Indian Statistical Institute, ISI BStat & BMath 2021 UGA Solutions & Discussions

ISI BMATH & BSTAT 2021 Multiple Choice Questions UGA, Objectives: Solutions and Discussions

Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Answer Keys: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Question Paper: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGB Question Paper: Click Here

Problem 1 

The number of ways one can express 2^{2} 3^{3} 5^{5} 7^{7} as a product of two numbers a and b, where g \operatorname{cd}(a, b)=1, and 1<a<b, is
(A) 5 .
(B) 6 .
(C) 7 .
(D) 8 .


Denote c=2^23^35^57^7. If c=ab, then each a=2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}7^{\alpha_4} and b=2^{\beta_1}3^{\beta_2}5^{\beta_3}7^{\beta_4}, where

    \[\alpha_1+\beta_1=2, \alpha_2+\beta_2=3, \alpha_3+\beta_3=5, \alpha_4+\beta_4=7.\]

Since gcd(a,b)=1 we necessarily have \alpha_j=0 or \beta_j=0 for each j. To construct the pair (a,b) we have to choose either \alpha_j=0 or \beta_j=0 for each j\in \{1,2,3,4\}. Hence, there are 2^4 choices. In the representation c=ab numbers a and b are distinct. Since c=ab=ba and because of the condition a<b only half of choices are satisfying. So we need to divide the number of choices by 2. Finally, we must exclude the choice where all \alpha_j=0 (because then a=1). The answer is \frac{2^4}{2}-1=7.

Answer (c).

Problem 2

The sum of all the solutions of 2+\log _{2}(x-2)=\log _{(x-2)} 8 in the interval (2, \infty) is

(a) \frac{35}{8}.

(b) 5.

(c) \frac{49}{8}.

(d) \frac{55}{8}.


Denote y=\log_2(x-2). Then \log_{(x-2)}8=3\log_{(x-2)}2=\frac{3}{y}. The equation becomes




Its solutions are y_1=-3, y_2=1. Corresponding values of x are

    \[x_1=2+2^{y_1}=2+\frac{1}{8}=\frac{17}{8}, x_2=2+2^{y_2}=4.\]

The answer is 4+\frac{17}{8}=\frac{49}{8}.

Answer (C)

Problem 3

Let f: \mathbb{R} \rightarrow \mathbb{R} be a continuous function such that

    \[f(x+1)=\frac{1}{2} f(x) \text { for all } x \in \mathbb{R} \text { , }\]

and let a_{n}=\int_{0}^{n} f(x) d x for all integers n \geq 1. Then:

(a) \lim _{n \rightarrow \infty} a_{n} exists and equals \int_{0}^{1} f(x) d x.

(b) \lim _{n \rightarrow \infty} a_{n} does not exist.

(c) \lim _{n \rightarrow \infty} a_{n} exists if and only if \left|\int_{0}^{1} f(x) d x\right|<1.

(d) \lim _{n \rightarrow \infty} a_{n} exists and equals 2 \int_{0}^{1} f(x) d x.


We verify by induction that f(x+k)=\frac{1}{2^k}f(x), k\geq 0. Indeed, for k=0 this is obvious. If the result is true for k, then



    \[a_n=\int^n_0 f(x)dx=\sum^{n-1}_{k=0} \int^{k+1}_k f(x)dx=\sum^{n-1}_{k=0} \int^{1}_0 f(x+k)dx=\]

    \[=\sum^{n-1}_{k=0} \frac{1}{2^k}\int^{1}_0 f(x)dx=\int^1_0 f(x)dx\sum^{n-1}_{k=0}\frac{1}{2^k}\to\]

    \[\to\int^1_0 f(x)dx \sum^\infty_{k=0}\frac{1}{2^k}=2\int^1_0 f(x)dx, \ n\to\infty.\]

Answer (D)

Problem 4

Consider the curves x^{2}+y^{2}-4 x-6 y-12=0,9 x^{2}+4 y^{2}-900=0 and y^{2}-6 y-6 x+51=0. The maximum number of disjoint regions into which these curves divide the X Y -plane (excluding the curves themselves), is

(a) 4;

(b) 5;

(c) 6;

(d) 7.


The equation of the first curve is


This is a circle with center (2,3) and radius 5.

The equation of the second curve is


This is an ellipse. The first curve is entirely inside this ellipse. Indeed, if


then |x-2|\leq 5, |y-3|\leq 5, hence |x|\leq 7, |y|\leq 8 and

    \[9x^2+4y^2\leq 697<900.\]

The first two curves divide the plane into 3 regions.

The equation of the third curve is


This is a parabola with vertex (7,3) and which is rightward opened. Observe that the first curve and the third curve touch each other at the point (3,7) which is the rightmost point of the first curve. Hence, the third curve does not divide the region of the circle inside the third curve, while it divides two other regions into two parts. Totally, we have 5 regions.


Answer  (B)

Problem 5

A box has 13 distinct pairs of socks. Let p_{r} denote the probability of having at least one matching pair among a bunch of r socks drawn at random from the box. If r_{0} is the maximum possible value of r such that p_{r}<1, then the value of p_{r_{0}} is

(a) 1-\frac{12}{{}^{26}C{12}}.

(b) 1-\frac{13}{{}^{26}C_{13}}.

(c) 1-\frac{2^{13}}{{}^{26}C_{13}}.

(d) 1-\frac{2^{12}}{{}^{26}C_{12}}.


We observe that p_r is increasing with r. If we choose 14 socks, then by the pigeonhole principle we neccesarily choose at least one matching pair. So, p_{14}=1. We compute p_{13}. There are {}^{26}C_{13} choices of a bunch of 13 socks. There are no matches if we choose exactly one sock from each pair, i.e. there are 2^{13} choices when there are no matching pairs. Correspondingly, p_{r_0}=p_{13}=1-\frac{2^{13}}{{}^{26}C_{13}}.

Answer (C)

Problem 6

Let a, b, c, d>0, be any real numbers. Then the maximum possible value of c x+d y, over all points on the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, must be

(a) \sqrt{a^{2} c^{2}+b^{2} d^{2}}.

(b) \sqrt{a^{2} b^{2}+c^{2} d^{2}}

(c) \sqrt{\frac{a^{2} c^{2}+b^{2} d^{2}}{a^{2}+b^{2}}}.

(d) \sqrt{\frac{a^{2} b^{2}+c^{2} d^{2}}{c^{2}+d^{2}}}.


By the Cauchy-Schwarz inequality

    \[cx+dy=ac\frac{x}{a}+bd\frac{y}{b}\leq \sqrt{a^2c^2+b^2d^2}\sqrt{\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2}=\sqrt{a^2c^2+b^2d^2}.\]

Moreover, if

    \[x=\frac{a^2c}{\sqrt{a^2c^2+b^2d^2}}, y=\frac{b^2d}{\sqrt{a^2c^2+b^2d^2}},\]



i.e. the maximum possible value is \sqrt{a^2c^2+b^2d^2}.

Answer (A)

Problem 7

Let f(x)=\sin x+\alpha x, x \in \mathbb{R}, where \alpha is a fixed real number. The function f is one-to-one if and only if


(a) \alpha>1 or \alpha<-1

(b) \alpha \geq 1 or \alpha \leq-1.

(c) \alpha \geq 1 or \alpha<-1.

(d) \alpha>1 or \alpha \leq-1


If the function f(x)=\sin x+\alpha x is one-to-one, then it is either increasing or decreasing. Its derivative is

    \[f'(x)=\cos x+\alpha.\]

If \alpha\geq 1, then f'(x)\geq 0 with f(x)=0 in a countable set of points. f is strcitly increasing.

If \alpha\leq -1, then f'(x)\leq 0 with f(x)=0 in a countable set of points. f is strcitly decreasing.

Answer (B)

Problem 8

The value of

    \[1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots 2021}\]



(a) \frac{2021}{1010};

(b) \frac{2021}{1011};

(c) \frac{2021}{1012};

(d) \frac{2021}{1013}.


We use the formula


The quantity of interest is




Answer (B)

Problem 9

The volume of the region S=\{(x, y, z):|x|+2|y|+3|z| \leq 6\} is

(a) 36;

(b) 48;

(c) 72;

(d) 6


At first we note that the region \{(x,y):|x|+2|y|\leq a\} is the rhombus with vertices (a,0), (0,\frac{a}{2}), (-a,0), (0,-\frac{a}{2}). Its area is 4\cdot \frac{1}{2}\cdot a\cdot \frac{a}{2}=a^2. When z_0\in [-2,2], the cross-section of the region S by the plane \{z=z_0\} is the rhombus \{(x,y):|x|+2|y|\leq 6-3|z_0|\}. Its area is (6-3|z_0|)^2. To get the volume of S we integrate in z:

    \[\int^2_{-2} (6-3|z|)^2dz=2\int^2_0 (6-3z)^2dz=18\int^2_0 (2-z)^2dz=18\int^2_0 z^2dz=18\cdot \frac{8}{3}=48\]

Answer (B)

Problem 10

Let f: \mathbb{R} \rightarrow \mathbb{R} be a twice differentiable function such that \frac{d^{2} f(x)}{d x^{2}} is positive for all x \in \mathbb{R}, and suppose f(0)=1, f(1)=4 . Which of the following is not a possible value of f(2) ?

(a) 7;

(b) 8;

(c) 9;

(d) 10.


The derivative f'(x) is strictly increasing. By the mid-point theorem, there exists x_0\in (0,1), such that


So, for all x\in [1,2] we have f'(x)>3 and


7 is not a possible value of f(2).

Answer (A)

Problem 11


    \[f(x)=e^{-|x|}, x \in \mathbb{R}\]


    \[g(\theta)=\int_{-1}^{1} f\left(\frac{x}{\theta}\right) d x, \theta \neq 0\]


    \[\lim _{\theta \rightarrow 0} \frac{g(\theta)}{\theta}\]


(a) equals 0;

(b) equals +\infty;

(c) equals 2;

(d) does not exist.


Consider the transformation

    \[g(\theta)=\int^1_{-1} e^{-\frac{|x|}{|\theta|}}dx=2\int^1_0 e^{-\frac{x}{|\theta|}}dx=2|\theta|\int^{\frac{1}{|\theta|}}_0 e^{-x}dx.\]


    \[\lim_{\theta\to 0+}\frac{g(\theta)}{\theta}=\lim_{\theta\to 0+}2\int^{\frac{1}{|\theta|}}_0 e^{-x}dx=2\int^\infty_0 e^{-x}dx=2;\]

    \[\lim_{\theta\to 0-}\frac{g(\theta)}{\theta}=-\lim_{\theta\to 0-}2\int^{\frac{1}{|\theta|}}_0 e^{-x}dx=-2\int^\infty_0 e^{-x}dx=-2.\]

Answer (D)

Problem 12

The number of different ways to colour the vertices of a square P Q R S using one or more colours from the set \{Red, Blue, Green, Yellow \}, such that no two adjacent vertices have the same colour is

(a) 36;

(b) 48;

(c) 72;

(d) 84.


There are 4 possibilities to choose a colour for the vertix P. Then there are 3 possibilities to choose a colour for the vertix Q. If the vertex S has the same colour as Q, then there are 3 possibilities for the vertex R. In this case we have 4\cdot 3\cdot 3=36 colourings. If the vertex S has different colour than the vertex Q, then there are 2 possibilities for the vertex S and 2 possibilities for the vertex S. In this case we have 4\cdot 3\cdot 2\cdot 2=48 colourings. Totally there are 84 colourings.

Answer (D)

Problem 13

Define a=p^{3}+p^{2}+p+11 and b=p^{2}+1, where p is any prime number. Let d=\operatorname{gcd}(a, b). Then the set of possible values of d is


(a) \{1,2,5\}.

(b) \{2,5,10\}.

(c) \{1,5,10\}.

(d) \{1,2,10\}.


Observe that


If d|a and d|b, then d|10. If p is odd, then 2=a and b are even, so d\geq 2. If p=2, then
b=5, a=25 and d=5. So, we never have d=1. Possible values of d are \{2,5,10\}. We check that all values appear.

If p=2, then d=5.

If p=3, then b=10, a=50 and d=10.

If p=5, then b=26, a=166 and d=2.

Answer (B)

Problem 14

Consider all 2 \times 2 matrices whose entries are distinct and taken from the set \{1,2,3,4\}. The sum of determinants of all such matrices is

(a) 24;

(b) 10;

(c) 12;

(d) 0


Let S be the set of all 2\times 2 matrices with all distinct values from the set \{1,2,3,4\}. For each such matrix A denote by \tilde{A} the matrix obtained by interchanging columns of the matrix A. The transformation A\to \tilde{A} is a bijection of the set S. Hence

    \[\sum_{A\in S}\det(A)=\sum_{A\in S}\det(\tilde{A})=-\sum_{A\in S}\det(A).\]

It follows that \sum_{A\in S}\det(A)=0.

Answer (D)

Problem 15

Let a, b, c and d be four non-negative real numbers where a+b+c+d=
1. The number of different ways one can choose these numbers such that a^{2}+b^{2}+c^{2}+d^{2}=\max \{a, b, c, d\} is

(a) 1;

(b) 5;

(c) 11;

(d) 15;


Denote M=\max\{a,b,c,d\}. Then 0<M\leq 1. Further,

    \[a^2+b^2+c^2+d^2=M=M\cdot 1=aM+bM+cM+dM.\]



All summands in the right-hand side are non-negative. The latter equality is possible if and only if all summands in the right-hand side are equal to zero. In every configuration (a,b,c,d) some numbers are equal to zero (not all), and some are equal to M. Moreover, if k numbers from (a,b,c,d) are equal to zero (0\leq k\leq 3), then M=\frac{1}{4-k}. There are 2^4-1=15 possibilities to choose \leq 3 numbers from (a,b,c,d) that are equal to zero.


Answer (D)

Problem 16

The polynomial x^{4}+4 x+c=0 has at least one real root if and only if

(a) c<2;

(b) c \leq 2;

(c) c<3;

(d) c \leq 3.


Consider equation x^4=a-4x (i.e. a=-c). Consider the value of a when the line y=a-4x touches the curve y=x^4. If x_0 is the common point of these curves, then derivatives of functions a-4x and x^4 are equal at x=x_0:


Hence, x_0=-1 and (-1)^4=a+4, a=-3. When a\geq -3 the curves y=x^4 and y=a-4x will intersect. In terms of c the answer is c\leq 3.

Answer  (D)

Problem 17

The number of all integer solutions of the equation x^{2}+y^{2}+x-y= 2021 is

(a) 5;

(b) 7;

(c) 1;

(d) 0;


Denote x-y=n. Then


The right-hand side is even, so there are no integer solutions of the given equation.

Answer (D)

Problem 18

The number of different values of a for which the equation x^{3}-x+a= 0 has two identical real roots is

(a) 0;

(b) 1;

(c) 2;

(d) 3.


Let c be the value of two identical real roots of the equation x^3-x+a=0. Let d be the value of the third root. Then


Compare coefficients:

    \[2c+d=0, d=-2c;\]

    \[2cd+c^2=-1, c^2=\frac{1}{3}, c=\pm \frac{1}{\sqrt{3}}, d=\mp\frac{2}{\sqrt{3}}.\]

    \[a=-c^2d=\pm \frac{2}{3\sqrt{3}}\]

Answer (C)

Problem 19

Suppose f(x) is a twice differentiable function on [a, b] such that



    \[x^{2} \frac{d^{2} f(x)}{d x^{2}}+4 x \frac{d f(x)}{d x}+2 f(x)>0 \text { for all } x \in(a, b) \text { . }\]

(a) f is negative for all x \in(a, b).

(b) f is positive for all x \in(a, b).

(c) f(x)=0 for exactly one x \in(a, b).

(d) f(x)=0 for at least two x \in(a, b).


Consider the function g(x)=x^2f(x). Then



The function g(x) is strictly convex. By assumption, g(a)=g(b)=0, so g(x)<0 for all x\in (a,b). Correspondingly, f(x)<0 for all x\in (a,b).

Answer  (A)

Problem 20

Consider the following two subsets of \mathbb{C} :

    \[A=\left\{\frac{1}{z}:|z|=2\right\} \text { and } B=\left\{\frac{1}{z}:|z-1|=2\right\} \text { . }\]


(a) A is a circle, but B is not a circle.

(b) B is a circle, but A is not a circle.

(c) A and B are both circles.

(d) Neither A nor B is a circle.


We check that A=\{z:|z|=1/2\}. Indeed, |z|=2\Leftrightarrow |\frac{1}{z}|=\frac{1}{|z|}=\frac{1}{2}. A is a circle.

We check that B is a circle. Consider points z_{1,2}=2\pm \sqrt{3}i. They belong to the circle |z-1|=2. Hence,

    \[\frac{1}{z_{1,2}}=\frac{2\mp \sqrt{3}i}{7}\in B.\]

Denote w=\frac{1}{z_1}, so that \bar{w}=\frac{1}{z_2}. If B is a circle of center x, then |w_1-x|=|w_2-x| and necessarialy, x\in \mathbb{R}. Further, |-1-1|=2, hence -1\in B and


Taking squares we find that

    \[1+2x+x^2=(2/7-x)^2+3/49=\frac{1}{7}-\frac{4x}{7}+x^2, x=-\frac{1}{3}.\]

The radius of B is then \frac{2}{3}. So, we will prove that B=\{z: |z+1/3|=2/3\}. Write z=x+iy. The relation


is equivalent to (x-1)^2+y^2=4, i.e. x^2+y^2=3+2x. The relation |\frac{1}{z}+\frac{1}{3}|=2/3 is equivalent to


and this is equivalent to


and to


and to


Finally, it means that


i.e. B is a circle.

Answer (C)

Problem 21


For a positive integer n, the equation

    \[x^{2}=n+y^{2}, \quad x, y \text { integers }\]

does not have a solution if and only if

(a) n=2.

(b) n is a prime number.

(c) n is an odd number.

(d) n is an even number not divisible by 4 .


Rewrite the equation as


Assume that n is odd, n=2k+1. Then we can find the solution

    \[x=k+1, y=k.\]


    \[(x-y)(x+y)=1\cdot (2k+1)=n.\]

Assume that n is divisible by 4, n=4k. Then we can find the solution

    \[x=k+1, y=k-1.\]


    \[(x-y)(x+y)=2\cdot 2k=n.\]

Assume that n=2k, where k is odd. If there are solutions, then


i.e. either x-y or x+y is even. But,


so both numbers x-y and x+y are even and n is divisible by 4. This shows that in the case when n=2k, k is odd, there are no solutions.

Answer (D)

Problem 22

Let f: \mathbb{R} \rightarrow \mathbb{R} be any twice differentiable function such that its second derivative is continuous and

    \[\frac{d f(x)}{d x} \neq 0 \text { for all } x \neq 0 \text { . }\]


    \[\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=\pi\]

(a) for all x \neq 0, \quad f(x)>f(0).

(b) for all x \neq 0, \quad f(x)<f(0).

(c) for all x, \quad \frac{d^{2} f(x)}{d x^{2}}>0.

(d) for all x, \frac{d^{2} f(x)}{d x^{2}}<0.


Observe that f(0)=0 (otherwise \lim_{x\to 0}\frac{f(x)}{x^2} does not exist in \mathbb{R}). Since \pi>1, there exists \epsilon>0 such that

    \[x\in (-\epsilon,\epsilon)\setminus \{0\}\Rightarrow f(x)>x^2\]

On the set (0,\infty) we have f'(x)\ne 0. So, the derivative f'(x) preserves the sign on (0,\infty). If x\in (0,\epsilon), then relations f(0)=0, f(x)>0 imply that f' is positive somewhere on (0,x). But then it is positive everywhere on (0,\infty) and f(x)>f(0) for x>0.

On the set (-\infty,0) we have f'(x)\ne 0. So, the derivative f'(x) preserves the sign on (-\infty,0). If x\in (-\epsilon,0), then relations f(x)>0, f(0)=0 imply that f' is negative somewhere on (x,0). But then it is negative everywhere on (-\infty,0) and f(x)>f(0) for x<0.

Answer (A)

Problem 23

Let us denote the fractional part of a real number x by \{x\} (note:
\{x\}=x-[x] where [x] is the integer part of x ). Then,

    \[\lim _{n \rightarrow \infty}\left\{(3+2 \sqrt{2})^{n}\right\}\]

(a) equals 0 .

(b) equals 1 .

(c) equals \frac{1}{2}.

(d) does not exist.


We observe that

    \[(3+2\sqrt{2})^n+(3-2\sqrt{2})^n=\sum^n_{k=0} {}^nC_{k}3^{n-k}(2\sqrt{2})^k(1+(-1)^k)\]

is an integer. Also, 0<3-2\sqrt{2}<1. It follows that

    \[\{(3+2\sqrt{2})^n\}=\{-(3-2\sqrt{2})^n\}=1-(3-2\sqrt{2})^n\to 1, \ n\to\infty.\]

Answer (B)

Problem 24


    \[\begin{gathered} p(x)=x^{3}-3 x^{2}+2 x, x \in \mathbb{R}, \\ f_{0}(x)= \begin{cases}\int_{0}^{x} p(t) d t, & x \geq 0, \\ -\int_{x}^{0} p(t) d t, & x<0,\end{cases} \\ f_{1}(x)=e^{f_{0}(x)}, \quad f_{2}(x)=e^{f_{1}(x)}, \quad \ldots \quad, f_{n}(x)=e^{f_{n-1}(x)} \end{gathered}\]

How many roots does the equation \frac{d f_{n}(x)}{d x}=0 have in the interval (-\infty, \infty) ?

(a) 1;

(b) 3;

(c) n+3;

(d) 3 n.


For all n\geq 1 we have

    \[f'_n(x)=f_n(x) f'_{n-1}(x),\]

and f_n(x)>0. So, f'_n(x)=0 if and only if f'_{n-1}(x)=0. The number of solutions of the equation f'_n(x)=0 coincides with the number of solutions of the equation f'_0(x)=0. f'_0(x)=p(x), so we need tofind the number of solutions of the equation


There are three solutions: x=0,1,2.

Answer (B)

Problem 25

For 0 \leq x<2 \pi, the number of solutions of the equation

    \[\sin ^{2} x+2 \cos ^{2} x+3 \sin x \cos x=0\]


(a) 1;

(b) 2;

(c) 3;

(d) 4.


\cos x=0\Rightarrow |\sin x|=1 and points x=\pi/2,3\pi/2 are not solutions of the equation. Divide the equation by \cos^2 x:

    \[\tan^2 x+3\tan x+2=0.\]

Either \tan x=-1, or \tan x=-2. \tan x takes each value twice on [0,2\pi). There are 4 solutions of the initial equation.

Answer (D)

Problem 26

Let f: \mathbb{R} \rightarrow[0, \infty) be a continuous function such that

    \[f(x+y)=f(x) f(y)\]

for all x, y \in \mathbb{R}. Suppose that f is differentiable at x=1 and

    \[\left.\frac{d f(x)}{d x}\right|_{x=1}=2\]

Then, the value of f(1) \log _{e} f(1) is

(a) e;

(b) 2;

(c) \log _{e} 2;

(d) 1.


If f(y)=0 for some y\in \mathbb{R}, then for all x\in \mathbb{R} we have


which contradicts the assumption f'(1)=2. So, f(y)>0 for all y\in\mathbb{R}. Denote g(x)=\log_e f(x). Then g is continuous and


It is well-known that g(x)=cx. So, f(x)=e^{cx}. By assumption,



    \[f(1)\log_e f(1)=e^c c=2.\]

Answer (B)

Problem 27

The expression

    \[\sum_{k=0}^{10} 2^{k} \tan \left(2^{k}\right)\]


(a) \cot 1+2^{11} \cot \left(2^{11}\right)

(b) \cot 1-2^{10} \cot \left(2^{10}\right).

(c) \cot 1+2^{10} \cot \left(2^{10}\right).

(d) \cot 1-2^{11} \cot \left(2^{11}\right).


We use the relation \cos(x)=\frac{\sin(2x)}{2\sin(x)} for x close to 1:


Take the \log_e and differentiate both sides of this equation:




Answer (D)

Problem 28

If the maximum and minimum values of \sin ^{6} x+\cos ^{6} x, as x takes all real values, are a and b, respectively, then a-b equals

(a) \frac{1}{2}.

(b) \frac{2}{3}.

(c) \frac{3}{4}

(d) 1 .


Denote t=\sin^2 x. Then t\in [0,1]. Further,

    \[\sin^6 x+\cos^6 x=t^3+(1-t)^3.\]

Let f(t)=t^3+(1-t)^3, t\in [0,1]. The derivative of f(t) equals


    \[f'(t)\Leftrightarrow t>1/2.\]

So, \min_{t\in[0,1]} f(t)=f(1/2)=\frac{1}{4}, \max_{t\in[0,1]} f(t)=f(0)=1. Correspondingly, a-b=1-\frac{1}{4}=\frac{3}{4}.

Answer (C)

Problem 29

If two real numbers x and y satisfy (x+5)^{2}+(y-10)^{2}=196, then the minimum possible value of x^{2}+2 x+y^{2}-4 y is

(a) 271-112 \sqrt{5}.

(b) 14-4 \sqrt{5}.

(c) 276-112 \sqrt{5}.

(d) 9-4 \sqrt{5}.


Denote by C the circle (x+5)^2+(y-10)^2=196.

If x^2+2x+y^2-4y=r, then


and the point (x,y) belongs to the circle of radius \sqrt{r+5} and with center (-1,2). The point (-1,2) is in the interior of the circle C. The smallest possible value of r taken by points (x,y) corresponds to the radius of a circle with center (-1,2) for which it touches the circle C. Consider the line the passes through centers of circles (-1,2) and (-5,10):

    \[\begin{cases} x=-5+4t \\ y= 10-8 t \end{cases}\]

where t\geq 0. The point (-5+4t,10-8t) belongs to the circle C when

    \[16t^2+64t^2=196, \ t=\sqrt{\frac{49}{20}}=\frac{7}{2\sqrt{5}}.\]

Hence, the minimal possible radius r satisfies



Hence r=271-112\sqrt{5}.

Answer (A)

Problem 30

Define f: \mathbb{R} \rightarrow \mathbb{R} by

    \[f(x)= \begin{cases}(1-\cos x) \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{cases}\]



(a) f is discontinuous.

(b) f is continuous but not differentiable.

(c) f is differentiable and its derivative is discontinuous.

(d) f is differentiable and its derivative is continuous.


We compute the derivative of f. If x\ne 0, then

    \[f'(x)=\sin x\sin\frac{1}{x}-\frac{1-\cos x}{x^2}\cos \frac{1}{x}.\]

If x=0, then

    \[f'(0)=\lim_{x\to 0}\frac{1-\cos x}{x}\sin\frac{1}{x}=0,\]

as \lim_{x\to 0}\frac{1-\cos x}{x}=0 and |\sin\frac{1}{x}|\leq 1. So,

    \[f'(x)=\begin{cases} \sin x\sin\frac{1}{x}-\frac{1-\cos x}{x^2}\cos \frac{1}{x}, \ x\ne 0 \\ 0, \ x=0 \end{cases}\]

The derivative f'(x) is not continuous at x=0, as for x_n=\frac{1}{2\pi n} we have

    \[f'(x_n)=-\frac{1-\cos x_n}{x^2_n}\to -\frac{1}{2}.\]

Answer (C)


  • Sir, in solution 22, isn’t it that f(x)<f(0) for all x<0 since f' is negative in (-inf, 0).
    Also as per me, answer should be option A- I arrived at the answer by checking for f(x)=(pi)*sin²x..hehe

  • Sir, any ideas about the cutoff for UGA and UGB B.Stat?

  • Sayan Kumar Mukhopadhyay

    Sir, Q 23, if I substitute (3+2sqrt2)^n as t , then as n approaches infinity, t also approaches infinity. So the limit changes to lim t tends to infinity { t }, which does not exist. Please explain Sir🙏, where am I getting wrong?

  • Sayan Kumar Mukhopadhyay

    Sir, in Q 23, on substituting (3+2sqrt2)^n as t , as n approaches infinity, t also approaches infinity. So the limit changes to lim t tends to infinity { t }, which does not exist. Please explain Sir, where am I going wrong?

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