I.S.I. 2014 B.Stat B.Math entrance Objective Paper Solution

(This is a work in progress. If you have alternative solution put as comment)

1. The systems of inequalities $a – b^2 \ge \frac {1}{4} , b- c^2 \ge \frac {1}{4}, c – d^2 \ge \frac {1}{4}, d – a^2 \ge \frac {1}{4}$ has

(A) no solution; (B) Exactly one solution; (C) exactly two solutions; (D) infinitely many solutions;

2. Let $\log_{12} 18 = a.$ Then $\log_{24} 16$ is equal to

(A) $\displaystyle { \frac {8 – 4a}{5 – a} }$ ; (B) $\displaystyle {\frac{1}{3+a}}$ ; (C) $\displaystyle { \frac {4a – 1}{2 +3a}}$ ; (D) $\displaystyle { \frac{8 – 4a}{5 + a} }$ ;

3. The number of solutions of the equation tan x + sec x = 2 cos x , where $0 \le x \le \pi$

(A) 0; (B) 1; (C) 2; (D) 3;

(4) Using the digits 2, 3, 9 how many six digit numbers can be formed which are divisible by 6?

(A) 41; (B) 80; (C) 81; (D) 161;

(16) Let $A_0 = \phi$ ( the empty set) . For each i = 1, 2, 3, … , define the set $A_i = A_{i-1} \cup { A_{i-1} }$ . The set $A_3$ is

(A) $\phi$ ; (B) { $\phi$ } ; (C) ${ \phi { \phi } }$ ; (D) ${ \phi , { \phi } , { \phi , { \phi } } }$ ;

(17) Let $\displaystyle { f(x) = \frac{1}{x-2} }$ . The graphs of the functions $f$ and $f^{-1}$ intersect at

(A) $(1+\sqrt {2} , 1 + \sqrt {2} )$ and $(1 – \sqrt {2} , 1 – \sqrt {2} )$ ;
(B) $(1+\sqrt {2} , 1 + \sqrt {2} )$ and $(\sqrt {2} , -1 – \frac{1}{ \sqrt {2}} )$ ;
(C) $(1-\sqrt {2} , 1 – \sqrt {2} )$ and $(- \sqrt {2} , -1 + \frac{1}{\sqrt {2}} )$ ;
(D) $(\sqrt {2} , -1 – \frac{1}{\sqrt {2}} )$ and $(\sqrt {2} , 1 + \frac{1}{\sqrt {2}} )$ ;

(18) Let N be a number such that whenever you take N consecutive positive integers, at least one of them is coprime to 374. What is the smallest possible value of N?

(A) 4; (B) 5; (C) 6; (D) 7;

(19) Let $A_1 , A_2 , … , A_{18}$ be the vertices of a regular polygon with 18 sides. How many of the triangles $\Delta A_i A_j A_k$ , 1 le i < j < k \le 18 , are isosceles but not equilateral?

(A) 63; (B) 70; (C) 126; (D) 144;

(20) The limit $\displaystyle { \lim _ {x to 0} \frac{\sin ^{\alpha} x } {x} }$ exists only when

(A) $\alpha \ge 1$ ; (B) (A) $\alpha = 1$ ; (C) $|\alpha| \le 1$ ; (D) $\alpha$ is a positive integer

(21) Consider the region $R = { (x, y) : x^2 + y^2 \le 100 , \sin (x +y) > 0 }$ . What is the area of R?

(A) $25 \pi$ ; (B) $50 \pi$ ; (C) $50$ ; (D) $100 \pi – 50$ ;

(22) Consider a cyclic trapezium whose circumcenter is one of the sides. If the ratio of the two parallel sides is 1:4, what is the ratio of the sum of the two oblique sides to the longer parallel side?

(A) $\sqrt 3 : \sqrt 2$ ; (B) 3:2 ; (C) $\sqrt 2 : 1$ ; (D) $\sqrt 5 : \sqrt 3$ ;

(23) Consider the function $\displaystyle { f(x) = \left \lbrace \log _ e \left (\frac{4 + \sqrt{2x}}{x} \right) \right \rbrace ^2 }$ for x > 0. Then

(A) f decreases upto some point and increases after that
(B) f increases upto some point and decreases after that
(C) f increases initially, then decreases and then again increases
(D) f decreases initially, then increases and then again decreases

(24) What is the number of ordered triplets (a, b, c), where a, b, c are positive integers (not necessarily distinct), such that abc = 1000?

(A) 64; (B) 100; (C) 200; (D) 560;

(25) Let $f : (0, \infty) to (0, \infty)$ be a function differentiable at 3 and satisfying f(3) = 3 f'(3) > 0 . Then the limit $\displaystyle { \lim_{x to \infty} \left ( \frac { f \left ( 3 + \frac{3}{x} \right ) }{f(3)} \right )^x }$

(A) exists and is equal to 3; (B) exists and is equal to e; (C) exists and is always equal to f(3) ; (D) need not always exist.

(26) Let z be a non – zero complex number such that $|z – \frac{1}{z}| = 2$ . What is the maximum value of |z|?

(A) 1; (B) $\sqrt {2}$ ; (C) 2; (D) $1 + \sqrt 2$ ;

(27) The minimum value of |sin x + cos x + tan x + cosec x + sec x + cot x| is

(A) 0; (B) $2 \sqrt 2 – 1$ ; (C) $2 \sqrt 2 + 1$ ; (D) 6;

28. For any function $f : X to Y$ and any subset A of Y, define $f^{-1} (A) = { x \in X : f(x) \in A }$.
Let $A^c$ denote the complement of A in Y. For subsets $A_1. A_2$ of Y, consider the following statements:

(i) $f^{-1} (A_1^c \cap A_2^c ) = (f^{-1} (A_1))^c \cup (f^{-1} (A_2))^c$;
(ii) Of $f^{-1}(A_1) = f^{-1} (A_2)$ then $A_1 = A_2$

Then,

(A) both (i) and (ii) are always true;
(B) (i) is always true, but (ii) may not always be true.
(C) (ii) is always true, but (i) may not always be true.
(D) neither (i) nor (ii) is always true.

(29) Let f be a function such that f”(x) exists, and f”(x) > 0 for all $x \in [a, b]$. For any point $c \in [a, b]$ , let A(c) denote the area of the region bounded by y = f(x) the tangent to the graph of f at x = c and the lines x = a and x = b. Then,

(A) A(c) attains it’s minimum at $c = \frac {1}{2} (a +b )$ for any such f;
(A) A(c) attains it’s maximum at $c = \frac {1}{2} (a +b )$ for any such f;
(A) A(c) attains it’s minimum at both $c = a , c = b$ for any such f;
(D) the points c where A(c) attains its minimum depend on f.

(30) In $\Delta ABC$ , the lines BP, BQ, trisect $\angle ABC$ and the lines CM , CN trisect $\angle ACB$ . Let BP and CM intersect at X and BQ and CN intersect at Y. If $\angle ABC = 45^o$ and $\angle ACB = 75^o$ , then $\angle BXY$ is

(A) $45^o$ ; (B) $47 \frac{1}{2} ^o$ ; (C) $50^o$ ; $55^o$

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1. The systems of inequalities ${a – b^2 \ge \frac {1}{4}}$, ${b – c^2 \ge \frac {1}{4}}$, ${c – d^2 \ge \frac {1}{4}}$ , ${d – a^2 \ge \frac {1}{4}}$ has

(A) no solution; (B) Exactly one solution; (C) exactly two solutions; (D) infinitely many solutions;

Discussion:

If we add all of these inequalities we get $0 \ge a^2 – a + \frac{1}{4} + b^2 – b + \frac{1}{4} + c^2 – c + \frac{1}{4} + d^2 – d + \frac{1}{4}$

This implies $\left ( a – \frac {1}{2} \right )^2 + \left ( b – \frac {1}{2} \right )^2 + \left ( c – \frac {1}{2} \right )^2 + \left ( d – \frac {1}{2} \right )^2 \le 0$

Sum of squares can never be less than 0. They can be equal to zero if and only if each of the squares are zero. Hence the only solution is $a = b = c = d = \frac {1}{2}$

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2. Let $\log_{12} 18 = a.$ Then $\log_{24} 16$ is equal to

(A) $\displaystyle { \frac {8 – 4a}{5 – a} }$ ; (B) $\displaystyle {\frac{1}{3+a}}$ ; (C) $\displaystyle { \frac {4a – 1}{2 +3a}}$ ; (D) $\displaystyle { \frac{8 – 4a}{5 + a} }$ ;

Discussion:

$\displaystyle { \frac {\log 18}{\log 12} = a \implies \frac{2\log 3 + \log 2}{\log 3 + 2\log 2} = a \implies 2\log 3 + \log 2 = a \log 3 + 2a \log 2 }$

Hence $\displaystyle { (2-a) \log 3 = (2a – 1) \log 2 \implies \log 3 = \frac {2a – 1}{2-a} log 2 }$

Now $\displaystyle { \log_{24} 16 = \frac {log 24}{\log 16} = \frac {4 \log 2}{3\log 2 + \log 3} }$

Replacing log 3 by $\displaystyle { \frac {2a – 1}{2-a} \log 2 }$ we get

${\displaystyle {\log_{24} 16 = \frac {4 \log 2}{3\log 2 + \frac {2a – 1}{2-a} \log 2} = \frac {4 (2-a)}{ 3(2-a) + 2a – 1} = \frac {8-4a)}{ 5-a}}}$

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3. The number of solutions of the equation tan x + sec x = 2 cos x , where $0 \le x \le \pi$

(A) 0; (B) 1; (C) 2; (D) 3;

Discussion: Assume f(x) = \tan x + \sec x – 2 \cos x

Taking derivative we have $\displaystyle { f'(x) = \sec^2 x + \sec x \tan x + 2 \sin x = \sec^2 x + \frac { \sin x}{\cos^2 x} + 2 \cos x }$

Now in the interval $0 \le x \le \pi$ f'(x) >0

Hence the function is always increasing in the given interval (except at point of discontinuities).

The function is discontinuous at $x = \frac{\pi}{2}$ So we separately consider two intervals $[ 0, \frac{\pi}{2} )$ and $(\frac{\pi}{2} , \pi ]$

Now f(0) = -1 and $\displaystyle { f(\frac{\pi}{3}) = \sqrt 3 + 2 – 1 = \sqrt 3 + 1 }$

Since the function is continuous in the interval $0 \le x \frac{\pi}{3}$ and it is negative at x = 0 and positive at $x = \frac {\pi}{3}$ hence by intermediate value property theorem, the function ‘cuts’ the x axis at least once.

Since the derivative is positive, hence it would ‘cross the x axis’ exactly once in the interval $[ 0, \frac{\pi}{2} )$

Similarly we can show that it has exactly one solution in the interval $(\frac{\pi}{2} , \pi ]$.

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(4) Using the digits 2, 3, 9 how many six digit numbers can be formed which are divisible by 6?

(A) 41; (B) 80; (C) 81; (D) 161;

Discussion:

Since the number is divisible by 6, it has to be even and divisible by 3.

Hence sum of it’s digits must be divisible by 3. Therefore number of 2’s present can be either 0, 3 or 6 (why? Suppose there are two 2’s present. Then sum of the digits will be (some 6) + (some 9) + 4 . But 6’s and 9’s are divisible by 3 and 4 is not. Hence the sum of the digits is not divisible by 3 therefore the number is not divisible by 3. In similar manner we can discuss the remaining cases).

Case 1: 6 two’s are use: 1 case (222222)
Case 2: 3 two’s are used: remaining three digits are a mix of 3’s and 9’s. Now the Last digit has to be 2 (the number must be even). So first 5 digits are filled up with two 2’s and a mix of 3’s and 9’s.

We choose 2 out of 5 spots (for the two’s) in $\dbinom {5}{2} = 10$ ways and for each of the remaining three spots we have 2 choices (3 or 9) hence $2 \times 2 \times 2$.

Hence total number of ways = $10 \times 8 = 80$

Case 3: 0 two’s are used: not possible as the number is not even.

Total number of numbers = 80 +1 = 81

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(5) What is the value of the following integral? $\displaystyle { \int_\frac{1}{2014}^{2014} \frac{\tan^{-1} x} {x} dx }$

(A) $\displaystyle { \frac {\pi}{4} \log 2014 }$ ; (B) $\displaystyle { \frac {\pi}{2} \log 2014 }$ ; (C) $\displaystyle { {\pi}\log 2014 }$ ; (D) $\displaystyle { \frac {1}{2} \log 2014 }$ ;

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(6) A light ray travelling along the line y = 1, is reflected by a mirror placed along the line x = 2y. The reflected ray passes along which line?

(A) 4x – 3y = 5; (B) 3x – 4y = 2; (C) x – y = 1; (D) 2x – 3y = 1;

Since BD bisects $\angle ABD$ hence by angle bisector theorem $\displaystyle { \frac {BA}{BC} = \frac {AD}{DC} \implies \frac{2}{\sqrt{4 + (y -1)^2 } } = \frac{1}{-y} }$ (assuming y is negative).

Squaring both sides we get $4y^2 = 4 + y^2 + 1 – 2y \implies 3y^2 + 2y -5 = 0$

Value of y by solving this equation is -5/3.

Hence equation of the reflected line is $\frac {y-1}{x -2} = \frac { -5/3 -1} {0 -2}$

This implies 6(y -1) = 8(x -2)  or  4x – 3y = 5

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(7) For a real number x, [x] denote the greatest integer less than or equal to x. Then the number of real solutions of |2x – [x]| = 4 is

(A) 1; (B) 2; (C) 3; (D) 4;

Discussion:

Let x = I + f (where I = [x] and f = x – [x] )

Hence |2I + 2f – I| = 4

or |I + 2f| = 4 or I + 2f = 4 or -4

Now 2f = 4 – I or 2f = -4 -I

Hence 2f must be an integer. There f = 1/2 or 0

Thus I = 4, -4, 3 or -5

Hence 4 solutions.

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(8) What is the ratio of areas of regular pentagons inscribed inside and circumscribed around a given circle?

(A) $\cos 36^o$ ; (B) $\cos^2 36^o$ ; (C) $\cos^2 54^o$ ; (D) $\cos^2 72^o$ ;

Discussion:

Note that ratio of the areas of two pentagons is $\displaystyle { \frac{5 \Delta OED} {5 \Delta OFG } }$= $\displaystyle { \frac {OE \times OF \times \sin \angle EOD }{OF \times OG \times \sin \angle FOG} }$= $\displaystyle { \frac {OD^2 \sin 72^o}{OF^2 s\in 72^o} }$ = $\displaystyle {\left ( \frac {OD}{OF} \right ) ^2 }$

But $\displaystyle { \frac{OD}{OF} = \cos 36^o }$ since $\angle DOF = 36^o$

Hence answer is $\cos^2 36^o$

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(9) Let $z_1$ and $z_2$ be nonzero complex numbers satisfying $|z_1 + z_2| = |z_1 – z_2|$. The circumcenter of the triangle with the points $z_1 , z_2$ and the origin as it’s vertices is given by

(A) $\frac {1}{2} (z_1 – z_2)$ (B) $\frac {1}{3} (z_1 + z_2)$ (C) $\frac{1}{2} (z_1 + z_2)$ (D) $\frac {1}{3} (z_1 – z_2)$

Discussion:

Suppose A is the origin.

Let us complete the parallelogram with $A, z_1$ and $z_2$ . By parallelogram law of vector addition the other vertex $z_4 = z_1 + z_2$ . So the length of $Az_4 = |z_1 + z_2|$  Also the length of $z_1 z_2 = |z_1 – z_2|$

So according to the problem length of two diagonals of the parallelogram are equal.

Hence it is a rectangle. Therefore $Az_1z_2$ is a right angled triangle implying circumcenter is in the middle of hypotenuse. It is $\displaystyle { \frac {z_1 + z_2} {2} }$

(10) In how many ways can 20 identical chocolates be distributed among 8 students so that each student gets at least one chocolate and exactly two students gets at least two chocolates each?

(A) 308; (B) 364; (C) 616; (D) $\displaystyle { \dbinom {8}{2} \dbinom{17}{2} }$

Discussion:

First we give one chocolate to each student. That leaves us with 12 more chocolates. Now we must give one chocolate more to each of two students. So we choose the two students in $\dbinom{8}{2} = 28$ ways.

Finally remaining 10 chocolates have to be distributed among those two students (as if we give a chocolate to any other student then he will have two or more chocolates which we do not want) . This is same as the number of non negative  integer solutions of the equation $s_1 + s_2 = 10$ which is $\dbinom {11}{1} = 11$ .

Hence the total number of ways is $28 \times 11 = 308$

(11) Two vertices of a square lie on a circle of radius r, and the other two vertices lie on a tangent to this circle. Then, each side of the square is

(A) $\displaystyle { \frac {3r}{2} }$ ;  (B) $\displaystyle { \frac {4r}{3} }$ ;  (C) $\displaystyle { \frac {6r}{5} }$ ;  (A) $\displaystyle { \frac {8r}{5} }$ ;

Discussion:

Teacher: This diagram is your clue. Side of the triangle is of length ‘s’. FEDC is the square. A is the center. FE is tangent to the circle at B. BA extended intersects CD at H. Suppose $\angle ABC = x , \angle ACH = y$

Student: Is B the midpoint of EF?

Teacher: Right guess! Can you prove this?

Student: Let us extend BA to meet CD at H. AB is perpendicular to EF at B as A is the center and EF is tangential to the circle at B. So AH is perpendicular to CD at H (as CD parallel EF).

Now H is the center of CD as A is the center of the circle and AH is perpendicular to the chord CD at H. As H is the midpoint of CD, so B is the midpoint of EF (as BH is parallel to CF as both are perpendicular to EF).

Teacher: Very nice. Now apply a little trigonometry. Alternatively you may use Pythagoras Theorem.

Student: (A solution due to Tiya Chakrabarty). Let us use Pythagoras Theorem first.

$AH = \sqrt { r^2 – \frac{s^2}{4} }$

Also AH + AB = s or

Hence $(s-r)^2 = \sqrt { r^2 – \frac{s^2}{4} }$

(12) Let P be the set of all numbers obtained by multiplying five distinct integers between 1 and 100. What is the largest integer n such that $2^n$ divides at least one element of P ?

(A) 8; (B) 20; (C) 24; (D) 25;

Discussion:

Teacher: We have to maximize the number of 2’s in the product of five numbers chosen.

Student: So we have to choose numbers from 1 to 100 which have the largest number of 2’s in their prime factorization. Like 64, 32, 16 etc.

Teacher: Right. So which five numbers will you choose?

Student: First 64 because it is the highest power of 2 (6th power) from 1 to 100. Next we choose 32. Then we can pick 96 because it also has 5 two’s. Finally we pick 16 and 48 (with four 2’s each).

So total 24 two’s in the product.

Teacher: Excellent! So ANSWER is (C) .

(13) Consider the function $f(x) = ax^3 + bx^2 + cx + d$ , where a, b, c, and d are real numbers with a > 0 . If f is strictly increasing, then the function g(x) = f'(x) – f”(x) + f”'(x) is

(A) zero for some $x \in setR$ ; (B) positive for all $x \in setR$ ; (C) negative for all $x \in setR$ ; (D) strictly increasing;

(14) Let A be the set of all points (h, k) such that the area of the triangle formed by (h, k), (5, 6) and (3, 2) is the 12 square units. What is the least possible length of a line segment joining (0, 0) to a point in A?

(A) $\displaystyle { \frac{4}{\sqrt 5} }$ ; (B) $\displaystyle { \frac{8}{\sqrt 5} }$ ; (A) $\displaystyle { \frac{12}{\sqrt 5} }$ ; (A) $\displaystyle { \frac{16}{\sqrt 5} }$ ;

Teacher: There is a formula to compute area of a triangle when coordinates of three vertices are given. Suppose $(x_1, y_1) , (x_2, y_2) , (x_3, y_3)$ are the three vertices then ${ \frac{|x_1 y_2 – x_1 y_3 + x_2 y_3 – x_2 y_1 + x_3 y_1 – x_3 y_2|}{2}}$ . You may apply it here.

Student: Okay. Applying it we have two cases : k = 2h + 8 or k = 2h – 16. Now distance of (h, k) from origin is $\sqrt {h^2 + k^2 }$ . Replacing k in terms of h we have two expressions for distance: $\sqrt {5h^2 – 64h + 256}$ ; $\sqrt {5h^2 + 32h + 64}$ . I think we can separately compute the lowest value of these two expressions and use the lower of the two.

Teacher: Precisely so. One may use calculus or normal ‘completing the square method’ here.

(15) Let P = { abc : a, b, c positive integers, $a^2 + b^2 = c^2$ , and 3 divides c } . What is the largest integer n such that $3^n$ divides every element of P ?

(A) 1; (B) 2; (C) 3; (D) 4;

Teacher: There is a little problem with this problem! The statement is not very clear. Assume that the highest power of 3 that divides c is 1. (Otherwise c could have any power of 3 in it’s prime factorization. So highest power of 3 dividing abc could be as large as you please).

Student: Okay. Now as 3 divides c hence it also divides $c^2$. Hence it divides $a^2 + b^2$. Now any square quantity can be either 0 or 1 mod 3. Here both $a^2$ and $b^2$ must be 0 mod 3 (because if one of them is 0 and one of them 1 mod 3 then their sum is 1 mod 3; if both are 1 mod 3 then their sum is 2 mod 3. Both are contradictions as the sum is 0 mod 3).

(16) Let $A_0 = \phi$ ( the empty set) . For each i = 1, 2, 3, … , define the set $A_i = A_{i-1} \cup { A_{i-1} }$ . The set $A_3$ is

(A) $\phi$ ; (B) { $\phi$ } ; (C) ${ \phi { \phi } }$ ; (D) ${ \phi , { \phi } , { \phi , { \phi } } }$ ;

(17) Let $\displaystyle { f(x) = \frac{1}{x-2} }$ . The graphs of the functions $f$ and $f^{-1}$ intersect at

(A) $(1+\sqrt {2} , 1 + \sqrt {2} )$ and $(1 – \sqrt {2} , 1 – \sqrt {2} )$ ;
(B) $(1+\sqrt {2} , 1 + \sqrt {2} )$ and $(\sqrt {2} , -1 – \frac{1}{ \sqrt {2}} )$ ;
(C) $(1-\sqrt {2} , 1 – \sqrt {2} )$ and $(- \sqrt {2} , -1 + \frac{1}{\sqrt {2}} )$ ;
(D) $(\sqrt {2} , -1 – \frac{1}{\sqrt {2}} )$ and $(\sqrt {2} , 1 + \frac{1}{\sqrt {2}} )$ ;

(18) Let N be a number such that whenever you take N consecutive positive integers, at least one of them is coprime to 374. What is the smallest possible value of N?

(A) 4; (B) 5; (C) 6; (D) 7;

(19) Let $A_1 , A_2 , … , A_{18}$ be the vertices of a regular polygon with 18 sides. How many of the triangles $\Delta A_i A_j A_k$ , 1 \le i < j < k \le 18 , are isosceles but not equilateral?

(A) 63; (B) 70; (C) 126; (D) 144;

(20) The limit $\displaystyle { \lim _ {x to 0} \frac{\sin ^{\alpha} x } {x} }$ exists only when

(A) $\alpha \ge 1$ ; (B) (A) $\alpha = 1$ ; (C) $|\alpha| \le 1$ ; (D) $\alpha$ is a positive integer

(21) Consider the region $R = { (x, y) : x^2 + y^2 \le 100 , \sin (x +y) > 0 }$ . What is the area of R?

(A) $25 \pi$ ; (B) $50 \pi$ ; (C) $50$ ; (D) $100 \pi – 50$ ;

(22) Consider a cyclic trapezium whose circumcenter is one of the sides. If the ratio of the two parallel sides is 1:4, what is the ratio of the sum of the two oblique sides to the longer parallel side?

(A) $\sqrt 3 : \sqrt 2$ ; (B) 3:2 ; (C) $\sqrt 2 : 1$ ; (D) $\sqrt 5 : \sqrt 3$ ;

(23) Consider the function $\displaystyle { f(x) = \left \lbrace \log _ e \left (\frac{4 + \sqrt{2x}}{x} \right) \right \rbrace ^2 }$ for x > 0. Then

(A) f decreases upto some point and increases after that
(B) f increases upto some point and decreases after that
(C) f increases initially, then decreases and then again increases
(D) f decreases initially, then increases and then again decreases

(24) What is the number of ordered triplets (a, b, c), where a, b, c are positive integers (not necessarily distinct), such that abc = 1000?

(A) 64; (B) 100; (C) 200; (D) 560;

(25) Let $f : (0, \infty) to (0, \infty)$ be a function differentiable at 3 and satisfying f(3) = 3 f'(3) > 0 . Then the limit $\displaystyle { \lim_{x to \infty} \left ( \frac { f \left ( 3 + \frac{3}{x} \right ) }{f(3)} \right )^x }$

(A) exists and is equal to 3; (B) exists and is equal to e; (C) exists and is always equal to f(3) ; (D) need not always exist.

(26) Let z be a non – zero complex number such that $|z – \frac{1}{z}| = 2$ . What is the maximum value of |z|?

(A) 1; (B) $\sqrt {2}$ ; (C) 2; (D) $1 + \sqrt 2$ ;

(27) The minimum value of |\sin x + \cos x + \tan x + \cosec x + \sec x + \cot x| is

(A) 0; (B) $2 \sqrt 2 – 1$ ; (C) $2 \sqrt 2 + 1$ ; (D) 6;

28. For any function $f : X to Y$ and any subset A of Y, define $f^{-1} (A) = { x \in X : f(x) \in A }$.
Let $A^c$ denote the complement of A in Y. For subsets $A_1. A_2$ of Y, consider the following statements:

(i) $f^{-1} (A_1^c \cap A_2^c ) = (f^{-1} (A_1))^c \cup (f^{-1} (A_2))^c$;
(ii) Of $f^{-1}(A_1) = f^{-1} (A_2)$ then $A_1 = A_2$

Then,

(A) both (i) and (ii) are always true;
(B) (i) is always true, but (ii) may not always be true.
(C) (ii) is always true, but (i) may not always be true.
(D) neither (i) nor (ii) is always true.

(29) Let f be a function such that f”(x) exists, and f”(x) > 0 for all $x \in [a, b]$. For any point $c \in [a, b]$ , let A(c) denote the area of the region bounded by y = f(x) the tangent to the graph of f at x = c and the lines x = a and x = b. Then,

(A) A(c) attains it’s minimum at $c = \frac {1}{2} (a +b )$ for any such f;
(A) A(c) attains it’s maximum at $c = \frac {1}{2} (a +b )$ for any such f;
(A) A(c) attains it’s minimum at both $c = a , c = b$ for any such f;
(D) the points c where A(c) attains its minimum depend on f.

(30) In $\Delta ABC$ , the lines BP, BQ, trisect $\angle ABC$ and the lines CM , CN trisect $\angle ACB$ . Let BP and CM intersect at X and BQ and CN intersect at Y. If $\angle ABC = 45^o$ and $\angle ACB = 75^o$ , then $\angle BXY$ is

(A) $45^o$ ; (B) $47 \frac{1}{2} ^o$ ; (C) $50^o$ ; $55^o$

Discussion:

Teacher: This can be done by simple angle chasing. Focus on $\Delta XBC$. Try to Show XY bisects $\angle XBC$

Student: In triangle $\Delta XBC$ , BQ bisects $\angle XBC$ and CN bisects $\angle XCB$ . So Y must be the incenter of $\Delta XBC$ implying XY bisects $\angle BXC$.

Now $\angle XBC = \frac{2}{3} \angle ABC = 30^o , \angle XCB = \frac {2}{3} \angle ACB = 50^o$ . Hence $\angle BXC = 100^0 \implies \angle BXY = 50^o$.

18 Replies to “I.S.I. 2014 B.Stat B.Math entrance Objective Paper Solution”

1. Vinod your solutions are nice. We welcome you to contribute to our blog as well if you wish. In that way we will be able to help many students.

If you wish to contribute articles// solutions to our website let us know your e mail address and we will send an invite

1. We are a group of teachers, who focus strictly on ISI and Olympaid students. We have functioned since 2010 and worked with 100s of students. If you wish to post solutions please be a contributor. Very few people can see if you post a link as a comment.

1. For Question numbr 27 just assume sinX=a and cosX=b , so cosecX=(1/a) and secX= (1/b) . tanX=(a/b) and cotX=(b/a). then put A.M.-G.M. Inequality-
[a+b+(1/a)+(1/b)+(a/b)+(b/a)] / 6> a.b.(a/b).(b/a).(1/a).(1/b)
so the ans is 6

1. Batman says:

no you are absolutely wrong . its correct answer is option B .
you cant use AM -GM inequality as they all are never positive simultaneously except in the first quadrant. AM GM is applied for positive Numbers only.

1. Batman says:

it is not that much easy as you think AM GM is applied to positive numbers only. answer will be option B

2. navien says:

for question 20 use lhospitals rule

3. navien says:

sol to q20 lhospitals law

4. ATUL says:

What is the answer of Question 22

5. abc = 1000,
=2^3.5^3
so, 3, 2s will be distributed in a,b, c.
Now, say there are three boxes and you have to put three balls on it. That can e done by (5!)|(3!.2!] ways.
For 5s also it is the same.
So such a,b,c can be found in (5!)|(3!.2!] * (5!)|(3!.2!] ways.
Here the a,b,c are ordered triplet.
So the ans is 100

6. For 22, the only possible case is the longer parallel side is passing through the centre of the circumscribed circle and the oblique sides are equal.
Now it can be solved by normal geometry and the ans is √3 :√2

7. Esha says:

Answers of ques. 16, 14, 17,20,23,

1. Ashani Dasgupta says:

What do you mean?

1. Esha says:

I Mean i want solutioms or answers to these questions

8. Ravi says:

Please can someone post solution of 2016 isi paper subjective and objective also for 2015