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 = 51, 25 = 52, 125 = 53, and 625 = 54.
∴ {5, 25, 125, 625} = {x: x = 5n, 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.
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
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 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
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 25x2 – 30x + 11 = 0.
Sol. 25x2 – 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
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 S1,S2and S3 respectively, show that : S3 = 3(S2 - S1).
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 x2 + y2 + 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
x2 + y2 + 6x + 2y - 90 = 0.
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
x2 + y2 + 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 = r2
Since the circle passes through (2, – 2) and (3, 4), we have
Since the circle passes through (2, – 2) and (3, 4), we have
(2 - h)2 + (-2 - k)2 = r2 ...(1) and
(3 - h)2 + (4 - k)2 = r2 ...(2)
(3 - h)2 + (4 - k)2 = r2 ...(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 r2 = 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 4C2 ways.
(a) Two boys can be selected in 2C2 ways.
(a) Two boys can be selected in 2C2 ways.
Hence, probability = 2C2 /4C2 = 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 2C1 ways. One girl out of 2 girls can be selected in 2C1 ways.
Hence, probability = 2C1 x 2C1/4C2 = 2 x 2/2 x 3 = 2/3.
(c) Two girls can be selected from 2 girls in 2C2 ways.
Hence, probability = 2C2/4C2 = 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
⇒ PA2 = PB2
⇒ (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
PB2 = PC2
⇒ (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(PA) = 720, n(PB) = 450
We know that :
n(PA∪PB) = n(PA)+n(PB) - n(PA∩PB)
⇒ n(PA∪PB) = 720 + 450 - n(PA∩PB)
⇒ n(PA∪PB) = 1170 - n(PA∪PB)
⇒ n(PA∪PB) = 1170 - n(PA∩PB) ....(1)
Since PA∪PB⊂U, therefore
n(PA∪PB) ≤ n(U)
⇒ -n(PA∪PB) ≥ n(U)
⇒ 1170 - n(PA∪PB) ≥ 1170 - n(U) .....(2)
From (1) and (2), we have
n(PA∩PB) ≥ 1170 - n(U) But PA∪PB⊂U
⇒ n(PA∩PB) ≥ 1170 - 1000 ⇒ n(PA∪PB) ≤ n(U) = 1000
⇒ n(PA ∩ PB) ≥ 170 ⇒ Maximum value of n(PA∪PB) = 1000
Thus, the least value of n(PA∩PB) is 170 and the maximum value of n(PA∩PB) is 1000.
It is given that :
n(U) = 1000, n(PA) = 720, n(PB) = 450
We know that :
n(PA∪PB) = n(PA)+n(PB) - n(PA∩PB)
⇒ n(PA∪PB) = 720 + 450 - n(PA∩PB)
⇒ n(PA∪PB) = 1170 - n(PA∪PB)
⇒ n(PA∪PB) = 1170 - n(PA∩PB) ....(1)
Since PA∪PB⊂U, therefore
n(PA∪PB) ≤ n(U)
⇒ -n(PA∪PB) ≥ n(U)
⇒ 1170 - n(PA∪PB) ≥ 1170 - n(U) .....(2)
From (1) and (2), we have
n(PA∩PB) ≥ 1170 - n(U) But PA∪PB⊂U
⇒ n(PA∩PB) ≥ 1170 - 1000 ⇒ n(PA∪PB) ≤ n(U) = 1000
⇒ n(PA ∩ PB) ≥ 170 ⇒ Maximum value of n(PA∪PB) = 1000
Thus, the least value of n(PA∩PB) is 170 and the maximum value of n(PA∩PB) is 1000.
24. Solve : 2 cos2 x+3 sin x = 0.
Sol. We have
2 cos2 x+3 sin x = 0
⇒ 2(1-sin2 x)+3 sin x = 0 [sin2 x+cos2 x=1]
⇒ 2-2 sin2 x+3 sin x = 0
⇒ 2 sin2 x-3 sin x-2 = 0
⇒ 2 sin2 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
⇒ 2(1-sin2 x)+3 sin x = 0 [sin2 x+cos2 x=1]
⇒ 2-2 sin2 x+3 sin x = 0
⇒ 2 sin2 x-3 sin x-2 = 0
⇒ 2 sin2 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(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) 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
(x1 + x2)/2 = 1 ...(1)
(y1 + y2)/2 =2 ...(2)
(z1 + z2)/2 =- 3 ...(3)
(y1 + y2)/2 =2 ...(2)
(z1 + z2)/2 =- 3 ...(3)
Q is the mid-point of BC, therefore
(x2 + x3)/2 =3 ...(4)
(y2 + y3)/2 =0 ...(5)
(z2 + z3)/2 =1 ...(6)
(y2 + y3)/2 =0 ...(5)
(z2 + z3)/2 =1 ...(6)
R is the mid-point of AC, therefore
(x1 + x3)/2 =- 1 ...(7)
(y1 + y3)/2 =1 ...(8)
(z1 + z3)/2 =- 4 ...(9)
(y1 + y3)/2 =1 ...(8)
(z1 + z3)/2 =- 4 ...(9)
Adding (1), (4) and (7), we get
(x1 + x2)/2 + (x2 + x3)/2 + (x1 + x3)/2 = 1 +3-1
⇒ (x1 + x2 + x3) =3
Adding (2), (5) and (8), we get
(y1 + y2)/2 + (y2 + y3)/2 + (y1 + y3)/2 = 2 + 0 + 1
⇒ (y1 + y2 + y3) =3
Adding (3), (6) and (9), we get
(z1 + z2)/2 + (z2 + z3)/2 + (z1 + z3)/2 = - 3 + 1 -4
⇒ (z1 + z2 + z3) =- 6
We have centroid O of triangle is given by
{(x1 + x2 + x3)/3, (y1 + y2 + y3)/3, (z1 + z2 + z3)/3}= {3/3, 3/3, -6/3}
= (1, 1, - 2)
= (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.
is 1:2 then find the value of n.
Sol. Given expansion is 
T3 = nC2 xn-2 
T4 = nC3 xn-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.
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
⇒ a2n+2 + 2an+1 bn+1 + b2n+2 = (ab)(a2n + 2anbn + b2n)
⇒ a2n+2 + 2an+1 bn+1 + b2n+2 = (a2n+1 b + 2an+1 bn+1 +ab2n+1)
⇒ a2n+2 + b2n+2 = a2n+1 b + ab2n+1
⇒ a2n+2 - a2n+1 b = ab2n+1 - b2n+2
⇒ a2n+1(a –b) = b2n+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 x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.
If each observation is multiplied by 3 and the resulting observations are yi, then
yi = 3x, i.e., 
, for i = 1 to 6
, for i = 1 to 6
From (1) and (2), it can be observed that,
Substituting the values of xi and 
in (2), we obtain
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 E1, E2 and E3 be independent events that P, Q and R hit the target respectively. Then
P(E1) = 3/4, P(E2) = 2/3and P(E3) = 4/5
P(E1c) = 1 - 3/4 = 1/4, P(E2c) = 1 - 2/3 = 1/3and P(E3c) = 1 - 4/5 = 1/5 .
P(E1) = 3/4, P(E2) = 2/3and P(E3) = 4/5
P(E1c) = 1 - 3/4 = 1/4, P(E2c) = 1 - 2/3 = 1/3and P(E3c) = 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., E1 ∩ E2 ∩ E3c
(b) P hits, Q does not hit and R hits i.e., E1 ∩ E2c ∩ E3
(c) P does not hit, Q hits and R hits i.e., E1c ∩ E2 ∩ E3
(d) P hits, Q hits and R hits i.e., E1 ∩ E2 ∩ E3
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.
(b) P hits, Q does not hit and R hits i.e., E1 ∩ E2c ∩ E3
(c) P does not hit, Q hits and R hits i.e., E1c ∩ E2 ∩ E3
(d) P hits, Q hits and R hits i.e., E1 ∩ E2 ∩ E3
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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