AP EAPCET 2025 23rd May Evening Shift Question Paper with Solutions

$\begin{array}{cc}\because [2𝑥-2]& =\left[\right(2𝑥-1)-1]\\ & =[2𝑥-1]-1\\ \text{ let, }[2𝑥-1]& =𝑎\\ \Rightarrow [2𝑥-2]& =𝑎-1\\ \text{ so, }3(𝑎-1)+1& =2𝑎-1\\ \Rightarrow 3𝑎-3+1& =2𝑎-1\Rightarrow 𝑎=1\end{array}$

thus $[2𝑥-1]=1$

$\begin{array}{cc}& \Rightarrow 1\le (2𝑥-1)<2\Rightarrow 2\le 2𝑥<3\\ & \Rightarrow 1\le 𝑥<1.5\end{array}$

so, $𝑥\in [1,1.5)$

Now, $𝑘=2[2𝑥-1]-1=2\left(1\right)-1=1$

for $𝑥\in [1,1.5)$

$\begin{array}{ccc}𝑓\left(𝑥\right)& =[𝑘+5𝑥]=[1+5𝑥]& \\ \text{ at }& 𝑥=1,𝑓\left(𝑥\right)=\left[6\right]=6\left(\text{ minimum }\right)\\ \text{ at }& 𝑥=1.5,𝑓\left(𝑥\right)=[1+5\times 1.5]\\ & =\left[8.5\right]=8\text{ (maximum) }\end{array}$

so, $𝑓\left(𝑥\right)$ can be $6,7,8$

Hence, the range of $𝑓\left(𝑥\right)=\{6,7,8\}$

$\because 𝑦=𝑓\left(𝑥\right)=(𝑥+1{)}^2-1$

$\begin{array}{cc}& \Rightarrow 𝑦+1=(𝑥+1{)}^2\\ & \Rightarrow 𝑥=\sqrt{\surd 𝑦+1}-1\end{array}$

thus, ${𝑓}^{-1}\left(𝑥\right)=\sqrt{\surd 𝑥+1}-1$, defined for

$𝑥\ge -1$

Now, $𝑓\left(𝑥\right)={𝑓}^{-1}\left(𝑥\right)$

$\Rightarrow (𝑥+1{)}^2=\sqrt{\surd 𝑥+1}$

let $𝑥+1=𝑡$

$\Rightarrow {𝑡}^4=𝑡\Rightarrow 𝑡({𝑡}^3-1)=0$

therefore, $𝑡=0$

$\Rightarrow 𝑥+1=0\Rightarrow 𝑥=-1$

or $𝑡=1$

$\begin{array}{cc}& \Rightarrow 𝑥+1=1\Rightarrow 𝑥=0\\ & \text{ Hence, }\{𝑥\mid 𝑓(𝑥)={𝑓}^{-1}(𝑥\left)\right\}\\ & =\{0,-1\}\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}{11}^{12}-{11}^2& =𝑘(5\times {10}^9+6\times {10}^9+33\times {10}^8+110\times {10}^7+\dots +33)\\ {11}^{12}-{11}^2& =(10+1{)}^{12}-(10+1{)}^2\\ & ={10}^{12}+12\times {10}^{11}+66\times {10}^{10}+220\times {10}^9\cdots +66\times {10}^2+120+1-121\end{array}\end{array}$

$\begin{array}{cc}& ={10}^{12}+12\times {10}^{11}+66\times {10}^{10}+220\times {10}^9\dots +66\times {10}^2\\ & =2\times {10}^2(5\times {10}^9+6\times {10}^9+33\times {10}^8+110\times {10}^7+\dots +33)\\ & =200(5\times {10}^9+6\times {10}^9+33\times {10}^8+110\times {10}^7+\dots +33)\\ & \therefore 𝑘=200\end{array}$

$\because 𝑃=⎡⎣⎢\begin{array}{ccc}1& 𝛼& 3\\ 1& 3& 3\\ 2& 4& 4\end{array}⎤⎦⎥$

$\begin{array}{cc}& \because \mathrm{adj}\left(𝐴\right)=(\det 𝐴)\cdot {𝐴}^{-1}\\ & \Rightarrow \det (\mathrm{adj}(𝐴\left)\right)=(\det 𝐴{)}^{𝑛-1}\end{array}$

So, for $𝑛=3$

$\begin{array}{cc}\det (\mathrm{adj}(𝐴\left)\right)& =(\det 𝐴{)}^{3-1}\\ & =4^2=16...\left(𝑖\right)\end{array}$

$\begin{array}{cc}\text{ and }\det \left(𝐴\right)& =1(12-12)-𝛼(4-6)+3(4-6)\\ & =2𝛼-6...\left(𝑖𝑖\right)\end{array}$

Hence, $2𝛼-6=16$

$\begin{array}{cc}\Rightarrow & 2𝛼=22\\ \Rightarrow & 𝛼=11\end{array}$

We are given the equation ${𝑥}^3+6{𝑥}^2+5𝑥-42=0$.

Let's check if $𝑥=2$ is a root by plugging it in:

$(2{)}^3+6\times (2{)}^2+5\times 2-42=8+24+10-42=42-42=0$

Since the equation equals zero, $𝑥=2$ is a real root. So, $𝛼=2$.

Now, we substitute $𝛼=2$ into the matrix:

$⎡⎣⎢\begin{array}{ccc}𝛼-1& 𝛼+1& 𝛼+2\\ 𝛼-2& 𝛼+3& 𝛼-3\\ 𝛼+4& 𝛼-4& 𝛼+5\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 3& 4\\ 0& 5& -1\\ 6& -2& 7\end{array}⎤⎦⎥$

Next, let's find the determinant of this matrix:

$\begin{array}{cc}& =\mid \begin{array}{ccc}1& 3& 4\\ 0& 5& -1\\ 6& -2& 7\end{array}\mid \\ & =1\times (5\times 7-(-1)\times (-2\left)\right)-3\times (0\times 7-(-1)\times 6)+4\times (0\times (-2)-5\times 6)\\ & =1\times (35-2)-3\times (0+6)+4\times (0-30)\\ & =33-18-120\\ & =-105\end{array}$

Given matrix

$\begin{array}{cc}& ⎡⎣⎢\begin{array}{cccc}2& -3& 4& 0\\ 5& -4& 2& 1\\ 1& -3& 5& -4\end{array}⎤⎦⎥\\ & {𝑅}_2\to {𝑅}_2-\frac{5}{2}{𝑅}_1\text{ and }{𝑅}_3\to {𝑅}_3-\frac{1}{2}{𝑅}_1\\ & =⎡⎣⎢\begin{array}{cccc}2& -3& 4& 0\\ 0& 3.5& -8& 1\\ 0& -1.5& 3& -4\end{array}⎤⎦⎥\\ & {𝑅}_3\to {𝑅}_3+\frac{1.5}{3.5}{𝑅}_2={𝑅}_3+\frac{3}{7}{𝑅}_2\\ & =⎡⎣⎢\begin{array}{cccc}2& -3& 4& 0\\ 0& 3.5& -8& 1\\ 0& 0& -3/7& 25/7\end{array}⎤⎦⎥\end{array}$

$\because$ All 3 rows are non-zero

Hence, rank of the matrix $=3$

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

$=\frac{\left[𝑥\right(𝑥-1)+𝑦(𝑦-1\left)\right]+𝑖[𝑥𝑦-(𝑥-1\left)\right(𝑦-1\left)\right]}{{𝑥}^2+(𝑦-1{)}^2}$

$\because$ The fraction is purely imaginary.

So, ${𝑥}^2-𝑥+{𝑦}^2-𝑦=0$ {real part should be zero)

$\begin{array}{cc}& \Rightarrow {𝑥}^2-𝑥+\frac{1}{4}+{𝑦}^2-𝑦+\frac{1}{4}=\frac{1}{4}+\frac{1}{4}\\ & \Rightarrow {(𝑥-\frac{1}{2})}^2+{(𝑦-\frac{1}{2})}^2=\frac{1}{2}={\left(\frac{1}{\sqrt{\surd 2}}\right)}^2\end{array}$

Center of circle $(𝛼,𝛽)=(\frac{1}{2},\frac{1}{2})$

and Radius $\left(𝑟\right)=\frac{1}{\sqrt{\surd 2}}$

Thus, $\frac{𝛼}{𝛽}+\frac{𝛽}{𝛼}=\frac{(1/2)}{(1/2)}+\frac{(1/2)}{(1/2)}=2$

and $𝑟=\frac{1}{\sqrt{\surd 2}}$

$\Rightarrow {𝑟}^2=\frac{1}{2}$

So, $4{𝑟}^2=4\times \frac{1}{2}=2$

Hence, $\frac{𝛼}{𝛽}+\frac{𝛽}{𝛼}=4{𝑟}^2$

${\left(\frac{\sqrt{\surd 3}+𝑖}{\sqrt{\surd 3}-𝑖}\right)}^{𝑛}=-1$

$\because$ Modulus of numerator $=\sqrt{\surd (\sqrt{\surd 3}{)}^2+1^2}$

$=2𝜃={\tan }^{-1}\left(\frac{1}{\sqrt{\surd 3}}\right)=\frac{𝜋}{6}$

and modulus of denominator $=2$

$\begin{array}{cc}& 𝜃={\tan }^{-1}(-\frac{1}{\sqrt{\surd 3}})=-\frac{𝜋}{6}\\ & \begin{array}{cc}\therefore {\left(\frac{\sqrt{\surd 3}+𝑖}{\sqrt{\surd 3}-𝑖}\right)}^{𝑛}& ={\left(\frac{2\mathrm{cis}\left(\frac{𝜋}{6}\right)}{2\mathrm{cis}(-\frac{𝜋}{6})}\right)}^{𝑛}\\ & ={[\mathrm{cis}(\frac{𝜋}{6}-(-\frac{𝜋}{6}))]}^{𝑛}\\ & ={[\mathrm{cis}\left(\frac{𝜋}{3}\right)]}^{𝑛}=\mathrm{cis}\left(\frac{𝑛𝜋}{3}\right)\end{array}\end{array}$

$\begin{array}{cc}& \text{ and }\mathrm{cis}\left(\frac{𝑛𝜋}{3}\right)=-1\\ & =\mathrm{cis}\left(𝜋\right)\\ & \therefore \frac{𝑛𝜋}{3}=𝜋\\ & \Rightarrow 𝑛=3\end{array}$

thus, $𝑝=3$

Similarly, second expression

$\begin{array}{cc}& \begin{array}{cc}& {\left(\frac{1-\sqrt{\surd 3}𝑖}{1+\sqrt{\surd 3}𝑖}\right)}^{𝑚}=\mathrm{cis}\left(\frac{2𝜋}{3}\right)\\ \Rightarrow & {\left[\frac{2(\mathrm{cis}(-\frac{𝜋}{3}))}{2(\mathrm{cis}\left(\frac{𝜋}{3}\right))}\right]}^{𝑚}=\mathrm{cis}\left(\frac{2𝜋}{3}\right)\\ \Rightarrow & {[\mathrm{cis}(-\frac{𝜋}{3}-\frac{𝜋}{3})]}^{𝑚}=\mathrm{cis}\left(\frac{2𝜋}{3}\right)\\ \Rightarrow & \mathrm{cis}(-\frac{2𝜋}{3}\cdot 𝑚)=\mathrm{cis}\left(\frac{2𝜋}{3}\right)\end{array}\\ & \text{ Therefore, }-\frac{2𝜋}{3}\cdot 𝑚=\frac{2𝜋}{3}\end{array}$

$\begin{array}{cc}\Rightarrow & -2𝜋𝑚\equiv 2𝜋\cdot mod\left(6𝜋\right)\\ \Rightarrow & -2𝑚\equiv 2mod\left(6\right)\\ \Rightarrow & 2𝑚\equiv -2\equiv 4mod6\\ \Rightarrow & 𝑚=2mod3\end{array}$

$\therefore$ Least positive integer $=𝑚=2$ thus, $𝑞=2$

Hence, $\sqrt{\surd {𝑞}^2+{𝑝}^2}=\sqrt{\surd 3^2+2^2}=\sqrt{\surd 13}$

$\begin{array}{cc}& {𝑧}^8-20{𝑧}^4+64=0\\ & \Rightarrow {\left({𝑧}^4\right)}^2-16{𝑧}^4-4{𝑧}^4+64=0\\ & \Rightarrow ({𝑧}^4-16)({𝑧}^4-4)=0\\ & {𝑧}^4=16\text{ or }{𝑧}^4=4\\ & \Rightarrow 𝑧=\pm 2\pm 2𝑖,\pm \sqrt{\surd 2},\pm \sqrt{\surd 2}𝑖\end{array}$

imaginary roots are $2𝑖,-2𝑖,\sqrt{\surd 2}𝑖,-\sqrt{\surd 2}𝑖$ sum of square the imaginary roots

$\begin{array}{cc}& =(2𝑖{)}^2+(-2𝑖{)}^2+(\sqrt{\surd 2}𝑖{)}^2+(-\sqrt{\surd 2}𝑖{)}^2\\ & =-4-4-2-2=-12\end{array}$

Let $𝑓\left(𝑥\right)=(𝑎-3){𝑥}^2+12𝑥+(𝑎+6)$ to quadratic always be positive

$\begin{array}{cc}& 𝑎-3>0\\ \Rightarrow & 𝑎>3\\ \text{ and }& 𝐷=(12{)}^2-4(𝑎-3\left)\right(𝑎+6)<0\\ & =144-4({𝑎}^2+3𝑎-18)<0\\ \Rightarrow & {𝑎}^2+3𝑎-54>0\\ \Rightarrow & {𝑎}^2+9𝑎-6𝑎-54>0\\ \Rightarrow & (𝑎+9)(𝑎-6)>0\end{array}$

$𝑎\in (-\infty ,-9)\cup (6,\infty )$

Combine with $𝑎>3$

$\Rightarrow$ We take $𝑎\in (6,\infty )$

Least positive integeral value

$𝛼=7,𝑙=6$

put these values

$\begin{array}{cc}& (𝛼-3){𝑥}^2+12𝑥+(𝑙+2)\\ & =4{𝑥}^2+12𝑥+8=0\\ \Rightarrow & {𝑥}^2+3𝑥+2=0\\ \Rightarrow & (𝑥+1)(𝑥+2)=0\end{array}$

Hence, $𝑥=-1,-2$

$\because x^2+2 a x+b=0$

$$\begin{aligned} \Rightarrow \quad x & =\frac{-2 a \pm \sqrt{(2 a)^2-4 b}}{2} \\ & =-a \pm \sqrt{a^2-b} \end{aligned}$$

let,

$$\begin{aligned} & \alpha=-a+\sqrt{a^2-b} \\ & \beta=-a-\sqrt{a^2-b} \end{aligned}$$

so, $|\alpha-\beta|=2 \sqrt{a^2-b}$

given that $2 \sqrt{a^2-b} \leq 2 m$

$$\begin{array}{ll} \Rightarrow & a^2-b \leq m^2 \\ \Rightarrow & b \geq a^2-m^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array}$$

Also, for real and distinct roots

$$\begin{array}{cc} & D=4 a^2-4 b > 0 \\ \Rightarrow & b < a^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array}$$

Hence, $a^2-m^2 \leq b<a^2$

$\Rightarrow \quad b \in\left[a^2-m^2, a^2\right)$

$\begin{array}{cc}& {𝑥}^3-2{𝑥}^2+3𝑥-4=0\\ & \begin{array}{cc}& \text{ So, }𝛼+𝛽+𝛾=2𝛼𝛽+𝛽𝛾+𝛾𝛼=+3\text{, }\\ & 𝛼𝛽𝛾=4\\ & \text{ thus, }{𝛼}^2+{𝛽}^2+{𝛾}^2=(𝛼+𝛽+𝛾{)}^2\\ & \begin{array}{cc}& -2(𝛼𝛽+𝛽𝛾+𝛾𝛼)\\ & =4-2\left(3\right)=-2\\ \text{ and }& {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2\\ & =(𝛼𝛽{)}^2+(𝛽𝛾{)}^2+(𝛾𝛼{)}^2\end{array}\\ & \because {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2\end{array}\end{array}$

$\begin{array}{cc}& =(𝛼𝛽+𝛽𝛾+𝛾𝛼{)}^2-2𝛼𝛽𝛾(𝛼+𝛽+𝛾)\\ & =(3{)}^2-2(4\left)\right(2)=9-16=-7\end{array}$

and ${𝛼}^2{𝛽}^2{𝛾}^2=(𝛼𝛽𝛾{)}^2=4^2=16$

Hence, the cubic equation with roots ${𝛼}^2,{𝛽}^2,{𝛾}^2$ is

$\begin{array}{cc}& {𝑥}^3-({𝛼}^2+{𝛽}^2+{𝛾}^2){𝑥}^2+({𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2)𝑥-{𝛼}^2{𝛽}^2{𝛾}^2\\ & ={𝑥}^3-(-2){𝑥}^2+(-7)𝑥-16𝑥-{𝛼}^2{𝛽}^2{𝛾}^2\\ & ={𝑥}^3+2{𝑥}^2-7𝑥-16=0\end{array}$

$\begin{array}{cc}& \because {𝑥}^3+𝑝{𝑥}^2+𝑞𝑥+𝑟=0\\ & 𝛼+𝛽+𝛾=-𝑝,𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑞,\\ & 𝛼𝛽𝛾=-𝑟\end{array}$

$\begin{array}{cc}𝑆𝑜,{𝛼}^3& +{𝛽}^3+{𝛾}^3=(𝛼+𝛽+𝛾{)}^3\\ & -3(𝛼+𝛽+𝛾)(𝛼𝛽+𝛽𝛾+𝛾𝛼)+3𝛼𝛽𝛾\\ =& (-𝑝{)}^3-3(-𝑝\left)\right(𝑞)+3(-𝑟)\\ =& -{𝑝}^3+3𝑝𝑞-3𝑟\end{array}$

Number of combinations of choosing 4 digits out of 4

$={}^5{𝐶}_4=5$

Each 4-digit set can be arranged in 4! ways $=24$ ways

So, total number of 4 -digit numbers

$=5\times 24=120$

In each group, each digit appears

$\frac{24}{4}=6\text{ times }$

So, contribution per group

$=\text{ sum of digits }\times 6\times (1000+100+10+1)$

Now, compute total sum of digits for all 5 groups

$\begin{array}{c}\begin{array}{c}(2+3+5+8)+(1+3+5+8)\\ +(1+2+5+8)+(1+2+3+8)\\ +(1+2+3+5)\end{array}\\ =18+17+16+14+11=76\end{array}$

$\text{ Hence, total sum }\begin{array}{cc}& =76\times 6666\\ & =506616\end{array}$

Total possible ordered pairs $(𝑝,𝑞)$ of first 50 natural numbers

$=50\times 50=2500$

Out of these, exactly half will satisfy $𝑝>𝑞$ and there are 50 pairs where $𝑝=𝑞\left\{\right(1,1),(2,2)\dots (50,50\left)\right\}$

So, number of pairs with $𝑝\ne 𝑞$

$=2500-50=2450$

and for $𝑝>𝑞$

$=\frac{2450}{2}=1225$

Fathers $\left(𝐹\right)=5$, Teachers $\left(𝑇\right)=6$, students $\left(𝑆\right)=4$

for, $𝑇+𝐹+𝑆=7,𝑇\ge 0$, thus $𝐹\ge 1,𝑆\ge 1$ total combination ( $𝑇,𝐹,𝑆$ )

$(4,1,2),(4,2,1),(5,1,1),(3,2,2)$

Case $𝐈(𝑇=4,𝐹=1,𝑆=2)$

$\begin{array}{cc}& ={}^6{𝐶}_4\cdot {}^5{𝐶}_1\cdot {}^4{𝐶}_2\\ & =15\times 5\times 6=450\end{array}$

$\begin{array}{cc}\mathrm{𝐂𝐚𝐬𝐞}𝐈𝐈(𝑇& =4,𝐹=2,𝑆=1)\\ & ={}^6{𝐶}_4\cdot {}^5{𝐶}_2\cdot {}^4{𝐶}_1\\ & =15\times 10\times 4=600\end{array}$

Case III ( $𝑇=5,𝐹=1,𝑆=1$ )

$\begin{array}{cc}& ={}^6{𝐶}_5\cdot {}^5{𝐶}_1\cdot {}^4{𝐶}_1\\ & =6\times 5\times 4=120\end{array}$

Case IV ( $𝑇=3,𝐹=2,𝑆=2$ )

$\begin{array}{cc}& ={}^6{𝐶}_3\cdot {}^5{𝐶}_2\cdot {}^4{𝐶}_2\\ & =20\times 10\times 6=1200\end{array}$

Hence, total number of ways

$\begin{array}{cc}& =450+600+120+1200\\ & =2370\end{array}$

$\because (1+𝑥{)}^{𝑛}=\sum _{𝑛}^{𝑘=0}{𝐶}_{𝑘}{𝑥}^{𝑘}$

and

$({𝐶}_0+{𝐶}_1)-({𝐶}_2+{𝐶}_3)+({𝐶}_4+{𝐶}_5)-\cdots$

can be grouped as

$\begin{array}{c}\sum _{𝑘=0}^{𝑛/2}(-1{)}^{𝑘}({𝐶}_{2𝑘}+{𝐶}_{2𝑘+1})\\ \text{ and }(1+𝑖{)}^{𝑛}=\sum _{𝑘=0}^{𝑛}{𝐶}_{𝑘}{𝑖}^{𝑘}\\ \mathrm{Re}\left[\right(1+𝑖{)}^{𝑛}]={𝐶}_0-{𝐶}_2+{𝐶}_4-{𝐶}_6+\cdots \\ \mathrm{Im}\left[\right(1+𝑖{)}^{𝑛}]={𝐶}_1-{𝐶}_3+{𝐶}_5-{𝐶}_7+\cdots \end{array}$

Adding both

$\begin{array}{cc}& \mathrm{Re}\left[\right(1+𝑖{)}^{𝑛}]+\mathrm{Im}\left[\right(1+𝑖{)}^{𝑛}]\\ & =({𝐶}_0+{𝐶}_1)-({𝐶}_2+{𝐶}_3)+({𝐶}_4+{𝐶}_5)-\dots \end{array}$

Now, $(1+𝑖{)}^{𝑛}$

$=(\sqrt{\surd 2}{)}^{𝑛}(\cos \left(\frac{𝑛𝜋}{4}\right)+𝑖\sin \left(\frac{𝑛𝜋}{4}\right))$

Hence, $({𝐶}_0+{𝐶}_1)-({𝐶}_2+{𝐶}_3)+({𝐶}_4+{𝐶}_5)$

$\begin{array}{cc}-\dots & =\mathrm{Re}\left[\right(1+𝑖{)}^{𝑛}]+\mathrm{Im}\left[\right(1+𝑖{)}^{𝑛}]\\ & =2^{𝑛/2}(\cos \left(\frac{𝑛𝜋}{4}\right)+\sin \left(\frac{𝑛𝜋}{4}\right))\end{array}$

$1+\frac{4}{15}+\frac{4\cdot 10}{15\cdot 30}+\frac{4\cdot 10\cdot 16}{15\cdot 30\cdot 45}+\dots +\infty$

We know that

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

On comparing with given expansion, we get

$𝑛𝑥=\frac{4}{15}\Rightarrow 𝑥=\frac{4}{15𝑛}...\left(𝑖\right)$

and $\frac{𝑛(𝑛-1)}{2!}{𝑥}^2=\frac{4\cdot 10}{15\cdot 30}=\frac{40}{450}=\frac{4}{45}$

Using Eq. (i), $\frac{𝑛(𝑛-1)}{2!}{\left(\frac{4}{15𝑛}\right)}^2=\frac{4}{45}$

$\begin{array}{ccc}\Rightarrow & \frac{𝑛(𝑛-1)}{2}\times \frac{4\times 4}{15𝑛\times 15𝑛}& =\frac{4}{45}\\ \Rightarrow & \frac{𝑛(𝑛-1)}{2}\times \frac{16}{225{𝑛}^2}& =\frac{4}{45}\\ \Rightarrow & & 𝑛=\frac{-2}{3}\end{array}$

$\begin{array}{cc}& \text{ Also, }𝑥=\frac{4}{15}\left(\frac{-3}{2}\right)=\frac{-2}{5}\\ & \begin{array}{cc}\therefore (1+𝑥{)}^{𝑛}& ={(1-\frac{2}{5})}^{-2/3}\\ & ={\left(\frac{3}{5}\right)}^{-2/3}={\left(\frac{5}{3}\right)}^{2/3}\end{array}\end{array}$

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

Comparing both sides, we get

$\begin{array}{cccc}& & 𝐴+𝐵& =0\\ \Rightarrow & & 𝐴& =-𝐵...\left(𝑖\right)\end{array}$