TG EAPCET 2023 14th May Morning Shift Question Paper with Solutions

As, we know in ${}^{𝑛}{𝐶}_{𝑟},𝑛\ge 𝑟$ and $𝑛\in 𝐼$, $𝑟\ge 0$

So, to define, ${}^{(16-𝑥)}{𝐶}_{(2𝑥-1)}$

$\begin{array}{cc}& 16-𝑥\ge 2𝑥-1\\ \Rightarrow & 3𝑥\le 17\\ \Rightarrow & 𝑥\le \frac{17}{3}\text{ and }2𝑥-1\ge 0\\ \Rightarrow & 𝑥\ge \frac{1}{2}\end{array}$

So, $\frac{1}{2}\le 𝑥\le \frac{17}{3}$

Hence, $𝑥=\{1,2,3,4,5\}$

$\begin{array}{cc}& 𝑋=\{\left[\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right]\mid 𝑎,𝑏,𝑐,𝑑\in 𝑅\}\\ & 𝑓:𝑋\to 𝑅\\ & 𝑓\left(𝐴\right)=\det \left(𝐴\right),\forall 𝐴\in 𝑋\end{array}$

Now, by taking examples

$\begin{array}{cc}& \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right],\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]\in 𝑋\\ & \det \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\det \left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]=0\\ & 𝑓\left(\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\right)=𝑓\left(\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]\right)\\ & \Rightarrow \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\ne \left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right]\end{array}$

So, this function is not one-one.

Let $𝑓:𝑋\to 𝑌$

to show onto for each such that $𝑓\left(𝑥\right)={𝑦}_0$; ${𝑦}_0\in 𝑌$

there exists $𝑥\in 𝑋$

$\begin{array}{cc}& {𝑦}_0\in 𝑅\\ & \left[\begin{array}{cc}{𝑦}_0& 0\\ 0& 1\end{array}\right]\text{ such that }\mid \begin{array}{cc}{𝑦}_0& 0\\ 0& 1\end{array}\mid ={𝑦}_0\\ & 𝑓\left(\left[\begin{array}{cc}{𝑦}_0& 0\\ 0& 1\end{array}\right]\right)={𝑦}_0\end{array}$

This function is onto function.

$\because 𝑓(𝑥+𝑦)=𝑓\left(𝑥\right)+𝑓\left(𝑦\right)$

and $𝑓\left(1\right)=7$

Put $𝑥=𝑦=1$

Then, $𝑓\left(2\right)<text>=</text><text>\mid </text>2𝑓\left(1\right)$

$\Rightarrow 𝑓\left(2\right)=2\times 7=14$

On putting $𝑥=2,𝑦=1$ in Eq. (i), we get $𝑓\left(3\right)=𝑓\left(2\right)+𝑓\left(1\right)=14+7=21$

On putting $𝑥=3,𝑦=1$ in Eq. (i), we get $𝑓\left(4\right)=𝑓\left(3\right)+𝑓\left(1\right)=21+7=28$

i.e., $𝑓\left(1\right),𝑓\left(2\right),𝑓\left(3\right),\dots$, are in AP .

First term of $\mathrm{AP}=𝑓\left(l\right)=7$

and common difference $=𝑓\left(2\right)-𝑓\left(1\right)=7$

$\begin{array}{cc}𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,& \sum _{𝑟=1}^{𝑛}𝑓\left(𝑟\right)=𝑓\left(1\right)+𝑓\left(2\right)+\dots +𝑓\left(𝑛\right)\\ & =7+14+21+\dots +7𝑛={𝑆}_{𝑛}\end{array}$

$𝑆𝑜,{𝑆}_{𝑛}=\frac{𝑛}{2}(7+7𝑛)=\frac{7𝑛}{2}(𝑛+1)$

As we know,

If $𝑛$ is the order of a square matrix.

Then, $|\mathrm{adj}(\mathrm{adj}𝐴\left)\right|=|𝐴{|}^{(𝑛-1{)}^2}$

So, $\mid \mathrm{adj}(\mathrm{adj}{𝐴}^2)\mid ={\mid {𝐴}^2\mid }^{(𝑛-1{)}^2}$

∵ Order of $𝐴=𝑛=3$

$\begin{array}{cc}𝑇ℎ𝑒𝑛,\mid \mathrm{adj}(\mathrm{adj}{𝐴}^2)\mid & ={\mid {𝐴}^2\mid }^{(3-1{)}^2}\\ & ={\mid {𝐴}^2\mid }^4=|𝐴{|}^8\end{array}$

We have,

$𝐴$ and $𝐵$ are two square matrices

$\begin{array}{cc}(𝐴𝐵+𝐵𝐴{)}^{𝑇}+(𝐴𝐵-𝐵𝐴{)}^{𝑇}& =2𝐵𝐴\\ (𝐴𝐵{)}^{𝑇}+(𝐵𝐴{)}^{𝑇}+(𝐴𝐵{)}^{𝑇}-(𝐵𝐴{)}^{𝑇}& =2𝐵𝐴\\ 2(𝐴𝐵{)}^{𝑇}& =2𝐵𝐴\\ (𝐴𝐵{)}^{𝑇}& =𝐵𝐴\\ {𝐴}^{𝑇}{𝐵}^{𝑇}& =𝐵𝐴\end{array}$

If $𝐴$ is symmetric matrix and $𝐵$ is symmetric matrix.

$\begin{array}{cc}\because {𝐴}^{𝑇}& =𝐴\text{ and }{𝐵}^{𝑇}=𝐵\\ 𝐴\left(𝐵\right)& =𝐵𝐴\\ 𝐴𝐵& =𝐵𝐴\text{ (which is not possible) }\end{array}$

Now, if $𝐴$ and $𝐵$ are skew-symmetric matrix.

$\begin{array}{cc}\because {𝐴}^{𝑇}& =-𝐴\text{ and }{𝐵}^{𝑇}=-𝐵\\ -𝐴-𝐵& =𝐵𝐴\\ 𝐴𝐵& =𝐵𝐴\text{ (which is not possible) }\end{array}$

Thus, $𝐴$ and $𝐵$ are neither symmetric nor skew-symmetric matrix.

Given,

$\mathrm{adj}⎡⎣⎢\begin{array}{ccc}1& 0& 2\\ -1& 1& -2\\ 0& 2& 1\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}5& 𝑚& -2\\ 1& 1& 0\\ -2& -2& 𝑛\end{array}⎤⎦⎥$

Adjoint of matrix $=$ transpose of cofactor matrix

$\begin{array}{cc}& {𝐶}_{11}=\mid \begin{array}{cc}1& -2\\ 2& 1\end{array}\mid =5\\ & {𝐶}_{12}=-\mid \begin{array}{cc}-1& -2\\ 0& 1\end{array}\mid =1\\ & {𝐶}_{13}=\mid \begin{array}{cc}-1& 1\\ 0& 2\end{array}\mid =-2\end{array}$

$\begin{array}{cc}{𝐶}_{21}& =-\mid \begin{array}{cc}0& 2\\ 2& 1\end{array}\mid =4\\ {𝐶}_{22}& =\mid \begin{array}{cc}1& 2\\ 0& 1\end{array}\mid =1\\ {𝐶}_{23}& =-\mid \begin{array}{cc}1& 0\\ 0& 2\end{array}\mid =-2\\ {𝐶}_{31}& =\mid \begin{array}{cc}0& 2\\ 1& -2\end{array}\mid =-2\\ {𝐶}_{32}& =-\mid \begin{array}{cc}1& 2\\ -1& -2\end{array}\mid =0\\ {𝐶}_{33}=\mid \begin{array}{cc}1& 0\\ -1& 1\end{array}\mid & =1\\ \text{ So, Adj }⎡⎣⎢\begin{array}{ccc}1& 0& 2\\ -1& 1& -2\\ 0& 2& 1\end{array}⎤⎦⎥& =⎡⎣⎢\begin{array}{ccc}5& 1& -2\\ 4& 1& -2\\ -2& 0& 1\end{array}⎤⎦⎥\\ ⎡⎣⎢\begin{array}{ccc}5& 𝑚& -2\\ 1& 1& 0\\ -2& -2& 𝑛\end{array}⎤⎦⎥& =⎡⎣⎢\begin{array}{ccc}5& 4& -2\\ 1& 1& 0\\ -2& -2& 1\end{array}⎤⎦⎥\end{array}$

Hence, $𝑚=4,𝑛=1$

$\therefore 𝑚+𝑛=4+1=5$

We have,

$\begin{array}{c}𝐴=\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]\text{ and }\\ 𝑓\left(𝑥\right)=𝑥+{𝑥}^2+{𝑥}^3+\dots +{𝑥}^{2023}\\ \text{ Then, }{𝐴}^2=𝐴\cdot 𝐴=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ {𝐴}^3=𝐴{𝐴}^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ {𝐴}^4=𝐴{𝐴}^3=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ {𝐴}^{𝑛}=𝐴{𝐴}^{𝑛-1}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

So, $𝑓\left(𝐴\right)=𝐴+{𝐴}^2+{𝐴}^3+\dots +{𝐴}^{2023}$

$\begin{array}{cc}& =\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]+\dots +\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ & =\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]\end{array}$

Hence, $𝑓\left(𝐴\right)+𝐼=\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

Given, $𝑖=\sqrt{\surd -1}$

Then,

$\begin{array}{c}\sum _{\infty }^{𝑛=0}{\left(\frac{𝑖}{3}\right)}^{𝑛}={\left(\frac{𝑖}{3}\right)}^0+{\left(\frac{𝑖}{3}\right)}^1+{\left(\frac{𝑖}{3}\right)}^2+{\left(\frac{𝑖}{3}\right)}^3\\ +\dots +{\left(\frac{𝑖}{3}\right)}^{\infty }\end{array}$

$=1+\frac{𝑖}{3}+{\left(\frac{𝑖}{3}\right)}^2+{\left(\frac{𝑖}{3}\right)}^3+\dots$

Clearly, it forms a GP with infinite terms and first term as 1 and common ratio as $\left(\frac{𝑖}{3}\right)$.

$\begin{array}{cc}\Rightarrow {𝑆}_{\infty }\frac{1}{1-\frac{𝑖}{3}}& =\frac{3}{3-𝑖}\left[\begin{array}{c}\text{ sum of a GP with }\\ \text{ infinite term }=\frac{𝑎}{1-𝑟}\end{array}\right]\\ & =\frac{3(3+𝑖)}{(3-𝑖)(3+𝑖)}\\ & =\frac{9+3𝑖}{9-(-1)}=\frac{9+3𝑖}{10}\end{array}$

$\text{ Given, }𝑖=\sqrt{\surd -1}$

$\begin{array}{cc}\text{ Consider, }𝑧=& \frac{(1+𝑖{)}^{2025}}{(1-𝑖{)}^{2022}}=\frac{(1+𝑖{)}^{2022}}{(1-𝑖{)}^{2022}}(1+𝑖{)}^3\\ & =\frac{{\left[\right(1+𝑖{)}^2]}^{1011}(1+𝑖{)}^3}{{\left[\right(1-𝑖{)}^2]}^{1011}}\\ & =\frac{[1-1+2𝑖{]}^{1011}}{[1-1-2𝑖{]}^{1011}}(1+𝑖{)}^3\\ & ={\left(\frac{2𝑖}{-2𝑖}\right)}^{1011}(1+𝑖{)}^2(1+𝑖)\\ & =(-1{)}^{1011}[1-1+2𝑖\left]\right(1+𝑖)\\ & =-1\left(2𝑖\right)(1+𝑖)=-[2𝑖+2(𝑖{)}^2]\\ & =-2𝑖-(-2)=2-2𝑖\end{array}$

We have,

$\begin{array}{cc}\frac{|𝑧-𝑖|}{|𝑧+𝑖|}& =2,\text{ where }𝑧=𝑥+\\ |𝑧-𝑖|& =2|𝑧+𝑖|\\ |𝑥+(𝑦-1\left)𝑖\right|& =2|𝑥+(𝑦+1\left)𝑖\right|\end{array}$

Squaring on both sides, we get

$\begin{array}{c}|𝑥+(𝑦-1)𝑖{|}^2=(2|𝑥+(𝑦+1\left)𝑖\right|{)}^2\\ {𝑥}^2+(𝑦-1{)}^2=4({𝑥}^2+(𝑦+1{)}^2)\\ {𝑥}^2+{𝑦}^2+1-2𝑦=4{𝑥}^2+4{𝑦}^2+4+8𝑦\\ 3{𝑥}^2+3{𝑦}^2+8𝑦+2𝑦+4-1=0\\ 3{𝑥}^2+3{𝑦}^2+10𝑦+3=0\end{array}$

Hence, the locus of $𝑧$ is

$3{𝑥}^2+3{𝑦}^2+10𝑦+3=0$

$\begin{array}{cc}& \text{ Given, }{𝑥}_{𝑛}=\cos \frac{𝜋}{2^{𝑛}}+𝑖\sin \frac{𝜋}{2^{𝑛}}\\ & \prod _{𝑛=1}^{\infty }{𝑥}_{𝑛}={𝑥}_1\cdot {𝑥}_2\cdot {𝑥}_3\dots \end{array}$

$\begin{array}{cc}& =(\cos \frac{𝜋}{2}+𝑖\sin \frac{𝜋}{2})(\cos \frac{𝜋}{4}+𝑖\sin \frac{𝜋}{4})(\cos \frac{𝜋}{8}+𝑖\sin \frac{𝜋}{8})\\ & =\cos (\frac{𝜋}{2}+\frac{𝜋}{4}+\frac{𝜋}{8}+\dots )+𝑖\sin (\frac{𝜋}{2}+\frac{𝜋}{4}+\frac{𝜋}{8}+\dots )\end{array}$

$\begin{array}{cc}& =\cos \left(\frac{𝜋}{2}(1+\frac{1}{2}+\frac{1}{4}+\dots )\right)+𝑖\sin \left(\frac{𝜋}{2}(1+\frac{1}{2}+\frac{1}{4}+\dots )\right)\\ & =\cos (\frac{𝜋}{2}\times \frac{1}{1-\frac{1}{2}})+𝑖\sin (\frac{𝜋}{2}\times \frac{1}{1-\frac{1}{2}})\\ & =\cos 𝜋+𝑖\sin 𝜋\\ & =-1+𝑖\left(0\right)=-1+0=-1\end{array}$

Given, the roots of the equation ${𝑧}^2-𝑖=0$ are $𝛼$ and $𝛽$.

Now, ${𝑧}^2-𝑖=0$

$\begin{array}{cc}\Rightarrow & {𝑧}^2=𝑖\\ \Rightarrow & 𝑧=\pm \sqrt{\surd 𝑖}\\ \Rightarrow & 𝑧=\pm \sqrt{\surd \cos \frac{𝜋}{2}+𝑖\sin \frac{𝜋}{2}}\\ \Rightarrow & 𝑧=\pm {(\cos \frac{𝜋}{2}+𝑖\sin \frac{𝜋}{2})}^{1/2}\\ \Rightarrow & 𝑧=\pm (\cos \frac{𝜋}{4}+𝑖\sin \frac{𝜋}{4})\end{array}$

$\begin{array}{cc}& \text{ Thus, }𝛼=\cos \frac{𝜋}{4}+𝑖\sin \frac{𝜋}{4}\text{ and }\\ & \begin{array}{cc}& 𝛽=-\cos \frac{𝜋}{4}-𝑖\sin \frac{𝜋}{4}\\ & 𝛼=\cos \frac{𝜋}{4}+𝑖\sin \frac{𝜋}{4}\text{ and }\\ & 𝛽=\cos \frac{5𝜋}{4}+𝑖\sin \frac{5𝜋}{4}\\ & \therefore \left|\right(\arg 𝛽-\arg 𝛼\left)\right|=\mid \frac{5𝜋}{4}-\frac{𝜋}{4}\mid =\mid \frac{4𝜋}{4}\mid =𝜋\end{array}\end{array}$

Given, ${𝑥}^2+2𝑝𝑥-2𝑝+8>0$ for all real values of $𝑥$.

$\begin{array}{c}\text{ So, }(2𝑝{)}^2-4(1\left)\right(-2𝑝+8)<0\\ \Rightarrow & 4{𝑝}^2+8𝑝-32<0\\ \Rightarrow & {𝑝}^2+2𝑝-8<0\\ \Rightarrow & (𝑝+4)(𝑝-2)<0\\ \Rightarrow & -4<𝑝<2\\ \Rightarrow & 𝑝\in (-4,2)\end{array}$

The expression $\frac{𝑥+3}{(𝑥-1)(𝑥+2)}$ is not defined only at $𝑥=1$ and $𝑥\ne 2$

Let $𝑦=\frac{𝑥+3}{(𝑥-1)(𝑥+2)}$

$\Rightarrow (𝑥-1)(𝑥+2)𝑦=𝑥+3$

$\begin{array}{cc}& \Rightarrow 𝑦{𝑥}^2+𝑦𝑥-2𝑦-𝑥-3=0\\ & \Rightarrow 𝑦{𝑥}^2+(𝑦-1)𝑥-(2𝑦+3)=0\\ & \Rightarrow 𝑥=\frac{-(𝑦-1)\pm \sqrt{\surd \begin{array}{c}(𝑦-1{)}^2-4(𝑦)\\ (-2𝑦-3)\end{array}}}{2𝑦}\end{array}$

Now, the range set of $\frac{𝑥+3}{(𝑥-1)(𝑥+2)}$ will be all real values of $𝑦$ except those value for which

$\begin{array}{cc}& (𝑦-1{)}^2-4𝑦(-2𝑦-3)<0\\ \Rightarrow & {𝑦}^2-2𝑦+1+8{𝑦}^2+12𝑦<0\\ \Rightarrow & 9{𝑦}^2+10𝑦+1<0\\ \Rightarrow & (9𝑦+1)(𝑦+1)<0\\ \Rightarrow -1& <𝑦<-\frac{1}{9}\end{array}$

Thus, range of $\frac{𝑥+3}{(𝑥-1)(𝑥+2)}$ be

$\begin{array}{cc}& 𝑅-(-1,-\frac{1}{9})\\ & \therefore 𝛼=-1\text{ and }𝛽=-\frac{1}{9}\end{array}$

The intercepts of the equation

$(-1)𝑥+(-\frac{1}{9})𝑦+1=0\text{ or }\frac{𝑥}{1}+\frac{𝑦}{9}=1\text{ be }1$

and 9.

Thus, the sum of intercept be $1+9=10$.

We know that, $\sin {18}^{\circ }=\frac{\sqrt{\surd 5}-1}{4}$ and $\cos {36}^{\circ }=\frac{\sqrt{\surd 5}+1}{4}$

$\begin{array}{cc}\therefore {\sin }^2{18}^{\circ }& ={\left(\frac{\sqrt{\surd 5}-1}{4}\right)}^2\\ & =\frac{5+1-2\sqrt{\surd 5}}{16}=\frac{6-2\sqrt{\surd 5}}{16}\\ \text{ and }{\cos }^2{36}^{\circ }& ={\left(\frac{\sqrt{\surd 5}+1}{4}\right)}^2=\frac{5+1+2\sqrt{\surd 5}}{16}\\ & =\frac{6+2\sqrt{\surd 5}}{16}\end{array}$

Now, the quadratic equation whose roots are ${\sin }^2{18}^{\circ }$ and ${\cos }^2{36}^{\circ }$ be

$\begin{array}{cc}(𝑥-\frac{6-2\sqrt{\surd 5}}{16})(𝑥-\frac{6+2\sqrt{\surd 5}}{16})& =0\\ \Rightarrow ((𝑥-\frac{3}{8})+\frac{\sqrt{\surd 5}}{8})((𝑥-\frac{3}{8})-\frac{\sqrt{\surd 5}}{8})& =0\end{array}$

$\begin{array}{cc}\Rightarrow & {(𝑥-\frac{3}{8})}^2-{\left(\frac{\sqrt{\surd 5}}{8}\right)}^2=0\\ \Rightarrow & {𝑥}^2-\frac{3}{4}𝑥+\frac{9}{64}-\frac{5}{64}=0\\ \Rightarrow & {𝑥}^2-\frac{3}{4}𝑥+\frac{1}{16}=0\\ \Rightarrow & 16{𝑥}^2-12𝑥+1=0\end{array}$

Given equation is,

${𝑥}^4+{𝑥}^3-4{𝑥}^2+𝑥+1=0$

On putting $𝑥=𝑦+ℎ$

$\begin{array}{c}\therefore (𝑦+ℎ{)}^4+(𝑦+ℎ{)}^3-4(𝑦+ℎ{)}^2\\ +(𝑦+ℎ)+1=0\\ \Rightarrow {({𝑦}^2+{ℎ}^2+2𝑦ℎ)}^2+{𝑦}^3+{ℎ}^3+3𝑦{ℎ}^2+3{𝑦}^2ℎ\\ -4{𝑦}^2-4{ℎ}^2-8𝑦ℎ+𝑦+ℎ+1=0\\ \Rightarrow {𝑦}^4+{ℎ}^4+4{𝑦}^2{ℎ}^2+2{𝑦}^2{ℎ}^2+4𝑦{ℎ}^3\\ +4{𝑦}^3ℎ+{𝑦}^3+{ℎ}^3+3𝑦{ℎ}^2+3{𝑦}^2ℎ\\ -4{𝑦}^2-4{ℎ}^2-8𝑦ℎ+𝑦+ℎ+1=0\\ \Rightarrow {𝑦}^4+(4ℎ+1){𝑦}^3+(6{ℎ}^2+3ℎ-4){𝑦}^2\\ +(4{ℎ}^3+3{ℎ}^2-8ℎ+1)𝑦\\ +{ℎ}^4+{ℎ}^3-4{ℎ}^2+ℎ+1=0\end{array}$

Given, transformed equation does not contain ${𝑥}^2$ term.

$\begin{array}{cc}& \therefore (6{ℎ}^2+3ℎ-4)=0\\ & \Rightarrow ℎ=\frac{-3\pm \sqrt{\surd 105}}{12}\end{array}$

So, $𝛼=\frac{-3+\sqrt{\surd 105}}{12}$ and $𝛽=\frac{-3-\sqrt{\surd 105}}{12}$

$\begin{array}{cc}& \text{ Now, }12(𝛼-𝛽{)}^2=12[\left(\frac{-3+\sqrt{\surd 105}}{12}\right)\\ & {-\left(\frac{-3-\sqrt{\surd 105}}{12}\right)]}^2\\ & =12{\left[\frac{-3+\sqrt{\surd 105}+3+\sqrt{\surd 105}}{12}\right]}^2\\ & =12{\left(\frac{2\sqrt{\surd 105}}{12}\right)}^2=12\times \frac{4\times 105}{12\times 12}=35\end{array}$

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

On comparing the given equation with $𝑎{𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+𝑑=0$, we get

$\begin{array}{cc}& 𝑎=1,𝑏=2,𝑐=-1\text{ and }𝑑=-2\\ & \therefore 𝛼+𝛽+𝛾=-\frac{𝑏}{𝑎}=-2\\ & 𝛼𝛽𝛾=\frac{-𝑑}{𝑎}=2\text{ and }𝛼𝛽+𝛽𝛾+𝛼𝛾=\frac{𝑐}{𝑎}=-1\end{array}$

Now, $(𝛼+𝛽+𝛾{)}^2={𝛼}^2+{𝛽}^2+{𝛾}^2+2$

$\begin{array}{cc}& (𝛼𝛽+𝛽𝛾+𝛼𝛾)\\ \Rightarrow & (-2{)}^2={𝛼}^2+{𝛽}^2+{𝛾}^2+2(-1)\\ \Rightarrow & {𝛼}^2+{𝛽}^2+{𝛾}^2=4+2=6\\ \therefore & {({𝛼}^2+{𝛽}^2+{𝛾}^2)}^3=(6{)}^3\\ \Rightarrow & ({𝛼}^6+{𝛽}^6+{𝛾}^6)+3[({𝛼}^2+{𝛽}^2+{𝛾}^2)\\ & ({𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛼}^2{𝛾}^2)-{𝛼}^2{𝛽}^2{𝛾}^2]=216\\ \Rightarrow & {𝛼}^6+{𝛽}^6+{𝛾}^6+3\left[\right(6\left)\right(9)-4]=216\\ & [\because {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛼}^2{𝛾}^2=9]\\ \Rightarrow & {𝛼}^6+{𝛽}^6+{𝛾}^6=216-150\\ \Rightarrow & {𝛼}^6+{𝛽}^6+{𝛾}^6=66\end{array}$

If polygon has $𝑛$ sides, then number of diagonals is

$\begin{array}{cccc}& & {}^{𝑛}{𝐶}_2-𝑛& =35\\ \Rightarrow & & \frac{𝑛(𝑛-1)}{2}-𝑛& =35\\ \Rightarrow & & {𝑛}^2-𝑛-2𝑛& =70\\ \Rightarrow & & {𝑛}^2-3𝑛-70& =0\\ \Rightarrow & & {𝑛}^2-10𝑛+7𝑛-70& =0\\ \Rightarrow & & 𝑛(𝑛-10)+7(𝑛-10)& =0\\ & \Rightarrow & 𝑛=10\text{ or }𝑛=-7\Rightarrow 𝑛& =10\end{array}$

Thus, there are 10 vertices.

Now, if $𝐴$ and $𝐵$ are two vertices, then the number of triangle formed by $𝐴𝐵$ as one of its side is equal to 8 .

We know that, any triangle can be obtained by joining any 3 points not in the same straight line. Thus, number of triangles obtained from 10 different point, number 3 of which are collinear $={}^{10}{𝐶}_3=120$

Triangles obtained from 4 points $={}^4{𝐶}_3=4$ and triangle obtained from remaining 6 points $={}^6{𝐶}_3=20$

So, required number of triangles $=120-20-4=96$

Case I A student chooses 4 out of first five questions and six out of the remaining 8 questions.

$\text{ ∴ Total number of ways }={}^5{𝐶}_4\times {}^8{𝐶}_6$

Case II A student chooses 5 out of first five questions and 5 out of remaining 8 questions.

$\text{ ∴ Total number of ways }={}^5{𝐶}_5\times {}^8{𝐶}_5$

∴ Total number of ways

$\begin{array}{cc}& =({}^5{𝐶}_4\times {}^8{𝐶}_6)+({}^5{𝐶}_5\times {}^8{𝐶}_5)\\ & =(5\times 28)+(1\times 56)=140+56=196\end{array}$

The given expression $(7-5𝑥{)}^{-2/3}$ can be written as, $7^{-2/3}{(1-\frac{5}{7}𝑥)}^{-2/3}$ Now, the expansion of ${(1-\frac{5}{7}𝑥)}^{-2/3}$ is valid, if

$\begin{array}{cc}& \mid \frac{5}{7}𝑥\mid <1\Rightarrow \frac{5}{7}\left|𝑥\right|<1\\ & \Rightarrow \left|𝑥\right|<\frac{7}{5}\Rightarrow -\frac{7}{5}<𝑥<\frac{7}{5}\end{array}$

Now, it can be concluded that $𝑐=7/5$

$\therefore 5𝑐+7=5\left(\frac{7}{5}\right)+7=7+7=14$