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Past Year Exams with Solutions
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∗† Econschool
1 ISI PEA 2006
1+x 1 −x
, 0 < x < 1 , then f
2 x 1+x^2
equals
A. 2 f(x)
f (x) 2
C. (f (x))
2 ;
D. none of these.
∂u ∂x
∂u ∂y
∂u ∂z
equals
C. u
D. none of these.
The number of subsets S of C that contain k elements and that also have the property
that S ∩ A contains i elements is
m
i
n
i
m
k − i
n
i
m
i
n
k − i
2 −| x | | is decreasing
is
A. one;
B. two;
C. three;
D. none of these.
∑n
i=1 x
2 i
is a measure of central tendency. If AM denotes the arithmetic mean of
the set of numbers, then which of the following statements is correct?
A. RMS < AM always;
B. RMS > AM always;
C. RMS < AM when the numbers are not all equal;
D. RMS > AM when numbers are not all equal.
For k > 1 , put ∆
k f = ∆
k− 1 f
. Then ∆
k f (x) equals
∑k
j=0(−1)
j
k
j
f (x + j)
B. p + q
1 p
1 q
D. none of these.
S = (x 1 y 2 + x 2 y 1 ) − 2 x 1 x 2. The number S is
A. always a negative integer;
B. can be a negative fraction;
C. always a positive number;
D. none of these.
2 +12x− 5 y = 11
is
D. none of these.
is incorrect?
A. f always has at least one maximum in the interval [0,1]
B. f always has at least one minimum in the interval [0,1]
C. ∃x ∈ [0, 1] such that f (x) = x
D. the function f must always have the property that f (0) ∈ { 0 , 1 } f (1) ∈
{ 0 , 1 } and f (0) + f (1) = I
2 ISI PEA 2007
(1+x)a− 1 (1+x)β^ − 1
equals
α β
α+ β+
α− 1 β− 1
2 − 1 is
A. odd;
B. not prime;
C. necessarily positive;
D. none of the above.
kx is a probability density function on the
interval [0,1] is
A. k = log 2;
B. k = 2 log 2;
C. k = 3 log 3;
D. k = 3 log 4
2 − q
2 is a prime number. Then, p − q is
A. a prime number;
B. an even number greater than 2
C. an odd number greater than 1 but not prime;
D. none of these.
A. is differentiable on (a, b)
B. is continuous in [a, b] but not differentiable;
C. has a continuous inverse;
D. none of these.
x 3 4
= 0 is satisfied by
A. x = 1;
B. x = 3
C. x = 4
D. none of these.
x +
x +
x +
x +.. ., then f
′ (x) is
x 2 f (x)− 1
A. no solution;
B. a unique non-degenerate solution;
C. a corner solution;
D. infinitely many solutions.
both f (x; 1) and f (x; 0) are probability density functions (p · d, f.). Then
A. f (x; θ) is a p.d.f. for all values of θ
B. f (x; θ) is a p.d.f. only for 0 < θ <
1 2
C. f (x; θ) is a p.d.f. only for
1 2
≤ θ < 1
D. f (x; θ) is not a p.d.f. for any value of θ.
x 5 1 4 3 2
y 0 4 2 0 − 1
A. r > 0;
B. r < − 0 .5;
C. − 0. 5 < r < 0;
D. r = 0.
transformation of a homogeneous function of degree one. Let f 1 (x) =
∑n
i=1 x
r ∑ i^ ,^ and^ f^2 (x) = n i=
aixi + b, where xi > 0 for all i, 0 < r < 1 , ai > 0 and b are constants. Then
A. f 1 is not homothetic but f 2 is;
B. f 2 is not homothetic but f 1 is;
C. both f 1 and f 2 are homothetic;
D. none of the above.
1 1 −x
, then h(h(h(x)) equals
1 1 −x
B. x
1 x
D. 1 − x
|x| x
is
A. continuous but not differentiable at x = 0
B. differentiable at x = 0
C. not continuous at x = 0;
D. continuously differentiable at x = 0.
2 dx (x−2)(x−1)x
equals
A. log
x(x−2) (x−1)^2
∣ + constant;
B. log
(x−2) x(x−1)^2
∣ + constant;
C. log
x^2 (x−1)(x−2)
∣ + constant;
D. log
(x−2)^2 x(x−1)
∣ + constant.
show up. If the restaurant has 50 tables and takes 52 reservations, then the probability
that it will be able to accommodate everyone is
209 552
4 5
4 5
1 5
example, [0.5] = 0, [1] = 1 and [1.5] = 1. Let I =
0
[x] + [x
2 ] dx. Then I equals
5 − 2
√ 2 2
D. none of these.
3 − n) (n
2 − 4) ( for n = 3, 4 ,.. .) is
A. divisible by 6 but not always divisible by 12
B. divisible by 12 but not always divisible by 24
C. divisible by 24 but not always divisible by 120
D. divisible by 120 but not always divisible by 720.
respectively. One buyer buys 5 kg. of A and 10 kg. of B and another buyer spends Rs 100
on A and Rs. 150 on B. If the average expenditure per mango (irrespective of variety) is the
same for the two buyers, then which of the following statements is the most appropriate?
A. p 1 = p 2
B. p 2 =
3 4
p 1
C. p 1 = p 2 or p 2 =
3 4
p 1
3 4
p 2 p 1
(r
2 ) between x
2 and y is found to be 1. Which of the following statements is the most
appropriate?
A. In the (x, y) scatter diagram, all points lie on a straight line.
B. In the (x, y) scatter diagram, all points lie on the curve y = x
2 .
C. In the (x, y) scatter diagram, all points lie on the curve y = a + bx
2 , a > 0 , b > 0
3 ISI PEA 2008
dx x+x log x
equals
A. log |x + x log x|+ constant
B. log |1 + x log x|+ constant
C. log | log x|+ constant
D. log |1 + log x|+ constant.
−1 + x is
√^1 x− 1
B. x
2
x − 1 ,
D. none of these.
x +
x+ x− 1
x+ x^2 +
is
D. none of these
x + 3y ≥ 3
3 x + y ≥ 3
x + y ≤ 3
An optimal solution of the above programme is given by
A. x =
3 4
, y =
3 4
B. x = 0, y = 3
C. x = − 1 , y = 3
D. none of the above.
The function f 1 is defined by f 1 (a 1 ) = b 1 , f 1 (a 2 ) = b 2 , f 1 (a 3 ) = b 3 and the function
f 2 is defined by f 2 (b 1 ) = c 1 , f 2 (b 2 ) = c 2 , f 2 (b 3 ) = f 2 (b 4 ) = c 3. Then the mapping
f 2 ◦ f 1 : {a 1 , a 2 , a 3 } → {c 1 , c 2 , c 3 } is
A. a composite and one − to − one function but not an onto function.
B. a composite and onto function but not a one − to − one function.
C. a composite, one − to − one and onto function.
D. not a function.
1 t− (^1) and y = t
t t− (^1) , t > 0 , t 6 = 1 then the relation between x and y is
A. y
x = x
1 y (^) ,
B. x
1 y (^) = y
1 x (^) ,
C. x
y = y
x ,
D. x
y = y
1 x (^).
xB ≥ 0 and xS ≥ 0 , is
D. none of these.
0
[x]
n f
′ (x)dx, where [x] stands for the integral part of x, n is a positive integer
and f
′ is the derivative of the function f, is
A. (n + 2
n ) (f (2) − f (0)),
n ) (f (2) − f (1))
n f (2) − (
n − 1) f (1) − f (0),
D. none of these.
42% completed university education and remaining 37% completed only school education.
Of those who went to college 61% reads newspapers regularly, 35% of those who went to
the university and 70% of those who completed only school education are regular readers of
newspapers. Then the percentage of those who read newspapers regularly completed only
school education is
D. none of these.
−x defined on the real line is
A. continuous but not differentiable at zero,
B. differentiable only at zero,
C. differentiable everywhere,
D. differentiable only at finitely many points.
team to win the World Cup. The team has equal odds for winning or losing any match.
What is the probability that they will win in odd number of matches?
unities. Each one did his own simplification.
′ s method: Divide the set of number into 5 equal parts, calculate the mean for each part
and then take the mean of these.
′ s method: Divide the set into 2000 and 3000 numbers and follow the procedure of A.
′ s method: Calculate the mean of 4500 numbers (which are 6 = 1 ) and then add 1. Then
3 5
2 x + 3y − 1 = 0 and 5x − 2 y + 3 = 0 are
A. equal but have opposite signs,
2 3
and
2 5
1 2
and −
3 5
D. Cannot say.
a b
b c
c d
d a
is always
A. less than
B. less than 2 but greater than or equal to
C. less than 4 but greater than or equal 2
D. greater than or equal to 4.
A. 2 < x < 3 ,
B. x > 3 ,
C. x < 2 ,
D. no such x exists.
3 − 5 x
2
A. all roots between 1 and 2 ,
B. all negative roots,
C. a root between 0 and 1 ,
D. all roots greater than 2.
f (x) = ax
2 e
−kx (k > 0 , 0 ≤ x ≤ ∞)
Then, a equals
k^3 2
k 2
k^2 2
D. k
n
x
p
x (1−
p)
n−x
. Then which of the following inequalities hold.
A. (n + 1)p − 1 < r < (n + 1)p,
B. r < (n + 1)p − 1
C. r > (n + 1)p
D. r < np.
(y 1 , y 2 ,... , yj + δ,... , yk − δ,... , yn) where yk − δ > yk− 1 >... > yj+1 > yj + δ δ > 0.
Let σ : standard deviation of y and σ
′ : standard deviation of y
′
. Then
A. σ < σ
′ ,
B. σ
′ < σ,
C. σ
′ = σ,
D. nothing can be said.
xf (x) 1 −F (x)
Then for x < e
μ and f (x) =
e−^
(log x−μ)^2 2
x
√ 2 π
, the function r(x) is
A. increasing in x,
B. decreasing in x
C. constant,
D. none of the above.
negative and each of its rows and columns sum to 1. Let yn× 1 = Pn×nxn× 1 where elements
of y are some rearrangements of the elements of x. Then
A. P is bistochastic with diagonal elements 1 ,
B. P cannot be bistochastic,
C. P is bistochastic with elements 0 and 1 ,
D. P is a unit matrix.
x x+
. Define fn(x) = f 1 (fn− 1 (x)) , where n ≥ 2. Then fn(x) is
A. decreasing in n,
B. increasing in n,
C. initially decreasing in n and then increasing in n,
D. initially increasing in n and then decreasing n.
1 −x−^2 n 1+x−^2 n^
, x > 0 equals
D. The limit does not exist.
∗ 1 , x
∗ 2 ) to the opti-
mization problem minimize f (x 1 , x 2 ) subject to x 1 + x 2 = 21 is
A. (x
∗ 1 = 10.^5 , x
∗ 2 = 10.5),
B. (x
∗ 1 = 11, x
∗ 2 = 10)
C. (x
∗ 1 = 10, x
∗ 2 = 11),
D. None of these.
1 9
1 2
2 9
1 3
Y denote the number of white and red balls respectively. The correlation between X and
Y is
D. some real number between −
1 2
and
1 2
x 2
and
h(x) = min{f (x), g(x)} with 0 ≤ x ≤ 1. Then h is
A. continuous and differentiable
B. differentiable but not continuous
C. continuous but not differentiable
D. neither continuous nor differentiable
2 f (x) + 3f (−x) = 55 − 7 x
for every x ∈ R, then f (3) equals
and 5 P.M.. They agree that each is to wait not more than 15 minutes for the other.
Assuming that each is independently equally likely to arrive at any point during the hour,
find the probability that they meet.
15 16
7 16
5 24
22 175
x 1
x 2
x 2
x 3
x 3
x 1
is always
1 3
12 +2^2 +...+n^2 n^3
equals
1 3
1 6
root. x
2 − 7 x + 12 = 0 and x
2 − 8 x + b = 0 The root of x
2 − 8 x + b = 0 that is not a root
of x
2 − 7 x + 12 = 0 is
1 (^2) +n
1 (^3) +n
1 (^4). Then, which of the following relationships
between n and μ is correct?
A. n = μ
B. n > μ
C. n < μ
D. None of the above.
A. f (z) = z
2
B. f (z) = az for some real number a
C. f (z) = log z
D. f (z) = e
z
∣ ∣ ∣ ∣ ∣ ∣ x a 2
2 x 0
2 1 1
for all values of x.
f (x) =
Ax − B if x ≤ − 1
2 x
2
4 if x > 1
3 4
1 4
1 4
3 4
1 2
1 2
x
2 x )
1 x (^) is
C. e
Y from 25 pairs of observations obtained the following results:
650 ,
2 = 460,
XY = 508. It was later discovered that at the time of
inputing, the pair (X = 8, Y = 12) had been wrongly input as (X = 6, Y = 14) and the
pair (X = 6, Y = 8) had been wrongly input as (X = 8, Y = 6). Calculate the value of the
correlation coefficient with the correct data.
4 5
2 3
5 6
2 − 1 which is nearest to the point (2,-0.5) is
D. None of the above
x ≤ 2 , then mean of X is
1 2
1 5
3 4
f (x + y) = f (xy) for all x, y ≥ 4. If f (8) = 9, then f (9) =
about f?
A. It decreases on the interval
−
1 (^2) , 2 + 3−^
1 2
B. It increases on the interval
− (^12) , 2 + 3
− (^12)
C. It decreases on the interval
−
1 2
D. It decreases on the interval [2,3]
by 5 metres by cutting squares of side x metres out of each corner and folding up the sides.
The largest possible volume in cubic metres of such a box is
