PUNE MOCK-1 (2017) SOLUTIONS
1 INFOMATHS/MCA/MATHS/
1.Ans. (b)Clearly, f : [4, ) [4, ) is a bijection. So, it is invertible.
Let f(x) = y. Then,
5x(x – 4) = y
x2 – 4x = log5y
x2 – 4x – log5y = 0
Hence,
We know that if g(x) is inverse of a bijection f(x), then
fog (x) = x f(g(x)) = x.
This relation suggests the following method for finding the inverse of a bijection.
Method II
Step I Obtain the bijection f(x)
Step II In the formula for f(x) replace f(x) by x and x by f-1 (x).
Step III Simplify the relation obtained in step II to get f-1 (x).
2.Ans. (a)Clearly, f : A B is a bijection. This fact can also be observed from the graph of f(x) as it represents an arc of the parabola y = x2 – x + 1 lying on the right side of the vertex (1/2, 3/4).
We know that the curves y = f(x) and y = f-1 (x) are mirror images of each other in the line mirror y = x. This means that the two curves intersect at points lying on the line y = x.
f(x) = f-1(x)
f(x) = x
x2 – x + 1 = x (x – 1)2 = 0 x = 1
Hence, the solution set of the equation f(x) = f-1 (x) is {1}
3.Ans. (b)We have,
[Using triangle inequality]
Thus, the maximum value of |z| is
4.Ans. (b)
Let f(x) = (x – a)3 + (x – b)3 + (x – c)3. Then,
f '(x) = 3{(x – a)3 + (x – b)3 + (x – c)3}
Clearly, f'(x) > 0 for all x.
So, f' (x) = 0 has no real roots.
Hence, f(x) = 0 has two imaginary and one real root.
5.Ans. (d)The total number of ordered pairs of disjoint subsets of S, except ordered pair (,) is
(4C024+4C123 + 4C2 22 + 4C3 21 + 4C4 20) – 1
= (1 + 2)4 – 1 = 80
Total number of unordered pairs of disjoint subsets of S equal to
6.Ans. (b)Following cases arise:
Case I When Ms. Radha is a speaker
In this case, Ms. Rani refuses to speak. So, we have to select 7 girls out of remaining 8 girls and 6 boys from 8 boys. The number of ways of doing this is 8C78C6 = 224.
Case II When Radha is not a speaker.
In this case, two subcases arise.
(i) When Rani is a speaker (ii) When Rani is not a speaker.
If Rani is a speaker, Mr Ravi refused to speak. So, we have to select 7 girls from remaining 8 girls and 6 boys from remaining 7 boys which can be done in 8C77C6 ways = 56 ways.
If Rani is not a speaker, then 8 girls are to be chosen out of remaining 8 girls and 6 boys from 8 boys which can be done in 8C68C8 = 28 ways.
Hence, required number of ways = 224 + 56 + 28 = 308.
7.Ans. (c)
We have,
(1 – t)12 (t + t12) (1 – t24)
Coefficient of t24 = Coefficient of t12 in
8.Ans. (a)
If A is an n n matrix, then |adj A | = |A|n-1
|P|2 = 1(3 – 7) – 4 (6 – 7) + 4 (2 – 1) = 4
|P| = 2
9.Ans. (b)
Given linear equations will passes a non-zero solution, if
10.Ans. (c)
We have,
a = 1 + 2 + 4 + ….. to n terms = 2n – 1
b = 1 + 3 + 9 + ….. to n terms
c = 1 + 5 + 25 + ….. to n terms
Applying R3 R3 – R1 and taking 2 common R2
= 2 2 = 0
11.Ans. (b)
Applying R2 R2 – 2R1, R3 R3 – R1
p(q – b) (r – c) – (a – p) b(r – c) – (a – p) c(q – b) = 0
p(q – b) (r – c) + b(p – a) (r – c) + c(p – a) (q – b) = 0
[Dividing throughout by (p – a) (q – b) (r – c)]
12.Ans. (a)
Consider the following events:
A = Sum of the digits on the selected ticket is 8.
B = product of the digits on the selected ticket is zero.
There are 14 tickets having product of digits appearing on them as zero. The numbers on such tickets are 00, 01, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40.
and
Required probability
13.Ans. (a)
We have,
P(A) + P(B) – 2P (A B) = P …(i)
P(B) + P(C) – 2P (B C) = P …(ii)
P(C) + P(A) – 2P (C A) = P …(iii)
and P(A B C) = P2
Adding (i), (ii) and (iii), we get
2[P(A) + P(B) + P(C) + P(A B) – P(B C) – P(A C)] = 3p
P(A) + P(B) + P(C) – P(A B) – P(B C) – P(A C) = 3p/2
Required probability
= P(A B C)
= P(A) + P(B) + P(C)– P(A B) – P(B C) – P(A C) + P(A B C)
14.Ans. (a)
Let AB be a straight line segment of length a and let P and Q be two points on it such that AP = x and AQ = y. Then, we have to find the probability for |PQ| > c i.e. |x – y| > c.
Clearly, 0 x a and 0 y a.
We observe that there are two variables x and y satisfying 0 x a and 0 y a. In the Cartesian plane the inequalities 0 x a and 0 y a represent the region enclosed by the square having x = 0, y = 0, x = a and y = a as its sides. Clearly, area enclosed by this square is a2.
Now, |x – y| > c
x – y > c and – (x – y) > c
x – y > and y – x > c
In order to find P(|x – y| > c), we require the area of the region enclosed by x – y > c, y – x > c, 0 x a and 0 y a. Clearly, shaded region is Fig. 1 represents the area enclosed.
We have,
Area of the shaded region = Area APQ + Area BRS
Hence, required probability
15.Ans. (c)
The number of ways of choosing a and b from the given set of 5n integers is 5nC2.
Let us divide the given set of 5n integers in 5 groups as follows:
G1 : 1, 6, 11, …, 5n – 4
G2 : 2, 7, 12, …, 5n – 3
G3 : 3, 8, 13, …, 5n – 2
G4 : 4, 9, 14, …, 5n – 1
G5 : 5, 10, 15, …, 5n
We have, a4 – b4 = (a – b) (a + b) (a2 + b2)
So, we observe that a4 – b4 will be divisible by 5, if both a and b belong to the last group or if they belong to any of the remaining four groups. Thus, the number of favourable elementary events is nC2 + 4nC2.
Hence, required probability
16.Ans. (d)
Since a biscuit can be given to any one of N beggars. Therefore, each biscuit can be distributed in N ways. So, the total number of ways of distributing n biscuits among N beggars is
Now, r biscuits can be given to a particular beggar in nCr ways and the remaining (n – r) biscuits can be distributed to (N – 1) beggars in (N – 1)n-r ways. Thus, the number of ways in which a particular beggar receives r biscuits is
Hence, required probability
17.Ans. (b)
We have,
and 2P(A) = P(B) = p
Now,
18.Ans. (b)
Let Ai denote the event of getting 1, first time, in ith throw.
Clearly,
p = Probability of getting 1 in a throw
q = Probability of not getting 1 in a throw
Required probability = P(A2 A4 A6 ….)
= P(A2) +P(A4)+P(A6)+P(A8) + …..
= pq + q3p + q5p + q7p + …..
19.Ans. (d)
We have,
Suppose B has n outcomes. Then,
P(B)
Since A and B are independent events.
P(A B) = P(A) P(B)
is an integer between 0 and 10
is an integer between 0 and 5
n = 5, 10.
20.Ans. (c)
Let A denote the event that each American man is seated adjacent to his wife and B denote the event that Indian man is seated adjacent to his wife. Then,
Required probability = P(B/A)
21.Ans. (d)
From multiplication Theorem on probability, we have
22.Ans. (a)
We have,
and
and
and
where P(E) = x and P(F) = y
[On eliminating xy]
Substituting in we get
When and for
Hence or
23.Ans. (a)
For i = 1, 2, 3, let Wi (Bi, Ri) denote the event that the ball drawn from ith bag is white (black, red).
Required probability
= P{(W1 W2 W3) (B1 B2 B3) (R1 R2 R3)}
= P(W1 W2 W3) + P (B1 B2 B3) + (R1 R2 R3)
= P(W1)P(W2)P(W3)+ P(B1)P(B2)P(B3) +P(R1)P(R2)P(R3)
24.Ans. (d)
Let Ei(i = 1, 2, 3) denote the event of selecting ith bag and A denote the event of drawing one white and one red ball. Then,
Required probability = P(E2/A)
25.Ans. (b)
Clearly,
Required probability
26.Ans. (c)
Clearly, point (a, a2) lies on the opposite sides of the two lines.
27.Ans. (b)
It is given that a, b, c are in H.P.
passes through the fixed point (1, -2).
28.Ans. (b)
The line cuts y-axis at P(0, 1). Clearly, its image Q(0, -1) lies on the reflected ray AR produced backward.
So, the equation of the reflected ray AR is
or
29.Ans. (a)
Let (h, k) be the coordinates of the centroid of ABC whose third vertex C(x1, y1) moves on the line 2x + 3y = 9.
We have,
and and
As lies on
Hence, the locus of is
30.Ans. (b)
We have,
S1 = 102 + 72 – 4 10 – 2 7 – 20 > 0.
So, P lies outside the circle.
Joint P with the centre C(2, 1) of the given circle. Suppose PC cuts the circle at A and B. Then, PB is the greatest distance of P from the circle.
Now, we have
and, CB = Radius
PB = PC + CB = (10 + 5) = 15
31.Ans. (b)
Let P(t2, t) be a point on the curve x = y2 and S be the distance between P and the line y – x – 1 = 0. Then,
Clearly, S is minimum when
For this value of t, we get
32.Ans. (a)
We have,
for all x R.
Hence, f(x) is periodic with period /2.
Students are advised to learn the periods of following standard functions.
Function / Periodsin x
cos x
tan
cosec x
sec x
cot x
x-[x] / 2
2
2
2
1
33.Ans. (a)
We have,
34.Ans. (b)
We have,
[Using L’ Hospital’s Rule]
f(a) = g(a) = k]
k = 4
35.Ans. (b)
We have,
36.Ans. (c)
Not given solution
37.Ans. (c)
Since f(x) is an odd differentiable function defined on R. Therefore,
f(-x) = - f(x) for all x R
Differentiable both sides w.r.t.x, we get
-f'(-x) = - f'(x) for all x R
f'(-x) = f'(x) for all x R
f'(-3) = f'(3) = - 2
ALITERWe know that the derivative of a differentiable odd function is an even function. Therefore, f ' (x) is an even function. Hence,
f ' (-3) = f ' (3) = - 2
38.Ans. (a)
We have,
39.Ans. (b)
We have,
Putting y = vx and we get
[On integrating]
which is the required solution.
40.Ans. (a)
We have,
…(i)
Putting x = 0 and y = 1 in (i), we get
Putting C = 4 in (i), we get
Putting we get
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