TG EAPCET 2023 12th May Evening Shift Question Paper with Solutions

The given function is

$𝑓\left(𝑥\right)={\sin }^{-1}({\log }_2\left(\frac{{𝑥}^2}{2}\right))$

To find the domain of $𝑓\left(𝑥\right)$

Domain $\left(𝑓\right(𝑥\left)\right)=$ Domain of

${\sin }^{-1}({\log }_2\left(\frac{{𝑥}^2}{2}\right))$

where, ${\log }_2\left(\frac{{𝑥}^2}{2}\right)\in [-1,1]$

$\begin{array}{cc}& \Rightarrow -1\le {\log }_2\left(\frac{{𝑥}^2}{2}\right)\le 1\\ & \Rightarrow \frac{1}{2}\le \frac{{𝑥}^2}{2}\le 2\\ & \Rightarrow 1\le {𝑥}^2\le 4\end{array}$

For, ${𝑥}^2\le 4\Rightarrow -2\le 𝑥\le 2$

$\Rightarrow 𝑥\in [-2,2]...\left(𝑖\right)$

and also, ${𝑥}^2\ge 1$

$\Rightarrow 𝑥\in (-\infty ,-1]\cup [1,\infty )...\left(𝑖𝑖\right)$

Now, from Eqs. (i) and (ii), taking intersection, we get

$𝑥\in [-2,-1]\cup [1,2]$

Given function,

$\begin{array}{cc}𝑓\left(𝑥\right)=& -\sqrt{\surd -{𝑥}^2-6𝑥-5}\\ & =-\sqrt{\surd -({𝑥}^2+6𝑥+5)}\end{array}$

Let $𝑦=𝑓\left(𝑥\right)=-\sqrt{\surd -(𝑥+5)(𝑥+1)}$

Now, squaring on both sides, we get

$\begin{array}{cc}& {𝑦}^2=-({𝑥}^2+6𝑥+5)\\ \Rightarrow & {𝑥}^2+6𝑥+(5+{𝑦}^2)=0\end{array}$

Applying quadratic formula, we get

$\begin{array}{cc}& 𝑥=\frac{-6\pm \sqrt{\surd 36-4(5+{𝑦}^2)}}{2}\\ & =\frac{-6\pm \sqrt{\surd 16-4{𝑦}^2}}{2}\end{array}$

Clearly, $16-4{𝑦}^2\ge 0\Rightarrow {𝑦}^2\le \frac{16}{4}=4$

$\begin{array}{cc}\Rightarrow & {𝑦}^2\le 4\\ \Rightarrow & -2\le 𝑦\le 2\end{array}$

Clearly, $𝑦<0$ because $\sqrt{\surd -({𝑥}^2-6𝑥-5)}$ always positive.

So, $𝑦\in [-2,0]$

Thus, range of given function is $[-2,0]$

Given function is, $𝑓\left(𝑥\right)=2𝑥+\sin 𝑥$, $𝑥\in 𝑅$

For one one $𝑓\left({𝑥}_1\right)=𝑓\left({𝑥}_2\right)$

$\Rightarrow 2{𝑥}_1+\sin {𝑥}_1=2{𝑥}_2+\sin {𝑥}_2$

This can be possible only, when ${𝑥}_1={𝑥}_2$

So, $𝑓\left(𝑥\right)$ is one-one.

For onto Range of

$𝑓\left(𝑥\right)=𝑅=\text{ Co-domain }$

So, Range $=$ Co-domain

Thus, $𝑓\left(𝑥\right)$ is onto.

$\mid \begin{array}{ccc}\sqrt{\surd 3}& 2\sqrt{\surd 5}& \sqrt{\surd 5}\\ \sqrt{\surd 15}& 5& \sqrt{\surd 10}\\ 3& \sqrt{\surd 15}& 5\end{array}\mid$

Taking $\sqrt{\surd 3},\sqrt{\surd 5}$ and $\sqrt{\surd 5}$ from ${𝐶}_1,{𝐶}_2$ and ${𝐶}_3$, we get

$\begin{array}{cc}& \sqrt{\surd 3}\times \sqrt{\surd 5}\times \sqrt{\surd 5}\mid \begin{array}{ccc}1& 2& 1\\ \sqrt{\surd 5}& \sqrt{\surd 5}& \sqrt{\surd 2}\\ \sqrt{\surd 3}& \sqrt{\surd 3}& \sqrt{\surd 5}\end{array}\mid \\ & =5\sqrt{\surd 3}\mid \begin{array}{ccc}1& 1& 1\\ \sqrt{\surd 5}& 0& \sqrt{\surd 2}\\ \sqrt{\surd 3}& 0& \sqrt{\surd 5}\end{array}\mid \end{array}$

$\begin{array}{cc}& \text{ (Applying }{𝐶}_2\to {𝐶}_2-{𝐶}_1,\\ & -5\sqrt{\surd 3}(5-\sqrt{\surd 6})\\ & =-25\sqrt{\surd 3}+15\sqrt{\surd 2}=15\sqrt{\surd 2}-25\sqrt{\surd 3}\end{array}$

Given that $𝐴$ is a non-singular matrix, and it satisfies the equation:

$(𝐴-2𝐼)(𝐴-3𝐼)=0$

Since $𝐴$ is non-singular, it implies that $\left|𝐴\right|\ne 0$, which means ${𝐴}^{-1}$ exists.

Expanding the given equation:

${𝐴}^2-2𝐴-3𝐴+6𝐼=0$

Simplifies to:

${𝐴}^2-5𝐴+6𝐼=0$

Multiplying both sides of this equation by ${𝐴}^{-1}$ yields:

$𝐴-5𝐼+6{𝐴}^{-1}=0$

Dividing through by 5 gives:

$\frac{𝐴}{5}+\frac{6}{5}{𝐴}^{-1}=𝐼$

Given that $𝐴𝐵=$ Scalar matrix

$=\left[\begin{array}{cc}𝜆& 0\\ 0& 𝜆\end{array}\right]$

where $𝜆=$ scalar

Also, given that $𝐵=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ and

$\begin{array}{cc}& \det \left(3𝐴\right)=27\\ & \Rightarrow \det 𝐴=\frac{27}{9}=3\end{array}$

Now, let $𝐴=\left[\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right]$

$\therefore 𝐴𝐵=\left[\begin{array}{cc}𝜆& 0\\ 0& 𝜆\end{array}\right]$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow \left[\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}𝜆& 0\\ 0& 𝜆\end{array}\right]\\ & \Rightarrow \left[\begin{array}{cc}𝑎& 2𝑎+3𝑏\\ 𝑐& 2𝑐+3𝑑\end{array}\right]=\left[\begin{array}{cc}𝜆& 0\\ 0& 𝜆\end{array}\right]\end{array}\\ & \text{ Comparing on both sides, we get }\\ & \begin{array}{cc}& 𝑎=𝜆,𝑐=0,2𝑎+3𝑏=0,2𝑐+3𝑑=𝜆\\ & \therefore 𝑏=-\frac{2𝜆}{3}\text{ and }𝑑=\frac{𝜆}{3}\\ & \text{ Thus, }𝐴=\left[\begin{array}{cc}𝜆& -\frac{2𝜆}{3}\\ 0& \frac{{𝜆}^3}{3}\end{array}\right]\\ & \left|𝐴\right|=\mid \begin{array}{cc}𝜆& -2𝜆\\ 0& \frac{𝜆}{3}\end{array}\mid =\frac{{𝜆}^2}{3}=3\text{ (given) }\\ & \Rightarrow {𝜆}^2=9\Rightarrow 𝜆=\pm 3\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \therefore {𝐴}^2=\left[\begin{array}{cc}𝜆& -\frac{2𝜆}{3}\\ 0& \frac{{𝜆}^2}{3}\end{array}\right]\left[\begin{array}{cc}𝜆& -\frac{2𝜆}{3}\\ 0& \frac{{𝜆}^2}{3}\end{array}\right]\\ & =⎡⎣\begin{array}{cc}{𝜆}^2& -\frac{2{𝜆}^2}{3}-\frac{2{𝜆}^2}{9}\\ 0& \frac{{𝜆}^2}{9}\end{array}⎤⎦=⎡⎣\begin{array}{cc}{𝜆}^2& -\frac{8{𝜆}^2}{9}\\ 0& \frac{{𝜆}^2}{9}\end{array}⎤⎦\\ & {𝐴}^2=\left[\begin{array}{cc}9& -8\\ 0& 1\end{array}\right]\\ & \text{ Clearly, }{𝐴}^{-1}=\frac{1}{3}\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\\ & \Rightarrow 3{𝐴}^{-1}=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]\end{array}\\ & \text{ Thus, }\\ & 3{𝐴}^{-1}+{𝐴}^2=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]+\left[\begin{array}{cc}9& -8\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}10& -6\\ 0& 4\end{array}\right]\end{array}$

If $𝐴$ is a symmetric matrix with real entries, it means ${𝐴}^{𝑇}=𝐴$.

Now, suppose the inverse ${𝐴}^{-1}$ exists.

For symmetric matrices, the transpose of the inverse is equal to the inverse of the transpose, so:

$({𝐴}^{-1}{)}^{𝑇}=({𝐴}^{𝑇}{)}^{-1}={𝐴}^{-1}$

This equation shows that ${𝐴}^{-1}$ is also symmetric. Therefore, if the inverse of a symmetric matrix exists, it is necessarily symmetric as well.

Given that

$(𝑧=\sin \left(\frac{6𝜋}{5}\right)+𝑖(1+\cos \frac{6𝜋}{5}))$

Thus, clearly $(\sin \frac{6𝜋}{5},1+(\cos \frac{6𝜋}{5}))$ lies in the 2 nd quadrant of complex plane. Hence, its argument is

$\begin{array}{cc}& \arg \left(𝑧\right)=𝜋-{\tan }^{-1}\mid \left(\frac{𝑦}{𝑧}\right)\mid \\ & (\because 𝑧=𝑥+𝑖𝑦,\forall 𝑥<0\text{ and }{𝑦}^{}⩾0)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& =𝜋-{\tan }^{-1}\mid \frac{1+\cos \frac{6𝜋}{5}}{\sin \frac{6𝜋}{5}}\mid \\ & =𝜋-{\tan }^{-1}\mid \frac{1-\cos \frac{𝜋}{5}}{\sin \frac{𝜋}{5}}\mid \\ & =𝜋-{\tan }^{-1}\mid \frac{2{\sin }^2\frac{𝜋}{10}}{2\sin \frac{𝜋}{10}\cos \frac{𝜋}{10}}\mid \\ & =𝜋-{\tan }^{-1}\mid \frac{\sin \frac{𝜋}{10}}{\cos \frac{𝜋}{10}}\mid \\ & =𝜋-{\tan }^{-1}(\tan \frac{𝜋}{10})\\ & =𝜋-\frac{𝜋}{10}(\because -\frac{𝜋}{2}\le {\tan }^{-1}𝑥\le \frac{1}{2})\\ & =\frac{10𝜋-𝜋}{10}=\frac{9𝜋}{10}\end{array}\\ & \text{ Thus, }\arg \left(𝑧\right)=\frac{9𝜋}{10}\end{array}$

Given,

$\begin{array}{cc}& 𝑥+𝑖𝑦=\sqrt{\surd \frac{3+𝑖}{1+3𝑖}}...\left(𝑖\right)\\ \Rightarrow & \overline{𝑥+𝑖𝑦}=\sqrt{\surd \frac{\overline{3+𝑖}}{1+3𝑖}}\\ \Rightarrow & 𝑥-𝑖𝑦=\sqrt{\surd \frac{3-𝑖}{1-3𝑖}}...\left(𝑖𝑖\right)\end{array}$

On multiplying Eqs. (i) and (ii), we get

$\begin{array}{cc}& {𝑥}^2+{𝑦}^2=\sqrt{\surd \frac{(3+𝑖)(3-𝑖)}{(1+3𝑖)(1-3𝑖)}}=\sqrt{\surd \frac{9+1}{1+9}}=1\\ \Rightarrow & {({𝑥}^2+{𝑦}^2)}^2=(1{)}^2=1\end{array}$

Now, Img $\left(\frac{2𝑧+1}{𝑖𝑧+1}\right)=-2$

Now, let $𝑧=𝑥+𝑖𝑦$

$\begin{array}{cc}& \therefore \frac{2𝑧+1}{𝑖𝑧+1}=\frac{2(𝑥+𝑖𝑦)+1}{𝑖(𝑥+𝑖𝑦)+1}=\frac{(2𝑥+1)+𝑖2𝑦}{(1-𝑦)+𝑖𝑥}\\ & =\frac{\left\{\right(2𝑥+1)+𝑖2𝑦\}\left\{\right(1-𝑦)-𝑖𝑥\}}{(1-𝑦{)}^2+{𝑥}^2}\end{array}$

$=\frac{\begin{array}{c}(2𝑥+1)(1-𝑦)+𝑖(2𝑦-2{𝑦}^2)-𝑖(2{𝑥}^2+𝑥)+2𝑥𝑦\end{array}}{(1-𝑦{)}^2+{𝑥}^2}$

From Eq. (i), we get

$\begin{array}{cc}& \frac{2𝑦-2{𝑦}^2-2{𝑥}^2-𝑥}{1+{𝑦}^2-2𝑦+{𝑥}^2}=-2\\ & \Rightarrow 2𝑦-2{𝑦}^2-2{𝑥}^2-𝑥=-2-2{𝑦}^2+4𝑦-2{𝑥}^2\\ & \Rightarrow 2𝑦-𝑥+2-4𝑦=0\\ & \Rightarrow 𝑥+2𝑦-2=0\end{array}$

which is a straight line.

$\begin{array}{cc}& \mid \begin{array}{ccc}𝜔+{𝜔}^2& {𝜔}^2+{𝜔}^9& {𝜔}^9+𝜔\\ {𝜔}^{27}+{𝜔}^{31}& {𝜔}^{31}+{𝜔}^{17}& {𝜔}^{17}+{𝜔}^{27}\\ {𝜔}^{30}+{𝜔}^{41}& {𝜔}^{41}+{𝜔}^{19}& {𝜔}^{19}+{𝜔}^{30}\end{array}\mid \\ & \begin{array}{cc}=& \mid \begin{array}{ccc}𝜔& {𝜔}^2& {𝜔}^9\\ {𝜔}^{27}& {𝜔}^{31}& {𝜔}^{17}\\ {𝜔}^{30}& {𝜔}^{41}& {𝜔}^{19}\end{array}\mid +\mid \begin{array}{ccc}{𝜔}^2& {𝜔}^9& 𝜔\\ {𝜔}^{31}& {𝜔}^{17}& {𝜔}^{27}\\ {𝜔}^{41}& {𝜔}^{19}& {𝜔}^{30}\end{array}\mid \\ =& 𝜔\times {𝜔}^{17}\times {𝜔}^{19}\mid \begin{array}{ccc}1& 𝜔& {𝜔}^8\\ {𝜔}^{10}& {𝜔}^{14}& 1\\ {𝜔}^{11}& {𝜔}^{22}& 1\end{array}\mid +𝜔\times {𝜔}^{17}\times {𝜔}^{19}\mid \begin{array}{ccc}𝜔& {𝜔}^8& 1\\ {𝜔}^{14}& 1& {𝜔}^{10}\\ {𝜔}^{22}& 1& {𝜔}^{11}\end{array}\mid \\ =& {𝜔}^{37}\mid \begin{array}{ccc}1& 𝜔& {𝜔}^2\\ 𝜔& {𝜔}^2& 1\\ {𝜔}^2& 𝜔& 1\end{array}\mid +{𝜔}^{37}\mid \begin{array}{ccc}𝜔& {𝜔}^2& 1\\ {𝜔}^2& 1& 𝜔\\ 𝜔& 1& {𝜔}^2\end{array}\mid \\ =& {𝜔}^{37}\times 0+{𝜔}^{37}\times 0\\ =& 0+0=0\end{array}\end{array}$

If $𝑖=\sqrt{\surd -1}$, then consider evaluating $(1+𝑖{)}^{10}+(1-𝑖{)}^{10}$.

Solution

We know that ${𝑖}^2=-1$.

First, let's compute $(1+𝑖{)}^2$ and $(1-𝑖{)}^2$:

Calculate $(1+𝑖{)}^2$:

$(1+𝑖{)}^2=(1^2+2\cdot 1\cdot 𝑖+{𝑖}^2)=1+2𝑖+{𝑖}^2=1+2𝑖-1=2𝑖$

Calculate $(1-𝑖{)}^2$:

$(1-𝑖{)}^2=(1^2-2\cdot 1\cdot 𝑖+{𝑖}^2)=1-2𝑖+{𝑖}^2=1-2𝑖-1=-2𝑖$

Now, raise each result to the 5th power:

$\left\{\right(1+𝑖{)}^2{\}}^5=(2𝑖{)}^5$

$(2𝑖{)}^5=2^5\cdot {𝑖}^5=32\cdot {𝑖}^5$

Since ${𝑖}^5={𝑖}^4\cdot 𝑖=1\cdot 𝑖=𝑖$, it follows that:

$(2𝑖{)}^5=32𝑖$

$\left\{\right(1-𝑖{)}^2{\}}^5=(-2𝑖{)}^5$

$(-2𝑖{)}^5=(-2{)}^5\cdot {𝑖}^5=-32\cdot {𝑖}^5$

Similarly, ${𝑖}^5=𝑖$, thus:

$(-2𝑖{)}^5=-32𝑖$

Finally, sum these results:

$(1+𝑖{)}^{10}+(1-𝑖{)}^{10}=32𝑖-32𝑖=0$

Thus, the final answer is $0$.

Given that

$\begin{array}{cc}& {𝑥}^2-1\le 0...\left(𝑖\right)\\ & \Rightarrow {𝑥}^2\le 1\\ & \Rightarrow -1\le 𝑥\le 1\\ & \text{ and also }{𝑥}^2-𝑥-2\ge 0...\left(𝑖𝑖\right)\\ & \Rightarrow (𝑥-2)(𝑥+1)\ge 0\\ & \Rightarrow 𝑥\in (-\infty ,-1]\cup [2,\infty )\end{array}$

Thus, the set of all value of $𝑥$ which satisfy both the inequalities ${𝑥}^2-1\le 0$ and ${𝑥}^2-𝑥-2\ge 0$ simultaneously is $\{-1\}$.

Given the function $𝑓\left(𝑥\right)=\frac{1-𝑥+{𝑥}^2}{1+𝑥+{𝑥}^2}$, we want to find its minimum value for all real values of $𝑥$.

First, differentiate $𝑓\left(𝑥\right)$ with respect to $𝑥$:

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

Simplify the expression:

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

To find the critical points, set ${𝑓}^{\prime }\left(𝑥\right)=0$:

$2{𝑥}^2-2=0{𝑥}^2=1𝑥=\pm 1$

Evaluate $𝑓\left(𝑥\right)$ at $𝑥=1$ and $𝑥=-1$:

At $𝑥=1$:

$𝑓\left(1\right)=\frac{1-1+1^2}{1+1+1^2}=\frac{1}{3}$

At $𝑥=-1$:

$𝑓(-1)=\frac{1-(-1)+1^2}{1-1+1^2}=3$

Hence, the minimum value of $𝑓\left(𝑥\right)$ is $\frac{1}{3}$.

The given equations are

${𝑥}^2-6𝑥+𝑎=0...\left(𝑖\right)$

$𝑎𝑛𝑑{𝑥}^2-𝑐𝑥+6=0...\left(𝑖𝑖\right)$

Since, $𝛼$ is a common roots in both the equations

Thus, the other roots of first and second equations are $4𝛽$ and $3𝛽$.

Hence, $(𝛼,4𝛽)$ and $(𝛼,3𝛽)$ are the roots of first and second equations.

Thus, $4𝛼𝛽=𝑎$ and $3𝛼𝛽=6$

$𝛼𝛽=\frac{𝑎}{4}\Rightarrow 𝑎=8$

So, from Eq. (i), we get

${𝑥}^2-6𝑥+8=0\Rightarrow 𝑥=2\text{ and }4$

If $𝑥=2$ is a common roots, then

From Eq. (ii), we get $4-2𝑐+6=0$

$\begin{array}{cc}& \Rightarrow 𝑐=5\\ & \therefore {𝑥}^2-5𝑥+6=0\Rightarrow 𝑥=3,2\end{array}$

Thus, their common root is 2 .

To find ${𝛼}^{15}+{𝛽}^{15}$ for the roots $𝛼$ and $𝛽$ of the equation ${𝑥}^2+2𝑥+2=0$, let's solve the quadratic equation:

${𝑥}^2+2𝑥+2=0$

First, we complete the square:

$(𝑥+1{)}^2=-1$

This implies:

$𝑥+1=\pm 𝑖$

Thus, the roots are:

$𝑥=𝑖-1\text{or}𝑥=-𝑖-1$

Assign $𝛼=𝑖-1$ and $𝛽=-𝑖-1$.

Next, calculate ${𝛼}^{15}+{𝛽}^{15}$:

${𝛼}^2={𝑖}^2-2𝑖+1=-1-2𝑖+1=-2𝑖$

${𝛽}^2=(-𝑖{)}^2-2(-𝑖)+1=-1+2𝑖+1=2𝑖$

Now express ${𝛼}^{15}+{𝛽}^{15}$ as:

${𝛼}^{15}+{𝛽}^{15}=({𝛼}^2{)}^7\cdot 𝛼+({𝛽}^2{)}^7\cdot 𝛽$

Substitute ${𝛼}^2$ and ${𝛽}^2$:

$=(-2𝑖{)}^7\cdot (𝑖-1)+(2𝑖{)}^7\cdot (-𝑖-1)$

Simplify:

Calculate $(-2𝑖{)}^7$:

$(-2𝑖{)}^7=(-2{)}^7\cdot {𝑖}^7=-128\cdot (-𝑖)=128𝑖$

Calculate $(2𝑖{)}^7$:

$(2𝑖{)}^7=2^7\cdot {𝑖}^7=128\cdot (-𝑖)=-128𝑖$

Substitute back:

$=128𝑖(𝑖-1)-128𝑖(-𝑖-1)$

Simplify the expression:

$=128𝑖\cdot 𝑖-128𝑖\cdot 1-128𝑖\cdot (-𝑖)-128𝑖\cdot (-1)$

$=128(-1)-128𝑖+128-128𝑖$

Combine terms:

$=128(-1+1)+128𝑖-128𝑖$

$=256(-1)=-256$

Thus, ${𝛼}^{15}+{𝛽}^{15}=-256$.

Consider the equation:

${𝑥}^4-2𝑎{𝑥}^3+4𝑏{𝑥}^2+8𝑎𝑥+16=0\left(𝑖\right)$

Let the roots of equation (i) be $𝛼$, $𝛽$, $𝛾$, and $𝛿$.

The product of the roots for polynomial equation (i) is given by:

$𝛼𝛽𝛾𝛿=\frac{\text{Constant term}}{\text{Coefficient of }{𝑥}^4}=\frac{16}{1}=16\left(𝑖𝑖\right)$

Suppose we have a new equation whose roots are $𝑃$ times the roots of equation (i), thus the roots become $𝑃𝛼$, $𝑃𝛽$, $𝑃𝛾$, and $𝑃𝛿$.

Now, the product of these new roots is:

$𝑃𝛼\cdot 𝑃𝛽\cdot 𝑃𝛾\cdot 𝑃𝛿={𝑃}^4\times 𝛼𝛽𝛾𝛿=1$

So,

${𝑃}^4=\frac{1}{𝛼𝛽𝛾𝛿}$

Substituting the value from equation (ii), we get:

${𝑃}^4=\frac{1}{16}$

Thus,

$\left|𝑃\right|=\frac{1}{(16{)}^{\frac{1}{4}}}=\frac{1}{2}$

Therefore, the absolute value of $𝑃$ is $\frac{1}{2}$.

Let three digits are $𝑎,𝑏$ and $𝑐$ such that

$𝑎+𝑏+𝑐=10...\left(𝑖\right)$

where, $𝑎\ge 1,𝑏\ge 0,𝑐\ge 0$

Let $𝑑=𝑎-1\Rightarrow 𝑎=𝑑+1$, where $𝑑\ge 0$

$\Rightarrow 𝑑+𝑏+𝑐=9$, where $0\le 𝑑\le 8$,

$0\le 𝑏\le 9,0\le 𝑐\le 9$

Thus, total number of 3 -digits number is

$\begin{array}{cc}{}^{9+3-1}{𝐶}_{3-1}-1& ={}^{11}{𝐶}_2-1\\ & =\frac{11\times 10\times 9!}{9!\times 2}-1\\ & =55-1=54\end{array}$

The number of words start with E , $H,M,O$ and $R=5\times 5!=5\times 120=600$

The number of words start with TE $=4!=24$

The number of words start with

THE, THM and THO $=3\times 3!=3\times 6=18$

The number of words start with

THRE, THRM $=2\times 2!=2\times 2=4$

Next word will be "THROEM".

Hence, the position of word "THROEM" is $600+24+18+4+1=647$

A student is allowed to select at most $𝑛$ books from a collection of $(2𝑛+1)$ books.

The number of ways, he select at least one book is 255.

This selection is given by combination concepts, we get

$\begin{array}{cc}& {}^{2𝑛+1}{𝐶}_1+{}^{2𝑛+1}{𝐶}_2+{}^{2𝑛+1}{𝐶}_3...\left(𝑖\right)\\ & +\dots +{}^{2𝑛+1}{𝐶}_{𝑛}=255\end{array}$

We know that, ${}^{𝑛}{𝐶}_{𝑟}={}^{𝑛}{𝐶}_{𝑛-𝑟}$ and

$\begin{array}{cc}& {}^{2𝑛+1}{𝐶}_0+{}^{2𝑛+1}{𝐶}_1+{}^{2𝑛+1}{𝐶}_2+\dots +\\ & {}^{2𝑛+1}{𝐶}_{𝑛}=\frac{2^{2𝑛+1}}{2}\end{array}$

Thus, $2({}^{2𝑛+1}{𝐶}_0+{}^{2𝑛+1}{𝐶}_1+\dots +{}^{2𝑛+1}{𝐶}_{𝑛})$

$\begin{array}{cc}& \Rightarrow ({}^{2𝑛+1}{𝐶}_0+{}^{2𝑛+1}{𝐶}_1+\dots +{}^{2𝑛+1}{𝐶}_{𝑛})\\ & =\frac{2^{2𝑛}\cdot 2}{2}=2^{2𝑛}\end{array}$

$\begin{array}{cc}\Rightarrow & (1+{}^{2𝑛+1}{𝐶}_1+{}^{2𝑛+1}{𝐶}_2+\dots +{}^{2𝑛+1}{𝐶}_{𝑛})\\ & =2^{2𝑛}\\ & [\because {}^{2𝑛+1}{𝐶}_0=1]\end{array}$

$\begin{array}{cc}\Rightarrow & {}^{2𝑛+1}{𝐶}_1+{}^{2𝑛+1}{𝐶}_2+\dots +{}^{2𝑛+1}{𝐶}_{𝑛}\\ & =2^{2𝑛}-1...\left(𝑖𝑖\right)\end{array}$

Now, comparing the Eqs. (i) and (ii), we get

$\begin{array}{cc}& \begin{array}{cc}{2}^{2𝜋}-1& =255\\ \Rightarrow 2^{2𝜋}& =256=2^8\Rightarrow 2𝑛=8\end{array}\\ & (\because {𝑎}^{𝑚}={𝑎}^{𝑛}\Rightarrow 𝑎𝑚=𝑛,\text{ where }𝑎>0)\\ & \Rightarrow 𝑛=4\\ & \text{ Then, the value of }𝑛\text{ is }4\text{. }\end{array}$

To determine the number of integral terms in the expansion of $(\sqrt{\surd 3}+\text{exception 25:}{)}^{256}$, consider the general term in the expansion:

${𝑇}_{𝑟+1}=(\frac{256}{𝑟})\left(\sqrt{\surd 3}{)}^{256-𝑟}\right(\text{exception 25:}{)}^{𝑟}$

This expression simplifies to:

${𝑇}_{𝑟+1}=(\frac{256}{𝑟})\left(3{)}^{\frac{256-𝑟}{2}}\right(5{)}^{\frac{𝑟}{8}}$