Indian Statistical Institute, ISI MMATH 2021 PMA Solutions & Discussions

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};

(c) 0;

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


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


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:


The sum 9 appears in 4 cases:


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


We find that


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


We observe that



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

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 !}


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





On the other hand,

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


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


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]


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}


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.


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)


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




    \[\sqrt{a}<2; \ 0<a<4.\]

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.


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}}


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

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





Taking square roots we find


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

    \[\prod_{k=1}^{5} k !\]

\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.


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

(c) 2 \ln 2;

(d) \ln 2


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}


Consider the characteristic equation of the recurrence:

    \[k^2-\alpha k -(1-\alpha)=0.\]

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


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,


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


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


We find b:

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

So, the equation of the line is




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.


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




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




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


We compute the trace of the matrix:


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.


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.


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.


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


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).


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}


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}

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

(d) none of the above


The gradient of f equals

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


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

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


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


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}


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


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 !


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


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.


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


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


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


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}


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


Differentiate this equation in x:


Differentiate in y:

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



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


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

The number of elements in the set

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

(a) 2021

(b) 2020

(c) 2021^{2}-1

(d) 2020 \times 2021


The determinant of A must be zero:


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)


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