In 2026, the following Cheenta students have been successful for Indian Statistical Institute's M.Stat Entrance. They ranked within the first 50 in the entire country in these entrances.
I.S.I. M.Stat Entrance
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's B.Stat Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 200 in the entire country in these entrances. Most of these students attended the problem solving workshops regularly, which happen 5 days every week.
CMI B.Sc. Math Entrance
I.S.I. B.Stat Entrance
I.S.I. B.Math Entrance
| 1. (A) | 2. (C) | 3. (B) | 4. (B) | 5. (A) | 6. (D) |
| 7. (D) | 8. (D) | 9. (B) | 10. (B) | 11. (C) | 12. (D) |
| 13. (C) | 14. (D) | 15. (B) | 16. (D) | 17. (D) | 18. (C) |
| 19. (C) | 20. (A) | 21. (A) | 22. (C) | 23. (B) | 24. (A) |
| 25. (C) | 26. (B) | 27. (A) | 28. (B) | 29. (D) | 30. (B) |
In the \(x y\)-plane, the curve \(3 x^3 y+6 x y+2 x y^3=0\) represents
(A) a pair of straight lines
(B) an ellipse
(C) a pair of straight lines and an ellipse
(D) a hyperbola
Solution
\(3 x^3 y+6 x y+2 x y^3=0\)
\(x y\left(3 x^2+2 y^2+6\right)=0\)
\(\Rightarrow x y=0\)
OR
\(3 x^2+2 y^2+6=0\)
no solution as \(3 x^2+2 y^2 \geqslant 0\)
\(\therefore \quad x y=0\)
\(\Rightarrow x=0\) or \(y=0\)
Answer: Pair of Straight Lines.
Let \(I=\int_3^5 \frac{1}{1+x^3} d x\). Then
(A) \(I<\frac{1}{64}\)
(B) \(I>\frac{1}{13}\)
(C) \(\frac{1}{63}<I<\frac{1}{14}\)
(D) \(I>\frac{1}{2}\left(\frac{1}{14}+\frac{1}{63}\right)\)
Solution
We know that
\(m(a-b) \leqslant \int_b^a p(x) d x \leqslant M(a-b)\)
Where \(m=\min . f(x)\) and
\[
B \leqslant x \leqslant a
\]
\(M=m o x \quad f(x)\)
\(b \leqslant x<a\)
So, \(2 x \frac{1}{126}<\int_3^5 f(x) d x \quad \leqslant 2 x \frac{1}{28}\)
\(\Rightarrow \frac{1}{ 63}<\int_3^5 f(x) d x<\frac{1}{14}\)
The coefficient of \(x^8\) in \((1-3 x)^6\left(1+9 x^2\right)^6(1+3 x)^6\) is
(A) \(-3^9 \times 5\)
(B) \(3^9 \times 5\)
(C) \(-3^8 \times 5\)
(D) \(3^8 \times 5\)
Solution:
\((1-3 x)^6\left(1+9 x^2\right)^6(1+3 x)^6\)
=\(\left(1-9 x^2\right)^6\left(1+9 x^2\right)^6\)
=\(\left(1-81 x^4\right)^6\)
\(\therefore\binom{6}{0} 1^0 \cdot\left(-81 x^4\right)^6+\cdots+\binom{6}{4} 1^4\left(-81 x^4\right)^2+\cdots\)
=\(5 \times 3^9 \times x^8\)
Consider two events \(A\) and \(B\) with probabilities \(P(A)\) and \(P(B)\) respectively such that \(0<P(A), P(B)<1\). Define
\[
P(A \mid B)=\frac{P(A \cap B)}{P(B)}
\]
Consider the following statements.
(I) \(P\left(A \mid B^c\right)+P(A \mid B)=1\).
(II) \(P\left(A^c \mid B\right)+P(A \mid B)=1\).
Then, in general,
(A) (I) is true and (II) is false
(B) (I) is false and (II) is true
(C) both (I) and (II) are true
(D) both (I) and (II) are false
Solution
Let \(f(x)=7 x^{11}+4 x^3-3\). Then \(f\) has
(A) exactly 1 real root
(B) exactly 3 real roots
(C) exactly 5 real roots
(D) 11 real roots
Solution
\(f(x)=7 x^{11}+4 x^5-3\)
\(f^{\prime}(x)=77 x^{10}+20 x^4\)
\(\therefore f^{\prime}(x)>0 \quad) if (\quad x \neq 0\)
\[
\lim_{x \to \infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to -\infty} f(x) = -\infty
\]
By intermediate value theorem there is at least 1 root
\(\therefore f\) is (almost) monotone hence there is exactly 1 root.
Let \(A\) be an \(m \times n\) matrix with the \((i, j)\) th entry given by the real number \(a_{i j}, 1 \leq i \leq m, 1 \leq j \leq n\). Let
\[
a = \max_{1 \leq j \leq n} \left( \min_{1 \leq i \leq m} a_{ij} \right)
\quad \text{and} \quad
\beta = \min_{1 \leq j \leq n} \left( \max_{1 \leq i \leq m} a_{ij} \right).
\]
Then
(A) \(\alpha \leq \beta\) but not necessarily \(\alpha=\beta\)
(B) \(\beta \leq \alpha\) but not necessarily \(\alpha=\beta\)
(C) \(\alpha=\beta\)
(D) nothing can be said in general
Solution
Consider the cyclic quadrilateral \(A B C D\) given below.
Assume that \(A B=B C, A D=C D\), and \(\frac{A B}{A D}=\frac{1}{3}\). Let \(\theta=\angle A D C\). Then \(\cos \theta\) is equal to
(A) \(\frac{1}{5}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{3}{5}\)
(D) \(\frac{4}{5}\)
Solution
\(x^2+x^2-2 x \cdot x \cdot \cos (180-\theta)\)
=\((3 x)^2+(3 x)^2-2 \cdot 3 x \cdot 3 x \cos \theta\)
\(\Rightarrow 2 x^2+2 x^2 \cos \theta=18 x^2-18 x^2 \cos \theta\)
\(\Rightarrow 20 x^2 \cos \theta=16 x^2\)
\(\cos \theta=\frac{16 x^2}{20 x^2}\)
=\(\frac{4}{5}\)
Let \(A={(x, y): x, y \in[0,1]}\) and \(B={(x, y): x, y \in[0,2]}\). Define \(f: A \rightarrow B\) by \(f(x, y)=\left(x^2+y, x+y^2\right)\). Then \(f\) is
(A) one-to-one but not onto
(B) onto but not one-to-one
(C) both one-to-one and onto
(D) neither one-to-one nor onto
Solution
\(A={(x, y): x, y \in[0,1]}\)
\(B={(x, y): x, y \in[0,2]}\)
\(f: A \rightarrow B\)
\(f(x, y)=\left(x^2+y ; x+y^2\right)\)
Not one -on - one
as \(\begin{array}{r}(1,0) \ (0,1)\end{array}\) --> both map to \((1,1)\)
Onto, Examina the presimage of \([0,2]\)
\(x^2+y=0\)
\(x+y^2=2\).
\(\because \quad 0 \leqslant x, y \leqslant 1\)
\(\therefore x^2+y=0 \Rightarrow \quad x=0 \quad y=0\)
But then \(x+y^2=2\) is not true hence no solution.
The number of ordered pair \((a, b)\) of positive integers with \(a<b\) satisfying \(a^2+b^2=2025\) is
(A) 0
(B) 1
(C) 2
(D) 6
Solution
\(a^2+b^2=2.25\)
\(a<b\)
\(\therefore \quad 2025=5^2 \cdot 3^4=45^2\)
\(0,1,4,7 \quad \bmod 9\)
None of these work except \(0+0\)
\(\therefore \quad a^2 \equiv 0 \mathrm{mod} 9\)
\(b^2 \equiv 0\) mod 9
\(\therefore a \leq 0 \bmod a\)
or \(a \equiv 6 \bmod 9\).
\(\therefore\) Both a and 6 are divisible by 3
\(\left(3 a_1\right)^2+\left(3 b_1\right)^2=25 \times 81\)
\(\Rightarrow a_1^2+b_1^2=9 \times 25\)
\(\Rightarrow \quad a_2{ }^2+b_2{ }^2=25\)
\(\therefore a_2=3, b_2=9\)
\(\therefore a=27 \quad b=36\)
Twelve boxes are placed along a circle. In each box, \(1,2,3\) or 4 balls are put such that the total number of balls in any 4 consecutive boxes is same. The number of ways this can be done is
(A) 4 !
(B) \(4^4\)
(C) \((4!)^3\)
(D) \((4!)^4\)
Solution
once we choose number of balls in first 4 boxes, the remaining choices become fixed.
For each of the four boxes, we have four choices,
\(\therefore 4^4\) cases in total
Problem 11
Let \(a_0=0, a_1=1\) and \(a_n=5 a_{n-1}+a_{n-2}\) for \(n \geq 2\). Then the value of the determinant
is
(A) -1
(B) \(-5^{101}\)
(C) 1
(D) \(5^{101}\)
Solution
The lengths of the three sides of a right angled triangle are geometric progression. The smallest angle of the triangle is
(A) \(\tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\)
(B) \(\cos ^{-1}\left(\frac{2}{\sqrt{5}-1}\right)\)
(C) \(\sin ^{-1}\left(\frac{2}{\sqrt{5}-1}\right)\)
(D) \(\sin ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\)
Solution
Consider the following statements about two similar triangles \(\Delta_1), and (\Delta_2\).
\(S_1\): Lengths of the sides of \(\Delta_1\) are in arithmetic progression
\(S_2\). Lengths of the sides of \(\Delta_1\) are in geometric progression.
\(S_3\). Lengths of the sides of \(\Delta_2\) are in arithmetic progression.
\(S_4\). Lengths of the sides of \(\Delta_2\) are in geometric progression.
Then
(A) \(S_1\) implies \(S_3\), but \(S_2\) does not imply \(S_4\)
(B) \(S_1\) does not imply \(S_3\), but \(S_2\) implies \(S_4\)
(C) \(S_1\) implies \(S_3\), and \(S_2\) implies \(S_4\)
(D) \(S_1\) does not imply \(S_3\), and \(S_2\) does not imply \(S_4\)
Solution
Let \((\sigma, b, c)\) and \(\left(a^{\prime}, b^{\prime}, c^{\prime}\right)\) are
Sides of two similar triangles sit.
\(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}\)
and \(b=a r\) and \(a-a x^2\)
Ther, \(\frac{b^{\prime}}{a^{\prime}}=\frac{c\prime}{b\prime}=r\) also
So, \(S_2\) implips \(S_4\)
Now, let
\(b=a+d\) \(c=a+d\)
so, \(a^{\prime}: a k, b^{\prime}:(a+d) k\).
So, \(a^{\prime}, b^{\prime}, c^{\prime}\), are also in AP
so, \(S_2\) implies \(S_4\)
For each \(n \geq 1\), let \(a_n\) and \(b_n\) be real numbers such that \(a_n \neq 0\) \(\frac{a n d}{c} b_n \neq 0\). Let
\(\left(a_n+i b_n\right)^n=n\left(a_n+i b_n\right) \quad\) for all \(n \geq 6\).
Then
(A) no such (a_n, b_n) exist
(B) \(x_n=n^{\frac{1}{n+1}} \cos \frac{2 \pi}{n+1}, b_n=n^{\frac{1}{n+1}} \sin \frac{2 \pi}{n+1}\)
(C) \(a_n=n^{\frac{1}{n}} \cos \frac{2 \pi}{n}, b_n=n^{\frac{1}{n}} \sin \frac{2 \pi}{n}\)
(D) \(a_n=n^{\frac{1}{n-1}} \cos \frac{2 \pi}{n-1}, b_n=n^{\frac{1}{n-1}} \sin \frac{2 \pi}{n-1}\)
Solution
\(\left(a_n+i \cdot b_n\right)^n=n\left(a_n+i \cdot b_n\right)\).
\(\Rightarrow\left(a_n+i \cdot b_n^m\right)^{n-1}=n\).
As, \(\quad a_n+i b_n \neq 0\)
So, \(a_n+i b_n=\)
\(\frac{1}{n-1}\left[\cos \left(\frac{2 \pi k}{n-1}\right)\right.\) \(fi (\sin \left(\frac{2 \pi k}{n-1}]\right)\)
where \(0 \leqslant K<(n-1)\)
So, option (D)
The lengths of the two adjacent sides of a parallelogram are 2 cm and 3 cm . The length of one diagonal is \(\sqrt{19} \mathrm{~cm}\). Then the length of the other diagonal is
(A) \(\sqrt{5} \mathrm{~cm}\)
(B) \(\sqrt{7} \mathrm{~cm}\)
(C) \(\sqrt{15} \mathrm{~cm}\)
(D) \(\sqrt{21} \mathrm{~cm}\)
Solution
\(d_1^2=a_b^2+b_b^2-2 a b \cos \theta\)
\(d_2^2=a^2+b^2-2 a b \cos \left(180^{\circ}-\theta\right)\)
\(\therefore d_1^2+d_2^2=2\left(a^2+b^2\right)\)
\(\Rightarrow 19+d_2^2=2(4+9)\)
\(\Rightarrow \quad d_2^2=26-19 \Rightarrow d_2^2=7\)
\(=d_2=\sqrt{7}\)
Let \(f\) \((x) = \frac{1}{1+|x-1|}+\frac{1}{1+|x+1|}\). Then the function \(f\) has
(A) neither a local maximum nor a local minimum
(B) a local minimum at \(x=0\), but no local maximum
(C) local maxima at \(x= \pm 1\), but no local minimum
(D) a local minimum at \(x=0\) and local maxima at \(x= \pm 1\)
Solution
\(f(x)=\frac{1}{1+|x-1|}+\frac{1}{1+|x+1|}\)
when, \(x<-1\),
\(f(x)=\frac{1}{2-x}-\frac{1}{x} \Rightarrow f^{\prime}(x)=\frac{1}{(2-x)^2}+\frac{1}{x^2}>0\)
Hence, (f(x)) is increasing
\(f(x)=\frac{1}{x+2}+\frac{1}{2-x} \Rightarrow f^{\prime}(w)=\frac{1}{(2-w)^2}-\frac{1}{(x+2)^2}\)
Critical point, at \(x=0\)
\(f^{\prime}(n)>0) for (0 \leqslant n<1\).
\(f^{\prime}(x)<0) for (-1<x<0\)
when, \(n>1\).
\(f(n)=\frac{1}{n}+\frac{1}{n+2}\)
\(=-\frac{1}{n^2}-\frac{1}{(n+2)^2}<0\)
\(f(x)\) is decreasing.
at \(x= \pm 1 \quad f(x)\) is changing it's sign from positive to
& at \(x=0\) it is changing from negative to positive.
Hence, option (D)
\(f(x) =
\begin{cases}
-1, & \text{if } x < 0,
\ 0, & \text{if } x = 0,
\ 1, & \text{if } x > 0.
\end{cases}\)
Then the function \(F\) defined by \(F(x)=\int_{-5}^x f(t) d t\) is
(A) not continuous
(B) continuous, but nowhere differentiable
(C) differentiable everywhere
(D) differentiable everywhere except at 0
Solution
Let
\(L=\lim _{n \rightarrow \infty}(n+100)^{\frac{5}{5 g_e(n-50)}}\).
Then
(A) \(2 \leq L \leq 16\)
(B) \(16 \leq L \leq 32\)
(C) \(32 \leq L<243\)
(D) \(L>243\)
Solution
\(L=\lim _{n \rightarrow \infty}(n+100) \frac{5}{\ln (n-\infty)}\)
\(\Rightarrow \ln L=\frac{5 \ln (n+100)}{\left.\lim _{n \rightarrow \infty} \frac{\ln (n-50)}{\ln (n-5}\right)}=5\)
\(L=e^5\)
\(2<e<3\)
\(32<L<243\)
Let \(a, b, c, d\) be positive integers such that the product abcd (=999). Then the number of different ordered 4 -tuples \((a, b, c, d)\) is
(A) 20
(B) 48
(C) 80
(D) 84
Solution
\(999=3^3 \times 27\)
Hence the number of non-negetive integer solution
\(\binom{3+4-1}{4-1}\binom{1+4-1}{4-1}\)
\(=20 \times 4\)
\(=80\)
Let \(|x|\) denote the greatest integer less than or equal to \(x \in \mathbb{R}\) and \(|x|\) has its usual meaning, that is, \(|x|=x\) if \(x \geq 0\), and \(|x|=-x\), if \(x<0\). Then the value of the integral
\(\int_{-2}^1([x]+2)^{|x|} d x\)
is
(A) \(1+\frac{1}{\log _e 2}\)
(B) \(1+\log _e \frac{1}{2}\)
(C) \(2-\log _e 2\)
(D) none of the above
Solution
\(\int_{-2}^1([x]+2)^{|x|} d x\)
\(=\int_0^1 2^x d x+\int_{-1}^0 1^{-x} d x\)
\(f \int_{-2}^{-1} 0 \cdot d x\)
\(=\left[\frac{2 x}{\ln 2}\right]_0^1+1.1\)
\(\left(\frac{1}{\ln^2}+1\right)\)
For a real number \(x\), let \(f(x)=\int_{-20}^{20} g(t) g(x-t) d t\), where
\(g(x)= \begin{cases}1, & \text { if } x \in[0,1] \ 0, & \text { otherwise }\end{cases}\)
Then \(f(x)\) is equal to
(A) \(\begin{cases}x, & \text { if } x \in[0,1], \ 2-x, & \text { if } x \in[1,2], \ 0, & \text { otherwise }\end{cases}\)
(B) \(\begin{cases}1+x, & \text { if } x \in[0,1), \ 1-x, & \text { if } x \in[1,2), \ 0, & \text { otherwise }\end{cases}\)
(C) \(\begin{cases}1, & \text { if } x \in(-20,20), \ 0, & \text { otherwise }\end{cases}\)
(D) none of the above
Solution
Now, if \(x<0\) or \(x>2\)then the integration becomes 0 .
if \(\quad 0 \leq x \leq 1\)
Then \(f(x)\)
\(=\int_{x-1}^x g(t) d t\)
\(=\int_0^x d t=x\)
\(=\int_{-20}^{20} \theta(t) g(x-t) d t\)
\(=\int_0^1 g(x-t) d t\)
\(=\int_0^1 g(t+x-1) d t\)
\(=\int_{x-1}^x g(t) d t\)
it \(10 \leqslant x \leqslant 2\)
\(f(x): \int_{x-1}^x g(t) d t\)
\(=\int_{x-1}^1 d t=(2-x)\)
Let \(n \geq 3\). There are \(n\) straight lines in a plane, no two of which are parallel and no three pass through a common point. Their points of intersection are joined. Then the number of fresh line segments thus created is
(A) \(\frac{n(n-1)(n-2)}{8}\)
(B) \(\frac{n(n-1)(n-2)(n-3)}{6}\)
(C) \(\frac{n(n-1)(n-2)(n-3)}{8}\)
(D) none of the above
Solution
The lines intersect at \(\binom{n}{2}\) different points But there are also \((n-1)\) points in part of line
So, Total fresh line
\(=\binom{\binom{n}{2}}{2}-n\binom{n-1}{2}\)
\(=\left(\frac{n(n-1)}{2}\right)-n \cdot \frac{(n-1)(n-2)}{2}\)
\(=\frac{n(n-1)(n-2)(n+1)}{8}-\frac{n(n-1)(n-2)}{2}\)
\(=\frac{n(n-1)}{8} \cdot\left(n^2-n-2-4 n+\theta\right)\)
\(=\frac{n(n-1)(n-2)(n-3)}{8}\)
In a certain test there are \(n\) questions. At least \(i\) questions were wrongly answered by \(2^{n-i}\) students, where \(i=1,2, \ldots, n\). If the total number of wrong answers given by all students is 2047 , then (n) is equal to
(A) 10
(B) 11
(C) 12
(D) 13
Solution
At least \(i\) questions were wrongly answered by \(2^{n-i}\) students.
\(\therefore\) At least wrong answers
\(=\) Exactly \(n-3\) wrong answers +
Exactly \(n-2\) wrong answers+
Exactly \(n-1\) wrong answers+
Exactly \(n\) wrong answers
\(\therefore\) Exactly: questions wrong \(=2^{n-1}-2^{n-2}\) Exactly 2 questions wrong \(=2^{n-2}-2^{n-3}\)
\(\therefore\) Total number of wrongs
\[
\begin{aligned}
& 1\left(2^{n-1}-2^{n-2}\right)+2\left(2^{n-2}-2^{n-3}\right)+\cdots+n\left(2^1-2^0\right) \
& =2^{n-1}+2^{n-2}+2^{n-3}+\cdots+2^1+2^0
\end{aligned}
\]
\(\therefore 2^0+2^1+2^2+\cdots+2^{n-3}+2^{n-2}+2^{n-1}=2047\)
\(\Rightarrow \frac{2^0\left(2^n-1\right)}{2-1}=2047\)
\(\Rightarrow \quad 2^n-1=2047 \Rightarrow n=11\)
Let \(n\) be a positive integer. The value of \(\sum_{k=0}^n \tan ^{-1} \frac{1}{k^2+k+1}\) is
(A) \(\tan ^{-1}(n+1)\)
(B) \(\tan ^{-1}\left(\frac{1}{n+1}\right)\)
(C) \(\tan ^{-1} n\)
(D) \(\tan ^{-1}\left(\frac{1}{n}\right)\)
Solution
\(=\sum_{k=0}^n \tan ^{-1} \frac{1}{k^2+k+1}\)
\(=\sum_{k=0}^n \tan ^{-1} \frac{1}{k(k+1)+1}\)
\(=\sum_{k=0}^n \tan ^{-1} \frac{(k+1)-k}{k(k+1)+1}\)
\(=\sum_{k=0}^n\left[\tan ^{-1}(k+1)-\tan ^{-1}(k)\right]\)
\(=\tan ^{-1}(n+1)\)
Let \(d\) be the side length of the largest possible equilateral triangle that can be put inside a square of side length 1 . Then
(A) \(d<1\)
(B) \(d=1\)
(C) \(1<d<\frac{2}{3^{1 / 4}}\)
(D) \(d \geq \frac{2}{3^{1 / 4}}\)
Let \(f(x)=\left(x^2+18\right)(x-4) x(x+4)-2\). Then
(A) \(f\) has exactly one real root
(B) \(\int\) has exactly 3 distinct real roots
(C) \(f\) has 5 distinct real roots
(D) \(f\) has a repeated root
Solution
\(f(x)=\left(x^2+18\right) x(x-4)(x+4)-2\)
\(f^{\prime}(x)=5 x^4+6 x^2-288\)
\(f^{\prime}\) has two real roots
But \(f\) can have at most three real roots. But \(f(-4)<0 \quad f(-3)>0 \quad f(0)<0 \quad f(4)<0 \quad f(5)>0\)
\( \therefore \) it has 3 real roots (B).
Yet \(k\) be a positive integer and \(f(x)=e^x-1\). Then
\[\lim _{x \rightarrow 0} \frac{f(x)+f\left(\frac{x}{2}\right)+f\left(\frac{x}{2^2}\right)+\cdots+f\left(\frac{x}{2^k}\right)}{x}\]
(x) \(2-\frac{1}{2^k}\)
(B) \(2-\frac{1}{2^{k+1}}\)
(C) \(k\)
(D) \(2^{k+1}-1\)
Solution
\(f(x)=e^x-1\)
\(\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\frac{e^{x / 2}-1}{x}+\frac{e^{x / 2^2}-1}{x}+\cdots+\frac{e^{x / 2^x}-1}{x}\)
\(=\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\frac{e^{x / 2}-1}{x / 2} \times \frac{1}{2}+\frac{e^{x / 2^2}-1}{x / 2^2} \times \frac{1}{2^2}+\cdots+\frac{e^{x / 2^x}-1}{x / 2^x} \times \frac{1}{2^x}\)
=\(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^x}=-\frac{1\left(\frac{1}{2^{x+1}}-1\right)}{1 / 2}\)
\(=-2\left(\frac{1}{2^{x+1}}-1\right)\)
\(=-\left(\frac{1}{2^k}-2\right)=2-\frac{1}{2^k}\)
Let
\[a_n=\frac{n^2}{\sqrt{n^6+1}}+\frac{n^2}{\sqrt{n^6+2}}+\cdots+\frac{n^2}{\sqrt{n^6+n}}, \quad n \geq 1\]
Then \(\lim _{n \rightarrow \infty} a_n\)
(A) does not exist
(B) is equal to 1
(C) is equal to \(e\)
(D) is equal to \(\frac{1}{e}\)
Solution
\(a_n=\frac{n^2}{\sqrt{n^6+1}}+\frac{n^2}{\sqrt{n^6+2}}+\cdots+\frac{n^2}{\sqrt{n^6+n}}\)
\(\sqrt{n^6+0} \leq \sqrt{n^6+r} \leq \sqrt{n^6+n}\)\
\(\frac{1}{\sqrt{n^6}} \leqslant \frac{1}{\sqrt{n^6+r}} \leqslant \frac{1}{\sqrt{n^6+n}}\)
\(\frac{n^2}{n^3}-\leqslant \frac{n^2}{\sqrt{n^6+r}} \leqslant \frac{n^2}{\sqrt{n^6+n}}\)
\(\sum_{r=1}^n \frac{n^2}{n^3} \leqslant \sum_{r=1}^n \frac{n^2}{\sqrt{n^6+r}} \leqslant \sum_{n=1}^n \frac{n^2}{\sqrt{n^6+n}}\)
\(\frac{n^3}{n^3} \leqslant \sum_{r=1}^n \frac{n^2}{\sqrt{n^6+r}} \leqslant \frac{n^3}{\sqrt{n^6+n}}\)
\(1 \leqslant \sum_{r=1}^n \frac{n^2}{\sqrt{n^6+r}} \leqslant \frac{n^{B^3}}{\sqrt{n^6+n}}\)
\[
1 \leq \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n^2}{\sqrt{n^6 + r}} \leq 1 = \lim_{n \to \infty} \frac{n^3}{\sqrt{n^6 + n}}
\]
So, \(\lim _{n \rightarrow \infty} a_n=1\)
A subset ${u_1, u_2, u_3, u_4, u_5}$ of the first 90 positive integers can be selected in $\binom{90}{5}$ ways. Let $u_{\text{max}} = \max{u_1, u_2, u_3, u_4, u_5}$ and $u_{\text{min}} = \min{u_1, u_2, u_3, u_4, u_5}$. Then the arithmetic mean of $u_{\text{max}} + u_{\text{min}}$ over all such subsets is
(A) 45
(B) 46
(C) 89
(D) 91
Solution
No. of subsets whose max element 90 is \(\quad 90\binom{89}{4}\)
When \(u_{\text {max }}=89 \rightarrow 89\binom{88}{4}\)
When \(u_{\text {max }}^{\text {max }}=88 \rightarrow 88\binom{87}{4}\)
" " "
" " "
" " "
when \(u_{\max }=5 \rightarrow 5\binom{4}{4}\)
Now, when \(u_{\min }=1 \rightarrow 1\binom{89}{4}\).
when \(u_{\min }^{\min }=2 \rightarrow 2\binom{88}{4}\)
when \(u_{\text {min }}=3 \rightarrow 3\binom{87}{4}\)
" " "
" " "
when \(u_{\min }=86 \rightarrow 86\binom{4}{4}\)
\(\therefore\) Sum of all possible values
\(90\binom{89}{4}+89\binom{88}{4}+\left(88\binom{87}{4}+\cdots+5\binom{4}{4}\right.\)
\(+1\binom{89}{4}+2\binom{88}{4}+\cdots+86\binom{4}{4}\)
\(=91\left[\binom{89}{4}+\binom{88}{4}+\binom{87}{4}+\cdots+\binom{4}{4}\right]\)
\(=91\binom{90}{5} \quad\) [Hockey-Stick Identity]
\(\therefore A M\) of all \(u_{\text {max }}+u_{\text {min }}=\frac{91\binom{90}{5}}{\binom{90}{5}}\)
\(=91\)
Let
\[a_n=\frac{2^3-1}{2^3+1} \times \frac{3^3-1}{3^3+1} \times \cdots \times \frac{n^3-1}{n^3+1}, \quad n \geq 2\]
Then \(\lim _{n \rightarrow \infty} a_n\)
(A) does not exist
-(B) is equal to \(\frac{2}{3}\)
(C) is equal to 1
(D) is equal to \(\frac{1}{2}\)
Solution
\(\frac{r^3-1}{r^3+1}=\frac{r-1}{r+1} \times \frac{r^2+r+1}{r^2-r+1}\)
\(\left(\frac{r-1}{r+1}\right) \times \frac{(r+1)^2-(r+1)+1}{\left(r^2-r+1\right)}\)
\(\prod_{r=2}^n\left(\frac{r-1}{r+1}\right) \prod_{r=2}^n \frac{(r+1)^2-(r+1)+1}{r^2-r+1}\)
\(\prod_{r=2}^n\left(\frac{r-1}{r+1}\right)=\frac{1}{3} \times \frac{2}{4} \times \frac{8}{5} \times \frac{4}{6} \times \frac{5}{7} \times \frac{8}{8} \cdots=i\)
\(\prod_{r=2}^n \frac{(r+1)^2-(r+1)+1}{r^2-r+1}=\frac{3^2-3+1}{2^2-3+1} \times \frac{4^2-4+1}{3^2-3+1} \times \frac{5^2-5+1}{4^2-4+1} \times \cdots\)
= \(\frac{1}{4-2+1}=\frac{1}{3}\) (ii)
So from (i) and (ii) we get \(\frac{2}{3}\)
Problem 1
Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) is differentiable and \(\left|f^{\prime}(x)\right|<\frac{1}{2}\) for all \(x \in \mathbb{R}\). Show that for some \(x_0 \in \mathbb{R}, f\left(x_0\right)=x_0\).
Solution
Problem 2
If the interior angles of a triangle (A B C) satisfy the equality,
\[
\sin ^2 A+\sin ^2 B+\sin ^2 C=2\left(\cos ^2 A+\cos ^2 B+\cos ^2 C\right)
\]
prove that the triangle must have a right angle.
Problem 3
Suppose \(f:[0,1] \rightarrow \mathbb{R}\) is differentiable with \(f(0)=0\). If \(\left|f^{\prime}(x)\right| \leq f(x)\) for all \(x \in[0,1]\), then show that \(f(x)=0\) for all \(x\).
Problem 4
Let \(S^1={\{z \in \mathbb{C}| | z \mid=1}\}\) be the unit circle in the complex plane. Let Let \(f: S^1 \rightarrow S^1\) be the map given by \( f(z)=z^2 \). We define \(f^{(1)}:=f\) and \(f^{(k+1)}:=f \circ f^{(k)}\) for \(k \geq 1\). The smallest positive integer \(n\) such that \(f^{(n)}(z)=z\) is called the period of \(z\). Determine the total number of points in \(S^1\) of period 2025.
(Hint: \(2025=3^4 \times 5^2) \)
Problem 5
Let \(a, b, c\) be nonzero real numbers such that \(a+b+c \neq 0\). Assume that
\[
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}
\]
Show that for any odd integer \(k\),
\[
\frac{1}{a^k}+\frac{1}{b^k}+\frac{1}{c^k}=\frac{1}{a^k+b^k+c^k}
\]
Problem 6
Let \(\mathbb{N}\) denote the set of natural numbers, and let \(\left(a_i, b_i\right)\), \(1 \leq i \leq 9\), be nine distinct tuples in \(\mathbb{N} \times \mathbb{N}\). Show that there are three distinct elements in the set \({2^{a_i} 3^{b_i}: 1 \leq i \leq 9}\) whose product is a perfect cube.
Problem 7
Consider a ball that moves inside an acute-angled triangle along a straight line, until it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence (=) angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.
Problem 8
Let \(n \geq 2\) and let \(a_1 \leq a_2 \leq \cdots \leq a_n\) be positive integers such that \(\sum_{i=1}^n a_i=\Pi_{i=1}^n a_i\). Prove that \(\sum_{i=1}^n a_i \leq 2 n\) and determine when equality holds.
In today's discussion, we delve into a fascinating problem from the 2016 CMI B.Sc. entrance exam that draws on key concepts from number theory, specifically the Chinese Remainder Theorem (CRT), but applies them in the context of polynomials. The problem asks us to find a polynomial $P(x)$ that satisfies two conditions:
This setup mirrors the CRT for integers but applies it to the algebraic framework of polynomials.
The video walks through the similarities between integer and polynomial division and emphasizes how techniques like the Euclidean algorithm can be extended to polynomials. Using polynomial differentiation and integration, we solve the given conditions, ultimately arriving at a general form for $P(x)$ by adjusting constants.
A key takeaway is the parallel between solving congruences for integers using the Euclidean algorithm and doing the same for polynomials, underscoring the algebraic unity between these two domains.
$$
P(x) \equiv 1 \quad\left(\bmod x^{100}\right) \quad \text { and } \quad P(x) \equiv 2 \quad\left(\bmod (x-2)^3\right)
$$
This highlights the deep connection between the two fields.
$$
P^{\prime}(x) \text { must be divisible by } x^{99} \text { and } \quad(x-2)^2
$$
$$
P(x)=a \cdot \frac{x^{101}}{101}-4 \cdot \frac{x^{100}}{100}+4 \cdot \frac{x^{99}}{99}+b
$$
$$
P_1(x) \cdot x^{100}+P_2(x) \cdot(x-2)^3=1
$$
| 1.[A] | 2.[D] | 3.[D] | 4.[D] | 5.[B] |
| 6.[B] | 7.[A] | 8.[A] | 9.[A] | 10.[B] |
| 11.[B] | 12.[A] | 13.[C] | 14.[B] | 15.[C] |
| 16.[C] | 17.[A] | 18.[C] | 19.[B] | 20.[B] |
| 21.[B] | 22.[C] | 23.[B] | 24.[B] | 25.[C] |
| 26.[B] | 27.[A] | 28.[D] | 29.[D] | 30.[C] |
Let $0<x<\frac{1}{6}$ be a real number. When a certain biased dice is rolled, a particular face $F$ occurs with probability $\frac{1}{6}-x$ and its opposite face occurs with probability $\frac{1}{6}+x$; the other four faces occur with probability $\frac{1}{6}$. Recall that opposite faces sum to 7 in any dice. Assume that the probability of obtaining the sum 7 when two such dice are rolled is $\frac{13}{96}$. Then, the value of $x$ is:
(A) $\frac{1}{8}$
(B) $\frac{1}{12}$
(C) $\frac{1}{24}$
(D) $\frac{1}{27}$
An office has 8 officers including two who are twins. Two teams, Red and Blue, of 4 officers each are to be formed randomly. What is the probability that the twins would be together in the Red team?
(A) $\frac{1}{5}$
(B) $\frac{3}{7}$
(C) $\frac{1}{4}$
(D) $\frac{3}{14}$
Suppose Roger has 4 identical green tennis balls and 5 identical red tennis balls. In how many ways can Roger arrange these 9 balls in a line so that no two green balls are next to each other and no three red balls are together
(A) 8
(B) 9
(C) 11
(D) 12
The number of permutations $\sigma$ of $1,2,3,4$ such that $|\sigma(i)-i|<2$ for every $1 \leq i \leq 4$ is
(A) 2
(B) 3
(C) 4
(D) 5
Let $f(x)$ be a degree 4 polynomial with real coefficients. Let $z$ be the number of real zeroes of $f$. and $e$ be the number of local extrema (i.e., local maxima or Minima ) of $f$. Which of the following is a possible $(z, e)$ pairs?
(A) $(4,4)$
(B) $(3,3)$
(C) $(2,2)$
(D) $(0,0)$
A number is called a palindrome if it reads the same backward or forward. For example, 112211 is a palindrome. How many 6-digit palindromes are divisible by 495 ?
(A) 10
(B) 11
(C) 30
(D) 45
Let $A$ be a square matrix of real numbers such that $A^4=A$. Which of the following is true for every such A ?
(A) $\quad \operatorname{det}(A) \neq-1$
(B) $A$ must be invertiible.
(C) $A$ can not be invertiible.
(D) $A^2+A+I=0$ where $I$ denotes the identity matrix.
Consider the real-valued function $h:{0,1, \ldots, 100} \rightarrow R$ such that $h(0)=5, h(100)=20$ and satisfying $h(i)=\frac{1}{0}(h(i+1)+h(i-1))$, for every $i=1,2, \ldots, 99$. Then, the value of $h(1)$ is :
(A) 5.15
(B) 5.5
(C) 6
(D) 6.15
An up-right path is a sequence of points $a_0=\left(x_0, y_0\right), a_1=\left(x_1, y_1\right), \cdots$ such that $a_{i+1}-a_i$ is either $(1,0)$ or $(0,1)$. The number of up-right paths from $(0,0)$ to $(100,100)$ which pass through $(1,2)$ is
(A) $3\binom{197}{99}$
(B) $3\binom{100}{50}$
(C) $2\binom{197}{98}$
(D) $3\binom{197}{100}$
Let $f(x)=\frac{1}{2} x \sin x-(1-\cos x)$. The smallest positive integer $k$ such that $\lim _{x \rightarrow 0} \frac{f(x)}{x^k} \neq 0$ is :
(A) 3
(B) 4
(C) 5
(D) 6
Nine students in a class gave a test for 50 marks. Let $S_1 \leq S_2 \leq \ldots \leq S_5 \leq \ldots \leq S_8 \leq S_9$ denote their ordered scores. Given that $S_1=20$ and $\sum_{i=1}^9 S_1=250$, let $m$ be the smallest value that $S_5$ can take and $M$ be the largest value that $S_5$ can take. Then the pair $(m, M)$ is given by?
(A) $(20,35)$
(B) $(20,34)$
(C) $(25,34)$
(D) $(25,30)$
Let 10 red balls and 10 white balls be arranged in a straight line such that 10 each are on either side of a central mark. The number of such symmetrical arrangements about the central mark is
(A) $\frac{10!}{5!5!}$
(B) 10 !
(C) $\frac{10!}{5!}$
(D) 2.10 !
If $z=x+i y$ is a complex number such that $\left|\frac{z-i}{z+i}\right|<1$, then we must have
(A) $x>0$
(B) $x<0$
(C) $y>0$
(D) $y<0$
Let $S=\{x-y |x, y \text{ are real numbers with} x^2+y^2=1\}$. Then maximum number in the set $S$ is
(A) 1
(B) $\sqrt{2}$
(C) $2 \sqrt{2}$
(D) $1+\sqrt{2}$
In a factory, 20 workers start working on a project of packing consignments. They need exactly 5 hours to pack one consignment. Every hour 4 new workers joint the existing workforce. It is mandatory to would relive a worker after 10 hours. Then the number of consignments that would be packed in the initial 113 hours is
(A) 40
(B) 50
(C) 45
(D) 52
Let $A B C D$ be a rectangle with its shorter side $a>0$ units and perimeter $2 s$ units. Let $P Q R S$ be any rectangle such that vertices $A, B, C$ and $D$ respectively lie on the lines $P Q, Q R, R S$ and $S P$. Then the maximum area of such a rectangle $P Q R S$ in square units is given by
(A) $s^2$
(B) $2 a(s-a)$
(C) $\frac{s^2}{2}$
(D) $\frac{5}{2} a(s-a)$
The number of pairs of integers $(x, y)$ satisfying the equation $x y(x+y+1)=5^{2018}+1$ is
(A) 0
(B) 2
(C) 1009
(D) 2018
Let $p(n)$ be the number of digits when $8^n$ is written in base 6 , and let $q(n)$ be the number of digits when $6^n$ is written in base 4 . For example, $8^2$ in base 6 is 144 , hence $p(2)=3$. Then $\lim _{n \rightarrow \infty} \frac{p(n) q(n)}{n^2}$ equals:
(A) 1
(B) $\frac{4}{3}$
(C) $\frac{3}{2}$
(D) 2
For a real number $\alpha$, let $S_\alpha$ denote the set of those real numbers $\beta$ that satisfy $\alpha \sin (\beta)=\beta \sin (\alpha)$. Then which of the following statements is true?
(A) For any $\alpha, S_\alpha$ is an infinite set
(B) $\quad S_\alpha$ is finite set if and only if $\alpha$ is not an integer multiple of $\pi$
(C) There are infinitely many numbers $\alpha$ for which $S_\alpha$ is the set of all real numbers
(D) $\quad S_\alpha$ is always finite
If $A=\left(\begin{array}{ll}1 & 1 \ 0 & i\end{array}\right)$ and $A^{2018}=\left(\begin{array}{ll}a & b \ c & d\end{array}\right)$, then $a+d$ equals :
(A) $1+i$
(B) 0
(C) 2
(D) 2018
Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be two functions. Consider the following two statements :
$P(1)$ : If $\lim_{x \rightarrow 0} f(x)$ exists and $\lim{x \rightarrow 0} f(x) g(x)$ exists, then $\lim _{x \rightarrow 0} g(x)$ must exist.
$P(2)$ : If $f, g$ are differentiable with $f(x)<g(x)$ for every real number $x$, then $f^{\prime}(x)<g^{\prime}(x)$ for all $x$
Then, which one of the following is a correct statement?
(A) Both $P(1)$ and $P(2)$ are true.
(B) Both $P(1)$ and $P(2)$ are false.
(C) $\quad P(1)$ is true and $P(2)$ is false.
(D) $\quad P(1)$ is false and $P(2)$ is true.
The number of solutions of the equation $\sin (7 x)+\sin (3 x)=0$ with $0 \leq x \leq 2 \pi$ is :
(A) 9
(B) 12
(C) 15
(D) 18
A bag contains some candies, $\frac{2}{5}$ of them are made of white chocolate and remaining $\frac{3}{5}$ are made of dark chocolate. Out of the white chocolate candies, $\frac{1}{3}$ are wrapped in red paper, the rest are wrapped in blue paper. Out of the dark chocolate candles, $\frac{2}{3}$ are wrapped in red paper, the rest wrapped in blue paper. If a randomly selected candy from the bag is found to be wrapped in red paper, then what is the probability that it is made up of dark chocolate?
(A) $\frac{2}{3}$
(B) $\frac{3}{4}$
(C) $\frac{3}{5}$
(D) $\frac{1}{4}$
A party is attended by twenty people. In any subset of four people, there is at least one person who knows the other three (we assume that if $X$ knows $Y$, then $Y$ knows $X$ ). Suppose there are three people in the party who do not know each other. How many people in the party know everyone?
(A) 16
(B) 17
(C) 18
(D) Cannot be determined from the given data.
The sum of all natural numbers $a$ such that $a^2-16 a+67$ is a perfect square is :
(A) 10
(B) 12
(C) 16
(D) 22
The sides of a regular hexagon $A B C D E F$ are extended by doubling them (for example, $B A$ extends to $B A^{\prime}$ with $\left.B A^{\prime}=2 B A\right)$ to form a bigger regular hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} F^{\prime} F^{\prime}$ as in the figure
Then the ratio of the areas of the bigger to the smaller hexagon is:
(A) 2
(B) 3
(C) $2 \sqrt{3}$
(D) $\pi$
Between 12 noon and 1 PM, there are two instants when the hour hand and the minute hand of a clock are at right angles. The difference in minutes between these two instants is:
(A) $32 \frac{8}{11}$
(B) $30 \frac{8}{11}$
(C) $32 \frac{5}{11}$
(D) $30 \frac{5}{11}$
For which values of $\theta$, with $0<\theta<\frac{\pi}{2}$, does the quadratic polynomial in $t$ given by $t^2+4 t \cos \theta+\cot \theta$ have repeated roots?
(A) $\frac{\pi}{6}$ or $\frac{5 \pi}{18}$
(B) $\frac{\pi}{6}$ or $\frac{5 \pi}{12}$
(C) $\frac{\pi}{12}$ or $\frac{5 \pi}{18}$
(D) $\frac{\pi}{12}$ or $\frac{5 \pi}{12}$
Let $\alpha, \beta, \gamma$ be complex numbers which are the vertices of an equilateral triangle. Them, we must have :
(A) $\alpha+\beta+\gamma=0$
(B) $\alpha^2+\beta^2+\gamma^2=0$
(C) $\alpha^2+\beta^2+\gamma^2+\alpha \beta+\beta \gamma+\gamma \alpha=0$
(D) $\quad(\alpha-\beta)^2+(\beta-\gamma)^2+\left(\gamma-\alpha^2\right)=0$
Assume that $n$ copies of unit cubes are glued together side by side to form a rectangular solid block. If the number of unit cubes that are completely invisible is 30 , then the minimum possible value of $n$ is :
(A) 204
(B) 180
(C) 140
(D) 84
Welcome to a thrilling exploration of locus problems in geometry, a crucial concept for anyone preparing for math competitions like the IOQM, American Math Competition, and GMD. Whether you're aiming for ISI, CMI, or just looking to sharpen your mathematical skills, understanding loci will give you a solid edge.
In simpler terms, a locus is the path traced by a moving point that follows a specific rule.Imagine you have a fixed point, O, and a moving point, P. Point P doesn't move randomly; it follows a specific rule. Our goal is to find out the path that point P traces as it moves according to this rule. This path is known as the locus of point P.
Let’s start with a simple rule: point P is always 3 units away from point O. What happens then? P traces out a circle!
Both views are important and help you understand the circle in different ways.
Next, let’s try a different rule. Suppose you have two fixed points, O1 and O2, and a moving point, P. The rule is that the sum of the distances from P to O1 and O2 is always 5 units. What shape does P trace out? The answer is an ellipse!
To understand this, imagine P moving so that the distances to O1 and O2 always add up to 5. Visualizing this movement helps you see how the ellipse forms.
For our final example, imagine a big fixed circle with a diameter of 4 cm and a small moving circle with a diameter of 1 cm. If the small circle rolls around the big circle, what path does a point on the edge of the small circle trace? This path is called a hypocycloid.
As the small circle rolls, the point on its edge creates a unique and interesting path. Visualizing this helps you understand the movement and the resulting shape.
You have two fixed points, A and B, and a moving point, P. The sum of the distances from P to A and B is constant. What path does P trace out?
Share your answers in the comments!
In 2024, the following Cheenta students are successful for Indian Statistical Institute's BStat-BMath Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 100 in the entire country in these entrances.
Most of these students attended the problem solving workshops regularly, which happen 5 times every week.
CMI B.Sc. Math Entrance
I.S.I. B.Stat Entrance
I.S.I. B.Math Entrance
In this video we talk to the successful candidates and learn from their strategies. We also have a virtual award ceremony. The students presented with books and access to Cheenta Research Programs.
Find, with proof, all possible values of $t$ such that
$$\lim _{n \rightarrow \infty}{\frac{1+2^{1 / 3}+3^{1 / 3}+\cdots+n^{1 / 3}}{n^t}}=c$$
for some real number $c>0$. Also find the corresponding values of $c$.
Suppose $n \geq 2$. Consider the polynomial
$$
Q_n(x)=1-x^n-(1-x)^n .
$$
Show that the equation $Q_n(x)=0$ has only two real roots, namely 0 and 1.
Let $A B C D$ be a quadrilateral with all internal angles $<\pi$. Squares are drawn on each side as shown in the picture below. Let $\Delta_1, \Delta_2, \Delta_3$ and $\Delta_4$ denote the areas of the shaded triangles shown. Prove that
$$
\Delta_1-\Delta_2+\Delta_3-\Delta_4=0
$$
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function which is differentiable at 0 . Define another function $g: \mathbb{R} \rightarrow \mathbb{R}$ as follows:
$$
g(x)= \begin{cases}f(x) \sin \left(\frac{1}{x}\right) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}
$$
Suppose that $g$ is also differentiable at 0 . Prove that
$$
g^{\prime}(0)=f^{\prime}(0)=f(0)=g(0)=0 .
$$
Let $P(x)$ be a polynomial with real coefficients. Let $\alpha_1, \ldots, \alpha_k$ be the distinct real roots of $P(x)=0$. If $P^{\prime}$ is the derivative of $P$ show that for each $i=1,2, \ldots, k$,
$$
\lim _{x \rightarrow a_i} \frac{\left(x-\alpha_i\right) P^{\prime}(x)}{P(x)}=r_i,
$$
for some positive integer $r_i$.
Q6. Let $x_1, \ldots, x_{2024}$ be non-negative real numbers with $\sum_{i=1}^{2024} x_i=1$. Find, with proof, the minimum and maximum possible values of the expression
$$
\sum_{i=1}^{1012} x_i+\sum_{i=1013}^{2024} x_i^2 .
$$
Consider a container of the shape obtained by revolving a segment of the parabola $x=1+y^2$ around the $y$-axis as shown below. The container is initially empty. Water is poured at a constant rate of $1 \mathrm{~cm}^3 / \mathrm{s}$ into the container. Let $h(t)$ be the height of water inside the container at time $t$. Find the time $t$ when the rate of change of $h(t)$ is maximum.
In a sports tournament involving $N$ teams, each team plays every other team exactly once. At the end of every match, the winning team gets 1 point and the losing team gets 0 points. At the end of the tournament, the total points received by the individual teams are arranged in decreasing order as follows:
$$
x_1 \geq x_2 \geq \cdots \geq x_N .
$$
Prove that for any $1 \leq k \leq N$,
$$
\frac{N-k}{2} \leq x_k \leq N-\frac{k+1}{2}
$$
| 1) D | 7) D | 13) B | 19) C | 25) B |
| 2) B | 8) A | 14) D | 20) C | 26) D |
| 3) C | 9) D | 15) D | 21) A | 27) B |
| 4) A | 10) A | 16) D | 22) A | 28) C |
| 5) B | 11) B | 17) B | 23) C | 29) D |
| 6) D | 12) B | 18) A | 24) B | 30) A |
If $x=1+\sqrt[5]{2}+\sqrt[5]{4}+\sqrt[5]{8}+\sqrt[3]{16}$, then the value of $\left(1+\frac{1}{x}\right)^{30}$ is
(a) 2
(b) 5
(c) 32
(d) 64
$x=1+2^{1 / 5}+2^{2 / 5}+2^{3 / 5}+2^{4 / 5}$
Let $a=2^{1 / 5}$
$\therefore x=1+a+a^2+a^3+a^4$
$=\frac{a^5-1 i}{a-1}=\frac{1}{2^{1 / 5}-1}$
$\begin{aligned} & \therefore \frac{1}{x}=2^{1 / 5}-1 \ & \quad\left(1+\frac{1}{x}\right)^{30}=2^6=64\end{aligned}$
Let $j$ be a number selected at random from ${\{1,2, \ldots, 2024}\}$. What is the probability that $j$ is divisible by 9 and 15 ?
(a) $\frac{1}{23}$
(b) $\frac{1}{46}$
(c) $\frac{1}{44}$
(d) $\frac{1}{253}$
$\operatorname{lcm}(9,15)=45$
$\lfloor \frac{2024}{45} \rfloor,=44.$
So the set of favourable integers has 44 elements.
So, probability = $\frac{44}{2024}=\frac{1}{46}$
Let $S_n$ be the set of all $n$-digit numbers whose digits are all 1 or 2 and there are no consecutive 2's. (Example: 112 is in $S_3$ but 221 is not in $S_3$ ). Then the number of elements in $S_{10}$ is
(a) $512$
(b) $256$
(c) $144$
(d) $89$
12th Fibonacci Number
$\sum_{k=0}^n\binom{n-k+1}{k}$
$=F_{n+2}$
=$144$
There are 30 True or False questions in an examination. A student knows the answer to 20 questions and guesses the answers to the remaining 10 questions at random. What is the probability that the student gets exactly 24 answers correct?
(a) $\frac{105}{2^9}$
(b) $\frac{105}{2^8}$
(c) $\frac{105}{2^{10}}$
(d) $\frac{4}{2^{10}}$
Binomial Distribution
$x$ be the R.V. of selecting unknown questions. (answer is unknown)
Then $x \sim \operatorname{Bin}(10,1 / 2)$
So, $P(x=4)=\binom{10}{4} \frac{1}{2^{10}}=\frac{105}{2^9}$
Let $T$ be a right-angled triangle in the plane whose side lengths are in a geometric progression. Let $n(T)$ denote the number of sides of $T$ that have integer lengths. Then the maximum value of $n(T)$ over all such $T$ is
(a) $0$
(b) $1$
(c) $2$
(d) $3$
$a, a r, a r^2$
Now, $a^2+a^2 r^2=a^2 r^4$
$\Rightarrow 1+r^2=r^4$
But $r$ has a positive solution
So, only $a$ can be an integer $\max (n(T))=1$
Let $x_1, x_2, \ldots, x_n$ be non-negative real numbers such that $\sum_{i=1}^n x_i=1$. What is the maximum possible value of $\sum_{i=1}^n \sqrt{x_i}$ ?
(a) $1$
(b) $\sqrt{n}$
(c) $n^{3 / 4}$
(d) $n$
Cauchy Schwarz
$\left(\sum\left(\sqrt{\left.x_i-1\right)}\right)^2 \leq\left(2\left(\sqrt{x_i}\right)^2\right)\left(\sum 1^2\right)\right.$
$\therefore \sum \sqrt{x_1} \leqslant \sqrt{n}$
The precise interval on which the function $f(x)=\log _{1 / 2}\left(x^2-2 x-3\right)$ is monotonically decreasing, is
(a) $(-\infty,-1)$
(b) $(-\infty,1)$
(c) $(1, \infty)$
(d) $(3, \infty)$
$f(x)=-\log _2\left(x^2-2 x-3\right)$
$f^{\prime}(x)=-\frac{\ln^2}{\log _2(x-3)}-\frac{\ln^2}{\log _2(x+1)}$
in $(3, \infty)$
$f^{\prime}(x)<0 \quad \forall x$
So, in $(3, \infty), f(x)$ is monotonically decreasing.
The angle subtended at the origin by the common chord of the circles $x^2+y^2-6 x-6 y=0$ and $x^2+y^2=36$ is
(a) $\pi / 2$
(b) $\pi / 4$
(c) $\pi / 3$
(d) $2 \pi / 3$
The length of the chord
$=2 \sqrt{36-18}$
$=6 \sqrt{2}$
So, ABC is an isosceles triangle. $ (\frac{\pi}{2}) $
In $\triangle A B C, C D$ is the median and $B E$ is the altitude. Given that $\overline{C D}=\overline{B E}$, what is the value of $\angle A C D ?$
(a) $\pi / 3$
(b) $\pi / 4$
(c) $\pi / 5$
(d) $\pi / 6$
If the points $z_1$ and $z_2$ are on the circles $|z|=2$ and $|z|=3$, respectively, and the angle ineluded between these vectors is $60^{\circ}$, then the value of $\frac{\left|z_1+z_2\right|}{\left|z_1-z_2\right|}$ is
(a) $\sqrt{\frac{19}{7}}$
(b) $\sqrt{19}$
(c) $\sqrt{7}$
(d) $\sqrt{\frac{7}{19}}$
Let $n \geqslant 1$. The maximum possible number of primes in the set ${\{n+6, n+7, \ldots . n+34, n+35}\}$ is
(a) 7
(b) 8
(c) 12
(c) 13
Suppose 40 distinguishable balls are to be distributed into 4 different boxes such that each box gets exactly 10 balls. Out of these 40 balls, 10 are defective and 30 are non-defective. In how many ways can the balls be distributed such that all the defective balls go to the first two boxes?
(a) $\frac{40!}{(10!)^4}$
(b) $\frac{30!\cdot 20!}{(10!)^5}$
(c) $\frac{20!\cdot 20!}{(10!)^5}$
(d) $\frac{30!-10!}{(10!)^4}$
The number of elements in the set
${\{x: 0 \leqslant x \leqslant 2,\left|x-x^5\right|=\left|x^5-x^6\right|}\}$ is
(a) 2
(b) 3
(c) 4
(d) 5
For $0 \leq x \leq 1$
$x=0$ is two solutions.
$x=1$
if $0<x<1$ then,
$x-x^5=x^3-x^6$
$\Rightarrow x+x^6=2 x^5$
$\Rightarrow 1+x^3=2 x^4$
$\text { Consider }=x^5-2 x^4+1=f(x)$
$f^{\prime}(x)=5 x^4-8 x^3=x^3(5 x-8)=5 x^3\left(x-\frac{8}{5}\right)$
$\text { So, } f^{\prime}(x)<0 \text { for } x \in(0,1)$
$\text { So, Now sloution for } x \text { for (1) }$
$\text { in } x \in(0,1) \text {. }$
$\text { if } x \in(1,2) \text { then, }$
$x^5-x=x^6-x^5$
$x^6+x=2 x^5$
$x^5+1=2 x^4$
Consider,
$f(x)=x^5-2 x^4+1$
$f^{\prime}(x)=5 x^3\left(x-\frac{8}{5}\right) \quad \text { Here } x \in(1,2)$
So, at some point $f^{\prime}(x)$ changes sign for $x \in(1,2)$
$\text { Now, } f(1.5)<0$
$f(2)>0$
So, by $I VT$, there is a solution for $f$.
$\text { So, for } 0 x \in(1,2)$
there is a solution.
Total 3 solutions
In a room with $n \geqslant 2$ people, each pair shakes hands between themselves with probability $\frac{2}{n^2}$ and independently of other pairs. If $p_n$ is the probability that the total number of handshakes is at most 1 , then $\lim _{n \rightarrow \infty} p_n$ is equal to
(a) 0
(b) 1
(c) $e^{-1}$
(d) $2e^{-1}$
The number of positive solutions to the equation
$$e^x \sin x=\log x+e^{\sqrt{x}}+2$$
(a) 0
(b) 1
(c) 2
(d) $\infty$
$e^x \sin x=\log x+e^{\sqrt{x}}+2$
$f(x)=e^x \sin x$
$f^{\prime}(x)=e^x \cos x+e^x \sin x$
Now, $-e^x<e^x \sin x<R^x$
Now, $\begin{aligned} g(x)=\log +e^{\sqrt{x}}+2\end{aligned}$ is an increasing graph for $x>0$. So infinitely many.
Let $n>1$ be the smallest composite integer that is coprime to $\frac{10000}{9900 \%}$. Then
(a) $n \leqslant 100$
(b) $100<n \leqslant 9900$
(c) $9900<n \leqslant 10000$
(d) $n>10000$
Let $P={\{(x, y): x+1 \geqslant y, x \geqslant-1, y \geqslant 2 x}\}$. Then the minimum value of $(x+y)$ where $(x, y)$ varies over the set $P$ is
(a) $-1$
(b) $-3$
(c) $3$
(d) $0$
Let $A={1, \ldots, 5}$ and $B={1, \ldots, 10}$. Then the number of ordered pairs $(f, g)$ of functions $f: A \rightarrow B$ and $g: B \rightarrow A$ satisfying $(g \circ f)(a)=a$ for all $a \in A$ is
(a) $\frac{10!}{5!} \times 5^5$
(b) $5^{10} \times 5!$
(c) $10!\times 5!$
(d) $\binom{10}{5} \times 10^5$
Let
$$S=\frac{1}{\sqrt{10000}}+\frac{1}{\sqrt{10001}}+\cdots+\frac{1}{\sqrt{160000}}$$
Then the largest positive integer not exceeding $S$ is
(a) $200$
(b) $400$
(c) $600$
(d) $800$
The real number $x$ satisfies
$$
\frac{|x|^2-|x|-2}{2|x|-|x|^2-2}>2
$$
if and only if $x$ belongs to
(a) $(-2,-1) \cup(1,2)$
(b) $(-2 / 3,0) \cup(0,2 / 3)$
(c) $(-1,-2 / 3) \cup(2 / 3,1)$
(d) $(-1,0) \cup(0,1)$
Consider points of the form $\left(n, n^k\right)$, where $n$ and $k$ are integers with $n \geq 0, k \geq 1$. How many such points are strictly inside the circle of radius 10 with centre at the origin?
(a) $11$
(b) $12$
(c) $15$
(d) $17$
Let $n>1$, and let us arrange the expansion of $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$ in decreasing powers of $x$. Suppose the first three coefficients are in arithmetic progression. Then, the number of terms where $x$ appears with an integer power, is
(a) $3$
(b) $2$
(c) $1$
(d) $0$
The limit
$$
\lim _{n \rightarrow \infty} \frac{2 \log 2+3 \log 3+\cdots+n \log n}{n^2 \log n}
$$
is equals to
(a) $0$
(b) $1 / 4$
(c) $1 / 2$
(d) $1$
$\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1 \log 1+2 \log \alpha+\cdots+n \log n}{n \log n}\right]$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{r=1}^n$ $\frac{r \log r}{n \log n}$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{\gamma=1}^n .$ $\frac{\left(\frac{\gamma}{n}\right)\left[\log \left(\frac{\gamma}{n}\right)+\log n\right]}{\log n}$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{\gamma=1}^n .$ $\frac{\left(\frac{\gamma}{n}\right) \log \left(\frac{\gamma}{n}\right)}{\log (n)}$ $+\lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right)$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{\log (n)}$ $\frac{1}{n} \cdot \sum_{r=1}^n\left(\frac{r}{n}\right) \log \left(\frac{r}{n}\right)$ $+\lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2}$
$=\lim _{n \rightarrow \infty}$ $\frac{1}{\log (n)}$ $× \lim _{n \rightarrow \infty}$ $\frac{1}{n} \sum_{\gamma=1}^n\left(\frac{\gamma}{n}\right) \log \left(\frac{\gamma}{n}\right)+\frac{1}{2}$
$=\lim _{x \rightarrow \infty} \frac{1}{\log (x)} \times \int_0^0 x \log (x)+\frac{1}{2}$
$=0 \times 0+\frac{1}{2}$
$=\frac{1}{2}$
Let $p<q$ be prime numbers such that $p^2+q^2+7 p q$ is a perfect square. Then, the largest possible value of $q$ is:
(a) $7$
(b) $11$
(c) $23$
(d) $29$
$p^2+q^2+r p q=x^2 $ with $x>p+q \in \mathbb{N}$
$\Rightarrow \quad(p+q)^2+5 p q=x^2$
$\Rightarrow \quad 5 p q=x^2-(p+q)^2$
$\Rightarrow \quad 5 p q=(x+p+q)(x-p-q)$
Considering $x+p+q>x-p-q$ & $q>p$,
Case 1:
$x+p+q=5 p ; 5 q$
$x-p-q=q \text p$
$\Rightarrow 2 p+2 q=5 p-q$ $\text { (or) } 5 q-p$
$\Rightarrow p=q,$ absurd
Case 2:
$x+p+q=q$
$x-p-q=5 p$
$\quad \Rightarrow \quad 2 p+2 q=q-5 p$
$\quad \Rightarrow \quad q=7 p$ not possible
Case 3:
$x+p+q=5 p q$
$x-p-q=1$
$\quad \Rightarrow$ $2 p+2 q=5 p q-1$
$\quad \Rightarrow$ $(5 q-2)(5 p-2)=7 $,
but $5 q-2>7$, not possible
Case 4:
$x+p+q=p q$
$x-p-q=5$
$\Rightarrow 2 p+2 q=p q-5$
$\quad \Rightarrow \quad(p-2)(q-2)=9$
$\Rightarrow \quad q-2=q, \quad p-2=1$
$\Rightarrow \boxed {\quad q=11, p=3}$
The set of all real numbers $x$ for which $3^{2^{1-x^2}}$ is an integer has
(a) $3$ elements
(b) $15$ elements
(c) $24$ elements
(d) infinitely many elements.
The maximum possible value of $1-x^2$ is 1 $(when =0)$ and the minimum value is $(-\infty)$. So,
$-\infty<1-x^2 \leqslant 1$
and it wan take any value in the intervene So,
$0<2^{1-x^2} \leqslant 2$
Thus,
$1=3^0<3^{2^{1-2}} \leq 3^2=9$
Hence, $3^{2^{1-x^2}}$ an take all the putegors 2 to 9 and each of which has two corresponding valet for $x$ except when $x=0$. Thus, the number of such $x=2 \times 8-1=15$
Let $a, b, c$ be three complex numbers. The equation
$$a z+b \bar{z}+c=0$$
represents a straight line on the complex plane if and only if
(a) $a = b$
(b) $\tilde{a} c=b \vec{c}$
(c) $|a|=|b| \neq 0$
(d) $|a|=|b| \neq 0$ and $\bar{a} c=b \bar{c}$
In the adjoining figure, $C$ is the centre of the circle drawn, $A, F, E$ lie on the circle and $B C D F$ is a rectangle. If $\frac{D E}{A B}=2$, then $\frac{F E}{F A}$ equals
(a) $\sqrt{\frac{3}{2}}$
(b) $\sqrt{2}$
(c) $\sqrt{\frac{5}{2}}$
(d) $\sqrt{3}$
For every increasing function $b:[1, \infty) \rightarrow[1, \infty)$ such that
$$
\int_1^{\infty} \frac{\mathrm{d} x}{b(x)}<\infty
$$
we must have
(a) $\sum_{k=1}^{\infty} \frac{\sqrt{\log k}}{b(k)}<\infty$
(b) $\sum_{k=3}^{\infty} \frac{\log k}{b(\log k)}<\infty$
(c) $\sum_{k=1}^{\infty} \frac{e^k}{b\left(e^k\right)}<\infty$
(d) $\sum_{k=3}^{\infty} \frac{1}{\sqrt{b(\log k)}}<\infty$
Consider the following two statements:
(I) There exists a differentiable function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $g\left(x^3+x^5\right)=e^x-100$.
(II) There exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $g\left(e^x\right)=x^3+x^5 $
Then
(a) Only (I) is correct
(b) Only (II) is correct
(c) Both (I) and (II) are correct.
(d) Neither (I) nor (II) is correct.
(I)
$g\left(x^3+x^5\right)=e^x-100$
differentiating both sides, we get
$g^{\prime}\left(x^3+x^5\right)\left(3 x^2+5 x^4\right)=e^x$
Note that the LHS is well-defined if $g(x)$ is differentiable substitute
$x=0 \Rightarrow 0=1$, which is absurd.
Hence, such a fraction $g(x)$ cannot be differentiable (I) is incorrect.
(II)
$g\left(e^x\right)=x^3+x^5$
$g(0)=\lim _{x \rightarrow-\infty} g\left(e^x\right)$ because $\lim _{x \rightarrow-\infty} e^x=0$ and that $g$ is continuous.
But notice that $\operatorname{Lim}_{x \rightarrow-\infty} x^3+x^5=-\infty$
$\Rightarrow \quad \lim _{x \rightarrow-\infty} g\left(e^x\right)=-\infty=g(0)$, which is absurd if we assume "g" as continuous (bounded on compact sets)
(II) is incorrect
Neither (I) nor (II) is correct
Let $\psi: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $\int_{-1}^1 \psi(x) \mathrm{d} x=1$.
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function. Then
$$
\lim_ {\varepsilon \rightarrow 0} \frac{1}{\varepsilon} \int_{1-\varepsilon}^{1+\varepsilon} f(y) \psi\left(\frac{1-y}{\varepsilon}\right) \mathrm{dy}
$$
equals
(a) $f(1)$
(b) $f(1) \psi(0)$
(c) $f^{\prime}(1) \psi(0)$
(d) $f(1) \psi(1)$
$\lim_{\varepsilon \mapsto 0} \frac{1}{\varepsilon} \int_{1-\varepsilon}^{1+\varepsilon} f(y) \psi\left(\frac{1-y}{\varepsilon}\right) d y$
Substitute $\frac{1-y}{\varepsilon}=t$ $\Rightarrow 1-y=\varepsilon t$ $\Rightarrow 1-\varepsilon t=y$
$1-y=\varepsilon t$
$-d y=\varepsilon d t$
=$\lim_{\varepsilon \rightarrow 0} \int_{-1}^{1} f(1-\varepsilon t) \psi(t)$
=$f(1)$
$\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$
General Term, $\binom{n}{r}\left(x^{1 / 2}\right)^{n-r}\left(\frac{1}{2 x^{1 / a}}\right)^r$
$\binom{n}{r}x \frac{2 n-3 r}{4}$
Put, $r=0,1,2$
$\binom{n}{0},\binom{n}{1}\left(\frac{1}{2}\right),\binom{n}{2}\left(\frac{1}{2}\right)^2$ are in A.P.
So, $n=1$ or 8
$\therefore n=8$.
Now simple calculation, gives (A)
Quadratic diophantine equations can be hard to solve. However in this particular discussion, we will use a technique from elementary number theory to solve it. This problem is from the entrance of Indian Statistical Institute.
We use two facts:
Watch the video and then try the quiz that follows.