TG EAPCET 2024 9th May Morning Shift Question Paper with Solutions

We have,

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\sqrt{\surd \cos (\sin 𝑥)}+{\cos }^{-1}\left(\frac{1+{𝑥}^2}{2𝑥}\right)\\ \\ & \Rightarrow \cos (\sin 𝑥)>0\\ \\ & \because \sin 𝑥\in [-1,1]\\ \\ & \therefore \cos (\sin 𝑥)>0,\forall 𝑥\in 𝑅\\ \\ & \text{ and }-1\le \frac{1+{𝑥}^2}{2𝑥}\le 1\\ \\ & -1\le \frac{1+{𝑥}^2}{2𝑥}\text{ and }\frac{1+{𝑥}^2}{2𝑥}\le 1\\ \\ & \Rightarrow 𝑥\in \{-1,1\}\end{array}$

To find largest positive integer $𝐾$, that divides ${81}^{𝑛}+20𝑛-1$ for $𝑛\in 𝑁$, we start evaluating the expression for small values of $𝑛$

For $𝑛=1:{81}^1+20.1-1=100$

For $𝑛=2:{81}^2+20.2-1=6600$

Now,

The greatest common divisor of 100 and 6600 is 100 . Hence, $𝐾=100$ is the largest integer that divides

${81}^{𝑛}+20𝑛-1$ for all $𝑛\in 𝑁$

and the divisior of 100 are

$\begin{array}{cc}& 1,2,4,5,10,20,25,50,100\\ \\ & \begin{array}{c}\Rightarrow 𝑆=1+2+4+5+10+20+25+50+100=217\end{array}\end{array}$

Hence, $𝑆-𝐾=117$

$\because 𝐴+𝐵$ is symmecric

$\begin{array}{cc}& \Rightarrow (𝐴+𝐵{)}^{𝑇}=𝐴+𝐵\\ \\ & \Rightarrow {𝐴}^{𝑇}+{𝐵}^{𝑇}=𝐴+𝐵\dots \text{ (i) }\end{array}$

and $𝐴-𝐵$ is skew-symmetric

$\begin{array}{cc}\Rightarrow (𝐴-{𝐴}^{𝑇}& =-(𝐴-𝐵)\\ \\ {𝐴}^{𝑇}-{𝐵}^{𝑇}& =-𝐴+𝐵\dots \text{ (ii) }\end{array}$

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

$𝐵={𝐴}^{𝑇}$

$\Rightarrow 𝐵={⎡⎣⎢\begin{array}{ccc}-1& 2& 3\\ \\ 4& 3& -2\\ \\ 3& -4& 5\end{array}⎤⎦⎥}^{𝑟}=⎡⎣⎢\begin{array}{ccc}-1& 4& 3\\ \\ 2& 3& -4\\ \\ 3& -2& 5\end{array}⎤⎦⎥$

Given $𝐷$ is the transpose of $𝐶$

$\begin{array}{cc}& \Rightarrow ={𝐶}^{𝑟}={⎡⎣⎢\begin{array}{ccc}0& 1& -2\\ \\ 2& -1& 0\\ \\ 0& 2& 1\end{array}⎤⎦⎥}^{𝑇}\\ \\ & 𝐷=⎡⎣⎢\begin{array}{ccc}0& 2& 0\\ \\ 1& -1& 2\\ \\ -2& 0& 1\end{array}⎤⎦⎥\\ \\ & 50,𝐵+𝐷=⎡⎣⎢\begin{array}{ccc}-1& 4& 3\\ \\ 2& 3& -4\\ \\ 3& -2& 5\end{array}⎤⎦⎥+⎡⎣⎢\begin{array}{ccc}0& 2& 0\\ \\ 1& -1& 2\\ \\ -2& 0& 1\end{array}⎤⎦⎥\\ \\ & =⎡⎣⎢\begin{array}{ccc}-1& 6& 3\\ \\ 3& 2& -2\\ \\ 1& -2& 6\end{array}⎤⎦⎥\end{array}$

We have, ${𝐴}^2+𝐼=2𝐴\Rightarrow {𝐴}^2=2𝐴-𝐼$

$\begin{array}{cc}& \because {𝐴}^9={𝐴}^2\cdot {𝐴}^2{𝐴}^2{𝐴}^2𝐴\\ \\ & =(2𝐴-𝐼)(2𝐴-𝐼)(2𝐴-𝐼)(2𝐴-𝐼)𝐴\\ \\ & \begin{array}{cc}\Rightarrow {𝐴}^9=& (4{𝐴}^2+𝐼-4𝐴)(4{𝐴}^2+𝐼-4𝐴)𝐴\\ \\ & =(4𝐴-3𝐼)(4𝐴-3𝐼)𝐴\\ \\ & =(16{𝐴}^2+9𝐼-24𝐴)𝐴\\ \\ & =(8𝐴-7𝐼)𝐴\\ \\ & =8{𝐴}^2-7𝐴\\ \\ & =16𝐴-8𝐼-7𝐴\\ \\ & =9𝐴-8𝐼\end{array}\end{array}$

Let $\Delta =⎡⎣⎢\begin{array}{ccc}\frac{{𝑎}^2+{𝑏}^2}{𝑐}& 𝑐& 𝑐\\ \\ 𝑎& \frac{{𝑏}^2+{𝑐}^2}{𝑎}& 𝑎\\ \\ 𝑏& 𝑏& \frac{{𝑐}^2+{𝑎}^2}{𝑏}\end{array}⎤⎦⎥$

On multiply ${𝑅}_1,{𝑅}_2$ and ${𝑅}_3$ by $𝑐,𝑎$ and $𝑏$, respectively

$\Rightarrow \Delta =\frac{1}{𝑎𝑏𝑐}⎡⎣⎢\begin{array}{ccc}{𝑎}^2+{𝑏}^2& {𝑐}^2& {𝑐}^2\\ \\ {𝑎}^2& {𝑏}^2+{𝑐}^2& {𝑎}^2\\ \\ {𝑏}^2& {𝑏}^2& {𝑐}^2+{𝑎}^2\end{array}⎤⎦⎥$

Apply ${𝑅}_1\to {𝑅}_1-({𝑅}_2+{𝑅}_3)$

$\begin{array}{cc}\therefore \Delta =& \frac{1}{𝑎𝑏𝑐}⎡⎣⎢\begin{array}{ccc}0& -2{𝑏}^2& -2{𝑎}^2\\ \\ {𝑎}^2& {𝑏}^2+{𝑐}^2& {𝑎}^2\\ \\ {𝑏}^2& {𝑏}^2& {𝑐}^2+{𝑎}^2\end{array}⎤⎦⎥\\ \\ & =\frac{-2}{𝑎𝑏𝑐}⎡⎣⎢\begin{array}{ccc}0& {𝑏}^2& {𝑎}^2\\ \\ {𝑎}^2& {𝑏}^2+{𝑐}^2& {𝑎}^2\\ \\ {𝑏}^2& {𝑏}^2& {𝑐}^2+{𝑎}^2\end{array}⎤⎦⎥\end{array}$

Apply ${𝑅}_2\to {𝑅}_2-{𝑅}_1$ and ${𝑅}_3\to {𝑅}_3-{𝑅}_1$, we get

$\begin{array}{cc}& =\frac{-2}{𝑎𝑏𝑐}⎡⎣⎢\begin{array}{ccc}0& {𝑏}^2& {𝑎}^2\\ \\ {𝑎}^2& {𝑐}^2& 0\\ \\ {𝑏}^2& 0& {𝑐}^2\end{array}⎤⎦⎥\\ \\ & =\frac{-2}{𝑎𝑏𝑐}[0-{𝑏}^2({𝑎}^2{𝑐}^2-0)+{𝑎}^2(0-{𝑏}^2{𝑐}^2)]\\ \\ & =\frac{-2}{𝑎𝑏𝑐}(-2{𝑎}^2{𝑏}^2{𝑐}^2)=4𝑎𝑏𝑐\end{array}$

We have,

$\begin{array}{cc}& 𝑥-2𝑦+3𝑧=4\\ \\ & 3𝑥+𝑦-2𝑧=7\\ \\ & 2𝑥+3𝑦+𝑧=6\end{array}$

Here,

$\begin{array}{cc}& 𝐷=\mid \begin{array}{ccc}1& -2& 3\\ \\ 3& 1& -2\\ \\ 2& 3& 1\end{array}\mid \\ \\ & =1(1+6)+2(3+4)+3(9-2)=42\\ \\ & {𝐷}_{𝑥}=\mid \begin{array}{ccc}4& -2& 3\\ \\ 7& 1& -2\\ \\ 6& 3& 1\end{array}\mid \\ \\ & =4(1+6)+2(7+12)+3(21-6)=111\\ \\ & {𝐷}_{𝑦}=\mid \begin{array}{ccc}1& 4& 3\\ \\ 3& 7& -2\\ \\ 2& 6& 1\end{array}\mid =3\\ \\ & {𝐷}_{𝑧}=\mid \begin{array}{ccc}1& -2& 4\\ \\ 3& 1& 7\\ \\ 2& 3& 6\end{array}\mid =21\end{array}$

So, system of linear equations has unique solution and $𝑧=\frac{{𝐷}_{𝑧}}{𝐷}=\frac{21}{42}=\frac{1}{2}$

We have,

$\begin{array}{cc}& \sqrt{\surd 5}-𝑖\sqrt{\surd 15}=𝑟(\cos 𝜃+𝑖\sin 𝜃)\\ \\ & \begin{array}{cc}\Rightarrow 𝑟& =\sqrt{\surd (\sqrt{\surd 5}{)}^2+(-\sqrt{\surd 15}{)}^2}\\ \\ & =\sqrt{\surd 5+15}=2\sqrt{\surd 5}\end{array}\\ \\ & \text{ and }𝜃={\tan }^{-1}\left(\frac{\text{ Imaginary part }}{\text{ Real part }}\right)\\ \\ & ={\tan }^{-1}\left(\frac{-\sqrt{\surd 15}}{\sqrt{\surd 5}}\right)={\tan }^{-1}(-\sqrt{\surd 3})\end{array}$

$\begin{array}{cc}& \Rightarrow 𝜃=-\frac{𝜋}{3}\\ \\ & \text{ So, }{𝑟}^2(\sec 𝜃+3{\mathrm{cosec}}^2𝜃)\\ \\ & =(2\sqrt{\surd 5}{)}^2[\sec (-\frac{𝜋}{3})+3{\mathrm{cosec}}^2(-\frac{𝜋}{3})]\\ \\ & =20[2+3\times \frac{4}{3}]=20\times 6=120\end{array}$

We have, $𝑧=𝑥+𝑖𝑦$

$\begin{array}{cc}\therefore \frac{2𝑧-𝑖}{𝑧-2}=& \frac{2(𝑥+𝑖𝑦)-𝑖}{𝑥+𝑖𝑦-2}\\ \\ & =\frac{2𝑥+(2𝑦-1)𝑖}{𝑥-2+𝑖𝑦}\\ \\ & =\frac{2𝑥+(2𝑦-1)𝑖}{(𝑥-2)+𝑦𝑖}\times \left[\frac{(𝑥-2)-𝑦𝑖}{(𝑥-2)-𝑦𝑖}\right]\\ \\ & 2𝑥(𝑥-2)+𝑦(2𝑦-1)\\ \\ & =\frac{+\left[\right(𝑥-2\left)\right(2𝑦-1)-2𝑥𝑦]𝑖}{(𝑥-2{)}^2+{𝑦}^2}\\ \\ =& \frac{2{𝑥}^2-4𝑥+2{𝑦}^2-𝑦}{(𝑥-2{)}^2+{𝑦}^2}\\ \\ & +\frac{\left[\right(𝑥-2\left)\right(2𝑦-1)-2𝑥𝑦]𝑖}{(𝑥-2{)}^2+{𝑦}^2}\end{array}$

$\because \frac{2𝑧-𝑖}{𝑧-2}$ is purely real number

$\begin{array}{cc}& \Rightarrow \frac{(𝑥-2)(2𝑦-1)-2𝑥𝑦}{(𝑥-𝑧{)}^2+{𝑦}^2}=0\\ \\ & \Rightarrow 2𝑥𝑦-𝑥-4𝑦-2𝑥𝑦+2=0\\ \\ & \Rightarrow 𝑥+4𝑦=2\end{array}$

Given $𝑥$ and $𝑦$ are two complex number with $\left|𝑥\right|=\left|𝑦\right|=1$ and $𝑥={𝑒}^{𝑖2𝛼}$

$\begin{array}{cc}𝑦& ={𝑒}^{𝑖.3𝛽},\text{ where }𝛼+𝛽=\frac{𝜋}{36}\\ \\ \Rightarrow {𝑥}^6& ={\left({𝑒}^{𝑖2𝛼}\right)}^6={𝑒}^{𝑖.12𝛼}\\ \\ {𝑦}^4& ={\left({𝑒}^{𝑖.3𝛽}\right)}^4={𝑒}^{𝑖.12𝛽}\end{array}$

So, ${𝑥}^6\cdot {𝑦}^4={𝑒}^{𝑖.12𝛼}\cdot {𝑒}^{𝑖,12𝛽}={𝑒}^{𝑖.12(𝛼+𝛽)}$

Since, $𝛼+𝛽=\frac{𝜋}{36}$

$\Rightarrow {𝑥}^6\cdot {𝑦}^4={𝑒}^{𝑖.12\frac{𝜋}{36}={𝑒}^{𝑖\frac{𝜋}{3}}}$

and $\frac{1}{{𝑥}^6\cdot {𝑦}^4}={𝑒}^{-𝑖\cdot \frac{𝜋}{3}}$

$\begin{array}{cc}\Rightarrow {𝑥}^6\cdot {𝑦}^4+\frac{1}{{𝑥}^6\cdot {𝑦}^4}& ={𝑒}^{𝑖\cdot \frac{𝜋}{3}}+{𝑒}^{-𝑖\cdot \frac{𝜋}{3}}\\ \\ & =2\cos \frac{𝜋}{3}=1\end{array}$

We have,

$\begin{array}{cc}& {𝑥}^{14}+{𝑥}^9-{𝑥}^5-1=0\\ \\ & \Rightarrow ({𝑥}^9-1)({𝑥}^5+1)=0\\ \\ & {𝑥}^9=1\\ \\ & 𝑥=(1{)}^{1/9}\\ \\ & 𝑥=(\cos 0^{\circ }\\ \\ & {+𝑖\sin 0^{\circ })}^{1/9}\\ \\ & =(\cos 2𝐾𝜋\\ \\ & +𝑖\sin 2𝐾𝜋{)}^{1/9}\\ \\ & =\cos \frac{2𝐾𝜋}{9}+𝑖\sin \frac{2𝐾𝜋}{9}\\ \\ & \text{ using Demoivre’s theorm here, }𝐾=0\text{, }1,\dots .,8\\ \\ & \begin{array}{c}{𝑥}^5=-1\\ \\ 𝑥=(\cos 𝜋+𝑖\sin 𝜋{)}^{1/5}\\ \\ =[\cos (2𝐾𝜋+𝜋)\\ \\ +𝑖\sin (2𝐾𝜋+𝜋){]}^{1/5}\\ \\ =[\cos (2𝐾+1)𝜋\\ \\ +𝑖\sin (2𝐾+1)𝜋{]}^{1/5}\\ \\ =\cos \frac{(2𝐾+1)𝜋}{5}\\ \\ +𝑖\sin \frac{(2𝐾+1)𝜋}{5}\\ \\ \text{ where }𝐾=0,1,\dots .,5\end{array}\end{array}$

So, one of root of the given equation is

$\begin{array}{cc}& \text{ So, one of root }\frac{2𝜋}{9}+𝑖\sin \frac{2𝜋}{9}\text{ for }𝐾=1\\ \\ & =\frac{\sqrt{\surd 5}+1}{4}+𝑖\frac{\sqrt{\surd 10-2\sqrt{\surd 5}}}{4}\end{array}$

We have,

$\begin{array}{c}3{𝑥}^2+(2𝑘+1)𝑥-5𝑘=0\end{array}$

$\because$ roots are real and equal.

$\begin{array}{cc}\Rightarrow & 𝐷=0\\ \\ \Rightarrow & (2𝑘+1{)}^2-4(-5𝑘)\times 3=0\\ \\ \Rightarrow & 4{𝑘}^2+4𝑘+1+60𝑘=0\\ \\ \Rightarrow & 4{𝑘}^2+64𝑘+1=0\\ \\ & 𝑘=\frac{-64\pm \sqrt{\surd 4096-16}}{2\times 4}\\ \\ & =\frac{-64\pm \sqrt{\surd 4080}}{8}=\frac{-16\pm \sqrt{\surd 255}}{2}\end{array}$

$𝐻𝑒𝑛𝑐𝑒,𝑘=\frac{-16+\sqrt{\surd 255}}{2}(\because \frac{1}{2}<𝑘<0)$

Let $𝛼$ be the common root of

$\begin{array}{cc}& 2{𝑥}^2+𝑎𝑥-2=0\text{ and }{𝑥}^2+𝑥+0\to 2=0\\ \\ & \Rightarrow 2{𝛼}^2+𝑎𝛼-2=0\dots \text{ (i) }\\ \\ & {𝛼}^2+𝛼+2𝑎=0\dots \text{ (ii) }\\ \\ & \Rightarrow \frac{{𝛼}^2}{2{𝑎}^2+2}=\frac{𝛼}{-2-4𝑎}=\frac{1}{2-𝑎}[𝑎\ne 2,-\frac{1}{2}]\\ \\ & \Rightarrow 𝛼=\frac{4𝑎+2}{𝑎-2}\end{array}$

So, $\frac{(4𝑎+2{)}^2}{(𝑎-2{)}^2(2{𝑎}^2+2)}=\frac{-1}{𝑎-2}$

$\begin{array}{c}\frac{2(2𝑎+1{)}^2}{({𝑎}^2+1)}=2-𝑎\\ \\ 8{𝑎}^2+2+8𝑎=2{𝑎}^2+2-{𝑎}^3-𝑎\\ \\ {𝑎}^3+6{𝑎}^2+9𝑎=0\Rightarrow 𝑎({𝑎}^2+6𝑎+9)=0\\ \\ \Rightarrow 𝑎=-3[\because 𝑎\ne 0)\\ \\ \therefore 𝑎{𝑥}^2-4𝑥-2𝑎=-3{𝑥}^2-4𝑥+6=0\\ \\ 3{𝑥}^2+4𝑥-6=0\\ \\ 𝑥=\frac{-4\pm \sqrt{\surd 16+72}}{6}\\ \\ =\frac{-2\pm \sqrt{\surd 22}}{3}\end{array}$

So, one root of $𝑎{𝑥}^2-4𝑥-2𝑎=0$ is

$=\frac{-2+\sqrt{\surd 22}}{3}.$

We have,

$\begin{array}{c}2{𝑥}^3-3{𝑥}^2+5𝑥-7=0\\ \\ \therefore 𝛼+𝛽+𝛾=-(-\frac{3}{2})=\frac{3}{2}\dots \text{ (i) }\\ \\ 𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{5}{2}\dots \text{ (ii) }\\ \\ 𝛼𝛽𝛾=\frac{7}{2}\dots \text{ (iii) }\end{array}$

On squaring both side of Eq. (ii), we get

$\begin{array}{cc}& (𝛼𝛽+𝛽𝛾+𝛾𝛼{)}^2=\frac{25}{4}\\ \\ & {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2+2\\ \\ & (𝛼{𝛽}^2𝛾+𝛼𝛽{𝛾}^2+{𝛼}^2𝛽𝛾)=\frac{25}{4}\\ \\ & \Sigma {𝛼}^2{𝛽}^2+2𝛼𝛽𝛾(𝛼+𝛽+𝛾)=\frac{25}{4}\end{array}$

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

$\begin{array}{cc}& \Sigma {𝛼}^2{𝛽}^2+2\left(\frac{7}{2}\right)\left(\frac{3}{2}\right)=\frac{25}{4}\\ \\ & \Sigma {𝛼}^2{𝛽}^2=\frac{25}{4}-\frac{21}{2}=-\frac{17}{4}\end{array}$

We have,

${𝑥}^4-{𝑥}^3-16{𝑥}^2+4𝑥+48=0\dots \text{ (i) }$

By hit and trial, we get, ${𝑥}^2-4$ is the factor of

$\begin{array}{c}{𝑥}^4-{𝑥}^3-16{𝑥}^2+4𝑥+48=0\\ \\ \Rightarrow \frac{{𝑥}^4-{𝑥}^3-16{𝑥}^2+4𝑥+48}{{𝑥}^2-4}\\ \\ \Rightarrow {𝑥}^2-𝑥-12\\ \\ \Rightarrow {𝑥}^4-{𝑥}^3-16{𝑥}^2+4𝑥+48=({𝑥}^2-4)\\ \\ ({𝑥}^2-𝑥-12)\\ \\ \Rightarrow (𝑥-2)(𝑥+2)(𝑥-4)(𝑥+3)=0\end{array}$

$\Rightarrow$ Roots of above Eq. (i) are,

$-3,-2,2,4$

$\text{ So, }\begin{array}{cc}{𝛼}^4+{𝛽}^4+{𝛾}^4+{𝛿}^4& =(-3{)}^4+(-2^4\\ \\ +& (2^4+(4{)}^4=369\end{array}$

We have,

$2,3,5,7$

No repetition

Number of given digits $=𝑛=4$

Sum of digits = 17

Sum of all four digits formed

$\begin{array}{cc}& =(𝑛-1)!(\text{ Sum of digits) (11111 ......... n times) }\\ \\ & =3!\left(17\right)\left(1111\right)\\ \\ & =6\times 17\times 1111\\ \\ & =113322\end{array}$

Given that there are 15 identical gold coins to be distributed among 3 persons.

Also, each person gets at least 3 coins $\Rightarrow$ total $=3\times 3=9$ coins

$\therefore 9$ coins are already distributed in 1 ways. we now have 6 coins to be distributed between 3 people.

$\begin{array}{cc}& \Rightarrow {𝑥}_1+{𝑥}_2+{𝑥}_3=6,{𝑥}_{𝑖}\ge 0\\ \\ & \text{ Number of }\end{array}$

Number of solutions is given by

$\begin{array}{cc}& {}^{𝑛+𝑟-1}{𝐶}_{𝑟-1}\\ \\ & \Rightarrow 𝑛=6,𝑟=3\\ \\ & \therefore \text{ Reanired solution }\end{array}$

$\therefore$ Required Number $={}^{6+3-1}{𝐶}_{3-1}$

$={}^8{𝐶}_2=28$

4 letter words from

ACCOMMODATION

There are different letters

"ACCOMMODATION"

namely,

A, C, D, I, M, N, O, T

$\begin{array}{cccccccc}2& 2& 1& 1& 2& 1& 3& 1\end{array}$

Case (I) all 4 letters of the word are different numbers of combinations

$={}^8{𝑐}_4=70$

Case (II) 2 letters are same and 2 are different.

$={}^4{𝑐}_1\cdot {}^7{𝑐}_2-6=78$

( $\because$ when we select 2 same letter from 0 and 3rd letter also from, 0 then this need to be subtracted and such cases are 6)

Case (III) When 3 letters are same

$=1.{}^7{𝑐}_1=7$

Case (IV) When 2 letters are same and other 2 are also same.

$={}^4{𝑐}_1\times {}^3{𝑐}_1=12$

Hence, total number of combination

$=70+78+7+12=167$

We have,

$\begin{array}{cc}& 2\frac{{𝑐}_1}{{𝑐}_0}+4\frac{{𝑐}_2}{{𝑐}_1}+6\frac{{𝑐}_3}{{𝑐}_2}+\dots .+2𝑛\frac{{𝑐}_{𝑛}}{{𝑐}_{𝑛-1}}=650\\ \\ & \Rightarrow \sum _{𝑟=1}^{𝑛}2𝑟\frac{{}^{𝑛}{𝑐}_{𝑟}}{{}^{𝑛}{𝑐}_{𝑟-1}}=650\end{array}$

$\begin{array}{cc}& \Rightarrow 2\sum _{𝑟=1}^{𝑛}𝑟⎡⎣\frac{\frac{𝑛!}{𝑟!(𝑛-𝑟)!}}{\frac{𝑛!}{(𝑟-1)!(𝑛-𝑟+1)!}}⎤⎦=650\\ \\ & \Rightarrow 2\sum _{𝑟=1}^{𝑛}𝑟⎡⎣⎢\begin{array}{c}\frac{𝑛!}{𝑟\cdot (𝑟-1)!(𝑛-𝑟)!}\\ \\ \times \frac{(𝑟-1)!(𝑛-𝑟+1)\cdot (𝑛-𝑟)!}{𝑛!}\end{array}⎤⎦⎥\end{array}$

$\begin{array}{c}=650\\ \\ \Rightarrow & 2\Sigma (𝑛-𝑟+1)=650\\ \\ \Rightarrow & 2[{𝑛}^2+𝑛-\frac{𝑛(𝑛+1)}{2}]=650\\ \\ \Rightarrow & {𝑛}^2+𝑛-650=0\\ \\ \Rightarrow & (𝑛+26)(𝑛-25)=0\\ \\ \Rightarrow & 𝑛=25\end{array}$

Hence, ${}^{𝑛}{𝑐}_2={}^{25}{𝑐}_2=300$

We have,

$\frac{𝑥}{(𝑥-2)(𝑥-3)}=\frac{𝐴}{(𝑥-2)}+\frac{𝐵}{(𝑥-3)}$

On comparing, we get

$\begin{array}{cc}& 𝐴=-2𝐵=3\\ \\ & \Rightarrow \frac{𝑥}{(𝑥-2)(𝑥-3)}=\frac{-2}{𝑥-2}+\frac{3}{𝑥-3}\\ \\ & =\frac{1}{(1-\frac{𝑥}{2})}+\frac{-1}{(1-\frac{𝑥}{3})}\end{array}$

$\begin{array}{cc}& ={(1-\frac{𝑥}{2})}^{-1}-{(1-\frac{𝑥}{3})}^{-1}\\ \\ & =[1+\frac{𝑥}{2}+{\left(\frac{𝑥}{2}\right)}^2+{\left(\frac{𝑥}{2}\right)}^3+\dots .+{\left(\frac{𝑥}{2}\right)}^{𝑛}]-\\ \\ & [1+\frac{𝑥}{3}+{\left(\frac{𝑥}{3}\right)}^2+\dots .+{\left(\frac{𝑥}{3}\right)}^{𝑛}]\end{array}$

$\Rightarrow$ Coefficient of ${𝑥}^2=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}$

We have,

$\begin{array}{cc}& \frac{{𝑥}^4}{({𝑥}^2+1)(𝑥-2)}=𝑓\left(𝑥\right)+\frac{𝐴𝑥+𝐵}{{𝑥}^2+1}+\frac{𝐶}{𝑥-2}\\ \\ & \frac{{𝑥}^4}{({𝑥}^2+1)(𝑥-2)}\\ \\ & 𝑓\left(𝑥\right)\left[({𝑥}^2+1)\right(𝑥-2\left)\right]\\ \\ & =\frac{+(𝐴𝑥+𝐵)(𝑥-2)+𝐶({𝑥}^2+1)}{({𝑥}^2+1)(𝑥-2)}\\ \\ & \Rightarrow \frac{{𝑥}^4}{({𝑥}^2+1)(𝑥-2)}\end{array}$