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 100100 students. Let ai,11100a_i,1\le 1\le 100 , denote the number of friends the i-th student has in the class. For each 0j990\le j\le99 , let cjc_j denote the number of students having at least jj friends. Prove that :i=1100ai=j=099cj\displaystyle \sum_{i=1}^{100} a_i = \sum_{j=0}^{99} c_j

Solution :

ai=\displaystyle a_i = Number of friends of the ii-th student

cj=\displaystyle c_j = Number of students having at least jj friends

It is harmless to renumber each student so that their corresponding number of friends form an increasing sequence {ai}\{a_i\}. Since now cjc_j denotes the number of students having at least jj friends it follows that there is some kk for which akja_k\ge j , and since {ai}\{a_i\} is an increasing sequence for all n>kn\gt k ,ana_n will have at least jj friends. So cjc_j is exactly the number of elements of the set S={ak,ak+1,,a100}\displaystyle S = \{a_k,a_{k+1},\cdots,a_{100}\} where each of these students numbering from kk on wards have friends aka_k and number of these students is value of cjc_j .

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

f(x)=0x[0,1)=1x1\displaystyle \begin{aligned} f(x)&=0\quad x\in[0,1) \\ &=1\quad x\ge1\end{aligned}

Now note that beginning with 0j990\le j\le99 we find an nn such that ai>k  i>na_i\gt k\forall\; i>n and for all i<ni\lt n ,k>aik>a_i and we have k=ank=a_n .

So we need to count the number of entries for each jj from some ana_n where it has an=ja_n=j to a100a_{100} .Our function perfectly fits that requirement and we have ,

cj=i=1100f(aij)\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,

j=099cj=j=099i=1100f(aij)=i=1100j=099f(aij)=i=11000j<aif(aij)Since for jai  f(aij)=0=i=11000j<ai1=i=1100ai\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 0θπ20\le \theta \le \dfrac{\pi}{2} Prove that sinθ2θπ\sin \theta \ge \dfrac{2\theta}{\pi}

Solution :

Define f(x)=πsinx2xf(x)=\pi\sin x-2x . So f(x)f(x) has a critical point at x=cos1(2π)x=\cos^{-1}\left(\dfrac{2}{\pi}\right) . For x[0,cos1(2π)]x\in\left[0,\cos^{-1}\left(\dfrac{2}{\pi}\right)\right] ,f(x)0f'(x)\ge 0 and thus ff is increasing. So,

θ0f(θ)f(0)sinθ2θπ\displaystyle \begin{aligned} &\theta\ge 0 \\ &f(\theta)\ge f(0) \\ &\sin \theta \ge \dfrac{2\theta}{\pi} \end{aligned}

Similarly For x[cos1(2π),π2]x\in\left[\cos^{-1}\left(\dfrac{2}{\pi}\right),\dfrac{\pi}{2}\right] ,f(x)0f'(x)\le 0 and thus ff is decreasing we have,

θπ2f(θ)f(π/2)sinθ2θπ\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:RRf:\mathbb{R}\to \mathbb{R} be a function that is differentiable n+1n+1 times for some positive integer nn . The ii-th derivative is denoted by f(i)f^{(i)}. Suppose, f(1)=f(0)=f(1)(0)=f(2)(0)==f(n)(0)=0\displaystyle f(1)=f(0)=f^{(1)}(0)=f^{(2)}(0)=\cdots=f^{(n)}(0)=0. Prove that f(n+1)(x)=0f^{(n+1)}(x)=0 for some x(0,1)x\in(0,1).

Solution:

By Rolle's Theorem since f(1)=f(0)=0f(1)=f(0)=0 we have f(c1)=0f'(c_1)=0 for some c1(0,1)c_1\in(0,1) . Now if we define f1(x)=f(x)f_1(x)=f'(x) then we have by rolle's theorem that f(c2)=0f''(c_2)=0 for some c2(0,c1)c_2\in (0,c_1) since f(0)=f(c1)=0f'(0)=f'(c_1)=0 . Proceeding in this manner we will have f(n+1)(cn+1)=0f^{(n+1)}(c_{n+1})=0 for some cn+1(0,cn)c_{n+1}\in (0,c_n) since f(n)(0)=f(n)(cn)=0f^{(n)}(0)=f^{(n)}(c_n)=0 . Since 0<cn+1<cn<cn1<<10<c_{n+1}<c_n<c_{n-1}<\cdots<1 we have have proved the result that f(n+1)(x)=0f^{(n+1)}(x)=0 for some x(0,1)x\in (0,1) where x=cn+1x=c_{n+1}.

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

Solution:

It suffices to prove that P(P(X))=XP(P(X))=X is solvable whenever P(X)=XP(X)=X is solvable. Let x0x_0 be the unique solution to P(X)=XP(X)=X so that P(x0)=x0P(x_0)=x_0. So it follows that P(P(x0))=x0P(P(x_0))=x_0 is true and x0x_0 is the unique fixed point of PPP\circ P . Since it is given that P(X)=XP(X)=X has no solution so does PP(x)P\circ P(x) . Thus proved !

Problem: Let ff be a real-valued differentiable function on the real line R\mathbb{R} such that limx0f(x)x2\displaystyle \lim_{x\to 0}\dfrac{f(x)}{x^2} exists and is finite. Prove that f(0)=0f'(0)=0

Solution:

limx0f(x)x2=limx01x(f(x)f(0)x+f(x)x)=limx01x(f(0)+f(0)x)=limx0(f(0)x+f(0)x2)\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)=0f(0)=f'(0)=0 and thus proved.

Problem: Let p(x)=x7+x6+b5x5++b1x+b0p(x)=x^7+x^6+b_5 x^5+\cdots+b_1 x+b_0 and q(x)=x5+c4x4++cx+c0q(x)=x^5+c_4 x^4+\cdots+c_x+c_0 be polynomials with integer coefficients. Assume that p(i)=q(i)p(i)=q(i) for integers i[1,6]i\in[1,6] . Show that there exists a negative integer rr such that p(r)=q(r)p(r)=q(r)

Solution: Define a dummy polynomial g(x)=p(x)q(x)g(x)=p(x)-q(x) so that g(x)g(x) has roots 1,2,3,4,5,61,2,3,4,5,6 . Since it's a 77-th degree polynomial it must have 7 roots. Moreover since gg is monic we may write g(x)=(x1)(x2)(x3)(x4)(x5)(x6)(xα)g(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-\alpha) for some integer α\alpha which is the 77-th root.

Since g(x)=p(x)q(x)=x7+x6++(b0c0)g(x)=p(x)-q(x)=x^7+x^6+\cdots+(b_0-c_0) it follows that the sum of the roots is 1-1 . Therefore 1+2++6+α=11+2+\cdots+6+\alpha=-1 which makes α=22\alpha=-22 . Thus g(22)=0g(-22)=0 which implies p(22)=q(22)p(-22)=q(-22) and we are proved.

Problem: Given the polynomial f(x)=xn+a1xn1+a2xn1++anf(x)=x^n+a_1 x^{n-1}+a_2 x^{n-1}+\cdots+a_n with real coefficients such that a12<a2a_1^2\lt a_2 . Show that not all roots of f(x)f(x) can be real.

Solution: Let the roots of the polynomial be αi\alpha_i for i[1,n]i\in[1,n]. Since a1=αi\displaystyle a_1 = -\sum \alpha_i and a2=i<jnαiαj\displaystyle a_2=\sum_{i<j \le n} \alpha_i \alpha_j it follows that a122a2=αi2>0\displaystyle a_1^2-2a_2=\sum \alpha_i^2 \gt 0 which implies a12>2a2a_1^2\gt 2a_2 which is a contradiction with a12<a2a_1^2\lt a_2 and thus our assumption that all of αi\alpha_i's are real was false. Thus not all real roots are possible.

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

Solution: Let f(x)g(x)=px+qf(x)g(x)=px+q and by differentiating twice we have f(x)g(x)+2f(x)g(x)+f(x)g(x)=0f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)=0 . Since f,gf,g are non decreasing we have f(x)0f'(x)\not\lt 0 and g(x)0g'(x)\not\lt 0 for all x(a,b)x\in(a,b) . By putting f(x)=g(x)f''(x)=g(x) and g(x)=f(x)g''(x)=f(x) we have (f(x))2+(g(x))2=2f(x)g(x)0\left(f(x)\right)^2+\left(g(x)\right)^2=-2f'(x)g'(x)\le 0 . But since (f(x))20\left(f(x)\right)^2 \ge 0 and (g(x))20\left(g(x)\right)^2 \ge 0 and thus (f(x))2+(g(x))20\left(f(x)\right)^2+\left(g(x)\right)^2\ge 0 . So 0(f(x))2+(g(x))200\le \left(f(x)\right)^2+\left(g(x)\right)^2\le 0 implies (f(x))2+(g(x))2=0\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    x(a,b)f(x)= g(x)=0\;\forall\; x\in(a,b)

Problem: Let NN be a positive integer such that N(N101)N(N-101) is the square of a positive integer. Then determine all possible values of NN.

Solution: Let N(N101)=k2N(N-101)=k^2 for some kNk\in\mathbb{N} . Since NN takes positive integral values the discriminant of the quadratic N2101Nk2=0N^2-101N-k^2=0 must be a perfect square. Hence 1012+4k2=m2101^2+4k^2=m^2 for some positive integer mm . This factorizes to (m+2k)(m2k)=1012(m+2k)(m-2k)=101^2.

Since 101101 is a prime so RHS can be factored in exactly two ways as a product of two factors namely 101×101101\times 101 and 1012×1101^2 \times 1 . Since m+2k>m2km+2k \gt m-2k so we must have (m+2k)(m2k)=1012×1(m+2k)(m-2k)=101^2 \times 1 which implies m+2k=1012m+2k=101^2 and m2k=1m-2k=1 from which we get k=2550,m=5101k=2550,m=5101 and this gives N=2601N=\boxed{2601} as the only positive integer solution.

Problem(Subjective 173): Let P1,P2,,PnP_1,P_2,\cdots,P_n be polynomials in xx , each having all integer coefficients, such that P1=P12+P22++Pn2P_1=P_1^2+P_2^2+\cdots+P_n^2 . Assume that P1P_1 is not the zero polynomial. Show that P1=1P_1=1 and P2=P3==Pn=0P_2=P_3=\cdots=P_n=0

Solution: Since P1P_1 is not the zero polynomial with integer coefficients it's obvious that P12P10P_1^2-P_1\ge 0. This gives, P22+P32++Pn2=P1P120P_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 P22+P32++Pn20P_2^2+P_3^2+\cdots+P_n^2 \ge 0 . This together implies that P22+P32++Pn2=0P_2^2+P_3^2+\cdots+P_n^2 =0 and thus each of PiP_i's are zero that is P2=P3==Pn=0P_2=P_3=\cdots=P_n=0 . Substituting this we have P12=P1P_1^2=P_1 which makes P1=1P_1=1 since P10P_1 \not = 0. Thus proved.

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

Solution: Since S={1,2,,n}S=\{1,2,\cdots,n\} and also for any rSr\in S we have S={f(r,1),f(r,2),,f(r,n)}S=\{f(r,1),f(r,2),\cdots,f(r,n)\} and thus it follows that {f(r,s):sS}\{f(r,s):s\in S\} is just a permutation of {1,2,,n}\{1,2,\cdots,n\} . Say for some a1Sa_1\in S and any rSr\in S we have f(r,a1)=1f(r,a_1)=1 . Now it is safe to assume or rather substitute r=a1r=a_1 to have f(a1,a1)=1f(a_1,a_1)=1 . In a similar manner for some a2Sa_2\in S we have for any rSr\in S, f(r,a2)=2f(r,a_2)=2 and assuming r=a2r=a_2 we have f(a2,a2)=2f(a_2,a_2)=2 . This process repeated nn times will give f(an,an)=nf(a_n,a_n)=n for some anSa_n\in S where it must be noted that (a1,a2,,an)=σ(1,2,,n)(a_1,a_2,\cdots,a_n)=\sigma(1,2,\cdots,n). So this proves that S={f(a1,a1),f(a2,a2),,f(an,an)}={f(r,r):rS}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 x5+x=10x^5+x=10 . Prove that it has only one real root which lies between 11 and 22 and further the root must be irrational.

Solution: Let f(x)=x5+x10f(x)=x^5+x-10 and applying Descarte's rule of signs to f(x)f(x) we have only 11 sign change so it may have at most one positive real root. Applying same to f(x)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 11 and 22 since f(1)<0f(1)<0 and f(2)>0f(2)>0. Now had the root been rational it has to be an integer by the Rational Root Theorem. Clearly x=1x=1 isn't a root and f(x)>0f(x)\gt 0 for all integers x>1x\gt 1. So there must be an irrational root.

Problem: Let 0<a<b0<a<b. Consider two circles with radii aa and bb and centers (a,0)(a,0) and (b,0)(b,0) respectively. Let CC 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)A(a,0) to be the center of Γ1\Gamma_1 , B(b,0)B(b,0) to be the center of Γ\Gamma and CC to be the center of Γ2\Gamma_2 . Joining AC and BC we find that if rr be the radius of the variable circle at any instant then AC=a+r|AC|=a+r and BC=br|BC|=b-r respectively. Thus AC+BC=a+bAC+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 CC is ellipse with it's foci at (a,0)(a,0) and (b,0)(b,0) respectively.

To be continued....

Note by Aditya Narayan Sharma
2 years, 7 months ago

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Lovely problems!

Md Zuhair - 2 years, 7 months ago

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Glad you liked that !

Aditya Narayan Sharma - 2 years, 7 months ago

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Problem:

Suppose f:RRf : \mathbb{R} \mapsto \mathbb{R} is a function defined by

f(x):={1if x=1e(x101)+(x1)2sin(1x1)otherwise f(x) := \begin{cases} 1 & \text{if } x = 1 \\ e^{\left( x^{10} - 1 \right)} + {\left( x-1 \right)}^2 \sin \left( \dfrac{1}{x-1} \right) & \text{otherwise} \end{cases}

(a) Find f(1)f'(1).

(b) Evaluate limu[100uuk=1100f(1+ku)]\displaystyle \lim_{u \to \infty} \left[ 100u - u \sum_{k=1}^{100} f \left( 1 + \dfrac ku \right) \right].

Please solve this one.

Tapas Mazumdar - 2 years, 7 months ago

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Here is another interesting question from ISI's Test of Mathematics at the 10+2 level.Qno 173 subjective question. Let $P1,P2,...,Pn$ be polynomials in $x$ each having all integer coefficients such that $P1=P1^2+P2^2+...+Pn^2$.Assume that $P1$ is not the zero polynomial. Show that $P1=0$ and $P2=P3=...=Pn=0$I could not solve this one.

nikhil murarka - 2 years, 7 months ago

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@nikhil murarka I have added your problem with a solution, hope it helps

Aditya Narayan Sharma - 2 years, 7 months ago

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Hey can you tell me the max and min values for cos (sin x)+ sin (cosx) with full solution...

Md Zuhair - 2 years, 7 months ago

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This in fact involves numerical calculations , if you check the derivatives of f(x)f(x) in [0,π][0,\pi] which is ok since it's periodic we would get 0,π0,\pi as obvious stationary points. Also f(0)=2sin2(12)f''(0)=-2\sin^2 \left(\dfrac{1}{2}\right) and f(π)=2cos2(12)f''(\pi)=-2\cos^2 \left(\dfrac{1}{2}\right) so this would produce local maximums. It would thus involve finding another stationary point using numerical approximation such that it has an absolute minimum and maximum, which would involve solving transcendental equations. I think this are unlikely to appear at the exams

Aditya Narayan Sharma - 2 years, 7 months ago

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Add more!

Md Zuhair - 2 years, 6 months ago

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Great problems.Really enjoyed it.

Mayank Jha - 2 years, 6 months ago

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In problem number 5 from top how you did this.Can you please explain?especially 2nd step

Vedansh Priyadarshi - 2 years, 6 months ago

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In the year 2012, this was one of the questions, Let T1 and T2 be two circles centred at the points (a,0),(b,0);0<a<b and having radii a,b respectively.Let T be the circle touching T1 externally and T2 internally. Find the locus of the centre of of T. Can you help with this?

Kaustav Ghosh - 2 years, 7 months ago

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I have added that, hope it was helpful.

Aditya Narayan Sharma - 2 years, 7 months ago

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