Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 6 Application of Derivatives.

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

### Plus Two Maths Application of Derivatives 4 Marks Important Questions

Question 1.

(a) Find the equation of the tangent to the curve \(x^{\frac{2}{3}}+y^{\frac{2}{3}}=2\) at (1,1).

(b) Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. (May – 2015)

Answer:

Question 2.

(a) The slope of the tangent to the curve given

\(x=1-\cos \theta, y=\theta-\sin \theta \text { by at } \theta=\frac{\pi}{2}\)

(i) 0

(ii) – 1

(iii) 1

(iv) Not defined.

(b) Find the intervals in which the function *f*(x) = x^{2} – 4x + 6 is strictly decreasing.

(C) Find the minimum and maximum value, if any, of the function *f*(x) = (2x – 1)^{2} + 3 (March – 2016)

Answer:

(a) (iii) 1

(b) Given; *f*(x) = x^{2} – 4x + 6 ⇒ *f’*(x) = 2x – 4

For turning points; *f*’(x) = 2x – 4 0 ⇒ x = 2

So volurn.,e is niaxirnum when h = 2r

The intervals are (- ∞, 2); (2, ∞)

*f’*(0) = 2 x 0 – 4 = -4

Therefore *f*(x) is decreasing in (- ∞, 2)

(c) *f*(x) = (2 x 1)^{2} + 3

⇒ *f’*(x) 2(2x – 1) x 2 *f”*(x) = 8

For tuming points; *f’*(x) = 8x – 4 = 0 ⇒ x = 1/2

*f*(x) has minimum value at x = 1/2 minimum value is \(f\left(\frac{1}{2}\right)=3\)

2)

Question 3.

(a) Which of the following function has neither local maxima nor local minima?

(i) *f*(x) = x^{2} + x

(ii) *f*(x) = logx

(iii) *f*(x) = x^{3} – 3x + 3

(iv) *f*(x) = 3 + |x|

(b) Find the equation of the tangent to the curve y = 3x^{2 }at (1,1). (March – 2016)

Answer:

Question 4.

(i) The slope of the normal to the curve, y = x^{3} – x^{2} at (1, -1) is

(a) 1

(b) – 1

(c) 2

(d)0

(ii) Find the intervals in which the function *f*(x) = 2x^{3} – 24x + 25 is increasing or decreasing. (May – 2016)

Answer:

(i) (b) – 1

(ii) *f*(x) = 2x^{3} – 24x + 25

*f’*(x) = 6x^{2} – 24

*f’*(x) = O

⇒ 6x^{2} – 24 = 0 ⇒ x^{2 }= 4 ⇒ x = – 2,2

Therefore the intervals are (-∞, -2); (-2, 2); (2, ∞)

*f*(x) is increasing in the intervals (-∞, -2); (2, ∞)

f(x) is decreasing in the intervals (-2, 2)

Question 5.

(i) The slope of the normal to the curve, y^{2} – 4x at (1,2) is

(a) 1

(b) 1/2

(c) 2

(d) – 1

(ii) Find the intervals in which the function 2x^{3} + 9x^{2} + 12x – 1 is strictly increasing. (March – 2017)

Answer:

(i) (b) – 1

(ii) *f*(x) = 2x^{3} + 9x^{2} + 12x – 1

*f’*(x) = 6x^{2} + 18x + 12

= 6(x^{2} + 3x + 2) = 6(x + 1) (x + 2)

*f’*(x) = O

⇒ 6(x + 1)(x + 2) = 0 ⇒ x = – 1 – 2

Therefore the intervals are

(- ∞, – 2); (- 2, – 1); (- 1, ∞)

In the ¡nterval (- ∞, – 2)

*f’*( – 3) = 6(- 3 + 1) (- 3 + 2) > 0

Therefore increasing In the interval (- 2, – 1)

*f’*(- 1.5) = 6(- 1.5 + 1)(- 1.5 + 2) < 0

Therefore decreasing In the interval (- 1, ∞)

*f’*(0) = 6(0 + 1)(0 + 2) > 0

Therefore increasing

Question 6.

Find two positive numbers whose sum is 16 and sum of whose cubes is minimum. (March – 2017)

Answer:

Let the numbers be x and 16 – x. Then,

S = x^{3} + (16 – x)^{3}

= S’ = 3x^{2} + 3(16 – x)^{2}(- 1)

⇒ S” = 6x + 6(16 – x)………..(1)

For turning points S’ = 0 ⇒ 3 x^{2} – 3(16 – x)^{2} = 0

⇒ x^{2} – 16 + 32x – x^{2} =0

⇒ – 16^{2} + 32x = 0 = x^{2} = \(\frac{16 \times 16}{32}\) =8

(1) ⇒ S” = 6(8) + 6(16 – 8) > 0

TherefocemrnimumM x = 8

Thusthe numbers are8 and 16 – 8 = 8

### Plus Two Maths Application of Derivatives 6 Marks Important Questions

Question 1.

(i) Show that the function x^{3} – 6x^{2} + 15x + 4 is strictly increasing in R.

(ii) Find the approximate change in volume of a cube of side x meters caused by an increase in the side by 3%.

(iii) Find the equation of the tangent and normal at the point (1,2) on the parabola y^{2} =^{ }4x. (March – 2010)

Answer:

(i) Given; *f*(x) = x^{3} – 6x^{2 }+ 15x + 4

*f’*(x) = 3x^{2} – 12x + 15 = 3(x^{2} – 4x +5)

= 3(x^{2} – 4x + 4 + 1) = 3(x – 2)^{2 }+ 1) > 0

For any value of x, *f*(x) is a strKly ¡ncreasing.

(ii) We have; V = x^{3 }and Δx = 3% of x = 0.03x

\(d V=\frac{d V}{d x} \Delta x=3 x^{2} \Delta x\)

= 3x^{2} x 0.03x = 0.09x^{3} = 0.09V

\(\Rightarrow \frac{d V}{V}=0.09\)

Therefore 9% is the approximate increase In volume.

(iii) Given; y^{2}….4x ⇒ 2y \(\frac{d y}{d x}\) = 4 ⇒ \(\frac{d y}{d x}=\frac{2}{y}\)

Slope at (1,2) = \(\frac{2}{2}\) = 1

Equation of tangent at (1,2) is; y – 2 = 1(x – 1)

⇒ x – y + 1 = 0

Equation of normal at (1,2) is; y – 2 = – 1(x – 1)

⇒ x + y – 3 = 0

Question 2.

Consider the parametric forms

x = 1 + \(\frac{1}{t}\) – and y = t – \(\frac{1}{t}\) ofa curve

(i) Find \(\frac{d y}{d x}\)

(ii) Find the equation of the tangent at t = 2.

(iii) Find the equation of the normal at t = 2. (May – 2010)

Answer:

Question 3.

(i) The radius of a circle is increasing at the rate of 2cmls. Find the rate at which area of the circle is increasing when radius is

6cm.

(ii) Prove that the function *f*(x) = log sin x is strictly increasing in \(\left(0, \frac{\pi}{2}\right)\) and strictly decreasing in \(\left(\frac{\pi}{2}, \pi\right)\)

(iii) Find the maximum and minimum value of the function *f*(x) = x^{3} – 6x^{2} + 9x + 15. (March – 2011)

Answer:

Question 4.

(i) Find the approximate value of (82)^{1/4} up to three places of decimals using differentiation.

(ii) Find two positive numbers such that Their sum is 8 and the sum of their squares is minimum. (May – 2011)

Answer:

(ii) Let the numbers be x and 8 – x. Then,

S = x^{2} + (8 – x)^{2}

⇒ S’ = 2x + 2(8 – x)( – 1)

⇒ S” = 2 + 2 = 4 ………..(1)

For turning points S’ = 0 = 2x – 2(8 – x) = 0

⇒ 4x – 16 = 0 ⇒ x = 4

(1) ⇒ S” = 4 > 0

Therefore minimum at x = 4

Thus the numbers are 4 and 8 – 4 = 4.

Question 5.

(i) The slope of the tangent to the curve y = x^{3} – 1 at x = 2 is ……….

(ii) Use differentiation to approximate \(\sqrt{36.6}\)

(iii) Find two numbers whose sum is 24 and whose product as large as possible. (March – 2012, March – 2016)

Answer:

Therefore minimum at x =12

Thus the numbers are 12 and 24 – 12 = 12.

Question 6.

(i) Show that the function x^{3} – 3x^{2} + 6x – 5 is strictly increasing on R.

(ii) Find the interval in which the function *f*(x) = sin x + cosx; 0 < x < 2π is strictly increasing or strictly decreasing. (May – 2012)

Answer:

(i) Given; *f*(x) = x^{3} – 3x^{2} + 6x – 5

*f’*(x) = 3x^{2} – 6x + 6 = 3(x^{2} – 2x +2)

= 3(x^{2} – 2x + 1 + 1) 3(x – 1)^{2} + 1) > 0

For any value cit x, *f*(x) is a strictly increasing.

Question 7.

(i) Find the slope of the normal to the curve y = sinθ at θ = π/4

(ii) Show that the function *f*(x) = x^{3} – 6x^{2} + 15x + 4 is strictly increasing in R.

(iii) Show that all rectangles with a given perimeter, the square has the maximum area. (March – 2013)

Answer:

(ii) *f*(x) = x^{3} – 6x^{2 }+ 15x + 4

Differentiating w.r.t x;

*f*(x) = 3x^{2} – 12x + 15 = 3(x^{2} – 4x + 5)

= 3 (x^{2} – 4x + 4 + 1)

= 3 ((x – 2)^{2 }+ 1) > 0, ∀x∈R

Therefore fis strictly increasing in R.

(iii) Let x and ybe the length and breadth of a rectangle with area A and perimeter P.

Question 8.

A right circular cylinder is inscribed in a given cone of radius R cm and height H cm as shown in the figure.

(i) Find the Surface Area S of the circular cylinder as a function of x.

(ii) Find a relation connecting x and R when S is a maximum. (May – 2013)

Answer:

(i) There are two similar triangles ΔDJB and ΔDHF

Question 9.

(i) Which of the following function is Increasing for all values of x in its domain?

(a) sin x

(b) log x

(c) x^{2}

(d) |x|

(ii) Find a point on the curve y = (x – 2)^{2} at which the tangent is parallel to the chord joining the points (2,0) and (4,4).

(iii) Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 6x^{2}. (March – 2014)

Answer:

(i) (b) log x

(ii) Given; y = (x – 2)^{2} ⇒ \(\frac{d y}{d x}\) = 2(x – 2)

Slope of the chord = \(\frac{4-0}{4-2}=2\)

\(\Rightarrow 2=2(x-2) \Rightarrow x=3 \Rightarrow y=(3-2)^{2}=1\)

Therefore the required point is (3, 1)

(iii) Given; p(x) = 41 – 24x – 6x^{2}

p’(x) = – 24 – 12x

p”(x) = – 12

For turning points p’(x) = – 24 – 12x = 0

⇒ x = -2

Since p”(x) = – 12 always maximum Therefore maximum value p(- 2) = 41 – 24(- 2) 6(- 2)^{2} = 65

Question 10.

(a) Find the slope of the tangent to the parabola y^{2} = 4ax at (at^{2}, 2at).

(b) Find the intervals in which the function x^{2} – 2x + 5 is strictly increasing.

(c) A spherical bubble volume at the rate of which the diminishing when the is decreasing in 2cm^{3}/sec. Find the surface area is radius is 3cm. (May – 2014)

Answer:

Question 11.

(a) Which of the following function is always increasing?

(i) x + sin 2x

(ii) x – sin 2x

(ill) 2x + sin 3x

(iv) 2x – sin 2x

(b) The radius of a cylinder is increasing at a rate of 1cm/s and its height decreasing at a rate of 1cm/s. Find the rate of change of its volume when the radius is 5cm and the height is 5cm.

(c) Write the equation of tangent at (1,1) on the curve 2x^{2} + 3y^{2} = 5. (March – 2015)

Answer: