Indian Statistical Institute, ISI BStat & BMath 2018 Solutions & Hints

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ISI BMath & BSTAT 2018 Subjective Questions UGB along with hints and 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 even with the hints provided, then only look at the solution of 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 me on youtube to get more such videos in the future. 

Problem 1.

Find all pairs (x, y) with x, y real, satisfying the equations:

    \[ \sin\left(\frac{x+y}{2}\right)=0, \ |x|+|y|=1. \]

Topic: Triginometry
Difficulty level: Very Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 1 UGB Solution & Discussion Video:

Hint 1

Zeros of the sine function are of the form \pi n, n\in \mathbb{Z}.

Hint 2

Consider the cases n=0 and n\neq 0.

Hint 1

In the case n\neq 0 apply the triangle inequality.

Full Solution

Zeros of the sine function are of the form \pi n, n\in \mathbb{Z}. So,

    \[ \frac{x+y}{2}=\pi n, \ x+y=2\pi n. \]

Consider the case n=0. Then y=-x, |y|=|x| and

    \[ 1=|x|+|y|=2|x|, |x|=\frac{1}{2}. \]

We obtain two solutions: x=\frac{1}{2}, y=-\frac{1}{2} and x=-\frac{1}{2}, y=\frac{1}{2}.

 

 

Let n\ne 0, then |n|\geq 1. From the triangle inequality we obtain

    \[ 1=|x|+|y|\geq |x+y|=|2\pi n|=2\pi |n|\geq 2\pi \]

But 2\pi >1 and there are no solution (x,y) with x+y=2\pi n, n\ne 0.

 

 

So, there are two such pairs: (x,y)=(\frac{1}{2},-\frac{1}{2}) and (x,y)=(-\frac{1}{2},\frac{1}{2}).

 

Problem 2.

Suppose that PQ and RS are two chords of a circle intersecting at a point O. It is given that PO = 3 cm and SO = 4 cm. Moreover, the

area of the triangle POR is 7 cm^2. Find the area of the triangle QOS.

Topic: Calculus: Geometry
Difficulty Level: Very Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 2 UGB Solution & Discussion Video:

Diagram

Hint 1

Can you figure out that triangles triangles POR and QOS are similar?

Hint 2

Find out what happens to the ratio of the area of two similar triangle. Figure out the coefficient of similarity.

Hint 3

Did you really needed this hint?
If you have figured out the ratio as in hint 2 then what are you waiting for?
You know the area of triangle POR.

Full Solution

Consider triangles POR and QOS. Consider angles \angle PRO=\angle PRS and \angle OQS=\angle PQS. They are subtended by the same chord PS and are located on the same side of the chord (since PQ and RS intersect). So, \angle PRO=\angle OQS. Angles \angle POR and \angle QOS are vertical, so \angle POR =\angle QOS. Triangles POR and QOS are similar by the equality of two angles. Coefficiant of similarity is found from the ratio of sides: the side PO=3 is opposite to the angle \angle PRO, the side SO=4 is opposite to the angle \angle OQS. So, the coefficient of similarity is k=\frac{4}{3}. The area of the triangle QOS is k^2 times bigger than the area of the triangle POR: area of QOS =7\cdot (\frac{4}{3})^2=\frac{112}{9}.

Problem 3.

Let f : \mathbb{R}\to \mathbb{R} be a continuous function such that for all x\in\mathbb{R} and

for all t\geq 0,

    \[ f(x)=f(e^t x). \]

Show that f is a constant function.

Topic: Calculus: Functions, Limits
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 3 UGB Solution & Discussion Video:

Hint 1

Let x\ne 0. For any natural number n\geq 1 set t=\ln n\geq 0.

Hint 2

    \[ f\left(\frac{x}{n}\right)=f\left(e^t\frac{x}{n}\right)=f\left(n\frac{x}{n}\right)=f(x). \]

Hint 3

    \[ f(0)=\lim_{n\to \infty}f\left(\frac{x}{n}\right)=\lim_{n\to\infty}f(x)=f(x). \]

Full Solution

Let x\ne 0. For any natural number n\geq 1 set t=\ln n\geq 0. Then

    \[ f\left(\frac{x}{n}\right)=f\left(e^t\frac{x}{n}\right)=f\left(n\frac{x}{n}\right)=f(x). \]

Taking the limit as n\to \infty and using continuity of f, we get

    \[ f(0)=\lim_{n\to \infty}f\left(\frac{x}{n}\right)=\lim_{n\to\infty}f(x)=f(x). \]

For all x\in\mathbb{R}, f(x)=f(0) and f is a constant function.

Problem 4.

Let f : (0, \infty) \to\mathbb{R} be a continuous function such that for all x\in (0,\infty),

    \[ f(2x) = f(x). \]

Show that the function g defined by the equation

    \[ g(x)=\int^{2x}_xf(t)\frac{dt}{t} \mbox{ for } x>0 \]

is a constant function.

Topic: Calculus: Differentiation, Definite Integral
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 4 UGB Solution & Discussion Video:

Hint 1

Try to see if you have all the conditions satisfied to apply Leibniz integral rule.

If you don’t know what Leibniz integral rule is then read: Leibntz Integral rule.

Hint 2

The function t\to \frac{f(t)}{t} is continuous on (0,\infty).

Hints 3

If you are convinced that you can apply Leibntz integral rule then calculate the derivative of g(x).

Full Solution

The function t\to \frac{f(t)}{t} is continuous on (0,\infty). So, g is continuously differentiable on (0,\infty). We calculate the derivative of g:

    \[ g'(x)=\frac{d}{dx}\left(\int^{2x}_xf(t)\frac{dt}{t}\right)=2\frac{f(2x)}{2x}-\frac{f(x)}{x}=\frac{f(2x)-f(x)}{x}=0. \]

So, g is a constant function.

 

Problem 5.

Let f:\mathbb{R}\to\mathbb{R} be a differentiable function such that its derivative f' is a continuous function. Moreover, assume that for all x\in\mathbb{R},

    \[ 0\leq |f'(x)|\leq \frac{1}{2}. \]

Define a sequence of real numbers \{a_n\}_{n\in\mathbb{N}} by:

    \[ a_1=1, \]

    \[ a_{n+1}=f(a_n) \mbox{ for all } n\in \mathbb{N}. \]

Prove that there exists a positive real number M such that for all n\in\mathbb{N},

    \[ |a_n|\leq M. \]

Topic: Calculus: Differentiation, Continuity
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 5 UGB Solution & Discussion Video:

Hint 1

Try to apply Mean Value Theorem.

Let x<y. By the mean value theorem, there is c\in(x,y) such that

    \[ |f(y)-f(x)|=|f'(c)| (y-x). \]

So,

    \[ |f(y)-f(x)|\leq \frac{1}{2}|y-x|. \]

Hint 2

For any n\geq 2 we have

    \[ |a_{n+1}-a_n|=|f(a_n)-f(a_{n-1})|\leq \frac{1}{2}|a_n-a_{n-1}|. \]

Hint 3

By induction try to verify that for all n\geq 1

    \[ |a_{n+1}-a_n|\leq \frac{1}{2^{n-1}}|a_2-a_1|. \]

Full Solution

Let x<y. By the mean value theorem, there is c\in(x,y) such that

    \[ |f(y)-f(x)|=|f'(c)| (y-x). \]

So,

    \[ |f(y)-f(x)|\leq \frac{1}{2}|y-x|. \]

For any n\geq 2 we have

    \[ |a_{n+1}-a_n|=|f(a_n)-f(a_{n-1})|\leq \frac{1}{2}|a_n-a_{n-1}|. \]

By induction we verify that for all n\geq 1

    \[ |a_{n+1}-a_n|\leq \frac{1}{2^{n-1}}|a_2-a_1|. \]

For n=1 inequality turns into equality. Assume that

    \[ |a_{n+1}-a_n|\leq \frac{1}{2^{n-1}}|a_2-a_1|. \]

Then

    \[ |a_{n+2}-a_{n+1}|\leq \frac{1}{2}|a_{n+1}-a_{n}|\leq \frac{1}{2^n}|a_2-a_1|. \]

Hence, for all n\geq 1 we have

    \[ |a_{n+1}-a_n|\leq \frac{1}{2^{n-1}}|a_2-a_1|. \]

Now, by the triangle inequality

    \[ |a_{n+1}-a_1|\leq \sum^n_{k=1}|a_{k+1}-a_k|\leq \sum^n_{k=1} \frac{1}{2^{k-1}}|a_2-a_1|\leq |a_2-a_1|\sum^\infty_{k=1}\frac{1}{2^{k-1}}=2|a_2-a_1|. \]

So, for all n\geq 1 we have

    \[ |a_n|\leq |a_n-a_1|+|a_1|\leq |a_1|+2|a_2-a_1| \]

The needed statement holds with M=|a_1|+2|a_2-a_1|=1+2|f(1)-1|.

Problem 6.

Let a\geq b\geq c>0 be real numbers such that for all n\in\mathbb{N}, there

exist triangles of side lengths a^n, b^n, c^n. Prove that the triangles are

isosceles.

Topic: Geometry, Inequality, Limits
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 6 UGB Solution & Discussion Video:

Hint 1

Assume a>b. Then also a>c. Since a^n,b^n,c^n are the sides of a triangle, the triangle inequality holds:

    \[ a^n\leq b^n+c^n. \]

Hint 2

It follows that for all n\geq 1

    \[ 1\leq \left(\frac{b}{a}\right)^n+\left(\frac{c}{a}\right)^n. \eqno(1) \]

Hint 3

However, by assumption 0<\frac{b}{a}<1 and 0<\frac{c}{a}<1.

Now try to apply limits on \bigg(\frac{b}{a}\bigg)^n and \bigg(\frac{c}{a}\bigg)^n

Full Solution

Assume a>b. Then also a>c. Since a^n,b^n,c^n are the sides of a triangle, the triangle inequality holds:

    \[ a^n\leq b^n+c^n. \]

It follows that for all n\geq 1

    \[ 1\leq \left(\frac{b}{a}\right)^n+\left(\frac{c}{a}\right)^n. \eqno(1) \]

However, by assumption 0<\frac{b}{a}<1 and 0<\frac{c}{a}<1. So,

    \[ \lim_{n\to\infty} \left(\frac{b}{a}\right)^n=\lim_{n\to \infty} \left(\frac{c}{a}\right)^n=0. \]

Passing to the limit in (1) we obtain 1\leq 0, which is impossible. So, a=b and all triangles with sides a^n,b^n,c^n are isosceles.

Problem 7.

Let a, b, c\in \mathbb{N} be such that

    \[ a^2+b^2=c^2 \mbox{ and } c-b=1. \]

Prove that

 

 

(i) a is odd,

 

(ii) b is divisible by 4,

 

(iii) a^b+b^a is divisible by c.

Topic: Number Theory
Difficulty Level: First two parts: Very Easy, Third Part: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 7 UGB Solution & Discussion Video:

Hint 1

Substitute c=b+1 in the equation a^2+b^2=c^2:

Hint 2

Since a is odd, write a=2d+1. we get

    \[ 2b+1=a^2=(2d+1)^2=4d^2+4d+1. \]

Hint 3

Substitute b=c-1. We must prove that a^{c-1}+(c-1)^a is divisible by c. Expand

    \[ a^{c-1}+(c-1)^a=a^{c-1}+\sum^a_{j=0}{a\choose j} c^j (-1)^{a-j}= \]

a is odd, so (-1)^a=-1

    \[ =(a^{c-1}-1)+\sum^a_{j=1}{a\choose j} c^j (-1)^{a-j}= \]

    \[ =(a^{c-1}-1)+c\sum^a_{j=1}{a\choose j} c^{j-1} (-1)^{a-j}. \]

The second summand is divisible by c, so it is enough to check that a^{c-1}-1 is divisible by c.

Full Solution

(i) Substitute c=b+1 in the equation a^2+b^2=c^2:

    \[ a^2+b^2=(b+1)^2=b^2+2b+1 \]

It follows that a^2=2b+1. a^2 is odd, hence a is odd.

 

(ii) Since a is odd, write a=2d+1. we get

    \[ 2b+1=a^2=(2d+1)^2=4d^2+4d+1. \]

So,

    \[ b=2d^2+2d=2d(d+1). \]

One of the numbers d,d+1 is even, so the product d(d+1) is divisible by 2 and b is divisible by 4.

 

(iii) Substitute b=c-1. We must prove that a^{c-1}+(c-1)^a is divisible by c. Expand

    \[ a^{c-1}+(c-1)^a=a^{c-1}+\sum^a_{j=0}{a\choose j} c^j (-1)^{a-j}= \]

a is odd, so (-1)^a=-1

    \[ =(a^{c-1}-1)+\sum^a_{j=1}{a\choose j} c^j (-1)^{a-j}= \]

    \[ =(a^{c-1}-1)+c\sum^a_{j=1}{a\choose j} c^{j-1} (-1)^{a-j}. \]

The second summand is divisible by c, so it is enough to check that a^{c-1}-1 is divisible by c. From part (ii) we have b=4m, so c=b+1=4m+1. From the initial equation

    \[ a^2=c^2-b^2=(c-b)(c+b)=c+b=2c-1. \]

Then

    \[ a^{c-1}-1=a^{4m}-1=(a^2)^{2m}-1=(2c-1)^{2m}-1= \]

expand

    \[ =\sum^{2m}_{j=0}{2m\choose j}2^jc^j(-1)^{2m-j}-1=\sum^{2m}_{j=1}{2m\choose j}2^jc^j(-1)^{2m-j}+(-1)^{2m}-1= \]

2m is even, so (-1)^{2m}=1

    \[ =\sum^{2m}_{j=1}{2m\choose j}2^jc^j(-1)^{2m-j}=c\sum^{2m}_{j=1}{2m\choose j}2^jc^{j-1}(-1)^{2m-j} \]

and the expression a^{c-1}-1 is also divisible by c.

Problem 8.

Let n\geq 3. Let A = ((a_{ij}))_{1\leq i,j\leq n} be an n\times n matrix such that

a_{ij}\in\{1,-1\} for all 1\leq i,j\leq n. Suppose that

    \[ a_{k1}=1 \mbox{ for all } 1\leq k\leq n \mbox{ and} \]

    \[ \sum^n_{k=1}a_{ki}a_{kj}=0 \mbox{ for all } i\ne j. \]

Show that n is a multiple of 4.

Topic: Number Theory
Difficulty Level: Very Hard

Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 8 UGB Solution & Discussion Video:

Hint 1

Applying the definition of the matrix A with i=1 we get that for all j\ne 1

    \[ \sum^n_{k=1}a_{kj}=0. \]

 

Hint 2

Since n\geq 3 we can choose i,j such that 1<i<j. Then

    \[ \sum^n_{k=1}a_{ki}=0, \sum^n_{k=1}a_{kj}=0, \sum^n_{k=1}a_{ki}a_{kj}=0. \eqno(2) \]

 

 

 

Consider following quantities:

 

N_{1,1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=a_{kj}=1,

 

N_{1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=1, a_{kj}=-1,

 

N_{-1,1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=-1, a_{kj}=1,

 

N_{-1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=a_{kj}=-1.

 

Clearly, N_{1,1}+N_{1,-1}+N_{-1,1}+N_{-1,-1}=n.

Hint 3

Further,

 

N_{1,1}+N_{1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=1,

 

N_{1,1}+N_{-1,1} is the number of indices k\in \{1,\ldots,n\} such that a_{kj}=1,

 

N_{-1,1}+N_{-1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=-1,

 

N_{1,-1}+N_{-1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{kj}=-1.

 

Using these quantities we calculate sums in (2):

    \[ 0=\sum^n_{k=1}a_{ki}=\sum_{k:a_{ki}=1}1-\sum_{k:a_{ki}=-1}1=N_{1,1}+N_{1,-1}-N_{-1,1}-N_{-1,-1}; \]

Full Solution

Applying the definition of the matrix A with i=1 we get that for all j\ne 1

    \[ \sum^n_{k=1}a_{kj}=0. \]

 

 

 

Since n\geq 3 we can choose i,j such that 1<i<j. Then

    \[ \sum^n_{k=1}a_{ki}=0, \sum^n_{k=1}a_{kj}=0, \sum^n_{k=1}a_{ki}a_{kj}=0. \eqno(2) \]

 

 

 

Consider following quantities:

 

N_{1,1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=a_{kj}=1,

 

N_{1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=1, a_{kj}=-1,

 

N_{-1,1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=-1, a_{kj}=1,

 

N_{-1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=a_{kj}=-1.

 

Clearly, N_{1,1}+N_{1,-1}+N_{-1,1}+N_{-1,-1}=n.

 

 

Further,

 

N_{1,1}+N_{1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=1,

 

N_{1,1}+N_{-1,1} is the number of indices k\in \{1,\ldots,n\} such that a_{kj}=1,

 

N_{-1,1}+N_{-1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{ki}=-1,

 

N_{1,-1}+N_{-1,-1} is the number of indices k\in \{1,\ldots,n\} such that a_{kj}=-1.

 

Using these quantities we calculate sums in (2):

    \[ 0=\sum^n_{k=1}a_{ki}=\sum_{k:a_{ki}=1}1-\sum_{k:a_{ki}=-1}1=N_{1,1}+N_{1,-1}-N_{-1,1}-N_{-1,-1}; \]

    \[ 0=\sum^n_{k=1}a_{kj}=\sum_{k:a_{kj}=1}1-\sum_{k:a_{kj}=-1}1=N_{1,1}+N_{-1,1}-N_{1,-1}-N_{-1,-1}; \]

    \[ 0=\sum^n_{k=1}a_{ki}a_{kj}=\sum_{k:a_{kj}=a_{kj}=1}1+\sum_{k:a_{kj}=a_{kj}=-1}1-\sum_{k:a_{ki}=1,a_{kj}=-1}1-\sum_{k:a_{ki}=-1,a_{kj}=1}1= \]

    \[ =N_{1,1}+N_{-1,-1}-N_{1,-1}-N_{-1,1}. \]

We get following relations:

    \[ \begin{cases} N_{1,1}+N_{1,-1}-N_{-1,1}-N_{-1,-1}=0 \\ N_{1,1}+N_{-1,1}-N_{1,-1}-N_{-1,-1}=0\\ N_{1,1}+N_{-1,-1}-N_{1,-1}-N_{-1,1}=0 \end{cases} \]

 

Adding first and second we get 2N_{1,1}-2N_{-1,-1}=0, N_{-1,-1}=N_{1,1}.

 

 

Adding first and third we get 2N_{1,1}-2N_{-1,1}=0, N_{-1,1}=N_{1,1}.

 

 

Adding second and third we get 2N_{1,1}-2N_{1,-1}=0, N_{1,-1}=N_{1,1}.

 

 

It follows that n=N_{1,1}+N_{1,-1}+N_{-1,1}+N_{-1,-1}=4N_{1,1}. n is a multiple of 4.

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