TG EAPCET 2020 10th September Evening Shift Question Paper with Solutions

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ We have, }𝑓\left(𝑥\right)=𝑥-\frac{1}{𝑥}\mathrm{gl}\\ & \therefore \left(𝑓\right(𝑥){)}^3={𝑥}^3-\frac{1}{{𝑥}^3}-3(𝑥)\left(\frac{1}{𝑥}\right)(𝑥-\frac{1}{𝑥})\\ & =𝑓\left({𝑥}^3\right)-3(𝑥-\frac{1}{𝑥})\\ & \Rightarrow \left(𝑓\right(𝑥){)}^3=𝑓\left({𝑥}^3\right)-3𝑓(𝑥)\\ & \Rightarrow 3𝑓\left(𝑥\right)=𝑓\left({𝑥}^3\right)-\left[𝑓\right(𝑥){]}^3\end{array}\end{array}$

We have,

$𝑓\left(𝑥\right)=\left[𝑥\right]\text{ and }𝑔\left(𝑥\right)=3\left[\frac{𝑥}{3}\right]$

Given, $𝑓\left(𝑥\right)=𝑔\left(𝑥\right)$

$\therefore \left[𝑥\right]=3\left[\frac{𝑥}{3}\right]$

Here, $\left[𝑥\right]$ and $\left[\frac{𝑥}{3}\right]$ is an integers

Let $\left[\frac{𝑥}{3}\right]=𝑘$

$\begin{array}{c}\because \left[𝑥\right]=3𝑘\\ \because 𝑥\in [3𝑘,3𝑘+1)\\ \{𝑥\in 𝐑/3𝑘\le 𝑥<3𝑘+1,𝑘\in 𝐙\}\end{array}$

We have,

$\begin{array}{cc}& \begin{array}{cc}{𝑆}_{𝑛}=& 1^2+2\times 2^2+3^2+2\times 4^2+5^2\\ & +2\times 6^2+\dots \infty \end{array}\\ \therefore & {𝑇}_{𝑛}=(2𝑛-1{)}^2+2\times (2𝑛{)}^2\\ & =4{𝑛}^2+1-4𝑛+8{𝑛}^2\\ =& 12{𝑛}^2-4𝑛+1\\ & \Sigma {𝑇}_{𝑛}=12\Sigma {𝑛}^2-4\Sigma 𝑛+\Sigma 1\\ =& 12\frac{𝑛(𝑛+1)(2𝑛+1)}{6}-\frac{4𝑛(𝑛+1)}{2}+𝑛\\ =& 2𝑛(𝑛+1)(2𝑛+1)-2𝑛(𝑛+1)+𝑛\\ =& 2𝑛(𝑛+1)(2𝑛+1-1)+𝑛\\ =& 2𝑛(𝑛+1)\left(2𝑛\right)+𝑛\\ =& 4{𝑛}^2(𝑛+1)+𝑛\\ =& 𝑛\left[4𝑛\right(𝑛+1)+1]\\ =& 𝑛(4{𝑛}^2+4𝑛+1)=𝑛(2𝑛+1{)}^2\end{array}$

Here, ${𝑆}_{𝑛}$ is even

$\begin{array}{cc}& {𝑆}_{𝑚}=\frac{𝑚}{2}{(2\frac{𝑚}{2}+1)}^2𝑛=\frac{𝑚}{2}\\ \Rightarrow & {𝑆}_{𝑚}=\frac{𝑚}{2}(𝑚+1{)}^2\\ \therefore & {𝑆}_{𝑛}=\frac{𝑛}{2}(𝑛+1{)}^2\text{ when }𝑛\text{ is even }\end{array}$

We have, $𝐴=⎡⎣⎢\begin{array}{ccc}1& 4& 2\\ 2& -1& 4\\ -3& 7& -6\end{array}⎤⎦⎥$ and

$\begin{array}{cc}& 𝐵=⎡⎣⎢\begin{array}{ccc}{𝑏}_{11}& {𝑏}_{12}& {𝑏}_{13}\\ {𝑏}_{21}& {𝑏}_{22}& {𝑏}_{23}\\ {𝑏}_{31}& {𝑏}_{32}& {𝑏}_{33}\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}2& 0& -2\\ {𝑏}_{21}& {𝑏}_{22}& {𝑏}_{23}\\ {𝑏}_{31}& {𝑏}_{32}& {𝑏}_{33}\end{array}⎤⎦⎥\\ & \therefore 𝐴𝐵=⎡⎣⎢\begin{array}{ccc}1& 4& 2\\ 2& -1& 4\\ -3& 7& -6\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}2& 0& -2\\ {𝑏}_{21}& {𝑏}_{22}& {𝑏}_{23}\\ {𝑏}_{31}& {𝑏}_{32}& {𝑏}_{33}\end{array}⎤⎦⎥\end{array}$

$=⎡⎣⎢\begin{array}{cc}2+4{𝑏}_{21}+2{𝑏}_{31}& 0+4{𝑏}_{22}+2{𝑏}_{32}\\ 4-{𝑏}_{21}+4{𝑏}_{31}& 0+{𝑏}_{22}+4{𝑏}_{32}\\ -6+7{𝑏}_{21}-6{𝑏}_{31}& 0+7{𝑏}_{22}-6{𝑏}_{32}\\ & -2+4{𝑏}_{23}+2{𝑏}_{33}\\ & -4-{𝑏}_{23}+4{𝑏}_{33}\\ & 6+7{𝑏}_{23}-6{𝑏}_{33}\end{array}⎤⎦⎥$

$=⎡⎣⎢\begin{array}{ccc}2& 14& -4\\ 4& 1& -8\\ -6& 15& 12\end{array}⎤⎦⎥$

On solving above equal matrices with corresponding elements, we get

$\begin{array}{cc}& {𝑏}_{21}={𝑏}_{31}=0,{𝑏}_{22}=3,{𝑏}_{32}=1,{𝑏}_{23}=0\\ & \text{ and }{𝑏}_{33}=-1\\ & \therefore 𝐵=⎡⎣⎢\begin{array}{ccc}2& 0& -2\\ 0& 3& 0\\ 0& 1& -1\end{array}⎤⎦⎥\\ & \therefore \left|𝐵\right|=2(-3-0)+(-2)(0-0)=-6\\ & \text{ and Trace }\left(𝐵\right)=2+3-1=4\\ & \therefore \left|𝐵\right|+\mathrm{trace}\left(𝐵\right)=-6+4=-2\end{array}$

Given, $𝐴$ is $𝑚\times 𝑛$ matrix of rank 4

$𝐴$ contains $𝑚$ th order of non-singular sub matrix.

$𝐴$ is non-singular matrix

$\therefore 𝐴$ is a square matrix of order $𝑀$

∴ Rank of $𝐴$ is $𝑀$

∴ order of $𝐴=4$       [rank of $𝐴=4$ ]

Hence, rows of $𝐴=4$.

Given, $𝐶$ and $𝐷$ are non-singular matrix of order $𝑛$

$\begin{array}{cc}\because \left|𝐶\right|& \ne 0,\left|𝐷\right|\ne 0\\ 𝐶𝐷& =-𝐷𝐶\\ \left|𝐶𝐷\right|& =|-𝐷𝐶|\\ \left|𝐶\right|\left|𝐷\right|& =(-1{)}^{𝑛}|𝐷\left|\right|𝐶|\\ 1& =(-1{)}^{𝑛}\end{array}$

$\therefore 𝑛$ is even integer.

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}{𝑧}_1& ={𝑥}_1+𝑖{𝑦}_1\\ {𝑧}_2& ={𝑥}_2+𝑖{𝑦}_2\\ {𝑧}_3& ={𝑥}_1+\frac{𝑖{𝑥}_2}{2}\\ {𝑧}_4& =2{𝑦}_1+𝑖{𝑦}_2\\ \left|{𝑧}_1\right|& =1,\left|{𝑧}_2\right|=2\text{ and }\mathrm{Re}\\ \left({𝑧}_1{𝑧}_2\right)=0& \end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }{𝑧}_1={𝑒}^{𝑖𝛼}\text{ and }{𝑧}_2=2{𝑒}^{𝑖𝛽}\\ & {𝑧}_1{𝑧}_2=2{𝑒}^{𝑖(𝛼+𝛽)}\end{array}$

$\begin{array}{cc}& {𝑧}_1{𝑧}_2=2(\cos (𝛼+𝛽)+𝑖\sin (𝛼+𝛽\left)\right)\\ & \mathrm{Re}({𝑧}_1+{𝑧}_2)=2\cos (𝛼+𝛽)=0\\ & \therefore 𝛼+𝛽=\frac{𝜋}{2}\\ & 𝛽=\frac{𝜋}{2}-𝛼\end{array}$

$\begin{array}{cc}\therefore {𝑧}_1& =\cos 𝛼+𝑖\sin 𝛼\\ {𝑧}_2& =2(\cos 𝛽+𝑖\sin 𝛽)\\ {𝑧}_2& =2(\sin 𝛼+𝑖\cos 𝛼)\\ \therefore {𝑧}_3& =\cos 𝛼+𝑖\left(\frac{2\sin 𝛼}{2}\right)\\ {𝑧}_3& =\cos 𝛼+𝑖\sin 𝛼\\ \left|{𝑧}_3\right|& =1\\ {𝑧}_4& =2(\sin 𝛼+𝑖\cos 𝛼)\\ \left|{𝑧}_4\right|& =2\\ {𝑧}_3{𝑧}_4& =2(\cos 𝛼+𝑖\sin 𝛼)(\sin 𝛼+𝑖\cos 𝛼)\\ {𝑧}_3{𝑧}_4& =2\left[\right(\cos 𝛼\sin 𝛼-\cos 𝛼\sin 𝛼)\\ & +𝑖({\sin }^2𝛼+{\cos }^2𝛼)]\end{array}$

$\begin{array}{c}{𝑧}_3{𝑧}_4=2𝑖\\ \mathrm{Re}\left({𝑧}_3{𝑧}_4\right)=0\end{array}$

We have, $\left|𝑧\right|\ge 3$

Then, $\mid 𝑧+\frac{3}{𝑧}\mid \le \left|𝑧\right|+\mid \frac{3}{𝑧}\mid$

$=\mid 𝑧+\frac{3}{𝑧}\mid \le 3+\mid \frac{3}{𝑧}\mid$

∴ Least value of $\mid 𝑧+\frac{3}{𝑧}\mid$ is not 1

And $|{𝑧}_1-{𝑧}_2|\le \left|{𝑧}_1\right|+\left|{𝑧}_2\right|$ which is true.

$\begin{array}{cc}& \text{ At }𝜃=\frac{𝜋}{2}\\ & {\left(\frac{\cos 𝜃+𝑖\sin 𝜃}{\sin 𝜃+𝑖\cos 𝜃}\right)}^{2020}+{\left(\frac{1+\cos 𝜃+𝑖\sin 𝜃}{1-\cos 𝜃+𝑖\sin 𝜃}\right)}^{2021}\end{array}$

$\begin{array}{cc}& ={\left(\frac{0+𝑖\cdot 1}{1+𝑖\cdot 0}\right)}^{2020}+{\left(\frac{1+0+𝑖}{1-0+𝑖}\right)}^{2021}\\ & =(𝑖{)}^{2020}+{\left(\frac{1+𝑖}{1+𝑖}\right)}^{2021}\\ & =1+(1{)}^{2021}=1+1\\ & =2=𝑥+𝑖𝑦\Rightarrow 𝑥=2,𝑦=0\\ \therefore 𝑥& +𝑦=2+0=2\end{array}$

As, $𝜔$ is complex root of unity, then

$\begin{array}{c}1+𝜔+{𝜔}^2=0,{𝜔}^3=1\\ \therefore (𝜔𝑥+2)({𝜔}^2𝑥+2)-3\\ ={𝜔}^3{𝑥}^2+2𝜔𝑥+2{𝜔}^2𝑥+4-3\\ ={𝑥}^2+2𝑥(𝜔+{𝜔}^2)+1(\because {𝜔}^3=1)\\ ={𝑥}^2+2𝑥(-1)+1={𝑥}^2-2𝑥+1\\ \therefore \sum _{𝑥=1}^{10}\left(\right(𝜔𝑥+2)({𝜔}^2𝑥+2)-3)\\ =\sum _{𝑥=1}^{10}({𝑥}^2-2𝑥+1)=\sum _{𝑥=1}^{10}{𝑥}^2-2\sum _{𝑥=1}^{10}𝑥+\sum _{𝑥=1}^{10}1\\ =\frac{10(10+1)(2\times 10+1)}{6}-\frac{2\times 10(10+1)}{2}+10\\ =\frac{10\times 11\times 21}{6}-10\times 11+10\\ =385-100=285\end{array}$

We have,

$\sqrt{\surd \frac{5𝑥}{𝑥-2}}+\sqrt{\surd \frac{𝑥-2}{5𝑥}}=\frac{29}{10}$

Let $\sqrt{\surd \frac{5𝑥}{𝑥-2}}=𝑦$

Then, $𝑦+\frac{1}{𝑦}=\frac{29}{10}$

$\begin{array}{cc}\Rightarrow & \frac{{𝑦}^2+1}{𝑦}=\frac{29}{10}\\ \Rightarrow & 10{𝑦}^2+10=29𝑦\\ \Rightarrow & 10{𝑦}^2-29𝑦+10=0\\ \Rightarrow & 10{𝑦}^2-25𝑦-4𝑦+10=0\\ \Rightarrow & 5𝑦(2𝑦-5)-2(2𝑦-5)=0\\ \Rightarrow & (5𝑦-2)(2𝑦-5)=0\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑦=\frac{2}{5},\frac{5}{2}\\ & \text{ Now, if }𝑦=\frac{2}{5}\Rightarrow \sqrt{\surd \frac{5𝑥}{𝑥-2}}=\frac{2}{5}\end{array}$

$\begin{array}{cc}\Rightarrow & \frac{5𝑥}{𝑥-2}=\frac{4}{25}\\ \Rightarrow & 125𝑥=4𝑥-8\\ \Rightarrow & 121𝑥=-8\\ \Rightarrow & 𝑥=-8/121\end{array}$

$\text{ And if }𝑦=\frac{5}{2}\Rightarrow \sqrt{\surd \frac{5𝑥}{𝑥-2}}=\frac{5}{2}$

$\begin{array}{cc}\Rightarrow & \frac{5𝑥}{𝑥-2}=\frac{25}{4}\\ \Rightarrow & \frac{𝑥}{𝑥-2}=\frac{5}{4}\\ \Rightarrow & 5𝑥-10=4𝑥\\ \Rightarrow & 𝑥=10\end{array}$

$\begin{array}{cc}& \therefore \text{ Let }𝛼=10\text{ and }𝛽=\frac{-8}{121}(\because 𝛼>𝛽)\\ & \\ & \begin{array}{cc}\therefore & \sqrt{\surd {𝛼}^2-{11}^4{𝛽}^2}\\ =& \sqrt{\surd (10{)}^2-{11}^4{\left(\frac{-8}{121}\right)}^2}\\ =& \sqrt{\surd 100-64}\\ =& \sqrt{\surd 36}=6\end{array}\end{array}$

Let $𝑦=\frac{9\cdot 3^{2𝑥}+6\cdot 3^{𝑥}+4}{9\cdot 3^{2𝑥}-6\cdot 3^{𝑥}+4}$

Put, $3^{𝑥}=𝑡$

$\begin{array}{cc}& \therefore 𝑦=\frac{9\cdot {𝑡}^2+6𝑡+4}{9{𝑡}^2-6𝑡+4}\\ & \Rightarrow 𝑦(9{𝑡}^2-6𝑡+4)=9{𝑡}^2+6𝑡+4\\ & \Rightarrow 9{𝑡}^2(𝑦-1)-6𝑡(𝑦+1)+(4𝑦-4)=0\\ & \Rightarrow (𝑦-1){𝑡}^2\frac{-6}{9}(𝑦+1)𝑡+\frac{4}{9}(𝑦-1)=0\\ & \Rightarrow (𝑦-1){𝑡}^2-\frac{2}{3}(𝑦+1)𝑡+\frac{4}{9}(𝑦-1)=0\end{array}$

$\because 𝑡$ is real

$\therefore 𝐷\ge 0$

$\begin{array}{c}\Rightarrow \frac{4}{9}(𝑦+1{)}^2-4(𝑦-1\left)\frac{4}{9}\right(𝑦-1)\ge 0\\ \Rightarrow \frac{4}{9}(𝑦+1{)}^2-\frac{16}{9}(𝑦-1{)}^2\ge 0\\ \Rightarrow (𝑦+1{)}^2-4(𝑦-1{)}^2\ge 0\\ \Rightarrow {𝑦}^2+1+2𝑦-4{𝑦}^2-4+8𝑦\ge 0\\ \Rightarrow -3{𝑦}^2+10𝑦-3\ge 0\\ \Rightarrow 3{𝑦}^2-10𝑦+3\le 0\\ \Rightarrow (𝑦-3)(3𝑦-1)\le 0\\ \Rightarrow \frac{1}{3}\le 𝑦\le 3\\ \text{ Hence, minimum value }=\frac{1}{3}.\end{array}$

Let the roots of the given cubic equation ${𝑥}^3-𝑝{𝑥}^2+𝑝𝑥-1=0$ be $𝛼,𝛽,𝛾$.

From the standard relationships between coefficients and roots, we get:

$\begin{array}{cc}𝛼+𝛽+𝛾& =𝑝,\\ 𝛼𝛽+𝛽𝛾+𝛾𝛼& =𝑝,\\ 𝛼𝛽𝛾& =1.\end{array}$

The question says that the new equation, whose roots are the squares of the original roots (i.e., ${𝛼}^2,{𝛽}^2,{𝛾}^2$), is the same as the given equation itself. Therefore, the symmetric sums of these squared roots must also be:

$\begin{array}{cc}{𝛼}^2+{𝛽}^2+{𝛾}^2& =𝑝,\\ {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2& =𝑝,\\ {𝛼}^2{𝛽}^2{𝛾}^2& =1.\end{array}$

Now, using the identity:

$(𝛼+𝛽+𝛾{)}^2={𝛼}^2+{𝛽}^2+{𝛾}^2+2(𝛼𝛽+𝛽𝛾+𝛾𝛼)$

Substitute the known values:

${𝑝}^2=𝑝+2𝑝$

Simplify this to get:

${𝑝}^2=3𝑝$

Since $𝑝\ne 0$, divide both sides by $𝑝$:

$𝑝=3$

$\begin{array}{cc}& \text{ Let roots of the equation }\\ & \begin{array}{cc}& 8{𝑥}^3+6𝑝{𝑥}^2+3𝑞𝑥-27=0\text{ are }\frac{𝑎}{𝑟},𝑎,\text{ ar }.\\ & \therefore \frac{𝑎}{𝑟}+𝑎+𝑎𝑟=\frac{-6𝑝}{8}=\frac{-3𝑝}{4}\\ & \Rightarrow 𝑎(\frac{1}{𝑟}+1+𝑟)=\frac{-3𝑝}{4}...\left(𝑖\right)\\ & (\frac{𝑎}{𝑟}\cdot 𝑎)+(𝑎\cdot 𝑎𝑟)+(\frac{𝑎}{𝑟}\cdot 𝑎𝑟)=\frac{3𝑞}{8}\\ & \Rightarrow {𝑎}^2(\frac{1}{𝑟}+𝑟+1)=\frac{3𝑞}{8}...\left(𝑖𝑖\right)\\ & \text{ and }\frac{𝑎}{𝑟}\times 𝑎\times 𝑎𝑟=\frac{27}{8}\\ & \Rightarrow {𝑎}^3=\frac{27}{8}\Rightarrow 𝑎=\frac{3}{2}...\left(𝑖𝑖𝑖\right)\end{array}\end{array}$

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

$𝑎=\frac{3𝑞/8}{-3𝑝/4}=-\frac{𝑞}{2𝑝}...\left(𝑖𝑣\right)$

From Eqs. (iii)and (iv), we have

$\frac{-𝑞}{2𝑝}=\frac{3}{2}\Rightarrow -𝑞=3𝑝$

Now,

${𝑞}^2+9{𝑝}^2+6𝑝𝑞+\frac{𝑞}{𝑝}$

$\begin{array}{cc}\Rightarrow & 9{𝑝}^2+9{𝑝}^2+6𝑝(-3𝑝)+\frac{(-3𝑝)}{𝑝}\\ \Rightarrow & 18{𝑝}^2-18{𝑝}^2-3=-3\end{array}$

Given, word ATRAPATRAM

Here, $A\to 4$ times

$\begin{array}{cc}& T\to 2\text{ times }\\ & R\to 2\text{ times }\\ & P\to 1\text{ time }\\ & M\to 1\text{ time }\end{array}$

(i) All A's are together,

Then it will be consider as one letter and remaining 6 letters and $l^{\prime }{A}^{\prime }s$ (including 4 A's) will be consider as 7 letters.

∴ Number of arrangement

$=\frac{7!}{2!2!}=𝑥...\left(𝑖\right)$

(ii) no two A's are together. such that-T-R-P-T-R-M-

∴ Number of arrangement

$\begin{array}{cc}& ={}^7{𝐶}_4\times \frac{6!}{2!2!}=𝑦...\left(𝑖𝑖\right)\\ \therefore & 𝑥+𝑦=\frac{7!}{2!2!}+{}^7{𝐶}_4\times \frac{6!}{2!2!}\\ =& \frac{7\times 6!}{2!2!}+{}^7{𝐶}_3\times \frac{6!}{2!2!}=\frac{6!}{2!2!}(7+{}^7{𝐶}_3)\\ =& \frac{6!}{2!2!}(7+35)=\frac{6!}{2!2!}\left(42\right)\end{array}$

Number between 1 and 10000 are formed using digit 2 and 3 only one and digit 4 twice

$\text{ one- }\text{ digit }=\{2,3,4\}=3$

Two digit

$=\{23,24,32,34,42,43,44\}=7$

Three digit number

$\begin{array}{cc}& =\{234,243,244,324,342,344\\ & 423,432,424,434,442,443\}=12\end{array}$

Four digit number

$\begin{array}{c}=\{2344,2434,2443,3244,3424,\\ 3442,4234,4324\dots \}\end{array}$

Rank of $324=14$

Rank of $4324=3+7+12+9=31$

$\therefore 𝑥-𝑦=31-14=17$

Given, 9th and 10th terms are numerically greatest terms in the expansion of $(5𝑥-6𝑦{)}^{𝑛}$

$9\le \frac{(𝑛+1)\left(\frac{6𝑦}{5𝑥}\right)}{1+\left(\frac{6𝑦}{5𝑥}\right)}$

Here, $𝑦=\frac{1}{2}$ and $𝑥=\frac{2}{5}$

$\begin{array}{cc}\because & \frac{𝑦}{𝑥}=\frac{5}{4}\Rightarrow \frac{6𝑦}{5𝑥}=\frac{3}{2}\\ & 9\le \frac{(𝑛+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}\\ & 9\le \frac{3(𝑛+1)}{5}\\ & 𝑛+1\ge 15\\ & 𝑛\ge 14\\ \because & 𝑛=14\end{array}$

$\begin{array}{cc}& \text{ Middle term of }(5𝑥-6𝑦{)}^{14}\\ & =\mid {}^{14}{𝐶}_7\left(5𝑥{)}^7\right(-6𝑦{)}^7\mid \\ & =\mid {}^{14}{𝐶}_7{(5\times \frac{2}{5})}^7{(-6\times \frac{1}{2})}^7\mid \\ & ={}^{14}{𝐶}_7\times 6^7\end{array}$

$\begin{array}{cc}& \text{We have, }\\ & 1-\frac{3}{16}+\frac{1\cdot 4}{1\cdot 2}{\left(\frac{3}{16}\right)}^2-\frac{1\cdot 4\cdot 7}{1\cdot 2\cdot 3}{\left(\frac{3}{16}\right)}^3+\dots =\end{array}$

$\begin{array}{cc}& \text{ We know that, }\\ & \begin{array}{cc}(1-𝑥{)}^{𝑛}=1-𝑛𝑥+& \frac{𝑛(𝑛-1)}{2!}{𝑥}^2\\ & -\frac{𝑛(𝑛-1)(𝑛-2)}{3!}{𝑥}^3\end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }(1-𝑥{)}^{𝑛}=1-\frac{3}{16}+\frac{1\cdot 4}{1\cdot 2}{\left(\frac{3}{16}\right)}^2\\ & \begin{array}{cc}& -\frac{1\cdot 4\cdot 7}{1\cdot 2\cdot 3}{\left(\frac{3}{16}\right)}^3\dots \\ \therefore 𝑛𝑥=\frac{3}{16}& \text{ and }\frac{𝑛(𝑛-1)}{2!}{𝑥}^2=\frac{1\cdot 4}{2!}{\left(\frac{3}{16}\right)}^2\\ 𝑛𝑥& =\frac{3}{16}\text{ and }\\ 𝑛(𝑛-1){𝑥}^2& =4{\left(\frac{3}{16}\right)}^2\end{array}\end{array}$

$\begin{array}{c}{𝑛}^2\cdot {𝑥}^2=\frac{3^2}{16}\text{ and }𝑛(𝑛-1){𝑥}^2=4{\left(\frac{3}{16}\right)}^2\\ 𝑛(𝑛-1){𝑥}^2=4{𝑛}^2{𝑥}^2\\ {𝑛}^2-𝑛=4{𝑛}^2\\ 𝑛-1=4𝑛\Rightarrow 𝑛=-\frac{1}{3}\\ 𝑥=\frac{3}{16}\times -3=\frac{-9}{16}\\ \because (1-𝑥{)}^{𝑛}={(1+\frac{9}{16})}^{-1/3}\\ ={\left(\frac{25}{16}\right)}^{-1/3}={\left(\frac{4}{5}\right)}^{2/3}\end{array}$

$\begin{array}{cc}& \begin{array}{c}𝐼𝑓\frac{2𝑥+1}{(𝑥-1{)}^2({𝑥}^2+1)}=\frac{𝐴}{𝑥-1}+\frac{𝐵}{(𝑥-1{)}^2}+\frac{𝐶𝑥+𝐷}{{𝑥}^2+1}\end{array}\\ & \begin{array}{c}\Rightarrow 2𝑥+1=𝐴(𝑥-1)({𝑥}^2+1)+𝐵({𝑥}^2+1)+(𝐶𝑥+𝐷)(𝑥-1{)}^2\end{array}\\ & \Rightarrow 2𝑥+1={𝑥}^3(𝐴+𝐶)+{𝑥}^2(-𝐴+𝐵-2𝐶+𝐷)+𝑥(𝐴+𝐶-2𝐷)+(-𝐴+𝐵+𝐷)\end{array}$