This is a work in progress. Please come back for the solutions. You can also suggest your solutions in the comment section
Answer Key
| Problem 1 -> A | Problem 7 -> C | 13. | Problem 19 -> B | Problem 25 -> C |
| Problem 2 -> D | 8. | Problem 14 -> C | Problem 20 -> D | Problem 26 -> D |
| Problem 3 -> | 9. | Problem 15 -> B | Problem 21 -> B | Problem 27 -> A |
| Problem 4 -> | 10. | Problem 16 -> A | Problem 22 -> C | Problem 28 -> |
| Problem 5 -> C | 11. | Problem 17 -> C | Problem 23 -> B | Problem 29 -> B |
| Problem 6 -> D | 12 | Problem 18 -> A | Problem 24 -> C | Problem 30 -> |
Any positive real number $x$ can be expanded as
$x=a_{n} \cdot 2^{n}+a_{n-1} \cdot 2^{n-1}+ \\ \cdots+a_{1} \cdot 2^{1}+a_{0} \cdot 2^{0}+a_{-1} \\ \cdot 2^{-1}+a_{-2} \cdot 2^{-2}+\cdots$
for some $n \geq 0$, where each $a_{i} \in{0,1}$. In the above-described expansion of $21.1875$, the smallest positive integer $k$ such that $a_{-k} \neq 0$ is:
(A) 3 <—Answer
(B) 2
(C) 1
(D) 4
Solution
$x=a_{n} 2^{n}+a_{n-1} 2^{n-1}+\cdots+a_{1} 2^{1}+a_{1}$
Binary represation of $21.1875$ is $10101.0011$ (very standard Method to Convert Dercimal to Binary)
So, $-k=-3$
So, we got $k=3$ (A)
Suppose, for some $\theta \in\left[0, \frac{\pi}{2}\right], \frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $(\cot 3 \theta) \tan \theta$ equals
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{8}$
(D) $\frac{1}{7}$ <— Answer
Solution
$\frac{\cos 3 \theta}{ \cos \theta}=\frac{1}{3}$
$\Rightarrow \frac{4 \cos^{ 3} \theta-3 \cos \theta}{\cos \theta}=\frac{1}{3}$
$\Rightarrow 4 \cos ^{2} \theta-3=\frac{1}{3}$
$\Rightarrow \cos ^{2} \theta=\frac{10}{12}$
So, $\sin ^{2} \theta=\frac{2}{12}$
$\tan 3 \theta\cot \theta=\frac{8-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}=\frac{3-\frac{1}{5}}{1-\frac{3}{5}}=7$
So we got, $\cot 3 \theta \tan \theta=\frac{1}{7}$ (D)
The locus of points $z$ in the complex plane satisfying $z^{2}+|z|^{2}=0$ is
(A) a straight line
(B) a pair of straight lines
(C) a circle
(D) a parabola
Solution
Amongst all polynomials $p(x)=c_{0}+c_{1} x+\cdots+c_{10} x^{10}$ with real coefficients satisfying $|p(x)| \leq|x|$ for all $x \in[-1,1]$, what is the maximum possible value of $\left(2 c_{0}+c_{1}\right)^{10}$ ?
(A) $4^{10}$
(B) $3^{10}$
(C) $2^{10}$
(D) 1
Solution
Let $\mathbb{Z}$ denote the set of integers. Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be such that $f(x) f(y)=f(x+y)+f(x-y)$ for all $x, y \in \mathbb{Z}$. If $f(1)=3$, then $f(7)$ equals
(A) 840
(B) 844
(C) 843 <— Answer
(D) 842
Solution
$f:\mathbb{Z} \rightarrow \mathbb{Z}$
$f(x) f(y)=f(x+y)+f(x-y)$
$f(1) f(0)=f(1)+f(1)$
$\Rightarrow 3 f(0)=f(1)+f(3)$
$\Rightarrow 3 f(0)=6$
$\Rightarrow f(0)=2$
Now, $f(1) f(1)=f(2)+f(0)$
$\Rightarrow 9=f(2)+2$
$\Rightarrow f(2)=7$
Again, $f(2) f(1)=f(3)+f(1)$
$\Rightarrow 7 \times 3=f(3)+3$
$\Rightarrow f(3)=18$
Again, $f(3) f(1)=f(4)+f(2)$
$\Rightarrow 18 \times 3=f(4)+7$
$\Rightarrow f(4)=47$
$f(3) f(4)=f(4+3)+f(1)$
$\Rightarrow (18 \times 47)-3=f(7)$
$\Rightarrow f(3)=843$ (C)
Let $A$ and $B$ be two $3 \times 3$ matrices such that $(A+B)^{2}=A^{2}+B^{2}$. Which of the following must be true?
(A) $A$ and $B$ are zero matrices.
(B) $A B$ is the zero matrix.
(C) $(A-B)^{2}=A^{2}-B^{2}$
(D) $(A-B)^{2}=A^{2}+B^{2}$ <— Answer
Solution
$(A+B)(A+B)$
$=A^{2}+B A+A B+B^{2}$
So, $B A+A B=0$
$(A-B)(A-B)$
$=A^{2}-A B-B A+B^{2}$
$=A^{2}+B^{2}$ (D)
Let $\left(n_{1}, n_{2}, \cdots, n_{12}\right)$ be a permutation of the numbers $1,2, \cdots, 12$. The number of arrangements with
$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}$
and $n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$ equals:
(A) $^{12} C_{5}$
(B) $^{12} C_{6}$
(C) $^{11} C_{6}$
(D) $\frac{11!}{2}$
Solution
$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$
$n_{6}$ is the smallest so,$n_{6}=1$
So, if we just choose first 5 it is enough.
So, $^{11} C_{5}$ as $n_{6}$ is fixed
$= ^{11} C_{6}$ (C)
The sides of a regular hexagon $A B C D E F$ is extended by doubling them to form a bigger hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ as in the figure below.
Then the ratio of the areas of the bigger to the smaller hexagon is:
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) 4
In how many ways can we choose $a_{1}<a_{2}<a_{3}<a_{4}$ from the set ${1,2, \ldots, 30}$ such that $a_{1}, a_{2}, a_{3}, a_{4}$ are in arithmetic progression?
(A) 135
(B) 145
(C) 155
(D) 165
Suppose the numbers 71,104 and 159 leave the same remainder $r$ when divided by a certain number $N>1$. Then, the value of $3 N+4 r$ must equal:
(A) 53
(B) 48
(C) 37
(D) 23
If $x, y$ are positive real numbers such that $3 x+4 y<72$, then the maximum possible value of $12 x y(72-3 x-4 y)$ is:
(A) 12240
(D) 13824
(C) 10656
(D) 8640
What is the minimum value of the function $|x-3|+|x+2|+|x+1|+|x|$ for real $x$ ?
(A) 3
(B) 5
(C) 6
(D) 8
Consider a differentiable function $u:[0,1] \rightarrow \mathbb{R}$. Assume the function $u$ satisfies
$u(a)=\frac{1}{2 r} \int_{a-r}^{a+r} u(x) d x, \quad$ for all $a \in(0,1)$ and all $r<\min (a, 1-a)$.
(A) $u$ attains its maximum but not its minimum on the set ${0,1}$.
(B) $u$ attains its minimum but not maximum on the set ${0,1}$.
(C) If $u$ attains either its maximum or its minimum on the set ${0,1}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set ${0,1}$.
A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance $x$ below the ladders?
(A) 3
(B) $2 \sqrt{2}$
(C) $\frac{8}{3}$ <— Answer
(D) $2 \sqrt{3}$
Solution
$\frac{a}{a+b}=\frac{x}{4}$, in $\triangle A C D$
$\frac{b}{a+b}=\frac{x}{8}$, in $\triangle A B D$
So, from the previous relations we get, $a=2 b$.
So, $\frac{x}{4}=\frac{2 b}{3 b}$
$\Rightarrow x=\frac{8}{3}$ feet (C)
Let $y=x+c_{1}, y=x+c_{2}$ be the two tangents to the ellipse $x^{2}+4 y^{2}=1$. What is the value of $\left|c_{1}-c_{2}\right| ?$
(A) $\sqrt{2}$
(B) $\sqrt{5}$ <— Answer
(C) $\frac{\sqrt{5}}{2}$
(D) 1
Solution
$x^{2}+4 y^{2}=1 \rightarrow(1)$
We want $\frac{d y}{d x}=1$
So differentiating $(1)$ with respect of $x$ we get
$2 x+8 y \frac{d y}{d x}=0$
for $\frac{d y}{d y}$ to be 1 we get $x=-4y$
Substituting $(1)$
$16 y^{2}+ 4y^{2}=1$
$\Rightarrow y_{0}=\frac{1}{2 \sqrt{5}} \text{or}-\frac{1}{2 \sqrt{5}}$
Let $y_{0}=\frac{1}{2 \sqrt{5}}$,
$\Rightarrow x_{0}=\frac{2}{\sqrt{5}}$
$c_{2}=-\frac{\sqrt{5}}{2}$
Now, $|c_{1}-c_{2}|=\sqrt{5}$ (B)
In the figure below, $A B C D$ is a square and $\triangle C E F$ is a triangle with given sides inscribed as in the figure. Find the length $B E$.
(A) $\frac{13}{\sqrt{17}}$ <— Answer
(B) $\frac{14}{\sqrt{17}}$
(C) $\frac{15}{\sqrt{17}}$
(D) $\frac{16}{\sqrt{17}}$
Solution
$\cos \theta=\frac{a}{4}=\frac{A F}{3}$
$A F=\frac{3 a}{4}$
$\tan \theta=\frac{1}{4}=\frac{A E}{\frac{3 a}{4}}$
$\Rightarrow A E=\frac{3 a}{16}$
$a^{2}+\frac{a^{2}}{16}=16$
$\Rightarrow \quad \frac{17 a^{2}}{16}=16$
$\Rightarrow \quad a^{2}=\frac{16 \times 16}{17}$
$\Rightarrow \quad a=\frac{16}{\sqrt{17}}$
$B E =\frac{13 a}{16}$
$B E =\frac{13}{\sqrt{17}}$ (A)
Let $p$ and $q$ be two non-zero polynomials such that the degree of $p$ is less than or equal to the degree of $q$, and $p(a) q(a)=0$ for $a=0,1,2, \ldots, 10$. Which of the following must be true?
(A) degree of $q \neq 10$
(B) degree of $p \neq 10$
(C) degree of $q \neq 5$ <— Answer
(D) degree of $p \neq 5$
Solution
$p(a) q(a)=0$, for $a=0,1,2, \cdots, 10$
Hence the polynonial, $P(x) q(x)$ has at least 11 roots as so $deg(p(x) q(x)) \geq 11$
Now, if $deg(q(x))=5$, then $deg(p(x)) \leq 5 \Rightarrow deg(p(x) q(x)) \leq 10$ which is a contradiction.
Degree of $q \neq 5$
For $n \in \mathbb{N}$, let $a_{n}$ be defined
(A) equals 0 <— Answer
(B) equals $\frac{\pi}{4}$
(C) equals $\frac{\pi}{2}$
(D) does not exist
Solution
$a_{n}=\int_{0}^{n} \frac{1}{1+n x^{2}} d x$
$=\frac{1}{\sqrt{n}} \int_{0}^{n} \frac{1}{1+(\sqrt{n} x)^{2}} d(\sqrt{n} x)$
$=\frac{1}{\sqrt{n}}\left[\tan ^{-1}(\sqrt{n} x)\right]_{0}^{n}$
$a_{n}=\frac{1}{\sqrt{n}} \tan ^{-1}\left(n^{3 / 2}\right)$
$=\frac{1}{\sqrt{n}}\left(\frac{\pi}{2}-\cot ^{-1}\left(n^{3 / 2}\right)\right)$
$\Rightarrow a_{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{\sqrt{n}} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)$
$\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)=\frac{1}{n^{3 / 2}}$, as $\frac{1}{n^{3 / 2}} \rightarrow 0$
$\lim {n \rightarrow \infty} a{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{n^{2}}$
$\Rightarrow \lim {n \rightarrow \infty} a{n}=0$
A $3 \times 3$ magic square is a $3 \times 3$ rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square
(A) 90
(B) 96 <— Answer
(C) 94
(D) 99
Solution
Asrume the sum is $s$ and $x$ as shown. Fill the rest of the square setting the diagonals equal to $s$ we get.
$s-2x=40$
$s-x=68$
So we get that $s=96$
The number of positive integers $n$ less than or equal to 22 such that 7 divides $n^{5}+4 n^{4}+3 n^{3}+2022$ is
(A) 7
(B) 8
(C) 9
(D) 10 <— Answer
Solution
$n^{5}+4 n^{4}+3 n^{3}+2022$
$\left.\equiv n^{3}(n+1)(n+3)+6 (\bmod 7\right)$
checking $n \equiv \pm 0, \pm 1, \pm 2, \pm 3 (\bmod 7)$
Now we see that only $n=1, \pm 2$ is permitted
Thus, $n \in{1,2,5,8,9,12,15,16,10,22}$
In a class of 45 students, three students can write well using either hand. The number of students who can write well only with the right hand is 24 more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:
(A) 33
(B) 36 <— Answer
(C) 39
(D) 41
Solution
$R =$ Set of right handers
$L $= Set of left handers
$|R \cap L|=3, | R \cup L|=45$
$24+\left|R^{C} \cap L|=\right| R \cap L^{c}|=|\left(R^{c}\right)^{c} \cap L^{c} \mid \left|\left(R^{c} \cup L\right)^{c}\right|$
$=45-\mid R^{c} \cup L|$
$45-\left(\left|R^{c}\right|+|L|-| R^{c} \cap L \mid\right)$
$=\left|R^{c} \cap L\right|+\left(45-\left|R^{c}\right|\right)-|L|$
$=\left|R^{c} \cap L\right|+|R|-L \mid$
Thenefore, $|R|-|L|=24$
Moreover, $|R|+|L| =|R \cup L|+\mid R \cap L| =45+3=48$
Hence, $|R|=\frac{24+48}{2}=\frac{72}{2}=36$
Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
(B) $3^{10}$
(C) $2^{10} \omega$ <— Answer
(D) $3^{10} \omega^{2}$
Solution
$2 \equiv-1$ mod 3
$\Rightarrow 2^{k} \equiv(-1)^{k}$ mod 3
$\Rightarrow \omega^{2 k}=\omega^{(-1)^{k}}$
$\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1-\omega^{2}+\omega^{2^{{2}}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left(1-\omega^{2^{2}}+ \omega^{2^{3}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1-\omega^{2^{3}}+\omega^{2^{4}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left(1- \omega^{2^{4}} + \omega^{2{^5}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1- \omega^{2^{5}}+\omega^{2^{6}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left.(1-\ \omega^{2^{6}}+\omega^{2^{7}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1-\omega^{2^{7}}+\omega^{2^{8}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left(1-\omega^{2^{8}}+\omega^{2^{9}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1- \omega^{2^{9}}+\omega^{2^{10}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
Product $(-1)^{10} 2^{10} \cdot \omega^{10+5}=2^{10}$
The function $x^{2} \log {e} x$ in the interval $(0,2)$
(A) exactly one point of local maximum and no points of local minimum.
(B) exactly one point of local minimum and no points of local maximum. <— Answer
(C) points of local maximum as well as local minimum.
(D) neither a point of local maximum nor a point of local minimum.
Solution
$f(x) =x^{2} \log {e} x$ $\Rightarrow f^{\prime}(x)=x+2 x \log {e} x=x(1+2 \log _{e}^{x})$
The number of triples $(a, b, c)$ of positive integers satisfying the equation $(2,3,4) \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{2}{a b c}$
and such that $a<b<c$, equals:
(A) 3
(B) 2
(C) 1 <— Answer
(D) 0
Solution
$\Rightarrow a(b+c)+b c=a b c+2$
$\Rightarrow a(b+c-b c)=2-b c$
$\Rightarrow a(b c-b-c)=b c-2$
$b c-b-c \mid b c-2$
$\Rightarrow b c-b-c \mid b+c-2$
$b c-b-c \leq b+c-2$
$\Rightarrow b c-2 b-2 c \leq-2$
$\Rightarrow(b-2)(c-2) \leq 2$
$b-2=1$, $c-2=2$
So, $b=3, c=4$
So, $a=2$ (C)
An urn contains 30 balls out of which one is special. If 6 of these balls are taken out at random, what is the probability that the special ball is chosen?
(A) $\frac{1}{30}$
(B) $\frac{1}{6}$
(C) $\frac{1}{5}$ <— Answer
(D) $\frac{1}{15}$
Solution
A triangle has sides of lengths $\sqrt{5}, 2 \sqrt{2}, \sqrt{3}$ units. Then, the radius of its inscribed circle is :
(A) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{2}$
(B) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{3}$
(C) $\sqrt{5}+\sqrt{3}+2 \sqrt{2}$
(D) $\frac{\sqrt{5}+\sqrt{3}-2 \sqrt{2}}{2}$ <— Answer
Solution
$r_{\Delta}=\frac{2 \times {\text {Area in }} \Delta \triangle A B C}{\text { perimeter of } \triangle A B C}$
$\Rightarrow r_{\Delta}=2 \times \frac{\frac{1}{2} \times \sqrt{3} \times \sqrt{5}}{\sqrt{3}+\sqrt{5}+2 \sqrt{2}}$
$\Rightarrow r_{\Delta}=\frac{\sqrt{3} \times \sqrt{5}(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{(\sqrt{3}+\sqrt{5})^{2}-(2 \sqrt{2})^{2}}$
$\frac{\sqrt{3} \times \sqrt{5} \times(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{2 \times \sqrt{3} \times \sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}-2 \sqrt{2}}{2}(D)$
Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled $80 \mathrm{~km}$, the triangle became right-angled.
When the first ship reaches the port, the second ship was still $120 \mathrm{~km}$ from the port. Find the initial distance of the ships from the port.
(A) $240 \mathrm{~km}$ <— Answer
(B) $300 \mathrm{~km}$
(C) $360 \mathrm{~km}$
(D) $180 \mathrm{~km}$
Solution
Let $P=Port$
$S_{1}=$ Ship 1
$S_{2}=$ Ship 2
When $S_{1}$ travelled $x$ km, $S_{2}$ travelled $80 km$ \& when $S_{1}$ travelled $d$ km, $S_{2}$ travelled $d-120 km$. Since they we moving at constant velocites.
$\frac{x}{80}=\frac{d}{d-120} \Rightarrow d-x=\frac{d(d-200)}{(d-120)}$
$d-x =(d-80) \cos 60^{\circ}$
$=\frac{1}{2}(d-80)$
So, $\frac{d(d-200)}{(d-120)}=\frac{1}{2}(d-80)$
$d^{2}-200 d-120 \times 80=0 \Rightarrow(d-240)(d+40)=0$
So, $d=240$ (A)
If $x_{1}>x_{2}>\cdots>x_{10}$ are real numbers, what is the least possible value of $\left(\frac{x_{1}-x_{10}}{x_{1}-x_{2}}\right)\left(\frac{x_{1}-x_{10}}{x_{2}-x_{3}}\right) \cdots\left(\frac{x_{1}-x_{10}}{x_{9}-x_{10}}\right)$
(A) $10^{10}$
(B) $10^{9}$
(C) $9^{9}$
(D) $9^{10}$
Solution
$x_{1}>x_{2}>\ldots>x_{10} \in \mathbb{R}$
$x_{i}-x_{j}>0 \quad \forall 1 \leq j < i \leq 10$
$(x_{1}-x_{2})+(x_{2}-x_{3})+\ldots+(x_{9}-x_{10})=(x_{1}-x_{10})$
By AM-GM Inequality we get, $[(x_{1}-x_{2})+\cdots+(x_{4}-x_{0})]^{9} \geq 9^{9}(x_{1}-x_{2})(x_{2}-x_{3}) \cdots(x_{9}-x_{10})$
$(x_{1}-x_{10})^{9} \geq 9^{9}(x_{1}-x_{2}) \ldots(x_{1}-x_{10})$
$\frac{(x_{1}-x_{10})}{(x_{1}-x_{2})} \times \frac{(x_{1}-x_{10})}{(x_{2}-x_{3})} \cdots \frac{(x_{1}-x_{6})}{(x_{1}-x_{10})} \geq 9^{9}$
Equality is achieved when $x_{i}$ are in AP eg $x_{i}=i$(C)
The range of values that the function
Solution
$f(x)=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$
$=\frac{1}{2}-\frac{1}{2\left(2 x^{2}+4 x+9\right)}$
$=\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]}$
$2(x+1)^{2}+7 \geq 7$
$-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq-\frac{1}{14}$
$\frac{1}{2}>\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq \frac{1}{2}-\frac{1}{14}=\frac{3}{7}$
In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area $x$ of the remaining triangle.
(B) 13
(C) 14
orms a squole of
(A) 12
(D) 15
Note. In this question-paper, $\mathbb{R}$ denotes the set of real numbers.
Consider a board having 2 rows and $n$ columns. Thus there are $2 n$ cells in the board. Each cell is to be filled in by 0 or 1 .
(a) In how many ways can this be done such that each row sum and each column sum is even?
(b) In how many ways can this be done such that each row sum and each column sum is odd?
Solution
Consider the function
Solution
Consider the parabola $C: y^{2}=4 x$ and the straight line $L$ : $y=x+2$. Let $P$ be a variable point on $L$. Draw the two tangents from $P$ to $C$ and let $Q_{1}$ and $Q_{2}$ denote the two points of contact on $C$. Let $Q$ be the mid-point of the line segment joining $Q_{1}$ and $Q_{2}$. Find the locus of $Q$ as $P$ moves along $L$.
Solution
Let $P(x)$ be an odd degree polynomial in $x$ with real coefficients. Show that the equation $P(P(x))=0$ has at least as many distinct real roots as the equation $P(x)=0$.
Solution
Since $P(x)$ and $P(P(x))$ both are odd degree polynomial it must have at least one real root.
Suppose, the distinct real roots of $P(x)$ be $\alpha_{1}, \alpha_{2,}, \ldots, \alpha_{k}$
Roots of $P(P(x))$ happens for all $x$ such that $p(x)=\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$
$P(P(x))=\left(P(x)-\alpha_{1}\right)\left(P(x)-\alpha_{2}\right)\left(P(x)-\alpha_{3}\right) \cdots\left(P(x)-\alpha_{k}\right)Q(x)$
Now, $P(x)=\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are $k$ different odd degree polynomials therefore there shoul we at least $K$ distinct real roots of $P(P(x))$. They should be distinct because $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are themselves distinct.
For any positive integer $n$, and $i=1,2$, let $f_{i}(n)$ denote the number of divisors of $n$ of the form $3 k+i$ (including 1 and $n$ ). Define, for any positive integer $n$.
Solution
(i) Factors of $5^{2022}$ :
$1,5^{1}, 5^{2}, 5^{3}, \ldots, 5^{2021}, 5^{2022}$
$5^{\text {odd }}\equiv (-1)^{\text {odd }}(\bmod 3)$
$\quad\quad\equiv 2(\bmod 3)$
$5^{\text {even }} \equiv(-1)^{\text {even }} \equiv 1 (\bmod 3)$
$f_{1}\left(5^{2022}\right)=1012$
$f_{2}\left(5^{2022}\right)=1011$
$f\left(5^{2022}\right)=1$
(ii) $21^{2022}=3^{2022} \times 7^{2022}$
factors without 3 as a factor are only considered:
$1,7^{1}, 7^{2}, 7^{3}, \cdots, 7^{2021}, 7^{2022}$
$7^{k} \equiv 1^{k} \equiv 1(\bmod 3) \forall k \in N_{0}$
$f_{1}\left(21^{2022}\right)=2023$
$f_{2}\left(21^{2022}\right)=0$
$f\left(2^{2022}\right)=2023$
Consider a sequence $P_{1}, P_{2}, \ldots$ of points in the plane such that $P_{1}, P_{2}, P_{3}$ are non-collinear and for every $n \geq 4, P_{n}$ is the midpoint of the line segment joining $P_{n-2}$ and $P_{n-3}$. Let $L$ denote the line segment joining $P_{1}$ and $P_{5}$. Prove the following:
(a) The area of the triangle formed by the points $P_{n}, P_{n-1,} P_{n-2}$ converges to zero as $n$ goes to infinity.
(b) The point $P_{9}$ lies on $L$.
Solution
(a) Let us denole the complex no. connespida to $P_{n}$ by $p_{n}$.
Then
satisfies the recurrence relation
$2 p_{n}=p_{n-2}+p_{n-3}, n \geq 4 $
It suffices to find three independent solutions to $2 z_{n}=z_{n-2}+z_{n-3}, n \geq 4$ to find the genenal solution. Let's try to find solutions of the form $z_{n}=\lambda^{n}$.
$2 \lambda^{n}=\lambda^{n-2}+\lambda^{n-3}$
$\Rightarrow 2 \lambda^{3}=\lambda+1$
$\Rightarrow 2 \lambda^{3}-2 \lambda^{2}+2 \lambda^{2}-2 \lambda+\lambda-1=0$
$\Rightarrow (\lambda-1)\left(2 \lambda^{2}+2 \lambda+1\right)=0$
$\Rightarrow \lambda=1, \frac{-2 \pm \sqrt{4-8}}{4}$
$=1,-\frac{1}{2} \pm 1 / 2$ Let $\varepsilon=e^\frac{i2 \pi}{8}$
So the roots are $1,\frac{\varepsilon^{3}}{\sqrt{2}} \frac{\varepsilon^{5}}{\sqrt{2}}$
Thenefone, the general sol is
$$
Z_{n}=A+B \frac{\varepsilon^{3 n}}{(\sqrt{2})^{n}}+C \frac{\varepsilon^{5 n}}{(\sqrt{2})^{n}}
$$
The (complex) constants can be adjusted to make
$$
z_{1}=p_{1}, z_{2}=p_{2}, z_{3}=p_{3} .
$$
As $n \rightarrow \infty$
$$
\left|\left(\frac{\varepsilon^{3}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0, \left|\left(\frac{\varepsilon^{5}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0
$$
Thus $Z n \rightarrow A$
Thenefore, eventually all points are contained in a ball of radius $\varepsilon$ about $A$, no matter how small $\varepsilon>0$ might be. Thus, the anea of $z_{n} z_{n-1} z_{n-2}$ must go to zero.
(b) $p_{9}=\frac{p_{7}+p_{6}}{2}$
$=\frac{\frac{1}{2}\left(p_{5}+p_{4}\right)+\frac{1}{2}\left(p_{4}+p_{3}\right)}{2}$
$=\frac{2 p_{4}+p_{5}+p_{3}}{4}$
$=\frac{p_{2}+p_{1}+p_{5}+p_{3}}{4}$
$=\frac{p_{1}+p_{5}+\left(p_{2}+p_{3}\right)}{4}$
$=\frac{p_{1}+p_{5}+2 p_{5}}{4}$
$=\frac{p_{1}+3 p_{5}}{4}$
7. Let
Find the minimum value of
Find the smallest positive real number $k$ such that the following inequality holds
This is a work in progress. Please come back for the solutions. You can also suggest your solutions in the comment section
Answer Key
| Problem 1 -> A | Problem 7 -> C | 13. | Problem 19 -> B | Problem 25 -> C |
| Problem 2 -> D | 8. | Problem 14 -> C | Problem 20 -> D | Problem 26 -> D |
| Problem 3 -> | 9. | Problem 15 -> B | Problem 21 -> B | Problem 27 -> A |
| Problem 4 -> | 10. | Problem 16 -> A | Problem 22 -> C | Problem 28 -> |
| Problem 5 -> C | 11. | Problem 17 -> C | Problem 23 -> B | Problem 29 -> B |
| Problem 6 -> D | 12 | Problem 18 -> A | Problem 24 -> C | Problem 30 -> |
Any positive real number $x$ can be expanded as
$x=a_{n} \cdot 2^{n}+a_{n-1} \cdot 2^{n-1}+ \\ \cdots+a_{1} \cdot 2^{1}+a_{0} \cdot 2^{0}+a_{-1} \\ \cdot 2^{-1}+a_{-2} \cdot 2^{-2}+\cdots$
for some $n \geq 0$, where each $a_{i} \in{0,1}$. In the above-described expansion of $21.1875$, the smallest positive integer $k$ such that $a_{-k} \neq 0$ is:
(A) 3 <—Answer
(B) 2
(C) 1
(D) 4
Solution
$x=a_{n} 2^{n}+a_{n-1} 2^{n-1}+\cdots+a_{1} 2^{1}+a_{1}$
Binary represation of $21.1875$ is $10101.0011$ (very standard Method to Convert Dercimal to Binary)
So, $-k=-3$
So, we got $k=3$ (A)
Suppose, for some $\theta \in\left[0, \frac{\pi}{2}\right], \frac{\cos 3 \theta}{\cos \theta}=\frac{1}{3}$. Then $(\cot 3 \theta) \tan \theta$ equals
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{8}$
(D) $\frac{1}{7}$ <— Answer
Solution
$\frac{\cos 3 \theta}{ \cos \theta}=\frac{1}{3}$
$\Rightarrow \frac{4 \cos^{ 3} \theta-3 \cos \theta}{\cos \theta}=\frac{1}{3}$
$\Rightarrow 4 \cos ^{2} \theta-3=\frac{1}{3}$
$\Rightarrow \cos ^{2} \theta=\frac{10}{12}$
So, $\sin ^{2} \theta=\frac{2}{12}$
$\tan 3 \theta\cot \theta=\frac{8-\tan ^{2} \theta}{1-3 \tan ^{2} \theta}=\frac{3-\frac{1}{5}}{1-\frac{3}{5}}=7$
So we got, $\cot 3 \theta \tan \theta=\frac{1}{7}$ (D)
The locus of points $z$ in the complex plane satisfying $z^{2}+|z|^{2}=0$ is
(A) a straight line
(B) a pair of straight lines
(C) a circle
(D) a parabola
Solution
Amongst all polynomials $p(x)=c_{0}+c_{1} x+\cdots+c_{10} x^{10}$ with real coefficients satisfying $|p(x)| \leq|x|$ for all $x \in[-1,1]$, what is the maximum possible value of $\left(2 c_{0}+c_{1}\right)^{10}$ ?
(A) $4^{10}$
(B) $3^{10}$
(C) $2^{10}$
(D) 1
Solution
Let $\mathbb{Z}$ denote the set of integers. Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be such that $f(x) f(y)=f(x+y)+f(x-y)$ for all $x, y \in \mathbb{Z}$. If $f(1)=3$, then $f(7)$ equals
(A) 840
(B) 844
(C) 843 <— Answer
(D) 842
Solution
$f:\mathbb{Z} \rightarrow \mathbb{Z}$
$f(x) f(y)=f(x+y)+f(x-y)$
$f(1) f(0)=f(1)+f(1)$
$\Rightarrow 3 f(0)=f(1)+f(3)$
$\Rightarrow 3 f(0)=6$
$\Rightarrow f(0)=2$
Now, $f(1) f(1)=f(2)+f(0)$
$\Rightarrow 9=f(2)+2$
$\Rightarrow f(2)=7$
Again, $f(2) f(1)=f(3)+f(1)$
$\Rightarrow 7 \times 3=f(3)+3$
$\Rightarrow f(3)=18$
Again, $f(3) f(1)=f(4)+f(2)$
$\Rightarrow 18 \times 3=f(4)+7$
$\Rightarrow f(4)=47$
$f(3) f(4)=f(4+3)+f(1)$
$\Rightarrow (18 \times 47)-3=f(7)$
$\Rightarrow f(3)=843$ (C)
Let $A$ and $B$ be two $3 \times 3$ matrices such that $(A+B)^{2}=A^{2}+B^{2}$. Which of the following must be true?
(A) $A$ and $B$ are zero matrices.
(B) $A B$ is the zero matrix.
(C) $(A-B)^{2}=A^{2}-B^{2}$
(D) $(A-B)^{2}=A^{2}+B^{2}$ <— Answer
Solution
$(A+B)(A+B)$
$=A^{2}+B A+A B+B^{2}$
So, $B A+A B=0$
$(A-B)(A-B)$
$=A^{2}-A B-B A+B^{2}$
$=A^{2}+B^{2}$ (D)
Let $\left(n_{1}, n_{2}, \cdots, n_{12}\right)$ be a permutation of the numbers $1,2, \cdots, 12$. The number of arrangements with
$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}$
and $n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$ equals:
(A) $^{12} C_{5}$
(B) $^{12} C_{6}$
(C) $^{11} C_{6}$
(D) $\frac{11!}{2}$
Solution
$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$
$n_{6}$ is the smallest so,$n_{6}=1$
So, if we just choose first 5 it is enough.
So, $^{11} C_{5}$ as $n_{6}$ is fixed
$= ^{11} C_{6}$ (C)
The sides of a regular hexagon $A B C D E F$ is extended by doubling them to form a bigger hexagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime} F^{\prime}$ as in the figure below.
Then the ratio of the areas of the bigger to the smaller hexagon is:
(A) $\sqrt{3}$
(B) 3
(C) $2 \sqrt{3}$
(D) 4
In how many ways can we choose $a_{1}<a_{2}<a_{3}<a_{4}$ from the set ${1,2, \ldots, 30}$ such that $a_{1}, a_{2}, a_{3}, a_{4}$ are in arithmetic progression?
(A) 135
(B) 145
(C) 155
(D) 165
Suppose the numbers 71,104 and 159 leave the same remainder $r$ when divided by a certain number $N>1$. Then, the value of $3 N+4 r$ must equal:
(A) 53
(B) 48
(C) 37
(D) 23
If $x, y$ are positive real numbers such that $3 x+4 y<72$, then the maximum possible value of $12 x y(72-3 x-4 y)$ is:
(A) 12240
(D) 13824
(C) 10656
(D) 8640
What is the minimum value of the function $|x-3|+|x+2|+|x+1|+|x|$ for real $x$ ?
(A) 3
(B) 5
(C) 6
(D) 8
Consider a differentiable function $u:[0,1] \rightarrow \mathbb{R}$. Assume the function $u$ satisfies
$u(a)=\frac{1}{2 r} \int_{a-r}^{a+r} u(x) d x, \quad$ for all $a \in(0,1)$ and all $r<\min (a, 1-a)$.
(A) $u$ attains its maximum but not its minimum on the set ${0,1}$.
(B) $u$ attains its minimum but not maximum on the set ${0,1}$.
(C) If $u$ attains either its maximum or its minimum on the set ${0,1}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set ${0,1}$.
A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance $x$ below the ladders?
(A) 3
(B) $2 \sqrt{2}$
(C) $\frac{8}{3}$ <— Answer
(D) $2 \sqrt{3}$
Solution
$\frac{a}{a+b}=\frac{x}{4}$, in $\triangle A C D$
$\frac{b}{a+b}=\frac{x}{8}$, in $\triangle A B D$
So, from the previous relations we get, $a=2 b$.
So, $\frac{x}{4}=\frac{2 b}{3 b}$
$\Rightarrow x=\frac{8}{3}$ feet (C)
Let $y=x+c_{1}, y=x+c_{2}$ be the two tangents to the ellipse $x^{2}+4 y^{2}=1$. What is the value of $\left|c_{1}-c_{2}\right| ?$
(A) $\sqrt{2}$
(B) $\sqrt{5}$ <— Answer
(C) $\frac{\sqrt{5}}{2}$
(D) 1
Solution
$x^{2}+4 y^{2}=1 \rightarrow(1)$
We want $\frac{d y}{d x}=1$
So differentiating $(1)$ with respect of $x$ we get
$2 x+8 y \frac{d y}{d x}=0$
for $\frac{d y}{d y}$ to be 1 we get $x=-4y$
Substituting $(1)$
$16 y^{2}+ 4y^{2}=1$
$\Rightarrow y_{0}=\frac{1}{2 \sqrt{5}} \text{or}-\frac{1}{2 \sqrt{5}}$
Let $y_{0}=\frac{1}{2 \sqrt{5}}$,
$\Rightarrow x_{0}=\frac{2}{\sqrt{5}}$
$c_{2}=-\frac{\sqrt{5}}{2}$
Now, $|c_{1}-c_{2}|=\sqrt{5}$ (B)
In the figure below, $A B C D$ is a square and $\triangle C E F$ is a triangle with given sides inscribed as in the figure. Find the length $B E$.
(A) $\frac{13}{\sqrt{17}}$ <— Answer
(B) $\frac{14}{\sqrt{17}}$
(C) $\frac{15}{\sqrt{17}}$
(D) $\frac{16}{\sqrt{17}}$
Solution
$\cos \theta=\frac{a}{4}=\frac{A F}{3}$
$A F=\frac{3 a}{4}$
$\tan \theta=\frac{1}{4}=\frac{A E}{\frac{3 a}{4}}$
$\Rightarrow A E=\frac{3 a}{16}$
$a^{2}+\frac{a^{2}}{16}=16$
$\Rightarrow \quad \frac{17 a^{2}}{16}=16$
$\Rightarrow \quad a^{2}=\frac{16 \times 16}{17}$
$\Rightarrow \quad a=\frac{16}{\sqrt{17}}$
$B E =\frac{13 a}{16}$
$B E =\frac{13}{\sqrt{17}}$ (A)
Let $p$ and $q$ be two non-zero polynomials such that the degree of $p$ is less than or equal to the degree of $q$, and $p(a) q(a)=0$ for $a=0,1,2, \ldots, 10$. Which of the following must be true?
(A) degree of $q \neq 10$
(B) degree of $p \neq 10$
(C) degree of $q \neq 5$ <— Answer
(D) degree of $p \neq 5$
Solution
$p(a) q(a)=0$, for $a=0,1,2, \cdots, 10$
Hence the polynonial, $P(x) q(x)$ has at least 11 roots as so $deg(p(x) q(x)) \geq 11$
Now, if $deg(q(x))=5$, then $deg(p(x)) \leq 5 \Rightarrow deg(p(x) q(x)) \leq 10$ which is a contradiction.
Degree of $q \neq 5$
For $n \in \mathbb{N}$, let $a_{n}$ be defined
(A) equals 0 <— Answer
(B) equals $\frac{\pi}{4}$
(C) equals $\frac{\pi}{2}$
(D) does not exist
Solution
$a_{n}=\int_{0}^{n} \frac{1}{1+n x^{2}} d x$
$=\frac{1}{\sqrt{n}} \int_{0}^{n} \frac{1}{1+(\sqrt{n} x)^{2}} d(\sqrt{n} x)$
$=\frac{1}{\sqrt{n}}\left[\tan ^{-1}(\sqrt{n} x)\right]_{0}^{n}$
$a_{n}=\frac{1}{\sqrt{n}} \tan ^{-1}\left(n^{3 / 2}\right)$
$=\frac{1}{\sqrt{n}}\left(\frac{\pi}{2}-\cot ^{-1}\left(n^{3 / 2}\right)\right)$
$\Rightarrow a_{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{\sqrt{n}} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)$
$\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{1}{n^{3 / 2}}\right)=\frac{1}{n^{3 / 2}}$, as $\frac{1}{n^{3 / 2}} \rightarrow 0$
$\lim {n \rightarrow \infty} a{n}=\frac{\pi}{2 \sqrt{n}}-\frac{1}{n^{2}}$
$\Rightarrow \lim {n \rightarrow \infty} a{n}=0$
A $3 \times 3$ magic square is a $3 \times 3$ rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square
(A) 90
(B) 96 <— Answer
(C) 94
(D) 99
Solution
Asrume the sum is $s$ and $x$ as shown. Fill the rest of the square setting the diagonals equal to $s$ we get.
$s-2x=40$
$s-x=68$
So we get that $s=96$
The number of positive integers $n$ less than or equal to 22 such that 7 divides $n^{5}+4 n^{4}+3 n^{3}+2022$ is
(A) 7
(B) 8
(C) 9
(D) 10 <— Answer
Solution
$n^{5}+4 n^{4}+3 n^{3}+2022$
$\left.\equiv n^{3}(n+1)(n+3)+6 (\bmod 7\right)$
checking $n \equiv \pm 0, \pm 1, \pm 2, \pm 3 (\bmod 7)$
Now we see that only $n=1, \pm 2$ is permitted
Thus, $n \in{1,2,5,8,9,12,15,16,10,22}$
In a class of 45 students, three students can write well using either hand. The number of students who can write well only with the right hand is 24 more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:
(A) 33
(B) 36 <— Answer
(C) 39
(D) 41
Solution
$R =$ Set of right handers
$L $= Set of left handers
$|R \cap L|=3, | R \cup L|=45$
$24+\left|R^{C} \cap L|=\right| R \cap L^{c}|=|\left(R^{c}\right)^{c} \cap L^{c} \mid \left|\left(R^{c} \cup L\right)^{c}\right|$
$=45-\mid R^{c} \cup L|$
$45-\left(\left|R^{c}\right|+|L|-| R^{c} \cap L \mid\right)$
$=\left|R^{c} \cap L\right|+\left(45-\left|R^{c}\right|\right)-|L|$
$=\left|R^{c} \cap L\right|+|R|-L \mid$
Thenefore, $|R|-|L|=24$
Moreover, $|R|+|L| =|R \cup L|+\mid R \cap L| =45+3=48$
Hence, $|R|=\frac{24+48}{2}=\frac{72}{2}=36$
Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
(B) $3^{10}$
(C) $2^{10} \omega$ <— Answer
(D) $3^{10} \omega^{2}$
Solution
$2 \equiv-1$ mod 3
$\Rightarrow 2^{k} \equiv(-1)^{k}$ mod 3
$\Rightarrow \omega^{2 k}=\omega^{(-1)^{k}}$
$\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1-\omega^{2}+\omega^{2^{{2}}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left(1-\omega^{2^{2}}+ \omega^{2^{3}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1-\omega^{2^{3}}+\omega^{2^{4}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left(1- \omega^{2^{4}} + \omega^{2{^5}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1- \omega^{2^{5}}+\omega^{2^{6}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left.(1-\ \omega^{2^{6}}+\omega^{2^{7}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1-\omega^{2^{7}}+\omega^{2^{8}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
$\left(1-\omega^{2^{8}}+\omega^{2^{9}}\right)=\left(1-\omega+\omega^{2}\right)=-2 \omega$
$\left(1- \omega^{2^{9}}+\omega^{2^{10}}\right)=\left(1-\omega^{2}+\omega\right)=-2 \omega^{2}$
Product $(-1)^{10} 2^{10} \cdot \omega^{10+5}=2^{10}$
The function $x^{2} \log {e} x$ in the interval $(0,2)$
(A) exactly one point of local maximum and no points of local minimum.
(B) exactly one point of local minimum and no points of local maximum. <— Answer
(C) points of local maximum as well as local minimum.
(D) neither a point of local maximum nor a point of local minimum.
Solution
$f(x) =x^{2} \log {e} x$ $\Rightarrow f^{\prime}(x)=x+2 x \log {e} x=x(1+2 \log _{e}^{x})$
The number of triples $(a, b, c)$ of positive integers satisfying the equation $(2,3,4) \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{2}{a b c}$
and such that $a<b<c$, equals:
(A) 3
(B) 2
(C) 1 <— Answer
(D) 0
Solution
$\Rightarrow a(b+c)+b c=a b c+2$
$\Rightarrow a(b+c-b c)=2-b c$
$\Rightarrow a(b c-b-c)=b c-2$
$b c-b-c \mid b c-2$
$\Rightarrow b c-b-c \mid b+c-2$
$b c-b-c \leq b+c-2$
$\Rightarrow b c-2 b-2 c \leq-2$
$\Rightarrow(b-2)(c-2) \leq 2$
$b-2=1$, $c-2=2$
So, $b=3, c=4$
So, $a=2$ (C)
An urn contains 30 balls out of which one is special. If 6 of these balls are taken out at random, what is the probability that the special ball is chosen?
(A) $\frac{1}{30}$
(B) $\frac{1}{6}$
(C) $\frac{1}{5}$ <— Answer
(D) $\frac{1}{15}$
Solution
A triangle has sides of lengths $\sqrt{5}, 2 \sqrt{2}, \sqrt{3}$ units. Then, the radius of its inscribed circle is :
(A) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{2}$
(B) $\frac{\sqrt{5}+\sqrt{3}+2 \sqrt{2}}{3}$
(C) $\sqrt{5}+\sqrt{3}+2 \sqrt{2}$
(D) $\frac{\sqrt{5}+\sqrt{3}-2 \sqrt{2}}{2}$ <— Answer
Solution
$r_{\Delta}=\frac{2 \times {\text {Area in }} \Delta \triangle A B C}{\text { perimeter of } \triangle A B C}$
$\Rightarrow r_{\Delta}=2 \times \frac{\frac{1}{2} \times \sqrt{3} \times \sqrt{5}}{\sqrt{3}+\sqrt{5}+2 \sqrt{2}}$
$\Rightarrow r_{\Delta}=\frac{\sqrt{3} \times \sqrt{5}(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{(\sqrt{3}+\sqrt{5})^{2}-(2 \sqrt{2})^{2}}$
$\frac{\sqrt{3} \times \sqrt{5} \times(\sqrt{3}+\sqrt{5}-2 \sqrt{2})}{2 \times \sqrt{3} \times \sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}-2 \sqrt{2}}{2}(D)$
Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled $80 \mathrm{~km}$, the triangle became right-angled.
When the first ship reaches the port, the second ship was still $120 \mathrm{~km}$ from the port. Find the initial distance of the ships from the port.
(A) $240 \mathrm{~km}$ <— Answer
(B) $300 \mathrm{~km}$
(C) $360 \mathrm{~km}$
(D) $180 \mathrm{~km}$
Solution
Let $P=Port$
$S_{1}=$ Ship 1
$S_{2}=$ Ship 2
When $S_{1}$ travelled $x$ km, $S_{2}$ travelled $80 km$ \& when $S_{1}$ travelled $d$ km, $S_{2}$ travelled $d-120 km$. Since they we moving at constant velocites.
$\frac{x}{80}=\frac{d}{d-120} \Rightarrow d-x=\frac{d(d-200)}{(d-120)}$
$d-x =(d-80) \cos 60^{\circ}$
$=\frac{1}{2}(d-80)$
So, $\frac{d(d-200)}{(d-120)}=\frac{1}{2}(d-80)$
$d^{2}-200 d-120 \times 80=0 \Rightarrow(d-240)(d+40)=0$
So, $d=240$ (A)
If $x_{1}>x_{2}>\cdots>x_{10}$ are real numbers, what is the least possible value of $\left(\frac{x_{1}-x_{10}}{x_{1}-x_{2}}\right)\left(\frac{x_{1}-x_{10}}{x_{2}-x_{3}}\right) \cdots\left(\frac{x_{1}-x_{10}}{x_{9}-x_{10}}\right)$
(A) $10^{10}$
(B) $10^{9}$
(C) $9^{9}$
(D) $9^{10}$
Solution
$x_{1}>x_{2}>\ldots>x_{10} \in \mathbb{R}$
$x_{i}-x_{j}>0 \quad \forall 1 \leq j < i \leq 10$
$(x_{1}-x_{2})+(x_{2}-x_{3})+\ldots+(x_{9}-x_{10})=(x_{1}-x_{10})$
By AM-GM Inequality we get, $[(x_{1}-x_{2})+\cdots+(x_{4}-x_{0})]^{9} \geq 9^{9}(x_{1}-x_{2})(x_{2}-x_{3}) \cdots(x_{9}-x_{10})$
$(x_{1}-x_{10})^{9} \geq 9^{9}(x_{1}-x_{2}) \ldots(x_{1}-x_{10})$
$\frac{(x_{1}-x_{10})}{(x_{1}-x_{2})} \times \frac{(x_{1}-x_{10})}{(x_{2}-x_{3})} \cdots \frac{(x_{1}-x_{6})}{(x_{1}-x_{10})} \geq 9^{9}$
Equality is achieved when $x_{i}$ are in AP eg $x_{i}=i$(C)
The range of values that the function
Solution
$f(x)=\frac{x^{2}+2 x+4}{2 x^{2}+4 x+9}$
$=\frac{1}{2}-\frac{1}{2\left(2 x^{2}+4 x+9\right)}$
$=\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]}$
$2(x+1)^{2}+7 \geq 7$
$-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq-\frac{1}{14}$
$\frac{1}{2}>\frac{1}{2}-\frac{1}{2\left[2(x+1)^{2}+7\right]} \geq \frac{1}{2}-\frac{1}{14}=\frac{3}{7}$
In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area $x$ of the remaining triangle.
(B) 13
(C) 14
orms a squole of
(A) 12
(D) 15
Note. In this question-paper, $\mathbb{R}$ denotes the set of real numbers.
Consider a board having 2 rows and $n$ columns. Thus there are $2 n$ cells in the board. Each cell is to be filled in by 0 or 1 .
(a) In how many ways can this be done such that each row sum and each column sum is even?
(b) In how many ways can this be done such that each row sum and each column sum is odd?
Solution
Consider the function
Solution
Consider the parabola $C: y^{2}=4 x$ and the straight line $L$ : $y=x+2$. Let $P$ be a variable point on $L$. Draw the two tangents from $P$ to $C$ and let $Q_{1}$ and $Q_{2}$ denote the two points of contact on $C$. Let $Q$ be the mid-point of the line segment joining $Q_{1}$ and $Q_{2}$. Find the locus of $Q$ as $P$ moves along $L$.
Solution
Let $P(x)$ be an odd degree polynomial in $x$ with real coefficients. Show that the equation $P(P(x))=0$ has at least as many distinct real roots as the equation $P(x)=0$.
Solution
Since $P(x)$ and $P(P(x))$ both are odd degree polynomial it must have at least one real root.
Suppose, the distinct real roots of $P(x)$ be $\alpha_{1}, \alpha_{2,}, \ldots, \alpha_{k}$
Roots of $P(P(x))$ happens for all $x$ such that $p(x)=\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}$
$P(P(x))=\left(P(x)-\alpha_{1}\right)\left(P(x)-\alpha_{2}\right)\left(P(x)-\alpha_{3}\right) \cdots\left(P(x)-\alpha_{k}\right)Q(x)$
Now, $P(x)=\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are $k$ different odd degree polynomials therefore there shoul we at least $K$ distinct real roots of $P(P(x))$. They should be distinct because $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are themselves distinct.
For any positive integer $n$, and $i=1,2$, let $f_{i}(n)$ denote the number of divisors of $n$ of the form $3 k+i$ (including 1 and $n$ ). Define, for any positive integer $n$.
Solution
(i) Factors of $5^{2022}$ :
$1,5^{1}, 5^{2}, 5^{3}, \ldots, 5^{2021}, 5^{2022}$
$5^{\text {odd }}\equiv (-1)^{\text {odd }}(\bmod 3)$
$\quad\quad\equiv 2(\bmod 3)$
$5^{\text {even }} \equiv(-1)^{\text {even }} \equiv 1 (\bmod 3)$
$f_{1}\left(5^{2022}\right)=1012$
$f_{2}\left(5^{2022}\right)=1011$
$f\left(5^{2022}\right)=1$
(ii) $21^{2022}=3^{2022} \times 7^{2022}$
factors without 3 as a factor are only considered:
$1,7^{1}, 7^{2}, 7^{3}, \cdots, 7^{2021}, 7^{2022}$
$7^{k} \equiv 1^{k} \equiv 1(\bmod 3) \forall k \in N_{0}$
$f_{1}\left(21^{2022}\right)=2023$
$f_{2}\left(21^{2022}\right)=0$
$f\left(2^{2022}\right)=2023$
Consider a sequence $P_{1}, P_{2}, \ldots$ of points in the plane such that $P_{1}, P_{2}, P_{3}$ are non-collinear and for every $n \geq 4, P_{n}$ is the midpoint of the line segment joining $P_{n-2}$ and $P_{n-3}$. Let $L$ denote the line segment joining $P_{1}$ and $P_{5}$. Prove the following:
(a) The area of the triangle formed by the points $P_{n}, P_{n-1,} P_{n-2}$ converges to zero as $n$ goes to infinity.
(b) The point $P_{9}$ lies on $L$.
Solution
(a) Let us denole the complex no. connespida to $P_{n}$ by $p_{n}$.
Then
satisfies the recurrence relation
$2 p_{n}=p_{n-2}+p_{n-3}, n \geq 4 $
It suffices to find three independent solutions to $2 z_{n}=z_{n-2}+z_{n-3}, n \geq 4$ to find the genenal solution. Let's try to find solutions of the form $z_{n}=\lambda^{n}$.
$2 \lambda^{n}=\lambda^{n-2}+\lambda^{n-3}$
$\Rightarrow 2 \lambda^{3}=\lambda+1$
$\Rightarrow 2 \lambda^{3}-2 \lambda^{2}+2 \lambda^{2}-2 \lambda+\lambda-1=0$
$\Rightarrow (\lambda-1)\left(2 \lambda^{2}+2 \lambda+1\right)=0$
$\Rightarrow \lambda=1, \frac{-2 \pm \sqrt{4-8}}{4}$
$=1,-\frac{1}{2} \pm 1 / 2$ Let $\varepsilon=e^\frac{i2 \pi}{8}$
So the roots are $1,\frac{\varepsilon^{3}}{\sqrt{2}} \frac{\varepsilon^{5}}{\sqrt{2}}$
Thenefone, the general sol is
$$
Z_{n}=A+B \frac{\varepsilon^{3 n}}{(\sqrt{2})^{n}}+C \frac{\varepsilon^{5 n}}{(\sqrt{2})^{n}}
$$
The (complex) constants can be adjusted to make
$$
z_{1}=p_{1}, z_{2}=p_{2}, z_{3}=p_{3} .
$$
As $n \rightarrow \infty$
$$
\left|\left(\frac{\varepsilon^{3}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0, \left|\left(\frac{\varepsilon^{5}}{\sqrt{2}}\right)^{n}\right| \rightarrow 0
$$
Thus $Z n \rightarrow A$
Thenefore, eventually all points are contained in a ball of radius $\varepsilon$ about $A$, no matter how small $\varepsilon>0$ might be. Thus, the anea of $z_{n} z_{n-1} z_{n-2}$ must go to zero.
(b) $p_{9}=\frac{p_{7}+p_{6}}{2}$
$=\frac{\frac{1}{2}\left(p_{5}+p_{4}\right)+\frac{1}{2}\left(p_{4}+p_{3}\right)}{2}$
$=\frac{2 p_{4}+p_{5}+p_{3}}{4}$
$=\frac{p_{2}+p_{1}+p_{5}+p_{3}}{4}$
$=\frac{p_{1}+p_{5}+\left(p_{2}+p_{3}\right)}{4}$
$=\frac{p_{1}+p_{5}+2 p_{5}}{4}$
$=\frac{p_{1}+3 p_{5}}{4}$
7. Let
Find the minimum value of
Find the smallest positive real number $k$ such that the following inequality holds
