AP EAPCET 2021 19th August Evening Shift Question Paper with Solutions

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

Now, $𝑓(-𝑥)=\frac{-𝑥}{\frac{1}{{𝑒}^{𝑥}}-1}-\frac{𝑥}{2}+1=\frac{-𝑥{𝑒}^{𝑥}}{1-{𝑒}^{𝑥}}-\frac{𝑥}{2}+1$

$\begin{array}{cc}& =\frac{-2𝑥{𝑒}^{𝑥}-𝑥(1-{𝑒}^{𝑥})}{2(1-{𝑒}^{𝑥})}+1\\ & =\frac{-2𝑥{𝑒}^{𝑥}-𝑥+𝑥{𝑒}^{𝑥}}{2(1-{𝑒}^{𝑥})}+1\\ & =\frac{-𝑥{𝑒}^{𝑥}-2𝑥+𝑥}{2(1-{𝑒}^{𝑥})}+1\\ & =\frac{-2𝑥+𝑥(1-{𝑒}^{𝑥})}{2(1-{𝑒}^{𝑥})}+1\\ & =\frac{-𝑥}{1-{𝑒}^{𝑥}}+\frac{𝑥}{2}+1\\ & =\frac{𝑥}{{𝑒}^{𝑥}-1}+\frac{𝑥}{2}+1=𝑓\left(𝑥\right)\end{array}$

$\therefore 𝑓\left(𝑥\right)$ is an even function.

$𝑓\left(𝑥\right)=\frac{1}{\left[𝑥\right]-1}$

$𝑓\left(𝑥\right)$ is not defined when

$\begin{array}{cc}& \left[𝑥\right]-1=0\\ \Rightarrow & \left[𝑥\right]=1\Rightarrow 1\le 𝑥<2\\ \Rightarrow & 𝑥\in [1,2)\\ \text{ Domain }𝑓=𝑅-[1,2)\end{array}$

$𝑓\left(𝑥\right)=\frac{4^{𝑥}}{4^{𝑥}+2}$

$𝑓(1-𝑥)=\frac{4^{1-𝑥}}{4^{1-𝑥}+2}=\frac{\frac{4}{4^{𝑥}}}{\frac{4}{4^{𝑥}}+2}$

$=\frac{4}{4+2.4^{𝑥}}$

$𝑓(1-𝑥)=\frac{2}{4^{𝑥}+2}$ ...... (i)

and $1-𝑓\left(𝑥\right)=1-\frac{4^{𝑥}}{4^{𝑥}+2}$

$=\frac{4^{𝑥}+2-4^{𝑥}}{4^{𝑥}+2}=\frac{2}{4^{𝑥}+2}=𝑓(1-𝑥)$

$\Rightarrow 𝑓(1-𝑥)+𝑓\left(𝑥\right)=1$ ..... (ii)

Put $𝑥=\frac{1}{4}$ in Eq. (ii), we get

$𝑓\left(\frac{3}{4}\right)+𝑓\left(\frac{1}{4}\right)=1$ ...... (iii)

Now, put $𝑥=\frac{1}{2}$ in Eq. (ii), we get

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

or $2𝑓\left(\frac{1}{2}\right)=1$ ..... (iv)

Adding Eqs. (iii) and (iv), we have

$𝑓\left(\frac{1}{4}\right)+2𝑓\left(\frac{1}{2}\right)+𝑓\left(\frac{3}{4}\right)=2$

Clearly, $2^{𝑛}>𝑛+1$ is valid for, $\forall 𝑛\ge 2$.

$\begin{array}{cc}& \mid \begin{array}{ccc}𝑏+𝑐& 𝑎& 𝑎\\ 𝑏& 𝑐+𝑎& 𝑏\\ 𝑐& 𝑐& 𝑎+𝑏\end{array}\mid \\ & =(𝑏+𝑐)\left[\right(𝑐+𝑎\left)\right(𝑎+𝑏)-𝑏𝑐]-𝑎(𝑎𝑏+{𝑏}^2-𝑏𝑐)\\ & =(𝑏+𝑐)(𝑎𝑐+𝑏𝑐+{𝑎}^2+𝑎𝑏-𝑏𝑐)+𝑎(𝑏𝑐-{𝑐}^2-𝑎𝑐)\\ & -{𝑎}^2𝑏-𝑎{𝑏}^2+𝑎𝑏𝑐+𝑎𝑏𝑐-𝑎{𝑐}^2-{𝑎}^2𝑐\\ & =𝑎𝑏𝑐+{𝑎}^2𝑏+𝑎{𝑏}^2+𝑎{𝑐}^2+{𝑎}^2𝑐+𝑎𝑏𝑐-{𝑎}^2𝑏-𝑎{𝑏}^2+𝑎𝑏𝑐+𝑎𝑏𝑐-𝑎{𝑐}^2-{𝑎}^2𝑐\\ & =4𝑎𝑏𝑐\end{array}$

Option (c) is correct.

${𝐷}^{𝑇}=𝐴𝐵𝐶$

and $𝑆=𝐴𝐵𝐶𝐷$

${𝑆}^2=\left(𝐴𝐵𝐶𝐷\right)\left(𝐴𝐵𝐶𝐷\right)=({𝐷}^{𝑇}.𝐷)\left({𝐷}^{𝑇}𝐷\right)$

$=({𝐷}^{𝑇}𝐷{)}^2$ ..... (i)

$\because$ $𝑆=𝐴𝐵𝐶𝐷={𝐷}^{𝑇}𝐷$

${𝑆}^{𝑇}=({𝐷}^{𝑇}𝐷{)}^{𝑇}={𝐷}^{𝑇}𝐷$

$\therefore$ From Eq. (i), we get

${𝑆}^2=({𝑆}^{𝑇}{)}^2$

$\therefore$ ${𝑆}^2=({𝑆}^{𝑇}{)}^2=({𝑆}^2{)}^{𝑇}$

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}{𝑎}^2& 15& 31\\ 12& {𝑏}^2& 41\\ 35& 61& {𝑐}^2\end{array}⎤⎦⎥,\\ 𝐵& =⎡⎣⎢\begin{array}{ccc}2𝑎& 3& 5\\ 2& 2𝑏& 8\\ 1& 4& 2𝑐-3\end{array}⎤⎦⎥\end{array}$

Given, trace of $𝐴=\mathrm{trace}$ of $𝐵$

$\begin{array}{cc}& \Rightarrow {𝑎}^2+{𝑏}^2+{𝑐}^2=2𝑎+2𝑏+2𝑐-3\\ & \Rightarrow ({𝑎}^2-2𝑎+1)+({𝑏}^2-2𝑏+1)+({𝑐}^2-2𝑐+1)=0\\ & \Rightarrow (𝑎-1{)}^2+(𝑏-1{)}^2+(𝑐-1{)}^2=0\\ & \Rightarrow 𝑎=1,𝑏=1,𝑐=1\end{array}$

Then product of diagonal elements of $𝐵$

$=\left(2𝑎\right)\left(2𝑏\right)(2𝑐-3)=\left(2\right)\left(2\right)(2-3)=-4$

$\mid \begin{array}{ccc}𝑎& 𝑎+1& 𝑎-1\\ -𝑏& 𝑏+1& 𝑏-1\\ 𝑐& 𝑐-1& 𝑐+1\end{array}\mid$

$+\mid \begin{array}{ccc}𝑎+1& 𝑏+1& 𝑐-1\\ 𝑎-1& 𝑏-1& 𝑐+1\\ (-1{)}^{𝑛+2}𝑎& (-1{)}^{𝑛-1}𝑏& (-1{)}^{𝑛}𝑐\end{array}\mid =0$

$\begin{array}{cc}& \because \left|𝐴\right|<text>=</text><text>\mid </text>{𝐴}^{𝑇}]\\ & \Rightarrow \mid \begin{array}{ccc}𝑎& -𝑏& 𝑐\\ 𝑎+1& 𝑏+1& 𝑐-1\\ 𝑎-1& 𝑏-1& 𝑐+1\end{array}\mid \\ & +(-1{)}^{𝑛}\mid \begin{array}{ccc}𝑎+1& 𝑏+1& 𝑐-1\\ 𝑎-1& 𝑏-1& 𝑐+1\\ 𝑎& -𝑏& 𝑐\end{array}\mid =0\end{array}$

$\begin{array}{cc}& {𝑅}_2\leftrightarrow {𝑅}_3\text{ and }{𝑅}_3\leftrightarrow {𝑅}_1\text{ in 2nd determinant }\\ & \Rightarrow \mid \begin{array}{ccc}𝑎& -𝑏& 𝑐\\ 𝑎+1& 𝑏+1& 𝑐-1\\ 𝑎-1& 𝑏-1& 𝑐+1\end{array}\mid \\ & +(-1{)}^{𝑛}\mid \begin{array}{ccc}𝑎& -𝑏& 𝑐\\ 𝑎+1& 𝑏+1& 𝑐-1\\ 𝑎-1& 𝑏-1& 𝑐+1\end{array}\mid =0\end{array}$

Both determinants are equal and their sum is zero when $𝑛$ is an odd integer.

$\because |𝑧-2|=|𝑧-1|$

Let $𝑧=𝑥+𝑖𝑦$, then

$\begin{array}{cc}\Rightarrow & |𝑥+𝑖𝑦-2{|}^2=|𝑥+𝑖𝑦-1{|}^2\\ \Rightarrow & (𝑥-2{)}^2+{𝑦}^2=(𝑥-1{)}^2+{𝑦}^2\\ \Rightarrow & {𝑥}^2-4𝑥+4={𝑥}^2-2𝑥+1\\ \Rightarrow & -2𝑥+4𝑥=4-1\Rightarrow 𝑥=3/2\end{array}$

Which is parallel to $𝑌$-axis.

${\left(\frac{1+𝑖}{1-𝑖}\right)}^{𝑚}=1\Rightarrow {\left[\frac{(1+𝑖)(1+𝑖)}{(1-𝑖)(1+𝑖)}\right]}^{𝑚}=1$

$\Rightarrow \left(\frac{1-1+2𝑖}{1+1}\right)=1$

$\Rightarrow {𝑖}^{𝑚}=1$

$\Rightarrow 𝑚$ is multiple of $4[\because {𝑖}^{4𝑛}=1,\forall 𝑛\in 𝑁]$

Option (a) is not a multiple of 4 .

$\therefore 𝑚\ne 1934$

$\begin{array}{cc}& \text{ Given, }{𝑥}^2-\sqrt{\surd 3}𝑥+1=0\\ & \Rightarrow 𝑥=\frac{\sqrt{\surd 3}\pm \sqrt{\surd 3-4}}{2}\\ & 𝑥=\frac{\sqrt{\surd 3}}{2}\pm \frac{1}{2}𝑖=\cos \frac{𝜋}{6}\pm 𝑖\sin \frac{𝜋}{6}\\ & \Rightarrow {𝑥}^{2𝑛}=\cos \frac{2𝑛𝜋}{6}\pm 𝑖\sin \frac{2𝑛𝜋}{6}\\ & \Rightarrow {𝑥}^{2𝑛}=\cos \frac{𝑛𝜋}{3}\pm 𝑖\sin \frac{𝑛𝜋}{3}\\ & \text{ and }{𝑥}^{-2𝑛}=\cos \frac{𝑛𝜋}{3}\pm 𝑖\sin \frac{𝑛𝜋}{3}\\ & \mathrm{Now},{({𝑥}^{𝑛}-\frac{1}{{𝑥}^{𝑛}})}^2=({𝑥}^{2𝑛}+\frac{1}{{𝑥}^{2𝑛}}-2)\\ & =-2+2\cos \frac{𝑛𝜋}{3}\\ & \Rightarrow \sum _{𝑛=1}^{24}{({𝑥}^{𝑛}-\frac{1}{{𝑥}^{𝑛}})}^2=\sum _{𝑛=1}^{24}(-2+2\cos \frac{𝑛𝜋}{3})\\ & =-48+2[\cos \frac{𝜋}{3}+\cos \frac{2𝜋}{3}+\dots +\cos \frac{24𝜋}{3}]\\ & =-48+2\left[\frac{\cos (\frac{2𝜋}{3}+\frac{23𝜋}{6})\cdot \sin \frac{24𝜋}{6}}{\sin \frac{𝜋}{6}}\right]\\ & =-48+2\cdot 0=-48\\ & \end{array}$

Given equation, $11{𝑥}^2+12𝑥-13=0$

If $𝛼,𝛽$ are the roots, then

$𝛼+𝛽=\frac{-12}{11}\text{ and }𝛼𝛽=-\frac{13}{11}$

Now, $\frac{1}{{𝛼}^2}+\frac{1}{{𝛽}^2}=\frac{(𝛼+𝛽{)}^2-2𝛼𝛽}{(𝛼𝛽{)}^2}$

$\begin{array}{cc}& =\frac{\frac{144}{121}+\frac{26}{11}}{\frac{169}{121}}\\ & =\frac{144+286}{169}=2.54\end{array}$

Let $𝛼$ be the common root of

$\begin{array}{cc}& {𝑥}^3+𝑎𝑥+1=0\\ & \text{ and }{𝑥}^4+𝑎{𝑥}^2+1=0\\ & \text{ Then, }{𝛼}^3+𝑎𝛼+1=0\\ & \text{ and }{𝛼}^4+𝑎{𝛼}^2+1=0\end{array}$

subtracting both equations

$\begin{array}{cc}{𝛼}^4-{𝛼}^3+𝑎{𝛼}^2-𝑎𝛼& =0\\ 𝛼({𝛼}^3-{𝛼}^2+𝑎𝛼-𝑎)& =0\\ \Rightarrow {𝛼}^2(𝛼-1)+𝑎(𝛼-1)=0\\ [𝛼=0\text{ doesn’t give a root }]\\ \Rightarrow (𝛼-1)({𝛼}^2+𝑎)=0\\ 𝛼=1\text{ or }𝑎=-{𝛼}^2\end{array}$

Clearly, $𝛼=1$ gives $𝑎=-2$ in both equations.

Hence, for $𝑎=-2$ both equation have a common root.

Given, equation, $7{𝑥}^2-13𝑥+𝑎=0$

$\because$ Roots are rational $\Rightarrow$ Discriminate is a perfect square

$𝐷=169-28𝑎\Rightarrow 𝐷={𝑏}^2-4𝑎𝑐$

$\Rightarrow 𝐷$ is perfect square for $𝑎=6$

$\because$ $𝑝,𝑞,𝑟,𝑠$ are in AP.

Let $𝑝=𝑎,𝑞=𝑎+𝑑,𝑟=𝑎+2𝑑,𝑠=𝑎+3𝑑$

Given, $𝑝,𝑞$ are roots of

${𝑥}^2-2𝑥+𝐴=0$

Then, $𝑝+𝑞=2$ and $𝑝𝑞=𝐴$

$\Rightarrow 2𝑎+𝑑=2$ .... (i)

Also, r and s are roots of ${𝑥}^2-18𝑥+𝐵=0$

Then, $𝑟+𝑠=18$ and $𝑟𝑠=𝐵$

$\Rightarrow 2𝑎+5𝑑=18$ ..... (ii)

Solving Eqs. (i) and (ii), we get

$𝑎=-1,𝑑=4$

Now, $𝑝𝑞=𝐴$ (product of roots)

or $𝐴=𝑎(𝑎+𝑑)=(-1)(-1+4)$

$𝐴=-3$

and $𝑟𝑠=𝐵$

or $𝐵=(𝑎+2𝑑)(𝑎+3𝑑)$

$=(-1+8)(-1+12)=77$

$\therefore$ $𝐴=-3,𝐵=77$

$\begin{array}{cc}& \frac{1}{𝑟+1}\{{}^{𝑛}{𝑃}_{𝑟+1}-{}^{𝑛-1}{𝑃}_{𝑟+1}\}\\ & =\frac{1}{𝑟+1}[\frac{𝑛!}{(𝑛-𝑟-1)!}-\frac{(𝑛-1)!}{(𝑛-𝑟-2)!}]\\ & =\frac{1}{(𝑟+1)}\cdot \frac{(𝑛-1)!}{(𝑛-𝑟-2)!}(\frac{𝑛}{𝑛-𝑟-1}-1)\end{array}$

$\begin{array}{cc}& =\frac{1}{(𝑟+1)}\frac{(𝑛-1)!}{(𝑛-𝑟-2)!}\frac{(𝑟+1)}{(𝑛-𝑟-1)}\\ & =\frac{(𝑛-1)!}{(𝑛-𝑟-1)!}={}^{𝑛-1}{𝑃}_{𝑟}\end{array}$

Total number of balls $=10$

Total number of ways of selecting four balls $={}^{10}{𝐶}_4=210$

Number of ways of selecting all green balls $={}^4{𝐶}_4=1$

Number of ways of selecting 4 balls in which at least one black ball is selected $=210-1=209$

Total number of arrangements = 9!

Number of arrangements of best and worst paper together = 2! $\times$ 8!

Number of arrangements when best and worst are never together = 9! $-$ 2! $\times$ 8!

$\begin{array}{cc}& \frac{-{𝑥}^2+6𝑥+13}{(3𝑥+5)({𝑥}^2+4𝑥+4)}=\frac{-{𝑥}^2+6𝑥+13}{(3𝑥+5)(𝑥+2{)}^2}\\ & \Rightarrow \frac{-{𝑥}^2+6𝑥+13}{(3𝑥+5)(𝑥+2{)}^2}=\frac{𝐴}{3𝑥+5}+\frac{𝐵}{𝑥+2}+\frac{𝐶}{(𝑥+2{)}^2}\\ & \Rightarrow -{𝑥}^2+6𝑥+13=𝐴(𝑥+2{)}^2+𝐵(3𝑥+5\left)\right(𝑥+2)+𝐶(3𝑥+5)\dots (i)\end{array}$

It is an identity

$\therefore$ True for every value of $𝑥$.

Put $𝑥=-2$ in Eq. (i),

$\begin{array}{c}-(-2{)}^2+6(-2)+13=𝐴\cdot 0+𝐵\cdot 0+𝐶(3(-2)+5)\\ \Rightarrow -4-12+13=-𝐶\Rightarrow 𝐶& =3\end{array}$

Put $𝑥=\frac{-5}{3}$ in Eq. (i), we get

$\begin{array}{cc}-{\left(\frac{-5}{3}\right)}^2& +6\left(\frac{-5}{3}\right)+13\\ =𝐴{(\frac{-5}{3}+2)}^2+𝐵\cdot 0+𝐶\cdot 0\\ \Rightarrow \frac{-25}{9}-10+13& =𝐴\cdot \frac{1}{9}\\ \Rightarrow \frac{-25}{9}+3& =\frac{𝐴}{9}\Rightarrow 𝐴=2\end{array}$

Put $𝑥=0$ in Eq. (i), we get

$\begin{array}{c}13=4𝐴+10𝐵+5𝐶\\ \Rightarrow 13=8+10𝐵+15\\ \Rightarrow 10𝐵=13-23\text{ or }𝐵=-1\\ \therefore \frac{-{𝑥}^2+6𝑥+13}{(3𝑥+5)({𝑥}^2+4𝑥+4)}\\ =\frac{2}{3𝑥+5}+\frac{-1}{𝑥+2}+\frac{3}{(𝑥+2{)}^2}\end{array}$

$3\sin 𝐴+4\cos 𝐵=6$ ...... (i)

and $4\sin 𝐵+3\cos 𝐴=1$ .... (ii)

Squaring Eqs. (i) and (ii), we get

$9{\sin }^2𝐴+16{\cos }^2𝐵+24\sin 𝐴\cos 𝐵=36$ ..... (iii)

$16{\sin }^2𝐵+9{\cos }^2𝐴+24\sin 𝐵\cos 𝐴=1$ .... (iv)

Adding Eqs. (iii) and (iv), we get

$\begin{array}{cc}& 9({\sin }^2𝐴+{\cos }^2𝐴)+16({\sin }^2𝐵+{\cos }^2𝐵)\\ & +24(\sin 𝐴\cos 𝐵+\cos 𝐴\sin 𝐵)=37\\ & \Rightarrow 25+24\sin (𝐴+𝐵)=37\\ & \sin (𝐴+𝐵)=1/2\\ & \end{array}$

Let $\tan 𝐴+2\cot 2𝐴$

$=\frac{\sin 𝐴}{\cos 𝐴}+\frac{2\cos 2𝐴}{\sin 2𝐴}$

$=\frac{\sin 𝐴}{\cos 𝐴}+\frac{2(1-2{\sin }^2𝐴)}{2\sin 𝐴\cos 𝐴}$

$=\frac{{\sin }^2𝐴+1-2{\sin }^2𝐴}{\sin 𝐴\cos 𝐴}$

$=\frac{1-{\sin }^2𝐴}{\sin 𝐴\cos 𝐴}=\frac{{\cos }^2𝐴}{\sin 𝐴\cos 𝐴}=\cot 𝐴$

$\therefore \tan 𝛼+2\tan 2𝛼+4\tan 4𝛼+8\cot 8𝛼=\tan 𝛼+2\tan 2𝛼+4(\tan 4𝛼+2\cot 2(4𝛼\left)\right)$

$=\tan 𝛼+2\tan 2𝛼+4\cot 4𝛼$

$=\tan 𝛼+2(\tan 2𝛼+2\cot 2(2𝛼\left)\right)$

$=\tan 𝛼+2\cot 2𝛼=\cot 𝛼$

$\begin{array}{cc}& \text{ If }𝑓\left(𝑥\right)=\frac{\cot 𝑥}{1+\cot 𝑥}\text{ and }𝛼+𝛽=\frac{5𝜋}{4}\\ & \Rightarrow \cot (𝛼+𝛽)=\cot \frac{5𝜋}{4}=1\Rightarrow \frac{\cot 𝛼\cot 𝛽-1}{\cot 𝛽+\cot 𝛼}=1\\ & \Rightarrow \cot 𝛼\cot 𝛽=\cot 𝛽+\cot 𝛼+1....\left(i\right)\\ & \text{ Now, }𝑓\left(𝛼\right)𝑓\left(𝛽\right)=\frac{\cot 𝛼}{1+\cot 𝛼}\times \frac{\cot 𝛽}{1+\cot 𝛽}\\ & =\frac{\cot 𝛼\cot 𝛽}{1+\cot 𝛼+\cot 𝛽+\cot 𝛼\cot 𝛽}=\frac{\cot 𝛼\cot 𝛽}{2\cot 𝛼\cot 𝛽}=\frac{1}{2}\end{array}$

$\begin{array}{cc}& \cos 2𝜃=\cos 𝜃+\sin 𝜃\\ & \Rightarrow ({\cos }^2𝜃-{\sin }^2𝜃)-(\cos 𝜃+\sin 𝜃)=0\\ & \Rightarrow (\cos 𝜃+\sin 𝜃)(\cos 𝜃-\sin 𝜃-1)=0\\ & \Rightarrow \cos 𝜃+\sin 𝜃=0\\ & \text{ or }\cos 𝜃-\sin 𝜃=1\\ & \text{ When }\cos 𝜃+\sin 𝜃=0\text{, then }\tan 𝜃=-1\\ & \text{ then }𝜃=𝑛𝜋+\frac{3𝜋}{4}\Rightarrow 𝜃=\frac{3𝜋}{4},\frac{7𝜋}{4}\in [0,2𝜋]\\ & \text{ and when }\cos 𝜃-\sin 𝜃=1\text{, then }\sin 𝜃+(1-\cos 𝜃)=0\\ & \text{ and }2\sin \frac{𝜃}{2}(\cos \frac{𝜃}{2}+\sin \frac{𝜃}{2})=0\\ & \end{array}$

$\begin{array}{cc}& \text{ and }\sin \frac{𝜃}{2}=0,\cos \frac{𝜃}{2}+\sin \frac{𝜃}{2}=0\\ & \Rightarrow \frac{𝜃}{2}=𝑛𝜋,\tan \frac{𝜃}{2}=-1\\ & \Rightarrow 𝜃=0,2𝜋\in [0,2𝜋]\text{, }\\ & 𝜃=\frac{6𝜋}{4}\in [0,2𝜋]\end{array}$

Sum of all values of $𝜃$

$=\frac{3𝜋}{4}+\frac{7𝜋}{4}+0+2𝜋+\frac{6𝜋}{4}=\frac{24𝜋}{4}$

$\begin{array}{cc}& \sin [2{\cos }^{-1}\{\cot (2{\tan }^{-1}𝑥)\}]=0\\ & \Rightarrow 2{\cos }^{-1}\{\cot (2{\tan }^{-1}𝑥)\}={\sin }^{-1}\left(0\right)=0\\ & \Rightarrow {\cos }^{-1}(\cot (2{\tan }^{-1}𝑥))=0\\ & \Rightarrow \cot \left[(2{\tan }^{-1}𝑥)\right]=\cos 0=1\\ & \Rightarrow \cot [{\tan }^{-1}\left(\frac{2𝑥}{1-{𝑥}^2}\right)]=1\\ & \Rightarrow \cot [{\cot }^{-1}\left(\frac{1-{𝑥}^2}{2𝑥}\right)]=1\\ & \Rightarrow \frac{1-{𝑥}^2}{2𝑥}=1\\ & \Rightarrow 1-{𝑥}^2=2𝑥\\ & \Rightarrow {𝑥}^2+2𝑥-1=0\\ & 𝑥=\frac{-2\pm \sqrt{\surd 4+4}}{2}=\frac{-2\pm 2\sqrt{\surd 2}}{2}\\ & 𝑥=-1\pm \sqrt{\surd 2}\end{array}$

Given equation holds for 2 values of $𝑥$.

$\because 𝑟=\frac{\Delta }{𝑠},{𝑟}_1=\frac{\Delta }{𝑠-𝑎},{𝑟}_2=\frac{\Delta }{𝑠-𝑏}$,

${𝑟}_3=\frac{\Delta }{𝑠-𝑐}\text{ and }2𝑠=𝑎+𝑏+𝑐$

Then, ${𝑟}_1+{𝑟}_2+{𝑟}_3-𝑟=\frac{\Delta }{𝑠-𝑎}+\frac{\Delta }{𝑠-𝑏}+\frac{\Delta }{𝑠-𝑐}-\frac{\Delta }{𝑠}$

$\begin{array}{cc}& =\Delta \left(\frac{2𝑠-(𝑎+𝑏)}{(𝑠-𝑎)(𝑠-𝑏)}\right)+\Delta \left(\frac{𝑐}{𝑠(𝑠-𝑐)}\right)\\ & =\Delta \frac{(2𝑠-2𝑠+𝑐)}{(𝑠-𝑎)(𝑠-𝑏)}+\Delta \left(\frac{𝑐}{𝑠(𝑠-𝑐)}\right)\\ & =\Delta \left[\frac{𝑐𝑠(𝑠-𝑐)+𝑐(𝑠-𝑎)(𝑠-𝑏)}{𝑠(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)}\right]\\ & [\because {\Delta }^2=𝑠(𝑠-𝑎\left)\right(𝑠-𝑏\left)\right(𝑠-𝑐\left)\right]\\ & =\frac{\Delta 𝑐}{{\Delta }^2}\left(𝑠\right(𝑠-𝑐)+(𝑠-𝑎\left)\right(𝑠-𝑏\left)\right)\\ & =\frac{𝑐}{\Delta }({𝑠}^2-𝑠𝑐+{𝑠}^2-(𝑎+𝑏)𝑠+𝑎𝑏))\end{array}$