## Kerala Plus One Maths Model Question Paper 2

**Time Allowed: 2 1/2 hours**

**Cool off time: 15 Minutes**

**Maximum Marks: 80**

**General Instructions to Candidates :**

- There is a ‘cool off time’ of 15 minutes in addition to the writing time .
- Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
- Read instructions carefully .
- Read questions carefully before you answering.
- Calculations, figures and graphs should be shown in the answer sheet itself.
- Malayalam version of the questions is also provided.
- .Give equations wherever necessary.
- Electronic devices except non programmable calculators are not allowed in the Examination Hall.

**Answer any six from question numbers 1 to 7. Each carries three scores.**

Question 1.

Find 7+ 77 + 777 + 7777 +…. to n terms.

Question 2.

Question 3.

**a.** If A and B are two events in a random experiment, then

P(A) + P(B) – P(A ∩ B)

**b.** Given P(A) = 0.5, P(B) = 0.6 and P

(A ∩ B) = 0.3. Find P( A ∪ B) and P(A’).

Question 4.

Let A = {x:x is an integer, 1/2 < x < 7/2}

B = {2, 3, 4}

**a.** Write A in the roster form.

**b.** Find the power set of (A ∪ B)

**c.** Verify that (A – B )∪ (A ∩ B)=A

Question 5.

(i) Find the value of sin

\(\left( \frac { 3l\pi }{ 3 } \right) \)

(ii) Find the principal and general solutions of the equation cosx =

\(\frac { -\sqrt { 3 } }{ 2 } \)

Question 6.

A committee of 3 persons is to be constituted from a group of 2 men and 3 women.

**a.** In how many ways can this be done?

**b.** How many of these committees would consist of 1 man and 2 women.

Question 7.

**a.** Which one of the following points lies in the sixth octant?

i. (-4, 2, -5)

ii. (-4, -2, -5)

iii. (4, -2, -5)

iv. (4, 2, 5)

**b.** Find the ratio in which the YZ plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).

**Answer any six from question numbers 8 to 17. Each carries four scores.**

Question 8.

Consider the following statement :

**a.** Prove that P(1) is true.

**b.** Hence by using the principle of mathematical induction, prove that P(n) is true for all natural numbers n.

Question 9.

**a.** Write the negation of the statement: “Every natural number is greater than zero”

**b.** Verify by the method of contradiction : “P : √13 is irrational”

Question 10.

**a.** The 8th term in the expansion of (√2+√3)

is

i. 27 √2

ii. 27 √3

iii. 72 √2

iv. 72 √3

**b.** Find the term independent of x in the expansion of

Question 11.

**a.** In a random experiment, 6 coins are tossed simultaneously. Write the number of sample points in the sample space.

**b.** Given that P(A’)=0.5. P(B) = 0.6, P

(A ∩ B) = 0.3. Find P(A’) , P(A ∪ B), P(A ∩ B)and P(A ∪ B).

Question 12.

Let U={1,2, 3, 4, 5, 6, 7, 8, 9}, A= {1, 2, 4, 7} and B = {1, 3, 5, 7}

**a.** Find A ∪ B.

**b.** Find A’, B’ and hence show that (A ∪B)’ = (A’∩ B’)

**c.** The power set P(A ∪ B) has elements.

Question 13.

**a.** Let A= {7, 8} and B = {5, 4, 2} Find A x B.

**b.** Determine the domain and range of the

relation R defined by R = {(x, y): y = x + 1, x ∈ {0, 1,2, 3,4, 5}}

Question 14.

**a.** Find the distance between the points (2,-1, 3) and (-2, 1, 3)

**b.** Find the co-ordinates of the point which divides the line segment joining the points

(-2, 3, 5) and (1, -4,6) internally in the ratio of 2:3.

Question 15. A hyperbola whose transverse axis is X – axis, center (0,0) and the foci (±√10,0) passes through the point (3, 2)

**a.** Find the equation of the hyperbola.

**b.** Find its eccentricity.

Question 16.

**a.** Solve for the natural number n;

12. ^{(n+1)} p_{3} =5. ^{(n+1)} p_{3}

**b.** In how many ways can 7 athletes be chosen out of 12?

Question 17.

Consider the statement: “n(n +1) (2n + 1) is divisible by 6”

**a.** Verify the statement for n = 2.

**b.** By assuming that P(k) is true for a natural number k, verify that P(k + 1) is true.

**Answer any five from question numbers 18 to 24. Each carries six scores.**

Question 18.

**a.** Match the following:

**b.** Find the derivatives of tan x using the first principles.

Question 19.

**a.** Find the equation of the line passing through the points (3, -2) and (-1,4).

**b.** Reduce the equation √3x + y – 8 = 0 into normal form.

**c.** If the angle between two lines is π/4 and slope of the line is 1/2, find the slope of the other line.

Question 20.

**a.** If x is the mean and c is the standard deviation of a distribution, then the coefficient of variation is……..

**b.** Find the standard deviation for the following data:

x_{i} : 3 8 13 18 23

f_{i }: 7 10 15 10 6

Question 21.

**a.** Which one of the following pair of a. Which one of the following pair of straight lines are parallel?

i. x – 2xy-4 = 0; 2x – 3y – 4 = 0

ii. x – 2y – 4 = 0; x – 2y – 5 = 0

iii. 2x – 3y – 8=0; 3x – 3y – 8 = 0

iv. 2x – 3y – 8 = 0; 3x – 2y – 8 = 0

**b.** Equation of a straight line is 3x – 4y + 10 = 0.Convert it into the intercept form and write the x-intercept and y-intercept.

**c.** Find the equation of the line perpen dicular to the line x – 7y + 5 = 0 and having x-intercept 3.

Question 22.

**a.** If (x+1, y-2) = (3,1), write the values of x and y.

**b.** Let A={ 1, 2, 3, 4, 5} and B={4, 6, 9} be two sets. Define a relation R from A to B by R={(x, y): x-y is a positive integer}. Find A X B and hence write R in the Roster form.

**c.** Define the modulus function. What is its domain? Draw a rough sketch.

Question 23.

**a.** Which among the following is the interval corresponding to the inequality -2 < x ≤ 3 ?

i. [-2, 3]

ii. [-2, 3)

iii. (-2, 3]

iv. (-2, 3)

**b.** Solve the following inequalities graphically

2x + y ≥ 4

x + y ≤ 3

2x – 3y ≤ 6

Question 24.

**a.** In how many ways can the letters of the word, PERMUTATIONS be arranged if:

i. the words start with P and end with S?

ii. there are always 4 letters between P and S?

**b.** In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together.

**c.** How many chords can be drawn through 21 points?

**Answers**

Answer 1.

Answer 2.

Answer 3.

**a.** P(A) + P(B) – P (A ∩ B) = P (A ∪ B)

**b.** P(A ∪ B) = P(A) + P(B) – P (A ∩ B)

= 0.5+ 0.6=0.3 = 1.1-0.3 = 0.8

P(A)’= 1- P (A) = 1-0.5 = 0.5

Answer 4.

**a.** Let A = {1, 2,3}

**b.** A ∪ B = {1, 2, 3,4}

P(A ∪ B)= {1}, {2}, {3}, {4},

{1, 2}, {1, 3}, {1, 4}, {2, 3},

{2, 4}, {3, 4}, {1, 2 ,3}, {1, 2, 4},

{2, 3,4}, {1,3,4}, {1,2, 3, 4},

{(j)} = 24 = 16 elements

**c.** A-B= {l}, A ∩ B = {2, 3}

(A – B) ∪ (A ∩ B) = {1, 2, 3} = A

Answer 5.

Answer 6.

**a.** Required number of ways = 5C_{3} = 10

**b.** Number of committees =

2C_{1} x 3C_{2} = 2 x 3 = 6

Answer 7.

**a.** i. (-4, 2,-5)

**b.** Let YZ plane divides the line joining the points A (-2, 4, 7) and B (3, -5, 8) at R (x, y, z) in the ratio k: 1.

Then x coordinates of R = 0.

Answer 8.

Answer 9.

**a.** Let p : “Every natural number is greater than zero”. ~p: “ Every natural number is not greater than zero”.

**b.** Let us assume that √13 is a rational number.√13=a/b, where a and b are co-pnme, i.e., a and b have no common factors, which implies that 13b^{2}= a^{2} ⇒ 13 divides a. There exists an integer ‘k’ such that a= 13k a^{2}= 169 k^{2} ⇒ 13b^{2}= 169k^{2 ⇒} 13k^{2} ⇒ 13 divides b.

i.e., 13 dvides both a and b, which is contradiction to our assumption that a and b have no common factor. ∴ Our supposition is wrong. ∴ √13 is an irrational number.

Answer 10.

Answer 11.

**a.** 2^{6}

**b.** P(A) = 0.5; P(B) = 0.6; P(A ∩ B) = 0.3

P(A’ ) = 1 – P(A) = 1 – 0.5 = 0.5

P(A ∪ B)=P(A)+P(B)-P(A ∩ B)

= 0.5 + 0.6 – 0.3 = 0.8

P( A ∩ B)=l-P P(A ∪ B)= 1-0.8 = 0.2

P(A ∪ B)=l-P(A ∩ B)= 1-0.3 =0.7

Answer 12.

U = {1,2, 3, 4, 5,7},

A = {1,2, 4, 7} and B = {1,3, 5, 7}

**a.** A ∪ B = {1,2, 3, 4, 5, 7}

**b.** A’ = {3, 5, 6, 8, 9}

B’ = {2, 4, 6, 8, 9}

A ∪ B = {1,2, 3, 4, 5, 7}

(A ∪ B)’ = {6, 8, 9}

A’∩ B’ = {6, 8, 9},

So (A ∪ B)’ = A’∩B’

c. The power set P(A ∪B) has 26 or 64 elements.

Answer 13.

**a.** A = {7, 8} and B = {5, 4, 2}

A x B = {(7, 5), (7, 4), (7,2), (8, 5), (8, 4), (8, 2)}

**b.** Domain = {0,1, 2, 3, 4, 5}

Range = {1,2, 3, 4, 5, 6}

R = {(0,1). (1, 2), (2, 3), (3, 4),(4, 5), (5, 6)}

Answer 14.

Answer 15.

Equation of hyperbola is

Answer 16.

Answer 17.

**a.** When n = 2,

2 (2+1) (2×2 + 1) = 2 x 3 x 5 = 30 is divisible by 6

**b.** P(k) = k(k + l)(2k + 1) is divisible by 6

P (k + 1) = (k + l)(k + 2) (2k + 1)

+ (k + 1) (k + 2)2 = (k + 1) (2k +1) k + (k + 1)

(2k + 1) x 2 + (k + 1) (k + 2)

= k (k + 1) (2k + 1) + 2 (k + 1)

[2k + 1 + k + 2]

= k (k + 1) (2k + 1) + 6(k + 1)^{2}

is divisible by 6

∴ P(k + 1) is true

Answer 18.

Answer 19.

Answer 20.

Answer 21.

**a.** ii. x-2y-4 = 0; x-2y – 5 = 0

**b.** 3x – 4y + 10 = 0; 3x – 4y = -10

**c.** Slope of the given line is

\(\frac { -A }{ B } =-\frac { 1 }{ -7 } =\frac { 1 }{ 7 } \)

Slope of the required line is -7

Given x intercept of the required line is 3, the point is (3,0).

Hence equation of the required line is

y – 0= -7 (x – 3); y + 7x = 21 or 7x + y – 21 = 0

Answer 22.

**a.** (x+1, y-2) = (3, 1) x+l=3; x=2 y-2 = i; y=3

**b.** AxB = { (1,4),( 1,6),( 1,9), (2,4), (2,6), (2,9),(3,4),

(3,6), (3,9), (4,4), (4,6) ,(4,9), (5,4), (5,6), (5,9)}

R = {(5,4)}

**c.** A real function R is to be a modulus function, if f (x) = | x |, x ∈ R, is known as modulus function.

Answer 23.

**a.** ii (-2, 3)

**b.** 2x + y = 4

Solution region:

2x + y ≥ 4

⇒ 0 ≥ 4, which is false. || by putting x = 0, y = 0

Hence shade the half plane, which does not contain the origin. x + y < 3

⇒ 0 ≥ 3, which is true.

Hence shade the half plane, which contains the origin.

2x – 3y < 6

⇒ 0 ≤ 6, which is true.

Hence shade the half plane, which contains the origin. The common region shown in the figure ’ is the solution region.

Answer 24.

**a.** i. When word start with P and end with S, then there are 10 letters to be arranged of which T appears two times.

ii. When there are always 4 letters between P & S

P & S can be at

1st and 6^{th} place

2nd and 7^{th} place

3rd and 8^{th} place

4^{th} and 9^{th} place

5^{th }and 10^{th} place

6^{th} and 11^{th} place

7^{th} and 12^{th} place

So, P & S will placed in 7 ways & can be arranged in 7 x 2 ! = 14

The remaining 10 letters with 2T’s 10!

can be arranged in\(\frac { 10i }{ 2i } \)=1814400 ways

The required number of arrangements = 14 x 1814400 = 25401600

**b.** Can be done in 5 ! ways, such that xGxGxGxGxGx

where G represents the seats for girls and cross mark represents the seats for boys = Total number of ways