Indian Statistical Institute, ISI MMATH 2021 PMA Solutions & Discussions

Fractions > Indian Statistical Institute, ISI MMATH 2021 PMA Solutions & Discussions

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

Note: You must try the problems on your own before looking at the solutions. I have made the solutions so that you don’t have to join any paid program to get the solutions. After you have failed several times in your attempt to solve a question, then only look at the solution to that problem. If you cannot solve a single question on your own then this exam is not for you. And if you like my work then please do like my videos and subscribe to me on youtube to get more such videos in the future. 

Indian Statistical Institute, ISI, M.Math 2021 PMB Subjective Questions Solutions and Discussions: Click Here

To view ISI MMATH 2020 solutions, hints, and discussions: Click Here

To view ISI MMATH 2018 solutions, hints, and discussions: Click Here

To view ISI MMATH 2017 solutions, hints, and discussions: Click Here

To view ISI MMATH 2016 solutions, hints, and discussions: Click Here

To view previous year’s question papers: Click Here

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};

(c) 0;

(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

(c) 1 / 18

(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

(c) the system has infinitely many solutions

(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

(c) 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 !}

(c) \frac{(n+1)^{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)

(c) [0,1]

(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

(c) n !

(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;

(c) 3;

(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)

(c) (\mathrm{C})(0,2)

(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

(c) T is onto but not one-to-one

(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}}

(c) \frac{1}{2}-\frac{1}{\sqrt{3}}

(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;

(c) 64;

(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

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

is

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

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

(c) 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}

(c) \frac{\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

(c) 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

(c) converges for 0<x<\frac{1}{2} and diverges for x \geq \frac{1}{2}

(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

(c) must be 0

(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;

(c) 2;

(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.

(c) the smallest exponent k>1 such that g^{k}=a b for some g \in G is 4 .

(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;

(c) 2;

(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};

(c) \left[\begin{array}{c} t \\ 0 \\ -2 t \end{array}\right], 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

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

is

(a) \frac{3}{4}

(b) \frac{6}{13}

(c) -\frac{1}{2}

(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

(c) \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

(c) 1-(1 / 6)^{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

(c) 2^{n} n !

(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

(c) A is a straight line segment and B is a circle

(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}

(c) f^{-1} does not exist, but g^{-1} does

(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

(c) none of the two axes

(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

(c) \frac{p}{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

(c) 2021^{2}-1

(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

Leave a Reply to Ritu Cancel reply