UGA

Problem 1

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.

Problem 2

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, \(2 x \frac{1}{126}<\int_3^5 f(x) d x \quad \leqslant 2 x \frac{1}{28}\)

\(\Rightarrow \frac{1}{ 63}<\int_3^5 f(x) d x<\frac{1}{14}\)

Problem 3

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:

\((1-3 x)^6\left(1+9 x^2\right)^6(1+3 x)^6\)

=\(\left(1-9 x^2\right)^6\left(1+9 x^2\right)^6\)

=\(\left(1-81 x^4\right)^6\)

\(\therefore\binom{6}{0} 1^0 \cdot\left(-81 x^4\right)^6+\cdots+\binom{6}{4} 1^4\left(-81 x^4\right)^2+\cdots\)

=\(5 \times 3^9 \times x^8\)

Problem 4

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

Problem 5

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.

Problem 6

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

Problem 7

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

\(x^2+x^2-2 x \cdot x \cdot \cos (180-\theta)\)

=\((3 x)^2+(3 x)^2-2 \cdot 3 x \cdot 3 x \cos \theta\)

\(\Rightarrow 2 x^2+2 x^2 \cos \theta=18 x^2-18 x^2 \cos \theta\)

\(\Rightarrow 20 x^2 \cos \theta=16 x^2\)

\(\cos \theta=\frac{16 x^2}{20 x^2}\)

=\(\frac{4}{5}\)

Problem 8

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.

Problem 9

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

\(\left(3 a_1\right)^2+\left(3 b_1\right)^2=25 \times 81\)

\(\Rightarrow a_1^2+b_1^2=9 \times 25\)

\(\Rightarrow \quad a_2{ }^2+b_2{ }^2=25\)

\(\therefore a_2=3, b_2=9\)

\(\therefore a=27 \quad b=36\)

Problem 10

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

Problem 12

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

Problem 13

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 \(S_4\)

Problem 14

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)

Problem 15

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

\(d_1^2=a_b^2+b_b^2-2 a b \cos \theta\)

\(d_2^2=a^2+b^2-2 a b \cos \left(180^{\circ}-\theta\right)\)

\(\therefore d_1^2+d_2^2=2\left(a^2+b^2\right)\)

\(\Rightarrow 19+d_2^2=2(4+9)\)

\(\Rightarrow \quad d_2^2=26-19 \Rightarrow d_2^2=7\)

\(=d_2=\sqrt{7}\)

Problem 16

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)

Problem 17

\(f(x) =
\begin{cases}
-1, & \text{if } x < 0,
\ 0, & \text{if } x = 0,
\ 1, & \text{if } x > 0.
\end{cases}\)

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

Problem 18

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

\(L=\lim _{n \rightarrow \infty}(n+100) \frac{5}{\ln (n-\infty)}\)

\(\Rightarrow \ln L=\frac{5 \ln (n+100)}{\left.\lim _{n \rightarrow \infty} \frac{\ln (n-50)}{\ln (n-5}\right)}=5\)

\(L=e^5\)

\(2<e<3\)

\(32<L<243\)

Problem 19

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

\(\binom{3+4-1}{4-1}\binom{1+4-1}{4-1}\)

\(=20 \times 4\)

\(=80\)

Problem 20

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

\(\int_{-2}^1([x]+2)^{|x|} d x\)

\(=\int_0^1 2^x d x+\int_{-1}^0 1^{-x} d x\)

\(f \int_{-2}^{-1} 0 \cdot d x\)

\(=\left[\frac{2 x}{\ln 2}\right]_0^1+1.1\)

\(\left(\frac{1}{\ln^2}+1\right)\)

Problem 21

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 \(10 \leqslant x \leqslant 2\)

\(f(x): \int_{x-1}^x g(t) d t\)

\(=\int_{x-1}^1 d t=(2-x)\)

Problem 22

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

\(=\binom{\binom{n}{2}}{2}-n\binom{n-1}{2}\)

\(=\left(\frac{n(n-1)}{2}\right)-n \cdot \frac{(n-1)(n-2)}{2}\)

\(=\frac{n(n-1)(n-2)(n+1)}{8}-\frac{n(n-1)(n-2)}{2}\)

\(=\frac{n(n-1)}{8} \cdot\left(n^2-n-2-4 n+\theta\right)\)

\(=\frac{n(n-1)(n-2)(n-3)}{8}\)

Problem 23

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

\[
\begin{aligned}
& 1\left(2^{n-1}-2^{n-2}\right)+2\left(2^{n-2}-2^{n-3}\right)+\cdots+n\left(2^1-2^0\right) \
& =2^{n-1}+2^{n-2}+2^{n-3}+\cdots+2^1+2^0
\end{aligned}
\]

\(\therefore 2^0+2^1+2^2+\cdots+2^{n-3}+2^{n-2}+2^{n-1}=2047\)

\(\Rightarrow \frac{2^0\left(2^n-1\right)}{2-1}=2047\)

\(\Rightarrow \quad 2^n-1=2047 \Rightarrow n=11\)

Problem 24

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

\(=\sum_{k=0}^n \tan ^{-1} \frac{1}{k^2+k+1}\)

\(=\sum_{k=0}^n \tan ^{-1} \frac{1}{k(k+1)+1}\)

\(=\sum_{k=0}^n \tan ^{-1} \frac{(k+1)-k}{k(k+1)+1}\)

\(=\sum_{k=0}^n\left[\tan ^{-1}(k+1)-\tan ^{-1}(k)\right]\)

\(=\tan ^{-1}(n+1)\)

Problem 25

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) \(d \geq \frac{2}{3^{1 / 4}}\)

Problem 26

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).

Problem 27

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

\(f(x)=e^x-1\)

\(\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\frac{e^{x / 2}-1}{x}+\frac{e^{x / 2^2}-1}{x}+\cdots+\frac{e^{x / 2^x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\frac{e^{x / 2}-1}{x / 2} \times \frac{1}{2}+\frac{e^{x / 2^2}-1}{x / 2^2} \times \frac{1}{2^2}+\cdots+\frac{e^{x / 2^x}-1}{x / 2^x} \times \frac{1}{2^x}\)

=\(1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^x}=-\frac{1\left(\frac{1}{2^{x+1}}-1\right)}{1 / 2}\)

\(=-2\left(\frac{1}{2^{x+1}}-1\right)\)

\(=-\left(\frac{1}{2^k}-2\right)=2-\frac{1}{2^k}\)

Problem 28

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, \(\lim _{n \rightarrow \infty} a_n=1\)

Problem 29

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 \(u_{\text {max }}^{\text {max }}=88 \rightarrow 88\binom{87}{4}\)

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" " "

" " "

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 \(u_{\text {min }}=3 \rightarrow 3\binom{87}{4}\)

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" " "

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 \(u_{\text {max }}+u_{\text {min }}=\frac{91\binom{90}{5}}{\binom{90}{5}}\)

\(=91\)

Problem 30

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 \(\frac{2}{3}\)

UGB

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.