TG EAPCET 2024 10th May Morning Shift Question Paper with Solutions

Given,

$\begin{array}{c} f(x)=\sin ^{-1}\left(x^{2}-1\right)-3 \log _{3}\left(3^{x}-2\right) \\ -1 \leq x^{2}-1 \leq 1 \\ 0 \leq x^{2} \leq 2 \\ -\sqrt{2} < x < \sqrt{2} \end{array}$

$\therefore \sin ^{-1}\left(x^2-1\right)$ is not defined for $(-\infty,-\sqrt{2})$ and $(\sqrt{2}, 0)$

For $3 \log _3\left(3^x-2\right) \quad \Rightarrow \quad 3^x-2>0$

$$\begin{aligned} & 3^x>2 \\ & x>\log _3 2 \end{aligned}$$

$\therefore 3 \log _3\left(3^x-2\right)$ is not defined for $x \leq \log _3(2)$

So, intervel $[-\sqrt{2}, \sqrt{2}]$ combine with $x>\log _3(2)$. The domain of $f(x)$ is $\log _3(2)<x \leq \sqrt{2}$.

So, $f(x)$ is not defined for

$$\begin{aligned} & \left(-\infty, \log _3 2\right) \cup(\sqrt{2}, 0) \\ & a=\log _3 2, b=\sqrt{2} \\ & 3^a+b^2=2+2=4 \end{aligned}$$

Given,

$𝑓\left(𝑥\right)=\frac{1}{\sqrt{\surd |𝑥-|𝑥\left|\right|}}$

For

$\begin{array}{c}𝑥>0,\left|𝑥\right|=𝑥\\ \\ 𝑥-\left|𝑥\right|=0\\ \\ |𝑥-|𝑥\left|\right|=0\end{array}$

$\therefore 𝑓\left(𝑥\right)$ is not possible for $𝑥>0$.

For

$\begin{array}{cc}𝑥<0,\left|𝑥\right|& =-𝑥\\ \\ 𝑥-\left|𝑥\right|& =-2𝑥\\ \\ |𝑥-|𝑥\left|\right|& =2𝑥\\ \\ 𝑓\left(𝑥\right)& =\frac{1}{\sqrt{\surd -2𝑥}}\end{array}$

$\therefore 𝐴$ is set of negative numbers and $𝐵$ is a set of positive numbers.

$\therefore 𝐴\cap 𝐵=𝜙$

We have to check each option, From option (a), we have,

$\begin{array}{c}(2𝑛+7)<(𝑛+3{)}^2\\ \\ \text{ Let }𝑃\left(𝑛\right)\equiv (2𝑛+7)<(𝑛+3{)}^2\\ \\ 𝑃\left(1\right)=(2+7)=9<(1+3{)}^2=4^2=16\end{array}$

Let us assume that $𝑃\left(𝐾\right)$ is true.

We have to prove that $𝑃(𝐾+1)$ is true

$\begin{array}{c}(2𝐾+7)<(𝐾+3{)}^2\\ \\ 2𝐾+7<{𝐾}^2+9+6𝐾\end{array}$

Add 2 on LHS and $7+2𝐾$ on RHS

as $2<7+2𝐾$

$\begin{array}{c}\Rightarrow 2𝐾+9<{𝐾}^2+16+8𝐾\\ \\ 2(𝐾+1)+7<\left(\right(𝐾+1)+3{)}^2\end{array}$

$\therefore 𝑃(𝐾+1)$ is true.

Hence, $𝑃\left(𝑥\right)$ is true, $\forall 𝑛\in 𝑁$.

From option b,

$\begin{array}{c}\text{ Let }𝑄\left(𝑛\right)\equiv 1^2+2^2+\dots +{𝑛}^2>\frac{{𝑛}^3}{3}\\ \\ 𝑄\left(1\right)=1^2>\frac{1}{3}\end{array}$

Let $𝑄\left(𝐾\right)$ be true.

$\begin{array}{c}{1}^2+2^2+3^2+\dots +{𝐾}^2>\frac{{𝐾}^3}{3}\\ \\ 1^2+2^2+3^2+\dots .+{𝐾}^2\\ \\ +(𝐾+1{)}^2>\frac{{𝐾}^3}{3}+(𝐾+1{)}^2\\ \\ 1^2+2^2+3^2+\dots .+{𝐾}^2+(𝐾+1{)}^2\\ \\ >\frac{{𝐾}^3+3{𝐾}^2+6𝐾+3}{3}\\ \\ 1^2+2^2+3^2+\dots +{𝐾}^2+(𝐾+1{)}^2>\\ \\ \frac{{𝐾}^3+3{𝐾}^2+3𝐾+1+3𝐾+2}{3}\\ \\ =\frac{(𝐾+1{)}^3+3𝐾+2}{3}>\frac{(𝐾+1{)}^3}{3}\end{array}$

$\therefore 𝑄(𝐾+1{)}^2$ is true.

Hence, $𝑄\left(𝑛\right)$ is true, $\forall 𝑛\in 𝑁$

From option (c),

$\begin{array}{c}𝑅\left(𝑛\right)={35}^{2𝑛+1}+2^{3𝑛+1}\\ \\ 𝑅\left(1\right)=3\times 125+2^4=375+16=391\end{array}$

which is divisible by 23

Let us assume it is true for $𝑛=𝑘,𝑘\in 𝑁$

$\therefore 𝑆\left(𝑘\right)=3\cdot 5^{2𝑘+1}+2^{3𝑘+1}$ is divisible by

$23\forall 𝑘\in 𝑁$

Let $3\cdot 5^{2𝑘+1}+2^{3𝑘+1}=23𝑡,𝑡\in 𝑁$

$\therefore 3\cdot 5{𝑘}^{2𝑘+1}=23𝑡-2^{3𝑘+1}$

Now, we have to prove that for $𝑛=𝑘+1$

$\begin{array}{cc}\therefore 𝑆(𝑘+1)& =3\cdot 5^{2(𝑘+1)+1}+2^{3(𝑘+1)+1}\\ \\ & =3\cdot 5^{2𝑘+1}\cdot 5^2+2^{3𝑘+1}\cdot 2^3\\ \\ & =(23𝑡-2^{3𝑘+1})25+8\cdot 2^{3𝑘+1}\\ \\ & =23𝑡\times 25-25\times 2^{3𝑘+1}+8\cdot 2^{3𝑘+1}\\ \\ \therefore 𝑆(𝑘+1)& =23𝑡\times 25-{172}^{3𝑘+1}\end{array}$

Hence, $3\cdot 5^{2𝑛+1}+2^{3𝑛+1}$ is not divisible by 23.

Hence, $𝑅\left(𝑛\right)$ is not divisible by $23,\forall 𝑛\in 𝑁$. From option (d),

$2+7+12+\dots +(5𝑛-3)$ is an AP with common difference as 5 .

$\begin{array}{cc}{𝑆}_{𝑛}& =\frac{𝑛}{2}[2\times 2+(𝑛-1\left)5\right]\\ \\ & =\frac{𝑛}{2}(5𝑛-1)\end{array}$

Given, $5{𝐴}^{-1}=⎡⎣⎢\begin{array}{ccc}-3& 2& 2\\ \\ 2& -3& 2\\ \\ 2& 2& -3\end{array}⎤⎦⎥$

and

$𝐴=⎡⎣⎢\begin{array}{ccc}𝑥& 𝑦& 𝑦\\ \\ 𝑦& 𝑥& 𝑦\\ \\ 𝑦& 𝑦& 𝑥\end{array}⎤⎦⎥$

$\Rightarrow 5{𝐴}^{-1}𝐴=5𝐼$

$⎡⎣⎢\begin{array}{ccc}-3& 2& 2\\ \\ 2& -3& 2\\ \\ 2& 2& -3\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}𝑥& 𝑦& 𝑦\\ \\ 𝑦& 𝑥& 𝑦\\ \\ 𝑦& 𝑦& 𝑥\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}5& 0& 0\\ \\ 0& 5& 0\\ \\ 0& 0& 5\end{array}⎤⎦⎥$

$⎡⎣⎢\begin{array}{ccc}-3𝑥+4𝑦& -𝑦+2𝑥& -𝑦+2𝑥\\ \\ -𝑦+2𝑥& -3𝑥+4𝑦& -𝑦+2𝑥\\ \\ -𝑦+2𝑥& -𝑦+2𝑥& -3𝑥+4𝑦\end{array}⎤⎦⎥$

$=⎡⎣⎢\begin{array}{ccc}5& 0& 0\\ \\ 0& 5& 0\\ \\ 0& 0& 5\end{array}⎤⎦⎥$

$\begin{array}{cc}& \Rightarrow 𝑦=2𝑥\text{ and }-3𝑥+4𝑦=5\\ \\ & \Rightarrow 5𝑥=5\\ \\ & \Rightarrow 𝑥=1\end{array}$

Put $𝑥=1$, then $𝑦=2$

$𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ \\ 2& 1& 2\\ \\ 2& 2& 1\end{array}⎤⎦⎥$

$\begin{array}{cc}{𝐴}^2& =⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ \\ 2& 1& 2\\ \\ 2& 2& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ \\ 2& 1& 2\\ \\ 2& 2& 1\end{array}⎤⎦⎥\\ \\ & =⎡⎣⎢\begin{array}{ccc}9& 8& 8\\ \\ 8& 9& 8\\ \\ 8& 8& 9\end{array}⎤⎦⎥\\ \\ {𝐴}^2-4𝐴& =⎡⎣⎢\begin{array}{ccc}9& 8& 8\\ \\ 8& 9& 8\\ \\ 8& 8& 9\end{array}⎤⎦⎥-4⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ \\ 2& 1& 2\\ \\ 2& 2& 1\end{array}⎤⎦⎥\\ \\ & =⎡⎣⎢\begin{array}{ccc}5& 0& 0\\ \\ 0& 5& 0\\ \\ 0& 0& 5\end{array}⎤⎦⎥=5𝐼\end{array}$

$\begin{array}{cc}& \text{ Given, }𝐴=⎡⎣⎢\begin{array}{ccc}9& 3& 0\\ \\ 1& 5& 8\\ \\ 7& 6& 2\end{array}⎤⎦⎥\\ \\ & \text{ and }𝐴{𝐴}^{𝑇}-{𝐴}^2=⎡⎣⎢\begin{array}{ccc}{𝑎}_{11}& {𝑎}_{12}& {𝑎}_{13}\\ \\ {𝑎}_{21}& {𝑎}_{22}& {𝑎}_{23}\\ \\ {𝑎}_{31}& {𝑎}_{32}& {𝑎}_{33}\end{array}⎤⎦⎥\end{array}$

Now,

${𝐴}^2=⎡⎣⎢\begin{array}{ccc}9& 3& 0\\ \\ 1& 5& 8\\ \\ 7& 6& 2\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}9& 3& 0\\ \\ 1& 5& 8\\ \\ 7& 6& 2\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}84& 42& 24\\ \\ 70& 76& 56\\ \\ 83& 63& 52\end{array}⎤⎦⎥$

$\begin{array}{cc}& 𝐴{𝐴}^{𝑇}-{𝐴}^2=⎡⎣⎢\begin{array}{ccc}9& 3& 0\\ \\ 1& 5& 8\\ \\ 7& 6& 2\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}9& 1& 7\\ \\ 3& 5& 6\\ \\ 0& 8& 2\end{array}⎤⎦⎥\\ \\ & -⎡⎣⎢\begin{array}{ccc}84& 42& 24\\ \\ 70& 76& 56\\ \\ 83& 63& 52\end{array}⎤⎦⎥\\ \\ & =⎡⎣⎢\begin{array}{ccc}90& 24& 81\\ \\ 24& 90& 53\\ \\ 81& 53& 89\end{array}⎤⎦⎥-⎡⎣⎢\begin{array}{ccc}84& 42& 24\\ \\ 70& 76& 56\\ \\ 83& 63& 52\end{array}⎤⎦⎥\\ \\ & =⎡⎣⎢\begin{array}{ccc}6& -18& 57\\ \\ -46& 14& -3\\ \\ -2& -10& 37\end{array}⎤⎦⎥\\ \\ & \sum {𝑎}_{𝑖𝑗}=6-18+57-46+14\\ & 1\le 𝑖\le 3\\ & 1\le 𝑗\le 3\end{array}$

Given,

$\begin{array}{c}{\Delta }_1=\mid \begin{array}{ccc}1& {𝑎}^2& 𝑏𝑐\\ \\ 1& {𝑏}^2& 𝑐𝑎\\ \\ 1& {𝑐}^2& 𝑎𝑏\end{array}\mid \\ \\ {𝑅}_2\to {𝑅}_2-{𝑅}_1\text{ and }{𝑅}_3\to {𝑅}_3-{𝑅}_1\\ \\ {\Delta }_1=\mid \begin{array}{ccc}1& {𝑎}^2& 𝑏𝑐\\ \\ 0& {𝑏}^2-{𝑎}^2& 𝑐(𝑎-𝑏)\\ \\ 0& {𝑐}^2-{𝑎}^2& 𝑏(𝑎-𝑐)\end{array}\mid \\ \\ =(𝑎-𝑏)(𝑐-𝑎)\mid \begin{array}{ccc}1& {𝑎}^2& 𝑏𝑐\\ \\ 0& -(𝑎+𝑏)& 𝑐\\ \\ 0& 𝑐+𝑎& -𝑏\end{array}\mid \\ \\ =(𝑎-𝑏)(𝑐-𝑎)(𝑎𝑏+{𝑏}^2-{𝑐}^2-𝑎𝑐)\\ \\ =(𝑎-𝑏)(𝑐-𝑎)\left(𝑎\right(𝑏-𝑐)+(𝑏-𝑐\left)\right(𝑏+𝑐)\\ \\ {\Delta }_1=(𝑎-𝑏)(𝑏-𝑐)(𝑐-𝑎)(𝑎+𝑏+𝑐)\dots \\ \\ \text{ and }{\Delta }_2=\mid \begin{array}{ccc}1& 1& 1\\ \\ {𝑎}^2& {𝑏}^2& {𝑐}^2\\ \\ {𝑎}^3& {𝑏}^3& {𝑐}^3\end{array}\mid \\ \\ {𝐶}_2\to {𝐶}_2-{𝐶}_1\text{ and }{𝐶}_3\to {𝐶}_3-{𝐶}_1\\ \\ =\mid \begin{array}{ccc}1& 0& 0\\ \\ {𝑎}^2& {𝑏}^2-{𝑎}^2& {𝑐}^2-{𝑎}^2\\ \\ {𝑎}^3& {𝑏}^3-{𝑎}^3& {𝑐}^3-{𝑎}^3\end{array}\mid \\ \\ =-(𝑎-𝑏)(𝑐-𝑎)\\ \\ \mid \begin{array}{ccc}1& 0& 0\\ \\ {𝑎}^2& 𝑎+𝑏& 𝑐+𝑎\\ \\ {𝑎}^3& ({𝑎}^2+{𝑏}^2+𝑎𝑏)& {𝑐}^2+{𝑎}^2+𝑎𝑐\end{array}\mid \\ \\ {𝐶}_2\to {𝐶}_2-{𝐶}_3\\ \\ =-(𝑎-𝑏)(𝑐-𝑎)\\ \\ \mid \begin{array}{ccc}1& 0& 0\\ \\ {𝑎}^2& 𝑏-𝑐& 𝑐+𝑎\\ \\ {𝑎}^3& ({𝑏}^2-{𝑐}^2)+𝑎(𝑏-𝑐)& {𝑐}^2+{𝑎}^2+𝑎𝑐\end{array}\mid \\ \\ =-(𝑎-𝑏)(𝑏-𝑐)(𝑐-𝑎)\\ \\ \mid \begin{array}{ccc}1& 0& 0\\ \\ {𝑎}^2& 1& 𝑐+𝑎\\ \\ {𝑎}^3& 𝑏+𝑐+𝑎& {𝑐}^2+{𝑎}^2+𝑎𝑐\end{array}\mid \end{array}$

$\begin{array}{c}=-(𝑎-𝑏)(𝑏-𝑐)(𝑐-𝑎)\\ \\ [{𝑐}^2+{𝑎}^2+𝑎𝑐-𝑏𝑐-{𝑐}^2-𝑎𝑐-𝑎𝑏-𝑎𝑐-{𝑎}^2]\\ \\ =-(𝑎-𝑏)(𝑏-𝑐)(𝑐-𝑎)[-𝑏𝑐-𝑎𝑏-𝑎𝑐]\\ \\ {\Delta }_2=(𝑎-𝑏)(𝑏-𝑐)(𝑐-𝑎)[𝑎𝑏+𝑏𝑐+𝑐𝑎]\end{array}$

From Ist and 2nd equation replace ${\Delta }_1$ and ${\Delta }_2$

$\begin{array}{c}\frac{{\Delta }_1}{{\Delta }_2}=\frac{6}{11}\Rightarrow \frac{𝑎+𝑏+𝑐}{𝑎𝑏+𝑏𝑐+𝑐𝑎}=\frac{6}{11}\\ \\ \therefore 11(𝑎+𝑏+𝑐)=6(𝑎𝑏+𝑏𝑐+𝑐𝑎)\end{array}$

Given, $𝑥+3𝑦+7=0$

$\begin{array}{c}3𝑥+10𝑦-3𝑧+18=0\\ \\ 3𝑦-9𝑧+2=0\\ \\ 𝐷=\mid \begin{array}{ccc}1& 3& 0\\ \\ 3& 10& -3\\ \\ 0& 3& -9\end{array}\mid =0\\ \\ {𝐷}_{𝑥}=\mid \begin{array}{ccc}-7& 3& 0\\ \\ -18& 10& -3\\ \\ -2& 3& -9\end{array}\mid =99\\ \\ {𝐷}_{𝑦}=\mid \begin{array}{ccc}1& -7& 0\\ \\ 3& -18& -3\\ \\ 0& -2& -9\end{array}\mid =-33\\ \\ {𝐷}_{𝑧}=\mid \begin{array}{ccc}1& 3& -7\\ \\ 3& 10& -18\\ \\ 0& 3& -2\end{array}\mid =-11\end{array}$

As $𝐷=0$, and none ${𝐷}_{𝑥},{𝐷}_{𝑦}$ and ${𝐷}_{𝑧}$ is zero, the given system of equation has no solution.

Given, $\sqrt{\surd -5+12𝑖}=\sqrt{\surd \frac{13-5}{2}}+𝑖\sqrt{\surd \frac{13+5}{2}}$

$\begin{array}{cc}& =2+3𝑖\\ \\ \therefore 𝑥+𝑖𝑦& =\frac{13(2+3𝑖)}{(2-3𝑖)(3+2𝑖)}\\ \\ & =\frac{13(2+3𝑖)}{6-9𝑖+4𝑖-6{𝑖}^2}=\frac{13(2+3𝑖)}{12-5𝑖}\\ \\ & =\frac{13(2+3𝑖)(12+5𝑖)}{(12-5𝑖)(12+5𝑖)}\\ \\ & =\frac{13(24+36𝑖+10𝑖+15{𝑖}^2)}{169}\\ \\ & =\frac{13(9+46𝑖)}{169}=\frac{9}{13}+\frac{46𝑖}{13}\\ \\ \Rightarrow 𝑥& =\frac{9}{13}\text{ and }𝑦=\frac{46}{13}\\ \\ 13𝑦-26\dot{𝑥}& =46-18=28\end{array}$

Given, $𝑧=𝑥+𝑖𝑦$

$\begin{array}{c}|𝑥+𝑖𝑦-1|+|𝑥+𝑖𝑦+𝑖|=2\\ \\ \sqrt{\surd (𝑥-1{)}^2+{𝑦}^2}+\sqrt{\surd {𝑥}^2+(𝑦+1{)}^2}=2\\ \\ \sqrt{\surd (𝑥-1{)}^2+{𝑦}^2}=2-\sqrt{\surd {𝑥}^2+(𝑦+1{)}^2}\\ \\ {𝑥}^2+{𝑦}^2-2𝑥+1=4+{𝑥}^2+{𝑦}^2+2𝑦+1-4\sqrt{\surd {𝑥}^2+(𝑦+1{)}^2}\\ \\ 𝑥+𝑦+2=2\sqrt{\surd {𝑥}^2+{𝑦}^2+2𝑦+1}\end{array}$

$\begin{array}{c}{𝑥}^2+{𝑦}^2+4+2𝑥𝑦+4𝑥+4𝑦\\ \\ =4{𝑥}^2+4{𝑦}^2+8𝑦+4\\ \\ 3{𝑥}^2+3{𝑦}^2-2𝑥𝑦-4𝑥+4𝑦=0\end{array}$

To find one of the values of $(-64𝑖{)}^{5/6}$, let's proceed with the computation:

Let $𝑍=(-64𝑖{)}^{5/6}$.

We can express $𝑍$ as:

$𝑍={\left(\right(-𝑖{)}^5)}^{1/6}\times (64{)}^{5/6}$

Calculate $(64{)}^{5/6}$:

${64}^{5/6}=32$

Handle the imaginary unit:

Since $(-𝑖{)}^5=-𝑖$, we represent this in polar coordinates:

$-𝑖=\cos \left(\frac{3𝜋}{2}\right)+𝑖\sin \left(\frac{3𝜋}{2}\right)$

Find $(-𝑖{)}^{1/6}$:

Using De Moivre’s theorem, $(-𝑖{)}^{1/6}$ can be expressed as:

$(-𝑖{)}^{1/6}={[\cos \left(\frac{3𝜋}{2}\right)+𝑖\sin \left(\frac{3𝜋}{2}\right)]}^{1/6}$

$=\cos \left(\frac{3𝜋}{12}\right)+𝑖\sin \left(\frac{3𝜋}{12}\right)$

$=\frac{1}{\sqrt{\surd 2}}(1+𝑖)$

Combine the results:

Therefore, $𝑍$ becomes:

$𝑍=32\times \frac{1}{\sqrt{\surd 2}}(1+𝑖)=16\sqrt{\surd 2}(1+𝑖)$

The calculation yields one of the values of $(-64𝑖{)}^{5/6}$ as $16\sqrt{\surd 2}(1+𝑖)$.

Given, $𝑥+\frac{4}{𝑥}=2\sqrt{\surd 3}$

${𝑥}^2-2\sqrt{\surd 3}𝑥+4=0$

Roots, $𝑥=\sqrt{\surd 3}\pm 𝑖\Rightarrow 𝛼,𝛽=\sqrt{\surd 3}\pm 𝑖$

$\begin{array}{cc}& {𝛼}^{2024},{𝛽}^{2024}=(\sqrt{\surd 3}\pm 𝑖{)}^{2024}\\ \\ & =2^{2024}{[\frac{\sqrt{\surd 3}}{2}\pm \frac{1}{2}𝑖]}^{2024}\\ \\ & =2^{2024}{[\cos \frac{𝜋}{6}\pm 𝑖\sin \frac{𝜋}{6}]}^{2024}\\ \\ & =2^{2024}[\cos \frac{2024𝜋}{6}\pm 𝑖\sin \frac{2024𝜋}{6}]\\ \\ <text>\therefore </text><text>\mid </text>{𝛼}^{2024}& -{𝛽}^{2024}\mid =2^{2024}\times \mid 2𝑖\sin \left(\frac{2024𝜋}{6}\right)\mid \\ \\ & =2^{2024}\times 2\times \mid \sin (337𝜋+\frac{𝜋}{3})\mid \\ \\ & =2^{2025}\times \frac{\sqrt{\surd 3}}{2}\end{array}$

Hence, $\frac{2}{\sqrt{\surd 3}}\mid {𝛼}^{2024}-{𝛽}^{2024}\mid =2^{2025}$

Given, $12{𝑥}^{1/3}-25{𝑥}^{1/6}+12=0$

Let $𝑦={𝑥}^{1/6}$, then ${𝑦}^2={𝑥}^{1/3}$

So, $12{𝑦}^2-25𝑦+12=0$

$\begin{array}{cc}𝑦& =\frac{4}{3}\text{ or }\frac{3}{4}\\ \\ {𝑥}^{1/6}& =\frac{4}{3}\text{ or }\frac{3}{4}\\ \\ 𝑥& ={\left(\frac{4}{3}\right)}^6\text{ or }{\left(\frac{3}{4}\right)}^6\end{array}$

Here, $𝛼={\left(\frac{4}{3}\right)}^6$ and $𝛽={\left(\frac{3}{4}\right)}^6$

$\text{exception 25:}={(\frac{4}{3}\times \frac{4}{3})}^{6\times \frac{1}{6}}=\frac{16}{9}$

Given, $7+6𝑥-3{𝑥}^2$ or $-3{𝑥}^2+6𝑥+1$

As coefficient of ${𝑥}^2$ is less than 0 , then $𝑓\left(\frac{-𝑏}{2𝑎}\right)$ is minima or maxima.

$\frac{-𝑏}{2𝑎}=\frac{-6}{2\times (-3)}=1$

$𝑓\left(l\right)=-3+6+7=10$

Here, $𝛼=1,𝛽=10$

Let $𝑚$ and $𝑛$ be roots of ${𝑥}^2+𝑥-10=0$

$\begin{array}{c}𝑚+𝑛=-1\text{ and }𝑚𝑛=-10\\ \\ {𝑚}^2+{𝑛}^2=(𝑚+𝑛{)}^2-2𝑚𝑛=21\end{array}$

Given, ${𝑥}^3+3{𝑥}^2-10𝑥-24=0$

Factorise

$\Rightarrow (𝑥+4)(𝑥+2)(𝑥-3)=0$

Here, $𝛼=3,𝛽=-2𝛾=-4$

${𝛼}^3+3{𝛽}^2-10𝛾-24=11𝐾$

$27+3\times 4-10\times (-4)-24=11𝐾$

$27+12+40-24=11𝐾\Rightarrow 𝐾=\frac{55}{11}=5$

Given,

$8{𝑥}^3-42{𝑥}^2+63𝑥-27=0$

Factorise $(4𝑥-3)(2𝑥-3)(𝑥-3)=0$

Roots are $𝑥=\frac{3}{4},\frac{3}{2}$ or 3

So, $𝛽=\frac{3}{4},𝛾=\frac{3}{2}$ and $𝛼=3$

Let $𝑓\left(𝑥\right)=\frac{3}{2}{𝑥}^2+3𝑥+3$

Extreme value $=𝑓\left(\frac{-𝑏}{2𝑎}\right)$

$\frac{-𝑏}{2𝑎}=\frac{-3}{2\times \frac{3}{2}}\Rightarrow -1\Rightarrow 𝑓(-1)=\frac{3}{2}-3+3=\frac{3}{2}$

Alphabets present in the word 'COLLEGE' are C, O, L, E, G

Alphabets in ascending order are

C, E, G, L, O

Number of words starts with $\mathrm{CE}=\frac{5!}{2!}$

Number of words starts with CG $=\frac{5!}{2!2!}$

Number of words starts with $\mathrm{CL}=\frac{5!}{2!}$

Number of words starts with $\mathrm{COE}=\frac{4!}{2!}$

Number of words starts with COG $=\frac{4!}{2!2!}$

Number of words starts with COLE $=3$ !

Number of words starts with COLG $=\frac{3!}{2!}$

Number of words starts with

COLLEE $=1$ !

Next words in COLLEGE

$\therefore$ Required number of ways

$\begin{array}{c}=\frac{5!}{2!}+\frac{5!}{2!2!}+\frac{5!}{2!}+\frac{4!}{2!}+\frac{4!}{2!2!}+3!+\frac{3!}{2!}+1+1\\ \\ =60+30+60+12+6+6+3+2\\ \\ =179\end{array}$

Case I Let one digit be 1 or 7 , then sum of the other two digits be $2,5,8$, 11 or 14 or $3𝑥+2$ Therefore, the 2 digits can be $(3,5),(5,9)$.

$\therefore$ The possible number are 8 .

Case II Let one digit be 3 or 9 the sum of other two digits can be multiple of 3 . The other 2 digit can be $(1,5)$ or $(7,5)$ $\therefore$ The possible numbers are 8 .

Case III Let one digit can be 5 , then sum of other digit can be $1,4,7,10,13$ or 16.

$\therefore$ The other 2 digit numbers can be $(1,3),(1,9),(3,7),(7,6)$

$\therefore$ The possible number can be 8 in ${}^{-}$ number.

Hence, the required number are

$8+8+8=24$

We have, part A has 7 questions, Part B has 5 questions and part $𝐶$ has 3 questions.