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Students can Download Chapter 9 Differential Equations Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 9 Differential Equations

Plus Two Maths Differential Equations Three Mark Questions and Answers

Question 1.
y = e^2x(a + bx), a and b are arbitrary constants.
Answer:
y = e^2x(a + bx) ____(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = e^2xb + (a + bx)2e^2x
\(\frac{d y}{d x}\) = 2y + be^2x ⇒ \(\frac{d y}{d x}\) – 2y = be^2x ____(2)
Differentiating (2) with respect to x,

Question 2.
y = e^x(acosx + 6sinx), a and b are arbitrary constants.
Answer:
y = e^x(acosx + 6sinx) ___(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = e^x(-asinx + bcosx) + e^x(acosx + bsinx) \(\frac{d y}{d x}\) = e^x(-asin x + b cos x) + y
\(\frac{d y}{d x}\) – y = e^x(-a sin x + b cos x) ____(2)
Differentiating (2) with respec to x,

Question 3.
y = c₁e^x + c₂ e^-x , c₁ and c₁ are arbitrary constants.
Answer:
y = c₁e^x + c₂ e^-x ___(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = c₁e^x + c₂ e^-x __(2)
Differentiating (2) with respect to x,

Question 4.
(x – a)² + 2y² = a², a is a arbitrary constants.
Answer:
(x – a)² + 2y² = a² ___(1)
Differentiating with respect to x,

Question 5.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Answer:
y\(\frac{d y}{d x}\) = x ⇒ ydy = xdx
Integrating on both sides,
∫ydy = ∫xdx + c
⇒ \(\frac{y^{2}}{2}=\frac{x^{2}}{2}\) + c ____(1)
Since it passes through (0, -2),

Question 6.
Form the DE representing the family of parabolas having a vertex at origin and axis along positive direction of x-axis.
Answer:
Let (a, 0) be focus of the given family of parabolas.
y² = 4ax ____(1)
Differentiating with respect to x,

Question 7.
For the DE xy \(\frac{d y}{d x}\) = (x + 2)(y + 2), find the solution curve passing through the point(1,- 1).
Answer:
xy \(\frac{d y}{d x}\) = (x + 2)(y + 2)
⇒ \(\frac{y}{y+2} d x=\frac{x+2}{x} d x\)
Integrating on both sides,

⇒ y – 2 log|y + 2| = x + 2log|x| + c ____(1)
Since it passes through (1, -1),
⇒ -1 – 2log|-1 + 2| = 1 + 2log|l| + c
⇒ -2 = c
(1) ⇒ y – 2log|y + 2| = x + 2log|x| – 2.

Question 8.
Solve the initial value problem: \(\frac{d y}{d x}\) = y tan 2x; y(0) = 2.
Answer:
\(\frac{d y}{d x}\) = y tan 2x
⇒ \(\frac{d y}{y}\) tan 2xdx,
This is a variable type
∴∫\(\frac{d y}{y}\) = ∫tan 2xdx ⇒ log y = \(\frac{1}{2}\) log|sec 2x| + c
Given y(0) = 2 ⇒ log 2 = \(\frac{1}{2}\) log|sec 0| + c ⇒ c = log 2
log y = \(\frac{1}{2}\) log|sec 2x| + log 2 ⇒

Plus Two Maths Differential Equations Four Mark Questions and Answers

Question 1.
(i) Consider the differential equation given below. (1)
\(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\). Write the order and degree of the DE (if defined)
(ii) Find the Differential equation satisfying the family of curves y² = a(b² – x²), a and b are arbitrary constants. (3)
Answer:
(i) 4; degree is not defined

(ii) y² = a(b² – x²) ____(1)
Differentiating with respect to x,
2y \(\frac{d y}{d x}\) = -a2x ⇒ y\(\frac{d y}{d x}\) = -ax ____(2)
Differentiating (2) with respect to x,

Which is the differential equation.

Question 2.

1. Find the Differential equation satisfying the family of curves y = ae^3x + be^-2x, a and b are arbitrary constants. (3)
2. Hence write the degree and order of the DE. (1)

Answer:
1. y = ae^3x + be^-2x ____(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = ae^3x × 3 + be^-2x × -2 ____(2)
Differentiating (2) with respect to x,
⇒ \(\frac{d^{2} y}{d x^{2}}\) = 9ae^3x + 4be^-2x ____(3)
Now, (3) + 2 × (2) ⇒ \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=15 a e^{3 x}\)

Using (4), (5) in (1), we have,

2. Order: 2; degree: 1.

Question 3.
Consider the equation of all circles which pass through the origin and whose centres are on the x-axis.

1. Define the general equation of the circle.(1)
2. Find the DE corresponding to the above equation. (3)

Answer:
1. The general equation of the circle, passing through the origin and whose centers lies on x-axis can be taken as (x – h)² + y² = h² where h being an arbitrary constant.

2. Simplifying (x – h)² + y² = h² we get,
x² – 2hx + h² + y² = h² ⇒ x² – 2hx + h² = 0 _____(1)
Differentiating we get,
2x + 2y \(\frac{d y}{d x}\) – 2h = 0 ⇒ h = x + y \(\frac{d y}{d x}\)
Substituting in (1) we can eliminate h

Question 4.
Find a particular solution satisfying the given condition. (x³ + x² + x +1)\(\frac{d y}{d x}\) = 2x² + x, y = 1, when x = 0.
Answer:

Integrating on both sides,
∫dy = ∫\(\frac{2 x^{2}+x}{\left(x^{2}+1\right)(x+1)} d x\)
Splitting into partial fractions we have, (see Unit:7)

Question 5.

1. Write the degree of the DE y’ = 2xy. [0, 1, 2, 3] (1)
2. Express y’ = 2xy in the form Mdx = Ndy. Where M is a function of x and N is the function of y. (2)
3. Solve y’ = 2xy, y(0) = 1 (1)

Answer:

1. Degree = 1

2. We have, \(\frac{d y}{d x}\) = 2xy ⇒ \(\frac{d y}{y}\) = 2xdx, which is of the form Mdx = Ndy.

3. Solution is ∫\(\frac{d y}{y}\) = 2∫xdx ⇒ log|y| = x² + c
Given y(0) = 1 ⇒ log|1| = 0 + c ⇒ c = 0
⇒ log|y| = x² ⇒ y = e^x2.

Question 6.
Solve the following DE \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\).
Answer:
\(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\), this is a Homogeneous DE.
Therefore, put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) to convert it into variable separable form.
The DE becomes,


Therefore integrating we get,

Question 7.
Solve the linear differential equation \(x \frac{d y}{d x}-y=(x-1) e^{x}\).
Answer:
Given, x\(\frac{d y}{d x}\) – y = (x – 1)e^x, dividing both sides by x ,we get

Solution is
y × IF = ∫Q(IF)dx + c

Question 8.
(i) Choose the correct answer from the bracket. The solution of the differential equation xdy + ydx = 0 represents (1)
(a) a rectangular hyperbola
(b) a parabola whose centre is origin
(c) a straight line whose centre is origin
(d) a circle whose centre is origin.
(ii) From the DE of the family of circles touching the x-axis at origin. (3)
Answer:
(i) (c) a straight line whose centre is origin.

(ii) Let (0, a) be the centre of the circle. Therefore the equation of the circle is
x² + (y – a)² = a²
⇒ x² + y² = 2ay
⇒ \(\frac{x^{2}+y^{2}}{y}\) = 2a ____(1)

Differentiating with respect to x,

Question 9.
Solve the DE x²\(\frac{d y}{d x}\) = x² – 2y² + xy.
Answer:
x²\(\frac{d y}{d x}\) = x² – 2y² + xy

this is a Homogeneous DE.
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Integrating on both sides,

Question 10.
Choose the correct answer from the bracket

1. The DE \(\frac{d y}{d x}+\frac{y}{x}\) = e^x, x > 0 is of order …..[0,1,2,3] (1)
2. The integrating factor \(\frac{d y}{d x}+\frac{y}{x}\) = e^x, is……..[x, e^x, -x, e^-x] (1)
3. Solve \(\frac{d y}{d x}+\frac{y}{x}\) = e^x (2)

Answer:
1. Order = 1

2. \(\frac{d y}{d x}+\frac{y}{x}\) = e^x is of the form \(\frac{d y}{d x}\) + Py = Q,
where P = \(\frac{1}{x}\), Q = e^x
IF = e^∫Pdx = e^{∫\(\frac{1}{x}\)dx} = e^logx = x

3. Solution is y.IF = ∫e^x. IFdx
⇒ yx = ∫x.e^xdx ⇒ yx = x.e^x – ∫e^xdx
⇒ yx = x.e^x – e^x + c ⇒ yx = e^x(x – 1) + c.

Question 11.
Solve the DE \(\frac{d y}{d x}=\frac{x+y}{x-y}\).
Answer:
\(\frac{d y}{d x}=\frac{x+y}{x-y}\), this is a Homogeneous DE.
Put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\)

Integrating on both sides,

Plus Two Maths Differential Equations Six Mark Questions and Answers

Question 1.
Consider the DE \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)

1. Identify the DE ? Give reason. (1)
2. Explain the method of solving the DE. (1)
3. Solve the DE. (4)

Answer:
1. Given DE is a Homogeneous DE. Since y³ + 3x²y and x³ + 3xy² are Homogeneous functions of same degree (deg = 3).

2. By giving a substitution y = v x and \(\frac{d y}{d x}\) = v + x\(\frac{d y}{d x}\)
we can convert the DE into variable separable.

3. Now we have, \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)

∴ Integrating we get,
\(\int \frac{1+3 v^{2}}{2 v\left(1-v^{2}\right)} d v=\int \frac{d x}{x}\)

\(\frac{1}{2}\) log v – log(1 – v²) = log x + log c

Question 2.
Consider the D.E \(\frac{d y}{d x}+\frac{y}{x}=x^{2}\)

1. Find degree and order of DE . (1)
2. Solve the D.E. (4)
3. Find the particular solution when x = 1, y = 1. (1)

Answer:
1. Degree: 1, Order: 1.

2. The given D.E is first order linear DE of the form
\(\frac{d y}{d x}\) + Py = Q. Comparing we get, P = \(\frac{1}{x}\), Q = x²
∴ ∫Pdx = ∫\(\frac{1}{x}\)dx = logx
Integrating factor (I.F) = e^∫Pdx = e^logx = x
y.x = ∫x².xdx + c = ∫x³ dx + c
⇒ y.x = \(\frac{x^{4}}{4}\) + c ___(1)

3. Given, y = 1 when x = 1, then (1)
⇒ 1 × 1 = \(\frac{1}{4}\) + c ⇒ c = \(\frac{3}{4}\)
Therefore particular solution is
y.x = \(\frac{x^{4}}{4}\) + \(\frac{3}{4}\) ⇒ 4xy = x³ + 3.

Question 3.
Consider the equation.\(\frac{d y}{d x}\) + y = sin x

1. What is the order and degree of this equation? (1)
2. Find the integrating factor. (2)
3. Solve this equation. (3)

Answer:
1. Order = 1, Degree = 1

2. Given, \(\frac{d y}{d x}\) + y = sin x is of the form
\(\frac{d y}{d x}\) + Py = Q ⇒ P = 1, Q = sinx
Integrating factor = e^∫Pdx = e^∫1dx = e^x

3. Therefore solution is
y.IF = ∫Q.IFdx + c ⇒ ye^x = ∫e^x sinxdx + c ____(1)
∫sinx.e^xdx = e^x sinx – ∫cosx.e^xdx
= e^x sin x – cosx.e^x – ∫sinx.e^x dx
⇒ 2∫e^x sin xdx = e^x(sin x – cos x)
⇒ ∫e^x sinxdx = \(\frac{e^{x}}{2}\)(sinx – cosx)
(1) ⇒ ye^x = \(\frac{e^{x}}{2}\)(sinx – cosx) + c.

Question 4.
Considerthe D.E (x² – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)

1. Find \(\frac{d y}{d x}\), degree and order of the above differential equation. (1)
2. Find the integrating factor of the above differential equation. (2)
3. Solve the differential equation. (3)

Answer:
1. Given, (x² – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)

Here, Degree = 1, Order = 1.

2. The given DE is of the form \(\frac{d y}{d x}\) + Py = Q
Where,

Splitting it into partial fractions we get,

Put x = 1, ⇒ 6 = 2B ⇒ B = 3,
put x = -1, ⇒ 2 = -2A ⇒ A = -1

3. Solution is y × IF = ∫Q × IFdx + c

Question 5.
(i) The degree of the differential Equation \(\frac{d^{2} y}{d x^{2}}+\cos \left(\frac{d y}{d x}\right)=0\) is
(a) 2
(b) 1
(c) 0
(d) Not defined
(ii) Solve \(\frac{d y}{d x}\) + 2y tanx = sinx; y = 0, x = \(\frac{\pi}{3}\) (5)
Answer:
(i) (d) Not defined.

(ii) \(\frac{d y}{d x}\) + 2y tanx = sinx
Then, P = 2tanx, Q = sinx
IF = e^∫Pdx = e^∫2tanxdx = e^{2log sec x} = sec² x Solution is; y × IF = ∫Q(IF)dx + c
⇒ ysec² x = ∫sinx sec² xdx + c
⇒ ysec² x = ∫tanx secx dx + c
⇒ ysec²x = secx + c
Here; y = 0, x = \(\frac{\pi}{3}\)
⇒ 0 × sec² \(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + c ⇒ c = -2
⇒ ysec² x = secx – 2.

Question 6.
(i) The order of the differential equation \(x^{4} \frac{d^{2} y}{d x^{2}}=1+\left(\frac{d y}{d x}\right)^{3}\) is
(a) 1
(b) 3
(c) 4
(d) 2
(ii) Find the particular solution of the (1 + x²) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1 (5)
Answer:
(i) (d) 2

(ii) (1 + x²) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1

⇒ 0(1 + 1²) = tan^-11 + c ⇒ c = \(-\frac{\pi}{4}\)
⇒ y(1 + x²) = an^-1x – \(\frac{\pi}{4}\).

Students can Download Chapter 6 Rural Development Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 6 Rural Development

Plus One Economics Rural Development One Mark Questions and Answers

Question 1.
Rural developments include:
(i) education and skill developments
(ii) land reforms
(iii) measures for alleviation of poverty
(iv) all of the above
Answer:
(iv) all of the above

Question 2.
Shift of workforce from agriculture to other allied activities like livestock, poultry, fisheries, etc, is known as.
Answer:
Diversification.

Question 3.
Golden revolution is associated with?
(i) Horticulture
(ii) Livestock
(iii) Fisheries
(iv) Pulses
Answer:
(i) Horticulture

Question 4.
How many banks were nationalised in 1969?
(i) 6
(ii) 9
(iii) 14
(iv) 20
Answer:
(iii) 14

Question 5.
White revolution is related to
(i) Bamboo
(ii) Cash crop
(iii) Milk
(iv) Foodgrain
Answer:
(iii) Milk

Question 6.
Blue revolution is related to
(i) Milk
(ii) Fish
(iii) Egg
(iv) None of these
Answer:
(ii) Fish

Question 7.
Which among the following is an advantage of organic fanning
(i) It is labour intensive
(ii) It is environment-friendly
(iii) It uses locally produced organic inputs
(iv) All the above
Answer:
(iv) All the above

Question 8.
Complete the following statement.
The credit provisions of SHGs are generally referred to as ………………..
Answer:
Microcredit programs.

Question 9.
“Regulated Market yards benefit farmers as well as consumers”. Do you agree with this statement?
Answer:
Yes, I agree with this statement that the Regulated Market yields benefit farmers as well as consumers.

Plus One Economics Rural Development Two Mark Questions and Answers

Question 1.
What is TANWA?
Answer:
TANWA stands for Tamil Nadu Women in Agriculture. It is a project initiated in Tamil Naud to train women in latest agricultural techniques. It includes women to actively participate in raising agricultural productivity and family income. They promote small scale household activities like soap manufacturing, doll making etc.

Question 2.
Name few horticultural crops.
Answer:
Names of horticultural crops are:

• Fruits
• Vegetables
• Medicinal plants

Plus One Economics Rural Development Three Mark Questions and Answers

Question 1.
Complete the following.
a. Percentage of population living in the rural area is
(i) 25%
(ii) 50%
(iii) 75%
(iv) 90%
b. NABARD is the open bank providing credit to sector
(i) ombined sector
(ii) industrial sector
(iii) agricultural sector
(iv) tertiary sector
c. The Golden revolution was a period of very high productivity in
(i) food grain production
(ii) horticulture
(iii) organic farming
(iv) pisciculture
Answer:
a. (iii) 75%
b. (iii) agricultural sector
c. (ii) horticulture

Question 2.
Why the co-operative marketing could not gain momentum in India?
Answer:
Co-operative marketing has failed to gain adequate success in India because of inadequate coverage of farmer members and lack of appropriate link between marketing and processing management.

Question 3.
What do you mean by organic farming?
Answer:
Eco-friendly technology of cultivation is known as organic farming, it is the farming without using chemical fertilizers or pesticides. It uses natural mannure.

Question 4.
Give reasons for the declining share of agriculture in GDP of India.
Answer:
After the initiation of reforms, the growth rate of agriculture sector has declined to 2.3% during the 1990s which was lower than earlier years. The reasons for this decline are inadequate infrastructure, alternate employment opportunities in the industry or service sector, increase in casualisation of employment in agriculture, etc.

Question 5.
Explain the term ‘golden revolution’.
Answer:
The growth of horticulture sector at a spectacular speed is called ‘Gloden Revolution’. The period between 1991 – 2003 is known as the period of golden revolution. During this period, the investment in horticulture became highly productive and the sector emerged as a sustainable livelihood option. Thus India could become leading producer of mangoes, bananas, coconuts etc.

Question 6.
Name a few traditional industries in India.
Answer:
The traditional industries of India are:

• Pottery
• Crafts
• Handloom
• Textiles
• Bamboo

Question 7.
Prepare a note on ‘Kudumbashree’ in Kerala.
Answer:
Kudumbasree is a women-oriented community-based poverty reduction programme being implemented in kerala. The objective of this group is to encourage savings. The small savings in women’s group creates more employment and productive atmosphere in women’s groups.

Question 8.
Prepare a note on NABARD?
Answer:
NABARD stands for National Bank for Agriculture and Rural Development. It was established in 1982, July. It has been assigned the role performed by agricultural refinance and development corporation and also the role performed by RBI for rural credit. NABARD provides short term loans for agricultural operation.

Plus One Economics Rural Development Four Mark Questions and Answers

Question 1.
What are the alternative channels available for agricultural marketing? Give examples:
Answer:
Following are the alternative channels available for agricultural marketing.

1. Co-operative Credit Societies
2. Reserve Bank of India
3. Regional Rural Banks
4. Commercial Banks
5. NABARD
6. Self Help Groups (SHGs)

Question 2.
Critically appraise Rural banking operations in the light of recent farmers suicides in India.
Answer:
Operation of rural banking had a positive effect on rural farm and non-farm output, income, and employment. Farmers were given a variety of loans for meeting production needs. This helped India to achieve food security and to gain a huge buffer stock of grains.

However our banking system in the rural area could not meet all loan needs of the farmers. Some of the farmers still depend on local moneylenders. This system failed to develop a culture of deposit mobilization. Agriculture loan default was very high.

Question 3.
Mention some obstacles that hinder the mechanism of agricultural marketing.
Answer:
The farmers suffer from many problems relating to agricultural marketing. Some obstacles that hinder the mechanism of agricultural marketing are mentioned below.

• faulty weighing
• manipulation of accounts
• forced to sell at lower prices
• lack of proper storage facilities
• lack of infrastructure facilities
• inadequate coverage of farmers by cooperatives
• inefficient financial management

Question 4.
Match the following.

Answer:

Question 5.
‘Rainbow revolution is meant for overall agricultural development’, comment.
Answer:
The rainbow revolution is meant for overall development of Indian agriculture. Under rainbow revolution, the following revolutions are included.

Plus One Economics Rural Development Five Mark Questions and Answers

Question 1.
State whether true or false.

1. Organic food has higher nutritional value compared to non-organic food
2. Ten commercial banks were nationalized in 1969
3. Golden revolution happened in fisheries
4. SHG stands for Social Health Groups
5. Use of chemical fertilizers lead to pollution.

Answer:

1. True
2. False. Fourteen commercial banks were nationalized in 1969
3. False. Golden revolution happened in horticulture
4. False. SHG stands for Self Help Groups
5. True

Question 2.
“Organic farming and sustainable development are closely related to each other”. Substantiate.
Answer:
In recent years, awareness of the harmful effect of chemical-based fertilizers and pesticides on our health is on a rise. Conventional agriculture relies heavily on chemical fertilizers and toxic pesticides etc., which enter the food supply, penetrate the water sources, harm the livestock, deplete the soil and devastate natural eco-systems.

Efforts in evolving technologies which are ecofriendly are essential for sustainable development and one such technology which is eco-friendly is organic farming. In short, organic agriculture is a whole system of farming that restores, maintains and enhances the ecological balance.

There is an increasing demand for organically grown food to enhance food safety throughout the world. Thus it can be stated that organic farming and sustainable development are closely related to each other.

Question 3.
Identify the benefits and limitations of organic farming.
Answer:
A. Benefits of Organic Farming
1. Organic agriculture offers a means to substitute costlier agricultural inputs with locally produced organic inputs that are cheaper and thereby generate good returns on investment.

2. Organic agriculture also generates incomes through international exports as the demand for organically grown crops is on a rise. Studies across countries have shown that organically grown food has more nutritional value than chemical farming thus providing us with healthy foods.

3. Since organic farming requires more labour input than conventional farming, India will find organic farming an attractive proposition

4. the organic produce is pesticide-free and produced in an environmentally sustainable way

5. Organic fanning helps in sustainable development of agriculture and India has a clear advantage in producing organic products for both domestic and international markets.

B. Limitations of organic farming
1. Popularizing organic farming requires awareness and willingness on the part of farmers to adapt to new technology.

2. Inadequate infrastructure and the problem of marketing the products are major concerns which need to be addressed apart from appropriate agriculture policy to promote organic farming.

3. It has been observed that the yields from organic farming are less than modem agricultural farming in the initial years. Therefore, small and marginal farmers may find it difficult to adapt to large scale production.

4. Organic produce may also have more blemishes . and a shorter shelf life than sprayed produce. Moreover choice in production of off-season crops is quite limited inorganic farming.

Question 4.
Narrate the importance of agricultural diversification in promoting rural sector?
Answer:
Agricultural diversification is a system of fanning that encourages production of a variety of plant and animals and their products. Non-farm employment provides greater income stability, economies of size also reduces uncertainty of business. It involves maximum number of persons as it is labour intensive in nature.

The majority of small and marginal farmers cultivate mainly low-value subsistence crops. In the absence of farm and non-farm employment opportunities, they are forced to live below poverty line. So this problem is solved with the help of commercialisation and diversification of small farmers within and outside agriculture and their proper integration with local and global markets.

This is helpful not only in liberating the small and marginal farmers from the poverty trap, but also to meet the country’s growing demands for fruits, vegetables, milk and milk products, meat, fishes, eggs, etc. which generally show rising trends with increasing levels of per capita income in the economy.

Question 5.
“Rural development focuses on actions for the development of the areas that are lagging behind in the overall development of the development of the village economy” point out the areas of rural development.
Answer:
Rural development is quite a comprehensive term but it essentially means a plan of action for the development of areas which are lagging behind in socio-economic development. Some of the areas which are challenging and need fresh initiatives for development in India include

a. Development of human resources including:

1. Literacy, more specifically, female literacy, education and skill development
2. Health, addressing both sanitation and public health

b. Land reforms

c. Development of infrastructure development like electricity, irrigation, credit, marketing, transport facilities including construction of village roads and feeder roads to nearby highways, facilities for agriculture research and extension, and information dissemination

d. Introduction of special measures for alleviation of poverty and bringing about significant improvement in the living conditions of the weaker sections of the population emphasizing access to productive employment opportunities.

Question 6.
Explain some measures taken by the government to improve agricultural marketing.
Answer:
Agricultural marketing encompasses all activities in moving farm products from the producers to the final consumers such as storage, transport, processing, etc. In India, agricultural marketing was regarded as operationally inefficient and exploitative in character. Hence to improve the position of agricultural marketing following things are required :

1.  Facilities for storing goods.
2. Adequate and economical transport.
3. Eliminination of intermediaries.
4. Basic information of market conditions to get better prices for the produce.

The above mentioned requirements for efficient agricultural marketing can be also classified into the following.

1. Promotion of cooperative agricultural marketing societies.
2. Developing regulated markets where marketing practices have been standardized so as to provide encouragement to farmers to come to mandis to dispose off their produce and prevent them from exploitation in the hand of Dallas.
3. Granding and standardisation of produce to help the farmer to fetch better prices.
4. Developing warehousing facilities, road transport, etc.
5. Declaration of support price that offers a minimum price to the farmers for their produce.

Students can Download Chapter 8 Application of Integrals Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals

Plus Two Maths Application of Integrals Four Mark Questions and Answers

Question 1.
Consider the following figure.

1. Find the point of intersection (P) of the given parabola and the line. (2)
2. Find the area of the shaded region. (2)

Answer:
1. We have, y = x² and y = x ⇒ x = x² ⇒
⇒ x² – x = 0 ⇒ x(x – 1) = 0 ⇒ x = 0, 1
When x = 0, y =0 and x = 1, y = 1.
Therefore the points of intersections are (0, 0) and(1, 1).

2. Required area

Question 2.
1. Find the point of intersection ‘p’ of the given parabola and the line. (2)

2. Find the area of the shaded region. (2)
Answer:
1. Given, y = x², y = 2x
⇒ 2x = x² ⇒ x² – 2x = 0 ⇒ x(x – 2) = 0 ⇒ x = 0, 2
We have, y = 2x
⇒ when x = 0 ⇒ y = 0, when x = 2 ⇒ y = 4
‘P’ has co-ordinate (2, 4)

2. Area = \(\int_{0}^{2} 2 x d x-\int_{0}^{2} x^{2} d x=\left(x^{2}\right)^{2}-\left(\frac{x^{3}}{3}\right)_{0}^{2}=4-\frac{8}{3}=\frac{12-8}{3}=\frac{4}{3}\).

Question 3.
Consider the following figure.

1. Find the point of Intersection P of the circle x² + y² = 32 and the line y = x. (1)
2. Find the area of the shaded region. (3)

Answer:
1. x² + x² = 32 ⇒ 2x² = 32 ⇒ x² = 16 = 4
Therefore the point of intersection P is (4, 4).

2. We have x² + y² = 32 ⇒ y = \(\sqrt{32-x^{2}}\).
The required area =

= 8 + [8π – 8 – 4π] = 4π.

Question 4.

1. Shade the area enclosed by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant. (2)
2. Find the area of the region bound by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant. (2)

Answer:
1.

2. Area

Question 5.

1. Draw a rough sketch of the graph of the function y² = 4x. (2)
2. Find the area by the curve and the line x= 2. (2)

Answer:
1.

2. Area

Plus Two Maths Application of Integrals Six Mark Questions and Answers

Question 1.

1. Draw the graph of y² = 4x and y = x. (2)
2. Find the points of intersection of y² = 4x and y = x. (2)
3. Find the area bounded by the graphs.(2)

Answer:
1. y² = 4x is a parabola and y = x is a straight line passing through the origin.

2. x² = 4x ⇒ x² – 4x = 0 ⇒ x(x – 4) = 0 ⇒ x = 0, 4
When x = 0, y = 0 and when x = 4, y =4. Therefore the points of intersection are (0, 0) and (4, 4).

3. Area bounded by the graphs = Area under the parabola in the first quadrant – Area under the line.

Question 2.

1. Draw the graph of the function y = x² and x = y² in a coordinate axes. (2)
2. Find the point of intersection of the above graphs. (2)
3. Find the area of the region bounded by the above two curves. (2)

Answer:
1. The two function are parabolas as shown in the figure.

2. We have, y = x² and x = y²
x = (x²)² ⇒ x – x⁴ = 0 ⇒ x(1 – x³) = 0 ⇒ x = 0, 1
When x= 1, y= 1 and x = 0, y = 0.
Therefore the point is (0, 0) and (1, 1).

3. The required area = \(\int_{0}^{1} \sqrt{x} d x-\int_{0}^{1} x^{2} d x\)

Question 3.
Using the figure answer the following questions

1. Find the area of the shaded region as the sum of the area of two triangles. (2)
2. Define the function of the given graph. (2)
3. Verify the area of the shaded region using integration. (2)

Answer:
1. Area = Area of ∆OAD + Area of ∆ ABC
\(\frac{1}{2}\) × 3 × 3 + \(\frac{1}{2}\) × 2 × 2 = \(\frac{9}{2}\) + 2 = \(\frac{13}{2}\).

2. Area = \(\int_{0}^{5}\)f(x) dx = \(\int_{0}^{3}\)(-x + 3) dx + \(\int_{3}^{5}\)(x – 5) dx

Therefore verified.

Question 4.
The figure given below contains a straight line L with slope \(\sqrt{8}\) and a circle.

1. Find the equation of the line L and circle. (1)
2. Find the point of intersection P. (2)
3. Find the area of the shaded region. (3)

Answer:
1. The line L passes through origin and have slope 3, therefore its equation is y = \(\sqrt{8}\) x. The circle passes through origin and have radius 3, therefore its equation is x² + y² = 9.

2. We have, y =3x and x² + y² = 9
⇒ x² + (\(\sqrt{8}\)x)² = 9 ⇒ 9x² = 9
⇒ x = 1
∴ y = \(\sqrt{8}\) × 1 = \(\sqrt{8}\).
Therefore, coordinate of ‘P’ is (1, \(\sqrt{8}\)).

3. Area of the shaded region

Question 5.
Using the given figure answer the following

1. Define the equation of the circle and ellipse in the figure. (1)
2. Find the area of the ellipse using integration. (4)
3. Find the area of the shaded region. (Use formula to find the area of the circle.) (1)

Answer:
1. From the figure equation of the circle is x² + y² = 4 and that of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\).

2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y² = 1 ⇒ y = \(\frac{1}{2} \sqrt{4-x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx

3. Area of the circle of radius 2 = π (2)² = 4π
∴ Area of the shaded region = Area of the circle – Area of the ellipse
= 4π – 2π = 2π.

Question 6.

1. Find the area bounded by the curve y = sin x with X – axis, between x = 0 and x = 2π. (2)
2. Find the area of the region bounded by the curve y = x² and y = |x| (4)

Answer:
1. Area of y = sin x in each loop is same. Therefore;
2\(\int_{0}^{\pi}\)sin xdx = \(-2(\cos x)_{0}^{\pi}\) = -2 (cos π – cos0)
= -2(-1 – 1) = 4

2.

Area

Question 7.
Using integration, find the area of the region bounded by the triangle whose vertices are(-1, 1), (0, 5) and (3, 2). (6)
Answer:

Equation of BC is \(\frac{y-5}{1-5}=\frac{x-0}{-1-0}\)
⇒ y – 5 = 4x ⇒ 4x – y + 5 = 0 ⇒ y = 4x + 5
Equation of AB is x + y – 5 = 0 ⇒ y = 5 – x
Equation of AC is x – 4y + 5 = 0 ⇒ y = \(\frac{x}{4}+\frac{5}{2}\)
The required area = Area of the region PABCQP – Area of the region PACQP

Question 8.
Consider the functions f(x) = sin x and g(x) = cosx in the interval [0, 2π]

1. Find the x coordinates of the meeting points of the functions. (1)
2. Draw the rough sketch of the above functions. (2)
3. Find the area enclosed by these curves in the given interval. (3)

Answer:
1. f(x) = sin x and g(x) = cos x meet at multiples of \(\frac{\pi}{4}\)
x = \(\frac{\pi}{4}\), \(\frac{5 \pi}{4}\).

2.

3. Area = 2{Area under f(x) = sinx from \(\frac{\pi}{4}\) to π – Area under g(x) = cosx from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\)}

Question 9.

Evaluate \(\int_{0}^{r} \sqrt{r^{2}-x^{2}} d x\), where r is a fixed positive number. Hence prove the area of the circle of radius r is π r². (2)
Find the area of the circle, x² + y² = 16 which is exterior to parabola y² = 6x. (4)

Answer:
1.

2. Given, x² + y² = 16 and y² = 6x ⇒ x² + 6x = 16 ⇒ x² + 6x – 16 = 0 ⇒ (x + 8)(x – 2) = 0 ⇒ x = -8, 2.

Area = Area of the circle – Interior area of the parabola.

Question 10.
Using the figure answer the following questions

1. Define the equation of the ellipse and circle in the given figure. (1)
2. Find the area of the ellipse using integration. (4)
3. Find the area of the shaded region. (Area of the circle can be found using direct formula). (1)

Answer:
1. Equation of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\) and circle is x² + y² = 1.

2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y² = 1 ⇒ y = \(\frac{1}{2} \sqrt{4-x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx

3. Area of the circle = πr² = π × 1 = π
Required area = Area of ellipse – area of the circle = 2π – π = π.

Students can Download Chapter 5 Human Capital Formation in India Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 5 Human Capital Formation in India

Plus One Economics Human Capital Formation in India One Mark Questions and Answers

Question 1.
As per 2011 census, the literacy rate in India was.
(i) 74%
ii) 78%
(iii) 80%
(iv) 82%
Answer:
(i) 74%

Question 2.
Which among the following is related to health sector?
(i) UGC
(ii) NCERT
(iii) ICMR
(iv) AICTE
Answer:
(iii) ICMR

Question 3.
The state in which the literacy rate is the highest in the country is ……………..
(i) Delhi
(ii) Bihar
(iii) Mizoram
(iv) Kerala
Answer:
(iv) Kerala

Question 4.
The Adult literacy rate among females in India in the year2000 was ………………..
(i) 68.4%
(ii) 61.9%
(iii) 45.4%
(iv) 37.9%
Answer:
(iii) 45.4%

Question 5.
State true or false:
‘New Educational Policy was announced in 1987-88’.
Answer:
True.

Question 6.
Which state is having highest literacy rate in India according to 2011 census?
(i) Kerala
(ii) Punjab
(iii) Manipur
(iv) Hariyana
Answer:
(i) Kerala

Plus One Economics Human Capital Formation in India Two Mark Questions and Answers

Question 1.
Mention two government organizations that regulate the health and education sector.
Answer:
1. Organizations that regulate the education sector:

• The ministries of education at the union and state level
• National Council of Educational Research and Training (NCERT)
• University Grants Commission (UGC)
• All India Council of Technical Education (AICTE)

2. Organizations that regulate the health sector:

• The ministries of health at the union and state level
• Indian Council for Medical Research (ICMR)

Question 2.
Classify the following into preventive medicines, curative medicines, and social medicines.

1. Provision of clean drinking water.
2. Good sanitation
3. Medical intervention during illness
4. Vaccination

Answer:

• Preventive Medicine – Vaccination
• Curative Medicine – Medical intervention during illness.
• Social Medicine – Provision of clean drinking water, good sanitation

Question 3.
Classify the following into the features of human capital and human development

1. Education and health are considered as a means to increase human productivity.
2. Education and health are integral to human well-being.
3. Human beings are treated as a means to an end.
4. Human beings are treated as ends in themselves.

Answer:
a. Human Capital:
1. Education and health are considered as a means to increase human productively.
3. Human beings are treated as a means to an end.
b. Human development:
2. Education and health are integral to human well being.
4. Human beings are treated as ends in themselves.

Question 4.
Name the two major sources of human capital in a country.
Answer:
Two major sources of human capital in a country are:

1. Investment in education and
2. Investment in health.

Question 5.
Bring out the need for on-the-job training for a person.
Answer:
Firms give on-the-job-training to enhance the productive skills of the workers so as to enable them to absorb new technologies and modern ideas. It can be given in two ways:

1. Workers may be trained in the firm itself under the assistance of an experienced worker.
2. Workers may be sent off the firm campus for the training.

Plus One Economics Human Capital Formation in India Three Mark Questions and Answers

Question 1.
Give full form of the following

1. GER
2. SSA

Answer:

1. GER – Gross Enrolment Ratio
2. SSA – Sarva Shiksha Abhiyan

Question 2.
Find the odd one out

1. Roads, factory, teachers, power plants
2. Teachers, doctors, ports, scientists
3. UGC, NCERT, SCERT, NREGP

Answer:

1. Teachers. Others are physical capital
2. Ports. Others are human capital
3. NREGP. Others are educational regulatory institutions

Question 3.
“Adequate education and training to farmers can raise productivity in farms.”
Do you agree or not? Substantiate your answer.
Answer:
Yes, I do agree to the statement that adequate education and training to farmers can raise productivity in farms.
This is because the attainment of more information and training will help the farmers to attain more knowledge about new methods of farming and plant protection. The application of the new knowledge in the farms will enable them to raise productivity in the farms so that the overall productivity in the agricultural sector can be improved.

Question 4.
What do you mean by Human Development Index? How it is calculated?
Answer:
The quality of life index prepared and published by United Nations Development Programme is termed as Human Development Index. HDI studies the following three basic human capabilities:

1. Living a Long Life (Longevity)
2. Being Knowledgable (Educational Attainment)
3. Enjoying decent standard of living (Real Per Capita GDP)

In orderto calculate HDI we are required to construct the following three indices.
1. Life expectancy index: This index measures the degree of the achievement of the country regarding the expectancy of the people in a country. The value of index varies between 0-1.

2. Educational attainment index: This index measures the level of educational attainment of the people. The value of this index also varies between 0-1. Higher index shows higher level of educational attainment.

3. Real GDP per capita index: Real GDP is calculated at constant price. It shows the changes in physical production in real terms. Per capita, real GDP is GDP apanstant price divided by population. It is better measure of the quality of life of the people as compared to per capita income.

Plus One Economics Human Capital Formation in India Four Mark Questions and Answers

Question 1.
Point out the major problems of educational development in India.
Answer:
The major problems of educational development in India are pointed out below.

• Wastage of resources.
• Illiteracy
• Disparities in standard
• Lack of funds
• Problems of brain drain.

Question 2.
Discuss the following as a resource of human capital formation

1. Health infrastructure
2. Expenditure on migration

Answer:
1. Health infrastructure:
Expenditure on health is an important source of human capital formation. Health expenditure directly increases the supply of healthy labour force and is a source of human capital formation.

Preventive medicine, social medicine, curative medicine, and provision of clean drinking water and good sanitation are the various forms of health expenditure. In short provision of health facilities directly improve the efficiency of human capital.

2. Expenditure on migration:
Existence of unemployment is the main reason for the rural-urban migration in India. People migrate in search of jobs. Technically qualified persons migrate to other countries because of higher salaries that they may get in other countries.

Migration in any of these cases involves cost of transport, higher cost of living in the migrated places. The enhanced earnings in the new place outweigh the cost of migration. Hence expenditure on migration is a source of human capital formation.

Question 3.
Bring out the differences between human capital and human development.
Answer:

Question 4.
The following table shows the indicators of development in Education and health sectors. Analyze the table and make inference.

Answer:
1. The per capita income of the country was Rs.3687 in the year 1951. This value gradually increased over the years and in 2001, the real per capita income became Rs. 10306. this increase is a good sign of development as far a developing country like India is concerned.

2. Crude Death rate (Per 1000 population) of the country Was 25.1 in1951 which has fallen to 8.1 in 2001. Similarly, the infant mortality rate in India also declined significantly from 146 in 1951 to 63 in 2001. This shows the developments undergone in the health sector of the country.

3. Another important achievement in the field of education is rising literacy rate. It was 16.67% in 1951 whereas, the rate increased to 65.2% in 2001. The above table shows India’s achievements in the field of education, health and per capita income since independence. In all these three fields, country made very significant development during the last 50 years.

Plus One Economics Human Capital Formation in India Five Mark Questions and Answers

Question 1.
Trace the relationship between human capital and economic growth.
Answer:
Economic growth means the increase in real national income of a country. The contribution of the educated person to economic growth is more than that of an illiterate person. Education and health are important factors of economic growth.

Human capital contributes not only towards increasing labour productivity but also stimulates innovations and creates the ability to absorb new technologies. Education provides knowledge to understand changes in society and scientific advancements. Higher-income causes building of high level of human capital and also the high level of human capital causes growth of income.

Question 2.
“The percentage of ‘education expenditure of total government expenditure’ indicates the importance of education in the scheme of things before the government”. Critically analyze the changes in government spending on education.
Answer:
The government is supposed to make expenditure on education so as to improve the efficiency of human capital in the country. The percentage of ‘education expenditure of GDP’ expresses how much of our income is being committed to the development of education in the country.

During 1952-2002, education expenditure as percentage of total government expenditure increased from 7.92 to 13.17 and as percentage of GDP increased from0.64 to 4.02. Throughout this period the increase in education expenditure has not been uniform and there has been irregular rise and fall.

In addition to lowering expenditure on education, there exists a regional disparity in case of spending on education in various parts of the country. For example, the per capita education expenditure differs considerably across states from as high as Rs 3,440 in Lakshadweep to as low as Rs 386 in Bihar. This leads to differences in educational opportunities and attainments across states.

Thus it can be concluded that the government is not giving proper attention on the education sector. The relevance of education in the development of the country is to be recognized and therefore, given top priority to the expenditure in this sector.

Question 3.
“Physical capital is well distinguished from human capital”. Narrate them distinguishing points of physical capital and human capital.
Answer:
Physical capital is well distinguished from human capital. The major points of differences are given below.

1. Physical capital is tangible and can be easily sold in the market like any other commodity. Human capital is intangible; it is endogenously built in the body and mind of its owner.

2. Human capital is not sold in the market; only the services of human capital are sold and hence the necessity of the owner of the human capital to be present in the place of production. The physical capital is separable from its owner, whereas, human capital is inseparable from its owner.

3. The two forms of capital differ in terms of mobility across space. Physical capital is completely mobile between countries except for some artificial trade restrictions. Human capital is not perfectly mobile between countries as movement is restricted by nationality and culture.

4. In the case of human capital, depreciation takes place with ageing but can be reduced, to a large extent, through continuous investment in education, health, etc. This investment also facilitates the human capital to cope with change in technology which is not the case with physical capital.

5. Nature of benefits flowing from human capital are different from that of physical capital. Human capital benefits not only the owner but also society in general. This is called external benefit. An educated person can effectively take part in a democratic process and contribute to the socio-economic progress of a nation. A healthy person, by maintaining personal hygiene and sanitation, stops the spread of contagious diseases and epidemics.

6. Human capital creates both private and social benefits, whereas physical capital creates only private benefit. That is, benefits from a capital good flow to those who pay the price for the product and services produced by it.

Question 4.
Make some argument in favour of the need for different forms of government intervention in education and health sectors.
Answer:
Both education and health have been considered as important for human capital formation. The provision of basic education and health facilities has been accepted as a goal in All societies.

It is important that these facilities be made available free for the poor classes who can not afford them in order to ensure better skills and health of manpower. However, the private sector will not be forthcoming in making large investments in nonprofitable or free distribution of such services. Hence, the government has to play an important role.

Students can Download Chapter 7 Integrals Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals

Plus Two Maths Integrals Three Mark Questions and Answers

Question 1.
Integrate the following. (3 Score each)

1. ∫sin x sin 2x sin 3 xdx
2. ∫sec²x cos²2x dx

Answer:
1. We have sinxsin2xsin3x
= 1/2 (2sinxsin3x) sin2x
= 1/2 (cos2x – cos4x) sin2x
= 1/4 (2sin2xcos2x – 2cos4xsi n2x)
= 1/4 [sin4x – (sin6x – sin2x)]
= 1/4(sin4x + sin2x – sin6x)
∫sin x sin 2x sin 3 xdx
= \(\frac{1}{4}\) ∫(sin 4x + sin 2x – sin 6x)dx
= –\(\frac{1}{16}\) cos4x – \(\frac{1}{8}\) cos2x + \(\frac{1}{24}\) cos6x + c.

2. sec²x cos²2x = \(\frac{\left(2 \cos ^{2} x-1\right)^{2}}{\cos ^{2} x}\)
= \(\left(\frac{2 \cos ^{2} x}{\cos x}-\frac{1}{\cos x}\right)^{2}\) = (2cosx – secx)²
= 4cos²x + sec²x – 4
= 2(1 + cos2x) + sec²x – 4
= 2cos2x + sec²x – 2
∫sec² x cos² 2x dx = ∫(2 cos 2x + sec² x – 2)dx
= sin 2x + tan x – 2x + c.

Question 2.
Find \(\int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x\)?
Answer:

= ∫e^x [sec² x + tan x]dx
= ∫e^x[tanx + sec²x]dx = e^x tanx + c.

Question 3.
Evaluate \(\int \frac{\sec ^{2} x d x}{\sqrt{\tan ^{2} x+4}}\)?
Answer:
Put tanx = u, sec²xdx = dy

Question 4.
Find the following integrals.

Answer:
(i) I = \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Put cosx = t ⇒ -sin xdx = dt
When x = 0 ⇒ t = cos0 = 1,

(ii) I = \(\int_{0}^{1} x e^{x^{2}} d x\)
Put x² = t ⇒ 2xdx = dt
When x = 0 ⇒ t = 0,
x = 1 ⇒ t = 1
I = \(\frac{1}{2} \int_{0}^{1} e^{t} d t\) =

= [e¹ – e⁰] = e – 1.

Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,

(iv) I = \(\int_{0}^{2} x \sqrt{x+2} d x\)
Put x + 2 = t² ⇒ dx = 2tdt
When x = 0 ⇒ t = \(\sqrt{2}\), x = 2 ⇒ t = 2

(v) I = \(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin x} \cos x d x\)
Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,


Put tan x = t ⇒ sec² xdx = dt
When x = 0 ⇒ t = tan 0 = 0,

Question 5.
(i) If f (x) is an odd function, then \(\int_{-a}^{a} f(x)\) = ?
(a) 0
(b) 1
(c) 2\(\int_{0}^{a} f(x)\) dx
(d) 2a
Evaluate
(ii) \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x\)
(iii) \(\int_{-1}^{1} e^{|x|} d x\)
Answer:
(i) (a) 0.

(ii) Here, f(x) = sin⁹⁹x.cos¹⁰⁰x .then,
f(-x) = sin⁹⁹(- x).cos¹⁰⁰(- x) = – sin⁹⁹ x. cos¹⁰⁰ x = -f(x)
∴ odd function ⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x=0\).

(iii) Here, f(x) = e^|x|, f(-x) = e^|-x| = e^|x| = f(x)
∴ even function.

we have |x| = x, 0 ≤ x ≤ 1

Question 6.

1. Show that cos² x is an even function. (1)
2. Evaluate \(\int_{-\pi / 4}^{\pi / 4} \cos ^{2} x d x\) (2)

Answer:
1. Let f(x) = cos²x ⇒ f(-x) = cos² (-x) = cos² x = f(x) even.

2.

Question 7.
Find the following integrals.

Answer:

Question 8.
Find the following integrals.

Answer:

Add (1) and (2)

Question 9.
Find the following integrals.

1. \(\int \frac{1}{3+\cos x} d x\)
2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\)

Answer:
1. \(\int \frac{1}{3+\cos x} d x\)
Put t = tanx/2 ⇒ dt = 1/2 sec² x/2 dx

2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\) = \(\int \frac{2 x}{(x+2)(x+1)} d x\)

2x = A(x + 1) + B (x + 2)
when x = -1, -2 = B ; B = -2
when x = -2, -4 = -A ; A = 4

= 4log(x + 2) – 2log (x + 1) + C.

Plus Two Maths Integrals Four Mark Questions and Answers

Question 1.
Find the following integrals.

Answer:

x² + x +1 = A(x² + 1) + (Bx + C)(x + 2)
Put x = -2 ⇒ 4 – 2 + 1 = 5A ⇒ A = \(\frac{3}{5}\)
Equating the coefficients of x²
⇒ 1 = A + B ⇒ B = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
Equating the constants
⇒ 1 = A + 2C ⇒ 2C = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\) ⇒ C = \(\frac{1}{5}\)


⇒ 1 = A(x – 1) + B(x + 3)
Put x = 1 ⇒ 1 = 2A ⇒ A = \(\frac{1}{2}\)
Put x = -3 ⇒ 1 = -4B ⇒ B = – \(\frac{1}{4}\)


Equating the constants; ⇒ 1 = A
Equating the coefficients if t;
⇒ 0 = A + B ⇒ B = -1

Question 2.
Find the following integrals.

1. ∫ e^2x sin3xdx
2. ∫ x sin^-1xdx

Answer:
1. I = ∫e^2x sin3xdx = ∫ sin 3x × e^2xdx

2. ∫ x sin^-1xdx = ∫ sin^-1x × xdx

Question 3.
(i) Which of the following is the value of \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}\)? (1)

(ii) Evaluate \(\int \frac{2 x}{x^{2}+3 x+2} d x\) (3)
Answer:
(i) [sin^-1\(\frac{x}{a}\) + c]

(ii)

⇒ 2x = A(x + 1) + B(x + 2) ⇒
Put x = -2 and x = -1, we get A = 4, B = -2

Question 4.

1. Choose the correct answer from the bracket.
   ∫e^x dx = — (e^2x + c, e^-x + c, e^2x + c) (1)
2. Evaluate: ∫ e^x sinxdx

Answer:
1. e^x + c

2. I = ∫e^x sinxdx = sinx.e^x – ∫cos x.e^xdx
= sin x.e^x – (cos x.e^x – ∫(- sin x).e^x dx)
= sinx.e^x – cosxe^x – ∫sinx.e^xdx
= sin x.e^x – cos xe^x – I
2I = sin x.e^x – cos xe^x
I = \(\frac{1}{2}\)e^x(sinx – cosx) + c.

Question 5.
(i) f(x)∫g(x) dx – ∫(f'(x)∫g(x) dx)dx (1)
(a) ∫f'(x)g{x)dx
(b) ∫f(x)g'(x)dx
(c) ∫\(\frac{f(x)}{g(x)}\)dx
(d) ∫f(x)g(x)dx
(ii) Integrate sin^-1\(\sqrt{\frac{x}{a+x}}\)dx w.r.to x. (3)
Answer:
(i) (d) ∫f(x)g(x)dx

(ii) ∫sin^-1\(\sqrt{\frac{x}{a+x}}\)dx,
Put x = a tan²θ, θ = tan^-1\(\sqrt{\frac{x}{a}}\)
⇒ dx = 2a tanθ sec²θ dθ
I = ∫sin^-1\(\left(\frac{\tan \theta}{\sec \theta}\right)\) 2a tanθ sec²θ dθ
= ∫sin^-1(sinθ)2a tanθ sec²θ dθ
= 2a∫θ tanθ sec²θ dθ
Put tanθ = t, θ = tan^-1 t ⇒ sec²θ dθ = dt
= 2a ∫ tan^-1 t (t) dθ

= a[tan²θ.θ – tanθ + θ] + c
= a[θ(1 + tan²θ) – tanθ] + c

Question 6.
Match the following. (4)

Answer:

Question 7.
Evaluate \(\int \frac{x}{\sqrt{x+a}+\sqrt{x+b}} d x\)?
Answer:

Question 8.
Match the following.

Answer:
1.

2. ∫sec x(sec x + tan x)dx = ∫(sec² x + sec x. tan x)dx
= tanx + secx + c.

3. ∫e^3xdx = \(\frac{e^{3 x}}{3}\) + c.

4. ∫(sin x + cos x)dx = sin x – cosx + c.

Question 9.
Consider the integral I = \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)?

1. What substitution can be given for simplifying the above integral? (1)
2. Express I in terms of the above substitution. (1)
3. Evaluate I. (2)

Answer:
1. Substitute sin^-1 x = t.

2. We have, sin^-1 x = t ⇒ x = sint
Differentiating w.r.t. x; we get,
\(\frac{1}{\sqrt{1-x^{2}}}\)dx = dt
∴ I = ∫t sin t dt.

3. I = ∫t sin t dt = t.(-cost) -∫(-cost)dt = -t cost + sint + c
= -sin^-1 x. cos (sin^-1 x) + sin(sin^-1 x) + c
x – sin^-1 x.cos(sin^-1 x) + c.

Question 10.
Evaluate \(\int_{0}^{\pi / 4} \log (\tan x) d x\).
Answer:

Question 11.
Find the following integrals.

1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) (2)
2. \(\int \frac{1}{x^{2}-6 x+13} d x\) (2)

Answer:
1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) = \(\int \frac{\sin ^{2} x}{\cos ^{2} x} d x\) = ∫tan² xdx
= ∫(sec²x – 1)dx = tanx – x + c.

2. \(\int \frac{1}{x^{2}-6 x+13} d x\)

Question 12.
Match the following. Justify your answer.

Answer:

Question 13.
(i) ∫sin2x dx = ? (1)
(a) 2 cos x + c
(b) -2 sin x + c
(c) \(\frac{\cos 2 x}{2}\) + c
(d) \(-\frac{\cos 2 x}{2}\) + c
(ii) Evaluate ∫e^x sin 2x dx (3)
Answer:
(i) (d) \(-\frac{\cos 2 x}{2}\) + c.

(ii) Consider I = ∫e^x sin 2x dx
= ∫sin 2x. e^xdx = sinx.e^x – 2∫cos 2x. e^xdx
= sin 2x.e^x – 2 (cos 2x.e^x + 2∫sin 2x. e^xdx)
= sin 2x. e^x – 2 cos 2x e^x – 4 ∫sin 2x. e^xdx
= sin 2x. e^x – 2 cos 2x e^x – 4I
5 I = sin 2x. e^x – 2 cos 2x e^x
I = \(\frac{e^{x}}{5}\) (sin 2x – 2 cos 2x).

Question 14.

1. Resolve \(\frac{x^{2}+1}{x^{2}-5 x+6}\) into partial fractions. (2)
2. Hence evaluate ∫\(\frac{x^{2}+1}{x^{2}-5 x+6}\). (2)

Answer:
1.

2.

5x – 5 = A(x – 2) + B(x – 3)
x = 2, 5 = -B, B = -5
x = 3, 10 = A, A = 10
(1) ⇒ I = ∫ 1dx + ∫\(\frac{10}{x-3}\) dx – ∫\(\frac{5}{x-2}\) dx
= x + 10log(x – 3) – 5log(x – 2) + c.

Question 15.
Evaluate \(\int_{0}^{4}\) xdx as a limit of sum.
Answer:
By definition,
\(\int_{a}^{b}\) f(x) dx =
(b – a)\(\lim _{n \rightarrow \infty} \frac{1}{n}\){f(a) + f(a + h) +…….+f(a + {n – 1)h)}
Here, a = 0, b = 4, f(x) = x, h = \(\frac{4-0}{n}=\frac{4}{n}\) ⇒ nh = 4

Question 16.

1. Define the real valued function f(x) = |x² + 2x – 3| (2)
2. Evaluate \(\int_{0}^{2}\)|x² + 2x – 3|dx. (2)

Answer:
1. f(x) = |x² + 2x – 3| = |(x – 1) (x + 3)|
We have;

2. I = \(\int_{0}^{2}\)|x² + 2x – 3|dx

Question 17.
Consider the function f(x) = |x|+|x + 1|

1. Define the function f (x) in the interval [-2, 1]. (2)
2. Find the integral \(\int_{-2}^{1}\) f(x) dx (2)

Answer:
1. Given, f(x) = |x|+|x + 1|.
We have,

Combining these two functions, we get the function f(x).

2.

Question 18.
Evaluate \(\int_{\sqrt{6}}^{\sqrt{3}} \frac{d x}{1+\sqrt{\tan x}} d x\). (4)
Answer:

Plus Two Maths Integrals Six Mark Questions and Answers

Question 1.
(i) Fill in the blanks. (3)
(a) ∫ tan xdx = —
(b) ∫ cos xdx = —
(c) ∫\(\frac{1}{x}\)dx = —
(ii) Evaluate ∫sin³ xcos² xdx (3)
Answer:
(i) (a) log|secx| + c
(b) sinx + c
(c) log|x| + c.

(ii) ∫sin³ xcos² xdx = ∫sin² xcos² x sin xdx
= ∫(1 – cos² x)cos² x sin xdx
Put cos x = t ⇒ – sin xdx = dt
∴ ∫(1 – cos² x)cos² xsin xdx = -∫(1 – t² )t²dt
= ∫(t⁴ – t²)dt = \(\frac{t^{5}}{5}-\frac{t^{3}}{3}\) + c
= \(\frac{\cos ^{5} x}{5}-\frac{\cos ^{3} x}{3}\) + c.

Question 2.
Find the following integrals.

Answer:
(i) I = ∫(3x – 2)\(\sqrt{x^{2}+x+1} d x\)
Let 3x – 2 = A(2x + 1) + B
⇒ 3 = 2 A ⇒ A = \(\frac{3}{2}\)
⇒ -2 = A + B ⇒ -2 = \(\frac{3}{2}\) + B
⇒ B = -2 – \(\frac{3}{2}\) = – \(\frac{7}{2}\)


Using (2) and (3) in (1) we have;

(ii) I = \(\int \frac{2 x-3}{x^{2}+3 x-18} d x\)
Let 2x – 3 = A(2x + 3) + B
⇒ 2 = 2A ⇒ A = 1
⇒ -3 = 3A + B ⇒ -3 = 3 + B ⇒ B = -6

(iii) I = \(\int \frac{5 x+2}{1+2 x+3 x^{2}} d x\)
Let 5x + 2 = A{6x + 2) + B
⇒ 5 = 6 A ⇒ A = \(\frac{5}{6}\)
⇒ 2 = 2A + B ⇒ 2 = \(\frac{5}{3}\) + B ⇒ 2 – \(\frac{5}{3}\) = \(\frac{1}{3}\)

(iv) I = \(\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x\)
Let 5x + 3 = A(2x + 4) + B
⇒ 5 = 2A ⇒ A = \(\frac{5}{2}\)
⇒ 3 = 4A + B ⇒ 3 = 10 + B ⇒ B = -7


Using (2) and (3) in (1) we have;

Question 3.
Consider the expression \(\frac{1}{x^{3}-1}\)

1. Split it into partial fraction. (2)
2. Evaluate ∫ \(\frac{1}{x^{3}-1}\) dx (4)

Answer:
1.

1 = A (x² + x + 1) + (Bx + c)(x + 1),
Put x = -1 ⇒ 1 = A(1 + 1 + 1) ⇒ A= \(\frac{1}{3}\)
Equating like terms.
0 = A + B ⇒ B = – \(\frac{1}{3}\), 1 = A + C ⇒ C = \(\frac{2}{3}\)

2.

Put, x – 2 = D (2x – 1) + E ,
1 = 2 D ⇒ D = \(\frac{1}{2}\),
-2 = -D + E ⇒ E = –\(\frac{3}{2}\)

Question 4.
(i) Match the following (4)

(ii) Consider the function f(x) = \(\frac{x^{4}}{x+1}\) Evaluate ∫f(x)dx (2)
Answer:
(i)

(ii) Here the numerator is of degree 4 and denominator of degree 1. So to make it a proper fraction we have to divide Nr by Dr.

Question 5.

1. Evaluate the as \(\int_{0}^{2}\)x²dx the limit of a sum. (3)
2. Hence evaluate \(\int_{-2}^{2}\)x²dx (1)
3. If \(\int_{0}^{2}\) f(x)dx = 5 and \(\int_{-2}^{2}\) f(x)dx = 0, then \(\int_{-2}^{0}\) f(x)dx = …….. (2)

Answer:
1. Here the function is f(x) = x², a = 0, b = 2 and h = \(\frac{b-a}{n}=\frac{2}{n}\)
\(\int_{0}^{2}\)x²dx =

2. \(\int_{-2}^{2}\) x²dx = 2 \(\int_{0}^{2}\)x²dx = \(\frac{16}{3}\)

3.

Question 6.
Find ∫\(\sqrt{\tan x}\)xdx.
Answer:
Given;
I = ∫\(\sqrt{\tan x}\)xdx,
Put tanx = t² ⇒ sec²xdx = 2tdt ⇒ dx = \(\frac{2 t d t}{1+t^{4}}\)

Question 7.
(i) Match the following. (2)

(ii) Integrate \(\frac{\sec ^{2} x}{5 \tan ^{2} x-12 \tan x+14}\) w.r.to x. (4)
Answer:
(i)

Question 8.

1. Evaluate \(\int_{0}^{1} \sqrt{x} d x\) (1)
2. If \(\int_{0}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of a. (3)
3. Hence find \(\int_{a}^{a+1}\)x dx. (2)

Answer:
1.

2. Given;

3. When a = 0

When, a = 4

Question 9.
(i) Let f (x) be a function, then \(\int_{0}^{a}\) f(x) dx = ? (1)
(a) 2 \(\int_{0}^{a}\) f(x – a) dx
(b) \(\int_{0}^{a}\) f(a – x) dx
(c) f(a)
(d) 2\(\int_{0}^{a}\) f(a – x) dx
Evaluate

Answer:
(i) (b) \(\int_{0}^{a}\) f(a – x) dx

(ii)

(1) + (2)

⇒ I = 1.

(iii)

Question 10.
Find the following integrals.

1. ∫\(\frac{2 e^{x}}{e^{3 x}-6 e^{2 x}+11 e^{x}-6} d x\)
2. ∫\(\frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x\)

Answer:
1.

⇒ 1 = A(t – 2)(t – 3) + B(t – 1)(t – 3) + C(t – 1)(t – 2)
Put t = 1 ⇒ 1 = A(-1)(-2) ⇒ A = \(\frac{1}{2}\)
Put t = 2 ⇒ 1 = B(1)(-1) ⇒ B = -1
Put t = 3 ⇒ 1 = B(2)(1) ⇒ B = \(\frac{1}{2}\)

2. I = ∫\(\frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x\)dx
Put sin x = t ⇒ cosxdx = dt

⇒ 3t – 2 = A(t – 2) + B
Equating the coefficients if t; ⇒ 3 = A
Equating the constants
⇒ -2 = -2A + B ⇒ -2 = -6 + B ⇒ B = 4

Question 11.

1. Find ∫\(\frac{1}{x^{2}+a^{2}}\)dx (1)
2. Show that 3x + 1 = \(\frac{3}{4}\)(4x – 2) + \(\frac{5}{2}\) (2)
3. Evaluate \(\int \frac{3 x+1}{2 x^{2}-2 x+3} d x\) (3)

Answer:
1. ∫\(\frac{1}{x^{2}+a^{2}}\)dx = 1/a tan^-1 x/a + c.

2. 3x + 1 = A \(\frac{d}{d x}\)(2x² – 2x + 3) + B
= A(4x – 2) + B
3 = 4A; A = 3/4
1 = -2A + B
1 = -3/2 + B, B = 1 + 3/2 = 5/2
∴ 3x + 1 = 3/4(4x – 2) + 5/2

3.

Students can Download Chapter 12 ICT and Society Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 12 ICT and Society

Plus Two Computer Science ICT and Society One Mark Questions and Answers

Question 1.
IPR stands for ______.
Answer:
Intellectual Property Right.

Question 2.
WIPO stands for _____.
Answer:
World Intellectual Property Organisation

Question 3.
______ is the exclusive rights to prevent unauthorized copying of inventions by a Creator from the Unauthorised person or company.
Answer:
Patent

Question 4.
_____ is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company.
Answer:
Trademark

Question 5.
A product or article is designed so beautifully to attract customers. This type of design is called
Answer:
Industrial Design.

Question 6.
Aranmula Kannadi, Palakkadan Matta, Marayoor Sarkkara, etc are example of _______.
Answer:
Geographical indications.

Question 7.
_____ is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.
Answer:
Copyright

Question 8.
From the following which is the symbol for copyright.
(a) $
(b) ©
(c) ®
(d) ™
Answer:
(b) ©

Question 9.
From the following which is the symbol for Unregistered trademark.
(a) $
(b) ©
(c) ®
(d) ™
Answer:
(d) ™

Question 10.
From the following which is the symbol for Registered trademark.
(a) $
(b) ©
(c) ®
(d) ™
Answer:
(c) ®

Question 11.
Unauthorized copying or use of Intellectual property rights such as Patents, Copyrights and Trademarks are called ____.
Answer:
Intellectual Property Infringement.

Question 12.
_____ prevents others from the unauthorized or intentional copying or use of Patent without the permission of the creator.
Answer:
Patent Infringement.

Question 13.
______ is the illegal copying, distribution, or use of software.
Answer:
Piracy.

Question 14.
______ prevents others from the unauthorized or intentional copying or use of Trademark without the permission of the creator.
Answer:
Trademark Infringement

Question 15.
_____ prevents others from the unauthorized or intentional copying or use of Copy right without the permission of the creator.
Answer:
Copy right Infringement

Question 16.
______ is a virtual environment created by computer systems connected to the internet
Answer:
Cyberspace

Question 17.
A person committing crimes and illegal activities with the use of computers over Internet. This crime is included as _____ crime.
Answer:
Cybercrime

Question 18.
State True or False.
Cybercrimes can be classified into three categories such as against individual, property, and Government.
Answer:
True

Question 19.
Phishing, hacking, denial of service attacks, etc are ____ crimes.
Answer:
Cyber

Question 20.
Odd one out
(а) Identity theft
(b) Harassment
(c) violation of privacy
(d) credit card fraud
Answer:
(d) credit card fraud, it is a cybercrime against individual others are cyber crimes against property.

Question 21.
Odd one out
(a) Credit card theft
(b) Intellectual property theft
(c) Internet time theft
(d) Dissemination of obscene material
Answer:
(d) Dissemination of obscene material, It is cyber , crime against individual, the others are cyber against property.

Question 22.
Odd one out
(a) cyberterrorism
(b) Attacks against e-Governance websites
(c) Impersonation and cheating
(d) Website defacement
Answer:
(c) Impersonation and cheating, it is cybercrime against individual others are cyber crimes against Government.

Question 23.
IT Act amended in _____.
(a) 2015
(b) 2008
(c) 1900
(d) 1998
Answer:
(b) 2008

Question 24.
IT Act passed in Indian parliament is ____.
Answer:
2000.

Question 25.
The laws to prevent cyber crimes is termed as ____.
Answer:
Cyberlaw

Question 26.
_____ excessive enthusiasm for acquiring knowledge.
Answer:
Infomania

Question 27.
Phishing is an example of ______.
Answer:
Cybercrime.

Question 28.
ICT stands for _______.
(a) Internet and Communication Technology
(b) Information and Computer Technology
(c) Information and Communication Technology
(d) Integrated Communication Technology
Answer:
(c) Information and Communication Technology

Question 29.
Which of the following e-Governance helps citizens for interacting with the Government?
(a) G2E
(b) G2B
(c) G2C
(d) G2G
Answer:
(c) G2C

Question 30.
What are the different types of interactions in e-Governance?
Answer:
G2G, G2E.G2B, G2C.

Question 31.
The unauthorized use of intellectual property rights is termed as
Answer:
Infringement

Question 32.
Expand the term WIPO in connection with IPR.
Answer:
World Intellectual Property Organization.

Question 33.
The exclusive right granted to an invention is called
(a) Trademark
(b) Copy right
(c) Patent
(d) Design
Answer:
(c) Patent

Question 34.
The exclusive right given to a person over the creation of his/her mind for a period of time is called
Answer:
Patent / Intellectual Property Right

Question 35.
What is the name given to the process of using scientific knowledge for analyzing and presenting evidence of cyber related crimes before court?
Answer:
Cyber forensics

Question 36.
Which among the following are considered as violation to privacy?
1. Keeping hidden cameras in private places
2. Publishing private photos of individual in social media without their permission
3. Use of unauthorized software
4. Using simple password
(A) All the above are correct
(B) 1,2 and 3 only
(C) 1 and 4 only
(D) 1 and 2
Answer:
(D) 1 and 2

Plus Two Computer Science ICT and Society Two Mark Questions and Answers

Question 1.
“IPR (Intellectual Property Right) encourages innovation” Justify.
Some people spend lots of money,time body and mental power to create some products such as a classical movie, album, artistic work, discoveries, invention, software, etc. These type of Intellectual properties must be protected from unauthorized access by law. This is called Intellectual Property right(IPR). It enables to earn recognition, financial benefit, can sell the innovation, etc. It motivates further innovation.

Question 2.
Define the following terms.

1. Cyber space
2. Cyber crime

Answer:
1. CyberSpace:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too.

Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual(un real) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

2. Cyber crime:
Just like normal crimes (theft, trespassing private area, destroy, etc,) Cyber crimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc and some of them commit suicide.

Question 3.
Write a short note on

1. Trademark
2. Industrial design

Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 4.
Compare patent and Trademark.
1. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

2. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Question 5.
Write any one website for the following services.

1. e-Governance
2. e-Business
3. e-Banking
4. e-Learning

Answer:

1. e-Governance(any One) www.dhsekerala.gov.in, www.incometaxindia.gov.in, www.spark.gov.in,www.ceo.kerala.gov. in
2. e-Business www.indane.co.in, www.amazon.com,www.ebay.in
3. e-Banking www.onlinesbi.co.in
4. e-Learning www.ignouonline.ac.in,www.nptel.iitm.ac.in

Question 6.
Write a short note about EPS.
Answer:
Electronic Payment System(EPS): It is also called plastic money that is electronically exchange money between two individuals or firms(buyers and sellers) in an online environment.

Question 7.
What is cyberspace?
Answer:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too.

Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyberspace. The result for this integration is that tremendous speed and it is very cheap.

Question 8.
Why is cyberspace called a virtual world?
Answer:
The telephone system and computer system are integrated and create a virtual(un real) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap. This is an imaginary world. We can see persons with different behaviour. Because of good and bad people we can’t believe blindly. If we search a solution for a problem thousands of answers will get instantly and may confused us.

Question 9.
What is copyright? How does it differ from patent?
Answer:
1. Copyright:
The trademark is ©, copyright is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator for 60 years after the death of the author.

2. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

Question 10.
Explain the exclusive right given to the owner by IPR?
Answer:
The exclusive right given to the owner by I PR is owner can disclose their creations for money.

Question 11.
it is the unauthorized copying, distribution, and use of a creation without the permission of the creator. It is against the copyright act and hence the person committed deserve the punishment.

Question 12.
Match the following

Answer:
a – 2
b – 3
c – 4
d – 1

Question 13.
What do you meant by infringement?
Answer:
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called intellectual property lnfringement(violation). It is a punishable offence.

Plus Two Computer Science ICT and Society Three Mark Questions and Answers

Question 1.
Write a short note on the importance of IT Act 2000.
Answer:
Information Technology Act 2000(amended in 2008):
IT Act 2000 controls the use of Computer(client), Server, Computer Networks, data and Information in Electronic format and provide legal infrastructure for E-commerce, in India. This is developed to promote IT industry, control e-commerce also ensures the smooth functioning of E-Governance and it prevents cyber crimes.

The person those who violate this will be prosecuted. In India, IT bill introduced in the May 2000 Parliament Session and it is known as Information Technology Act 2000. Some exclusions and inclusions are introduced in December 2008.

Question 2.
“Infomania affects peoples’ lives and their loved ones.”
Comment on this statement.
Answer:
Info mania is excessive desire(infatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media. Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Question 3.
Define the term e-Business. What are the advantages and challenges of e-Business? Write any two e-Business websites.
Answer:
E-business(electronic Business): Providing ser¬vices or running business through internet is called E-business.
Advantages of e-business:

• It overcomes geographical limitations
• It reduces the operational cost
• It minimizes the time and cost
• It remains open all the time
• We can locate the product faster from a wider range of choices
  Challenges to E business
• Peoples are unaware of IT applications and its uses
• Most peoples don’t have plastic money(credit / debit card) and net banking
• It requires high security measurements otherwise you may lose money
• We can’t touch or smell products through online
• Some companies may not have proper Goods delivery service
  Useful e-Business websites

Question 4.
How do trademark and industrial design differ?
Answer:
Trademark:
This is a unique, sirhple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 5.
Why is Cyberlaw important?
Answer:
Just like normal crimes (theft, trespassing private area, destroy, etc.) Cybercrimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc.) also increased significantly. Due to cybercrime, the victims lose money, reputation, etc. and some of them commit suicide.

Cyberlaw ensures the use of computers and Internet by the people safely and legally. It consists of rules and regulations like Indian Penal Code (IPC) to stop crimes and for the smooth functions of Cyberworld. Two Acts are IT Act 2000 and IT Act Amended in 2008.

Question 6.
“Infomania has became a psychological problem”. Write your opinion.
Answer:
Info mania is the excessive desire(lnfatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Plus Two Computer Science ICT and Society Five Mark Questions and Answers

Question 1.
“Due to anonymous nature of Internet it is possible for the people to engage in variety of criminal activities.” Justify the statement with special reference to cyber crimes taking place against individual.
Answer:
Cyber crimes against individuals
i. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number etc) , Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

ii. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friend ship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

iii. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original ICT and Society one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias (Facebook, Twitter, etc), fake SMS, fake emails etc.

iv. Violation of privacy:
Trespassing into another person’s life and try to spoil life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

v. Dissemination of obscene material: With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

Question 2.
Explain different categories of cyber crimes in detail.
Answer:
Just like normal crimes( theft, trespassing private area, destroy, etc,) Cyber crimes(Virus, Trojan Horse, Phishing, Denial of Service, Pornography etc) also increased significantly . Due to cyber crime, the victims lose money, reputation,etc and some of them commit suicide.
Answer:
A. Cyber crimes against individuals:
1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number etc.), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friend ship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: F.ake accounts in Social Medias (Facebook, Twitter,etc), fake sms, fake emails, etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

B. Cyber crimes against property:
Stealing credit card details, hacking passwords of social media accounts or mail account or Net banking, uploading latest movies etc, are considered as cyber crimes against property.
1. Credit card fraud:
Stealing the details such as credit card number, company name, expiry date, cw number,password etc. and use these details to make payment for purchasing goods or transfer funds also.

2. Intellectual property theft:
The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself. Hence the revenue from the theatres are less significantly and undergoes huge loss.(Eg: Premam, Bahubali, etc) Copying a person’s creation and present as a new creation is called plagiarism. This can be identified some tools(programs) available in the Internet

3. Internet time theft:
This is deals with the misuse of WiFi Internet facility. If it is not protected by good password there is a chance of misuse our devices(Modem/Router) to access Internet without our consent by unauthorized persons. Hence our money and volume of data(Package) will lose and we may face the consequences if others make any crimes.

C. Cyber crimes against government:
The cyber crimes against Govt, websites is increased significantly. For example in 2015 the website of Registration Department of Kerala is hacked and destroys data from 2012 onwards.

1. Cyber terrorism:
It is deals with the attacks against very sensitive computer networks like computer controlled atomic energy power plants, air traffic controls, Gas line controls, telecom, Metro rail controls, Satellites, etc. This is a very serious matter and may lead to huge loss (money and life of citizens). So Govt, is very conscious and give tight security mechanism for their services.

2. Website defacement:
It means spoil or hacking websites and posting bad comments about the Govt.

3. Attacks against e-governance websites :
Its main target is a Web server. Due to this attack the Web server/ computer forced to restart and this results refusal of service to the genuine users. If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server. Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Question 3.
“For the implementation of e-Learning different tools.
Answer:
e Learning tools
1. Electronic books reader(eBooks): With the help of a tablet or portable computer or any other device we can read digital files by using an s/w is called electronic books reader.

2. e-text: The electronic format of textual data is called e-Text.

3. Online chat: Realtime exchange of text or audio or video messages between two or more person over the Internet.

4. e-Content: The data or information such as text, audio, video, presentations, images, animations, etc, are stored in electronic format.

5. Educational TV channels: TV channels dedicated only for the e-Learning purpose.
Eg. VICTERS (Virtual Classroom Technology on Edusat for Rural Schools OR Versatile ICT Enabled Resources for Students)

Students can Download Chapter 4 Graphs and Charts for Business Data Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Graphs and Charts for Business Data

Plus Two Accountancy Graphs and Charts for Business Data One Mark Questions and Answers

Question 1.
_____________ are the visual representation of numerical data
Answer:
Chart/ Graph

Question 2.
Chart / Graph has at least ____________ dimenstional relationship
(a) Two
(b) Three
(c) Four
(d) Five
Answer:
(a) Two

Question 3.
____________ chart is suitable for comparing multiple.
Answer:
Bar Chart

Question 4.
In column chart, the X-axis shows
(a) Value of each category
(b) Different categories
(c) Height of the chart
(d) Depth of the value
Answer:
(b) Different categories

Question 5.
________ Chart is similar to the column chart, with the difference being that the data series are displayed horizontally
(a) Line chart
(b) Pie chart
(c) Barchart
(d) Area chart
Answer:
(c) Bar chart

Question 6.
________ chart shows data changes for a certain period of time.
Answer:
Line chart

Question 7.
_______ chart contains only one data series
Answer:
Pie chart

Question 8.
_______chart shows values as circular sectors to the total circle
Answer:
Pie chart

Question 9.
Pie chart don’t have more than __________ categories.
(a) Ten
(b) Twenty Five
(c) Seven
(d) Three
Answer:
(c) Seven

Question 10.
____________ is a pictorial representation of data, which has at least two dimensional relationships.
(a) Graph
(b) Chart
(c) Diagram
(d) All the above
Answer:
(d) All the above

Question 11.
_________ Chart is used to compare values across categories.
(a) Column chart
(b) Line chart
(c) Pie chart
(d) Barchart
Answer:
(a) Column chart

Question 12.
_________ chart is used to display trends over time.
Answer:
Line chart

Question 13.
The entire chart including all elements is termed as ________
Answer:
Chart area

Question 14.
In 3D chart, the area is bounded by three axes ie _____, _____ & _____
Answer:
X, Y, Z

Question 15.
In ______ chart, both axes display values ie they have no category axis.
Answer:
XY Chart or Scatter diagram

Question 16.
_______ specifices the colour, symbol or pattern used to mark data series.
Answer:
Legends

Question 17.
The change the location of a chart, right click the chart and select.
(a) Chart Type
(b) Source Data
(c) Move here
(d) Chart Options
Answer:
(c) Move here

Question 18.
In 3D chart X, Y & Z axes are used to show
(a) Category, Value, Total
(b) Depth, Vertical, Horizontal
(c) Length, Breadth, Depth
(d) Category, Value, Series.
Answer:
(d) Category, Value, Series

Question 19.
Legand can be repositioned on the chart
(a) Anywhere
(b) On right side only
(c) On the corner only
(d) On the left side only
Answer:
(a) Anywhere

Question 20.
Which chart element details the data values and categories below the chart?
(a) Data table
(b) Data marker
(c) Data labels
(d) Datapoint
Answer:
(c) Data labels

Question 21.

Which type of chart is this?
Answer:
Radar chart

Question 22.
The intersection of both the axis (X-axis and Y-axis) is called __________ of the graph.
Answer:
Origin (0)

Question 23.
_________ Chart display in rings, where each ring represents a data series
Answer:
Doughnut chart.

Question 24.
Name the given chart

Answer:
Bar Chart

Question 25.
In _________ chart, the area below the plotted lines is solid
Answer:
Area chart.

Question 26.
Radar chart / Net chart is also known as ______
(a) Doughnut chart
(b) Pie chart
(c) Ara chart
(d) Star chart
Answer:
(d) Star chart

Question 27.
Which among the following is the special feature of 3D chart
(a) Chart area
(b) X & Y axes
(c) Chart wall
(d) Legend
Answer:
(c) Chart wall

Question 28.
Give a suitable name to the diagram

(a) Barchart
(b) Single line graph
(c) Pie chart
(d) Area Chart
Answer:
(c) Pie Chart

Plus Two Accountancy Graphs and Charts for Business Data Two Mark Questions and Answers

Question 1.
Name the different chart formats in Libre Office Calc
Answer:
Barchart, Column Chart, Pie chart, Line chart, Area chart, Doughnut chart, etc.

Question 2.
What is the importance of charts and graphs in business?
Answer:

1. Chart and graphs covey lots of business information in a visual format
2. Different business Data variables plotted in charts and graphs show the trend of the business in an easy way

Question 3.
Identify the type of chart

Answer:
Line chart

Question 4.
Give a short note on it.

1. Barchart
2. Pie chart

Answer:
1. Bar Chart
This type of chart shows a bar graph or column chart with horizontal bars. The Y-axis shows categories and the X-axis shows the value for each category. It is suitable for comparing multiple values.

2. Pie chart
A pie chart displays the contribution of each value to a total. It represents multiple subgroup of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

• Normal Pie chart
• Exploded Pie chart
• Doughnut chart or Donut chart
• Exploded Doughnut chart

Question 5.
What are the special features of graphs and charts?
Answer:

1. Graphs/charts are the pictorial representation of business data
2. A chart represents tabular numeric data
3. Dimensions in the data are often displayed on axes (X, Y, & Z)

Question 6.
Differentiate between Chart area and Chart wall?
Answer:
1. Chart area:
This is the total space that is enclosed by a chart. It is the background of the chart.

2. Chart wall:
In 2D chart, the wall or area is bounded by the X and Y-axis. In the 3D chart, the wall is bounded by three axes X, Y and Z

Question 7.
Quarterly sales of a business firm is used to create a bar graph. Identify the Data variables plotted on X and Y-axis
Answer:
X-axis – I^st Quarter, II^nd Quarter, III^rd Quarter, IV^th Quarter,
Y-axis – Sales in I^st Quarter, Sales in I^nd Quarter, Sales in III^rd Quarter, Sales in IV^th Quarter

Question 8.
Identify the type of chart

Answer:
(a) 2D Chart
(b) 3D Chart

Question 9.
What is the use of Auto shapes in LibreOffice Calc?
Answer:
Auto shapes tool bar allows drawing a number of geometrical shapes, arrows; flow chart elements, etc.

Question 10.
What is PIE chart? What are the specialties of PIE chart.
Answer:
Pie chart:
A pie chart displays the contribution of each value to a total. It represents multiple subgroup of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

1. Normal Pie chart
2. Exploded Pie chart
3. Doughnut chart or Donut chart
4. Exploded Doughnut chart

Question 11.
Choose the right statements from the following.

1. We can put on the right side of the origin positive values and on left side of the origin negative values of data on X-axis
2. The upward side of origin shows postiive values and downward side of the origin shows negative values of data on Y-axis.
3. We can put on the right side of the origin negative values and on left side of the origin positive values of data on X-axis.
4. The upward side of origin shows negative values and downward side of the origin shows positive values of data on Y-axis.

Answer:
Right statements a & b

Question 12.
Identify the type of given chart. List down its features.

Answer:

1. It is a Doughnut chart
2. Features of the doughnut chart
   • It displays data in rings
   • Each ring represents a data series
   • The first data series is displayed in the center of the chart

Question 13.
What is a 3-D chart?
Answer:
Charts can be prepared with three dimensional (3-D) effects. 3- D charts have a third axis. The third axis is called as Z-axis. So a 3-D chart has the fol¬lowing dimensions.

1. Horizontal axis – Indicate the category – known as X-axis
2. Vertical axis – Indicate the derived values – known as Y-axis
3. Depth axis – Indicate the series – known as Z-axis

Question 14
Differentiate between Data Marker and Data series.
Answer:
1. Data Marker:
Individual values plotted in a chart are called data marker or data point.

2. Data Series:
Data markers of the same colour or pattern is called data series.

Question 15.
Is there any difference between

1. A column chart and
2. A bar chart?

Substantiate your answer
Answer:
1. Column Chart:
It is the most commonly used chart type. It shows a bar chart or bar graph with vertical bars. The X-axis shows the categories and Y-axis shows the value for each category. Column chart are used to compare values across categories.

2. Bar Chart:
This type of chart shows a bar graph or column chart with horizontal bars. The Y-axis shows categories and the X-axis shows the value for each category. It is suitable for comparing multiple values.

Plus Two Accountancy Graphs and Charts for Business Data Three Mark Questions and Answers

Question 1.
Match the following

Answer:

Question 2.
What are the advantages of using Graph/ Chart?
Answer:
Advantages in using Graph/Chart:

1. It summarises a large data set in visual form
2. Charts or graphs can clarify trends better than do tables.
3. It helps to estimate key values at a glance
4. It shows each data category in a frequency distribution.
5. It permits a visual check of the accuracy and reasonableness of calculations
6. The charts and graphs allow the investigator to draw a valid conclusion.

Question 3.
What are the elements of a Chart/ Graph
Answer:

Question 4.
How to use word Art styles to format text.
Answer:

• Step 1: Click in the chart element that contains text to be changed.
• Step 2: Click on the format.
• Step 3: Click on word Art styles.
• Step 4: Choose suitable options related to text formating like text fill, text outlines, shadow, etc.

Plus Two Accountancy Graphs and Charts for Business Data Four Mark Questions and Answers

Question 1.
What are the difference between 2D charts and 3D charts
Answer:

Question 2.
List out the steps to Rotate a chart.
Answer:

• Step 1. Select the plot area of the chart.
• Step 2. Click on the format tab.
• Step 3. Click on format selection.
• Step 4. Click on 3D Rotation and type a value of angle between 0° to 360° and then click close
• Step 5. Click on the chart area of the chart and click on format tab.
• Step 6. Click on shape effects and then click on Bevel and select a bevel option.

Question 3.
What are the different types of charts?
Answer:

1. Column chart: column chart are used to compare values across categories
2. Line chart: Line charts are used to display trends over time
3. Pie chart: Pie charts display the contribution of each value to a total
4. Bar chart: Bar charts are best suited for comparing multiple values
5. Area chart: Area chart emphasis differences between several sets of data over a period of time.
6. Scatter chart: (XY chart) This chart compares pairs of values.
7. Radar chart: Display values relative to a centre point.
8. Doughnut chart: It shows the relationship of parts to a whole. This chart display data in rings, where each ring represents a data series.

1. Column Chart:
It is the most commonly used chart type. It shows a bar chart or bar graph with vertical bars. The X-axis shows the categories and Y-axis shows the value for each category. Column chart are used to compare values across categories.

2. Line Chart:
A line chart shows values in the Y-axis and categories in X-axis. The Y values of each data series is connected by a line. Line chart shows data changes for a certain period of time.

3. Pie chart:
A pie chart displays the contribution of each value to a total. It represents multiple subgroup of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

• Normal Pie chart
• Exploded Pie chart
• Doughnut chart or Donut chart
• Exploded Doughnut chart

4. Bar Chart:
This type of chart shows a bar graph or column chart with horizontal bars. The Y-axis shows categories and the X-axis shows the value for each category. It is suitable for comparing multiple values.

5. Area chart:
The chart shows values as points on the Y-axis. The X-axis shows categories. The Y values of each data series are connected by a line. The area between each two lines is filled with a colour.

6. Scatter chart:
Scatter chart is also known as XY chart. In this type of chart, both axes display values. This chart is used to show the relationship among two variables.

7. Radar chart:
It is also known as Net chart or Star chart. A radar chart has a separate axis for each category and the axes extend outward from the center of the chart. The value of each data point is plotted on the corresponding axis.

8. Doughnut chart:
Chart display in rings, where each ring represents a data series. The first data series is displayed in the centre of the chart.

Question 4.
Identify and explain the type of chart given below.
Answer:

1. This is scatter chart or XY chart.
2. Features:
   • Both axes display values (No category)
   • This chart is used to show the relationship among two variables
   • Generally this chart is used for scientific, statistical and engineering data

Question 5.
Match the following.

Answer:

• (a) – (ii);
• (b) – (iii);
• (c) – (iv);
• (d) – (i)

Question 6.
How can we change the format of a selected chart element?
Answer:

• Step 1. Click anywhere in the chart.
• Step 2. Click format
• Step 3. Click format selection
• Step 4. Select a category (Fill border, style, etc)
• Step 5. Select formatting options

Question 7.
List down any four advantages of charts/ Graphs
Answer:
Advantages in using Graph/Chart:

1. It summarises a large data set in visual form
2. Charts or graphs can clarify trends better than do tables.
3. It helps to estimate key values at a glance
4. It shows each data category in a frequency distribution.
5. It permits a visual check of the accuracy and reasonableness of calculations
6. The charts and graphs allow the investigator to draw a valid conclusion.

Question 8.
What are the features of Charts/ Graphs in Libre Office Calc?
Answer:

1. Chart is a graphical representation of data
2. They are visual representation of numerical data
3. Charts can be read more quickly than the raw data
4. A chart has at least two axes – X and Y

Plus Two Accountancy Graphs and Charts for Business Data Five Mark Questions and Answers

Question 1.
What Pie Chart? What are the different types of Pie Chart?
Answer:
Pie chart:
A pie chart displays the contribution of each value to a total. It represents multiple subgroups of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

1. Normal Pie chart
2. Exploded Pie chart
3. Doughnut chart or Donut chart
4. Exploded Doughnut chart

Question 2.
Name the different elements of given chart.

Answer:

• X-Axis Title
• Y-Axis Title
• Data label
• Main Title
• Legend
• X-Axis
• Y-Axis
• Data series

Question 3.
Write the steps of changing the chart type.
Answer:

1. First select the chart by double-clicking on it. The chart should now be surrounded by a gray bonder
2. Right-click on the chart and choose chart type.
3. Select the replacement chart type.
4. Click on [OK]

Question 4.
Write the steps for preparing a chart in Libre Office Calc.
Answer:

• Step 1: Enter the data in a worksheet with proper column and row titles
• Step 2: Select the range of data using the mouse
• Step 3: Click on Insert Tab → Object → Chart. Select a chart type from the “Choose a chart type” list in chart wizard window.
• Step 4: Naming chart, X-axis and Y-axis. Click on the chart → Right click → Insert titles (Names) → OK
• Step 5: Change the layout or styles of chart.
• Step 6: Show or hide a legend
• Step 7: Display or hide chart axes or gridlines
• Step 8: Move (resize) a chart
• Step 9: Save a chart

Plus Two Accountancy Graphs and Charts for Business Data Practical Lab Work Questions and Answers

Question 1.
Draw an Area Chart from the following

Procedure:
Step 1 – Open a new blanks worksheet in LibreOffice Calc

Step 2 – Enter the above data as follows.

Step 3 – Select the range A1: D6 which is to be shown in the chart.

Step 4 – Click on Insert menu → Click on Chart → Chart wizard → Select Area Chart → Finish
Output:

Question 2.
Quarterly sales of a product are given below. Draw a bar diagram/bar chart

Procedure:
Step 1 – Open a new blanks worksheet in LibreOffice Calc

Step 2 – Enter the above data as follows.

Step 3 – Select the range A1: B5 which is to be shown in the chart:

Step 4 – Click on Insert menu → Click on Chart → Chart wizard Click on Bar chart → Finish
Output:

Question 3.
Draw a 3D column chart from the following details.

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc.

Step 2 – Enter the following data in the respective cells.

Step 3-Select the range A1: B6, which is to be shown in the chart.

Step 4 – Click on Insert menu → Click on Chart → Chart wizard → Click on Column Chart → Tick 3D Look → Select Bar Charts → Finish.
Output:

Question 4.
The net profits of a firm for the last six years are given below. Draw a line chart.

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following data in the respective cells

Step 3 – Select the range A1: B7, which is to be shown in the chart

Step4- Click on Insert menu → Click on Chart → Chart Wizard → Click on Line Chart → Finish
Output:

Question 5.
Enter the following data into a LibreOffice Calc worksheet and draw a 3D Pie chart.

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc.

Step 2 – Enter the data in respective cells.

Step 3 – Select the range A2: B9, which is to be shown in the chart

Step 4 – Click on Insert menu → Click on Chart → Chart wizard → Click on Pie Chart Tick on 3D Look → Finish
Output:

Question 6.
Sales for the first six months in 2 years are given below. Draw a scatter chart in a LibreOffice Calc works sheet

Procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2 – Enter the data in the following cells.

Step 3 – Select the range A1: C7, which is to be shown in the chart

Step 4 – Click on → Insert menu Click on → Chart → Chart Wizard → Scatter Chart→ Finish
Output:

Question 7.
The production of different items in Oct. 2015 is listed below. Draw a Radar Chart

Procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2 – Enter the data in the following cells.

Step 3 – Select the range A1: C6, which is to be shown in the chart

Step 4 – Click on → Insert menu → Click on Chart → Chart Wizard → Click on Radar Chart → Finish
Output:

Question 8.
The following table shows the number of students passed in the higher secondary examination. Draw a doughnut chart.

Procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2- Enter the data in the following cells.

Step 3 – Select the range A1: D6, which is to be shown in the chart.

Step 4 – Click on → Insert menu → Click on Chart → Chart Wizard → Click on doughnut Chart → Finish.
Output:

Students can Download Chapter 6 Application of Derivatives Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives

Plus Two Maths Application of Derivatives Three Mark Questions and Answers

Question 1.
Find the equation of tangents and normals to the given curves x = cost, y = sin t at t = \(\frac{π}{4}\).
Answer:
Given; x = cost, y = sin t

Equation of tangent at t = \(\frac{π}{4}\) is;

Equation of normal at t = \(\frac{π}{4}\) is;

⇒ \(\sqrt{2}\)y + \(\sqrt{2}\)x = 0 ⇒ y + x = 0.

Question 2.
A ladder Sm long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the decreasing when the foot of the ladder is 4m away from the wall?
Answer:
From the figure we have;
x² + y² = 25 ____(1)

Differentiating w.r.t t;

From (1) when x = 4 ⇒ 16 + y² = 25 ⇒ y = 3
Given; \(\frac{d x}{d t}\) = 2cm/s = 0.02 m/s
(2) ⇒ 4(0.02) + 3 \(\frac{d x}{d t}\) = 0

Question 3.
Find the points on the curve y = x³, the tangents at which are inclined at an angle of 60° to x-axis?
Answer:
\(\frac{d y}{d x}\) = 3x²
Slope of the tangent = tan60°
i.e. 3x² = \(\sqrt{3}\)

Question 4.
Find the equation of the tangent to the parabola y² = 4x + 5 which is parallel to y = 2x + 7.
Answer:
y² = 4x + 5 _____(1)
2y \(\frac{d y}{d x}\) = 4
\(\frac{d y}{d x}\) = \(\frac{4}{2y}\) = \(\frac{2}{y}\)
Given tangent is parallel to y = 2x + 7
ie. Slope of the tangent is 2 ⇒ \(\frac{2}{y}\) = 2 ⇒ y = 1
∴ from (1) ⇒ 1 = 4x+ 5 ⇒ 4x = -4 ⇒ x = -1
So the point of contact is (-1, 1).
∴ Equation of tangent is
y -1 = 2(x + 1) ⇒ y = 2x + 3.

Question 5.
Find the intervals in which the function f given f(x) = 2x² – 3x is

1. Strictly increasing.
2. Strictly decreasing.

Answer:
Given; f(x) = 2x² – 3x ⇒ f'(x) = 4x – 3
For turning points; f'(x) = 0
⇒ 4x – 3 = 0 ⇒ x = \(\frac{3}{4}\)
The intervals are \(\left(-\infty, \frac{3}{4}\right),\left(\frac{3}{4}, \infty\right)\)
f'(0) = – 3 < 0
∴ Strictly decreasing in \(\left(-\infty, \frac{3}{4}\right)\)
f'(1) = 1 > 0
∴ Strictly increasing in \(\left(\frac{3}{4}, \infty\right)\).

Question 6.
Find the intervals in which the function f(x) = (x + 1)³ (x – 3)³ strictly increasing or decreasing.
Answer:
Given; f(x) = (x + 1)³ (x – 3)³
⇒ f'(x) = (x + 1)³ 3(x – 3)² + (x – 3)³3(x + 1)²
= 3(x + 1)²(x – 3)²(x + 1 + x – 3)
= 3(x + 1)²(x – 3)²(2x – 2)
= 6(x +1)² (x – 3)² (x -1)
⇒ 6(x +1)² (x – 3)² (x – 1) = 0
⇒ x = -1, 1, 3
The intervals are
(-∞, -1), (-1, 1), (1, 3), (3, ∞)
f'(-2) = (-2 – 1) < 0
∴ Strictly decreasing in (-∞, -1)
f'(0) = (0 – 1) < 0
∴ Strictly decreasing in (-1, 1)
f'(2) = (2 – 1) > 0
∴ Strictly increasing in (1, 3)
f'(4) = (4 – 1) > 0
∴ Strictly increasing in (3, ∞).

Question 7.
Find the intervals in which the function f(x) = x + \(\frac{1}{x}\) strictly increasing or decreasing.
Answer:

⇒ x = ±1
The intervals are (-∞, -1), (-1, 1), (1, ∞)
f'(-2) > 0
∴ Strictly increasing in (-∞, -1)
f'(0) < 0
∴ Strictly decreasing in (-1, 1)
f'(2) > 0
∴ Strictly increasing in (1, ∞).

Question 8.
Determine whether the f(x) = x² function is strictly monotonic on the indicated interval.

1. (-1, 1)
2. (-1, 0)
3. (0, 1)

Answer:
f(x) = x²
⇒ f'(x) = 2x
⇒ f'(x) = 0 ⇒ 2x = 0 ⇒ x = 0
This turning point divides the domain into the intervals (-∞, 0); (0, ∞).

1. Interval (-1,1) f'(x) < 0 and f'(x) > 0. So f(x) is not monotonic.
2. Interval (-1,0), f'(x) < 0. ∴ f(x) is strictly monotonic.
3. Interval (0, 1) f'(x) > 0 and f(x) is strictly monotonic.

Question 9.
Determine whether the f(x) = x³ – x function is strictly monotonic on the indicated interval.

1. (-1, 0)
2. (-1, -1/2)
3. (-1, 1)

Answer:
(x) = x³ -x ⇒ f'(x) = 3x² – 1
⇒ f'(x) = 0 ⇒ 3x² – 1 = 0 ⇒ x = ±\(\frac{1}{\sqrt{3}}\)
This turning point divides the domain into the intervals (-∞, \(\frac{1}{\sqrt{3}}\)); (-\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)); (\(\frac{1}{\sqrt{3}}\), ∞).

1. Interval (-1, 0), f'(x) changes sign. So not monotonic.
2. Interval (-1, -1/2), f'(x) > 0 strictly monotonic.
3. lnterval(-1, 1) not monotonic

Question 10.
Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 1%.
Answer:
We have; V = x³ and ∆x = 1% of x= 0.01x
dV = \(\frac{d V}{d x}\) ∆x = 3x²∆x
= 3x² × 0.01x = 0.03x³ = 0.03V
⇒ \(\frac{d V}{V}\) = 0.03
Therefore 3% is the approximate increase in volume.

Question 11.
If the radius of a sphere is measured as 7m with an error of 0.02m then find the approximate error in calculating its volume.
Answer:
Let r be the radius of the sphere and ∆r be the error in measuring the radius then r =7m and ∆r = 0.02 m
We have; V = \(\frac{4}{3}\) πr³
dV = \(\frac{d V}{dr}\) ∆r = \(\frac{4}{3}\) π3r² × ∆r
= 4π(7)² × 0.02 = 3.92 π m³.

Question 12.
The length of a rectangle is decreasing at the rate of 5 cm/min and the width is increasing at the rate of 4cm/min. When length is 8 cm and width is 6 cm, find the rate of change of its area.
Answer:
Let length = x and width = y

Question 13.
Find the equation of tangents and normals to the given curves y= x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
Answer:
Given; y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
⇒ \(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
Slope = \(\left(\frac{d y}{d x}\right)_{x=0}\) = -10
Equation of tangent at (0, 5) is;
y – 5 =(-10)(x – 0)
⇒ y – 5 = -10x ⇒ 10x + y – 5 = 0
Equation of normal at (0, 5) is;
y – 5 = \(\frac{1}{10}\)(x – 0)
⇒ 10y – 50 = x ⇒ x – 10y + 50 = 0.

Question 14.
Find the equation of tangents and normals to the given curves y = x³ at (1, 1)
Answer:
Given; y = x³
⇒ \(\frac{d y}{d x}\) = 3x²
Slope = \(\left(\frac{d y}{d x}\right)_{x=1}\) = 3
Equation of tangent at (1, 1) is; y -1 = (3)(x – 1)
⇒ y – 1 = 3x – 3 ⇒ 3x – y – 2 = 0
Equation of normal at (1, 1) is;
y – 1 = \(-\frac{1}{3}\)(x – 1)
⇒ 3y – 3 = -x + 1 ⇒ x+ 3y – 4 = 0.

Question 15.
The volume of a cube is increasing at the rate of 8cm³/s. How fast is the surface area increasing when the length of an edge is 12cm.
Answer:
Let V be the volume of the cube of side x.
We have volume = V = x³
Rate of change of volume with respect to time ‘t’ is;
ie; differentiating w.r.t t; \(\frac{d V}{d t}\) = 3x²\(\frac{d x}{d t}\)
Given; \(\frac{d V}{d t}\) = 8 and x = 12 ⇒ 8 = 3(12)²\(\frac{d x}{d t}\)

Now let Surface area = S = 6x²
Differentiating w.r.t t;

Question 16.
Find the intervals in which the function f(x) = -2x³ – 9x² – 12x + 1 strictly increasing or decreasing.
Answer:
Given; f(x) = -2x³ – 9x² – 12x + 1
⇒ f'(x) = -6x² – 18x – 12
= – 6(x² + 3x + 2)
= – 6(x + 2)(x +1)
⇒ f'(x) = 0 ⇒ -6(x + 2)(x +1) = 0
⇒ x = -2, -1
The intervals are (-∞, -2),(- 2, -1),(-1, ∞)
f'(-3) = -(-3 + 2)(-3 + 1) < 0
∴ Strictly decreasing in (-∞, -2)
f'(-1.5) = -(-1.5 + 2)(-1.5 + 1) > 0
∴ Strictly increasing in (- 2, -1)
f'(0) = -(0 + 2)(0 + 1) < 0
Strictly decreasing 1n(-1, ∞).

Question 17.
Find the local maxima and minima of the following functions. Also find the local maximum and minimum values. (each question carry 3 score)

1. f(x) = sin x + cosx, 0 < x < \(\frac{\pi}{2}\)
2. f(x) = x³ – 3x
3. f(x) = x³ – 6x² + 9x + 15
4. g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\), x > 0
5. g(x) = \(\frac{1}{x^{2}+2}\)

Answer:
1. Given; f(x) = sinx + cosx
⇒ f'(x) = cosx – sinx
For turning point f'(x) = 0
⇒ cosx – sinx = 0
⇒ cosx = sinx
⇒ x = \(\frac{\pi}{4}\)
f”(x) = -sin x – cosx
⇒ f (\(\frac{\pi}{4}\)) = -sin\(\frac{\pi}{4}\) – cos\(\frac{\pi}{4}\) < 0
Hence f(x) has a local maximum at x = \(\frac{\pi}{4}\) and local maximum value is

2. Given; f(x) = x³ – 3x
⇒ f'(x) = 3x² – 3
For turning point f'(x) = 0
⇒ 3x² – 3 = 0
⇒ x = ±1
f”(x) = 6x
When x = -1
⇒ f”(-1) = -6 < 0
Hence f(x) has a local maximum at x = -1 and local maximum value is
f(-1) = (-1)³ – 3(-1) = -1 + 3 = 2
When x = 1
⇒ f”(1) = 6 > 0
Hence f(x) has a local minimum at x = 1 and local minimum value is
f(1) = (1)³ – 3(1) = 1 – 3 = -2.

3. Given; f(x) = x³ – 6x² + 9x + 15
⇒ f'(x) = 3x² – 12x + 9
For turning point f'(x) = 0
⇒ 3x² – 12x + 9 = 0 ⇒ 3(x² – 4x + 3) = 0
⇒ 3(x – 1)(x – 3) = 0 ⇒ x = 1, 3
f”(x) = 6x – 12
When x = 1
⇒ f”( 1) = 6 – 12 < 0
Hence f(x) has a local maximum at x = 1 and local maximum value is
f(1) = (1)³ – 6(1)² + 9(1) + 15 = 19
When x = 3
⇒ f”(3) = 6(3) – 12 > 0
Hence f(x) has a local minimum at x = 3 and local minimum value is
f(3) = (3)³ – 6(3)² + 9(3) + 15 = 15.

4. Given; g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\)
⇒ g'(x) = \(\frac{1}{2}\) – \(\frac{2}{x^{2}}\)
For turning point g'(x) = 0

Since x > 0, the acceptable value of x = 2

Hence g(x) has a local maximum at x = 2 and local maximum value is g(2) = \(\frac{2}{2}\) + \(\frac{2}{2}\) = 2

5. Given; g(x) = \(\frac{1}{x^{2}+2}\)

For turning point g'(x) = 0


Hence g(x) has a local maximum at x = 2 and maximum value is g(2) = \(\frac{1}{0+2}=\frac{1}{2}\).

Question 18.
Find the absolute maximum value and minimum value of the following functions.

1. f(x) = x³, x ∈ [-2, 2]
2. f(x) = 4x – \(\frac{x^{2}}{2}\), x ∈ \(\left[-2, \frac{9}{2}\right]\)

Answer:
1. Given; f(x) = x³ ⇒ f'(x) = 3x²
For turning point f'(x) = 0 ⇒ 3x² = 0 ⇒ x = 0
f(- 2) = (-2 )³ = -8
f( 2) = (2)³ = 8
f(0) = (0)³ = 0
Absolute maximum = max{-8, 8, 0} = 8
Absolute minimum = min {-8, 8, 0} = – 8

2. Given; f(x) = 4x – \(\frac{x^{2}}{2}\) ⇒ f'(x) = 4 – x
For turning point f'(x) = 0 ⇒ 4 – x = 0
⇒ x = 4

Absolute maximum = max{-10, 8, 7.875} = 8
Absolute minimum = min {-10, 8, 7.87} = -10.

Question 19.
A television camera at ground level is filming the lift-off of a space shuttle that is rising vertically according to position equation S = 50t².
The camera is 2000 feet from the launch pad. Find the rate of change in the angle of elevation of the camera 10 seconds after lift-off.
Answer:

Question 20.
Determine whether the f(x) = sinx function is strictly monotonic on the indicated interval.

1. (0, 2π)
2. (0, π)
3. (-π/2, π/2)

Answer:
f(x) = sinx ⇒ f'(x) = cosx changes sign.

1. Interval (0, 2π). ∴ f(x) is not monotonic.
2. f'(x) changes sign in (0, π) not monotonic.
3. f'(x) > 0 in (-π/2, π/2), ie. f(x) is strictly monotonic.

Question 21.
Find the approximate change in the Surface Area of a cube of side x meters caused by decreasing the side by 1%.
Answer:
We have;
S = 6x² and ∆x = 1% of x = -0.01x
dS = \(\frac{d S}{d x}\) ∆x = 6 × 2x × ∆x
= 6 × 2x × -0.01x = -0.02 × 6x² = -0.02S
⇒ \(\frac{d S}{S}\) = -0.02
Therefore 2% is the approximate decrease in surface area.

Plus Two Maths Application of Derivatives Four Mark Questions and Answers

Question 1.
The length ‘x’ of a rectangle is decreasing at the rate of 2 cm/s and the width ‘y’ is increasing at the rate of 2 cm/s.

1. Find the rate of change of Perimeter.
2. Find \(\frac{d A}{dt}\) when x = 12 cm and y = 5 cm.

Answer:
Since the length ‘x’ is decreasing and the width ‘y’ is increasing, we have \(\frac{d x}{dt}\) = -2 cm/s and \(\frac{d y}{dt}\)
= 2 cm/sec.
1. The Perimeter ‘P’ of the rectangle is given by
P = 2 (x + y)

2. The area ‘A’ of the rectangle ‘A’ is given by
A = x.y

= 12(2) + 5(-2)
= 24 – 10 = 14 cm2/s.

Question 2.
Find the equation of all lines having slope -1. Which are tangents to the curve?
y = \(\frac{1}{x-1}\), x ≠ 1
Answer:

⇒ x² – 2x = 0 ⇒ x(x – 2) = 0 ⇒ x = 0, x = 2
At x = 0, y = -1
Equation of tangent at (0, -1) is;
At x = 2, y = \(\frac{1}{2-1}\) = 1
Equation of tangent at (2, 1) is; y – 1 = -1(x – 2)
⇒ y – 1 = -x + 2 ⇒ x + y – 3 = 0.

Question 3.
Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangent are parallel to x-axis.
Answer:
Given; x² + y² – 2x – 3 = 0
Differentiating with respect to x;

Since the tangent is parallel to x-axis \(\frac{d y}{d x}\) = 0
\(\frac{1-x}{y}\) = 0 ⇒ x = 1
We have; (1)² + y² – 2(1) – 3 = 0
⇒ y² = 4 ⇒ y = ±2
Hence the points are (1, 2), (1, -2).

Question 4.
Find the equation of the tangent to the curve y = \(\sqrt{3 x-2}\) which is parallel to the line 4x – 2y + 5 = 0.
Answer:
Slope of the line 4x – 2y + 5 = 0 is 2.

acceptable since y is positive.
Hence the point is \(\left(\frac{41}{48}, \frac{3}{4}\right)\)
Equation of tangent is;


⇒ 6(4y – 3) = (48x – 41)
⇒ 24y – 18 = 48x – 41
⇒ 48x – 24y – 23 = 0.

Question 5.
Prove that the curve x = y² and xy = k cut at right angles, if 8k² = 1.
Answer:
x = y² ___(1)
⇒ 1 = 2y \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2y}\)
xy = 2k ___(2)
⇒ x \(\frac{d y}{d x}\) + y.1 = 0 ⇒ \(\frac{d y}{d x}\) = \(-\frac{y}{x}\)
The product of the slopes will be – 1.

Question 6.
The gradient at any point (x, y) of a curve is 3x² – 12 and the curve through the point (2, -7).

1. Find the equation of the tangent at the point ( 2, -7 ). (2)
2. Find the equation to the curve. (2)

Answer:
1. Given gradient as 3x² – 12 ⇒ \(\frac{d y}{d x}\) = 3x² – 12
Slope at (2, -7) is given by
\(\left(\frac{d y}{d x}\right)_{x=2}\) = 3(2)² – 12 = 0
Since slope is zero, the tangent is parallel to x – axis.
Here y = – 7 is the equation of the tangent at (2, -7).

2. Given, \(\frac{d y}{d x}\) = 3x² – 12
⇒ ∫dy = ∫(3x² – 12 )dx
y = 3\(\frac{x^{3}}{3}\) – 12x + c ⇒ y = x³ – 12x + c ____(1)
Given (2, -7) is a point on the curve.
(1) ⇒ -7 = (2)³ – 12(2) + c ⇒ -7 = 8 – 24 + c ⇒ c = 9
∴ Curve is y = x³ – 12x + 9.

Question 7.
Consider the curve x^2/3 + y^2/3 = 2

1. Find the slope of the tangent to the curve at the point (1, 1). (2)
2. Find the equation of the normal at the point (1, 1). (2)

Answer:
1. Given, x^2/3 + y^2/3 = 2,
Differentiating w.r.t. x,

⇒ slope of tangent = – 1.

2. Slope of normal = \(-\frac{1}{-1}\) = 1.
Equation of the normal is
y – 1 = 1(x – 1) ⇒ y – x = 0.

Question 8.
Find the intervals in which the function f given by f(x) = 2x³ – 3x² – 36x + 7 is

1. Strictly increasing. (2)
2. Strictly decreasing. (2)

Answer:
Given; f(x) = 2x³ – 3x² – 36x + 7
⇒ f'(x) = 6x² – 6x – 36
f'(x) = 0 ⇒ 6x² – 6x – 36 = 0
⇒ 6(x² – x – 6) = 0
⇒ 6(x + 2)(x – 3) = 0 ⇒ x = -2, 3
The intervals are (-∞, -2),(- 2, 3), (3, ∞)
f'(-3) = 6(-3 + 2)(-3 – 3) > 0.
∴ Strictly increasing in (-∞, -2).
f'(0) = 6(2)(-3) < 0.
∴ Strictly decreasing in (- 2, 3).
f'(4) = 6(4 + 2)(4 – 3) > 0
∴ Strictly increasing in (3, ∞).

Question 9.
Use differentials to find the approximate value of \(\sqrt{0.6}\) up to 3 places of decimals.
Answer:
Take y = \(\sqrt{x}\), let x = 0.64 and ∆x = -0.04
Then; f(x) = y = \(\sqrt{x}\)
f(x + ∆x) = y + ∆y


(1) ⇒ \(\sqrt{0.6}\) = 0.8 – 0.025 = 0.775.

Question 10.
Use differentials to find the approximate value of (0.999)^{\(\frac{1}{10}\)} up to 3 places of decimals.
Answer:
Take = x^{\(\frac{1}{10}\)}, let x = 1 and ∆x = -0.001
Then; f(x) = y
f(x + ∆x) = y + ∆y
(0.999)^{\(\frac{1}{10}\)} = x^{\(\frac{1}{10}\)} + ∆y
(0.999)^{\(\frac{1}{10}\)} = (1)^{\(\frac{1}{10}\)} + ∆y
⇒ (0.999)^{\(\frac{1}{10}\)} = 1 + ∆y ______(1)

(1) ⇒ (0.999)^{\(\frac{1}{10}\)} = 1 – 0.0001 = 0.9999.

Question 11.
Use differentials to find the approximate value of (15)^{\(\frac{1}{4}\)} up to 3 places of decimals.
Answer:
Takey = x^{\(\frac{1}{4}\)}, let x = 16 and ∆x = -1
Then; f(x) = y
f(x + ∆x) = y + ∆y
(15)^{\(\frac{1}{4}\)} = x^{\(\frac{1}{4}\)} + ∆y
(15)^{\(\frac{1}{4}\)} = 16^{\(\frac{1}{4}\)} + ∆y
(15)^{\(\frac{1}{4}\)} = 2 + ∆y ____(1)
⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x

Question 12.
Use differentials to find the approximate value of (26.57)^{\(\frac{1}{3}\)} up to 3 places of decimals.
Answer:
Take y = x^{\(\frac{1}{3}\)}, let x = 27 and ∆x = -0.43
Then; f(x) = y
f(x + ∆x) = y + ∆y
f(x + ∆x) = f(x) + ∆y
(26.57)^{\(\frac{1}{3}\)} = 27^{\(\frac{1}{3}\)} + ∆y .
⇒ (26.57)^{\(\frac{1}{3}\)} = 3 + ∆y ____(1)

(1) ⇒ (26.57)^{\(\frac{1}{3}\)} = 3 – 0.016 = 2.984.

Question 13.
Find the approximate value of f(5.001) where f(x) = x³ – 7x² + 15
Answer:
Let x = 5 and ∆x = 0.001
Then; f(x) = y
f(x + ∆x) = y + ∆y
f (5.001) = f(x) + ∆y
f(5.001) = f(5) + ∆y
f(5.001) = 5³ – 7(5)² + 15 + ∆y
⇒ f(5.001) = -35 + ∆y …
⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x ⇒ dy = (3x² – 14x) × 0.001
= (3(5)² – 14(5)) × 0.001 = (75-70)0.001 = 0.005
(1) ⇒ f(5.001) = -35 + 0.005 = -34.995.

Question 14.
Find the approximate value of f(3.02) where f(x) = 3x² + 5x + 3
Answer:
Let x = 3 and ∆x = 0.02
Then; f(x) = y
f(x + ∆x) = y + ∆y
f(3.02) = f(x) + ∆y
f(3.02) = f(3) + ∆y
f(3.02) = 3(3)² + 5(2) + 3 + ∆y
⇒ f(3.02) = 45 + ∆y ____(1)
⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x ⇒ dy = (6x + 5) × 0.02
= (6(3) + 5) × 0.02 = (18 + 5)0.02 = 0.46
(1) ⇒ f(3.02) = 45 + 0.46 = 45.46.

Question 15.
Consider the function y = f\(\sqrt{x}\)

1. If x = 0.0036 and ∆x = 0.0001 find ∆y. (3)
2. Hence approximate \(\sqrt{.0037}\) using differentials. (1)

Answer:
1. Let x = .0036, ∆x = 0.0001

2. (1) ⇒ \(\sqrt{.0037}\) = .000833 + .06 = .060833.

Question 16.
Find the approximate value of \(\sqrt[3]{124}\).
Answer:
f(x) = \(\sqrt[3]{x}\) = x^1/3 ⇒ f¹(x) = 1/3x^-2/3 = \(\frac{1}{3 x^{2 / 3}}\)
Let x = 125, ∆x = -1
Then; f(x) = y

Question 17.
Find two numbers x and y such that their sum is 35 and the product is x² y⁵ a maximum.
Answer:
Given; x + y = 35 ⇒ y = 35 – x
P = x² y⁵ ⇒ P = x²(35 – x)⁵
⇒ p’ = 2x(3 5 – x)⁵ + x² 5(35 – x)⁴(-1)
⇒ P’ = x(35 – x)⁴[2(35- x) – 5x]
⇒ p’ = x(35 – x)⁴[70 – 7x]
⇒ p’ = 7x(35 – x)⁴[10 – x]
⇒ p” = 7[x(3 5 – x)⁴ [-1] + x(10 – x)4(35 – x)³ (-1) + (35 – x)⁴(10 – x)]
For turning points P’ = 0
⇒ 7x(35 – x)⁴[10 – x] = 0
⇒ x = 0, 35, 10
x = 0, 35 can be rejected since correspondingly y will be y = 35, 0
⇒ P” = 7[10(35 – 10)⁴[-1]] < 0
Therefore maximum at x = 10
Thus the numbers are 10 and 35 – 10= 10.

Question 18.
Using differentials, find the approximate value of (63)^1/3.
Answer:
Take y = x^{\(\frac{1}{3}\)}, let x = 64 and ∆x = 1
Then; f(x) = y
f(x + ∆x) = y + ∆y

Question 19.

1. Find the point on the curve y = x³ – 10x + 8 at which the tangent is parallel to the line y = 2x + 1. (2)
2. Is the given line tangent to the curve? Why?

Answer:
1. \(\frac{d y}{d x}\) = 3x² – 10
Slope of the line y = 2x +1 is 2
⇒ 3x² – 10 = 2 ⇒ 3x² = 12 ⇒ x = ±2
When x = 2
y = 2³ – 10 × 2 + 8 = 8 – 20 + 8 = -4
When x = – 2
y = (-2)³ -10 × (-2) + 8 = -8 + 20 + 8 = 20
Therefore the points are (2, -4); (-2, 20)

2. No. Since (2, -4); (-2, 20) does not satisfies the equation y = 2x +1.

Question 20.
Suppose that a spherical balloon is inflated and it has volume ‘v’ and radius ‘r’ at time ‘t’.

1. If the balloon is inflated by pumping 900c.c. of gas per second. Find the rate a which the radius of the balloon is increasing when the radius is 15 cm. (2)
2. Find the rate of change of its surface at the instant when it radius is 15 cm. (2)

Answer:
1. Let V be the volume of the sphere of radius r.
V = \(\frac{4}{3}\) πr³, given; \(\frac{d V}{d t}\) = 900, r = 15
Differentiating w.r.t t,

2. Let ‘s’ denote the surface area of the balloon, then
S = 4πr²
Differentiating, \(\frac{d s}{d t}\) = 4π.2r.\(\frac{d r}{d t}\) =8.π r .\(\frac{d r}{d t}\)
= 8π × 15 × \(\frac{1}{\pi}\) = 120 cm²/sec.

Question 21.
Use differentials to find the approximate value of (0.009)^{\(\frac{1}{3}\)} up to 3 places of decimals.
Answer:
Take y = x^{\(\frac{1}{3}\)}, let x = 0.008 and ∆x = 0.001
Then; F(x) = Y
f(x + ∆x) = y + ∆y

Question 22.
Find the approximate value of \(\sqrt{401}\).
Answer:
f(x) = \(\sqrt{x}\) = x^1/2
f'(x) = \(\frac{1}{2 \sqrt{x}}\)
Let x = 400 ∆x = 1
f(x) = y = \(\sqrt{x}\)
f(x + ∆x) = y + ∆y

Question 23.
Consider y = \(\frac{\log x}{x}\), in (0, ∞)

1. Find the value of x at which \(\frac{d y}{d x}\) = 0 (2)
2. Find the maximum value.

Answer:
1.

2.

∴ y is maximum when x = e.
The maximum value is \(\frac{1}{e}\).

Question 24.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.
Answer:
Slope of the line y = x – 11 is 1.
Given; y = x³ – 11x + 5 ⇒ \(\frac{d y}{d x}\) = 3x² – 11 = 1
⇒ 3x² = 12 ⇒ x = ±2
At x = 2, ⇒ y = x – 11 = 2- 11 = -9
⇒ (2, -9)
At x = -2, ⇒ y = x – 11 = -2 – 11 = -13
⇒ (-2, -13)
But the point (-2, -13) do not lie on the curve, hence the point is (2, -9).

Question 25.
Consider the curve y = x² – 2x + 7

1. Find the slope of the tangent of the curve at x = 2. (2)
2. Write down the equation of the tangent at x =2. (2)

Answer:
1. Given, y = x² – 2x + 7 ⇒ y’ = 2x – 2
(y’)_x=2 = 2(2) – 2 = 2.

2. At x = 2 , y = 2² – 2(2) + 7 = 7.
Equation of the tangent at (2, 7) is
y – 7 = 2(x – 2) ⇒ 2x – y + 3 = 0.

Question 26.
Find the absolute maximum value and minimum value of the following functions.

1. f(x) = 2x³ – 15x² + 36x + 1, x ∈ [1, 5]
2. f(x) = 12x^{\(\frac{4}{3}\)} – 6x^{\(\frac{1}{3}\)}, x ∈ [-1, 1]

Answer:
1. Given; f(x) = 2x³ – 15x² + 36x + 1, x ∈ [1, 5]
⇒ f'(x) = 6x² – 30x + 36
For turning point f'(x) = 0 ⇒ 6x² – 30x + 36 = 0
⇒ x² – 5x + 6 = 0 ⇒ (x -3)(x – 2) = 0
⇒ x = 3, 2
f(1) = 2(1)³ – 15(1)² + 36(1) + 1 = 24
f(2) = 2(2)³ – 15(2)² + 36(2) + 1 = 29
f(3) = 2(3)³ – 15(3)² + 36(3) + 1 = 28
f(5) = 2(5)³ – 15(5)² + 36(5) + 1 = 56
Absolute maximum = max {24, 29, 28, 56} = 56
Absolute minimum = min {24, 29, 28, 56} = 24

2. Given;

f'(x) = 0 at x = \(\frac{1}{8}\) and f'(x) is not defined at x = 0. Therefore;

Absolute maximum = max {18, 0, 6, \(-\frac{9}{4}\)} = 18
Absolute minimum = min {18, 0, 6, \(-\frac{9}{4}\)} = \(-\frac{9}{4}\).

Question 27.
Consider the function y = x³ – 6x² + 3x – 1

1. Find the slope at x= -1. (1)
2. Find the minimum gradient of the above curve. (3)

Answer:
1. Given,
y = x³ – 6x² + 3x – 1 ⇒ y’ = 3x² – 12x + 3
Gradient at (x = -1) = (y’)_x=1 = 3(-1)² – 12(-1) + 3 = 18.

2. Now for minimum gradient we have to apply maxima – minima condition to the function y’ .ie, y” = 6x – 12 , for turning points of y’ is given by y” = 0.
Therefore, 6x – 12 = 0 ⇒ x = 2
Now, y”’ = 6 > 0
∴ y’ is maximum at x = 2.
Minimum gradient at (x = 2) is
= 12 – 24 + 3 = – 9.

Plus Two Maths Application of Derivatives Six Mark Questions and Answers

Question 1.
A curve passes through the origin, and its gradient function is 2x – \(\frac{x^{2}}{2}\)

1. Find its y coordinate when x= 2. (4)
2. Find the equation of the tangent at x= 2. (2)

Answer:
Given;

Integrating we have; ∫dy = ∫(2x – \(\frac{x^{2}}{2}\))dx
⇒ y = x² – \(\frac{x^{3}}{6}\) + c ___(1)
Since the curve passes through (0, 0)
(1) ⇒ 0 = 0 + c ⇒ c = 0
∴ Equation of the curve is y = x² – \(\frac{x^{3}}{6}\)
When x = 2 ⇒ y = 2² – \(\frac{2^{3}}{6}\) = \(\frac{8}{3}\)
∴ coordinate is (2, \(\frac{8}{3}\)).

2. Slope at (2, \(\frac{8}{3}\)) = 2 × 2 – \(\frac{2^{2}}{2}\) = 2
∴ Equation of the tangent at (2, \(\frac{8}{3}\)) is given by
y – \(\frac{8}{3}\) = 2(x – 2) ⇒ 3y – 8 = 6x – 12 ⇒ 3y = 6x – 4.

Question 2.
(i) Choose the correct answer from the bracket. The slope of the tangent to the curve y = x³ – 2x + 3 at x = 1 is ____(1)
(a) 0
(b) 1
(c) 2
(d) 3
(ii) Find points on the curve \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1 at which the tangents are (2)
(a) Parallel to x-axis
(b) parallel to y – axis.
(iii) Use differential to approximate \(\sqrt{25.6}\). (3)
Answer:
(i) (b) 1, Since

(ii)

(a) \(\frac{d y}{d x}\) = 0, since tangents are parallel to x- axis.
\(\frac{-9x}{25}\) = 0, x = 0 ∴ y = ± 3;
The points are (0, 3) and (0, -3)

(b) \(\frac{-25 y}{9 x}\) = 0, since tangents are parallel to y-axis, slope of normal = 0; y = o
∴ x = ± 5
The points are (5, 0) and (-5, 0)

(iii) Take y = \(\sqrt{x}\) , let x = 25 and ∆x = 0.6
Then; f(x) = y = \(\sqrt{x}\)
f(x + ∆x) = y + ∆y

Question 3.
Let x and y be the length and breadth of the rectangle ABCD in a circle having radius r. Let ∠CAB = θ (Ref. figure). If ∆ represent area of the rectangle and r is a constant.

1. Write ∆ in terms of r and θ. (2)
2. Find \(\frac{d \Delta}{d \theta}\) and \(\frac{d^{2} \Delta}{d \theta^{2}}\). (1)
3. Hence find the maximum value of ∆. (2)
4. Show that the rectangle of maximum area that can be inscribed in a circle of radius r is a square of side \(\sqrt{2} r\). (1)

Answer:
1. Area of the rectangle is ∆ = xy
From the figure y = 2r sinθ, x = 2r cosθ
∆ = xy = 4r²sinθcosθ = 2r²sin2θ

2. \(\frac{d \Delta}{d \theta}\) = 4r² cos2θ ⇒ \(\frac{d^{2} \Delta}{d \theta^{2}}\) = -8r²sin2θ

3. For turning points

Therefore local maximum at θ = \(\frac{\pi}{4}\)

4. Then;

Hence the rectangle becomes a square.

Question 4.
The second derivative of the equation of a curve is given by the equation x \(\frac{d^{2} y}{d x^{2}}\) = 1, given y = 1, \(\frac{d y}{d x}\) = 0 when x= 1.

1. Find the slope at x = e. (2)
2. Find the equation of the curve. (2)
3. Find the equation of the normal at x= e. (2)

Answer:
1. Given;

Integrating we get,

∴ Slope of the curve at x = e is given by
\(\left(\frac{d y}{d x}\right)_{x=e}\) = loge = 1.

2. We have,
\(\frac{d y}{d x}\) = logx, ⇒ dy = logxdx
Integrating we get,
∫dy = ∫logx dx ⇒ y = logx.x – ∫\(\frac{1}{x}\).x dx + c₂
⇒ y = xlogx – x + c₂ ____(2)
Given; y = 1 when x = 1
(2) ⇒ 1 = 1log1 – 1 + c₂ ⇒ 1 = 0 – 1 + c₂ ⇒ c₂ = 2
Therefore the equation of the curve is
y = xlogx – x + 2

3. We have, y = xlogx – x + 2
When x = e
⇒ y = e log e – e + 2 ⇒ y = e – e + 2 = 2
So we have to find the slope at (e, 2),
We know; \(\left(\frac{d y}{d x}\right)_{x=e}\) = log e = 1
∴ Slope of the normal at (e, 2)= -1
∴ Equation of the normal at (e, 2) is.
y – 2 = (-1) (x – e)
y – 2 = – x + e ⇒ y + x = e + 2.

Question 5.
The given figure represents a cylinder Inscribed in a sphere.

1. Find an expression for the volume V of the cylinder. (2)
2. Find the height of the cylinder when its volume V is maximum. (2)
3. Find the volume and radius of the largest cylinder. (2)

Answer:
1. From the right triangle ∆OAB,
y² = R² – x² ⇒ y = \(\sqrt{R^{2}-x^{2}}\)
Which is the radius of the cylinder
Also height = 2 x
∴ Volume = V= π y² × 2x = 2π(R² – x²)x = 2π(R²x – x³).

2. Now,

For maximum or minimum,

3. Base radius =

Question 6.
If f(x) = x³ + 3x² – 9x + 4 is a real function

1. Find the intervals in which the function is increasing or decreasing. (3)
2. Find the points of local maxima or local minima of f(x) (2)
3. Graph of a function is given in the following figure:

Which among the following represents the graph of its derivative? (1)

Answer:
1. f'(x) = 3x² + 6x – 9
For turning points f'(x) = 3x² + 6x – 9 = 0
⇒ x = 1, -3
These turning point divide the domain of f(x) in the following intervals. (-∞, -3), (-3, 1), (1, ∞) in (-∞, -3)
⇒ f'(-4) = 3(-4)² + 6(-4) – 9 > 0
Hence increasing.
In (-3, 1) ⇒ f'(0) = 3(0)² + 6(0) – 9 < 0
Hence decreasing.
In (1, ∞) ⇒ f'(2) = 3(2)² + 6(2) – 9 > 0
Hence increasing.

2. x = -3 is a local maximum point and x = 1 is a local minimum point.

3. (a)

The function passes through origin and has a local maximum at x = 2.

Question 7.
Of all the Cylinders with given surface area, show that the volume is maximum when height is equal to the diameter of the base.
Answer:
Let r be the radius, h be the height, V be the volume and S be the surface area
S = 2πr² + 2 πrh

S – 6πr² = 0
2πr² + 2πrh – 6πr² = 0
So h = 2r
So volume is maximum when h = 2r.

Question 8.
Sand is pouring from a pipe. The falling sand forms a Cone on the ground in such a way that the height of the Cone is always one-sixth of the radius of the base.

1. Establish a relation between the volume ‘v’ and height ‘h’ of the Cone using the given condition. (2)
2. lf the sand is pouring at the rate of -12 cm/sec, Find the rate of change of height of the Cone. (2)
3. Find \(\frac{\mathrm{dh}}{\mathrm{dt}}\) when h = 4cm. (2)

Answer:
1. Given that the height of the Cone is one-sixth of the radius of the base, then h = \(\frac{r}{6}\) ⇒ r = 6h
Then Volume V = \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) π(6h)2.h
V = \(\frac{1}{3}\) π 36h².h = \(\frac{1}{3}\) π 36h³
V =12 πh³ ____(1)

2. Differentiating (1) we get

3. When h = 4 cm

Question 9.
(i) Choose the correct answer from the bracket. The rate of change of the area of a circle with respect to its radius r at r = 10cm is.
(a) 10π
(b) 20π
(c) 30π
(d) 40π (1)
(ii) Find the intervals in which the function f given by f(x) = x² – 6x + 5 is (2)
(a) Strictly increasing
(b) Strictly decreasing
(iii) Find the local minimum and local maximum value, if any, of the function f(x) = x³ – 6x² + 9x + 8 (3)
Answer:

(ii) f'(x) = 2x — 6; 2x – 6 = 0; x = 3
(-∞, 3 ) is strictly decreasing
(3, ∞) is strictly increasing.

(iii) f'(x) = 3x² – 12x + 9
f¹¹ = 6x — 12
For maxima, minima
f¹ = 0 → 3x² – 12x + 9 = 0
3(x – 3)(x — 1) = 0; x = 3, x = 1
At x = 3 f¹¹(x) = 6 × 3 – 12 = 18 – 12 = 6 > 0
f is minimum, the local minimum value of f = 8
At x = 1 f¹¹(x) = 6 × 1 – 12 = -6 < 0,
f is maximum, the local maximum value of f = 12.

Question 10.
A wire of length 28m is cut into two pieces. One of the pieces is be made into a square and the other into a circle. What should be the length of the two pieces so that combined area of the square and the circle is minimum using differentiation?
Answer:
Let the length of one piece be ‘x’ and other piece be ‘28 – x’. Let from the first piece we will make a circle of radius Y and from the second piece we will make a square of side y. Then,

Let A be the combined area of the circle and square, then

Question 11.
An open box of maximum volume is to be made from a square piece of tin sheet 24cm on a side by cutting equal squares from the corners and turning of the sides.
(i) Complete the following table. (2)

(ii) Using the above table, express V as a function of x and determine its domain. (1)
(iii) Find height (x. cm) of the box when volume V is maximum by differentiation. (3)
Answer:
(i)

(ii) Generalise the above table as a function.
V = x(24-2x)², 0 < x < 12.

(iii) \(\frac{d V}{d x}\) = x.2(24 – 2x)(-2) + (24 – 2x)²
= -4x(24 – 2x) + (24 – 2x)²
= -96x + 8x² + 576 + 4x² – 96x
= 12x² – 192x + 576
For maximum or minimum,

Therefore volume is maximum when x = 4 cm.

Question 12.
A square tank of capacity 250 m³ has to be dug out. The cost of land is Rs. 50 per m². The cost of digging increases with the depth and for the whole tank is Rs. 400 × (depth)².

1. Find an expression for the cost of digging the tank. (3)
2. Find the dimension of the tank when the total cost is least. (3)

Answer:
1. Let x, x and y be the length, breadth, and depth of the tank.
Then, V = x. x. y = 250 ⇒ y = \(\frac{250}{x^{2}}\).
Area of land = x²
⇒ Cost of land = 50 x²
(∵ cost of land is Rs.50/m²)
Cost of digging = 400 × (depth)² = 400 × (y)²
∴ Total cost = C = 50 x² + 400 × (y)²

2. We have, C = 50x² + \(\frac{400 \times(250)^{2}}{x^{4}}\).
Differentiating w.r.t.x, we get,


∴ Maximum at x = 10
m ⇒ when x = 10m and
y = \(\frac{250}{10^{2}}\) = 2.5m the total cost is least.

Question 13.
Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) times the radius of the base.
Answer:
Volume of the cone will be, V =\(\frac{1}{3}\)πr²h
h = \(\frac{3 V}{\pi r^{2}}\) ____(1)
Curved surface area will be, S = πrl
⇒ S² = π²r²l² = P
⇒ P = π²r²(h² + r²) ⇒ P = π²r²h² + π²r⁴)

⇒ 2r² = h² ⇒ h = \(\sqrt{2} r\).

Question 14.
Let ABC be an isosceles triangle inscribed in a circle having radius r. Then by figure, area of the triangle ABC is ∆

1. Find \(\frac{d \Delta}{d \theta}\) and \(\frac{d^{2} \Delta}{d \theta^{2}}\) (2)
2. Find the maximum value of ∆. (3)
3. Show that the isosceles triangle of maximum area that can be in scribed in a given circle is an equilateral triangle. (1)

Answer:
1. Area of the isosceles triangle is ∆ = \(\frac{1}{2}\)bh
From the figure b = r sin2θ, h = r + r cos2θ

2. For turning points

Therefore local maximum a θ = \(\frac{\pi}{6}\) which means the area of the isosceles triangle is maximum When θ = \(\frac{\pi}{6}\).

3. Then; ∠OCB = 30° ⇒ ∠ACB = 2∠OCB = 60°. Therefore the isosceles triangle is an equilateral triangle.

Question 15.
(i) Using the graph of the function f (x) in the interval [ a, h ] match the following.

(ii) Consider the function f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
(a) Find the turning points of f(x). (1)
(b) Explain the nature of the turning points (1)
(c) Find the absolute extreme values of f(x). (2)
Answer:
(i)

(ii) (a) f(x) = 12x³ – 24x² + 24x – 48
For turning points,
f'(x) = 0 ⇒ 12x³ – 24x² + 24x – 48 = 0
⇒ x³ – 2x² + 2x – 4 = 0
⇒ (x² + 2)(x – 2) = 0 ⇒ x = ± (\(\sqrt{-2}\), 2)
We admit only x = 2 as x = \(\sqrt{-2}\) is not a real number.
Therefore at x = 2 f (x) has a turning point.

(b) f”(x) = 36x² – 48x + 24
⇒ f”(2) = 36(2)² – 48 × 2 + 24 > 0
Therefore at x = 2 f(x) has a local minimum.

(c) f(0) = 25,
f(2) = 3(2)⁴ – 8(2)³ + 12(2)² – 48 × 2 + 25 = -39
f(3) = 3(3)⁴ – 8(3)³ + 12(3)² – 48 × 3 + 25 = 16
Consider the set { f (0), f{2), f (3)}
⇒ {25, -39, 16}
The maximum value of the above set is the absolute maximum and it is 25 at x = 0. The minimum value of the above set is the absolute minimum and it is -39 at x = 2.

Question 16.
An open box with a square base is to be made out of a given quantity of sheet of area a².

1. If the box has side x units, then show that volume V= \(\frac{a^{2} x-x^{3}}{4}\) (2)
2. Show that the maximum volume is \(\frac{a^{3}}{6 \sqrt{3}}\) (4)

Answer:
1. Area = a² = x² + 4xh,
h = height of the box.

2. We have,

Question 17.
For the function f(x) = sin2x, 0 < x < π
(i) Find the point between 0 and π that satisfies f'(x) = 0. (2)
(ii) Find the point of local maxima and local minima. (2)
(iii) Find the local maximum and local minimum value. (2)
Answer:

Question 18.
A cylindrical can with a volume of 125m3 (about 2 litres) is to be made by cutting its top and bottom from metal squares and forming its curved side by bending a rectangular sheet of metal to match its ends. What radius ‘r’ and height ‘h’ of the can will minimize the amount of material required.
Answer:

The circular top and bottom should be cut out from a square metal sheet of side 2r. Therefore the area of squares is 8r².
Area A = 8r² + 2πrh

∴ To minimize the amount of material, r = 2.5
\(h=\frac{125}{\pi(2.5)^{2}}=6.3\).

Question 19.
A rectangle sheet of tin with adjascent sides 45cm and 24cm is to be made into a box • without top, by cutting off equal squares from the comers and folding up the flaps

1. Taking the side of the square cut off as x, express the volume of the box as the function of x. (2)
2. For what value of x, the volume of the box will be maximum. (4)

Answer:
1. Length of the box = 45 – 2x
Breadth of the box = 24 – 2x
Height of the box = x
Volume; V = (45 – 2x)(24 – 2x)x
= (1080 – 138x + 4x²)x
= 4x³ – 138x² + 1080x.

2. \(\frac{d y}{d x}\) = 12x² – 276x + 1080

12x² – 276x + 1080 = 0
x² – 23x + 90 = 0
x = 18, 5
x = 18 is impossible
∴ x = 5 when x = 5, \(\frac{d^{2} y}{d x^{2}}\) < 0

The volume of the box is maximum at x = 5.