Indian Statistical Institute, ISI MMATH 2021 PMA Solutions & Discussions

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ISI MMATH 2021 Multiple Choice Questions PMA, Objectives: Solutions and Discussions

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Problem 1

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that $f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}$
for all $x \neq 0$ . Then the value of $f(0)$ is

(a) $-\frac{1}{8}$ ;

(b) $\frac{1}{8}$ ;

(d) $-\frac{1}{4}$ .

Solution:

By continuity,

$f(0)=\lim_{x\to 0}\frac{2-\sqrt{x+4}}{\sin 2x}=\lim_{x\to 0}\frac{4-x-4}{\sin 2x (2+\sqrt{x+4})}=$

$=\lim_{x\to 0}\left(-\frac{2x}{\sin 2x}\cdot \frac{1}{2(2+\sqrt{x+4})}\right)=-\frac{1}{2(2+\sqrt{4})}=-\frac{1}{8},$

where the relation $\lim_{x\to 0}\frac{\sin x}{x}=1$ was used.

Answer (a)

Problem 2

A person throws a pair of fair dice. If the sum of the numbers on the dice is a perfect square, then the probability that the number 3 appeared on at least one of the dice is

(a) $1 / 9$

(b) $4 / 7$

(d) $7 / 36$

Solution:

The sum of the numbers on the dice takes values from $2$ to $12.$ The only perfect squares that can appear are $4$ and $9.$ The sum $4$ appears in $3$ cases:

$(1,3),(2,2),(3,1).$

The sum $9$ appears in $4$ cases:

$(3,6),(4,5),(5,4),(6,3).$

So, the perfect square appears in $7$ cases. Among them in $4$ cases the number $3$ appears on one of the dice. The conditional probability of interest is $\frac{4}{7}.$

Answer (b)

Problem 3

Consider the system of linear equations: $x+y+z=5, \quad 2 x+2 y+3 z=4$ . Then

(a) the system is inconsistent

(b) the system has a unique solution

(d) none of the above is true

Solution:

We find that

$z=(2x+2y+3z)-2(x+y+z)=4-10=-6.$

Then the equations take the form

$\begin{cases} x+y=5-z=11 \\ 2x+2y=4-3z=22 \end{cases}$

These two equations are identical. So, for any $x$ we have a solution

$y=11-x, z=-6.$

Answer (c)

Problem 4

If $g^{\prime}(x)=f(x)$ then $\int x^{3} f\left(x^{2}\right) d x$ is given by

(a) $x^{2} g\left(x^{2}\right)-\int x g\left(x^{2}\right) d x+C$

(b) $\frac{1}{2} x^{2} g\left(x^{2}\right)-\int x g\left(x^{2}\right) d x+C$

(d) $x^{2} g\left(x^{2}\right)-\frac{1}{2} \int x g\left(x^{2}\right) d x+C$

Solution:

We observe that

$\frac{d}{dx}(g(x^2))=2xg'(x^2)=2xf(x^2).$

Hence,

$\int x^3 f(x^2)dx=\frac{1}{2}\int x^2 \frac{d}{dx}(g(x^2))dx=$

integrate by parts

$=\frac{1}{2}x^2 g(x^2)-\frac{1}{2}\int 2x g(x^2)dx+C=\frac{1}{2}x^2 g(x^2)-\int x g(x^2)dx+C$

Answer (b)

Problem 5

If $\left({ }^{n} C_{0}+{ }^{n} C_{1}\right)\left({ }^{n} C_{1}+{ }^{n} C_{2}\right) \cdots\left({ }^{n} C_{n-1}+{ }^{n} C_{n}\right)=k^{n} C_{0}{ }^{n} C_{1} \cdots{ }^{n} C_{n-1}$
then $k$ is equal to

(a) $\frac{(n+1)^{n}}{n !}$

(b) $\frac{n^{n}}{n !}$

(d) $\frac{(n+1)^{n+1}}{n !}$

Solution:

For $j\in \{0,\ldots,n-1\}$ we have the identity

${}^nC_j+{}^nC_{j+1}=\frac{n!}{j!(n-j)!}+\frac{n!}{(j+1)!(n-j-1)!}=\frac{n!}{j!(n-j-1)!}\left(\frac{1}{n-j}+\frac{1}{j+1}\right)=$

$=\frac{(n+1)!}{(j+1)!(n-j)!}={}^{n+1}C_{j+1}.$

Hence,

$\prod^{n-1}_{j=0}\left({}^nC_j+{}^nC_{j+1}\right)=\prod^{n-1}_{j=0}{}^{n+1}C_{j+1}=\frac{((n+1)!)^n}{(\prod^{n-1}_{j=0}(j+1)!)(\prod^{n-1}_{j=0}(n-j)!)}=\frac{((n+1)!)^n}{(\prod^n_{j=1}j!)^2}$

On the other hand,

$\prod^{n-1}_{j=0} {}^nC_j=\frac{(n!)^n}{(\prod^{n-1}_{j=0} j!)(\prod^{n-1}_{j=0} (n-j)!)}=\frac{(n!)^{n+1}}{(\prod^n_{j=1}j!)^2}.$

Hence,

$k=\frac{\prod^{n-1}_{j=0}\left({}^nC_j+{}^nC_{j+1}\right)}{\prod^{n-1}_{j=0} {}^nC_j}=\frac{((n+1)!)^n}{(n!)^{n+1}}=\frac{(n+1)^n}{n!}$

Answer (a)

Problem 6

Let $\left\{f_{n}\right\}$ be a sequence of functions defined as follows:

$f_{n}(x)=x^{n} \cos (2 \pi n x), \quad x \in[-1,1]$

Then $\lim _{n \rightarrow \infty} f_{n}(x)$ exists if and only if $x$ belongs to the interval

(a) $(-1,1)$

(b) $[-1,1)$

(d) $(-1,1]$

Solution:

If $|x|<1,$ then

$|f_n(x)|\leq |x|^n\to 0, \ n\to\infty;$

If $x=1,$ then

$f_n(x)=\cos(2\pi n)=1\to 1, \ n\to\infty;$

If $x=-1,$ then

$f_n(x)=(-1)^n\cos(-2\pi n)=(-1)^n,$

and this does not converge. The limit of $f_n(x)$ exists if and only if $-1<x\leq 1.$

Answer (d)

Problem 7

Let $S$ be a set of $n$ elements. The number of ways in which $n$ distinct non-empty subsets $X_{1}, \ldots, X_{n}$ of $S$ can be chosen such that $X_{1} \subseteq$
$X_{2} \cdots \subseteq X_{n}$ , is

(a) $\left(\begin{array}{l}n \\ 1\end{array}\right)\left(\begin{array}{l}n \\ 2\end{array}\right) \cdots\left(\begin{array}{l}n \\ n\end{array}\right)$

(b) $1$

(d) $2^{n}$

Solution:

Denote $X_0=\emptyset.$ Each difference $X_j\setminus X_{j-1},$ $1\leq j\leq n,$ contains at least one element. Denote this element by $s_j.$ If $1\leq i<j\leq n,$ then $s_i\in X_i\subset X_{j-1},$ so $s_i\ne s_j.$ All elements $s_1,\ldots,s_n$ are distinct. Hence, $(s_1,\ldots,s_n)$ is an ordering of the set $S$ . There are $n!$ ways to order elements of the set $S,$ and every ordering $(s_1,\ldots,s_n)$ defines a sequence of non-empty distinct subsets by the rule $X_j=\{s_1,\ldots,s_j\},$ $1\leq j\leq n.$

Answer (c)

Problem 8

Let $A$ be a $4 \times 4$ matrix such that both $A$ and $\operatorname{Adj}(A)$ are non-null.
If $\operatorname{det} A=0$ , then the rank of $A$ is

(a) $1$ ;

(b) $2$ ;

(d) $4$ .

Solution:

Elements of the adjugate matrix $Adj(A)$ are equal (up to signs) to cofactors of elements of the matrix $A$ (i.e. $3\times 3$ minors of $A$ ). Since $\det A=0,$ the rank of $A$ is $\leq 3.$ Since $Adj(A)\ne 0$ there exists at least one non-zero $3\times 3$ minor of $A$ and the rank of $A$ is $\geq 3.$ We deduce that the rank of $A$ is equal to $3$ .

Answer (c)

Problem 9

The set of all $a$ satisfying the inequality

$\frac{1}{\sqrt{a}} \int_{1}^{a}\left(\frac{3}{2} \sqrt{x}+1-\frac{1}{\sqrt{x}}\right) d x<4$

is equal to the interval

(a) $(-5,-2)$

(b) $(1,4)$

(d) $(0,4)$

Solution:

At first we observe that $a>0.$ Evaluate the integral:

$\int^a_1 \left(\frac{3}{2}\sqrt{x}+1-\frac{1}{\sqrt{x}}\right)dx=\left(x^{3/2}+x-2\sqrt{x}\right)\bigg|^a_1=a^{3/2}+a-2\sqrt{a}.$

The inequality transforms to

$\frac{a^{3/2}+a-2\sqrt{a}}{\sqrt{a}}<4;$

$a+\sqrt{a}-6<0;$

$(\sqrt{a}+3)(\sqrt{a}-2)<0;$

$\sqrt{a}<2; \ 0<a<4.$

Answer (d)

Problem 10

Let $C_{0}$ be the set of all continuous functions $f:[0,1] \rightarrow \mathbb{R}$ and $C_{1}$ be the set of all differentiable functions $g:[0,1] \rightarrow \mathbb{R}$ such that the derivative $g^{\prime}$ is continuous. (Here, differentiability at 0 means right differentiability and differentiability at 1 means left differentiability.) If $T: C_{1} \rightarrow C_{0}$ is defined by $T(g)=g^{\prime}$ , then

(a) $T$ is one-to-one and onto

(b) $T$ is one-to-one but not onto

(d) $T$ is neither one-to-one nor onto.

Solution:

Consider a constant function $g_a(x)=a.$ Then $T(g_a)=0.$ Since $g_a$ are distinct for different $a$ , the mapping $T$ is not one-to-one.

Given a continuous functions $f\in C_0,$ consider

$g(x)=\int^x_0 f(t)dt.$

By the Fundamental Theorem of Calculus, $g\in C_1$ and $T(g)=g'=f.$ The mapping $T$ is onto.

Answer (c)

Problem 11

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. If $a<b<c$ and $a+b+c=\frac{3}{2}$ , then the value of $a$ is

(a) $\frac{1}{2 \sqrt{2}}$

(b) $-\frac{1}{2 \sqrt{2}}$

(d) $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Solution:

Let $d$ be the difference of the arithmetic progression. Then $a=b-d,$ $c=b+d$ and we find

$a+b+c=3b=\frac{3}{2}, \ b=\frac{1}{2}.$

Further,

$\frac{b^2}{a^2}=\frac{c^2}{b^2},$

and

$b^4=a^2c^2=(b-d)^2(b+d)^2=(b^2-d^2)^2.$

Taking square roots we find

$b^2=|b^2-d^2|,$

i.e. either $d^2=0$ or $d^2=2b^2=\pm \sqrt{2}\frac{1}{2}=\pm\frac{1}{\sqrt{2}}.$ The case $d=0$ is impossible, as $a<b<c.$ Also, the condition $a<b<c$ implies that $d>0$ and $d=\frac{1}{\sqrt{2}}.$ Correspondingly, $a=\frac{1}{2}-\frac{1}{\sqrt{2}}.$

Answer (d)

Problem 12

The number of distinct even divisors of

$\prod_{k=1}^{5} k !$

is
$\begin{array}{llll}\text { (A) } 24 & \text { (B) } 32 & \text { (C) } 64 & \text { (D) } 72\end{array}$

(a) $24$ ;

(b) $32$ ;

(d) $72$ .

Solution:

We have

$\prod^5_{k=1} k!=2!3!4!5!=2\cdot (2\cdot 3) \cdot (2\cdot 3\cdot 4)\cdot (2\cdot 3\cdot 4\cdot 5)=2^8 3^35.$

Every even divisor of $\prod^5_{k=1} k!$ is of the type $2^a3^b5^c$ with $1\leq a\leq 8,$ $0\leq b\leq 3,$ $0\leq c\leq 1.$ There are $8\cdot 4\cdot 2=64$ even divisors.

Answer (c)

Problem 13

Let $D$ be the triangular region in the $x y$ -plane with vertices at $(0,0),(0,1)$ and $(1,1)$ . Then the value of

$\iint_{D} \frac{2}{1+x^{2}} d x d y$

(a) $\frac{\pi}{2}$

(b) $\frac{\pi}{2}-\ln 2$

(d) $\ln 2$

Solution:

The region $D$ is defined by inequalities $0\leq x\leq y\leq 1.$ Hence

$\iint_D \frac{2}{1+x^2}dxdy=\int^1_0 \frac{2}{1+x^2}\int^1_x dydy=\int^1_0 \frac{2(1-x)}{1+x^2}dx=$

$=2\int^1_0 \frac{1}{1+x^2}dx-\int^1_0 \frac{2x}{1+x^2}dx=\left(2\arctan(x) -\ln (1+x^2)\right)\bigg|^1_0=\frac{\pi}{2}-\ln(2).$

Answer (b)

Problem 14

Given a real number $\alpha \in(0,1)$ , define a sequence $\left\{x_{n}\right\}_{n \geq 0}$ by the following recurrence relation:

$x_{n+1}=\alpha x_{n}+(1-\alpha) x_{n-1}, \quad n \geq 1$

If $\lim _{n \rightarrow \infty} x_{n}=\ell$ then the value of $\ell$ is

(a) $\frac{\alpha x_{0}+x_{1}}{1-\alpha}$

(b) $\frac{(1-\alpha) x_{0}+x_{1}}{2-\alpha}$

(d) $\frac{(1-\alpha) x_{1}+x_{0}}{2-\alpha}$

Solution:

Consider the characteristic equation of the recurrence:

$k^2-\alpha k -(1-\alpha)=0.$

Roots are $k=1$ and $k=\alpha-1.$ Hence,

$x_n=A+B(\alpha-1)^n.$

Since $|\alpha-1|<1,$ we find that $l=\lim_{n\to\infty}x_n=A.$ To find $A$ we use first two values:

$\begin{cases} A+B=x_0 \\ A+B(\alpha-1)=x_1 \end{cases}$

Then $A(2-\alpha)=x_1-(\alpha-1)x_0.$ So,

$l=A=\frac{(1-\alpha)x_0+x_1}{2-\alpha}.$

Answer (b)

Problem 15

A straight line passes through the intersection of the lines given by $3 x-4 y+1=0$ and $5 x+y=1$ and makes equal intercepts of the same sign on the coordinate axes. The equation of the straight line is

(a) $23 x-23 y+11=0$

(b) $23 x-23 y-11=0$

(d) $23 x+23 y-11=0$

Solution:

We find the intersection of the lines:

$\begin{cases} 3x-4y=-1 \\ 5x+y=1 \end{cases} \Rightarrow x=\frac{3}{23}, y=\frac{8}{23}.$

The line we are looking for passes through the point $(\frac{3}{23},\frac{8}{23})$ and makes equal (positive) intercepts with the coordinate axes. The slope of such line is $-1$ and its equation is of the form

$x+y=b.$

We find $b:$

$\frac{3}{23}+\frac{8}{23}=b, \ \ b=\frac{11}{23}.$

So, the equation of the line is

$x+y=\frac{11}{23}.$

Equivalently,

$23x+23y-11=0.$

Answer (d)

Problem 16

The series

$\sum_{n} \frac{3 \cdot 6 \cdot 9 \cdots 3 n}{7 \cdot 10 \cdot 13 \cdots(3 n+4)} x^{n}, \quad x>0$

(a) converges for $0<x \leq 1$ and diverges for $x>1$

(b) converges for all $x>0$

(d) converges for $\frac{1}{2}<x<1$ and diverges for $0<x \leq \frac{1}{2}, x \geq 1$ .

Solution:

Denote $c_n=\frac{3\cdot 6\cdot \ldots \cdot 3n}{7\cdot 10\cdot \ldots \cdot (3n+4)}=\frac{\prod^n_{j=1}(3j)}{\prod^n_{j=1}(3j+4)}.$ Apply the ratio test:

$\frac{c_n}{c_{n+1}}=\frac{3(n+1)+4}{3(n+1)}\to 1, \ n\to\infty.$

Hence, the radius of convergence of the power series is equal to 1. The series is convergent for $|x|<1,$ and is divergent for $|x|>1.$

We study convergence at $x=1.$ Observe that

$n\frac{c_n}{c_{n+1}}-(n+1)\to \frac{1}{3}, \ n\to\infty.$

Hence, there exists $N$ such that for all $n\geq N$ we have

$n\frac{c_n}{c_{n+1}}-(n+1)>\frac{1}{4}.$

Equivalently,

$nc_n-(n+1)c_{n+1}>\frac{1}{4}c_{n+1}.$

Taking the sum from $n=N$ to $n=N+p-1$ we find

$Nc_N-(N+p)c_{N+p}>\frac{1}{4}\sum^{N+p}_{n=N+1}c_n.$

So,

$\sum^{N+p}_{n=N+1}c_n<4Nc_N.$

The sequence of partial sums $\sum^{k}_{n=1}c_n$ is bounded and the series is convergent.

Answer (a)

Problem 17

Suppose $A$ and $B$ are two square matrices such that the largest eigenvalue of $(A B-B A)$ is positive. Then the smallest eigen value of $(A B-B A)$

(a) must be positive

(b) must be negative

(d) is none of the above

Solution:

We compute the trace of the matrix:

$tr(AB-BA)=tr(AB)-tr(BA)=tr(AB)-tr(AB)=0.$

Since the trace is the sum of eigenvalues and the largest eigenvalue of $AB-BA$ is positive, the smallest eigenvalue of $AB-BA$ is negative.

Answer (b)

Problem 18

The number of saddle points of the function $f(x, y)=2 x^{4}-x^{2}+3 y^{2}$ is

(a) $1$ ;

(b) $0$ ;

(d) none of the above.

Solution:

To find stationary points we compute the gradient of $f:$

$\nabla f= (8x^3-2x \ \ 6y).$

Stationary point $(x,y)$ is found from the system

$\begin{cases} 8x^3-2x=0 \\ 6y=0 \end{cases}$

There are three stationary points: $(0,0),$ $(\frac{1}{2},0),$ $(-\frac{1}{2},0).$ The matrix of second derivatives of $f$ is

$\begin{pmatrix} 24x^2-2 & 0 \\ 0 & 6 \end{pmatrix}$

Eigenvalues are $24x^2-2$ and $6.$ The stationary point is a saddle point if the first eigenvalue is negative. It happens only for the point $(0,0).$

Answer (a)

Problem 19

Suppose $G$ is a cyclic group and $a, b \in G$ . There does not exist any $x \in G$ such that $x^{2}=a .$ Also, there does not exist an $y \in G$ such that $y^{2}=b$ . Then,

(a) there exists an element $g \in G$ such that $g^{2}=a b$ .

(b) there exists an element $g \in G$ such that $g^{3}=a b$ .

(d) none of the above is true.

Solution:

Let $f$ be the generator of $G.$ Then $a=f^k$ for some integer $k.$ Observe that $k$ is odd. Similarly, $b=f^l$ for some odd $l$ . But then $ab=f^{k+l}$ where $k+l$ is even. Define $g=f^{(k+l)/2}.$ Then $g^2=ab.$

Answer (a)

Problem 20

The number of real roots of the polynomial $x^{3}-2 x+7$ is

(a) $0$ ;

(b) $1$ ;

(d) $3$ .

Solution:

Denote $f(x)=x^3-2x+7.$ The derivative of $f$ is

$f'(x)=3x^2-2.$

$f'(x)>0\Leftrightarrow |x|>\sqrt{\frac{2}{3}}.$ The function $f$ is increasing on $(-\infty,-\sqrt{\frac{2}{3}}],$ decreasing on $[-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}]$ , increasing on $[\sqrt{\frac{2}{3}},\infty).$

$f(\sqrt{\frac{2}{3}})=\frac{2}{3}\sqrt{\frac{2}{3}}-2\sqrt{\frac{2}{3}}+7=7-\frac{4}{3}\sqrt{\frac{2}{3}}>0.$

The only real zero of $f$ is located in $(-\infty,\sqrt{\frac{2}{3}}).$

Answer (b)

Problem 21

Suppose that a $3 \times 3$ matrix $A$ has an eigen value $-1$ . If the matrix
$A+I$ is equal to

$\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

then the eigen vectors of $A$ corresponding to the eigenvalue $-1$ are in
the form,

(a) $\left[\begin{array}{c} 2 t \\ 0 \\ t \end{array}\right], t \in \mathbb{R}$ ;

(b) $\left[\begin{array}{c} 2 t \\ s \\ t \end{array}\right], s, t \in \mathbb{R}$ ;

(d) $\left[\begin{array}{c} t \\ s \\ 2 t \end{array}\right], s, t \in \mathbb{R}$

Solution:

Let $v$ be the eigenvector of $A$ corresponding to the eigenvalue $-1:$ $Av=-v.$ Then
$(A+I)v=Av+v=0.$ Denote $v=\begin{bmatrix} x \\ y \\ z \end{bmatrix}.$

$(A+I)v=\begin{bmatrix} x-2z \\ 0 \\ 0 \end{bmatrix}=0.$

So, $x=2z.$ Let $z=t,$ $y=s.$ The vector $v$ is of the form

$v=\begin{bmatrix} 2t \\ s \\ t \end{bmatrix}$

Answer (b)

Problem 22

Consider the function $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ defined by $f(x, y)=x^{2}(y-1)$ . For $\vec{u}=\left(\frac{1}{2}, \frac{1}{2}\right)$ and $\vec{v}=(3,4)$ , the value of the limit

$\lim _{t \rightarrow 0} \frac{f(\vec{u}+t \vec{v})-f(\vec{u})}{t}$

(a) $\frac{3}{4}$

(b) $\frac{6}{13}$

(d) none of the above

Solution:

The gradient of $f$ equals

$\nabla f= [2x(y-1) \ \ \ x^2].$

Hence

$\lim_{t\to 0}\frac{f(\vec{u}+t\vec{v})-f(\vec{u})}{t}=\nabla f(\vec{u}) \vec{v}=\begin{bmatrix} -\frac{1}{2} & \frac{1}{4} \end{bmatrix}\begin{bmatrix} 3 \\ 4 \end{bmatrix}=-\frac{1}{2}.$

Answer (c)

Problem 23

Suppose $\phi$ is a solution of the differential equation $y^{\prime \prime}-y^{\prime}-2 y=0$ such that $\phi(0)=1$ and $\phi^{\prime}(0)=5$ . Then

(a) $\phi(x) \rightarrow \infty$ as $|x| \rightarrow \infty$

(b) $\phi(x) \rightarrow-\infty$ as $|x| \rightarrow \infty$

(d) $\phi(x) \rightarrow \infty$ as $x \rightarrow-\infty$

Solution:

The characteristic equation is $k^2-k-2=0.$ Its roots are $k=2$ and $k=-1.$ The solution $\phi$ is of the form

$\phi(x)=Ae^{2x}+Be^{-x}.$

To find $A$ and $B$ we use initial conditions:

$\begin{cases} \phi(0)=A+B=1 \\ \phi'(0)=2A-B=5 \end{cases}$

Hence, $A=2, B=-1$ and the solution is $\phi(x)=2e^{2x}-e^{-x}.$ In particular, $\phi(x)\to -\infty$ as $x\to -\infty.$

Answer (c)

Problem 24

A fair die is rolled five times. What is the probability that the largest
number rolled is $5$ ?

(a) $5 / 6$

(b) $1 / 6$

(d) $(5 / 6)^{5}-(2 / 3)^{5}$

Solution:

Denote by $A_k$ the event that all numbers rolled are $\leq k,$ $1\leq k\leq 6.$ Then $P(A_k)=(\frac{k}{6})^5.$ The event of interest is $A_5\setminus A_4.$ Its probability is

$\left(\frac{5}{6}\right)^5-\left(\frac{4}{6}\right)^5=\left(\frac{5}{6}\right)^5-\left(\frac{2}{3}\right)^5$

Answer (d)

Problem 25

Two rows of $n$ chairs, facing each other, are laid out. The number of different ways that $n$ couples can sit on these chairs such that each person sits directly opposite to his/her partner is

(a) $n !$

(b) $n ! / 2$

(d) $2 n !$

Solution:

There are $n!$ choices how each couple can choose its pair of opposite chairs. When a pair of chairs is chosen, there are $2$ ways for each couple they can sit on these chairs, i.e. $2^n$ choices for all couples. Totally there are $2^nn!$ ways the couples can sit on the chairs.

Answer (c)

Problem 26

Consider the function $f: \mathbb{C} \rightarrow \mathbb{C}$ defined on the complex plane $\mathbb{C}$ by $f(z)=e^{z}$ . For a real number $c>0$ , let $A=\{f(z) \mid \operatorname{Re} z=c\}$ and $B=\{f(z) \mid \operatorname{Im} z=c\}$ . Then

(a) both $A$ and $B$ are straight line segments

(b) $A$ is a circle and $B$ is a straight line segment

(d) both $A$ and $B$ are circles

Solution:

Let $Re (z)=c.$ Then $z=c+iy,$ $y\in \mathbb{R},$ and the set $A$ consists of points of the form

$e^{c+iy}=e^c(\cos y +i\sin y), \ y\in\mathbb{R}.$

$A$ is a cricle of center $0$ nd radius $e^c.$

Let $Im(z)=c.$ Then $z=x+iy,$ $x\in \mathbb{R},$ and the set $B$ consists of point of the form

$e^{x+ic}=e^x e^{ic}, \ x\in\mathbb{R}.$

$B$ is a straight line connecting $0$ and $e^{ic}.$

Answer (b)

Problem 27

Consider two real valued functions $f$ and $g$ given by
$f(x)=\frac{x}{x-1}$ for $x>1$ , and $g(x)=7-x^{3}$ for $x \in \mathbb{R}$
Which of the following statements about inverse functions is true?

(a) Neither $f^{-1}$ nor $g^{-1}$ exists

(b) $f^{-1}$ exists, but not $g^{-1}$

(d) Both $f^{-1}$ and $g^{-1}$ exist.

Solution:

The function $f(x)$ is strictly decreasing: $f'(x)=-\frac{1}{(x-1)^2};$

$\lim_{x\to 1+}f(x)=+\infty, \ \ \lim_{x\to \infty}f(x)=1.$

So, $f$ is a bijection of the set $(1,\infty).$ $f^{-1}$ exists.

The function $g(x)$ is strictly decreasing: $g'(x)=-3x^2;$

$\lim_{x\to \infty}g(x)=-\infty, \ \ \lim_{x\to -\infty}g(x)=\infty.$

So, $g$ is a bijection of $\mathbb{R}.$ $g^{-1}$ exists.

Answer (d)

Problem 28

A circle is drawn with centre at $(-1,1)$ touching $x^{2}+y^{2}-4 x+6 y-3=0$ externally. Then the circle touches

(a) both the axes

(b) only the $x$ -axis

(d) only the $y$ -axis

Solution:

Let $r$ be the radius of the circle centered at $(-1,1).$ The equation of the second circle is

$(x-2)^2+(y+3)^2=16.$

It is centered at $(2,-3)$ and of radius $4.$ The distance between centers of circles is

$\sqrt{(2+1)^2+(-3-1)^2}=5.$

Hence, $r=5-4=1$ and the first circle touches both axes (as $(-1,0)$ and $(0,1)$ belong to the first circle).

Answer (a)

Problem 29

Let $f(x-y)=\frac{f(x)}{f(y)}$ for all $x, y \in \mathbb{R}$ and $f^{\prime}(0)=p, f^{\prime}(5)=q$ . Then the value of $f^{\prime}(-5)$ is

(a) $q$

(b) $-q$

(d) $\frac{p^{2}}{q}$

Solution:

Observe that $f(0)=\frac{f(x)}{f(x)}=1.$ By assumption,

$f(x)=f(x-y)f(y).$

Differentiate this equation in $x:$

$f'(x)=f'(x-y)f(y).$

Differentiate in $y$ :

$0=-f'(x-y)f(y)+f(x-y)f'(y), \ f'(x-y)f(y)=f(x-y)f'(y).$

Then

$f(y)=\frac{f(0)f'(y)}{f'(0)}=\frac{f'(y)}{p}.$

Take $x=0,$ $y=5:$

$p=f'(0)=f'(-5)f(5)=\frac{f'(-5)f'(5)}{p}=\frac{q}{p}f'(-5), \ f'(-5)=\frac{p^2}{q}.$

Answer (d)

Problem 30

Let

$A=\left[\begin{array}{lll} a & 1 & 1 \\ b & a & 1 \\ 1 & 1 & 1 \end{array}\right]$

The number of elements in the set

$\left\{(a, b) \in \mathbb{Z}^{2}: 0 \leq a, b \leq 2021, \operatorname{rank}(A)=2\right\}$

is
(a) $2021$

(b) $2020$

(d) $2020 \times 2021$

Solution:

The determinant of $A$ must be zero:

$\det(A)=a^2+1+b-a-b-a=(a-1)^2.$

Hence, $a=1$ and the matrix $A$ is of the form

$A=\begin{bmatrix} 1 & 1 & 1 \\ b & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}.$

The rank of $A$ is $2$ if and only if $b\ne 1.$ So, there are $2022-1=2021$ possible values for $b$

Answer (a)

4 Comments

Prasant

July 26, 2021 at 1:00 am Reply

Sir, what will be the expected cutoff for mmath 2021?
- FractionsHub
  
  July 26, 2021 at 9:29 am Reply
  
  Except cut off to go around 60 to 70 for PMA and 4 Questions for PMB.
PERURI MOULI SRI SAI SURYA PAVAN

July 28, 2021 at 11:12 pm Reply

Sir i got 77 marks in PMA and i solved 2 questions completely in PMB and just attempted another 3 questions in PMB.is there any possibility for getting inteview call
- Ritu
  
  February 18, 2024 at 1:02 pm Reply
  
  Are you selected for interview?

Indian Statistical Institute, ISI MMATH 2021 PMA Solutions & Discussions

ISI MMATH 2021 Multiple Choice Questions PMA, Objectives: Solutions and Discussions

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To view previous year’s question papers: Click Here

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