**Instructions**

1. Questions 1 to 10 carry 1 mark each.

2. Questions 11 to 22 carry 4 marks each.

3. Questions 23 to 29 carry 6 marks each.

2. Questions 11 to 22 carry 4 marks each.

3. Questions 23 to 29 carry 6 marks each.

**1.**

**Write the following set in the set-builder form: {5, 25, 125, 625}.**

**Sol.**It can be seen that 5 = 5

^{1}, 25 = 5

^{2}, 125 = 5

^{3}, and 625 = 5

^{4}.

∴ {5, 25, 125, 625} = {x: x = 5

^{n}, n ∈ N and 1 ≤ n ≤ 4}**2. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x,**

*y*and z are distinct elements.**Sol.**. It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}.

**3. Let f be the subset of Z ×Z defined by f={(ab,a+b):a,b∈Z} Is f a function from Z to Z? Justify your answer.**

**Sol.**We observe that

1 x 6 = 6 and 2 x 3 = 6

⇒ (1×6,1+6) ∈ f and (2×3,2+3) ∈ f

⇒ (6,7) ∈ f and (6,5) ∈ f.

We observe that an element 6 have appeared more than once as the first component of the ordered pairs in f. So f is not a function for Z to Z.

**4. Express the given complex number in the form**

*a*+*ib*: (1 – i) – (–1 + i6).**Sol.**(1 – i) – (–1 + i6) = 1 – i + 1 – 6i = 2 – 7i

**5. If 2x+ i (x - y) = 5, where x and y are real numbers, find the values of x and y.**

**Sol.**We have 2x+ i(x - y) = 5

or 2x+ i(x - y) = 5 + 0.i

Comparing the real and imaginary parts,we get

2x = 5 and x - y = 0

⇒ x = 5/2 and x = y

⇒ x = 5/2 and x = y

Thus x = y = 5/2.

**6. Solve the given inequality for real**

*x*: 4x + 3 < 5x + 7.**Sol.**

**4x + 3 < 5x + 7**

⇒ 4x + 3 – 7 < 5x + 7 – 7

⇒ 4x – 4 < 5x

⇒ 4x – 4 – 4x < 5x – 4x

⇒ –4 < x

Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–4, ∞).

**7. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?**

**Sol.**There will be as many ways as there are ways of filling 3 vacant places

In succession by the given six digits. In this case, the units place
can be filled by 2 or 4 or 6 only i.e., the units place can be filled in
3 ways. The tens place can be filled by any of the 6 digits in 6
different ways and also the hundreds place can be filled by any of the 6
digits in 6 different ways, as the digits can be repeated.

Therefore, by multiplication principle, the required number of three digit even numbers is 3 × 6 × 6 = 108.

**8. Find the sum of odd integer from 1 to 21.**

**Sol.**The odd integer from 1 to 21 are 11, namely 1, 3, 5, 7, 9, 11,13,15,17,19, 21.

**9. Write the equations for the**

*x*and*y*-axes.**Sol.**

**The y-coordinate of every point on the x-axis is 0.**

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is y = 0.

**10. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?**

**Sol.**When a die is rolled, the sample space is given by

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

Accordingly, E = {4} and F = {2, 4, 6}

It is observed that E ∩ F = {4} ≠ Φ

Therefore, E and F are not mutually exclusive events.

**11. A wheel makes 500 revolutions per minute. How many radians it turns in one second?**

**Sol.**In one revolution angle covered is 2 π radians

In one minute, number of revolutions = 500

In one second, number of revolutions = 500/60 = 50/6

∴ In one second angle covered = 50/6 X 2 π radians

or In one second angle covered = π radians

**12. Write the value of tan 75°.**

**Sol.**tan 75°

= tan(45°+30°)

**13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?**

**Sol.**Let U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T' ∩ C').

n(T' ∩ C') = n(T ∪ C)'

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

**14. Find the value of .**

**Sol.**

**15. Solve the quadratic equation 25x**

^{2}– 30x + 11 = 0.**Sol.**25x

^{2}– 30x + 11 = 0

Using formula for the roots of quadratic equations, we get

Hence, the roots are

**16. In how many of the distinct permutations of the letters in MISSISSIPPI do the four 1’s not come together?**

**Sol.**Total number of letters in “MISSISSIPPI” = 11

Out of these I is repeated four times

S is repeated four times

P is repeated two times

Thus, the total number of distinct permutations

There are 4 I’s viz, I, I, I, I. Considering these four I’s as one letter we have 8 letters (MSSSSPP and one letter obtained by combining all 1’s), out of which

S is repeated four times

P is repeated two times

and rest are different

Thus, the total number of distinct permutations in which 4 I’s are together

Hence, the total number of distinct permutations of the letters in MISSISSIPPI do the four I’s not come together

= 7×6×5×161

= 210 x 161

= 33810

**17. Let the sum of n, 2n, 3n terms of an A.P. be S**

_{1},S_{2}and S_{3}respectively, show that : S_{3}= 3(S_{2 }- S_{1}).**Sol.**

**18. Find coordinates of the foot of perpendicular from the point (3,-4) to line 4x – 15y + 17 = 0.**

**Sol.**The equation of the given line is 4x – 15y + 17 = 0 … (i)

The equation of a line perpendicular to the given line is 15x + 4y – k = 0, where k is a constant.

If this line passes through the point (3, -4), then

4 x 3 – 15 x (-4) – k = 0

12 + 60 – k = 0

k = 72

12 + 60 – k = 0

k = 72

Therefore the equation of a line passing through the point (3, -4) and perpendicular to the given line is

15x + 4y – 72 = 0 … (ii)

The required foot of the perpendicular is the point of intersection of lines (i) and (ii).

Solving equation (i) and (ii), we get

Therefore the required point is

**19. Find the equation of the circle which passes though the points (3,7), (5,5) and has its centre on the line x - 4y = 1.**

**Sol.**Let the equation of the required circle be x

^{2}+ y

^{2 }+ 2gx + 2fy + C =

^{ }0

^{ }..(i)

^{ }It passes through (3,7) and (5,5)

9 + 49 + 6g + 14f + C = 0 ...(ii)

25 + 25 + 10g + 10f + C = 0 ...(iii)

The centre (-g, -f) lies on x - 4y = 1

-g + 4f = 1 ...(iv)

Subtracting (ii) from (iii), we get

4g - 4f - 8 = 0. ...(v)

or g - f = 2 ...(vi)

Solving equation (iv) and (vi)

3f = 3 , g - 1 = 2

f = 1 , g = 3.

Substituting the values of g and f in (ii)

We get

9 + 49 + 18 + 14 + C = 0.

C = -90

Hence the equation of the circle

x

^{2}+ y

^{2}+ 6x + 2y - 90 = 0.

**20. Find the equation of the circle which passes through the points (2, –2), and (3, 4) and whose centre lies on the line**

*x*+*y*= 2.**Sol.**Let the equation of the circle be (x - h)

^{2}+ (y - k)

^{2}= r

^{2}

Since the circle passes through (2, – 2) and (3, 4), we have

(2 - h)

^{2}+ (-2 - k)^{2}= r^{2}...(1) and^{ }(3 - h)^{2}+ (4 - k)^{2}= r^{2}...(2)
Also, the centre lies on the line x + y = 2, we have

h + k = 2 ...(3)

Solving the equations (1), (2) and (3), we get

h = 0.7, k = 1.3 and r

^{2}= 12.58
Hence, the equation of the required circle is

(x - 0.7)

^{2}+ (y - 1.3)^{2}= 12.58.**21. A group of two persons is to be selected from two boys and two girls. What is the probability that the group will have**

**(a) two boys (b) one boy (c) only girls**

**Sol.**Total number of persons = 2 + 2 = 4. Two persons can be selected in

^{4}C

_{2}ways.

(a) Two boys can be selected in

^{2}C

_{2}ways.

Hence, probability =

^{2}C_{2}/^{4}C_{2 }= 1/6.
(b) It means that there is one boy and one girl in the group. One boy out of 2 boys can be selected in

^{2}C_{1}ways. One girl out of 2 girls can be selected in^{2}C_{1}ways.
Hence, probability =

^{2}C_{1 }x^{2}C_{1}/^{4}C_{2}= 2 x 2/2 x 3 = 2/3.
(c) Two girls can be selected from 2 girls in

^{2}C_{2}ways.
Hence, probability =

^{2}C_{2}/^{4}C_{2}= 1/6.**22. Find the point in XY-plane which is equidistant from three points A(2,0,3), B(0,3,2) and C(0,0,1).**

**Sol.**The z-coordinate = 0 on xy-plane.

Let P (x, y, 0)be a point on xy-plane such that PA = PB = PC

Now, PA = PB

⇒ PA

^{2}= PB^{2}
⇒ (x - 2)

^{2}+ (y - 0)^{2}+ (0 - 3)^{2}= (x -0)^{2}+ (y -3)^{2}+ (0 - 2)^{2}
⇒ 4x - 6y = 0 or 2x - 3y = 0 ...(1)

PB = PC

PB

^{2}= PC^{2}
⇒ (x - 0)

^{2}+ (y - 3)^{2}+ (0 - 2)^{2}= (x - 0)^{2}+ (y - 0)^{2}+ (0 - 1)^{2}
⇒ - 6y + 12= 0 ory =2 ...(2)

By (1), we get x = 3

Hence, the required point is (3, 2, 0).

**23. A market research group conducted a survey of 1000 consumers and reported that 720 consumers like product A and 450 consumers like product B. What is the leastnumber that must have liked both products?**

**Sol.**Let U be the set of all consumers who were questioned, PA be the set of consumers who liked product A and PB be the set of consumers who liked product B.

It is given that :

n(U) = 1000, n(P

_{A}) = 720, n(P

_{B}) = 450

We know that :

n(P

_{A}∪P

_{B}) = n(P

_{A})+n(P

_{B}) - n(P

_{A}∩P

_{B})

⇒ n(P

_{A}∪P

_{B}) = 720 + 450 - n(P

_{A}∩P

_{B})

⇒ n(P

_{A}∪P

_{B}) = 1170 - n(P

_{A}∪P

_{B})

⇒ n(P

_{A}∪P

_{B}) = 1170 - n(P

_{A}∩P

_{B}) ....(1)

Since P

_{A}∪P

_{B}⊂U, therefore

n(P

_{A}∪P

_{B}) ≤ n(U)

⇒ -n(P

_{A}∪P

_{B}) ≥ n(U)

⇒ 1170 - n(P

_{A}∪P

_{B}) ≥ 1170 - n(U) .....(2)

From (1) and (2), we have

n(P

_{A}∩P

_{B}) ≥ 1170 - n(U) But P

_{A}∪P

_{B}⊂U

⇒ n(P

_{A}∩P

_{B}) ≥ 1170 - 1000 ⇒ n(P

_{A}∪P

_{B}) ≤ n(U) = 1000

⇒ n(P

_{A}∩ P

_{B}) ≥ 170 ⇒ Maximum value of n(P

_{A}∪P

_{B}) = 1000

Thus, the least value of n(P

_{A}∩P

_{B}) is 170 and the maximum value of n(P

_{A}∩P

_{B}) is 1000.

**24. Solve : 2 cos**

^{2}x+3 sin x = 0.**Sol.**We have

2 cos

⇒ 2(1-sin

⇒ 2-2 sin

⇒ 2 sin

⇒ 2 sin

⇒ sin x(2 sin x + 1)-2(2 sin x+1) = 0

⇒ (2 sin x+1)(sin x-2) = 0

⇒ Either 2 sin x+1=0 or sin x-2 = 0

But sin x = 2 is not possible as sin x cannot be greater than 1.

∴ 2 sin x+1 = 0

⇒

⇒

Hence, the solution is given by

^{2}x+3 sin x = 0⇒ 2(1-sin

^{2}x)+3 sin x = 0 [sin^{2}x+cos^{2}x=1]⇒ 2-2 sin

^{2}x+3 sin x = 0⇒ 2 sin

^{2}x-3 sin x-2 = 0⇒ 2 sin

^{2}x+sin x-4 sin x-2 = 0⇒ sin x(2 sin x + 1)-2(2 sin x+1) = 0

⇒ (2 sin x+1)(sin x-2) = 0

⇒ Either 2 sin x+1=0 or sin x-2 = 0

But sin x = 2 is not possible as sin x cannot be greater than 1.

∴ 2 sin x+1 = 0

⇒

⇒

Hence, the solution is given by

**25. Find centroid of a triangle, mid-points of whose sides are (1, 2, - 3), (2, 0, 1) and (- 1, 1, - 4).**

**Sol.**Let A(x

_{1}, y

_{1}, z

_{1}), B(x

_{2}, y

_{2}, z

_{2}) and C(x

_{3}, y

_{3}, z

_{3}) are vertices of triangle ABC.

P(1, 2, -3), Q(3, 0, 1) and R(- 1, 1, - 4) are mid-points of sides AB, BC and AC respectively.

P is the mid-point of AB, therefore

(x

(y

(z

_{1}+ x_{2})/2 = 1 ...(1)(y

_{1}+ y_{2})/2 =2 ...(2)(z

_{1}+ z_{2})/2 =- 3 ...(3)
Q is the mid-point of BC, therefore

(x

(y

(z

_{2}+ x_{3})/2 =3 ...(4)(y

_{2}+ y_{3})/2 =0 ...(5)(z

_{2}+ z_{3})/2 =1 ...(6)
R is the mid-point of AC, therefore

(x

(y

(z

_{1}+ x_{3})/2 =- 1 ...(7)(y

_{1}+ y_{3})/2 =1 ...(8)(z

_{1}+ z_{3})/2 =- 4 ...(9)
Adding (1), (4) and (7), we get

(x

_{1}+ x_{2})/2 + (x_{2}+ x_{3})/2 + (x_{1}+ x_{3})/2 = 1 +3-1
⇒ (x

_{1}+ x_{2}+ x_{3}) =3
Adding (2), (5) and (8), we get

(y

_{1}+ y_{2})/2 + (y_{2}+ y_{3})/2 + (y_{1}+ y_{3})/2 = 2 + 0 + 1
⇒ (y

_{1}+ y_{2}+ y_{3}) =3
Adding (3), (6) and (9), we get

(z

_{1}+ z_{2})/2 + (z_{2}+ z_{3})/2 + (z_{1}+ z_{3})/2 = - 3 + 1 -4
⇒ (z

_{1}+ z_{2}+ z_{3}) =- 6
We have centroid O of triangle is given by

{(x

= (1, 1, - 2)

_{1}+ x_{2}+ x_{3})/3, (y_{1}+ y_{2}+ y_{3})/3, (z_{1}+ z_{2}+ z_{3})/3}= {3/3, 3/3, -6/3}= (1, 1, - 2)

Therefore, centroid is (1, 1, - 2).

**26. If the ratio of the coefficients of 3rd and 4th terms in the expansion of is 1:2 then find the value of n.**

**Sol.**Given expansion is

T

_{3}=^{n}C_{2}x^{n-2 }
T

_{4}=^{n}C_{3}x^{n-3 }
But we are given,

⇒ -n + 2 = 12

⇒ n = -10

**27. Find the value of n so that may be the geometric mean between a and b.**

**Sol.**G. M. of a and b is

By the given condition,

Squaring both sides, we obtain

⇒ a

^{2n+2}+ 2a^{n+1}b^{n+1}+ b^{2n+2}= (ab)(a^{2n}+ 2a^{n}b^{n}+ b^{2n})
⇒ a

^{2n+2}+ 2a^{n+1}b^{n+1}+ b^{2n+2}= (a^{2n+1}b + 2a^{n+1 }b^{n+1}+ab^{2n+1})
⇒ a

^{2n+2}+ b^{2n+2}= a^{2n+1}b + ab^{2n+1}
⇒ a

^{2n+2}- a^{2n+1}b = ab^{2n+1}- b^{2n+2}
⇒ a

^{2n+1}(a –b) = b^{2n+1 }(a – b)
⇒ 2n + 1 = 0

**28. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.**

**Sol.**Let the observations be x

_{1}, x

_{2}, x

_{3}, x

_{4}, x

_{5}, and x

_{6}.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are y

_{i}, then
y

_{i}= 3x, i.e., , for i = 1 to 6
From (1) and (2), it can be observed that,

Substituting the values of x

_{i}and in (2), we obtain
Therefore, variance of new observations

Hence, the standard deviation of new observations is

**29. P,Q,R shot to hit a target. If P hits it 3 times in 4 trials, Q hits it 2 times in 3 trials and R hits it 4 times in 5 trials, what is the probability that the target is hit by at least two persons?**

**Sol.**Let E

_{1}, E

_{2}and E

_{3}

_{ }be independent events that P, Q and R hit the target respectively. Then

P(E

_{1}) = 3/4, P(E

_{2}) = 2/3and P(E

_{3}) = 4/5

P(E

_{1}

^{c}) = 1 - 3/4 = 1/4, P(E

_{2}

^{c}) = 1 - 2/3 = 1/3and P(E

_{3}

^{c}) = 1 - 4/5 = 1/5 .

The target is hit by atleast two persons in the following mutually exclusive ways:

(a) P hits, Q hits and R does not hit i.e., E

(b) P hits, Q does not hit and R hits i.e., E

(c) P does not hit, Q hits and R hits i.e., E

(d) P hits, Q hits and R hits i.e., E

Required Probability = P[(a) U (b) U (c) U (d)] = P(a) + P(b) + P(c) + P(d)

= (3/4 x 2/3 x 1/5) + (1/4 x 2/3 x 4/5) + (3/4 x 2/3 x 4/5)

= 25/30 = 5/6.

_{1}^{ }∩ E_{2}∩ E_{3}^{c}(b) P hits, Q does not hit and R hits i.e., E

_{1}^{ }∩ E_{2}^{c}∩ E_{3}(c) P does not hit, Q hits and R hits i.e., E

_{1}^{c }∩ E_{2}∩ E_{3}(d) P hits, Q hits and R hits i.e., E

_{1}^{ }∩ E_{2}∩ E_{3}Required Probability = P[(a) U (b) U (c) U (d)] = P(a) + P(b) + P(c) + P(d)

= (3/4 x 2/3 x 1/5) + (1/4 x 2/3 x 4/5) + (3/4 x 2/3 x 4/5)

= 25/30 = 5/6.

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