TG EAPCET 2020 11th September Morning Shift Question Paper with Solutions

Given function $𝑓:[-3,2]\to [0,\text{exception 25:}]$, such that

$𝑓\left(𝑛\right)=\{\begin{array}{cc}2+\text{exception 25:},& -3\le 𝑛\le -1\\ {𝑛}^{2/3},& -1\le 𝑛\le 2\end{array}$

So, ${𝑓}^{\prime }\left(𝑛\right)=\{\begin{array}{cc}\frac{1}{3}{𝑛}^{-2/3}& -3<𝑛<-1\\ \frac{2}{3}{𝑛}^{-1/3},& -1<𝑛<2\end{array}$

$\because 𝑓\left(𝑛\right)$ is strictly increasing in $(-3,-1)$ and $(0,2)$ because ${𝑓}^{\prime }\left(𝑛\right)$ is positive, for $𝑛\in (-3,-1)\cup (0,2)$ and $𝑓$ is strictly decreasing in $(-1,0)$.

$\because 𝑓(-1)$ or $𝑓\left(2\right)$ is the maximum value of the function and $𝑓(-1)=1$ and

$\begin{array}{cc}& 𝑓\left(2\right)=2^{2/3}\\ & \therefore \text{exception 25:}=2^{2/3}\Rightarrow 𝑥=4\end{array}$

Given function, $𝑓\left(𝑥\right)=\frac{𝑥-\left[𝑥\right]}{\sqrt{\surd \left|𝑥\right|-𝑥}}$ will define, if

$\left|𝑥\right|>𝑥\Rightarrow 𝑥<0\Rightarrow 𝑥\in (-\infty ,0)...\left(𝑖\right)$

and the function, $𝑔\left(𝑥\right)=\frac{𝑥-\left[𝑥\right]}{\sqrt{\surd \left|𝑥\right|+𝑥}}$ will define, if

$\begin{array}{cc}& \left|𝑥\right|+𝑥>0\Rightarrow 𝑥>0\Rightarrow 𝑥\in (0,\infty )...\left(𝑖𝑖\right)\\ & \therefore 𝐴=\{𝑥\mid 𝑥\in 𝐑\text{ and }𝑥<0\}\\ & \text{ and }𝐵=\{𝑥\mid 𝑥\in 𝐑\text{ and }𝑥>0\}\\ & \therefore 𝐴\cap 𝐵=𝜙\end{array}$

Since, ${𝑛}^5-5{𝑛}^3+4𝑛$

$\begin{array}{cc}& =𝑛({𝑛}^4-5{𝑛}^2+4)\\ & =𝑛({𝑛}^4-4{𝑛}^2-{𝑛}^2+4)=𝑛({𝑛}^2-1)({𝑛}^2-4)\\ & =(𝑛-2)(𝑛-1)\left(𝑛\right)(𝑛+1)(𝑛+2)=𝑃\left(𝑛\right)\\ & \left(\text{ let }\right)\end{array}$

For $𝑛=1,2,𝑃\left(𝑛\right)$ is zero so it is divisible by 120 .

and for $𝑛=3,𝑃\left(3\right)=5!=120$ and so on, so the

$𝑃\left(𝑛\right)$ is divisible by 120 for all positive integers $𝑛$.

Given determinant

$\begin{array}{cc}\Delta & =\mid \begin{array}{ccc}1+{\sin }^2𝜃& {\cos }^2𝜃& 4\sin 4𝜃\\ {\sin }^2𝜃& 1+{\cos }^2𝜃& 4\sin 4𝜃\\ {\sin }^2𝜃& {\cos }^2𝜃& 1+4\sin 4𝜃\end{array}\mid =0\end{array}$

On applying ${𝐶}_1\to {𝐶}_1+{𝐶}_2+{𝐶}_3$, we get

$\begin{array}{cc}\Delta & =\mid \begin{array}{ccc}2+4\sin 4𝜃& {\cos }^2𝜃& 4\sin 4𝜃\\ 2+4\sin 4𝜃& 1+{\cos }^2𝜃& 4\sin 4𝜃\\ 2+4\sin 4𝜃& {\cos }^2𝜃& 1+4\sin 4𝜃\end{array}\mid =0\end{array}$

$\begin{array}{cc}\Rightarrow & \Delta =(2+4\sin 4𝜃)\\ & \mid \begin{array}{ccc}1& {\cos }^2𝜃& 4\sin 4𝜃\\ 1& 1+{\cos }^2𝜃& 4\sin 4𝜃\\ 1& {\cos }^2𝜃& 1+4\sin 4𝜃\end{array}\mid =0\end{array}$

On applying ${𝑅}_2\to {𝑅}_2-{𝑅}_1$ and ${𝑅}_3\to {𝑅}_3-{𝑅}_1$, we get

$\begin{array}{cc}\Rightarrow \Delta & =(1+2\sin 4𝜃)\mid \begin{array}{ccc}1& {\cos }^2𝜃& 4\sin 4𝜃\\ 0& 1& 0\\ 0& 0& 1\end{array}\mid =0\\ \Rightarrow \Delta & =2\sin 4𝜃+1=0\Rightarrow \sin 4𝜃=-\frac{1}{2}\\ \Rightarrow 4𝜃& =\frac{7𝜋}{6}\\ \Rightarrow 𝜃& =\frac{7𝜋}{24}\in (0,\frac{𝜋}{2})\end{array}$

For a non-singular matrix ${𝐴}_{3\times 3}$, it is given that

$\begin{array}{cc}& {𝐴}^{-1}=\frac{1}{3}({𝐴}^2-5𝐴+7𝐼)\\ \Rightarrow & {𝐴}^3-5{𝐴}^2+7𝐴-3𝐼=0...\left(𝑖\right)\end{array}$

$\begin{array}{cc}& \therefore 17{𝐴}^8-85{𝐴}^7+119{𝐴}^6-51{𝐴}^5-19{𝐴}^4-133{𝐴}^2+58𝐴+𝐼\\ & =({𝐴}^3-5{𝐴}^2+7𝐴-3𝐼)(17{𝐴}^5-19𝐴)+𝐴+𝐼\\ & =𝐴+𝐼\{\because {𝐴}^3-5{𝐴}^2+7𝐴-3𝐼=0\}\end{array}$

It is given that,

$\begin{array}{cc}& ⎡⎣⎢\begin{array}{ccc}2& 1& 1\\ 0& 3& -1\\ 1& -1& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}𝑥\\ 𝑦\\ 𝑧\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}1\\ 1\\ 0\end{array}⎤⎦⎥\\ \Rightarrow & 2𝑥+𝑦+𝑧=1...\left(𝑖\right)\\ & 3𝑦-𝑧=1...\left(𝑖𝑖\right)\end{array}$

and $𝑥-𝑦+𝑧=0...\left(𝑖𝑖𝑖\right)$

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

$\begin{array}{cc}2𝑥+𝑦+3𝑦-1& =1\Rightarrow 2𝑥+4𝑦=2\\ \Rightarrow 𝑥+2𝑦& =1...\left(𝑖𝑣\right)\end{array}$

Now, from Eqs. (ii) and (iv), we have

$\begin{array}{cc}& \frac{𝑥-1}{-2}=𝑦=\frac{𝑧+1}{3}=𝑘\left(𝑙𝑒𝑡\right)\\ \therefore ⎡⎣⎢\begin{array}{c}𝑥\\ 𝑦\\ 𝑧\end{array}⎤⎦⎥& =⎡⎣⎢\begin{array}{c}1\\ 0\\ -1\end{array}⎤⎦⎥+𝑘⎡⎣⎢\begin{array}{c}-2\\ 1\\ 3\end{array}⎤⎦⎥,𝑘\in 𝐑\end{array}$

$\begin{array}{cc}& \text{ Given, }\sin 5𝜃=𝑎{\cos }^4𝜃\sin 𝜃\\ & +𝑏{\cos }^2𝜃{\sin }^3𝜃+𝑐{\sin }^5𝜃\text{ at }𝜃=\frac{𝜋}{2},\end{array}$

we get

$𝑙=0+0+𝑐\Rightarrow 𝑐=𝑙...\left(𝑖\right)$

at   $𝜃=\frac{𝜋}{4},\text{ we get }$

$\begin{array}{cc}-\frac{1}{\sqrt{\surd 2}}& =𝑎\frac{1}{4}\times \frac{1}{\sqrt{\surd 2}}+𝑏\frac{1}{2}\frac{1}{2\sqrt{\surd 2}}+\frac{1}{4\sqrt{\surd 2}}\\ \Rightarrow -1& =\frac{𝑎}{4}+\frac{𝑏}{4}+\frac{1}{4}\\ \Rightarrow 𝑎+𝑏& =-5...\left(𝑖𝑖\right)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \begin{array}{cc}𝑎𝑡𝜃& =\frac{𝜋}{3},\text{ we get }\\ -\frac{\sqrt{\surd 3}}{2}& =𝑎\frac{1}{16}\times \frac{\sqrt{\surd 3}}{2}+𝑏\frac{1}{4}\frac{3\sqrt{\surd 3}}{8}+\frac{9\sqrt{\surd 3}}{32}\\ \Rightarrow -1& =\frac{𝑎}{16}+\frac{3𝑏}{16}+\frac{9}{16}\\ \Rightarrow 𝑎+3𝑏& =-25...\left(𝑖𝑖𝑖\right)\end{array}\end{array}\\ & \text{ From Eqs. (ii) and (iii), we get }\\ & 𝑏=-10\text{ and }𝑎=5;\therefore 𝑎𝑏𝑐=-50\end{array}$

According to given information $|𝑧-2𝑖|=2$ is a circle having centre is $(0,2)$ and radius is 2 .

If the area of $△𝐴𝐵𝐶$ is maximum the triangle must be equilateral triangle and point $𝑀$ is the mid point of $𝐵𝐶$, where $𝑀$ is the foot of perpendicular of point $𝐴({𝑧}_1=2+2𝑖)$ on $𝐵𝐶$.

∴ Sum of the imaginary parts of ${𝑧}_2$ and ${𝑧}_3$ is twice of the imaginary part of $𝑀=2\times 2=4$.

$\begin{array}{cc}& \text{ It is given that, }\\ & \begin{array}{cc}{𝐴}_{𝑛}& =\cos \left(\frac{𝜋}{2^{𝑛}}\right)+𝑖\sin \left(\frac{𝜋}{2^{𝑛}}\right),𝑛\in 𝐍\\ & ={𝑒}^{𝑖\left(\frac{𝜋}{2^{𝑛}}\right)}\end{array}\end{array}$

$\begin{array}{cc}\therefore {\left({𝐴}_1{𝐴}_2{𝐴}_3{𝐴}_4\right)}^4& ={𝑒}^{𝑖𝜋{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16})}^4}\\ & ={𝑒}^{𝑖\frac{15𝜋}{4}}={𝑒}^{𝑖(4𝜋-\frac{𝜋}{4})}\\ =\cos (4𝜋-\frac{𝜋}{4})& +𝑖\sin (4𝜋-\frac{𝜋}{4})\\ =\cos \frac{𝜋}{4}-𝑖\sin \frac{𝜋}{4}& =\frac{1-𝑖}{\sqrt{\surd 2}}\end{array}$

It is given that, if ${𝑥}^2+𝑥+1=0$, then

$\begin{array}{c}{𝐴}_{𝑟}={(𝑥+\frac{1}{𝑥})}^3{({𝑥}^2+\frac{1}{{𝑥}^2})}^3{({𝑥}^3+\frac{1}{{𝑥}^3})}^3\\ \dots {({𝑥}^{𝑟}+\frac{1}{{𝑥}^{𝑟}})}^3\end{array}$

As, we know that $𝑤$ and ${𝑤}^2$ are roots of quadratic equation ${𝑥}^2+𝑥+1=0$, so ${𝑤}^3=1$ and $1+𝑤+{𝑤}^2=0$

$\begin{array}{cc}& \therefore {𝐴}_{𝑟}={(𝑤+\frac{1}{𝑤})}^3{({𝑤}^2+\frac{1}{{𝑤}^2})}^3{({𝑤}^3+\frac{1}{{𝑤}^3})}^3\\ & \dots {({𝑤}^{𝑟}+\frac{1}{{𝑤}^{𝑟}})}^3\\ & \therefore {𝐴}_3=(-1)(-1)\left(8\right)=8\end{array}$

Similarly, ${𝐴}_6=8^2,{𝐴}_9=8^3,\dots$ and so on.

$\begin{array}{cc}& \therefore \frac{1}{{𝐴}_3}+\frac{1}{{𝐴}_6}+\frac{1}{{𝐴}_9}+\frac{1}{{𝐴}_{12}}+\dots \infty \\ & =\frac{1}{8}+\frac{1}{8^2}+\frac{1}{8^3}+\frac{1}{8^4}+\dots \infty =\frac{1/8}{1-\frac{1}{8}}=\frac{1}{7}\end{array}$

It is given that $𝑝$ and $𝑞$ are roots of quadratic equation ${𝑥}^2+7𝑥+3=0$

To find the quadratic equation whose roots are

$\begin{array}{cc}& \frac{3𝑝}{1-2𝑝}\text{ and }\frac{3𝑞}{1-2𝑞},\text{ put }\frac{3𝑝}{1-2𝑝}=𝑥\\ & 3𝑝=𝑥-2𝑝𝑥\Rightarrow 𝑝=\frac{𝑥}{3+2𝑥}\end{array}$

$\because 𝑝$ is the root of equation

$\begin{array}{cc}& {𝑥}^2+7𝑥+3=0,\text{ so }\\ & {\left(\frac{𝑥}{3+2𝑥}\right)}^2+7\left(\frac{𝑥}{3+2𝑥}\right)+3=0\\ & \Rightarrow {𝑥}^2+7𝑥(3+2𝑥)+3(3+2𝑥{)}^2=0\\ & \Rightarrow 27{𝑥}^2+57𝑥+27=0\\ & \Rightarrow 9{𝑥}^2+19𝑥+9=0\end{array}$

So, on comparing with

$\begin{array}{cc}& 𝑙{𝑥}^2+𝑚𝑥+𝑛=0\text{, we get }\\ & 𝑙=9,𝑚=19\text{ and }𝑛=9\\ & \therefore 𝑙-𝑚+𝑛=9-19+9=-1\end{array}$

Let $𝛼$ be the common root of the equation

$3{𝑥}^2-7𝑥+2=0$ and $𝑘{𝑥}^2+7𝑥-3=0$, so

$3{𝛼}^2-7𝛼+2=0...\left(𝑖\right)$

and $𝑘{𝛼}^2+7𝛼-3=0...\left(𝑖𝑖\right)$

By cross multiplication, we get

$\begin{array}{cc}& \frac{{𝛼}^2}{21-14}=\frac{-𝛼}{-9-2𝑘}=\frac{1}{21+7𝑘}\\ \Rightarrow & \frac{{𝛼}^2}{7}=\frac{𝛼}{2𝑘+9}=\frac{1}{7(𝑘+3)}\\ \therefore & \frac{1}{𝑘+3}=\frac{(2𝑘+9{)}^2}{49(𝑘+3{)}^2}\end{array}$

$\because 𝑘$ is positive, so

$\begin{array}{cc}& 49(𝑘+3)=4{𝑘}^2+36𝑘+81\\ \Rightarrow & 4{𝑘}^2-13𝑘-66=0\\ \Rightarrow & 4{𝑘}^2-24𝑘+11𝑘-66=0\\ \Rightarrow & 4𝑘(𝑘-6)+11(𝑘-6)=0\\ \Rightarrow & 𝑘=6[\therefore 𝑘=6\text{ is positive }]\end{array}$

Let the roots of

${𝑥}^3+𝑎{𝑥}^2+𝑏𝑥+𝑐=0\text{ are }𝛼-1,𝛼,𝛼+1$

as roots are in AP with common

difference 1.

Now, sum of roots $=3𝛼=-𝑎...\left(𝑖\right)$

Sum of product of roots taking two at a time

$\begin{array}{c}=(𝛼-1)𝛼+(𝛼+1)𝛼+(𝛼-1)(𝛼+1)=𝑏\\ \Rightarrow 3{𝛼}^2-1=𝑏& \dots \left(\text{ ii }\right)\end{array}$

and product of roots

$=𝛼({𝛼}^2-1)=-𝑐...\left(𝑖𝑖𝑖\right)$

From Eqs. (i), (ii) and (iii), we have

$\begin{array}{cc}& (-\frac{𝑎}{3})[\frac{𝑏+1}{3}-1]=-𝑐\\ \Rightarrow & \frac{𝑎}{3}\left(\frac{𝑏-2}{3}\right)=𝑐\\ \Rightarrow & 9𝑐=𝑎(𝑏-2)\end{array}$

Given equation,

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

So, roots $𝛼=-3,𝛽=-1$ and $𝛾=1$

Therefore, $(1+{𝛼}^2)(1+{𝛽}^2)(1+{𝛾}^2)$

$=(1+9)(1+1)(1+1)=40$

It is given that, $𝑥,𝑦,𝑧\ge -1$ and $𝑤\ge 1$

Now, let $𝑎=𝑥+1\ge 0$

$\begin{array}{cc}& 𝑏=𝑦+1\ge 0\\ & 𝑐=𝑧+1\ge 0\\ & 𝑑=𝑤-1\ge 0\end{array}$

∴ The given equation

$\begin{array}{cc}& 𝑥+𝑦+𝑧+𝑤=25\text{ reduce to }\\ & (𝑎-1)+(𝑏-1)+(𝑐-1)+(𝑑+1)=25\\ & \Rightarrow 𝑎+𝑏+𝑐+𝑑=27\end{array}$

∴ The number of required solution is same as the number of non-negative integral solutions of equation

$𝑎+𝑏+𝑐+𝑑=27$

and it is equal to ${}^{27+4-1}{𝐶}_{4-1}={}^{30}{𝐶}_3$

The total number of ways that all girls seated around a circle is 10 !.

The number of ways the three sisters seated together along with the other girls is $8!\times 3!$.

So, required number of ways

$\begin{array}{cc}& =10!-8!3!\\ & =8!(10\times 9-6)\\ & =8!\times 84\end{array}$

Given binomial ${(𝑎{𝑥}^2+\frac{𝑏}{{𝑥}^3})}^{15}$, so the general term,

${𝑇}_{𝑟+1}={}^{15}{𝐶}_{𝑟}{𝑎}^{15-𝑟}{𝑏}^{𝑟}{𝑥}^{30-5𝑟}$

∴ The coefficient of ${𝑥}^{10}$ is (at $𝑟=4$ )

$={}^{15}{𝐶}_4{𝑎}^{11}{𝑏}^4=𝑙\text{ (given) }$

Similarly, $𝑚=$ the constant term (at $𝑟=6$ )

$={}^{15}{𝐶}_6{𝑎}^9{𝑏}^6$

and $𝑛=$ coefficient of ${𝑥}^{-10}$ is (at $𝑟=8$ )

$={}^{15}{𝐶}_8{𝑎}^7{𝑏}^8$

$\begin{array}{cc}\because & \frac{𝑙}{𝑚}+\frac{𝑚}{𝑛}=\frac{26}{11}\\ \Rightarrow & \frac{{}^{15}{𝐶}_4{𝑎}^{11}{𝑏}^4}{{}^{15}{𝐶}_6{𝑎}^9{𝑏}^6}+\frac{{}^{15}{𝐶}_6{𝑎}^9{𝑏}^6}{{}^{15}{𝐶}_8{𝑎}^7{𝑏}^8}=\frac{26}{11}\\ \Rightarrow & \frac{6!9!{𝑎}^2}{4!11!{𝑏}^2}+\frac{8!7!{𝑎}^2}{6!9!{𝑏}^2}=\frac{26}{11}\\ \Rightarrow & \frac{6\times 5}{11\times 10}\left(\frac{{𝑎}^2}{{𝑏}^2}\right)+\frac{7}{9}\left(\frac{{𝑎}^2}{{𝑏}^2}\right)=\frac{26}{11}\\ \Rightarrow & \frac{{𝑎}^2}{{𝑏}^2}(\frac{3}{11}+\frac{7}{9})=\frac{26}{11}\\ \Rightarrow & \frac{{𝑎}^2}{{𝑏}^2}\left(\frac{27+77}{99}\right)=\frac{26}{11}\\ \Rightarrow & \frac{{𝑎}^2}{{𝑏}^2}=\frac{9\times 26}{104}=\frac{9}{4}\end{array}$

It is given that for a complex number $𝑧$,

$\begin{array}{cc}& (1+𝑧{)}^{𝑛}=1+{}^{𝑛}{𝐶}_1𝑧+{}^{𝑛}{𝐶}_2{𝑧}^2+{}^{𝑛}{𝐶}_3{𝑧}^3\\ & +\dots +{}^{𝑛}{𝐶}_{𝑛}{𝑧}^{𝑛}\end{array}$

Put $𝑧=\cos 𝑥+𝑖\sin 𝑥$

Then, $(1+\cos 𝑥+𝑖\sin 𝑥{)}^{𝑛}=1+{}^{𝑛}{𝐶}_1$

$\begin{array}{cc}& (\cos 𝑥+𝑖\sin 𝑥)+{}^{𝑛}{𝐶}_2(\cos 𝑥+𝑖\sin 𝑥{)}^2+\\ & \dots +{}^{𝑛}{𝐶}_{𝑛}(\cos 𝑥+𝑖\sin 𝑥{)}^{𝑛}\\ & \Rightarrow 2^{𝑛}{(\cos \frac{𝑥}{2})}^{𝑛}{(\cos \frac{𝑥}{2}+𝑖\sin \frac{𝑥}{2})}^{𝑛}\\ & =[1+{}^{𝑛}{𝐶}_1\cos 𝑥+{}^{𝑛}{𝐶}_2\cos (2𝑥)+{}^{𝑛}{𝐶}_3\cos (3𝑥)\\ & +\dots +{}^{𝑛}{𝐶}_{𝑛}\cos \left(𝑛𝑥\right)]+𝑖[{}^{𝑛}{𝐶}_1\sin 𝑥+{}^{𝑛}{𝐶}_2\sin (2𝑥)\\ & +{}^{𝑛}{𝐶}_3\sin \left(3𝑥\right)+\dots +{}^{𝑛}{𝐶}_{𝑛}\sin \left(𝑛𝑥\right)]\\ & \Rightarrow 2^{𝑛}{(\cos \frac{𝑥}{2})}^{𝑛}(\cos \frac{𝑛𝑥}{2}+𝑖\sin \frac{𝑛𝑥}{2})\\ & =[1+{}^{𝑛}{𝐶}_1\cos 𝑥+{}^{𝑛}{𝐶}_2\cos (2𝑥)+{}^{𝑛}{𝐶}_3\cos (3𝑥)\\ & +\dots +{}^{𝑛}{𝐶}_{𝑛}\cos \left(𝑛𝑥\right)]+𝑖[{}^{𝑛}{𝐶}_1\sin 𝑥+{}^{𝑛}{𝐶}_2\sin (2𝑥)\\ & +{}^{𝑛}{𝐶}_3\sin \left(3𝑥\right)+\dots +{}^{𝑛}{𝐶}_{𝑛}\sin \left(𝑛𝑥\right)]\end{array}$

On comparing imaginary parts, we get

$\begin{array}{cc}& {}^{𝑛}{𝐶}_1\sin 𝑥+{}^{𝑛}{𝐶}_2\sin \left(2𝑥\right)+{}^{𝑛}{𝐶}_3\sin \left(23𝑥\right)\\ & +\dots +{}^{𝑛}{𝐶}_{𝑛}\sin \left(𝑛𝑥\right)\\ & =2^{𝑛}{(\cos \frac{𝑥}{2})}^{𝑛}\sin \left(\frac{𝑛𝑥}{2}\right)\\ & \therefore \sum _{𝑟=0}^{100}100{𝐶}_{𝑟}\sin \left(𝑟𝑥\right)=2^{100}{(\cos \frac{𝑥}{2})}^{100}\\ & \sin \left(\frac{100𝑥}{2}\right)\\ & ={(2\cos \frac{𝑥}{2})}^{100}\sin \left(𝑘𝑥\right)\text{ (given) }\end{array}$

Therefore, $𝑘=50$.

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}& \frac{1}{{𝑥}^4+{𝑥}^2+1}=\frac{𝐴𝑥+𝐵}{{𝑥}^2+𝑥+1}+\frac{𝐶𝑥+𝐷}{{𝑥}^2-𝑥+1}\\ & \Rightarrow \frac{1}{{𝑥}^4+{𝑥}^2+1}=\frac{(𝐴𝑥+𝐵)({𝑥}^2-𝑥+1)+(𝐶𝑥+𝐷)({𝑥}^2+𝑥+1)}{{𝑥}^4+{𝑥}^2+1}\end{array}\end{array}$

$\begin{array}{cc}\Rightarrow l=(& 𝐴+𝐶){𝑥}^3+(𝐵-𝐴+𝐶+𝐷){𝑥}^2+(𝐴-𝐵+𝐶+𝐷)𝑥+(𝐵+𝐷)\end{array}$

On comparing the terms, we get

$\begin{array}{cc}& 𝐴+𝐶=0,𝐴=𝐵+𝐶+𝐷,\\ & 𝐴+𝐶+𝐷=𝐵\text{ and }𝐵+𝐷=1\\ & \therefore 𝐴+𝐵+𝐶+𝐷=1\end{array}$

Therefore, ${\cos }^{-1}(𝐴+𝐵+𝐶+𝐷)$

$={\cos }^{-1}\left(l\right)=0$