TG EAPCET 2025 2nd May Evening Shift Question Paper with Solutions

We have, $3𝑓\left(𝑥\right)+4𝑓\left(\frac{1}{𝑥}\right)=\frac{2}{𝑥}-1$

$\therefore 3𝑓\left(3\right)+4𝑓\left(\frac{1}{3}\right)=\frac{2}{3}-1=\frac{-1}{3}...\left(𝑖\right)$

Also, $3𝑓\left(\frac{1}{3}\right)+4𝑓\left(3\right)=6-1=5...\left(𝑖𝑖\right)$

On solving Eqs. (i) and (ii), we get

$𝑓\left(3\right)=3$

$\begin{array}{cc}& \text{ We have, }𝑦=\frac{{10}^{𝑥}-{10}^{-𝑥}}{{10}^{𝑥}+{10}^{-𝑥}}+1\\ & \Rightarrow 𝑦-1=\frac{{10}^{𝑥}-{10}^{-𝑥}}{{10}^{𝑥}+{10}^{-𝑥}}\\ & \text{ Let }𝑎={10}^{𝑥}\\ & \therefore {10}^{-𝑥}=\frac{1}{𝑎}\\ & \therefore 𝑦-1=\frac{𝑎-\frac{1}{𝑎}}{𝑎+\frac{1}{𝑎}}=\frac{{𝑎}^2-1}{{𝑎}^2+1}\\ & \Rightarrow (𝑦-1)({𝑎}^2+1)={𝑎}^2-1\Rightarrow (𝑦-1){𝑎}^2+(𝑦-1)={𝑎}^2-1\\ & \Rightarrow {𝑎}^2(𝑦-1-1)=-𝑦\Rightarrow {𝑎}^2(𝑦-2)=-𝑦\\ & \Rightarrow {𝑎}^2=\frac{-𝑦}{𝑦-2}\Rightarrow {10}^{2𝑥}=\frac{-𝑦}{𝑦-2}\\ & \Rightarrow 2𝑥={\log }_{10}\left(\frac{-𝑦}{𝑦-2}\right)\Rightarrow 2𝑥={\log }_{10}\left(\frac{𝑦}{2-𝑦}\right)\\ & \Rightarrow 𝑥=\frac{1}{2}{\log }_{10}\left(\frac{𝑦}{2-𝑦}\right)\end{array}$

We have, ${49}^{𝑘}+1$ is a factor of

$\begin{array}{cc}& 48({49}^{125}+{49}^{124}+\dots +{49}^2+49+1)\\ & \text{ Let }𝑆={49}^{125}+{49}^{124}+\dots +49+1\\ & \Rightarrow 𝑆=\frac{{49}^{126}-1}{49-1}=\frac{{49}^{126}-1}{48}\Rightarrow 48𝑆={49}^{126}-1\end{array}$

Since, ${49}^{𝑘}+1$ divides ${49}^{126}-1$

$\therefore 𝑘$ divides 126 .

$\Rightarrow 2𝑘$ divides 126

$\Rightarrow$ Greatest possible integer $𝑘=63$

We have, $\mid \begin{array}{ccc}1& 2& 3-𝜆\\ 0& -1-𝜆& 2\\ 1-𝜆& 1& 3\end{array}\mid =𝐴{𝜆}^3+𝐵{𝜆}^2+𝐶𝜆+𝐷$

Now, $\mid \begin{array}{ccc}1& 2& 3-𝜆\\ 0& -1-𝜆& 2\\ 1-𝜆& 1& 3\end{array}\mid$

$\begin{array}{cc}& =1(-3-3𝜆-2)+(1-𝜆)(4+(1+𝜆\left)\right(3-𝜆\left)\right)\\ & =(-3𝜆-5)+(1-𝜆)(4+3-𝜆+3𝜆-{𝜆}^2)\\ & =(-3𝜆-5)+(1-𝜆)(7+2𝜆-{𝜆}^2)\\ & =(-3𝜆-5)+(7+2𝜆-{𝜆}^2-7𝜆-2{𝜆}^2+{𝜆}^3)\\ & ={𝜆}^3-3{𝜆}^2-8𝜆+2\end{array}$

$\begin{array}{cc}& \therefore {𝜆}^3-3{𝜆}^2-8𝜆+2=𝐴{𝜆}^3+𝐵{𝜆}^2+𝐶𝜆+𝐷\\ & \text{ On comparing both sides, we get }\\ & \begin{array}{cc}𝐴& =1,𝐵=-3,𝐶=-8,𝐷=2\\ \therefore 𝐷+𝐴& =2+1=3\end{array}\end{array}$

We have

$𝐴+2𝐵=⎡⎣⎢\begin{array}{ccc}1& 2& 0\\ 6& -3& 3\\ -5& 3& 1\end{array}⎤⎦⎥...\left(𝑖\right)$

And $2𝐴-𝐵=⎡⎣⎢\begin{array}{ccc}2& -1& 5\\ 2& -1& 6\\ 0& 1& 2\end{array}⎤⎦⎥...\left(𝑖𝑖\right)$

Now, $2𝐴+4𝐵=⎡⎣⎢\begin{array}{ccc}2& 4& 0\\ 12& -6& 6\\ -10& 6& 2\end{array}⎤⎦⎥...\left(𝑖𝑖𝑖\right)$

Substract Eqs. (ii) from (iii), we get

$\begin{array}{cc}& \begin{array}{cc}& 5𝐵=⎡⎣⎢\begin{array}{ccc}2& 4& 0\\ 12& -6& 6\\ -10& 6& 2\end{array}⎤⎦⎥-⎡⎣⎢\begin{array}{ccc}2& -1& 5\\ 2& -1& 6\\ 0& 1& 2\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}0& 5& -5\\ 10& -5& 0\\ -10& 5& 0\end{array}⎤⎦⎥\\ \Rightarrow & 𝐵=⎡⎣⎢\begin{array}{ccc}0& 1& -1\\ 2& -1& 0\\ -2& 1& 0\end{array}⎤⎦⎥\end{array}\\ & \therefore \text{ From Eqs. (ii), we get }\\ & \begin{array}{cc}& 2𝐴=⎡⎣⎢\begin{array}{ccc}2& -1& 5\\ 2& -1& 6\\ 0& 1& 2\end{array}⎤⎦⎥+⎡⎣⎢\begin{array}{ccc}0& 1& -1\\ 2& -1& 0\\ -2& 1& 0\end{array}⎤⎦⎥\\ \Rightarrow & 2𝐴=⎡⎣⎢\begin{array}{ccc}2& 0& 4\\ 4& -2& 6\\ -2& 2& 2\end{array}⎤⎦⎥\Rightarrow 𝐴=⎡⎣⎢\begin{array}{ccc}1& 0& 2\\ 2& -1& 3\\ -1& 1& 1\end{array}⎤⎦⎥\\ \therefore & \mathrm{tr}\left(𝐴\right)-\mathrm{tr}\left(𝐵\right)\\ & =(1-1+1)-(0-1+0)=1+1=2\end{array}\end{array}$

We have,

$𝐴,𝐶$ are $3\times 3$ matrices

$𝐵\times 𝐷$ are $3\times 1$ matrices

$𝐴𝑋=𝐵$ has unique solution

$𝐶𝑋=𝐷$ has infinite solution.

Since, $𝐴𝑋=𝐵$ have a unique solution.

$\Rightarrow \mathrm{rank}\left(𝐴\right)=$ number of unknowns $=3$

$\therefore \mathrm{rank}\left(\right[𝐴:𝐵\left]\right)=\mathrm{rank}\left(𝐴\right)=3...\left(𝑖\right)$

Since, $𝐶𝑋=𝐷$ have infinite many solutions

$\therefore \mathrm{Rank}\left(𝐶\right)=$ not unique $\Rightarrow \mathrm{rank}\left(𝐶\right)<3$

$\Rightarrow \mathrm{Rank}\left(\right[𝐶:𝐷\left]\right)<3...\left(𝑖𝑖\right)$

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

$\mathrm{Rank}\left(\right[𝐴:𝐷\left]\right)\ge \mathrm{rank}\left[\right(𝐶:𝐵\left)\right]$

We have, $𝐴$ and $𝐵$ are two non-square matrices such that

$𝑃=𝐴+𝐵,𝑄={𝐴}^{𝑇}𝐵,𝑅=𝐴{𝐵}^{𝑇}$

Since, $𝐴+𝐵$ is defined,

So order of $𝐴$ and $𝐵$ are same.

Let order of $𝐴$ and $𝐵$ be $𝑚\times 𝑛$

$\therefore$ Order of ${𝐴}^{𝑇}$ is $𝑛\times 𝑚$ and order of ${𝐵}^{𝑇}$ is $𝑛\times 𝑚$

Now, order of $𝑃=$ order of $𝐴+𝐵=𝑚\times 𝑛$

Order of $𝑄=$ order of ${𝐴}^{𝑇}𝐵=𝑛\times 𝑛$

And order of $𝑅=$ order of $𝐴{𝐵}^{𝑇}=𝑚\times 𝑚$

Now, order of $𝑃𝑄=𝑚\times 𝑛$

Order of $𝑄𝑅=$ not defined

Order of $𝑅𝑄=$ not defined

Order of $𝑄𝑃=$ not defined

Order of $𝑅𝑃=𝑚\times 𝑛$

$\therefore$ Order of $𝐴=$ order of $𝑃𝑄$ and $𝑅𝑃$

We have, $|𝑧-1|\le 2$

$\begin{array}{cc}𝐴𝑛𝑑& \mid 𝜔𝑧-1-{𝜔}^2\mid =𝑟\\ & \left|𝜔\right|\mid 𝑧-{𝜔}^2-𝜔\mid =𝑟\\ & |𝑧+1|=𝑟\\ & [\because |𝜔|=1,1+𝜔+{𝜔}^2=0]\\ & |𝑧-1|\le 2\Rightarrow |𝑧+1-2|\le 2\\ & |𝑧+1|-2\le 2\Rightarrow |𝑧+1|\le 4\end{array}$

For no solution

$\therefore |𝑧+1|>4\Rightarrow 𝑟>4$

$\begin{array}{cc}& \text{ Given, }\left|𝑧\right|=2,{𝑧}_1=\frac{𝑧}{2}{𝑒}^{𝑖𝛼}\\ & \therefore 𝑧=2{𝑒}^{𝑖𝜃}\end{array}$

( $𝜃$ is the argument of $𝑧$ )

And  ${𝑧}_1=\frac{2{𝑒}^{𝑖𝜃}}{2}{𝑒}^{𝑖𝛼}={𝑒}^{𝑖(𝜃+𝛼)}$

Now, ${𝑧}_{𝑛}^1={𝑒}^{𝑖𝑛(𝜃+𝛼)}$

$\Rightarrow {𝑧}_{-𝑛}^1={𝑒}^{-𝑖𝑛(𝜃+𝛼)}$

We know that $\frac{{𝑒}^{𝑖𝑥}-{𝑒}^{-𝑖𝑥}}{{𝑒}^{𝑖𝑥}+{𝑒}^{-𝑖𝑥}}=𝑖\tan 𝑥$

$\begin{array}{cc}\therefore \frac{{𝑧}_{𝑛}^1-{𝑧}_{-𝑛}^1}{{𝑧}_{𝑛}^1+{𝑧}_{-𝑛}^1}& =\frac{{𝑒}^{𝑖𝑛(𝜃+𝛼)}-{𝑒}^{-𝑖𝑛(𝜃+𝛼)}}{{𝑒}^{𝑖𝑛(𝜃+𝛼)}+{𝑒}^{-𝑖𝑛(𝜃+𝛼)}}\\ & =𝑖\tan \left(𝑛\right(𝜃+𝛼\left)\right)\\ & =𝑖\tan (𝑛𝜃+𝑛𝛼)\end{array}$

Given, $𝑛,𝐾\in 𝑁$ such that $𝑛\ne 3𝐾$

Let $𝑧=\sqrt{\surd 3}+𝑖$, then $\overline{𝑧}=\sqrt{\surd 3}-𝑖$

$\begin{array}{cc}& \left|𝑧\right|=\left|\overline{𝑧}\right|=\sqrt{\surd 3+1}=2\\ & \arg \left(𝑧\right)={\tan }^{-1}\left(\frac{1}{\sqrt{\surd 3}}\right)=\frac{𝜋}{6}\\ & \arg \left(𝑧\right)={\tan }^{-1}(-\frac{1}{\sqrt{\surd 3}})=\frac{-𝜋}{6}\\ & \therefore 𝑧=2(\cos \frac{𝜋}{6}+𝑖\sin \frac{𝜋}{6})\\ & \Rightarrow {𝑧}^{2𝑛}=2^{2𝑛}(\cos \frac{2𝑛𝜋}{6}+𝑖\sin \frac{2𝑛𝜋}{6})\\ & \text{ And }\overline{𝑧}=2(\cos \frac{𝜋}{6}-𝑖\sin \frac{𝜋}{6})\\ & \Rightarrow {\overline{𝑧}}^{2𝑛}=2^{2𝑛}(\cos \frac{2𝑛𝜋}{6}-𝑖\sin \frac{2𝑛𝜋}{6})\\ & \therefore {𝑧}^{2𝑛}+{\overline{𝑧}}^{2𝑛}=2\cdot 2^{2𝑛}\cos \left(\frac{𝑛𝜋}{3}\right)\end{array}$

$=2^{2𝑛+1}\cos \left(\frac{𝑛𝜋}{3}\right)$

Since, $𝑛\ne 3𝐾\Rightarrow 𝑛$ is not a multiple of 3 $\cos \left(\frac{𝑛𝜋}{3}\right)\ne \cos \left(𝑘𝜋\right)=\pm 1=(-1{)}^{𝑛+1}{2}^{2𝑛}$

Let $𝑧=1-𝑖\sqrt{\surd 3}$

$\begin{array}{cc}& \therefore \left|𝑧\right|=\sqrt{\surd 1+(\sqrt{\surd 3}{)}^2}=2\\ & \arg \left(𝑧\right)={\tan }^{-1}\left(\frac{-\sqrt{\surd 3}}{1}\right)=-\tan \frac{𝜋}{3}\\ & \therefore 𝑧=2(\cos \left(\frac{-𝜋}{3}\right)+𝑖\sin (-\frac{𝜋}{3}))\\ & =2{𝑒}^{-𝑖𝜋/3}\\ & \therefore \text{exception 25:}=2^{1/3}\cdot {𝑒}^{4-𝑖𝜋/3+2𝑘𝜋𝑖}\\ & =2^{1/3}{𝑒}^{𝑖(-\frac{𝜋}{9}+\frac{2𝑘𝜋}{3})}\end{array}$

For any integer $𝑘\in \{0,1,2\}$

${𝜔}_0=$ lies in 4th quadrant

${𝜔}_1=$ lies in 2nd quadrant

${𝜔}_2=$ lies in 3rd quadrant

Given, $𝑙$ is the maximum value of $-3{𝑥}^2+4𝑥+1$ and $𝑚$ is the minimum value of $3{𝑥}^2+4𝑥+1$

For $-3{𝑥}^2+4𝑥+1,\frac{-𝑏}{2𝑎}=\frac{-4}{2(-3)}=\frac{2}{3}$

$\begin{array}{ccc}\therefore \text{ Maximum value }& =& -3{\left(\frac{2}{3}\right)}^2+4\left(\frac{2}{3}\right)+1\\ & & =-3\times \frac{4}{9}+\frac{8}{3}+1\\ \Rightarrow & & 𝑙=\frac{-4}{3}+\frac{8}{3}+1=\frac{7}{3}\end{array}$

For $3{𝑥}^2+4𝑥+1,\frac{-𝑏}{2𝑎}=\frac{-4}{2\times 3}=\frac{-2}{3}$

$\therefore$ Minimum value

$\begin{array}{cc}& =3{\left(\frac{-2}{3}\right)}^2+4\left(\frac{-2}{3}\right)+1\\ & =3\times \frac{4}{9}-\frac{8}{3}+1\\ \Rightarrow 𝑚& =\frac{4}{3}-\frac{8}{3}+1=\frac{-1}{3}\end{array}$

Now, foci of hyperbola $=(𝑙,0)$

$=(\frac{7}{3},0)$

And $(7𝑚,0)=(\frac{-7}{3},0)$

$\begin{array}{cc}& \therefore \text{ Distance between foci }=\\ & \Rightarrow 2𝑎\times 2=2\times \frac{7}{3}\Rightarrow 𝑎=\frac{7}{6}\\ & \text{ Also, }{𝑏}^2={𝑎}^2({𝑒}^2-1)\\ & \Rightarrow {𝑏}^2=\frac{49}{36}(4-1)\\ & \Rightarrow {𝑏}^2=\frac{49}{36}\times 3=\frac{49}{12}\end{array}$

$\begin{array}{cc}& \therefore \text{ Equation of hyperbola, }\frac{{𝑥}^2}{\frac{49}{36}}-\frac{{𝑦}^2}{\frac{49}{12}}=1\\ & \Rightarrow 36{𝑥}^2-12{𝑦}^2=49\end{array}$

Given equation is

${𝑥}^2-3𝑎𝑥+{𝑎}^2-2𝑎-𝑘=0$

For distinct real roots, $𝐷>0$

$\begin{array}{cc}& =(-3𝑎{)}^2-4\times 1\times ({𝑎}^2-2𝑎-𝑘)>0\\ & =9{𝑎}^2-4{𝑎}^2+8𝑎+4𝑘>0\\ & =5{𝑎}^2+8𝑎+4𝑘>0\end{array}$

Since, leading coefficient of ${𝑎}^2$ is $5>0$

So, $(8{)}^2-4\times (5\left)\right(4𝑘)<0=64-80𝑘<0$

$\begin{array}{cc}\Rightarrow 64& <80𝑘\\ & =𝑘>\frac{4}{5}\Rightarrow 𝑘\in (\frac{4}{5},\infty )\end{array}$

Let

$\begin{array}{cc}& 𝑓\left(𝑥\right)={𝑥}^4-10{𝑥}^3+37{𝑥}^2-60𝑥+36=0\\ & \text{ At }𝑥=1,𝑓\left(1\right)=1-10+37-60+36\ne 0\\ & \text{ At }𝑥=2,𝑓\left(2\right)=(2^4-10(2^3+37(2^2\\ & -60(2+36=0\\ & =16-80+148-120+36\\ & =200-200=0\end{array}$

So, $𝑥=2$ is a factor of $𝑓\left(𝑥\right)$

Similarly $𝑥=3$ is a factor of $𝑓\left(𝑥\right)$

$\therefore 𝑓\left(𝑥\right)=(𝑥-2{)}^2(𝑥-3{)}^2$

$\therefore$ Original roots are $2,2,3,3$

The distinct roots are 2 and 3

There are three possible pairs of distinct roots

(i) Increase the two $2𝑠$ by 1

(ii) Increase the two $3𝑠$ by 1

(iii) Increase one 2 and one 3 by 1

Let us assume we choose one root 2 and one root 3 to increase by 1

Since, these distinct roots.

Increase one ' 2 ' by $1:2+1=3$

Increase one ' 3 ' by $1:3+1=4$

Other two roots ( 2 and 3 ) are kept fixed

$\therefore$ New sets of roots of he transformed equation is $3,4,2,3$

Let the transformed equation be $𝑔\left(𝑥\right)$. its roots are $2,3,3,4$

$\begin{array}{cc}\therefore 𝑔\left(𝑥\right)& =(𝑥-2)(𝑥-3)(𝑥-3)(𝑥-4)\\ & =(𝑥-2)(𝑥-3{)}^2(𝑥-4)\end{array}$

The common roots are value of $𝑥$ for which $𝑓\left(𝑥\right)=0$ and $𝑔\left(𝑥\right)=0$

$\therefore$ Common roots are $2,3,3$

$\therefore$ Number of common roots $=3$

Given, $𝛼,𝛽,𝛾$ are the roots of the equations ${𝑥}^3-𝑃{𝑥}^2+𝑄𝑥-𝑅=0$

$\therefore \begin{array}{c}𝛼+𝛽+𝛾=𝑃\\ 𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑄\end{array}$

And $𝛼𝛽𝛾=𝑅$

Also, $(𝛼-2{)}^2,(𝛽-2{)}^2,(𝛾-2{)}^2$ are the roots of equations, ${𝑥}^3-5{𝑥}^2+4𝑥=0$

Now, ${𝑥}^2-5{𝑥}^2+4𝑥=0$

$\begin{array}{cc}\Rightarrow & 𝑥({𝑥}^2-5𝑥+4)=0\\ \Rightarrow & 𝑥(𝑥-1)(𝑥-4)=0\\ \Rightarrow & 𝑥=0,1,4\\ \therefore & (𝛼-2{)}^2=0\Rightarrow 𝛼=2\\ & (𝛽-2{)}^2=1\Rightarrow 𝛽-2=\pm 1\Rightarrow 𝛽=1,3\end{array}$

And $(𝛾-2{)}^2=4\Rightarrow 𝛾-2=\pm 2\Rightarrow 𝛾=0,4$

Now,

$\begin{array}{cc}𝑃+𝑄+𝑅=(𝛼+𝛽+𝛾)+(𝛼𝛽+𝛽𝛾& +𝛾𝛼)+𝛼𝛽𝛾\end{array}$

Case I when $𝛼=2,𝛽=1,𝛾=0$

then,

$\begin{array}{cc}𝑃+𝑄+𝑅& =(2+1+0)+(2+0+0)\\ & =5\end{array}$

Case II when $𝛼=2,𝛽=1,𝛾=4$

then,

$\begin{array}{cc}𝑃+𝑄+𝑅=(2+1+4)+(2+4+8)& +(2\times 1\times 4)=29\end{array}$

Case III when $𝛼=2,𝛽=3,𝛾=0$ then

$\begin{array}{cc}𝑃+𝑄+𝑅& =(2+3+0)+(6+0+0)+0\\ & =11\end{array}$

Case IV when $𝛼=2,𝛽=3,𝛾=4$

then

$\begin{array}{cc}𝑃+𝑄+𝑅& =(2+3+4)+(6+12+8)+24\\ & =59\end{array}$

Clearly, least value of $𝑃+𝑄+𝑅$ is 5 .

We have, $𝑥+𝑦+𝑧+𝑡=10...\left(𝑖\right)$

When $𝑥\ge 2,𝑧\ge 5$

let ${𝑥}^{\prime }=𝑥-2$

$\Rightarrow 𝑥={𝑥}^{\prime }+2$

${𝑧}^{\prime }=𝑧-5$

$\Rightarrow 𝑧={𝑧}^{\prime }+5$

Here, ${𝑥}^{\prime },{𝑧}^{\prime },𝑦\ge 0$ and $𝑡\ge 0$ are non-negative integers.

$\therefore$ Substitude these value in Eq. (i), we get

$\begin{array}{cc}& ({𝑥}^{\prime }+2)+𝑦+({𝑧}^{\prime }+5)+𝑡=10\\ \Rightarrow & {𝑥}^{\prime }+𝑦+{𝑧}^{\prime }+𝑡=3...\left(𝑖𝑖\right)\end{array}$

$\therefore$ Number of non-negative integral solution for Eq. (ii)

$\begin{array}{cc}& ={}^{3+4-1}{𝐶}_{4-1}={}^6{𝐶}_3\\ & =\frac{6\times 5\times 4}{3\times 2\times 1}=20\end{array}$

Integers lying between 1000 and 10000 such that the sum of all the digits in each of those numbers becomes 30 .

$\therefore$ Coefficient of ${𝑥}^{30}$ in the expansion of

$\begin{array}{cc}({𝑥}^1+{𝑥}^2& +\dots +{𝑥}^9){({𝑥}^0+{𝑥}^1+\dots +{𝑥}^9)}^3\\ & =𝑥\left(\frac{1-{𝑥}^9}{1-𝑥}\right){\left(\frac{1-{𝑥}^{10}}{1-𝑥}\right)}^3\\ & =𝑥(1-{𝑥}^9){(1-{𝑥}^{10})}^3(1-𝑥{)}^{-4}\end{array}$

$\Rightarrow$ Coefficient of ${𝑥}^{29}$ in

$(1-{𝑥}^9){(1-{𝑥}^{10})}^3(1-𝑥{)}^{-4}$

$\begin{array}{cc}& =(1-{𝑥}^9)(1-3{𝑥}^{10}+3{𝑥}^{20}-{𝑥}^{30})(1-𝑥{)}^{-4}\\ & =(1-{𝑥}^9-3{𝑥}^{10}+3{𝑥}^{19}+3{𝑥}^{20}-3{𝑥}^{29}\\ & -{𝑥}^{30}+{𝑥}^{39})(1-𝑥{)}^{-4}\end{array}$

Here, $(1-𝑥{)}^{-4}=\sum _{\infty }^{𝑘=0}𝑘+{}^3{𝐶}_3{𝑥}^{𝑘}$

$\therefore$ Coefficient of ${𝑥}^{29}$ from $1\cdot (1-𝑥{)}^{-4}$

$={}^{29+3}{𝐶}_3={}^{32}{𝐶}_3=4960$

Coefficient of ${𝑥}^{20}$ from $-{𝑥}^9\cdot (1-𝑥{)}^{-4}$

$=-{}^{20+3}{𝐶}_3=-{}^{23}{𝐶}_3=-1771$

Coefficient of ${𝑥}^{19}$ from $-3{𝑥}^{10}\cdot (1-𝑥{)}^{-4}$

$\begin{array}{cc}& =-3\left({}^{19+3}{𝐶}_3\right)\\ & =-3\left(1540\right)=-4620\end{array}$

Coefficient of ${𝑥}^{10}$ from $3{𝑥}^{19}(1-𝑥{)}^{-4}$

(Means form is $3{𝑥}^{19}{\Sigma }^{𝑘+3}{𝐶}_3{𝑥}^{𝑘}$ need

$\begin{array}{cc}{𝑥}^{29},𝑘& =10)\\ & =3^{10+{}^3}{𝐶}_3=3\left({}^{13}{𝐶}_3\right)=858\end{array}$

Coefficient of ${𝑥}^9$ from $3{𝑥}^{20}\cdot (1-𝑥{)}^{-4}$

$=3\left({}^{9+3}{𝐶}_3\right)=3\left(220\right)=660$

Coefficient of ${𝑥}^0$ from $-3{𝑥}^{29}\cdot (1-𝑥{)}^{-4}$

$=-3\left({}^{0+3}{𝐶}_3\right)=-3\left(1\right)=-3$

$\therefore$ Total numbers

$\begin{array}{cc}& =4960-1771-4620+858+660-3\\ & =84\end{array}$

Given word in most

Words start with $𝑀=3!=6$

Words start with $O=3!=6$

Words start with $\mathrm{SM}=2!=2$

Words start with $\mathrm{SO}=2!=2$

Words start with STM = 1 ! = 1

Words start with STO is STOM

Rank of STOM is $(6+6+2+2+1)+1=18$

Let the given expression,

$𝐸={(1+\frac{1}{𝑥})}^{20}\left[30𝑥\right(1+𝑥{)}^{29}+(1+𝑥{)}^{30}]$

Here, $30𝑥(1+𝑥{)}^{29}+(1+𝑥{)}^{30}$

$\begin{array}{cc}& =(1+𝑥{)}^{29}[30𝑥+(1+𝑥)]\\ & =(1+𝑥{)}^{29}(31𝑥+1)\\ \therefore 𝐸& ={(1+\frac{1}{𝑥})}^{20}(1+𝑥{)}^{29}(31𝑥+1)\\ & =\frac{(𝑥+1{)}^{20}}{{𝑥}^{20}}(1+𝑥{)}^{29}(31𝑥+1)\\ & =\frac{(𝑥+1{)}^{49}}{{𝑥}^{20}}(31𝑥+1)\\ & =(𝑥+1{)}^{49}(31{𝑥}^{-19}+{𝑥}^{-20})\\ & =31{𝑥}^{-19}(𝑥+1{)}^{49}+{𝑥}^{-20}(𝑥+1{)}^{49}\end{array}$