TG EAPCET 2020 14th September Evening Shift Question Paper with Solutions

We have,

$𝑓\left(𝑛\right)=⎧⎩⎨\begin{array}{cc}2𝑛,& \text{ if }𝑛>0\\ 1,& \text{ if }𝑛=0\\ -2𝑛-1,& \text{ if }𝑛<0\end{array}$

When, $𝑛>0$

$𝑓\left(𝑛\right)=2,4,6,8\dots \dots$

When, $𝑛<0$

$𝑓\left(𝑛\right)=1,3,5,7\dots ..$

∴ Range of $𝑓\left(𝑛\right)$ is $\{1,2,3,4,\dots \dots \}$.

$\therefore$ Codomain $=$ Range

Hence, $𝑓\left(𝑛\right)$ is onto.

Here, $𝑓\left(0\right)-𝑓(-1)=1$

$\therefore 𝑓$ is not one one.

We have,

$𝑓\left(𝑥\right)={\cos }^{-1}[{\log }_5({𝑥}^2+7𝑥+15)]$

$𝑓\left(𝑥\right)$ is defined of

$\begin{array}{cc}& {\log }_5({𝑥}^2+7𝑥+15)\in [-1,1]\\ & -1\le {\log }_5({𝑥}^2+7𝑥+15)\le 1\\ & \frac{1}{5}\le {𝑥}^2+7𝑥+15\le 5\\ \because & {𝑥}^2+7𝑥+15\le 5\\ \Rightarrow & {𝑥}^2+7𝑥+10\le 0\\ & (𝑥+5)(𝑥+2)\le 0\\ \Rightarrow & 𝑥\in [-5,-2]\end{array}$

We have, $𝑓\left(𝑛\right)=𝐴(-2{)}^{𝑛}+𝐵(-3{)}^{𝑛}$

$\begin{array}{cc}𝑓\left(𝑛\right)+𝑎𝑓(𝑛-1)+𝑏𝑓(𝑛-2)& =0\\ \therefore 𝐴(-2{)}^{𝑛}+𝐵(-3{)}^{𝑛}+𝑎\left(𝐴\right(-2{)}^{𝑛-1}& \\ +𝐵(-3{)}^{𝑛-1})+𝑏\left(𝐴\right(-2{)}^{𝑛-2}+𝐵(-3{)}^{𝑛-2}& =0\\ \Rightarrow 𝐴(-2{)}^{𝑛-2}[4-2𝑎+𝑏]+(-3{)}^{𝑛-2}& \\ 𝐵[9-3𝑎+𝑏]& =0\end{array}$

∴ It is possible only

$4-2𝑎+𝑏=0\text{ and }9-3𝑎+𝑏=0$

Solving, we get $𝑎=5,𝑏=6$

$\therefore (𝑎+𝑏)(𝑏-𝑎)=(5+6)(6-5)=11$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \mid \begin{array}{ccc}2𝑎-2𝑏-4& 4𝑎& 4𝑎\\ 4& 2-𝑏-𝑎& 4\\ 2𝑏& 2𝑏& 𝑏-𝑎-2\end{array}\mid \\ & =2\mid \begin{array}{ccc}𝑎-𝑏-2& 2𝑎& 2𝑎\\ 4& 2-𝑏-𝑎& 4\\ 2𝑏& 2𝑏& 𝑏-𝑎-2\end{array}\mid \\ & =2\mid \begin{array}{ccc}-(𝑎+𝑏+2)& 0& 2𝑎\\ 𝑎+𝑏+2& -(𝑎+𝑏+2)& 4\\ 0& 𝑎+𝑏+2& 𝑏-𝑎-2\end{array}\mid \\ & {𝐶}_1\to {𝐶}_1-{𝐶}_2\text{ and }{𝐶}_2\to {𝐶}_2-{𝐶}_3\end{array}\end{array}$

$\begin{array}{cc}& =2(𝑎+𝑏+2{)}^2\mid \begin{array}{ccc}-1& 0& 2𝑎\\ 1& -1& 4\\ 0& 1& 𝑏-𝑎-2\end{array}\mid \\ & =2(𝑎+𝑏+2{)}^2(-1(-𝑏+𝑎+2-4)+2𝑎\left(1\right)\\ & =2(𝑎+𝑏+2{)}^2(𝑎+𝑏+2)=2(𝑎+𝑏+2{)}^3\\ & =2\left[\right(𝑎+𝑏{)}^3+6(𝑎+𝑏{)}^2+12(𝑎+𝑏)+8]\end{array}$

We have,

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& 𝑥\end{array}⎤⎦⎥\\ {𝐴}^2& =⎡⎣⎢\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& 𝑥\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& 𝑥\end{array}⎤⎦⎥\\ {𝐴}^2& =⎡⎣⎢\begin{array}{ccc}2& -2& -16-4𝑥\\ -1& 3& 16+4𝑥\\ 4+𝑥& -8-2𝑥& -12+{𝑥}^2\end{array}⎤⎦⎥\end{array}$

Given, ${𝐴}^2=𝐴$

$\begin{array}{cc}\because & 4+𝑥=1\Rightarrow 𝑥=-3\\ \therefore & 𝐴=⎡⎣⎢\begin{array}{ccc}2& -2& -4\\ -1& 3& 4\\ 1& -2& -3\end{array}⎤⎦⎥\\ & \left|𝐴\right|=2(-9+8)+2(3-4)-4(2-3)\\ & =-2-2+4=0\end{array}$

Rank of $𝐴=2\Rightarrow 𝑟+𝑥=2-3=-1$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}{\Delta }_1& =\mid \begin{array}{cc}𝑚& 𝑏\\ 𝑛& 𝑑\end{array}\mid =𝑚𝑑-𝑏𝑛\\ {\Delta }_2& =\mid \begin{array}{cc}𝑎& 𝑚\\ 𝑐& 𝑛\end{array}\mid =𝑎𝑛-𝑚𝑐\\ {\Delta }_3& =\mid \begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\mid =𝑎𝑑-𝑏𝑐\\ {𝑥}^{𝑎}{𝑦}^{𝑏}& ={𝑒}^{𝑚},{𝑥}^{𝑐}{𝑦}^{𝑑}={𝑒}^{𝑛}\\ 𝑦& ={\left(\frac{{𝑒}^{𝑚}}{{𝑥}^{𝑎}}\right)}^{1/𝑏}\text{ and }𝑦={\left(\frac{{𝑒}^{𝑛}}{{𝑥}^{𝑐}}\right)}^{1/𝑑}\end{array}\end{array}$

$\begin{array}{cccc}& \therefore & \frac{{𝑒}^{𝑚/𝑏}}{{𝑥}^{𝑎/𝑏}}& =\frac{{𝑒}^{𝑛/𝑑}}{{𝑥}^{𝑐/𝑑}}\\ \Rightarrow & & {𝑥}^{𝑐/𝑑-𝑎/𝑏}& ={𝑒}^{𝑛/𝑑-𝑚/𝑏}\\ \Rightarrow & & 𝑥& ={\left({𝑒}^{𝑛/𝑑-𝑚/𝑏}\right)}^{\frac{1}{𝑐}-\frac{𝑎}{𝑏}}\\ \Rightarrow & & 𝑥& ={𝑒}^{(\frac{𝑏𝑛-𝑚𝑑}{𝑏𝑑}\times \frac{𝑏𝑑}{𝑏𝑐-𝑎𝑑})}\\ \Rightarrow & & 𝑥& ={𝑒}^{\frac{𝑏𝑛-𝑚𝑑}{𝑏𝑐-𝑎𝑑}}\\ \Rightarrow & & 𝑥& ={𝑒}^{\frac{𝑚𝑑-𝑏𝑛}{𝑎𝑑-𝑏𝑐}}={𝑒}^{\frac{{\Delta }_1}{{\Delta }_3}}\end{array}$

Putting the value of $𝑥$ in $𝑦=\frac{{𝑒}^{𝑚/𝑏}}{{𝑥}^{𝑎/𝑏}}$ we get $𝑦={𝑒}^{\frac{𝑎𝑛-𝑚𝑐}{𝑎𝑑-𝑏𝑐}}={𝑒}^{\frac{{\Delta }_2}{{\Delta }_3}}$

We have, $\left|𝑧\right|-𝑧=2+𝑖$

$\begin{array}{cc}& \text{ Let }𝑧=𝑥+𝑖𝑦\\ & \therefore \sqrt{\surd {𝑥}^2+{𝑦}^2}-𝑥-𝑖𝑦=2+𝑖\end{array}$

Equating real part and imaginary part, we get

$\begin{array}{cccc}& & \sqrt{\surd {𝑥}^2+{𝑦}^2}-𝑥& =2\text{ and }𝑦=-1\\ & \therefore & \sqrt{\surd {𝑥}^2+1}& =𝑥+2\end{array}$

$\begin{array}{cc}\Rightarrow & {𝑥}^2+1={𝑥}^2+4𝑥+4\\ \Rightarrow & 𝑥=-\frac{3}{4}\\ \Rightarrow & 𝑧=-\frac{3}{4}-𝑖\\ \therefore & \left|𝑧\right|=\sqrt{\surd \frac{9}{16}+1}\\ & =\sqrt{\surd \frac{25}{16}}=\frac{5}{4}\end{array}$

Given, $\arg (𝑧-2-3𝑖)=\frac{𝜋}{4}$

$\begin{array}{cc}\Rightarrow 𝑧& =𝑥+𝑖𝑦\\ 𝑧-2-3𝑖& =𝑥+𝑖𝑦-2-3𝑖\\ & =(𝑥-2)+(𝑦-3)𝑖\\ \arg (𝑧-2-3𝑖)& ={\tan }^{-1}\left(\frac{𝑦-3}{𝑥-2}\right)=\frac{𝜋}{4}\\ \Rightarrow \frac{𝑦-3}{𝑥-2}& =\tan \frac{𝜋}{4}\\ 𝑦-3& =𝑥-2\\ \Rightarrow 𝑥-𝑦+1& =0\end{array}$

∴ Locus of $𝑧$ is $𝑥-𝑦+1=0$.

We have,

$\begin{array}{c}1+\frac{\cos 𝜃}{2}+\frac{\cos 2𝜃}{4}+\frac{\cos 3𝜃}{8}+\dots .\\ =\frac{𝑎-2\cos 𝜃}{5+𝑏\cos 𝜃}\end{array}$

Put, $𝜃=0$

$\begin{array}{cccc}& \therefore & 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots .& =\frac{𝑎-2}{5+𝑏}\\ & \Rightarrow & \frac{1}{1-\frac{1}{2}}& =\frac{𝑎-2}{5+𝑏}\\ & \Rightarrow & 2& =\frac{𝑎-2}{5+𝑏}\\ & \Rightarrow & 10+2𝑏& =𝑎-2\\ & \Rightarrow & 𝑎-2𝑏& =12...\left(𝑖\right)\end{array}$

Put, $𝜃=𝜋$

$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\dots =\frac{𝑎+2}{5-𝑏}$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow & \frac{1}{1+\frac{1}{2}}=\frac{𝑎+2}{5-𝑏}\\ \Rightarrow & \frac{2}{3}=\frac{𝑎+2}{5-𝑏}\\ \Rightarrow & 10-2𝑏=3𝑎+6\\ \Rightarrow & 3𝑎+2𝑏=4...\left(𝑖𝑖\right)\end{array}\\ & \text{ From Eq. (i) and (ii), we get }\\ & \begin{array}{c}𝑎=4,𝑏=-4\\ \therefore (𝑎-𝑏{)}^2=(4+4{)}^2=8^2=64\end{array}\end{array}$

Given, ${𝑧}_1,{𝑧}_2,{𝑧}_3,\dots ,{𝑧}_{𝑛}$ are roots of equation

$\begin{array}{cc}(𝑧+1{)}^{𝑛}& ={𝑧}^{𝑛}\Rightarrow \left(\frac{𝑧+1}{𝑧}\right)=(1{)}^{1/𝑛}\\ \frac{𝑧+1}{𝑧}& =(\cos \frac{2𝑘𝜋}{𝑛}+𝑖\sin \frac{2𝑘𝜋}{𝑛})\\ 𝑧+1& =𝑍(\cos \frac{2𝑘𝜋}{𝑛}+𝑖\sin \frac{2𝑘𝜋}{𝑛})\\ 𝑧& =\frac{1}{\cos \frac{2𝑘𝜋}{𝑛}+𝑖\sin \frac{2𝑘𝜋}{𝑛}-1}\\ 𝑧& =\frac{1}{2}(1-𝑖\cot \frac{𝑘𝜋}{𝑛})\\ \mathrm{Re}𝑧& =\frac{1}{2^{\prime }}\mathrm{Im}\left(𝑧\right)=(\frac{-1}{2}\cot \frac{𝑘𝜋}{𝑛})\\ 2\mathrm{Re}\left(𝑧𝑖\right)& =1,{\cot }^{-1}\left[2\right|\mathrm{Im}𝑧𝑖\left|\right]=\frac{𝑘𝜋}{𝑛}\end{array}$

$\begin{array}{cc}& \sum _{𝑖=1}^{𝑣}\frac{{\cot }^{-1}\left(2\right(\mathrm{Im}𝑧𝑖)-1}{2\mathrm{Re}\left(𝑧𝑖\right)}=\sum _{𝑖=1}^{𝑣}(\frac{𝑘𝜋}{𝑛}-1)\\ & =\frac{𝜋}{𝑛}(1+2+3+\dots ..𝑛)-𝑛\\ & =𝜋\frac{\left(𝑛\right)(𝑛+1)}{2\left(𝑛\right)}-𝑛\\ & =𝜋\frac{(𝑛+1)}{2}-𝑛\\ & =\frac{1}{2}[𝑛𝜋+𝜋-2𝑛]\\ & =\frac{1}{2}[𝜋-(𝜋-2\left)𝑛\right]\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Let }𝑦=\frac{{𝑥}^2+14𝑥+9}{{𝑥}^2+2𝑥+3}\\ & \Rightarrow {𝑥}^2𝑦+2𝑥𝑦+3𝑦={𝑥}^2+14𝑥+9\\ & \Rightarrow {𝑥}^2(𝑦-1)+2𝑥(𝑦-7)+3𝑦-9=0\end{array}\\ & \text{ Here, }𝑥\in 𝐑\end{array}$

$\begin{array}{cc}\therefore 4(𝑦-7{)}^2-4(𝑦-1\left)\right(3𝑦-9)& \ge 0\\ (𝑦-7{)}^2-(3{𝑦}^2-12𝑦+9)& \ge 0\\ {𝑦}^2-14𝑦+49-3{𝑦}^2+12𝑦-9& \ge 0\end{array}$

$\begin{array}{cc}\Rightarrow & 2{𝑦}^2+2𝑦-40\le 0\\ \Rightarrow & {𝑦}^2+𝑦-20\le 0\\ & (𝑦+5)(𝑦-4)\le 0\end{array}$

$\therefore 𝑦\in [-5,4]$

∴ Maximum and minimum value of $\frac{{𝑥}^2+14𝑥+9}{{𝑥}^2+2𝑥+3}$ are 4 and -5 respectively.

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \frac{\sqrt{\surd 12-𝑥-{𝑥}^2}}{𝑥+10}\le \frac{\sqrt{\surd 12-𝑥-{𝑥}^2}}{2𝑥+9}\\ \Rightarrow & \sqrt{\surd 12-𝑥-{𝑥}^2}(2𝑥+9-𝑥-10)\le 0\\ \Rightarrow & \sqrt{\surd 12-𝑥-{𝑥}^2}(𝑥-1)\le 0\\ \therefore & 12-𝑥-{𝑥}^2\ge 0\text{ and }𝑥\le 1\\ \Rightarrow & {𝑥}^2+𝑥-12\le 0\text{ and }𝑥\le 1\\ \Rightarrow & (𝑥+4)(𝑥-3)\le 0\text{ and }𝑥\le 1\\ \Rightarrow & 𝑥\in [-4,3]\text{ and }𝑥\in (-\infty ,1]\\ \therefore & 𝑥\in [-4,1]\cup \left\{3\right\}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given equation, }\\ & \begin{array}{c}6{𝑥}^6-25{𝑥}^5+31{𝑥}^4-31{𝑥}^2+25𝑥-6=0\\ 6({𝑥}^6-1)-25𝑥({𝑥}^4-1)+31({𝑥}^4-{𝑥}^2)=0\\ ({𝑥}^2-1)[6{𝑥}^4+6{𝑥}^2+6-25{𝑥}^3-25𝑥+31{𝑥}^2]=0\\ 6{𝑥}^4+6-25({𝑥}^3+𝑥)+37{𝑥}^2=0\\ 6({𝑥}^2+\frac{1}{{𝑥}^2})-25(𝑥+\frac{1}{𝑥})+37=0\\ 6{(𝑥+\frac{1}{𝑥})}^2-12-25(𝑥+\frac{1}{𝑥})+37=0\\ 6{(𝑥+\frac{1}{𝑥})}^2-25(𝑥+\frac{1}{𝑥})+25=0\end{array}\end{array}$

$\begin{array}{cc}\therefore 6{𝑦}^2-25𝑦+25& =0[\because 𝑥+\frac{1}{𝑥}=𝑦]𝑥\\ (6𝑦-15)(6𝑦-10)& =0\\ 𝑦& =\frac{5}{2},\frac{5}{3}\end{array}$

$\begin{array}{cccc}& \Rightarrow & 𝑥+\frac{1}{𝑥}& =\frac{5}{2}\\ & \Rightarrow & 𝑥& =2,\frac{1}{2}\\ & & 𝑥+\frac{1}{𝑥}& =\frac{5}{3}\text{ has no real roots }\\ & \therefore & 𝛼+𝛽& =\frac{5}{3}\end{array}$

We have,

${𝑥}^5-8{𝑥}^4+25{𝑥}^3-38{𝑥}^2+28𝑥-8=0$

Let

$\begin{array}{cc}𝑓\left(𝑥\right)=& {𝑥}^5-8{𝑥}^4+25{𝑥}^3-38{𝑥}^2+28𝑥-8\\ & 𝑓\left(2\right)=32-128+200-152+56-8\\ & 𝑓\left(2\right)=0\\ \Rightarrow & {𝑓}^{\prime }\left(𝑥\right)=5{𝑥}^4-32{𝑥}^3+75{𝑥}^2-76𝑥+28\\ & {𝑓}^{\prime }\left(2\right)=80-256+300-152+28\\ \Rightarrow & {𝑓}^{\prime }\left(2\right)=0\\ & {𝑓}^{\prime \prime }\left(𝑥\right)=20{𝑥}^3-96{𝑥}^2+150𝑥-76\\ & {𝑓}^{\prime \prime }\left(2\right)=160-384+300-76\\ \Rightarrow & {𝑓}^{\prime \prime }\left(2\right)=0\\ & {𝑓}^{\prime \prime \prime }\left(𝑥\right)=60{𝑥}^2-192𝑥+150\\ & {𝑓}^{\prime \prime \prime }\left(2\right)=240-384+150=6\\ \Rightarrow & {𝑓}^{\prime \prime \prime }\left(2\right)\ne 0\end{array}$

∴ Given, $𝛼$ is a root of multiplicity 3

$\begin{array}{ccc}\therefore & 𝛼& =2\\ \Rightarrow & {𝛼}^2-5𝛼+6& =4-10+6=0\end{array}$

The number of positive integral solution of ${𝑥}_1+{𝑥}_2+{𝑥}_3+{𝑥}_4=10$ is ${}^9{𝐶}_3=\frac{9\times 8\times 7}{1\times 2\times 3}=84$

∴ Statement I is false.

The exponent of 2 in 25! is

$\begin{array}{cc}\left[\frac{25}{2}\right]+\left[\frac{25}{4}\right]& +\left[\frac{25}{8}\right]+\left[\frac{25}{16}\right]\\ & =12+6+3+1\\ & =22\end{array}$

The exponent of 5 in (25)! is

$\left[\frac{25}{5}\right]+\left[\frac{25}{25}\right]=5+1=6$

∴ Exponent of 10 in $25!=6$

Hence, (25)! $={10}^6\times 𝑘$

$\Rightarrow 𝑘\in 𝐍$

∴ Statement II is true.

The total number of collection of atleast $(𝑛+1)$ books from $(2𝑛+1)$ books is,

${}^{2𝑛+1}{𝐶}_{𝑛+1}+{}^{2𝑛+1}{𝐶}_{𝑛+2}+\dots {}^{2𝑛+1}{𝐶}_{2𝑛+1}=255$

$\begin{array}{cc}& \begin{array}{cccc}\Rightarrow & & \frac{2^{2𝑛+1}}{2}-1& =255\\ \Rightarrow & & 2^{2𝑛}& =256\\ \Rightarrow & & 2^{2𝑛}& =2^8\\ \Rightarrow & & 𝑛& =4\end{array}\\ & \text{ ∴ Total number of books }\\ & \begin{array}{cc}& =2𝑛+1\\ & =2\times 4+1=9\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \frac{{(1+\frac{2𝑥}{3})}^{-4}(4+5𝑥{)}^{1/2}}{(9+𝑥{)}^{3/2}}\\ & ={(1+\frac{2𝑥}{3})}^{-4}\times 2{(1+\frac{5}{4}𝑥)}^{1/2}\times (9+𝑥{)}^{-3/2}\\ & ={(1+\frac{2𝑥}{3})}^{-4}\times 2\times {(1+\frac{5}{4}𝑥)}^{1/2}\times \frac{1}{27}{(1+\frac{𝑥}{9})}^{-3/2}\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& =\frac{2}{27}(1-\frac{8𝑥}{3})(1+\frac{5}{8}𝑥)(1-\frac{1}{6}𝑥)\\ & =\frac{2}{27}(1-\frac{8}{3}𝑥+\frac{5}{8}𝑥-\frac{1}{6}𝑥)\\ & =\frac{2}{27}(1-\frac{53}{24}𝑥)\end{array}\\ & \text{ Put, }𝑥=\frac{6}{371}\\ & \begin{array}{cc}& =\frac{2}{27}(1-\frac{53\times 6}{24\times 371})\\ & =\frac{2}{27}(1-\frac{53}{1484})\\ & =\frac{2}{27}\times \frac{1431}{1484}=\frac{1}{14}\end{array}\end{array}$

We have, $\sqrt{\surd 3+𝑥}+\sqrt{\surd 5+𝑥}$

Here, $3<𝑥<5$

$\begin{array}{cc}\therefore & {𝑥}^{1/2}{(1+\frac{3}{𝑥})}^{1/2}+5^{1/2}{(1+\frac{𝑥}{5})}^{1/2}\\ & {𝑥}^{1/2}[1+\frac{1}{2}\left(\frac{3}{𝑥}\right)+\frac{1}{2}(\frac{1}{2}-1)\left(\frac{9}{{𝑥}^2}\right)\dots ]\end{array}$

$+5^{1/2}⎡⎣⎢\begin{array}{c}1+\frac{1}{2}\left(\frac{𝑥}{5}\right)+\frac{1}{2}(\frac{1}{2}-1)\left(\frac{{𝑥}^2}{5^2}\right)\\ +\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\left(\frac{{𝑥}^3}{5^3}\right)\end{array}⎤⎦⎥$

∴ Coefficient of ${𝑥}^{-3/2}$ is,

$\frac{1}{2}(\frac{1}{2}-1)9=\frac{1}{2}\times -\frac{1}{2}\times 9=-\frac{9}{4}$

and coefficient of ${𝑥}^3$ is,

$\begin{array}{cc}{5}^{1/2}& \times \frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\frac{1}{5^3}\\ & =5^{\frac{1}{2}-3}\times \frac{1}{2}\times -\frac{1}{2}\times -\frac{3}{2}=5^{\frac{-5}{2}}\times \frac{3}{8}\end{array}$

Sum of coefficient of ${𝑥}^{-3/2}$ and ${𝑥}^3=\frac{-9}{4}+\frac{3}{8}{5}^{-5/2}$

$=\frac{-18+3(5{)}^{-5/2}}{8}$

We have,

$\begin{array}{cc}\frac{{𝑥}^5-5}{{𝑥}^5+{𝑥}^2}& =𝑓\left(𝑥\right)+\frac{𝐴}{𝑥}+\frac{𝐵}{{𝑥}^2}+\frac{𝐶}{𝑥+1}\\ \frac{{𝑥}^5-5}{{𝑥}^3+{𝑥}^2}& ={𝑥}^2-𝑥+1+\frac{-{𝑥}^2-5}{{𝑥}^3+{𝑥}^2}\\ \therefore \frac{-{𝑥}^2-5}{{𝑥}^3+{𝑥}^2}& =\frac{𝐴}{𝑥}+\frac{𝐵}{{𝑥}^2}+\frac{𝐶}{𝑥+1}\\ -{𝑥}^2-5& =𝐴𝑥(𝑥+1)+𝐵(𝑥+1)+𝐶\left({𝑥}^2\right)\\ -{𝑥}^2-5& =(𝐴+𝐶){𝑥}^2+(𝐴+𝐵)𝑥+𝐵\\ 𝐴+𝐶& =-1,𝐴+𝐵=0,𝐵=-5\end{array}$

Solving we get, $𝐴=5,𝐵=-5,𝐶=-6$

$\begin{array}{cc}\therefore & 𝑓\left(𝑥\right)={𝑥}^2-𝑥+1\\ \therefore & 𝑓\left(𝐾\right)={𝐾}^2-𝐾+1\end{array}$

Given, $𝑓\left(𝐾\right)+𝐴+𝐵+𝐶=1$

$\begin{array}{cccc}& & {𝐾}^2-𝐾+1+5-5-6& =1\\ \Rightarrow & & {𝐾}^2-𝐾-6& =0\\ \Rightarrow & & (𝐾+2)(𝐾-3)& =0\\ \Rightarrow & & 𝐾& =3,-2\end{array}$