Indian Statistical Institute, ISI BStat & BMath 2020 UGA Solutions & Discussions

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ISI BMATH & BSTAT 2020 Multiple Choice Questions UGA, Objectives: Solutions and Discussions

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Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGA Objective Solutions: Click Here

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

Problem 1

For any real number $x$ , let $[x]$ be the greatest integer $m$ such that $m\leq x$ . Then the number of points of discontinuity of the function $g(x)=[x^2-2]$ on the interval $(-3,3)$ is

(a) $5$ ;

(b) $9$ ;

(d) $16$ .

Solution:

$f(x)=|x|+|x+1|+e^x.$

$|x|$ is differentiable everywhere except at $x=0.$

$|x+1|$ is differentiable everywhere except at $x=-1.$

$e^x$ is differentiable everywhere.

$f(x)$ is differentiable everywhere except at $x=0$ and $x=-1.$

Answer: (b)

Problem 2

If $f,g$ are real-valued differentiable functions on the real line $\mathbb{R}$ such that $f(g(x))=x$ and $f'(x)=1+(f(x))^2$ , then $g'(x)$ equals

(a) $\frac{1}{1+x^2}$ ;

(b) $1+x^2$ ;

(d) $1+x^4$ .

Solution:

Differentiate equality $f(g(x))=x:$

$f'(g(x))g'(x)=1$

$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{1+(f(g(x))^2}=\frac{1}{1+x^2}$

Answer: (a)

Problem 3

The number of subsets of $\{1,2,3,\dots,10\}$ having an odd number of elements is

(a) $1024$ ;

(b) $512$ ;

(d) $50$ .

Solution:

Let $A_n$ be the number of subsets of $\{1,2,\ldots,n\}$ having odd number of elements. Let $S\subset \{1,2,\ldots,n\}$ be a subset having odd number of elements. If $n\not \in S,$ then $S\subset \{1,2,\ldots,n-1\}.$ There are $A_{n-1}$ such subsets. If $n\in S,$ then $S\setminus \{n\}\subset \{1,2,\ldots,n-1\}$ has even number of elements. There are $2^{n-1}-A_{n-1}$ such subsets.

$A_n=A_{n-1}+2^{n-1}-A_{n-1}=2^{n-1}.$

$A_{10}=2^9=512.$

Answer: (b)

Problem 4

A group of $64$ players in a chess tournament needs to be divided into $32$ groups of $2$ players each. In how many ways can this be done $?$

(a) $\frac{64!}{32!2^{32}}$ ;

(b) $\binom{64}{2}\binom{62}{2}\dots \binom{4}{2}\binom{2}{2}$

(d) $\frac{64!}{2^{64}}$

Solution:

First, we permute all the $64$ players in $64!$ ways and then group the first two, second two, and so on and make $32$ groups, but inside each group the two players can get permuted in $2!=2$ ways and so we divide $64!$ by 2^{32} and since the groups can get interchanged we divide it by $32!$ .

Answer: (a)

Problem 5

The number of real solution of $e^x=\sin(x)$ is

(a) $0$ ;

(b) $1$ ;

(d) infinity.

Solution:

$e^x\in (0,1)$ for all $x<0.$ On each segment $[2\pi n, \frac{\pi}{2}+2\pi n],$ $n\leq -1,$ the function $\sin(x)$ increases from $0$ to $1$ , hence it intersects $e^x.$

Answer: (d)

Problem 6

What is the limit of $\sum^n_{k=1}\frac{e^{-k/n}}{n}$ as $n$ tends to $\infty ?$

(a)The limit dose not exist;

(b) $\infty$ ;

(d) $e^{-0.5}$ .

Solution:

Consider partition $0<\frac{1}{n}<\frac{2}{n}<\ldots<1$ of the segment $[0,1].$ On each segment $[\frac{k-1}{n},\frac{k}{n}]$ of the partition choose the value of $e^{-x}$ at the right end-point. Corresponding Riemann sum is

$\frac{1}{n}\sum^n_{k=1}e^{-k/n}\to \int^1_0 e^{-x}dx=1-e^{-1}, \ n\to\infty.$

Answer: (c)

Problem 7

Let $f,g$ be differentiable functions on the real line $\mathbb{R}$ with $f(0)>g(0)$ .
Assume that the set $M=\{t\in\mathbb{R}|f(t)=g(t)\}$ is non-empty and that $f'(t)\geq g'(t)$ for all $t\in M$ . Then which of the following is necessarily true ?

(a) If $t\in M$ , then $t<0$ .

(b) For any $t\in M,f'(t)>g'(t)$ .

(d) none of the above.

Solution:

Let $f(t)=(t+1)(t-1)^2,$ $g(t)=0.$ Then $f(0)=1>g(0).$ The set

$M=\{t\in \mathbb{R}:f(t)=g(t)\}=\{t\in \mathbb{R}:(t+1)(t-1)^2=0\}=\{-1,1\}$

is not empty. $f'(t)=(t-1)^2+2(t+1)(t-1)=(t-1)(2t+1),$ $g'(t)=0.$

$f'(-1)=2>0=g'(-1), f'(1)=0=g'(1).$

$f$ and $g$ satisfy the conditions of the problem. However, neither of (A),(B),(C) is true.

Answer: (d)

Problem 8

Consider the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5\dots$ obtained by writing one $1$ , two $2's$ , three $3's$ and so on. What is the $2020^{th}$ term in the sequence $?$

(a) $62$ ;

(b) $63$ ;

(d) $65$ .

Solution:

The last position where the integer $n$ appears is $1+2+\ldots+n=\frac{n(n+1)}{2}.$ If $n=63,$ then $\frac{63\cdot 64}{2}=2016.$ The number $64$ appears on $64$ positions
$2017,2018,2019,2020,\ldots,2080.$

Answer: (c)

Problem 9

Let $A=\{x_1,x_2,\dots,x_{50}\}$ and $B=\{y_1,y_2,\dots,y_{20}\}$ be two sets of real numbers. What is the total number of function $f:A\to B$ such that $f$ is onto and $f(x_1)\leq f(x_2)\leq\dots\leq f(x_{50})?$

(a) $\binom{49}{19}$ ;

(b) $\binom{49}{20}$ ;

(d) $\binom{50}{20}$ .

Solution:

To define an increasing mapping $f$ of $A=\{x_1,x_2,\ldots,x_{50}\}$ onto $B=\{y_1,y_2,\ldots,y_{20}\}$ we must split elements of $A$ into $20$ nonempty groups of consecutive integers (first group will be mapped to $y_1,$ second — to $y_2$ and so on), i.e. to place $19$ barriers on $49$ available places between $x_j'$ s.

Answer: (a)

Problem 10

The number of complex roots of the polynomial $z^5-z^4-1$ which have modulus $1$ is

(a) $0$ ;

(b) $1$ ;

(d) more that $2$

Solution:

Let $z$ be a solution to $z^5-z^4-1=0$ such that $|z|=1.$ Denote $w=z^5.$ Then $|w|=1,$

$|w-1|=|z^5-1|=|z^4|=1.$

Write $w=e^{i\theta},$ $-\pi<\theta\leq \pi.$

$1=|e^{i\theta}-1|^2=(1-\cos(\theta))^2+\sin(\theta)^2=2-2\cos(\theta), \ \cos(\theta)=\frac{1}{2}, \theta=\pm\frac{\pi}{3}.$

$z^5=w=e^{\pm\frac{i\pi}{3}}, z=e^{\pm \frac{i\pi}{15}+\frac{2\pi i n}{5}}, 0\leq n\leq 4.$

$z^4=z^5-1=e^{\pm \frac{i\pi }{3}}-1=e^{\pm\frac{2\pi i }{3}}.$

$e^{\pm\frac{2\pi i }{3}}=e^{\pm \frac{4\pi i}{15}+\frac{8\pi i n}{5}}$

$\pm\frac{2\pi i}{3}=\pm \frac{4\pi i}{15}+\frac{8\pi i n}{5}+2\pi k i.$

$\pm 10=\pm 4+24n+30k, \pm 6=24n+30k$

$\pm 1=4n+5k.$

Only two possibilities for $n:$

$n=1, k=-1: -1=4-5, z=e^{-\frac{\pi i}{15}+\frac{2\pi i}{5}}=e^{\frac{\pi i}{3}}$

$n=4, k=-3: 1=16-15, z=e^{\frac{\pi i}{15}+\frac{8\pi i}{5}}=e^{\frac{5\pi i}{3}}$

Answer: (c)

Problem 11

The number of real roots of the polynomial

$p(x)=(x^{2020}+2020x^2+2020)(x^3-2020)(x^2-2020)$

(a) $2$ ;

(b) $3$ ;

(d) $2025$ .

Solution:

Equation $x^{2020}+2020x^2+2020=0$ has no real roots.

Equation $x^3-2020=0$ has one real root $x=\sqrt[3]{2020}.$

Equation $x^2-2020=0$ has two real roots $x=\pm\sqrt{2020}.$

Answer: (b)

Problem 12

Which of the following is the sum of an infinite geometric sequence whose terms come from the set $\{1,\frac{1}{2},\frac{1}{4},\dots,\frac{1}{2^n},\dots\}?$

(a) $\frac{1}{5}$ ;

(b) $\frac{1}{7}$ ;

(d) $\frac{1}{11}$ .

Solution:

Let $a_1=\frac{1}{2^n}$ be the first term, and $a_2=\frac{1}{2^m}$ be the second term of the progression, $0\leq n<m.$ Then the common ratio of the progression is $\frac{a_2}{a_1}=2^{n-m}.$ The sum is equal to

$S=\frac{2^{-n}}{1-2^{n-m}}=\frac{1}{2^n-2^{2n-m}}.$

The difference in the denominator is odd, so we take $m=2n.$ Let $n=3.$ Then

$S=\frac{1}{8-1}=\frac{1}{7}.$

This is the sum of the progression $\frac{1}{2^3}+\frac{1}{2^9}+\frac{1}{2^{15}}+\ldots$

Answer: (b)

Problem 13

The integral part of $\sum^{9999}_{n=2}\frac{1}{\sqrt{n}}$ equals

(a) $196$ ;

(b) $197$ ;

(d) $199$ .

Solution:

Compare the sum with the integral of $\frac{1}{\sqrt{x}}:$

$\sum^{9999}_{n=2}\frac{1}{\sqrt{n}}=\sum^{9999}_{n=2}\int^n_{n-1}\frac{1}{\sqrt{n}}dx<\int^{9999}_1\frac{dx}{\sqrt{x}}=2\sqrt{9999}-2<2\cdot 100-2=198.$

$\sum^{9999}_{n=2}\frac{1}{\sqrt{n}}=\sum^{9999}_{n=2}\int^{n+1}_{n}\frac{1}{\sqrt{n}}dx>\int^{10000}_2\frac{dx}{\sqrt{x}}=2\sqrt{10000}-2\sqrt{2}=200-2\sqrt{2}>197.$

Hence, the integer part of the sum is $197.$

Answer: (b)

Problem 14

Let $a_n$ be the number of subsets of $\{1,2,\dots,n\}$ that do not contain any two consecutive numbers. Then

(a) $a_n=a_{n-1}+a_{n-2}$ ;

(b) $a_n=2a_{n-1}$ ;

(d) $a_n=a_{n-1}+2a_{n-2}$ .

Solution:

Consider a subset $S\subset \{1,2,\ldots,n\}$ that does not contain any two consecutive numbers. If $n\not \in S,$ then $S\subset \{1,2,\ldots,n-1\}$ (there are $a_{n-1}$ such subsets). If $n\in S,$ then $n-1\not\in S$ and $n\setminus \{n\}\subset \{1,2,\ldots,n-2\}$ (there are $a_{n-2}$ such subsets).

Answer: (a)

Problem 15

There are $128$ numbers $1,2,\dots,128$ which are arranged in a circular pattern in clockwise order. We start deleting numbers from this set in a clockwise fashion as follows. First delete the number $2$ , then skip the next available number (which is $3$ ) and delete $4$ . Continue in this manner, that is after deleting a number, skip the next available number clockwise and delete the number available after that, till only one number remains. What is the last number left $?$

(a) $1$ ;

(b) $63$ ;

(d) None of the above.

Solution:

Consider the same algorithm but with $2^n$ numbers arrange in the clockwise order. If $n=1,$ then we will delete $2$ and the only number remained will be $1.$ By induction we prove that the last number always be $1.$ Assume the statement is proved for $n-1$ and consider numbers $1,2,\ldots,2^n$ arranged circularly in clockwise order. After the first $2^{n-1}$ steps we will delete all even numbers and will get $2^{n-1}$ numbers with the first available element $1.$ By the inductive assumption, the last number left will be $1.$

Answer: (a)

Problem 16

Let $z$ and $w$ be complex numbers lying on the circles of radii $2$ and $3$ respectively, with centre $(0,0)$ . If the angle between the corresponding vectors is $60$ degrees, then the value of $|z+w|/|z-w|$ is:

(a) $\frac{\sqrt{19}}{\sqrt{7}}$ ;

(b) $\frac{\sqrt{7}}{\sqrt{19}}$ ;

(d) $\frac{\sqrt{7}}{\sqrt{12}}$ .

Solution:

$z=2e^{i\theta},$ $w=3e^{i\theta+i\frac{\pi}{3}}.$

$\frac{|z+w|}{|z-w|}=\frac{|2+3e^{i\pi/3}|}{|2-3e^{i\pi/3}|}=\sqrt{\frac{(2+\frac{3}{2})^2+\frac{27}{4}}{(2-\frac{3}{2})^2+\frac{27}{4}}}=\sqrt{\frac{76}{28}}=\sqrt{\frac{19}{7}}$

Answer: (a)

Problem 17

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 the length of the side of the square is

(a) $\frac{3r}{2}$ ;

(b) $\frac{4r}{3}$ ;

(d) $\frac{8r}{5}$ .

Solution:

Let $s$ be the side of the square. The sides of the right-angled triangle $OAB$

are equal

$OA=\frac{s}{2}, OB=r, AB=s-r.$

Hence,

$r^2=\frac{s^2}{4}+(s-r)^2=\frac{5s^2}{4}-2sr+r^2, s=\frac{8r}{5}.$

Answer: (d)

Problem 18

For a real number $x$ , let $[x]$ denote the greatest integer less than or equal to $x$ . Then the number of real solutions of $|2x-[x]|=4$ is

(a) $4$ ;

(b) $3$ ;

(d) $1$ .

Solution:

$2x-[x]=\pm 4.$ Let $x=n+r,$ $n\in \mathbb{Z},$ $0\leq r<1.$ Then $[x]=n$ and

$2n+2r-n=\pm 4,$

$n+2r=\pm 4, 2r=\pm 4-n.$

Either $r=0$ and $x=n=\pm 4,$ or $r=\frac{1}{2},$ $n=3$ or $-5,$ and $x=3.5$ or $-4.5.$

Answer: (a)

Problem 19

If $\sin(\tan^{-1}(x))=\cot(\sin^{-1}(\sqrt{\frac{13}{17}}))$ then $x$ is

(a) $\frac{4}{17}$ ;

(b) $\frac{2}{3}$ ;

(d) $\sqrt{\frac{17^2-13^2}{17\times 13}}$

Solution:

Let $\alpha=\tan^{-1}(x),$ $\beta=\sin^{-1}(\sqrt{\frac{13}{17}}).$ Then

$-\frac{\pi}{2}<\alpha,\beta <\frac{\pi}{2},$

$\tan\alpha=x, \sin\beta=\sqrt{\frac{13}{17}}.$

$\sin\alpha=\cot\beta=\frac{\cos\beta}{\sin\beta}.$

$\cos\beta=\sqrt{1-\sin(\beta)^2}=\frac{2}{\sqrt{17}}$

$\sin\alpha=\cot\beta=\frac{2}{\sqrt{13}}, \cos\alpha=\sqrt{1-\sin(\alpha)^2}=\frac{3}{\sqrt{13}},$

$x=\tan\alpha=\frac{2}{3}.$

Answer: (b)

Problem 20

If the word PERMUTE is permuted is all possible ways and the different resulting words are written down in alphabetical order (also known as dictionary order), irrespective of whether the word has meaning or not then the $720^{th}$ word would be:

(a) $EEMPRTU$ ;

(b) $EUTRPME$ ;

(d) $MEETPUR$ .

Solution:

Put the letters of the word PERMUTE in the dictionary order: EEMPRTU. Since the last six letters are distinct, the first $6!=720$ permutations do not change the position of the first letter and include all possible permutations of the last six letters. Then the 720th word is defined by the last (in the dictionary order) permutation of EMPRTU, i.e. it is EUTRPME

Answer:(b)

Problem 21

The points $(4,7,-1),(1,2,-1),(-1,-2,-1)$ and $(2,3,-1)$ in $\mathbb{R}^3$ are the vertices of a

(a) rectangle which is not a square;

(b) rhombus;

(d) trapezium which is not a parallelogram.

Solution:

All four points belong to the plane $z=-1.$ Denote $A=(4,7,-1),$ $B=(1,2,-1),$ $C=(-1,-2,-1)$ and $D=(2,3,-1).$ Observe equality of vectors

$\vec{AB}=(-3,-5,0)=\vec{DC}$

$\vec{AD}=(-2,-4,0)= \vec{BC}.$

Vectors $\vec{AB}$ and $\vec{AD}$ are not orthogonal and of different lengths.

Answer: (c)

Problem 22

Let $f(x),g(x)$ be function on the real line $\mathbb{R}$ such that both $f(x)+g(x)$ and $f(x)g(x)$ are, differentiable. Which of the following is FALSE ?

(a) $f(x)^2+g(x)^2$ is necessarily differentiable.

(b) f(x) is differentiable if and only if $g(x)$ is differentiable.

(d) If $f(x)>g(x)$ for all $x\in \mathbb{R}$ , then $f(x)$ is differentiable.

Solution:

Let

$f(x)=\begin{cases} 1, \ x\geq 0 \\ 0, \ x<0 \end{cases}, g(x)=\begin{cases} 0, \ x\geq 0 \\ 1, \ x<0 \end{cases}$

Then $f(x)+g(x)=1,$ $f(x)g(x)=0$ are differentiable, while both $f(x)$ and $g(x)$ are discontinuous.

Answer: (c)

Problem 23

Let $S$ be the set consisting of all those real numbers that can be written as $p-2a$ where $p$ and $a$ are the perimeter and area of a right-angled triangle having base length $1$ . Then $S$ is

(a) $(2,\infty)$ ;

(b) $(1,\infty)$ ;

(d) the real line $\mathbb{R}$ .

Solution:

Let $x$ be the height of a triangle. Then $p=1+x+\sqrt{x^2+1},$ $a=\frac{x}{2}.$ $S$ is the set of all real numbers of the form

$1+x+\sqrt{x^2+1}-x=1+\sqrt{x^2+1}>2.$

Answer: (a)

Problem 24

Let $S=\{1,2,\dots,n\}$ . For any non-empty subset $A$ of $S$ , let $l(A)$ denote the largest number in $A$ . If $f(n)=\sum_{A\subseteq S}l(A)$ , that is, $f(n)$ is the sum of the numbers $l(A)$ while $A$ ranges over all nonempty subsets of $S$ , then $f(n)$ is

(a) $2^n(n+1)$ ;

(b) $2^n(n+1)-1$ ;

(d) $2^n(n-1)+1$ .

Solution:

If $l(A)=k,$ then all other elements of $A$ can be arbitrary elements of $\{1,2,\ldots,k-1\}.$ It means that there are $2^{k-1}$ subsets of $\{1,2,\ldots,n\}$ with $l(A)=k.$

$f(n)=\sum^n_{k=1}k2^{k-1}.$

We compute

$\sum^n_{k=1}z^k=\frac{z^{n+1}-z}{z-1},$

$\sum^n_{k=1}kz^{k-1}=\frac{d}{dz}\sum^n_{k=1}z^k=\frac{d}{dz}\frac{z^{n+1}-z}{z-1}=\frac{(n+1)z^n-1}{z-1}-\frac{z^{n+1}-z}{(z-1)^2}.$

At $z=2$ we get $(n+1)2^n-1-2^{n+1}+2=2^n(n-1)+1.$

Answer:(d)

Problem 25

If $a,b,c$ are distinct odd natural numbers, then the number of rational roots of the polynomial $ax^2+bx+c$

(a) must be $0$ .

(b) must be $1$ ;

(d) cannot be determined from the given data.

Solution:

There are rational roots of the polynomial $ax^2+bx+c,$ if $\sqrt{b^2-4ac}$ is rational. It is possible only when the discriminant is a perfect square:

$b^2-4ac=d^2.$

Then $4ac=b^2-d^2=(b-d)(b+d).$ Observe that $d$ is odd, hence $b-d$ and $b+d$ are even.

$ac=\frac{b-d}{2}\frac{b+d}{2}$

and both $\frac{b-d}{2}$ and $\frac{b+d}{2}$ are odd:

$\frac{b-d}{2}=2k+1, \frac{b+d}{2}=2l+1.$

But then $b=2k+2l+2$ is even. This is impossible.

Answer: (a)

Problem 26

Let $A,B,C$ be finite subsets of the plan such that $A\cap B,B\cap C$ and $C\cap A$ are all empty. Let $S=A\cup B\cup C$ . Assume that no three points of $S$ are collinear and also assume that each of $A,B$ and $C$ has at least $3$ points. Which of the following statements is always true $?$

(a) There exists a triangle having a vertex from each of $A,B,C$ that does not contain any point of $S$ in its interior;

(b) Any triangle having a vertex from each of $A,B,C$ must contain a point of $S$ in its interior;

(c) There exists a triangle, having a vertex from each of $A,B,C$ that contains all the remaining points of $S$ in its interior;

(d) There exist $2$ triangles, both having a vertex from each of $A,B,C$ such that two triangles do not intersect.

Solution:

Let $n$ be the minimal number of points of $S$ that can be in the interior of a triangle that has a vertex from each $A,$ $B,$ $C.$ Consider such triangle $abc,$ where $a\in A,$ $b\in B,$ $c\in C.$ Let $n\geq 1$ and $x\in S.$ be inside the interior of $abc.$ Without loss of generality assume that $x=a'\in A.$ Then $a'bc$ has $<n$ points of $S$ in its interior, which is impossible. So, $n=0.$

Answer: (a)

Problem 27

Shubhaangi thinks she may be allergic to Bengal gram and takes a test that is known to give the following result:

For people who really do have the allergy, the test says “Yes” $90\%$ of the time.
For people who do not have the allergy, the test says “Yes” $15\%$ of the time

If $2\%$ of the population has the allergy and Shubhaangi’s test says “Yes”, then the chance that Shubhaangi does really have the allergy are
(a) $1/9$ ;

(b) $6/55$ ;

(d) cannot be determined from the given data

Solution:

Let $A$ denote the event that the test says “Yes”. Consider two hypotheses:

$H_0$ — Shubhaangi has the allergy;

$H_1$ — Shubhaangi does not have the allergy.

Then

$P(H_0)=0.02, P(H_1)=0.98, P(A|H_0)=0.9, P(A|H_1)=0.15.$

By the Bayes rule

$P(H_0|A)=\frac{P(A|H_0)P(H_0)}{P(A|H_0)P(H_0)+P(A|H_1)P(H_1)}=\frac{0.9\cdot 0.02}{0.9\cdot 0.02+ 0.15\cdot 0.98}=\frac{18}{165}=\frac{6}{55}$

Answer: (b)

Problem 28

For any real number $x$ , let $[x]$ be the greatest integer $m$ such that $m\leq x$ . Then the number of points of discontinuity of the function $g(x)=[x^2-2]$ on the interval $(-3,3)$ is

(a) $5$ ;

(b) $9$ ;

(d) $16$ .

Solution:

We observe that $g(x)$ is even and $g(x)=-2$ for $-1< x< 1.$ It is enough to find positive points of discontinuity and multiply the result by 2. The function $x\to x^2-2$ is increasing on from $-2$ to $7$ when $0<x<3.$ $x\in (0,3)$ is a point of discontinuity of $g(x)$ if and only if $x^2-2$ is integer. There are $8$ integers in the interval $(-2,7).$

Answer: (d)

Problem 29

The area of the region in the plane $\mathbb{R}^2$ given by points $(x,y)$ satisfying $|y|\leq 1$ and $x^2+y^2\leq 2$ is

(a) $\pi+1$ ;

(b) $2\pi-2$ ;

(d) $2\pi-1$ .

Solution:

Parametrize the region as follows $-1\leq y\leq 1,$ $-\sqrt{2-y^2}\leq x\leq \sqrt{2-y^2}.$ Then the area is

$\int^1_{-1} 2\sqrt{2-y^2}dy=4\sqrt{2}\int^{1}_0\sqrt{1-(\frac{y}{\sqrt{2}})^2}dy=8\int^{1/\sqrt{2}}_0\sqrt{1-y^2}dy=$

$y=\sin\theta$

$=8\int^{\pi/4}_0\cos^2\theta d\theta=8\left(\frac{\theta}{2}+\frac{\sin(2\theta)}{4}\right)\bigg|^{\theta=\pi/4}_{\theta=0}=\pi+2.$

Answer: (c)

Problem 30

Let $n$ be a positive integer and $t\in(0,1)$ . Then $\sum^n_{r=0}r\binom{n}{r}t^r(1-t)^{n-r}$
equals

(a) $nt$ ;

(b) $(n-1)(1-t)$ ;

(d) $(n^2-2n+2)t$

Solution:

$\sum^n_{r=0}r{n\choose r}t^r(1-t^{n-r}=t\sum^n_{r=1}\frac{n!}{(r-1)!(n-r)!}t^{r-1}(1-t)^{n-r}=$

$=nt\sum^{n-1}_{r=0}{n-1\choose r}t^r(1-t)^{n-r}=nt.$

Answer: (a)

5 Comments

fayaque ali

September 29, 2020 at 5:44 am Reply

For problem 22, option B is also a correct answer right? That is to say that the statement is false. Both f(x) and g(x) are independent of each other. The differentiability of g does not effect the differentiability of f.
- FractionsHub
  
  September 29, 2020 at 6:02 am Reply
  
  “Both f(x) and g(x) are independent of each other” How do you prove your claim? Do you have an example where $f(x)$ is differentiable but $g(x)$ is not or vice versa?
  - Soumil
    
    July 15, 2021 at 1:49 pm Reply
    
    for q22, option B should also be the answer right?
    Example : f(x)= 100 + |x-1|
    g(x) = 100 – |x-1|
    - Soumil
      
      July 15, 2021 at 1:51 pm Reply
      
      Sorry, I meant in q22, option *d* should also be the answer
Soumil

July 15, 2021 at 1:48 pm Reply

Yes for q22, option B should be the answer.
Example : f(x)= 100 + |x-1|
g(x) = 100 – |x-1|

Indian Statistical Institute, ISI BStat & BMath 2020 UGA Solutions & Discussions

ISI BMATH & BSTAT 2020 Multiple Choice Questions UGA, Objectives: Solutions and Discussions

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

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