Archive Of Best Indian Statistical Institute entrance problems

Indian Statistical Institute , the leading organisation in India in mathematics and research works conducts entrance tests every year. This test are based solely on mathematics on a higher level than grade school.

Here I present in this note the best problems so far available from the Institute's entrance test of Undergraduate admission. It must be noted that ISI never publishes any solution to it's given problems, the solutions presented are thus my own and might have faults which you are welcome to correct.

Comment problems from ISI that you have I would add to the note if possible with the solution. So here we go !

Problem : A class has $100$ students. Let $a_i,1\le 1\le 100$ , denote the number of friends the i-th student has in the class. For each $0\le j\le99$ , let $c_j$ denote the number of students having at least $j$ friends. Prove that :$\displaystyle \sum_{i=1}^{100} a_i = \sum_{j=0}^{99} c_j$

Solution :

$\displaystyle a_i =$ Number of friends of the $i$-th student

$\displaystyle c_j =$ Number of students having at least $j$ friends

It is harmless to renumber each student so that their corresponding number of friends form an increasing sequence $\{a_i\}$. Since now $c_j$ denotes the number of students having at least $j$ friends it follows that there is some $k$ for which $a_k\ge j$ , and since $\{a_i\}$ is an increasing sequence for all $n\gt k$ ,$a_n$ will have at least $j$ friends. So $c_j$ is exactly the number of elements of the set $\displaystyle S = \{a_k,a_{k+1},\cdots,a_{100}\}$ where each of these students numbering from $k$ on wards have friends $a_k$ and number of these students is value of $c_j$ .

Now we need to put these in a more formal way to help establish a general formula to get our $c_j$ . If we define a function by :

$\displaystyle \begin{aligned} f(x)&=0\quad x\in[0,1) \\ &=1\quad x\ge1\end{aligned}$

Now note that beginning with $0\le j\le99$ we find an $n$ such that $a_i\gt k\forall\; i>n$ and for all $i\lt n$ ,$k>a_i$ and we have $k=a_n$ .

So we need to count the number of entries for each $j$ from some $a_n$ where it has $a_n=j$ to $a_{100}$ .Our function perfectly fits that requirement and we have ,

$\displaystyle c_j=\sum_{i=1}^{100} f\left(\dfrac{a_i}{j}\right)$

Now to prove the given we apply summation again on both sides to get,

$\displaystyle \begin{aligned} \sum_{j=0}^{99} c_j &= \sum_{j=0}^{99} \sum_{i=1}^{100} f \left(\dfrac{a_i}{j}\right) \\ &= \sum_{i=1}^{100} \sum_{j=0}^{99} f \left(\dfrac{a_i}{j}\right) \\ &= \sum_{i=1}^{100} \sum_{0\le j\lt a_i} f \left(\dfrac{a_i}{j}\right) \quad \text{Since for } j\ge a_i\; f \left(\dfrac{a_i}{j}\right)=0 \\ & = \sum_{i=1}^{100} \sum_{0\le j\lt a_i} 1 \\ &= \sum_{i=1}^{100} a_i \end{aligned}$

Thus proved !

Problem : For $0\le \theta \le \dfrac{\pi}{2}$ Prove that $\sin \theta \ge \dfrac{2\theta}{\pi}$

Solution :

Define $f(x)=\pi\sin x-2x$ . So $f(x)$ has a critical point at $x=\cos^{-1}\left(\dfrac{2}{\pi}\right)$ . For $x\in\left[0,\cos^{-1}\left(\dfrac{2}{\pi}\right)\right]$ ,$f'(x)\ge 0$ and thus $f$ is increasing. So,

$\displaystyle \begin{aligned} &\theta\ge 0 \\ &f(\theta)\ge f(0) \\ &\sin \theta \ge \dfrac{2\theta}{\pi} \end{aligned}$

Similarly For $x\in\left[\cos^{-1}\left(\dfrac{2}{\pi}\right),\dfrac{\pi}{2}\right]$ ,$f'(x)\le 0$ and thus $f$ is decreasing we have,

$\displaystyle \begin{aligned} &\theta\le \dfrac{\pi}{2} \\ &f(\theta)\ge f(\pi/2) \\ &\sin \theta \ge \dfrac{2\theta}{\pi} \end{aligned}$

Thus combining we get the result !

Problem : Let $f:\mathbb{R}\to \mathbb{R}$ be a function that is differentiable $n+1$ times for some positive integer $n$ . The $i$-th derivative is denoted by $f^{(i)}$. Suppose, $\displaystyle f(1)=f(0)=f^{(1)}(0)=f^{(2)}(0)=\cdots=f^{(n)}(0)=0$. Prove that $f^{(n+1)}(x)=0$ for some $x\in(0,1)$.

Solution:

By Rolle's Theorem since $f(1)=f(0)=0$ we have $f'(c_1)=0$ for some $c_1\in(0,1)$ . Now if we define $f_1(x)=f'(x)$ then we have by rolle's theorem that $f''(c_2)=0$ for some $c_2\in (0,c_1)$ since $f'(0)=f'(c_1)=0$ . Proceeding in this manner we will have $f^{(n+1)}(c_{n+1})=0$ for some $c_{n+1}\in (0,c_n)$ since $f^{(n)}(0)=f^{(n)}(c_n)=0$ . Since $0<c_{n+1}<c_n<c_{n-1}<\cdots<1$ we have have proved the result that $f^{(n+1)}(x)=0$ for some $x\in (0,1)$ where $x=c_{n+1}$.

Problem: Let $P:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $P(X)=X$ has no real solution. Prove that $P(P(X))=X$ has no real solution.

Solution:

It suffices to prove that $P(P(X))=X$ is solvable whenever $P(X)=X$ is solvable. Let $x_0$ be the unique solution to $P(X)=X$ so that $P(x_0)=x_0$. So it follows that $P(P(x_0))=x_0$ is true and $x_0$ is the unique fixed point of $P\circ P$ . Since it is given that $P(X)=X$ has no solution so does $P\circ P(x)$ . Thus proved !

Problem: Let $f$ be a real-valued differentiable function on the real line $\mathbb{R}$ such that $\displaystyle \lim_{x\to 0}\dfrac{f(x)}{x^2}$ exists and is finite. Prove that $f'(0)=0$

Solution:

$\displaystyle \begin{aligned} \lim_{x\to 0} \dfrac{f(x)}{x^2} &= \lim_{x\to 0} \dfrac{1}{x}\left(\dfrac{f(x)-f(0)}{x}+\dfrac{f(x)}{x}\right) \\ &= \lim_{x\to 0} \dfrac{1}{x}\left(f'(0)+\dfrac{f(0)}{x}\right) \\ & = \lim_{x\to 0} \left(\dfrac{f'(0)}{x}+\dfrac{f(0)}{x^2}\right) \end{aligned}$

Hence for the limit to exist and be finite it follows that $f(0)=f'(0)=0$ and thus proved.

Problem: Let $p(x)=x^7+x^6+b_5 x^5+\cdots+b_1 x+b_0$ and $q(x)=x^5+c_4 x^4+\cdots+c_x+c_0$ be polynomials with integer coefficients. Assume that $p(i)=q(i)$ for integers $i\in[1,6]$ . Show that there exists a negative integer $r$ such that $p(r)=q(r)$

Solution: Define a dummy polynomial $g(x)=p(x)-q(x)$ so that $g(x)$ has roots $1,2,3,4,5,6$ . Since it's a $7$-th degree polynomial it must have 7 roots. Moreover since $g$ is monic we may write $g(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-\alpha)$ for some integer $\alpha$ which is the $7$-th root.

Since $g(x)=p(x)-q(x)=x^7+x^6+\cdots+(b_0-c_0)$ it follows that the sum of the roots is $-1$ . Therefore $1+2+\cdots+6+\alpha=-1$ which makes $\alpha=-22$ . Thus $g(-22)=0$ which implies $p(-22)=q(-22)$ and we are proved.

Problem: Given the polynomial $f(x)=x^n+a_1 x^{n-1}+a_2 x^{n-1}+\cdots+a_n$ with real coefficients such that $a_1^2\lt a_2$ . Show that not all roots of $f(x)$ can be real.

Solution: Let the roots of the polynomial be $\alpha_i$ for $i\in[1,n]$. Since $\displaystyle a_1 = -\sum \alpha_i$ and $\displaystyle a_2=\sum_{i<j \le n} \alpha_i \alpha_j$ it follows that $\displaystyle a_1^2-2a_2=\sum \alpha_i^2 \gt 0$ which implies $a_1^2\gt 2a_2$ which is a contradiction with $a_1^2\lt a_2$ and thus our assumption that all of $\alpha_i$'s are real was false. Thus not all real roots are possible.

Problem:Let $f$ and $g$ be two non-decreasing twice differentiable functions defined on an interval $(a,b)$ such that for each $x\in (a,b)$, $f''(x)=g(x)$ and $g''(x)=f(x)$. Suppose that $f(x)g(x)$ is linear in $x$ on $(a,b)$ . Show that we must have $f(x)=g(x)=0$ for all $x\in(a,b)$

Solution: Let $f(x)g(x)=px+q$ and by differentiating twice we have $f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)=0$ . Since $f,g$ are non decreasing we have $f'(x)\not\lt 0$ and $g'(x)\not\lt 0$ for all $x\in(a,b)$ . By putting $f''(x)=g(x)$ and $g''(x)=f(x)$ we have $\left(f(x)\right)^2+\left(g(x)\right)^2=-2f'(x)g'(x)\le 0$ . But since $\left(f(x)\right)^2 \ge 0$ and $\left(g(x)\right)^2 \ge 0$ and thus $\left(f(x)\right)^2+\left(g(x)\right)^2\ge 0$ . So $0\le \left(f(x)\right)^2+\left(g(x)\right)^2\le 0$ implies $\left(f(x)\right)^2+\left(g(x)\right)^2= 0$ and if sum of two or more squares are zero they must be individually zero. Thus $f(x)= g(x)=0\;\forall\; x\in(a,b)$

Problem: Let $N$ be a positive integer such that $N(N-101)$ is the square of a positive integer. Then determine all possible values of $N$.

Solution: Let $N(N-101)=k^2$ for some $k\in\mathbb{N}$ . Since $N$ takes positive integral values the discriminant of the quadratic $N^2-101N-k^2=0$ must be a perfect square. Hence $101^2+4k^2=m^2$ for some positive integer $m$ . This factorizes to $(m+2k)(m-2k)=101^2$.

Since $101$ is a prime so RHS can be factored in exactly two ways as a product of two factors namely $101\times 101$ and $101^2 \times 1$ . Since $m+2k \gt m-2k$ so we must have $(m+2k)(m-2k)=101^2 \times 1$ which implies $m+2k=101^2$ and $m-2k=1$ from which we get $k=2550,m=5101$ and this gives $N=\boxed{2601}$ as the only positive integer solution.

Problem(Subjective 173): Let $P_1,P_2,\cdots,P_n$ be polynomials in $x$ , each having all integer coefficients, such that $P_1=P_1^2+P_2^2+\cdots+P_n^2$ . Assume that $P_1$ is not the zero polynomial. Show that $P_1=1$ and $P_2=P_3=\cdots=P_n=0$

Solution: Since $P_1$ is not the zero polynomial with integer coefficients it's obvious that $P_1^2-P_1\ge 0$. This gives, $P_2^2+P_3^2+\cdots+P_n^2 = P_1-P_1^2 \le 0$ . But since a perfect square is always non-negative we have $P_2^2+P_3^2+\cdots+P_n^2 \ge 0$ . This together implies that $P_2^2+P_3^2+\cdots+P_n^2 =0$ and thus each of $P_i$'s are zero that is $P_2=P_3=\cdots=P_n=0$ . Substituting this we have $P_1^2=P_1$ which makes $P_1=1$ since $P_1 \not = 0$. Thus proved.

Problem: Let $S=\left\{1,2,\cdots,n\right\}$ where $n$ is an odd integer. Let $f$ be a function defined on $\left\{(i,j):i\in S\; j\in S\right\}$ taking values in $S$ such that : $f(r,s)=f(s,r)$ & $\left\{f(r,s):s\in S\right\}=S$ for all $r,s\in S$. Show that $\left\{f(r,r):r\in S\right\}=S$

Solution: Since $S=\{1,2,\cdots,n\}$ and also for any $r\in S$ we have $S=\{f(r,1),f(r,2),\cdots,f(r,n)\}$ and thus it follows that $\{f(r,s):s\in S\}$ is just a permutation of $\{1,2,\cdots,n\}$ . Say for some $a_1\in S$ and any $r\in S$ we have $f(r,a_1)=1$ . Now it is safe to assume or rather substitute $r=a_1$ to have $f(a_1,a_1)=1$ . In a similar manner for some $a_2\in S$ we have for any $r\in S$, $f(r,a_2)=2$ and assuming $r=a_2$ we have $f(a_2,a_2)=2$ . This process repeated $n$ times will give $f(a_n,a_n)=n$ for some $a_n\in S$ where it must be noted that $(a_1,a_2,\cdots,a_n)=\sigma(1,2,\cdots,n)$. So this proves that $S=\{f(a_1,a_1),f(a_2,a_2),\cdots,f(a_n,a_n)\}=\{f(r,r):r\in S\}$.

Problem: Consider the equation $x^5+x=10$ . Prove that it has only one real root which lies between $1$ and $2$ and further the root must be irrational.

Solution: Let $f(x)=x^5+x-10$ and applying Descarte's rule of signs to $f(x)$ we have only $1$ sign change so it may have at most one positive real root. Applying same to $f(-x)$ produces no change so there are no negative real roots. Since an equation of odd degree always has a real root(SInce complex roots occur in pairs) so it has 1 real root.

Next it is easy to show by Intermediate Value Theorem that the root lies between $1$ and $2$ since $f(1)<0$ and $f(2)>0$. Now had the root been rational it has to be an integer by the Rational Root Theorem. Clearly $x=1$ isn't a root and $f(x)\gt 0$ for all integers $x\gt 1$. So there must be an irrational root.

Problem: Let $0<a<b$. Consider two circles with radii $a$ and $b$ and centers $(a,0)$ and $(b,0)$ respectively. Let $C$ be the center of any circle in between the two circles and tangent to both. Determine the locus of the center of any such circle.

Solution:

We have assumed $A(a,0)$ to be the center of $\Gamma_1$ , $B(b,0)$ to be the center of $\Gamma$ and $C$ to be the center of $\Gamma_2$ . Joining AC and BC we find that if $r$ be the radius of the variable circle at any instant then $|AC|=a+r$ and $|BC|=b-r$ respectively. Thus $AC+BC=a+b$ is constant for any such circle. The sum of distances of the center of the smaller circle from two points (the center of the other two circles respectively) is constant which is a property of the ellipse. Thus the locus of $C$ is ellipse with it's foci at $(a,0)$ and $(b,0)$ respectively.

To be continued....