TG EAPCET 2022 19th July Evening Shift Question Paper with Solutions

$\begin{array}{cc}& \text{ Given, }𝑓\left(𝑥\right)=\frac{1}{2}-\tan \left(\frac{𝜋𝑥}{2}\right)\\ & \begin{array}{cc}& \text{ and }𝑔\left(𝑥\right)=\sqrt{\surd 3+4𝑥-4{𝑥}^2}\\ & \Rightarrow 3+4𝑥-4{𝑥}^2\ge 0\\ & \Rightarrow 3-4({𝑥}^2-𝑥)\ge 0\\ & \Rightarrow 3-4({𝑥}^2-𝑥+\frac{1}{4})+1\ge 0\\ & \Rightarrow 4-4{(𝑥-\frac{1}{2})}^2\ge 0\\ & \Rightarrow {(𝑥-\frac{1}{2})}^2\le 1\\ & \Rightarrow -1\le 𝑥-\frac{1}{2}\le 1\\ & \Rightarrow -1+\frac{1}{2}\le 𝑥\le 1+\frac{1}{2}\\ & \Rightarrow -\frac{1}{2}\le 𝑥\le \frac{3}{2}\Rightarrow 𝑥\in [-\frac{1}{2},\frac{3}{2}]\end{array}\end{array}$

Now, $𝑓\left(𝑥\right)=\frac{1}{2}-\tan \left(\frac{𝜋𝑥}{2}\right)$

As $𝑓:𝐴\to 𝐵$ and $𝑔:𝐵\to 𝐶$, so domain of $𝑔\left(𝑥\right)=$ range of $𝑓\left(𝑥\right)$

∴ Range of $𝑓\left(𝑥\right)=[-\frac{1}{2},\frac{3}{2}]$.

$\begin{array}{cc}& \text{ Given, }𝑓\left(𝑥\right)=\sqrt{\surd \frac{1-{𝑥}^2}{1+{𝑥}^2}}\\ & \begin{array}{cc}& \frac{1-{𝑥}^2}{1+{𝑥}^2}\ge 0\\ \Rightarrow & 1-{𝑥}^2\ge 0[\because 1+{𝑥}^2>0]\\ \Rightarrow & {𝑥}^2\le 1\\ \Rightarrow & 𝑥\in [-1,1]\\ \therefore & 𝐷=[-1,1]\\ \text{ Let }𝑦=\frac{1-{𝑥}^2}{1+{𝑥}^2}\\ & 𝑦+{𝑥}^2𝑦=1-{𝑥}^2\end{array}\end{array}$

          $\begin{array}{cc}& \begin{array}{cc}\Rightarrow & {𝑥}^2(𝑦+1)=1-𝑦\\ \Rightarrow & {𝑥}^2=\frac{1-𝑦}{1+𝑦}\\ \Rightarrow & 𝑥=\pm \sqrt{\surd \frac{1-𝑦}{1+𝑦}}\\ \Rightarrow & \frac{1-𝑦}{1+𝑦}\ge 0\end{array}\\ & \text{ As we know that }𝑦\ge 0,1+𝑦>0\\ & \begin{array}{cc}\Rightarrow 1-𝑦& \ge 0\\ 𝑦& \le 1\\ 𝑦& \in [0,1]\\ 𝐺& =[0,1]\\ 𝐷\cap 𝐺& =[0,1]\end{array}\end{array}$

$\begin{array}{cc}𝐺𝑖𝑣𝑒𝑛,𝐴& =\left[\begin{array}{cc}0& 1\\ 1& 𝑘\end{array}\right]\\ {𝐴}^2& =\left[\begin{array}{cc}0& 1\\ 1& 𝑘\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 𝑘\end{array}\right]=\left[\begin{array}{cc}1& 𝑘\\ 𝑘& 1+{𝑘}^2\end{array}\right]\end{array}$

$\begin{array}{cc}{𝐴}^3={𝐴}^2\times 𝐴& =\left[\begin{array}{cc}1& 𝑘\\ 𝑘& 1+{𝑘}^2\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 𝑘\end{array}\right]\\ & =\left[\begin{array}{cc}𝑘& 1+{𝑘}^2\\ 1+{𝑘}^2& 2𝑘+{𝑘}^3\end{array}\right]\end{array}$

After comparing with given value of ${𝐴}^3$, we get

$\begin{array}{cc}& 𝑎=𝑘,𝑏=1+{𝑘}^2,𝑐=1+{𝑘}^2\text{ and }\\ & 𝑑=2𝑘+{𝑘}^3\end{array}$

Given, $𝑑=228$

$\Rightarrow 2𝑘+{𝑘}^3=228\Rightarrow {𝑘}^3+2𝑘-228=0$

By hit and trial method, $𝑘=6$

$\begin{array}{cc}𝑏+𝑐& =1+{𝑘}^2+1+{𝑘}^2\\ & =2(1+{𝑘}^2)\\ & =2(1+36)=74\end{array}$

$𝐴$ and $𝐵$ are squae matrices of order 3.

$𝐶$ is a unit matrix of order 3.

$𝐷=(𝐴𝐵-𝐶{)}^{-1}$

$\Rightarrow 𝐷$ and $𝐴𝐵-𝐶$ are inverse of each other

$\begin{array}{cccc}& & 𝐷\times (𝐴𝐵-𝐶)& =𝐼\\ \Rightarrow & & \left|𝐷\right|\times |𝐴𝐵-𝐶|& =1\end{array}$

$\begin{array}{cc}\Rightarrow |𝐴𝐵-𝐶|=\frac{1}{\left|𝐷\right|}\text{ or }\left|𝐷\right|& =\frac{1}{|𝐴𝐵-𝐶|}\\ \text{ Now, }|𝐵𝐶-𝐶|\times \left|𝐵𝐷𝐴\right|& =\frac{1}{\left|𝐷\right|}\times \left|𝐵𝐷𝐴\right|\\ & =\frac{\left|𝐵\right|\times \left|𝐷\right|\times \left|𝐴\right|}{\left|𝐷\right|}\\ & =\left|𝐵\right|\times \left|𝐴\right|=\left|𝐵𝐴\right|\end{array}$

If $𝐷$ and $𝐴𝐵-𝐶$ are inverse of each other,

$\begin{array}{cc}𝑡ℎ𝑒𝑛(𝐴𝐵-𝐶)\times 𝐷& =𝐷\times (𝐴𝐵-𝐶)=𝐼\\ 𝐴𝐵𝐷-𝐷& =𝐷𝐴𝐵-𝐷=𝐼\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow & 𝐴𝐵𝐷=𝐷𝐴𝐵=𝐷+𝐼\\ \therefore & 𝐴𝐵𝐷=𝐷𝐴𝐵\end{array}\\ & \text{ Hence, both statements are true. }\end{array}$

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}0& 0& -1\\ 0& -1& 0\\ -1& 0& 0\end{array}⎤⎦⎥,𝐵=⎡⎣⎢\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}⎤⎦⎥\\ \left|𝐴\right|& =⎡⎣⎢\begin{array}{ccc}0& 0& -1\\ 0& -1& 0\\ -1& 0& 0\end{array}\mid =-1(0-1)=1\\ \mathrm{adj}𝐴& ={⎡⎣⎢\begin{array}{ccc}0& 0& -1\\ 0& -1& 0\\ -1& 0& 0\end{array}⎤⎦⎥}^{𝑇}\\ {𝐴}^{-1}& =\frac{1}{\left|𝐴\right|}\mathrm{adj}𝐴=𝐴\\ \left|𝐵\right|& =-1\left(1\right)=-1\\ \mathrm{adj}𝐵& =⎡⎣⎢\begin{array}{ccc}0& -1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}\therefore {𝐵}^{-1}=& \frac{1}{\left|𝐵\right|}\mathrm{adj}𝐵=𝐵\\ & {\left({𝐴}^{-1}𝐵\right)}^{-1}+{\left(𝐴{𝐵}^{-1}\right)}^{-1}\\ & (𝐴𝐵{)}^{-1}+(𝐴𝐵{)}^{-1}\\ & [\because {𝐴}^{-1}=𝐴\text{ and }{𝐵}^{-1}=𝐵]\\ & =2(𝐴𝐵{)}^{-1}...(𝑖)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝐴𝐵& =⎡⎣⎢\begin{array}{ccc}0& 0& -1\\ 0& -1& 0\\ -1& 0& 0\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}0& 0& -1\\ -1& 0& 0\\ 0& -1& 0\end{array}⎤⎦⎥\\ \left|𝐴𝐵\right|& =-1\left(1\right)=-1\\ \mathrm{adj}𝐴𝐵& ={⎡⎣⎢\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}⎤⎦⎥}^{𝑇}=⎡⎣⎢\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}⎤⎦⎥\end{array}\\ & \text{ From Eq. (i), }\\ & \begin{array}{cc}2𝐴{𝐵}^{-1}& =2\frac{1}{\left|𝐴𝐵\right|}\mathrm{adj}\left(𝐴𝐵\right)\\ & =-2⎡⎣⎢\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}0& -2& 0\\ 0& 0& -2\\ -2& 0& 0\end{array}⎤⎦⎥\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}& ⎡⎣⎢\begin{array}{ccc}7& 5& 𝛼\\ 𝛽& 2& 11\\ 3& 𝛾& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}1\\ 3\\ 2\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}𝛼+𝛽\\ -2𝛼+𝛽-2𝛾\\ 𝛼+2𝛽+3𝛾\end{array}⎤⎦⎥\\ & \Rightarrow ⎡⎣⎢\begin{array}{c}22+2𝛼\\ 𝛽+28\\ 5+3𝛾\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}𝛼+𝛽\\ -2𝛼+𝛽-2𝛾\\ 𝛼+2𝛽+3𝛾\end{array}⎤⎦⎥\end{array}\end{array}$

$\Rightarrow \begin{array}{cc}22+2𝛼& =𝛼+𝛽\\ -𝛼+𝛽& =22...\left(𝑖\right)\end{array}$

Also, $-2𝛼+𝛽-2𝛾=𝛽+28$

$\Rightarrow -2(𝛼+𝛾)=28$

or $𝛼+𝛾=-14...\left(𝑖𝑖\right)$

Also, $5+3𝛾=𝛼+2𝛽+3𝛾$

$𝛼+2𝛽=5...\left(𝑖𝑖𝑖\right)$

On adding Eqs. (i) and (iii),

$3𝛽=27\Rightarrow 𝛽=9$

From Eq. (i), $-𝛼+9=22$

$𝛼=-13$

From Eq. (ii), $-13+𝛾=-14$

$\begin{array}{cc}& 𝛾=-1\\ & \begin{array}{cc}\therefore 100+\frac{2𝛼+11𝛽}{𝛾}& =100+\left[\frac{(-26)+99}{-1}\right]\\ & =100-73=27\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }𝑧=𝛼+𝑖𝛽\text{, }\\ & \begin{array}{c}\left|𝑧\right|-𝑧=1+2𝑖;\left|𝑧\right|=\sqrt{\surd {𝛼}^2+{𝛽}^2}\\ \sqrt{\surd {𝛼}^2+{𝛽}^2}-𝛼-𝑖𝛽=1+2𝑖\\ \Rightarrow 𝛽=-2\\ \text{ and }\sqrt{\surd {𝛼}^2+4}-𝛼=1\end{array}\end{array}$

$\begin{array}{cc}\Rightarrow & {𝛼}^2+4=(1+𝛼{)}^2\\ & [\because \text{ on squaring both sides }]\\ \Rightarrow & {𝛼}^2+4={𝛼}^2+1+2𝛼\\ \Rightarrow & 2𝛼=3\\ \Rightarrow & 𝛼=\frac{3}{2}\end{array}$

$\begin{array}{cc}\therefore 𝑧\overline{𝑧}& =|𝑧{|}^2={𝛼}^2+{𝛽}^2\\ & ={\left(\frac{3}{2}\right)}^2+(-2{)}^2\\ & =\frac{9}{4}+4=\frac{25}{4}\end{array}$

Given, $𝑖{𝑧}^2-2(𝑖+1)𝑧+(2-𝑖)=0$

$\therefore -𝑖$ and $𝛼$ are the roots of above equation.

$\begin{array}{cccc}& & -𝑖\times 𝛼& =\frac{2-𝑖}{𝑖}\\ \Rightarrow & & 𝛼& =\frac{2-𝑖}{-{𝑖}^2}\\ \Rightarrow & & 𝛼& =2-𝑖\end{array}$

Clearly, $\tan 𝜃=-\frac{1}{2},𝛾=\sqrt{\surd 5}$

$\sin 𝜃=\frac{-1}{\sqrt{\surd 5}},\cos 𝜃=\frac{2}{\sqrt{\surd 5}}$

[ ∵ from the polar form]

$\begin{array}{cc}& \cos 2𝜃={\cos }^2𝜃-{\sin }^2𝜃=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}\\ & \cos 6𝜃=4{\cos }^{3}2𝜃-3\cos 2𝜃\\ & =4{\left(\frac{3}{5}\right)}^3-3\times \frac{3}{5}=\frac{108}{125}-\frac{9}{5}\\ & =\frac{108-225}{125}=-\frac{117}{125}\\ & \therefore 5^3\cos 6𝜃=5^3\times -\left(\frac{117}{125}\right)=-117\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ Let }{𝑧}^{𝑛}=1\\ & {𝑧}^{𝑛}-1=(𝑧-1)(𝑧-{𝛼}_1)(𝑧-{𝛼}_2)(𝑧-{𝛼}_3)\\ & \dots (𝑧-{𝛼}_{𝑛-1})\end{array}\end{array}$

Sum of $𝑛$th roots of unity $=0$

$\begin{array}{cc}1+{𝛼}_1+{𝛼}_2+{𝛼}_3+\dots +{𝛼}_{𝑛-1}& =0\\ \Rightarrow {𝛼}_1+{𝛼}_2+{𝛼}_3+\dots +{𝛼}_{𝑛-1}& =-1\end{array}$

Also, the coefficient of ${𝑧}^{𝑛-1}$ is zero, so sum of product of zeroes taken two at a time $=0$

$\begin{array}{cc}\Rightarrow {𝛼}_1+{𝛼}_2+\dots & +{𝛼}_{𝑛-1}\\ & +\sum _{1\le 𝑖<𝑗\le 𝑛-1}{𝛼}_{𝑖}{𝛼}_{𝑗}=0\end{array}$

$\Rightarrow -1+\sum _{1\le 𝑖<𝑗\le 𝑛-1}^{}{𝑎}_{𝑖}{𝑎}_{𝑗}=0$

$\Rightarrow \sum _{1\le 𝑖<𝑗\le 𝑛-1}^{}{𝑎}_{𝑖}{𝑎}_{𝑗}=1$

$\begin{array}{cc}𝑓\left(𝑥\right)& =𝐴{𝑥}^2+𝐵𝑥\\ 𝑔\left(𝑥\right)& =𝐿{𝑥}^2+𝑀𝑥+𝑁\\ 𝑓\left(𝑥\right)-𝑔\left(𝑥\right)& =𝐴{𝑥}^2+𝐵𝑥-𝐿{𝑥}^2-𝑀𝑥-𝑁\\ & =(𝐴-𝐿){𝑥}^2+(𝐵-𝑀)𝑥-𝑁\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝐿𝑒𝑡& 𝐴-𝐿=𝑃\\ & 𝐵-𝑀=𝑄\\ & 𝑓\left(𝑥\right)-𝑔\left(𝑥\right)=𝑃{𝑥}^2+𝑄𝑥-𝑁\\ & 𝑓\left(2\right)-𝑔\left(2\right)=1\\ & 4𝑃+2𝑄-𝑁=1...\left(𝑖\right)\end{array}\end{array}$

Also, $𝑓(3-𝑔(3)=4$

$9𝑃+3𝑄-𝑁=4...\left(𝑖𝑖\right)$

Also, $𝑓\left(4\right)-𝑔\left(4\right)=9$

$16𝑃+4𝑄-𝑁=9...\left(𝑖𝑖𝑖\right)$

Now, Eq. (i) $\times 4$-Eq. (iii), we get

$4𝑄-3𝑁=-5...\left(𝑖𝑣\right)$

Eq. (i) $\times 9-$ Eq. (ii) $\times 4$, we get

$6𝑄-5𝑁=-7...\left(𝑣\right)$

Eq. (iv) $\times 3$-Eq. (v) $\times 2$, we get

$\begin{array}{ccc}& -9𝑁+10𝑁& =-15+14\\ & \Rightarrow & 𝑁& =-1\end{array}$

From Eq. (iv),

$\begin{array}{cccc}& & 4𝑄-3(-1)& =-5\\ \Rightarrow & & 4𝑄+3& =-5\\ \Rightarrow & & 4𝑄& =-8\\ \Rightarrow & & 𝑄& =-2\end{array}$

From Eq. (i),

$\begin{array}{cccc}& & 4𝑃+2(-2)-(-1)& =1\\ & & 4𝑃-4+1& =1\\ & & 𝑃=1\\ & \therefore & 𝑓\left(𝑥\right)-𝑔\left(𝑥\right)& ={𝑥}^2-2𝑥+1\\ & & =(𝑥-1{)}^2\\ & & 𝑓\left(𝑥\right)-𝑔\left(𝑥\right)& =0\\ & \Rightarrow & (𝑥-1{)}^2& =0\\ & \Rightarrow & 𝑥& =1\end{array}$

∴ The root of $𝑓\left(𝑥\right)-𝑔\left(𝑥\right)=0$ is 1 .

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{2𝑥-3}{(𝑥-2)(𝑥-3)}\\ & \text{ Let }𝑦=\frac{2𝑥-3}{{𝑥}^2-5𝑥+6}\\ & \Rightarrow {𝑥}^2𝑦-5𝑥𝑦+6𝑦-2𝑥+3=0\\ & \Rightarrow {𝑥}^2𝑦-𝑥(5𝑦+2)+6𝑦+3=0\\ & 𝐷\ge 0\end{array}$

$\begin{array}{cc}& (5𝑦+2{)}^2-4𝑦(6𝑦+3)\ge 0\\ \Rightarrow & 25{𝑦}^2+4+20𝑦-24{𝑦}^2-12𝑦\ge 0\\ \Rightarrow & {𝑦}^2+8𝑦+4\ge 0\\ \therefore & {𝑦}^2+8𝑦+16-12\ge 0\\ \Rightarrow & (𝑦+4{)}^2\ge 12\end{array}$

From option (a),

$(-10+4{)}^2=36\ge 12$

From option (b),

$(2+4{)}^2=36\ge 12$

From option (c),

$(1+4{)}^2=25\ge 12$

From option (d),

$(-2+4{)}^2=4<12$

∴ Option (d) is correct.

Given, $𝛼,𝛽$ and $𝛾$ are the roots of the equation

$5{𝑥}^3-3{𝑥}^2+2𝑥-4=0$

Then, $𝛼+𝛽+𝛾=\frac{3}{5}$

$𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{2}{5}$

and   $𝛼𝛽𝛾=\frac{4}{5}$

Now, $𝛼+𝛽+𝛾=\frac{3}{5}$

$\begin{array}{cc}& \text{ On squaring both sides, we get }\\ & \begin{array}{cc}& (𝛼+𝛽+𝛾{)}^2=\frac{9}{25}\\ \Rightarrow & {𝛼}^2+{𝛽}^2+{𝛾}^2\\ & +2(𝛼𝛽+𝛽𝛼+𝛾𝛼)=\frac{9}{25}\\ \Rightarrow & {𝛼}^2+{𝛽}^2+{𝛾}^2=\frac{9}{25}-\frac{4}{5}=-\frac{11}{25}\\ \therefore & 𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{2}{5}\end{array}\end{array}$

$\begin{array}{cc}& \text{ On squaring both sides, we get }\\ & \begin{array}{cc}& {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛼}^2{𝛾}^2+2𝛼{𝛽}^2𝛾+2𝛼𝛽{𝛾}^2+2{𝛼}^2𝛽𝛾=\frac{4}{25}\\ & \Rightarrow \Sigma {𝛼}^2{𝛽}^2+2𝛼𝛽𝛾(𝛼+𝛽+𝛾)=4/25\\ & \Rightarrow \Sigma {𝛼}^2{𝛽}^2=\frac{4}{25}-2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)=\frac{4}{25}-\frac{24}{25}\\ & =-\frac{20}{25}=-\frac{4}{5}\end{array}\end{array}$

Given, $𝛼,𝛽$ and $𝛾$ are the roots of equation ${𝑥}^3+4{𝑥}^2-9𝑥-36=0$ and

$\begin{array}{cccc}& 𝛼+𝛽=0& & \\ \therefore & & 𝛼+𝛽+𝛾& =-4\\ \Rightarrow & & 𝛾& =-4\\ & & 𝛼𝛽+𝛽𝛾+𝛾𝛼& =-9\\ \Rightarrow & & 𝛼𝛽+𝛾(𝛼+𝛽)& =-9\\ \Rightarrow & & 𝛼𝛽& =-9\\ & & 𝛼(-𝛼)& =-9\\ & & & [\because 𝛼+𝛽=0]\end{array}$

So, $-{𝛼}^2=-9\Rightarrow {𝛼}^2=9.$

Since $𝛼+𝛽=0$, we get $𝛽=-𝛼$. Therefore, ${𝛽}^2={𝛼}^2=9$.

Also, from above, $𝛾=-4$, so ${𝛾}^2=16$.

Now substitute these values:

$\begin{array}{cc}{𝛼}^2+2{𝛽}^2+3{𝛾}^2& =9+2\times 9+3\times 16\\ & =9+18+48\\ & =75\end{array}$

Given, 2- $𝑖$ is one of the roots of the equation

${𝑥}^4-9{𝑥}^3+31{𝑥}^2-49𝑥+30=0$

As coefficient of ${𝑥}^4,{𝑥}^3,{𝑥}^2,𝑥$ and constant are real numbers and $2-𝑖$ is one of the root, then $2+𝑖$ is the another root of the equation.

$\begin{array}{cccc}& & (𝑥-2-𝑖)(𝑥-2+𝑖)& =0\\ \Rightarrow & & (𝑥-2{)}^2-{𝑖}^2& =0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow {𝑥}^2-4𝑥+4+1=0\\ & \Rightarrow {𝑥}^2-4𝑥+5=0\end{array}\\ & \text{ Now, }\end{array}$

$\begin{array}{cc}\therefore & {𝑥}^2-5𝑥+6=0\\ \Rightarrow & {𝑥}^2-3𝑥-2𝑥+6=0\\ \Rightarrow & 𝑥(𝑥-3)-2(𝑥-3)=0\\ \Rightarrow & (𝑥-2)(𝑥-3)=0\\ \Rightarrow & 𝑥=2,3\\ \therefore & 𝛼=2,𝛽=3\\ & 2𝛼-𝛽=2\times 2-3=1\end{array}$

Given, equation

$6{𝑥}^6-25{𝑥}^5+31{𝑥}^4-31{𝑥}^2+25𝑥-6=0$

which is of reciprocal equation of even degree with opposite sign.

$\therefore {𝑥}^2-1$ is the root of the equation.

$\therefore 𝑥=\pm 1$ are the zeroes of the equation.

$\begin{array}{cc}& \Rightarrow 6{𝑥}^4-25{𝑥}^3+37{𝑥}^2-25𝑥+6=0\\ & \text{ Dividing by }{𝑥}^2\text{, we get }\\ & \begin{array}{cc}6{𝑥}^2-25𝑥+37-\frac{25}{𝑥}+\frac{6}{{𝑥}^2}& =0\\ \Rightarrow 6({𝑥}^2+\frac{1}{{𝑥}^2})-25(𝑥+\frac{1}{𝑥})+37& =0\end{array}\end{array}$

$\begin{array}{cc}& 6({(𝑥+\frac{1}{𝑥})}^2-2)-25(𝑥+\frac{1}{𝑥})+37=0\\ & \Rightarrow 6{(𝑥+\frac{1}{𝑥})}^2-25(𝑥+\frac{1}{𝑥})+25=0\\ & \therefore 6{𝑡}^2-25𝑡+25=0[\text{ Let }𝑥+\frac{1}{𝑥}=𝑡]\\ & \Rightarrow 6{𝑡}^2-15𝑡-10𝑡+25=0\\ & \Rightarrow 3𝑡(2𝑡-5)-5(2𝑡-5)=0\\ & \Rightarrow 𝑡=\frac{5}{3},𝑡=\frac{5}{2}\\ & \therefore 𝑥+\frac{1}{𝑥}=\frac{5}{3}\text{ or }𝑥+\frac{1}{𝑥}=\frac{5}{2}\\ & \Rightarrow 3{𝑥}^2+3=5𝑥\text{ or }2{𝑥}^2+2=5𝑥\\ & \Rightarrow 3{𝑥}^2-5𝑥+3=0\\ & \text{ or }2{𝑥}^2-5𝑥+2=0\end{array}$

The roots of $3{𝑥}^2-5𝑥+3=0$ are not real

$\begin{array}{cccc}& & 2{𝑥}^2-5𝑥+2& =0\\ \Rightarrow & & 2{𝑥}^2-4𝑥-𝑥+2& =0\\ \Rightarrow & & 2𝑥(𝑥-2)-1(𝑥-2)& =0\\ \Rightarrow & & 𝑥& =1/2,2\end{array}$

∴ The rational roots $\frac{1}{2},2,-1,\frac{5}{3},\frac{5}{2}$.

The value of $𝑚=-1$ and $𝑀=2$

$𝑀-𝑚=2-(-1)=3$

The total possible arrangement of the word 'LINEAR' is $6!=720$.

Number of possible arrangement of the word 'LINEAR' in which $𝐸$ and $𝐴$ come together is $5!2!=120\times 2=240$

Number of possible arrangement in the word 'LINEAR' in which $𝐸$ and $𝐴$ come together and $𝑁$ and $𝑅$ come together $=2!2!4!=2\times 2\times 24=96$

Number of possible arrangement in which $𝐸$ and $𝐴$ come together but $𝑁$ and $𝑅$ do not come together $=240-96=144$

Given, 15 lines are concurrent at a point $𝑃$.

Now, if a line is not passing through a point $𝑃$, then the triangle can be formed if those 15 times, concurrent at 2 points, so the required number of ways of forming a triangle $={}^{15}{𝐶}_2$

$=\frac{15\times 14}{1\times 2}=15\times 7=105$

Given, the expansion $(𝑎+𝑥{)}^{𝑛}$ contains 15 terms $\Rightarrow 𝑛=14$

Middle term of expansion $(𝑎+1{)}^{𝑛}$ is

${𝑇}_8={}^{14}{𝐶}_7{𝑎}^7$

Ratio of the neighbouring term in the expansion $(𝑎+𝑥{)}^{𝑛}=16$

$\begin{array}{cc}\therefore & \frac{{𝑇}_8}{{𝑇}_6}=16\Rightarrow \frac{{}^{14}{𝐶}_8{𝑎}^8}{{}^{14}{𝐶}_6{𝑎}^6}=16\\ \Rightarrow & {𝑎}^2=16[\because {}^{14}{𝐶}_8={}^{14}{𝐶}_6]\\ \Rightarrow & 𝑎=4\end{array}$