AP EAPCET 2022 4th July Evening Shift Question Paper with Solutions

$\begin{array}{cc}& \{𝑥\in 𝑅/\frac{\sqrt{\surd |𝑥{|}^2-2|𝑥|-8}}{\log (2-𝑥-{𝑥}^2)}\text{ is a real number }\}\\ & \{𝑥\in 𝑅/\frac{\sqrt{\surd {𝑥}^2-2𝑥-8}}{\log (2+𝑥)(1-𝑥)}\text{ is a real number }\}\\ & =\{𝑥\in 𝑅/\frac{\sqrt{\surd (𝑥-4)(𝑥+2)}}{\log \left\{\right(2+𝑥\left)\right(1-𝑥\left)\right\}}\text{ is a real number }\}\\ & =𝜙\end{array}$

Given, function $𝑓\left(𝑥\right)=\sin (\log \left(\frac{\sqrt{\surd 4-{𝑥}^2}}{1-𝑥}\right))$ is defined for $𝑥\ne 1$ and $𝑥>-2$ and also $𝑥<1$. Thus, the domain of above function is $𝐷\left(𝑓\right)=\{𝑥\in 𝑅;𝑥\in (-2,1\left)\right\}$

Given, $𝐴=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$

$\begin{array}{cc}{\left({𝐴}^{𝑇}\right)}^2& +(12𝐴{)}^{𝑇}\\ & ={\left[\begin{array}{cc}2& -4\\ -3& 1\end{array}\right]}^2+{\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]}^{𝑇}\\ & =\left[\begin{array}{cc}16& -12\\ -9& 13\end{array}\right]+\left[\begin{array}{cc}24& -48\\ -36& 12\end{array}\right]\\ & =\left[\begin{array}{cc}16+24& -12-48\\ -9-36& 13+12\end{array}\right]=\left[\begin{array}{cc}40& -60\\ -45& 25\end{array}\right]\end{array}$

Given, $𝑎,𝑏$ and $𝑐$ are respectively 5th, 8th, and 13th terms of arithmetic progression.

$\begin{array}{cc}& \therefore 𝑎=𝑃+(5-1)𝑑\\ & \Rightarrow 𝑎=𝑃+4𝑑\\ & 𝑏=𝑃+(8-1)𝑑\\ & \Rightarrow 𝑏=𝑃+7𝑑\\ & 𝑐=𝑃+(13-1)𝑑\\ & \Rightarrow 𝑐=𝑃+12𝑑\\ & \therefore \mid \begin{array}{ccc}𝑎& 5& 1\\ 𝑏& 8& 1\\ 𝑐& 13& 1\end{array}\mid =\mid \begin{array}{ccc}𝑃+4𝑑& 5& 1\\ 𝑃+7𝑑& 8& 1\\ 𝑃+12𝑑& 13& 1\end{array}\mid \\ & =\mid \begin{array}{ccc}𝑃& 5& 1\\ 𝑃& 8& 1\\ 𝑃& 13& 1\end{array}\mid +\mid \begin{array}{ccc}4𝑑& 5& 1\\ 7𝑑& 8& 1\\ 12𝑑& 13& 1\end{array}\mid \end{array}$

$\begin{array}{cc}& =𝑃\mid \begin{array}{ccc}1& 5& 1\\ 1& 8& 1\\ 1& 13& 1\end{array}\mid +𝑑\mid \begin{array}{ccc}4+1& 5& 1\\ 7+1& 8& 1\\ 12+1& 13& 1\end{array}\mid \\ & =𝑃\cdot 0+𝑑\mid \begin{array}{ccc}5& 5& 1\\ 8& 8& 1\\ 13& 13& 1\end{array}\mid =𝑃\cdot 0+𝑑\cdot 0=0\end{array}$

Given that,

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 𝑎& -1& 0\\ 𝑏& 𝑐& 1\end{array}⎤⎦⎥\\ {𝐴}^2& =⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 𝑎& -1& 0\\ 𝑏& 𝑐& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 𝑎& -1& 0\\ 𝑏& 𝑐& 1\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 𝑏+𝑐𝑎+𝑏& 0& 1\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& \text{ Since, }{𝐴}^2=𝐼\text{ (given) }\\ & \Rightarrow 𝑏+𝑐𝑎+𝑏=0\\ & \Rightarrow 𝑏=\frac{-𝑐𝑎}{2}\end{array}$

Given,

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

Given, $\det 𝐴=\mid \begin{array}{ccc}-2& 𝑥& 1\\ 𝑥& 1& 1\\ 2& 3& -1\end{array}\mid =0$

$\begin{array}{cc}\Rightarrow & -2(-1-3)+𝑥(2+𝑥)+1(3𝑥-2)=0\\ \Rightarrow & 8+2𝑥+{𝑥}^2+3𝑥-2=0\\ \Rightarrow & {𝑥}^2+5𝑥+6=0\Rightarrow (𝑥+2)(𝑥+3)=0\\ \Rightarrow & 𝑥=-2\text{ and }-3\\ \therefore & 𝑙=-2\text{ and }𝑚=-3\\ & {𝑙}^3-{𝑚}^3=(-2{)}^3-(-3{)}^3=-8+27=19\end{array}$

Multiplicative inverse of complex number $(\sin 𝜃,\cos 𝜃)$ is

$\begin{array}{cc}𝑧& =\sin 𝜃+𝑖\cos 𝜃\\ \Rightarrow \frac{1}{𝑧}& =\frac{(\sin 𝜃-𝑖\cos 𝜃)}{(\sin 𝜃+𝑖\cos 𝜃)(\sin 𝜃-𝑖\cos 𝜃)}\\ & =\frac{(\sin 𝜃-𝑖\cos 𝜃)}{{\sin }^2𝜃+{\cos }^2𝜃}=(\sin 𝜃-𝑖\cos 𝜃)\end{array}$

Thus, multiplicative inverse of $(\sin 𝜃,\cos 𝜃)$ is $(\sin 𝜃,-\cos 𝜃)$.

$\begin{array}{cc}& \sum _{𝑘=0}^{440}{𝑖}^{𝑘}=𝑥+𝑖𝑦\\ & \Rightarrow 𝑥+𝑖𝑦=1+𝑖+{𝑖}^2+{𝑖}^3+{𝑖}^4+\dots .+{𝑖}^{440}=-1\\ & \therefore 𝑥=-1,𝑦=0\\ & \text{ Then, }{𝑥}^{100}+{𝑥}^{99}\cdot 𝑦+{𝑥}^{242}\cdot {𝑦}^2+{𝑥}^{97}\cdot {𝑦}^3\\ & \Rightarrow (-1{)}^{100}+0=1\end{array}$

$\begin{array}{cc}& \sin 5𝜃=\sin (3𝜃+2𝜃)\\ & =\sin 3𝜃\cos 2𝜃+\cos 3𝜃\cdot \sin 2𝜃\\ & =(3\sin 𝜃-4{\sin }^3𝜃)\cdot (1-2{\sin }^2𝜃)\\ & +(4{\cos }^3𝜃-3\cos 𝜃)\cdot (2\sin 𝜃\cos 𝜃)\\ & =3\sin 𝜃-6{\sin }^3𝜃-4{\sin }^3𝜃+8{\sin }^5𝜃+8\sin 𝜃{\cos }^4𝜃\\ & -6\sin 𝜃{\cos }^2𝜃\\ & =3\sin 𝜃-10{\sin }^3𝜃+8{\sin }^5𝜃+8\sin 𝜃{\cos }^4𝜃\\ & -6\sin 𝜃{\cos }^2𝜃\end{array}$

$\begin{array}{cc}& =3\sin 𝜃-10{\sin }^3𝜃+8{\sin }^5𝜃+8\sin 𝜃{(1-{\sin }^2𝜃)}^2\\ & -6\sin 𝜃(1-{\sin }^2𝜃)\\ & =3\sin 𝜃-10{\sin }^3𝜃+8{\sin }^5𝜃+8\sin 𝜃(1+{\sin }^4𝜃-2{\sin }^2𝜃)\\ & -6\sin 𝜃+6{\sin }^3𝜃\\ & =-3\sin 𝜃-4{\sin }^3𝜃+8{\sin }^5𝜃+8\sin 𝜃\\ & +8{\sin }^5𝜃-16{\sin }^3𝜃\\ & =5\sin 𝜃-20{\sin }^3𝜃+16{\sin }^5𝜃\\ & =5\sin 𝜃-20\sin 𝜃(1-{\cos }^2𝜃)+16{(1-{\cos }^2𝜃)}^2\sin 𝜃\\ & =5\sin 𝜃-20\sin 𝜃+20\sin 𝜃{\cos }^2𝜃+16\sin 𝜃\\ & (1+{\cos }^4𝜃-2{\cos }^2𝜃)\\ & =-15\sin 𝜃+20\sin 𝜃{\cos }^2𝜃+16\sin 𝜃+16{\cos }^4𝜃\sin 𝜃\\ & -32{\cos }^2𝜃\sin 𝜃\\ & =16{\cos }^4𝜃\sin 𝜃-12{\cos }^2𝜃\sin 𝜃+\sin 𝜃\\ & \end{array}$

$\begin{array}{cc}& {𝑒}^{𝑖𝜃}=\mathrm{cis}𝜃=\cos 𝜃+𝑖\sin 𝜃\\ & \sum _{𝑛=0}^{\infty }\frac{\cos \left(𝑛𝜃\right)}{2^{𝑛}}\\ & =\mathrm{Re}\left[\sum _{𝑛=0}^{\infty }\frac{\cos 𝑛𝜃}{2^{𝑛}}\right]\\ & =\mathrm{Re}\left[\sum _{𝑛=0}^{\infty }{\left(\frac{\cos 𝜃}{2}\right)}^{𝑛}\right]\text{ [Using De-moivre theorem] }\\ & =\mathrm{Re}\left[\sum _{𝑛=0}^{\infty }{\left(\frac{{𝑒}^{𝑖𝜃}}{2}\right)}^{𝑛}\right]\\ & =\mathrm{Re}[1+\frac{{𝑒}^{𝑖𝜃}}{2}+{\left(\frac{{𝑒}^{𝑖𝜃}}{2}\right)}^2+\dots .\infty ]\\ & 1+\frac{{𝑒}^{𝑖𝜃}}{2}+{\left(\frac{{𝑒}^{𝑖𝜃}}{2}\right)}^2+\dots ..\text{ is an G P }\\ & =\mathrm{Re}\left[\frac{1}{1-\frac{{𝑒}^{𝑖𝜃}}{2}}\right]=\mathrm{Re}\left[\frac{2}{2-{𝑒}^{𝑖𝜃}}\right]\end{array}$

$\begin{array}{cc}& =\mathrm{Re}\left[\frac{2}{2-\cos 𝜃-𝑖\sin 𝜃}\right]\\ & =\mathrm{Re}\left[\frac{2(2-\cos 𝜃-𝑖\sin 𝜃)}{(2-\cos 𝜃{)}^2-{𝑖}^2{\sin }^2𝜃}\right]\\ & =\mathrm{Re}[\frac{4-2\cos 𝜃}{4+{\cos }^2𝜃-4\cos 𝜃+{\sin }^2𝜃}\\ & +\frac{𝑖\sin 𝜃}{4+{\cos }^2𝜃-4\cos 𝜃+{\sin }^2𝜃}]\\ & =\mathrm{Re}[\frac{4-2\cos 𝜃}{5-4\cos 𝜃}+\frac{𝑖\sin 𝜃}{5-4\cos 𝜃}]\\ & =\frac{4-2\cos 𝜃}{5-4\cos 𝜃}\end{array}$

Given, $𝑓\left(𝑥\right)={𝑥}^2+𝑎𝑥+𝑏(𝑏\ne 0)$

$\Rightarrow 𝑓\left(𝑓\right(𝑥\left)\right)={({𝑥}^2+𝑎𝑥+𝑏)}^2+𝑎({𝑥}^2+𝑎𝑥+𝑏)+𝑏$

Since, $𝑓\left(𝑓\right(0\left)\right)=0$

Now, ${𝑏}^2+𝑎𝑏+𝑏=0$ ..... (i)

Take $𝑎=0$ and $𝑏=-1$, then

Eq. (i) is satisfied.

Hence, $𝑎+𝑏=0+(-1)=-1$

$\begin{array}{cc}& \text{Let }𝑓\left(𝑥\right)=|𝑥-2|\\ & \{\begin{array}{cc}𝑥-2,& 𝑥>2\\ -(𝑥-2),& 𝑥<2\end{array}\end{array}$

Then, for $𝑥>2$ the equation becomes

$\begin{array}{cccc}\Rightarrow & & (𝑥-2{)}^2+(𝑥-2)-2& =0\\ \Rightarrow & & {𝑥}^2-3𝑥& =0\\ \Rightarrow & & 𝑥& =0,3\end{array}$

Thus, the root of the equation for $𝑥>2$ is 3.

For $𝑥<0$, the equation becomes

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

The root which is less than 2 is 1.

Thus, the roots of given equations are 3, 1. Sum will be $3+1=4$

Let ${𝛼}_1$ and ${𝛼}_2$ be roots of the equation ${𝑥}^2+𝑎𝑥+𝑏=0$.

Therefore, ${𝛼}_1+{𝛼}_2=-𝑎$ and ${𝛼}_1{𝛼}_2=𝑏$ and ${𝛽}_1$ and ${𝛽}_2$ be the roots of the equation ${𝑥}^2+𝑏𝑥+𝑎=0$

Therefore, ${𝛽}_1+{𝛽}_2=-𝑏$ and ${𝛽}_1,{𝛽}_2=𝑎$

$\begin{array}{cc}& \text{ Given that }{𝛼}_1-{𝛼}_2={𝛽}_1-{𝛽}_2\\ & \Rightarrow {({𝛼}_1+{𝛼}_2)}^2-4{𝛼}_1{𝛼}_2={({𝛽}_1+{𝛽}_2)}^2-4{𝛽}_1{𝛽}_2\\ & \Rightarrow (-𝑎{)}^2-4𝑏=𝑡-4𝑎\\ & \Rightarrow {𝑎}^2-{𝑏}^2+4𝑎-4𝑏=0\end{array}$

$\begin{array}{cc}& \Rightarrow (𝑎-𝑏)(𝑎+𝑏)+4(𝑎-𝑏)=0\\ & \Rightarrow (𝑎-𝑏)(𝑎+𝑏+4)=0\end{array}$

Therefore, $𝑎-𝑏=0$ and $𝑎+𝑏+4=0$

$\begin{array}{cc}& \text{ Given, }(𝑥+𝑎)(𝑥+1991)+1\\ & \therefore (𝑥+𝑎)(𝑥+1991)-1=0\\ & \Rightarrow (𝑥+𝑎)(𝑥+1991)=-1\end{array}$

Multiplication of two integers is $-1$, this implies that either $(𝑥+𝑎)$ is +1 and $(𝑥+1991)$ is $-1$ or $(𝑥+𝑎)$ is $-1$ and $(𝑥+1991)$ is +1.

Let $(𝑥+𝑎)=1$, then $𝑥+1991=-1$

$\begin{array}{cc}& \Rightarrow 𝑥=-1-1991\\ & 𝑥=-1992\\ & \therefore 𝑎=1-𝑥=1-(-1992)=1+1992=1993\end{array}$

Let $(𝑥+𝑎)=-1$, then $𝑥+1991=+1$

$\begin{array}{cc}\Rightarrow 𝑥& =1-1991\\ & =-1990\end{array}$

$\begin{array}{cc}\therefore 𝑎& =-1-𝑥=-1-(-1990)\\ & =-1+1990=1989\end{array}$

So, $𝑎$ can be 1993 or 1989.

If set $𝐴$ contains $𝑚$ elements and set $𝐵$ comprises $𝑛$ elements, the number of injections from $𝐴$ to $𝐵$ can be defined as follows:

$\{\begin{array}{cc}0,& \text{if }𝑛<𝑚\\ {}^{𝑛}{𝑃}_{𝑚},& \text{if }𝑛\ge 𝑚\end{array}$

This means that if the number of elements in set $𝐵$ is less than the number in set $𝐴$ ($𝑛<𝑚$), there are no injections possible. Conversely, if $𝑛\ge 𝑚$, the number of injections is given by the permutation notation ${}^{𝑛}{𝑃}_{𝑚}$.

MULTIPLE has 8 letters, with 5 consonant (L repeated twice) and 3 vowels.

The vowels in 2nd, 5th and 8th places, these can be arranged in 3 ! ways $=6$ ways

The remaining 5 consonants can be arranged in $\frac{5!}{2!}$ ways $=\frac{120}{2}=60$ ways

Thus, the total number of ways $=6\times 60=360$

Given the inequality $(\frac{𝑛-1}{3})+(\frac{𝑛-1}{4})>(\frac{𝑛}{3})$:

We start with the given expression:

$(\frac{𝑛-1}{3})+(\frac{𝑛-1}{4})>(\frac{𝑛}{3})$

Using the properties of combinations, we know that:

$(\frac{𝑛}{𝑘})=(\frac{𝑛-1}{𝑘-1})+(\frac{𝑛-1}{𝑘})$

Applying this, we get:

$(\frac{𝑛-1}{3})+(\frac{𝑛-1}{4})>(\frac{𝑛}{3})$

This simplifies to:

$\frac{(𝑛-1)!}{3!(𝑛-4)!}+\frac{(𝑛-1)!}{4!(𝑛-5)!}>\frac{𝑛!}{3!(𝑛-3)!}$

Rewriting and dividing both sides by common terms:

$\frac{1}{4!}+\frac{1}{(𝑛-4)}>\frac{1}{𝑛-3\cdot 𝑛-4}$

Simplifying further:

$\frac{1}{4}>\frac{1}{(𝑛-3)(𝑛-4)}$

Taking the reciprocal, we have:

$4<(𝑛-3)(𝑛-4)$

To find the minimum value of $𝑛$:

$(𝑛-3)(𝑛-4)>4$

Solving this inequality, we find:

$𝑛-3>4$

Thus:

$𝑛>7$

The smallest integer $𝑛$ satisfying the inequality is $𝑛=8$.

Hence, the least value of $𝑛$ is 8.

Number of zeroes at the end of 4000!

$\begin{array}{cc}& =\left[\frac{4000}{5}\right]+\left[\frac{4000}{5^2}\right]+\left[\frac{4000}{5^{-3}}\right]+\left[\frac{4000}{5^4}\right]+\left[\frac{4000}{5^5}\right]\\ & =800+160+32+6+1=999\end{array}$

Now, we need one more pair of 5 and 2 so that the number ends in exactly 1000 zeroes.

$4005\le 𝑁<4009$

$\therefore$ From the option (c), $𝑁=4009$

Given, $\frac{13𝑥+43}{2{𝑥}^2+17𝑥+30}=\frac{𝐴}{2𝑥+5}+\frac{𝐵}{𝑥+6}$

$\Rightarrow \frac{13𝑥+43}{2{𝑥}^2+17𝑥+30}=\frac{𝐴(𝑥+6)+𝐵(2𝑥+5)}{(2𝑥+5)(𝑥+6)}$

Now, $13𝑥+43=(𝐴+2𝐵)𝑥+(6𝐴+5𝐵)$

$\therefore 𝐴+2𝐵=13$ and $6𝐴+5𝐵=43$

Solving these equations, we get

$𝐵=\frac{78-43}{7}=\frac{35}{7}=5$ and $𝐴=3$

$\therefore {𝐴}^2+{𝐵}^2=(3{)}^2+(5{)}^2=9+25=34$

Given $𝐴+𝐵+𝐶=𝜋$

$\begin{array}{cc}& \therefore \cos 𝐵=\cos 𝐴\cdot \cos 𝐶\\ & \Rightarrow \cos \{𝜋-(𝐴+𝐶\left)\right\}=\cos 𝐴\cdot \cos 𝐶\\ & \Rightarrow -\cos (𝐴+𝐶)=\cos 𝐴\cdot \cos 𝐶\\ & \Rightarrow -\cos 𝐴\cdot \cos 𝐶+\sin 𝐴\cdot \sin 𝐶=\cos 𝐴\cdot \cos 𝐶\\ & \Rightarrow \sin 𝐴\cdot \sin 𝐶=2\cos 𝐴\cos 𝐶\\ & \Rightarrow \frac{\sin 𝐴}{\cos 𝐴}\cdot \frac{\sin 𝐶}{\cos 𝐶}=2\\ & \Rightarrow \tan 𝐴\cdot \tan 𝐶=2\end{array}$

Within the context of triangle $△𝐴𝐵𝐶$, the following relationship holds:

We know that:

$(\tan \frac{𝐴}{2}\cdot \tan \frac{𝐵}{2}\cdot \tan \frac{𝐶}{2})\le \frac{1}{3\sqrt{\surd 3}}$

Thus, squaring both sides of the inequality, we get:

$\begin{array}{cc}{(\tan \frac{𝐴}{2}\cdot \tan \frac{𝐵}{2}\cdot \tan \frac{𝐶}{2})}^2& \le {\left(\frac{1}{3\sqrt{\surd 3}}\right)}^2\\ & =\frac{1}{27}\end{array}$

Observe that $\begin{array}{cc}\tan \left(\frac{7𝜋}{8}\right)& =\tan (𝜋-\frac{𝜋}{8})\\ & =-\tan \left(\frac{𝜋}{8}\right)\end{array}$

Now, using $\tan \left(2𝜃\right)=\frac{2\tan 𝜃}{1-{\tan }^2𝜃}$

Let $𝜃=\frac{𝜋}{8}$

So, $\tan (2\cdot \frac{𝜋}{8})=\frac{2\tan \frac{𝜋}{8}}{1-{\tan }^2\frac{𝜋}{8}}$

$\Rightarrow \tan \frac{𝜋}{4}=\frac{2\tan \frac{𝜋}{8}}{1-{\tan }^2\frac{𝜋}{8}}$

On putting $\tan \frac{𝜋}{8}=𝑥$,

$\begin{array}{cc}1& =\frac{2𝑥}{1-{𝑥}^2}\\ \Rightarrow & \\ \Rightarrow 1-{𝑥}^2-2𝑥& =0\Rightarrow {𝑥}^2+2𝑥-1=0\\ 𝑥& =-1-\sqrt{\surd 2},\sqrt{\surd 2}-1\end{array}$

But, $\tan \frac{𝜋}{8}>0$ so, only take $\tan \frac{𝜋}{8}=\sqrt{\surd 2}-1$

Thus, $\tan \left(\frac{7𝜋}{8}\right)=-\tan \left(\frac{𝜋}{8}\right)=-(\sqrt{\surd 2}-1)=1-\sqrt{\surd 2}$

Given,

$\begin{array}{cc}& 1+{\sec }^2𝑥{\sin }^2𝑥\\ & =1+\left(\frac{1}{{\cos }^2𝑥}\right){\sin }^2𝑥\\ & =1+{\tan }^2𝑥={\sec }^2𝑥\end{array}$

Given that $𝐴𝐵𝐶$ be triangle.

Using cosine rule, $\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}$

$\cos 𝐵=\frac{{𝑐}^2+{𝑎}^2-{𝑏}^2}{2𝑐𝑎},\cos 𝐶=\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}$

Now, $2(𝑏𝑐\cos 𝐴+𝑐𝑎\cos 𝐵+𝑎𝑏\cos 𝐶)$

$=2(\frac{𝑏𝑐\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}+𝑐𝑎\frac{{𝑐}^2+{𝑎}^2-{𝑏}^2}{2𝑐𝑎}}{+𝑎𝑏\cdot \frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}})$

$=2(\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2}+\frac{{𝑐}^2+{𝑎}^2-{𝑏}^2}{2}+\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2})$

$=2\left(\frac{{𝑏}^2+{𝑐}^2+{𝑎}^2}{2}\right)={𝑎}^2+{𝑏}^2+{𝑐}^2$

Given,

$\begin{array}{cc}& 4+6({𝑒}^{2𝑥}+1)\tanh 𝑥=11\cosh 𝑥+11\sinh 𝑥\\ & \Rightarrow 4+6({𝑒}^{2𝑥}+1)\cdot \frac{({𝑒}^{2𝑥}-1)}{({𝑒}^{2𝑥}+1)}\\ & =11\{\frac{{𝑒}^{2𝑥}+1}{2{𝑒}^{𝑥}}+\frac{{𝑒}^{2𝑥}-1}{2{𝑒}^{𝑥}}\}\\ & =11\left\{\frac{2{𝑒}^{2𝑥}}{2{𝑒}^{𝑥}}\right\}=11{𝑒}^{𝑥}\\ & \Rightarrow 4+6({𝑒}^{2𝑥}-1)=11{𝑒}^{𝑥}\\ & \text{ Let }{𝑒}^{𝑥}=𝑃\\ & \Rightarrow 4+6({𝑃}^2-1)=11𝑃\\ & \Rightarrow 6{𝑃}^2-11𝑃-2=0\\ & \Rightarrow 6{𝑃}^2-12𝑃+𝑃-2=0\\ & \Rightarrow (𝑃-2)(6𝑃+1)=0\end{array}$