AP EAPCET 2025 24th May Morning Shift Question Paper with Solutions

$𝐴=\{𝑥\in 𝑅/{\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right)\in [-\frac{𝜋}{2},\frac{𝜋}{2}]\}$

and

$𝐵=\{𝑦\in 𝑅/𝑦={\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right),𝑥\in 𝐴\}$

Since, domain of ${\sin }^{-1}\left(𝑥\right)$ is $[-1,1]$

So, domain of ${\sin }^{-1}\sqrt{\surd {𝑥}^2+𝑥+1}$ is real, so

so

$\begin{array}{cc}& \sqrt{\surd {𝑥}^2+𝑥+1}\in [-1,1]\\ \Rightarrow & \sqrt{\surd {𝑥}^2+𝑥+1}\in [0,1]\end{array}$

( $\because$ square root is always non-negative)

$\begin{array}{cc}\Rightarrow & {𝑥}^2+𝑥+1\le 1\\ \Rightarrow & {𝑥}^2+𝑥\le 0\\ \Rightarrow & 𝑥(𝑥+1)\le 0\end{array}$

So, $𝑥\in [-1,0]$ for set $𝐴$

Thus, $𝐴=[-1,0]$

Now, for set $𝐵$, since $𝑥\in 𝐴=[-1,0]$, the range values for $\sqrt{\surd {𝑥}^2+𝑥+1}$ for $𝑥\in [-1,0]$ we will find.

$\begin{array}{cc}\text{ At }𝑥& =-1,\sqrt{\surd (-1{)}^2-1+1}\\ & =\sqrt{\surd 1-1+1}=1\\ \text{ At }𝑥& =0,\sqrt{\surd 0^2+0+1}=\sqrt{\surd 0+1}=1\\ \text{ At }𝑥& =-\frac{1}{2},\sqrt{\surd {\left(\frac{-1}{2}\right)}^2-\frac{1}{2}+1}\\ & =\sqrt{\surd \frac{1}{4}-\frac{1}{2}+1}\\ & =\sqrt{\surd \frac{3}{4}}\end{array}$

So, $\sqrt{\surd {𝑥}^2+𝑥+1}\in [\sqrt{\surd \frac{3}{4}},1]=[\frac{\sqrt{\surd 3}}{2},1]$

Thus, $𝑦={\sin }^{-1}\left(\sqrt{\surd {𝑥}^2+𝑥+1}\right)$

$\begin{array}{cc}& \in [{\sin }^{-1}\left(\frac{\sqrt{\surd 3}}{2}\right),{\sin }^{-1}(1\left)\right]\\ & =[\frac{𝜋}{3},\frac{𝜋}{2}]\end{array}$

Hence, $𝐵=[\frac{𝜋}{3},\frac{𝜋}{2}]$

$\Rightarrow$ Now, $𝐴\cap 𝐵=[-1,0]\cap [\frac{𝜋}{3},\frac{𝜋}{2}]=𝜙$

So, $𝐴\cap 𝐵\ne 𝜙$ is false.

$\Rightarrow 𝐴\cap {𝐵}^{𝐶}=[-1,0]\cap {[\frac{𝜋}{3},\frac{𝜋}{2}]}^{𝐶}$ and the intersection with any set outside this cannot be $[0,1]$,

So, $𝐴\cap {𝐵}^{𝐶}=[0,1]$ is false.

$\Rightarrow {𝐴}^{𝐶}$ is everything except $[-1,0]$.

Since $𝐵=(𝜋/3,𝜋/2)\subset 𝑅/(-1,0)$

$\Rightarrow {𝐴}^{𝐶}\cap 𝐵$ is true.

$\Rightarrow 𝐴\cup 𝐵=𝑅-\left\{\right[-1,0]\cup [\frac{𝜋}{3},\frac{𝜋}{2}]\}$

So, union of $𝐴$ and $𝐵$ is the complement of their own union.

That's impossible.

Thus, ${𝐴}^{𝐶}\cap 𝐵=[\frac{𝜋}{3},\frac{𝜋}{2}]$, option (c) is correct.

Given function,

$𝑓\left(𝑥\right)=\sqrt{\surd {\log }_{𝑒}\left(\frac{1}{{𝑥}^2-4𝑥+4}\right)}+{\sin }^{-1}({𝑥}^2-2)$

Since, $\log$ is defined for positive values only.

So, $\frac{1}{{𝑥}^2-4𝑥+4}>0$

Also since, ${𝑥}^2-4𝑥+4=(𝑥-2{)}^2\ge 0$ this holds for $𝑥\ne 2$

$\Rightarrow \ln \left(\frac{1}{(𝑥-2{)}^2}\right)\ge 0$

Now, we know that $\ln \left(𝑡\right)$ is strictly increasing for $𝑡>0$ and $\ln \left(1\right)=0$, the inequality

$\ln \frac{1}{(𝑥-2{)}^2}\ge 0$ is equivalent to

$\frac{1}{(𝑥-2{)}^2}\ge 1\iff (𝑥-2{)}^2\le 1$

$\Rightarrow |𝑥-2|\le 1$

i.e., $1\le 𝑥\le 3$

But $\log$ is undefined for $𝑥=2$

$\therefore 𝑥\in [1,3]-\left\{2\right\}$

For ${\sin }^{-1}({𝑥}^2-2)$ to be real, ${𝑥}^2-2\in [-1,1]$

$\begin{array}{cc}\Rightarrow & -1\le {𝑥}^2-2\le 1\\ \Rightarrow & 1\le {𝑥}^2\le 3\Rightarrow 1\le \left|𝑥\right|\le \sqrt{\surd 3}\\ \Rightarrow & 𝑥\in [-\sqrt{\surd 3},-1]\cup [1,\sqrt{\surd 3}]\Rightarrow 𝑥\in [1,\sqrt{\surd 3}]\end{array}$

Given that $𝑛({𝑛}^2+3)$ is divisible by $𝑘$, for all $𝑛\in 𝑁$

For $𝑛=1,1(1^2+3)=1.4=4$ is true, when $𝑘$ is 2 and 4 .

Let assume for $𝑛=2,2(2^2+3)=14$ is true for 2 but not 4 .

$\therefore$ Maximum value of $𝑘$ is 2 .

Let $𝐴=⎡⎣⎢\begin{array}{ccc}1& 1& 2\\ 1& 2& 3\\ 2& 3& 3\end{array}⎤⎦⎥$ and $𝐵=⎡⎣⎢\begin{array}{ccc}1& 2& 3\\ 4& -3& -1\\ 2& 1& -4\end{array}⎤⎦⎥$

Also, $𝑎=\det (\mathrm{adj}(𝐴\left)\right)$ and $𝑏=\det \left({𝐵}^{-1}\right)$

Now, $𝑎=\det (\mathrm{adj}(𝐴\left)\right)=(\det (𝐴){)}^{𝑛-1}$ for an $𝑛\times 𝑛$ matrix.

Since $𝐴$ is $3\times 3$ matrix, so $𝑎=(\det (𝐴){)}^{3-1}=(\det \left(𝐴\right){)}^2$

$\begin{array}{cc}\det 𝐴& =1\cdot \mid \begin{array}{cc}2& 3\\ 3& 3\end{array}\mid -1\cdot \mid \begin{array}{cc}1& 3\\ 2& 3\end{array}\mid +2\cdot \mid \begin{array}{cc}1& 2\\ 2& 3\end{array}\mid \\ & =1\cdot (6-9)-1(3-6)+2(3-4)\\ & =-3+3-2=-2\\ 𝑎& =(\det (𝐴){)}^2=(-2{)}^2=4...\left(𝑖\right)\end{array}$

Now, $\det \left(𝐵\right)=\mid \begin{array}{ccc}1& 2& 3\\ 4& -3& -1\\ 2& 1& -4\end{array}\mid$

$\begin{array}{cc}& =1\cdot \mid \begin{array}{cc}-3& -1\\ 1& -4\end{array}\mid -2\mid \begin{array}{cc}4& -1\\ 2& -4\end{array}\mid +3\cdot \mid \begin{array}{cc}4& -3\\ 2& 1\end{array}\mid \\ & =(12+1)-2(-16+2)+3(4+6)\\ & =13+28+30=71\end{array}$

$\begin{array}{cc}& \text{ So, }𝑏=\det \left({𝐵}^{-1}\right)=\frac{1}{\det \left(𝐵\right)}=\frac{1}{71}\\ & \begin{array}{cc}& 𝑏=\frac{1}{71}\\ & \Rightarrow 𝑏+1=\frac{1+71}{71}=\frac{72}{71}\\ & \text{ Then, }\frac{𝑏+1}{18𝑏}=\frac{72/71}{18\times \frac{1}{71}}\\ & =\frac{72}{71}\times \frac{71}{18}=\frac{72}{18}=4=𝑎\text{ [using Eq. (i)] }\\ & \therefore \frac{𝑏+1}{18𝑏}=𝑎\end{array}\end{array}$

Given, $𝐴𝑋=𝐵$ and $𝐶𝑋=𝐷$ and $𝐴𝑋=𝐵$ has unique solution $𝐷$ and $𝐶𝑋=𝐷$ has unique solution $𝐵$.

Since, $𝑋=𝐷$ is the unique solution of $𝐴𝑋=𝐵$

$\therefore 𝐴𝐷=𝐵...\left(𝑖\right)$

And, $𝑋=𝐵$ is the unique solution of $𝐶𝑋=𝐷$,

$\therefore 𝐶𝐵=𝐷\Rightarrow 𝐵={𝐶}^{-1}𝐷...\left(𝑖𝑖\right)$

From Eq. (i) and (ii),

$\begin{array}{c}𝐴𝐷={𝐶}^{-1}𝐷\\ (𝐴-{𝐶}^{-1})𝐷=0\end{array}$

So, the solution of $(𝐴-{𝐶}^{-1})𝑋=0$ is $𝐷$

Given,

$𝑓\left(𝑥\right)=\frac{1}{2}\mid \begin{array}{cc}𝑓\left(𝑥\right)& 𝑓\left(\frac{1}{𝑥}\right)-𝑓\left(𝑥\right)\\ 1& 𝑓\left(\frac{1}{𝑥}\right)\end{array}\mid$

and $𝑓\left(2\right)=33$

Here, $𝑓\left(𝑥\right)=\frac{1}{2}\left[𝑓\right(𝑥)\cdot 𝑓\left(\frac{1}{𝑥}\right)-𝑓\left(\frac{1}{𝑥}\right)+𝑓(𝑥\left)\right]...\left(𝑖\right)$

Let

$\begin{array}{c}𝑓\left(𝑥\right)={𝑎}_0{𝑥}^{𝑛}+{𝑎}_1{𝑥}^{𝑛-1}+\dots +{𝑎}_{𝑛-1}𝑥+{𝑎}_{𝑛}\\ ({𝑎}_0\ne 0)\end{array}$

Putting this value of $𝑓\left(𝑥\right)$ in Eq. (i),

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{1}{2}\left[𝑓\right(𝑥)\cdot 𝑓\left(\frac{1}{𝑥}\right)-𝑓\left(\frac{1}{𝑥}\right)+𝑓(𝑥\left)\right]\\ \Rightarrow & 𝑓\left(𝑥\right)\cdot 𝑓\left(\frac{1}{𝑥}\right)=𝑓\left(\frac{1}{𝑥}\right)+𝑓\left(𝑥\right)\end{array}$

Hence, $𝑓\left(𝑥\right)={𝑥}^{𝑛}+1$ or $-{𝑥}^{𝑛}+1$

Now, $𝑓\left(2\right)=2^{𝑛}+1=33$

$\begin{array}{cc}\Rightarrow & 2^{𝑛}=32\\ \Rightarrow & 2^{𝑛}=2^5\\ \Rightarrow & 𝑛=5\end{array}$

And $𝑓\left(2\right)=-2^{𝑛}+1=33$

$\Rightarrow -2^{𝑛}=33-1=32=2^5$

which is not possible.

$\therefore 𝑓\left(𝑥\right)={𝑥}^{𝑛}+1={𝑥}^5+1$

So, $𝑓\left(3\right)=3^5+1=243+1=244$

We note that the quotient $\frac{𝑧-(2-𝑖)}{𝑧+(1+2𝑖)}$ is not defined if $𝑧=-(1+2𝑖)\iff 𝑥=-1$ and $𝑦=-2$

Since $𝑥=𝑥+𝑖𝑦$, then

$\begin{array}{cc}& \frac{𝑧-(2-𝑖)}{𝑧+(1+2𝑖)}=\frac{𝑥+𝑖𝑦-(2-𝑖)}{𝑥+𝑖𝑦+(1+2𝑖)}\\ & =\frac{(𝑥-2)+𝑖(𝑦+1)}{(𝑥+1)+𝑖(𝑦+2)}\\ & =\frac{(𝑥-2)+𝑖(𝑦+1)}{(𝑥+1)+𝑖(𝑦+2)}\times \frac{(𝑥+1)-𝑖(𝑦+2)}{(𝑥+1)-𝑖(𝑦+2)}\\ & =\frac{(𝑥-2)(𝑥+1)-𝑖(𝑥-2)(𝑦+2)}{+𝑖(𝑦+1)(𝑥+1)+(𝑦+1)(𝑦+2)}\\ & =\frac{\left[\right(𝑥-2\left)\right(𝑥+1)+(𝑦+1\left)\right(𝑦+2\left)\right]}{(𝑥+1{)}^2+(𝑦+2{)}^2}+\frac{𝑖\left[\right(𝑥+1\left)\right(𝑦+1)-(𝑥-2\left)\right(𝑦+2\left)\right]}{(𝑥+1{)}^2+(𝑦+2{)}^2}\end{array}$

Since, $\frac{𝑧-(2-𝑖)}{𝑧+(1+2𝑖)}$ is purely imaginary,

So real part $=0$

$\begin{array}{cc}& \therefore \frac{(𝑥-2)(𝑥+1)+(𝑦+1)(𝑦+2)}{(𝑥+1{)}^2+(𝑦+2)}=0\\ & \iff {𝑥}^2+𝑥-2𝑥-2+{𝑦}^2+2𝑦+𝑦+2=0\\ & {𝑥}^2+{𝑦}^2-𝑥+3𝑦=0\end{array}$

And $𝑥\ne -1$ and $𝑦\ne -2$

This is the locus of point $𝑃$. This is a circle.

$\begin{array}{cc}& \text{ Given, }(\sqrt{\surd 3}-𝑖{)}^{𝑛},𝑛\in 𝑁\\ & \text{ Modulus }\begin{array}{cc}|\sqrt{\surd 3}-1|& =\sqrt{\surd (\sqrt{\surd 3}{)}^2+(-1{)}^2}\\ & =\sqrt{\surd 3+1}=\sqrt{\surd 4}=2\end{array}\end{array}$

$𝜃=\arg (\sqrt{\surd 3}-𝑖)={\tan }^{-1}\left(\frac{-1}{\sqrt{\surd 3}}\right)=\frac{-𝜋}{6}$

Argument :

$𝜃=\arg (\sqrt{\surd 3}-𝑖)={\tan }^{-1}\left(\frac{-1}{\sqrt{\surd 3}}\right)=\frac{-𝜋}{6}$

So, in polar form,

$\begin{array}{cc}\sqrt{\surd 3}-𝑖& =2\cdot (\cos 𝜃+𝑖\sin 𝜃)\\ & =2[\cos \left(\frac{-𝜋}{6}\right)+\sin \left(\frac{-𝜋}{6}\right)]\\ & =2[\cos \frac{𝜋}{6}-𝑖\sin \frac{𝜋}{6}]\end{array}$

$\begin{array}{cc}& \Rightarrow (\sqrt{\surd 3}-𝑖{)}^{𝑛}={\{2\cdot [\cos \left(\frac{𝜋}{6}\right)-𝑖\sin \left(\frac{𝜋}{6}\right)]\}}^{𝑛}\\ & =2^{𝑛}\cdot [\cos \left(\frac{𝑛𝜋}{6}\right)-𝑖\sin \left(\frac{𝑛𝜋}{6}\right)]...\left(𝑖\right)\end{array}$

But given that $(\sqrt{\surd 3}-1{)}^{𝑛}=2^{𝑛}...(𝑖𝑖)$

From Eqs. (i) and (ii), we get

$\begin{array}{cc}& 2^{𝑛}[\cos \left(\frac{𝑛𝜋}{6}\right)-𝑖\sin \left(\frac{𝑛𝜋}{6}\right)]=2^{𝑛}\\ \Rightarrow & \cos \left(\frac{𝑛𝜋}{6}\right)-𝑖\sin \left(\frac{𝑛𝜋}{6}\right)=1\\ \Rightarrow & \cos \left(\frac{𝑛𝜋}{6}\right)=1\text{ and }\sin (-\frac{𝑛𝜋}{6})=1\\ \Rightarrow & -\frac{𝑛𝜋}{6}=2𝑘𝜋\\ \Rightarrow & 𝑛=\frac{-12𝑘𝜋}{𝜋}=-12𝑘\end{array}$

Since, $𝑛\in 𝑁$, for the smallest positive $𝑛$, let $𝑘=-1$

$\Rightarrow 𝑛=-12(-1)=12$

$\therefore$ The least positive value of $𝑛$ is 12 .

Let $𝑧=1+\sqrt{\surd 5}+𝑖\sqrt{\surd 10-2\sqrt{\surd 5}}$

$\begin{array}{cc}& \begin{array}{cc}𝑆𝑜,\left|𝑧\right|& =\sqrt{\surd (1+\sqrt{\surd 5}{)}^2+(\sqrt{\surd 10-2\sqrt{\surd 5}}{)}^2}\\ & =\sqrt{\surd 1+5+2\sqrt{\surd 5}+10-2\sqrt{\surd 5}}\\ & =\sqrt{\surd 16+2\sqrt{\surd 5}-2\sqrt{\surd 5}}=\sqrt{\surd 16}=4\\ \text{ And }𝜃& =\arg \left(𝑧\right)={\tan }^{-1}\left(\frac{\sqrt{\surd 10-2\sqrt{\surd 5}}}{1+\sqrt{\surd 5}}\right)\\ & ={\tan }^{-1}(\tan \left(\frac{𝜋}{5}\right))=\frac{𝜋}{5}\end{array}\end{array}$

So, $𝑧=\left|𝑧\right|\cdot (\cos 𝜃+𝑖\sin 𝜃)$

$\begin{array}{cc}& =4[\cos \left(\frac{𝜋}{5}\right)+𝑖\sin \left(\frac{𝜋}{5}\right)]\\ \therefore {𝑧}^5& =4^5{[\cos \left(\frac{𝜋}{5}\right)+𝑖\sin \left(\frac{𝜋}{5}\right)]}^5\\ & =1024[\cos (5\cdot \frac{𝜋}{5})+𝑖\sin (5\cdot \frac{𝜋}{5})]\\ & =1024[\cos 𝜋+𝑖\sin 𝜋]\\ & =1024[-1+0]=-1024\end{array}$

Given, equation is

$\sqrt{\surd 3{𝑥}^2+𝑥+5}=𝑥-3$

Here, $3{𝑥}^2+𝑥+5\ge 0$ (since square root is non-negative)

$\Rightarrow 𝑥-3\ge 0$

$\Rightarrow 𝑥\ge 3$

Now, squaring to both sides of given equation, we get

$\begin{array}{cc}& {\left(\sqrt{\surd 3{𝑥}^2+𝑥+5}\right)}^2=(𝑥-3{)}^2\\ \Rightarrow & 3{𝑥}^2+𝑥+5={𝑥}^2-6𝑥+9\\ \Rightarrow & 2{𝑥}^2+7𝑥-4=0\end{array}$

Using quadratic formula,

$\begin{array}{cc}𝑥& =\frac{-7\pm \sqrt{\surd 7^2-4\cdot 2\cdot (-4)}}{2\cdot 2}\\ & =\frac{-7\pm \sqrt{\surd 49+32}}{4}=\frac{-7\pm 9}{4}\\ \Rightarrow 𝑥& =\frac{2}{4}\text{ and }𝑥=\frac{-16}{4}\\ \Rightarrow 𝑥& =\frac{1}{2}\text{ and }𝑥=-4\end{array}$

Since domain, $𝑥\ge 3$, so $𝑥=-4$

And $𝑥=\frac{1}{2}$ are not possible.

$\therefore$ There is zero solution.

Given, $\frac{{𝑥}^2-1}{(𝑥-4)(𝑥-3)}\ge 1$

$\begin{array}{cc}& \Rightarrow \frac{{𝑥}^2-1}{(𝑥-4)(𝑥-3)}-1\ge 0\\ & \Rightarrow \frac{{𝑥}^2-1-(𝑥-4)(𝑥-3)}{(𝑥-4)(𝑥-3)}\ge 0\end{array}$

Since, the inequality is $\frac{7𝑥-13}{{𝑥}^2-7𝑥+12}\ge 0$

So, the critical points at $7𝑥-13=0$

$\Rightarrow 𝑥=\frac{13}{7}$

and at $𝑥=4$ and $𝑥=3$, the expression is undefined.

$\therefore 𝑥\in [\frac{13}{7},3)\cup (4,\infty )$

Given, equation is

$2{𝑥}^3+3{𝑥}^2-5𝑥-7=0$

Since, $𝛼,𝛽$ and $𝛾$ are the roots of the given equation.

$\therefore$ Sum of roots $=𝛼+𝛽+𝛾=\frac{-3}{2}$

Sum of products of roots

$=𝛼𝛽+𝛽𝛾+𝛾𝛼=-\frac{5}{2}$

Products of roots $=𝛼𝛽𝛾=-\left(\frac{-7}{2}\right)=\frac{7}{2}$

$\begin{array}{c}\frac{1}{{𝛼}^2}+\frac{1}{{𝛽}^2}+\frac{1}{{𝛾}^2}=\frac{{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2+{𝛼}^2{𝛽}^2}{{𝛼}^2{𝛽}^2{𝛾}^2}\\ =\frac{\left(𝛽{𝛾}^2\right)+(𝛾𝛼{)}^2+(𝛼𝛽{)}^2}{(𝛼𝛽𝛾{)}^2}\end{array}$

$\begin{array}{cc}& =\frac{(𝛼𝛽+𝛽𝛾+𝛾𝛼{)}^2-2𝛼𝛽𝛾(𝛼+𝛽+𝛾)}{(𝛼𝛽𝛾{)}^2}\\ & =\frac{{\left(\frac{-5}{2}\right)}^2-2\left(\frac{7}{2}\right)\left(\frac{-3}{2}\right)}{{\left(\frac{7}{2}\right)}^2}\\ & =\frac{\frac{25}{4}+\frac{21}{2}}{\frac{49}{4}}=\frac{\frac{67}{4}}{\frac{49}{4}}\\ & =\frac{67}{4}\times \frac{4}{49}=\frac{67}{49}\\ & \therefore \frac{1}{{𝛼}^2}+\frac{1}{{𝛽}^2}+\frac{1}{{𝛾}^2}=\frac{67}{49}\end{array}$

Given equation

$𝑎{𝑥}^4+𝑏{𝑥}^3+𝑐{𝑥}^2+𝑑𝑥+𝑒=0$

Let $𝛼,𝛼,𝛽,\frac{1}{𝛽}$ be root of equation

$\begin{array}{cc}& \therefore 𝛼+𝛼+𝛽+\frac{1}{𝛽}=-𝑏/𝑎\\ & \Rightarrow (𝛽+\frac{1}{𝛽})=-\frac{𝑏}{𝑎}-2𝛼\\ & \Rightarrow 𝛼\cdot 𝛼\cdot 𝛽\cdot \frac{1}{𝛽}=\frac{𝑒}{𝑎}\Rightarrow {𝛼}^2=𝑒/𝑎\\ & \Rightarrow {𝛼}^2+2\left(𝛼𝛽\right)+\frac{2𝛼}{𝛽}+1=𝑐/𝑎\\ & {𝛼}^2+2𝛼(𝛽+\frac{1}{𝛽})+1=𝑐/𝑎\\ & \Rightarrow {𝛼}^2+2𝛼(-\frac{𝑏}{𝑎}-2𝛼)+1=𝑐/𝑎\\ & \Rightarrow -3{𝛼}^2-\frac{2𝛼𝑏}{𝑎}+1=𝑐/𝑎\\ & \Rightarrow \frac{𝑐}{𝑎}+3{𝛼}^2+\frac{2𝛼𝑏}{𝑎}=1\\ & \Rightarrow 3\left(\frac{𝑒}{𝑎}\right)+2\sqrt{\surd \frac{𝑒}{𝑎}}\times \frac{𝑏}{𝑐}+\frac{𝑐}{𝑎}=1\\ & \Rightarrow 3𝑒+2𝑏\sqrt{\surd \frac{𝑒}{𝑎}}+𝑐=𝑎\end{array}$

The digit are 1 to 9 . Each such number is strictly increasing sequence of digits, i.e., digits don't repeat and are in order.

e.g., 12, 135, 1239 and soon.

So, total such numbers in the number of increasing digit sequence of length 2 to 4.

We choose subsets of digits from 1 to 9 of lengths 2,3 and 4 in increasing order.

$\therefore 2$-digits numbers

$={}^9{𝐶}_2=\frac{9!}{2!7!}=\frac{9\times 8}{2}=36$

3-digits numbers

$={}^9{𝐶}_3=\frac{9!}{3!6!}=\frac{9\times 8\times 7\times 6!}{6\times 6!}=84$

$\begin{array}{cc}& \begin{array}{cc}4\text{-digits numbers }& ={}^9{𝐶}_4=\frac{9!}{4!5!}\\ & =\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}=126\end{array}\\ & \text{ So, total such digits }\\ & =36+84+126=246\end{array}$

Given, word is AGAIN,

So the letters in alphabatical order : A, A, G, I, N

Words starting with $𝐴=\frac{4!}{2!}=12$

Words starting with $𝐺=\frac{4!}{2!}=12$

Words starting with $𝐼=\frac{4!}{2!}=12$

Words starting with $𝑁=\frac{4!}{2!}=12$

So, total words $=12+12+12+12=48$

Thus, the 50th word starts with $𝑁$.

49th word = NAAGI

$\therefore 50$ th word $=$ NAAIG.

We select exactly I wicket keeper from 4 .

So, number of ways $={}^4{𝐶}_1=4$

Now, we have 10 players left to select from batsmen, bowlers and all-rounders with the condition.

Batsmen $\ge 4$, Bowlers $\ge 3$, All-rounders $\ge 2$

So, total of these three $=10$

So, number of batsmen, $𝐵$ from 4 to 6 , number of bowlers, $𝐿$ (say)from 3 to 6 and number of all-rounder, $𝐴$ (say) from 2 to 4 .

$\therefore 𝐵+𝐿+𝐴=10$ (after 1 wicket keeper)

Now, the possible combinations for, (B, L, A) are

$(4,4,2),(4,3,3),(5,3,2)$

Total number of ways for 10 players

$\begin{array}{cc}& =({}^6{𝐶}_4{}^6{𝐶}_4{}^4{𝐶}_2+{}^6{𝐶}_4{}^6{𝐶}_3{}^4{𝐶}_3+{}^6{𝐶}_5{}^6{𝐶}_3{}^4{𝐶}_2)\\ & =(15\times 15\times 6+15\times 20\times 4+6\times 20\times 6)\\ & =(1350+1200+720)=3270\end{array}$

$\therefore$ Total number of ways for 11 players

$\begin{array}{cc}& =4\times 3270\\ & =13080\end{array}$

Given,

$\begin{array}{cc}𝑦& =\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12}+\dots +\infty \\ & =\frac{3}{1!}\cdot \frac{1}{4}+\frac{3\cdot 5}{2!}\cdot \frac{1}{4^2}+\frac{3\cdot 5\cdot 7}{3!}\cdot \frac{1}{4^3}+\dots +\infty \end{array}$

Since, using binomial expansion

$\begin{array}{cc}(1-𝑥{)}^{-𝑛}=1+𝑛𝑥+& \frac{𝑛(𝑛+1)}{2!}{𝑥}^2+\frac{𝑛(𝑛+1)(𝑛+2)}{3!}{𝑥}^3+\dots \end{array}$

Here, $𝑛𝑥=\frac{3}{4}...\left(𝑖\right)$

$\begin{array}{cc}& \begin{array}{cc}& \frac{𝑛(𝑛+1)}{2}{𝑥}^2=\frac{3\cdot 5}{2!}\cdot \frac{1}{4^2}\\ \Rightarrow & 𝑛(𝑛+1){𝑥}^2=\frac{15}{16}...\left(𝑖𝑖\right)\end{array}\\ & \text{ From Eq. (i) and (ii), we get }\\ & \begin{array}{cc}\frac{𝑛(𝑛+1){𝑥}^2}{{𝑛}^2{𝑥}^2}& =\frac{15/16}{(3/4{)}^2}\\ & =\frac{15/16}{9/16}=\frac{5}{3}\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow & \frac{𝑛+1}{𝑛}=\frac{5}{3}\\ \Rightarrow & 3𝑛+3=5𝑛\\ \Rightarrow & 𝑛=\frac{3}{2}\end{array}\\ & \text{ So, }𝑥=\frac{3}{4}\cdot \frac{1}{𝑛}=\frac{3}{4}\times \frac{2}{3}=\frac{1}{2}\\ & \begin{array}{cc}\therefore 𝑦& =(1-𝑥{)}^{-𝑛}-1={(1-\frac{1}{2})}^{-3/2}-1\\ & ={\left(\frac{1}{2}\right)}^{-3/2}-1=2^{3/2}-1=2\sqrt{\surd 2}-1\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ So, }𝑦+1=2\sqrt{\surd 2}-1+1=2\sqrt{\surd 2}\\ & (𝑦+1{)}^2=(2\sqrt{\surd 2}{)}^2=8\\ & \Rightarrow {𝑦}^2+2𝑦+1-8=0\\ & \Rightarrow {𝑦}^2+2𝑦-7=0\end{array}\\ & \text{ Thus, the quadratic equation is }\\ & {𝑦}^2+2𝑦-7=0\end{array}$