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

We have,

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\text{exception 25:}+\log ({𝑥}^2-𝑥-2)\\ \therefore & 2{𝑥}^2-7𝑥+5\ne 0\text{ and }{𝑥}^2-𝑥-2>0\\ \Rightarrow & (2𝑥-5)(𝑥-1)\ne 0\text{ and }(𝑥+1)(𝑥-2)>0\\ \Rightarrow & 𝑥\ne 1,\frac{5}{2}\text{ and }𝑥\in (-\infty ,-1)\cup (2,\infty )\\ \Rightarrow & 𝑥\in (-\infty ,-1)\cup (2,\frac{5}{2})\cup (\frac{5}{2},\infty )\end{array}$

$\therefore$ Domain of $𝑓$ is

$(-\infty ,-1)\cup (2,\frac{5}{2})\cup (\frac{5}{2},\infty )$

Given

$𝑓(3𝑥+\frac{1}{2𝑥})=9{𝑥}^2+\frac{1}{4{𝑥}^2}$

We want to use this to find $𝑓(𝑥+\frac{1}{𝑥})$.

Let

$𝑡=3𝑥+\frac{1}{2𝑥}$

Now square $𝑡$:

${𝑡}^2={(3𝑥+\frac{1}{2𝑥})}^2$

${𝑡}^2=9{𝑥}^2+\frac{1}{4{𝑥}^2}+2\cdot 3𝑥\cdot \frac{1}{2𝑥}$

${𝑡}^2=9{𝑥}^2+\frac{1}{4{𝑥}^2}+3$

So,

$9{𝑥}^2+\frac{1}{4{𝑥}^2}={𝑡}^2-3$

But from the given relation,

$𝑓\left(𝑡\right)=9{𝑥}^2+\frac{1}{4{𝑥}^2}$

Hence,

$𝑓\left(𝑡\right)={𝑡}^2-3$

Therefore,

$𝑓\left(𝑦\right)={𝑦}^2-3$

for the required input.

Now given

$𝑓(𝑥+\frac{1}{𝑥})=1$

Using $𝑓\left(𝑦\right)={𝑦}^2-3$, put

$𝑦=𝑥+\frac{1}{𝑥}$

Then

${𝑦}^2-3=1$

${𝑦}^2=4$

$𝑦=\pm 2$

So,

$𝑥+\frac{1}{𝑥}=\pm 2$

Now solve both cases.

For

$𝑥+\frac{1}{𝑥}=2$

${𝑥}^2+1=2𝑥$

${𝑥}^2-2𝑥+1=0$

$(𝑥-1{)}^2=0$

$𝑥=1$

For

$𝑥+\frac{1}{𝑥}=-2$

${𝑥}^2+1=-2𝑥$

${𝑥}^2+2𝑥+1=0$

$(𝑥+1{)}^2=0$

$𝑥=-1$

Therefore,

$𝑥=\pm 1$

So the correct option is B.

$\begin{array}{cc}\frac{1}{3\cdot 6}& +\frac{1}{6\cdot 9}+\frac{1}{9\cdot 12}+\dots \text{ to }9\text{ terms }\\ =& \frac{1}{9}(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots +\frac{1}{9\cdot 10})\\ =& \frac{1}{9}((1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\dots +(\frac{1}{9}-\frac{1}{10}))\\ & =(1-\frac{1}{10})=\frac{1}{9}\left(\frac{9}{10}\right)=\frac{1}{10}\end{array}$

We have, $𝛼,𝛽,𝛾$ are the roots of the equation $\mid \begin{array}{ccc}𝑥& 2& 2\\ \\ 2& 𝑥& 2\\ \\ 2& 2& 𝑥\end{array}\mid =0$ and

$min(𝛼,𝛽,𝛾)=𝛼$

Now, $\mid \begin{array}{ccc}𝑥& 2& 2\\ \\ 2& 𝑥& 2\\ \\ 2& 2& 𝑥\end{array}\mid =0$

$\Rightarrow \mid \begin{array}{ccc}𝑥-2& 0& 2\\ \\ 0& 𝑥-2& 2\\ \\ 2-𝑥& 2-𝑥& 𝑥\end{array}\mid =0$

$\begin{array}{c}[{𝐶}_1\to {𝐶}_1-{𝐶}_3\text{ and }{𝐶}_2\to {𝐶}_2-{𝐶}_3]\\ \\ \Rightarrow (𝑥-2{)}^2\mid \begin{array}{ccc}1& 0& 2\\ \\ 0& 1& 2\\ \\ -1& -1& 𝑥\end{array}\mid =0\\ \\ [{𝐶}_1\to \frac{1}{𝑥-2}{𝐶}_1\text{ and }{𝐶}_2\to \frac{1}{𝑥-2}{𝐶}_2]\\ \\ \Rightarrow (𝑥-2{)}^2[1(𝑥+2)-0+(-1)(0-2)]=0\\ \\ \Rightarrow (𝑥-2{)}^2(𝑥+4)=0\Rightarrow 𝑥=-4,2,2\\ \\ \therefore 𝛼=-4,𝛽=2\text{ and }𝛾=2\\ \\ \text{ Now, }2𝛼+3𝛽+4𝛾\\ \\ =2(-4)+3\left(2\right)+4\left(2\right)\\ \\ =-8+6+8=6\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ \\ & \begin{array}{cc}& \begin{array}{c}𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ \\ 3& 2& 3\\ \\ 1& 1& 2\end{array}⎤⎦⎥\text{ and }{𝐴}^{-1}=⎡⎣⎢\begin{array}{ccc}{𝑎}_{11}& {𝑎}_{12}& {𝑎}_{13}\\ \\ {𝑎}_{21}& {𝑎}_{22}& {𝑎}_{23}\\ \\ {𝑎}_{31}& {𝑎}_{32}& {𝑎}_{33}\end{array}⎤⎦⎥\\ \\ \left|𝐴\right|=1(4-3)-2(6-3)+2(3-2)=1-6+2=-3\begin{array}{}\end{array}\\ \\ {𝐶}_{11}=\mid \begin{array}{cc}2& 3\\ \\ 1& 2\end{array}\mid =1\\ \\ {𝐶}_{12}=-\mid \begin{array}{cc}3& 3\\ \\ 1& 2\end{array}\mid =-3\\ \\ {𝐶}_{13}=\mid \begin{array}{cc}3& 2\\ \\ 1& 1\end{array}\mid =1\\ \\ {𝐶}_{21}=-\mid \begin{array}{cc}2& 2\\ \\ 1& 2\end{array}\mid =-2\\ \\ {𝐶}_{22}=\mid \begin{array}{cc}1& 2\\ \\ 1& 2\end{array}\mid =0\\ \\ {𝐶}_{23}=-\mid \begin{array}{cc}1& 2\\ \\ 1& 1\end{array}\mid =1\end{array}\end{array}\end{array}$

$\begin{array}{cc}& {𝐶}_{31}=\mid \begin{array}{cc}2& 2\\ \\ 2& 3\end{array}\mid =2,{𝐶}_{32}=-\mid \begin{array}{cc}1& 2\\ \\ 3& 3\end{array}\mid =3\\ \\ & {𝐶}_{33}=\mid \begin{array}{cc}1& 2\\ \\ 3& 2\end{array}\mid =-4\\ \\ & \mathrm{adj}𝐴={⎡⎣⎢\begin{array}{ccc}1& -3& 1\\ \\ -2& 0& 1\\ \\ 2& 3& -4\end{array}⎤⎦⎥}^{𝑇}=⎡⎣⎢\begin{array}{ccc}1& -2& 2\\ \\ -3& 0& 3\\ \\ 1& 1& -4\end{array}⎤⎦⎥\\ \\ & \therefore {𝐴}^{-1}=\frac{1}{\left|𝐴\right|}(\mathrm{adj}𝐴)=\frac{1}{-3}⎡⎣⎢\begin{array}{ccc}1& -2& 2\\ \\ -3& 0& 3\\ \\ 1& 1& -4\end{array}⎤⎦⎥\end{array}$

Now,

$\sum _{\begin{array}{c}1\le 𝑖\le 3\\ \\ 1\le 𝑗\le 3\end{array}}{𝑎}_{𝑖𝑗}=(-\frac{1}{3})$

$\begin{array}{cc}& [1+(-2)+2+(-3)+0+3+1+1+(-4\left)\right]\\ \\ & =(-\frac{1}{3})(-1)=\frac{1}{3}\end{array}$

We have,

$𝐴𝑋=𝐷$, where

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}3& -4& 7\\ \\ 5& 2& -4\\ \\ 8& -6& -1\end{array}⎤⎦⎥,𝑋=⎡⎣⎢\begin{array}{c}𝑥\\ \\ 𝑦\\ \\ 𝑧\end{array}⎤⎦⎥\text{ and }𝐷=⎡⎣⎢\begin{array}{c}-6\\ \\ -9\\ \\ -5\end{array}⎤⎦⎥\\ \\ \left|𝐴\right|& =3(-2-24)+4(-5+32)+7(-30-16)\\ \\ & =-78+108-322\\ \\ & =-292\ne 0\end{array}$

$\therefore$ Rank $\left(𝐴\right)=$ Order of $𝐴=3$

$𝐴𝐷=⎡⎣⎢\begin{array}{ccc}3& -4& 7\\ \\ 5& 2& -4\\ \\ 8& -6& -1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}-6\\ \\ -9\\ \\ -5\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}-17\\ \\ -28\\ \\ 11\end{array}⎤⎦⎥$

Clearly, Rank $\left(\right[𝐴𝐷\left]\right)=3$

Hence, Rank $\left(𝐴\right)=\mathrm{Rank}\left(\right[𝐴𝐷\left]\right)=3$

We have,

$(𝑥,𝑦,𝑧)=(𝛼,𝛽,𝛾)$ is a unique solution of the system of simultaneous linear equations

$\begin{array}{cc}3𝑥-4𝑦+𝑧& =-7\\ \\ 2𝑥+3𝑦-𝑧& =10\\ \\ 𝑥-2𝑦-3𝑧& =3\end{array}$

so, by cramer's rule,

$\begin{array}{cc}& 𝑥=𝛼=\frac{\mid \begin{array}{ccc}-7& -4& 1\\ \\ 10& 3& -1\\ \\ 3& -2& -3\end{array}\mid }{\mid \begin{array}{ccc}3& -4& 1\\ \\ 2& 3& -1\\ \\ 1& -2& -3\end{array}\mid }\\ \\ & =\frac{-7(-9-2)+4(-30+3)+(-20-9)}{3(-9-2)+4(-6+1)+(-4-3)}\\ \\ & =\frac{77-108-29}{-33-20-7}\\ \\ & =\frac{-60}{-60}=1\end{array}$

We have,

$\begin{array}{cc}& \frac{(2-𝑖)𝑥+(1+𝑖)}{2+𝑖}+\frac{(1-2𝑖)𝑦+(1-𝑖)}{1+2𝑖}=1-2𝑖\\ \\ & \Rightarrow \frac{(2𝑥+1)+(1-𝑥)𝑖}{2+𝑖}+\frac{(𝑦+1)+(-2𝑦-1)𝑖}{1+2𝑖}=1-2𝑖\\ \\ & \Rightarrow \frac{\left[\right(2𝑥+1)+(1-𝑥\left)𝑖\right](2-𝑖)}{4-{𝑖}^2}+\frac{\left[\right(𝑦+1)+(-2𝑦-1\left)𝑖\right](1-2𝑖)}{1-4{𝑖}^2}=1-2𝑖\\ \\ & \Rightarrow \left[2\right(2𝑥+1)+(1-𝑥)+(𝑦+1)+2(-2𝑦-1\left)\right]\\ \\ & +𝑖[-2𝑥-1+2(1-𝑥)-2(𝑦+1)+(-2𝑦-1\left)\right]\\ \\ & =5(1-2𝑖)\\ \\ & \Rightarrow (3𝑥-3𝑦+2)+𝑖(-4𝑥-4𝑦-2)=5-10𝑖\end{array}$

We have,

$𝑧=1-\sqrt{\surd 3}𝑖=2{𝑒}^{-𝑖\frac{𝜋}{3}}$

Now, ${𝑧}^3-3{𝑧}^2+3𝑧$

$\begin{array}{cc}& \begin{array}{c}={\left(2{𝑒}^{-\frac{𝜋}{3}𝑖}\right)}^3-3{\left(2{𝑒}^{-\frac{𝜋}{3}𝑖}\right)}^2+3\left(2{𝑒}^{-\frac{𝜋}{3}𝑖}\right)\\ \\ \begin{array}{cc}& =8{𝑒}^{-𝜋𝑖}-12{𝑒}^{-\frac{2𝜋}{3}𝑖}+6{𝑒}^{-\frac{𝜋}{3}𝑖}\\ \\ & =8(\cos (-𝜋)+𝑖\sin (-𝜋\left)\right)\end{array}\\ \\ -12(\cos (-\frac{2𝜋}{3})+𝑖\sin (-\frac{2𝜋}{3}))\\ \\ +6(\cos (-\frac{𝜋}{3})+𝑖\sin (-\frac{𝜋}{3}))\\ \\ =8(-1+𝑖(0\left)\right)-12(-\frac{1}{2}+𝑖(-\frac{\sqrt{\surd 3}}{2}))\\ \\ +6(\frac{1}{2}+𝑖(-\frac{\sqrt{\surd 3}}{2}))\end{array}\\ \\ & =(-8+6+3)+𝑖(0+6\sqrt{\surd 3}-3\sqrt{\surd 3})\\ \\ & =1+3\sqrt{\surd 3𝑖}\end{array}$

Let, $(\sqrt{\surd 3}-𝑖{)}^{2/5}=𝑥$

$\begin{array}{cc}\Rightarrow & (\sqrt{\surd 3}-𝑖{)}^2={𝑥}^5\\ \\ \Rightarrow & {𝑥}^5=(\sqrt{\surd 3}{)}^2+(𝑖{)}^2-2\sqrt{\surd 3}𝑖\\ \\ \Rightarrow & {𝑥}^5=3+(-1)-2\sqrt{\surd 3}𝑖\\ \\ \Rightarrow & {𝑥}^5=2-2\sqrt{\surd 3}𝑖=2(1-\sqrt{\surd 3}𝑖)\end{array}$

Clearly, the product of all the values of $(\sqrt{\surd 3}-𝑖{)}^{2/5}$ is $2(1-\sqrt{\surd 3}𝑖)$

12 th roots of unity are $𝑙,{𝑒}^{𝑖\frac{𝜋}{6}},{𝑒}^{𝑖\frac{2𝜋}{6}}$, ${𝑒}^{𝑖\frac{3𝜋}{6}},{𝑒}^{𝑖\frac{4𝜋}{6}},\dots ,{𝑒}^{𝑖\frac{11𝜋}{6}}$.

30th roots of unity are $1,{𝑒}^{𝑖\frac{𝜋}{15}},{𝑒}^{𝑖\frac{2𝜋}{15}}$, ${𝑒}^{𝑖\frac{3𝜋}{15}},{𝑒}^{𝑖\frac{4𝜋}{15}},\dots ,{𝑒}^{𝑖\frac{29𝜋}{15}}$.

Clearly, common roots are $1,{𝑒}^{𝑖\frac{𝜋}{3}},{𝑒}^{𝑖\frac{2𝜋}{3}}$, ${𝑒}^{𝑖𝜋},{𝑒}^{𝑖\frac{4𝜋}{3}}$ and ${𝑒}^{𝑖\frac{5𝜋}{3}}$

Hence, the number of common roots is 6.

Given $𝛼$ is a root of the equation

$\begin{array}{cc}& \frac{𝑥-1}{\sqrt{\surd 2{𝑥}^2-5𝑥+2}}=\frac{41}{60},\text{ where }-\frac{1}{2}<𝛼<0\\ \\ \therefore & \frac{𝛼-1}{\sqrt{\surd 2{𝛼}^2-5𝛼+2}}=\frac{41}{60}\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{(𝛼-1{)}^2}{2{𝛼}^2-5𝛼+2}={\left(\frac{41}{60}\right)}^2\\ \\ & \Rightarrow 3600({𝛼}^2-2𝛼+1)=1681(2{𝛼}^2-5𝛼+2)\\ \\ & \Rightarrow 238{𝛼}^2+1205𝛼+238=0\\ \\ & \Rightarrow 238{𝛼}^2+1156𝛼+49𝛼+238=0\\ \\ & \Rightarrow 34𝛼(7𝛼+34)+7(7𝛼+34)=0\\ \\ & \Rightarrow (34𝛼+7)(7𝛼+34)=0\\ \\ & \Rightarrow 𝛼=-\frac{7}{34},-\frac{34}{7}\\ \\ & \because -\frac{1}{2}<𝛼<0\\ \\ & \therefore 𝛼=-\frac{7}{34}\end{array}$

We have,

$4+3𝑥-7{𝑥}^2$ attains its maximum value $𝑀$ at $𝑥=𝛼$

$\therefore 𝛼=\frac{-\left(3\right)}{2(-7)}=\frac{3}{14}$

Now, $𝑀=4+3\left(\frac{3}{14}\right)-7{\left(\frac{3}{14}\right)}^2=\frac{121}{28}$

Also, we have $5{𝑥}^2-2𝑥+1$ attains its minimum value $𝑀$ at $𝑥=𝛽$

$\therefore 𝛽=\frac{-(-2)}{2\left(5\right)}=\frac{2}{10}=\frac{1}{5}$

Now, $𝑚=5{\left(\frac{1}{5}\right)}^2-2\left(\frac{1}{5}\right)+1=\frac{4}{5}$

$\begin{array}{cc}\therefore \frac{28(𝑀-𝛼)}{5(𝑚+𝛽)}& =\frac{28(\frac{121}{28}-\frac{3}{14})}{5(\frac{4}{5}+\frac{1}{5})}\\ \\ & =\frac{115}{5}=23\end{array}$

Given,

$𝛼,𝛽,𝛾$ are the roots of the equation

$\begin{array}{c}2{𝑥}^3-5{𝑥}^2+4𝑥-3=0\\ \\ \therefore 𝛼+𝛽+𝛾=\frac{-(-5)}{2}=\frac{5}{2}\dots \text{ (i) }\\ \\ 𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{4}{2}=2\dots \text{ (ii) }\\ \\ 𝛼𝛽𝛾=\frac{-(-3)}{2}=\frac{3}{2}\dots \text{ (iii) }\end{array}$

Now, $\Sigma 𝛼𝛽(𝛼+𝛽)$

$\begin{array}{cc}& =𝛼𝛽(𝛼+𝛽)+𝛽𝛾(𝛽+𝛾)+𝛾𝛼(𝛼+𝛾)\\ \\ & ={𝛼}^2𝛽+𝛼{𝛽}^2+{𝛽}^2𝛾+𝛽{𝛾}^2+{𝛼}^2𝛾+𝛼{𝛾}^2\\ \\ & =(𝛼+𝛽+𝛾)(𝛼𝛽+𝛽𝛾+𝛾𝛼)-3𝛼𝛽𝛾\\ \\ & =\left(\frac{5}{2}\right)\left(2\right)-3\left(\frac{3}{2}\right)\end{array}$

[using Eqs. (i), (ii) and (iii)]

$=\frac{10}{2}-\frac{9}{2}=\frac{1}{2}$

Given, roots of the equation

${𝑥}^5+4{𝑥}^4-13{𝑥}^3-52{𝑥}^2+36𝑥+144=0$

are ${𝛼}_1𝛽,𝛾,2$, and $𝜀$, where

$𝛼<𝛽<𝛾<2<𝜀$

$\begin{array}{cc}& \text{ Now, }{𝑥}^5+4{𝑥}^4-13{𝑥}^3-52{𝑥}^2+36𝑥+144=0\\ \\ & \Rightarrow (𝑥+4)(𝑥+3)(𝑥+2)(𝑥-2)(𝑥-3)=0\\ \\ & \Rightarrow 𝑥=-4,-3,-2,2,3\\ \\ & \therefore 𝛼=-4,𝛽=-3,𝛾=-2\text{ and }𝜀=3\\ \\ & \text{ Now, }𝛼+2𝛽+3𝛾+5𝜀\\ \\ & =(-4)+2(-3)+3(-2)+5\left(3\right)\\ \\ & =-4-6-6+15\\ \\ & =-16+15=-1\end{array}$

The numbers will be divisible by 6 if and only if the unit digit of numbers is an even number and the sum of digits is divisible by 3 .

Collection of 4 -digits whose sum is divisible by 3 .

(1236) $\to \dots \dots \dots ..2$ or $6\to 3\times 2\times 1\times 2=12$

$\left(1245\right)\to \dots \dots \dots 2$ or $4\to 3\times 2\times 1\times 2=12$

$\left(1356\right)\to \dots \dots \dots 6\to 3\times 2\times 1\times 1=6$

$\left(2346\right)\to \dots \dots \dots .2$ or 4 or $6\to$

$3\times 2\times 1\times 3=18$

(3456) $\to \dots \dots ..4$ or $6\to 3\times 2\times 1\times 2=12$

$\text{ Total }=12+12+6+18+12=60$

$\therefore$ The number of numbers which are divisible by 4 are 60.

Given that the number of circular permutation of 9 distinct things taken 5 at a time is ${𝑛}_1$.

$\begin{array}{cc}\therefore {𝑛}_1& ={}^9{𝐶}_5(5-1)!\\ \\ & =\frac{9!}{5!(9-5)!}\times 4!=9\times 8\times 7\times 6\end{array}$

Also, given that the number of linear permutation of 8 distinct things taken 4 at a time is ${𝑛}_2$.

$\begin{array}{cc}\therefore {𝑛}_2={}^8{𝐶}_4(4!)& =\frac{8!}{4!(8-4)!}\times 4!\\ \\ & =8\times 7\times 6\times 5\end{array}$

Now, $\frac{{𝑛}_1}{{𝑛}_2}=\frac{9\times 8\times 7\times 6}{8\times 7\times 6\times 5}=\frac{9}{5}$

We know that

The number of ways in which $𝑛$ different things can be distributed to $𝑚$ person so that no person gets all the things is same as the number of functions (whose range is not a singleton set) from set $𝐴$ (containing $𝑛$ elements) to set $𝐵$ (containing $𝑚$ elements).

$\therefore$ The number of ways in which 4 different things can distributed to 6 person so that no person gets all the things

$\begin{array}{cc}& =6^4-6=6(216-1)\\ \\ & =6\left(215\right)=1290\end{array}$

Let 3 consecutive terms in the expansion of $(1+𝑥{)}^{23}$ be

${𝑇}_{𝑟+1},{𝑇}_{𝑟+2}\text{ and }{𝑇}_{𝑟+3}$

Coefficient of ${𝑇}_{𝑟+1}={}^{23}{𝐶}_{𝑟}$

Coefficient of ${𝑇}_{𝑟+2}={}^{23}{𝐶}_{𝑟+1}$

Coefficient of ${𝑇}_{𝑟+3}={}^{23}{𝐶}_{𝑟+2}$

According to the question,

$\begin{array}{cc}{}^{23}{𝐶}_{𝑟+1}-{}^{23}{𝐶}_{𝑟}& ={}^{23}{𝐶}_{𝑟+2}-{}^{23}{𝐶}_{𝑟+1}\\ \\ \Rightarrow & \frac{23!}{(𝑟+1)!(22-𝑟)!}-\frac{23!}{𝑟!(23-𝑟)!}\\ \\ & =\frac{23!}{(𝑟+2)!(21-𝑟)!}-\frac{23!}{(𝑟+1)!(22-𝑟)!}\\ \\ \Rightarrow & \frac{23-𝑟}{(𝑟+1)!(23-𝑟)!}-\frac{(𝑟+1)}{(𝑟+1)!(23-𝑟)!}\\ \\ & =\frac{22-𝑟}{(𝑟+2)!(22-𝑟)!}-\frac{𝑟+2}{(𝑟+2)!(22-𝑟)!}\\ \\ \Rightarrow & (𝑟+2)\left[\right(23-𝑟)-(𝑟+1\left)\right]\\ \\ \Rightarrow & =(23-𝑟)\left[\right(22-𝑟)-(𝑟+2\left)\right]\\ \\ \Rightarrow & (𝑟+2)(22-2𝑟)=(23-𝑟)(20-2𝑟)\\ \\ \Rightarrow & {𝑟}^2-84𝑟+416=0\\ \\ \Rightarrow & (𝑟-8)(𝑟-13)=0\\ \\ \Rightarrow 𝑟=8,13\end{array}$

Hence, 3 consecutive terms are either ${𝑇}_9,{𝑇}_{10}$ and ${𝑇}_{11}$ or ${𝑇}_{14},{𝑇}_{15}$ and ${𝑇}_{16}$.

Let the numerically greatest term in the expansion of $(3𝑥-16𝑦{)}^{15}$ when $𝑥=\frac{2}{3}$ and $𝑦=\frac{3}{2}$ is ${𝑇}_{𝑟+1}$

$\begin{array}{cc}{𝑇}_{𝑟+1}& ={}^{15}{𝐶}_{𝑟}{\left(3\left(\frac{2}{3}\right)\right)}^{15-𝑟}{(-16\left(\frac{3}{2}\right))}^{𝑟}\\ \\ & ={}^{15}{𝐶}_{𝑟}\left(2{)}^{15-𝑟}\right(-24{)}^{𝑟}\end{array}$

Clearly, the value the expression will be maximum only when the value $𝑟$ is largest possible even integer.

$\text{ So, }𝑟=14[\because 0\le 𝑟\le 15\text{ and }𝑟\in 𝑍]$

$\therefore 15$ th term will be the numerically greatest term.

$\begin{array}{cc}& \text{ Given, }\frac{3{𝑥}^4-2{𝑥}^2+1}{(𝑥-2{)}^4}\\ \\ & =𝐴+\frac{𝐵}{𝑥-2}+\frac{𝐶}{(𝑥-2{)}^2}+\frac{𝐷}{(𝑥-2{)}^3}+\frac{𝐸}{(𝑥-2{)}^4}\\ \\ & \Rightarrow 3{𝑥}^4-2{𝑥}^2+1=𝐴(𝑥-2{)}^4\\ \\ & +𝐵(𝑥-2{)}^3+𝐶(𝑥-2{)}^2+𝐷(𝑥-2)+𝐸\\ \\ & \Rightarrow 3{𝑥}^4-2{𝑥}^2+1=𝐴{𝑥}^4+(-8𝐴+𝐵){𝑥}^3\\ \\ & +(24𝐴-6𝐵+𝐶){𝑥}^2\\ \\ & +(-32𝐴+12𝐵-4𝐶+𝐷)𝑥\\ \\ & +(16𝐴-8𝐵+4𝐶-2𝐷+𝐸)\end{array}$

On comparing coefficients, we get

$\begin{array}{cc}& 𝐴=3,-8𝐴+𝐵=0\\ \\ \Rightarrow & 𝐵=24\\ \\ & 24𝐴-6𝐵+𝐶=-2\\ \\ \Rightarrow & 𝐶=70,-32𝐴+12𝐵-4𝐶+𝐷=0\\ \\ \Rightarrow & 𝐷=88\\ \\ & 16𝐴-8𝐵+4𝐶-2𝐷+𝐸=1\\ \\ \Rightarrow & 𝐸=41\\ \\ \therefore & 2𝐴+3𝐵-𝐶-𝐷+𝐸\end{array}$