TG EAPCET 2020 10th September Morning Shift Question Paper with Solutions

We have,

$𝑓\left(𝑥\right)=⎧⎩⎨⎪⎪\begin{array}{cc}\frac{60-5𝑥}{3},& 0\le 𝑥\le 6\\ 10,& 6\le 𝑥\le 7\\ 31-3𝑥,& 7\le 𝑥\le 10\end{array}$

Now, $𝑓\left(𝑥\right)=10\forall 𝑥\in [6,7]$

So, $𝑓\left(𝑥\right)$ is not one-one.

Again,

$\begin{array}{cc}& 0\le 𝑥\le 6\\ \Rightarrow & -30\le -5𝑥\le 0\\ \Rightarrow & 60-30\le 60-5𝑥\le 60\\ \Rightarrow & \frac{30}{3}\le \frac{60-5𝑥}{3}\le \frac{60}{3}\\ \Rightarrow & 10\le \frac{60-5𝑥}{3}\le 20\text{ and }7\le 𝑥\le 10\\ \Rightarrow & -30\le -3𝑥\le -21\\ \Rightarrow & 1\le 31-3𝑥\le 10\end{array}$

∴ Range of $𝑓\left(𝑥\right)=[1,20]$

It is given that co-domain of $𝑓\left(𝑥\right)=[1,20]$

∴ Range of $𝑓\left(𝑥\right)=$ co-domain of $𝑓\left(𝑥\right)$

So, $𝑓\left(𝑥\right)$ is onto.

We have,

$𝑓\left(𝑥\right)=\sqrt{\surd {\log }_{10}\left(\frac{5𝑥-{𝑥}^2}{4}\right)}$

For $𝑓\left(𝑥\right)$ to be defined

$\begin{array}{cc}& {\log }_{10}\left(\frac{5𝑥-{𝑥}^2}{4}\right)\ge 0\text{ and }\frac{5𝑥-{𝑥}^2}{4}>0\\ \Rightarrow & \frac{5𝑥-{𝑥}^2}{4}\ge 1\text{ and }\frac{5𝑥-{𝑥}^2}{4}>0\\ \Rightarrow & \frac{5𝑥-{𝑥}^2}{4}\ge 1\\ \Rightarrow & 5𝑥-{𝑥}^2\ge 4\\ \Rightarrow & {𝑥}^2-5𝑥+4\le 0\\ \Rightarrow & (𝑥-4)(𝑥-1)\le 0\end{array}$

$\begin{array}{cc}& \therefore 𝑥\in [1,4]\\ & \text{ So, domain of }𝑓\left(𝑥\right)\text{ is }[1,4]\text{. }\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ We have, }\frac{𝑚}{𝑛}=\frac{1}{1\cdot 7}+\frac{1}{7\cdot 13}+\frac{1}{13\cdot 19}\\ & +\dots +\frac{1}{[1+(20-1\left)6\right][7+(20-1\left)6\right]}\\ & =\frac{1}{1\cdot 7}+\frac{1}{7\cdot 13}+\frac{1}{13\cdot 19}+\dots +\frac{1}{115\times 121}\\ & =\frac{1}{6}[\frac{6}{1\cdot 7}+\frac{6}{7\cdot 13}+\frac{6}{13\cdot 19}+\dots +\frac{6}{115\times 121}]\end{array}\end{array}$

$\begin{array}{cc}& =\frac{1}{6}[\frac{7-1}{1\times 7}+\frac{13-7}{7\times 13}+\frac{19-13}{13\times 19}+\dots +\frac{121-115}{115\times 121}]\\ & =\frac{1}{6}[(\frac{1}{1}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{13})+(\frac{1}{13}-\frac{1}{19})\\ & +\dots +(\frac{1}{115}-\frac{1}{121})]\\ & =\frac{1}{6}[1-\frac{1}{121}]=\frac{1}{6}\times \frac{120}{121}=\frac{20}{121}\\ & \begin{array}{cc}\therefore & 𝑚=20,𝑛=121\\ \therefore & 5𝑚+2𝑛=5\times 20+2\times 121\\ & =100+242=342\end{array}\end{array}$

It is given that the matrices $𝐴$ and $𝐵$ are non-singular of order 3 and $\left|𝐵\right|=𝑘,𝑎$ positive integer, so

$\begin{array}{cc}\mid {𝑘}^{-1}{𝐴}^{-1}\mid ={\left({𝑘}^{-1}\right)}^3\mid {𝐴}^{-1}\mid & =\frac{1}{{𝑘}^3\left|𝐴\right|}\\ & [\because \mid {𝐴}^{-1}\mid =\frac{1}{\left|𝐴\right|}]\end{array}$

$\begin{array}{c}=\frac{1}{\left|𝐵{|}^3\right|𝐴|}\\ \because \mid \mathrm{adj}\left({𝐴}^{-1}\right)\mid ={\mid {𝐴}^{-1}\mid }^{3-1}={\mid {𝐴}^{-1}\mid }^2=\frac{1}{|𝐴{|}^2}\end{array}$

$[\because |\mathrm{adj}\left(𝐴\right)|=|𝐴{|}^{𝑛-1}$, where $𝑛$ is the order of matrix $𝐴$ ]

Now, since it is given that $𝐵𝐴{𝐵}^{-1}=𝐼$

$\begin{array}{cc}& \Rightarrow 𝐴{𝐵}^{-1}={𝐵}^{-1}\\ & \therefore 𝐵{𝐴}^{𝑘}{𝐵}^{-1}=𝐵{𝐴}^{𝑘-1}\left(𝐴{𝐵}^{-1}\right)=𝐵{𝐴}^{𝑘-1}{𝐵}^{-1}\\ & =𝐵{𝐴}^{𝑘-2}\left(𝐴{𝐵}^{-1}\right)=𝐵{𝐴}^{𝑘-2}{𝐵}^{-1}\\ & \therefore 𝐵{𝐴}^{𝑘}{𝐵}^{-1}=𝐵𝐴{𝐵}^{-1}=𝐼\\ & \text{ and },𝐵\frac{\mathrm{adj}\left(𝐵\right)}{\left|𝐵\right|}=𝐵{𝐵}^{-1}=𝐼\end{array}$

Therefore, $𝐵{𝐴}^{𝑘}{𝐵}^{-1}=𝐵\frac{\mathrm{adj}\left(𝐵\right)}{\left|𝐵\right|}$, if

$𝐵𝐴{𝐵}^{-1}=𝐼$

Now, the adj $(\mathrm{adj}\left({𝐴}^{-1}\right))$

$={\mid {𝐴}^{-1}\mid }^{3-2}\left({𝐴}^{-1}\right)$

$[\because \mathrm{adj}(\mathrm{adj}𝐴)=|𝐴{|}^{𝑛-2}𝐴$ where $𝑛$ is the order of matrix $𝐴$ ]

$=\frac{1}{\left|𝐴\right|}\left({𝐴}^{-1}\right)$

We have,

$\begin{array}{c}2𝑥+𝑝𝑦+6𝑧=8\\ 𝑥+2𝑦+𝑞𝑧=5\\ 𝑥+𝑦+3𝑧=4\end{array}$

Now, if we express above equation in matrix form, We have,

$𝐴𝑋=𝐵$

Where,

$𝐴=⎡⎣⎢\begin{array}{ccc}2& 𝑝& 6\\ 1& 2& 𝑞\\ 1& 1& 3\end{array}⎤⎦⎥,𝐵=⎡⎣⎢\begin{array}{c}8\\ 5\\ 4\end{array}⎤⎦⎥,𝑋=⎡⎣⎢\begin{array}{c}𝑥\\ 𝑦\\ 𝑧\end{array}⎤⎦⎥$

For no solution $\left|𝐴\right|=0$

$\begin{array}{cc}& \Rightarrow \mid \begin{array}{ccc}2& 𝑝& 6\\ 1& 2& 𝑞\\ 1& 1& 3\end{array}\mid =0\\ & \Rightarrow 2(6-𝑞)-𝑝(3-𝑞)+6(1-2)=0\\ & \Rightarrow 12-2𝑞-3𝑝+𝑝𝑞-6=0\\ & \Rightarrow 3𝑝+2𝑞-𝑝𝑞-6=0\\ & \Rightarrow 3𝑝-6+2𝑞-𝑝𝑞=0\\ & \Rightarrow 3(𝑝-2)-𝑞(𝑝-2)=0\\ & \Rightarrow (𝑝-2)(3-𝑞)=0\Rightarrow \\ & \overrightarrow{𝑝}=2,3\end{array}$

But for $𝑝=2$ equations $2𝑥+𝑝𝑦+6𝑧=8$ and $𝑥+𝑦+3𝑧=4$ are same.

So, $𝑝\ne 2,𝑞=3$

We have,

$\begin{array}{c}(𝜆-1)𝑥+(3𝜆+1)𝑦+2𝜆𝑧=0\\ (𝜆-1)𝑥+(4𝜆-2)𝑦+(𝜆+3)𝑧=0\\ 2𝑥+(3𝜆+1)𝑦+3(𝜆-1)𝑧=0\end{array}$

Now, it can be express as $𝐴𝑋=0$

where, $𝐴=⎡⎣⎢\begin{array}{ccc}𝜆-1& 3𝜆+1& 2𝜆\\ 𝜆-1& 4𝜆-2& 𝜆+3\\ 2& 3𝜆+1& 3(𝜆-1)\end{array}⎤⎦⎥$ and $𝑋=⎡⎣⎢\begin{array}{c}𝑥\\ 𝑦\\ 𝑧\end{array}⎤⎦⎥$

For non-zero solution $\left|𝐴\right|=0$

$⎡⎣⎢\begin{array}{ccc}𝜆-1& 3𝜆+1& 2𝜆\\ 𝜆-1& 4𝜆-2& 𝜆+3\\ 2& 3𝜆+1& 3(𝜆-1)\end{array}⎤⎦⎥=0$

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

$⎡⎣⎢\begin{array}{ccc}𝜆-1& 3𝜆+1& 2𝜆\\ 0& 𝜆-3& -𝜆+3\\ -𝜆+3& 0& 𝜆-3\end{array}⎤⎦⎥=0$

$\begin{array}{cc}& \Rightarrow (𝜆-1)\left[\right(𝜆-3{)}^2]+(-𝜆+3)\left(\right(3𝜆+1)\\ & (-𝜆+3)-2𝜆(𝜆-3))=0\\ & \Rightarrow (𝜆-1)(𝜆-3{)}^2-(𝜆-3)\\ & \left[\right(𝜆-3\left)\right(-3𝜆-1-2𝜆\left)\right)=0\\ & \Rightarrow (𝜆-3{)}^2[𝜆-1+5𝜆+1]=0\\ & \Rightarrow (𝜆-3{)}^2(6𝜆)=0\Rightarrow 𝜆=3,0\\ & \therefore 𝑝=3,𝑞=0\\ & \therefore {𝑝}^2+{𝑞}^2-𝑝𝑞=9+0-0=9\end{array}$

We have, $\left|𝑧\right|\le 2$

$\Rightarrow \sqrt{\surd {𝑥}^2+{𝑦}^2}\le 2\Rightarrow {𝑥}^2+{𝑦}^2\le 4...\left(𝑖\right)$

Again, $(1-𝑖)𝑧+(1+𝑖)\overline{𝑧}\ge 4$

$\begin{array}{cc}& \Rightarrow (1-𝑖)(𝑥+𝑖𝑦)+(1+𝑖)(𝑥-𝑖𝑦)\ge 4\\ & \Rightarrow 𝑥+𝑖𝑦-𝑖𝑥+𝑦+𝑥-𝑖𝑦+𝑖𝑥+𝑦\ge 4\\ & \Rightarrow 2𝑥+2𝑦\ge 4\Rightarrow 𝑥+𝑦\ge 2...\left(𝑖𝑖\right)\end{array}$

So, $𝐴\cap 𝐵=\{{𝑥}^2+{𝑦}^2\le 4\}\cap \{𝑥+𝑦\ge 2\}$

$\therefore 𝑧=\sqrt{\surd 3}+\frac{1}{2}𝑖$

Given equation, ${𝑧}^2(1-{𝑧}^2)=16,𝑧\in 𝐂$

Now, let ${𝑧}^2=𝑤=𝑟(\cos 𝜃+𝑖\sin 𝜃)$

where $𝑟>0$.

$\begin{array}{cc}& \therefore 1-{𝑧}^2=\frac{16}{{𝑧}^2}\Rightarrow {𝑧}^2+\frac{16}{{𝑧}^2}=1\\ & \Rightarrow (𝑟+\frac{16}{𝑟})\cos 𝜃+𝑖(𝑟-\frac{16}{𝑟})\sin 𝜃=1\end{array}$

On comparing real and imaginary parts, we get

$(𝑟+\frac{16}{𝑟})\cos 𝜃=1\text{ and }(𝑟-\frac{16}{𝑟})\sin 𝜃=0$

∴ Either $\sin 𝜃=0$

$\Rightarrow \cos 𝜃=1\Rightarrow 𝑟+\frac{16}{𝑟}=1$

(not possible)

or $𝑟-\frac{16}{𝑟}=0\Rightarrow {𝑟}^2=16$

$\begin{array}{cc}\Rightarrow & 𝑟=4\\ \Rightarrow & |𝑧{|}^2=4\Rightarrow |𝑧|=2\end{array}$

Let $𝑧=𝑥+𝑖𝑦$

Then, vertices of rectangle are $(𝑥,𝑦)$, $(𝑥,-𝑦),(-𝑥,-𝑦),(-𝑥,𝑦)$.

Now, area of rectangle $=\left(2𝑥\right)\left(2𝑦\right)=4𝑥𝑦$

It is given that,

$\begin{array}{cccc}& & 4𝑥𝑦& =2\sqrt{\surd 3}\Rightarrow 2𝑥𝑦=\sqrt{\surd 3}\\ & \therefore & & 𝑥=\frac{1}{2},𝑦=\sqrt{\surd 3}\\ & \therefore & & 𝑧=\frac{1}{2}+\sqrt{\surd 3}𝑖\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& {\left(\frac{\cos 𝜃+𝑖\sin 𝜃}{\sin 𝜃+𝑖\cos 𝜃}\right)}^8+{\left(\frac{1+\cos 𝜃-𝑖\sin 𝜃}{1+\cos 𝜃+𝑖\sin 𝜃}\right)}^{16}\\ & ={\left(\frac{\cos 𝜃+𝑖\sin 𝜃}{𝑖(\cos 𝜃-𝑖\sin 𝜃)}\right)}^8+{\left(\frac{2{\cos }^2\frac{𝜃}{2}-𝑖2\sin \frac{𝜃}{2}\cos \frac{𝜃}{2}}{2{\cos }^2\frac{𝜃}{2}+𝑖2\sin \frac{𝜃}{2}\cos \frac{𝜃}{2}}\right)}^{16}\end{array}\end{array}$

$\begin{array}{cc}=& \frac{1}{{𝑖}^8}{\left(\frac{\cos 𝜃+𝑖\sin 𝜃}{\cos 𝜃-𝑖\sin 𝜃}\right)}^8+{\left(\frac{\cos \frac{𝜃}{2}-𝑖\sin \frac{𝜃}{2}}{\cos \frac{𝜃}{2}+𝑖\sin \frac{𝜃}{2}}\right)}^{16}\\ =& \frac{\cos 8𝜃+𝑖\sin 8𝜃}{\cos 8𝜃-𝑖\sin 8𝜃}+\frac{\cos 8𝜃-𝑖\sin 8𝜃}{\cos 8𝜃+𝑖\sin 8𝜃}\\ =& (\cos 8𝜃+𝑖\sin 8𝜃)(\cos 8𝜃-𝑖\sin 8𝜃{)}^{-1}\\ & +(\cos 8𝜃-𝑖\sin 8𝜃(\cos 8𝜃+𝑖\sin 8𝜃{)}^{-1}\\ =& (\cos 8𝜃+𝑖\sin 8𝜃)(\cos 8𝜃+𝑖(\sin 8𝜃)+\\ & (\cos 8𝜃-𝑖\sin 8𝜃)(\cos 8𝜃-𝑖\sin 8𝜃)\\ =& (\cos 8𝜃+𝑖\sin 8𝜃{)}^2+(\cos 8𝜃-𝑖\sin 8𝜃{)}^2\\ =& {\cos }^{2}8𝜃+{𝑖}^2{\sin }^{2}8𝜃+2𝑖\cos 8𝜃\sin 8𝜃\\ & +{\cos }^{2}8𝜃+{𝑖}^2{\sin }^{2}8𝜃-2𝑖\cos 8𝜃\sin 8𝜃\\ =& 2({\cos }^{2}8𝜃-{\sin }^{2}8𝜃)=2\cos 16𝜃\end{array}$

We have,

$\begin{array}{cc}& 𝜆{𝑥}^2+13𝑥+7=0\\ \therefore & 𝐷=(13{)}^2-4(𝜆\left)\right(7)=169-28𝜆\end{array}$

For rational roots, $𝐷=$ perfect square

So, $𝜆\in (-3,7)$ has values $-2,0,6$ so that $𝐷$ become perfect square.

∴ Required sum $=-2+0+6=4$

We have, $𝑓\left(𝑥\right)=1-2𝑥-5{𝑥}^2$

$\begin{array}{cc}& =-5[{𝑥}^2+\frac{2}{5}𝑥-\frac{1}{5}]\\ & =-5[{(𝑥+\frac{1}{5})}^2-\frac{1}{5}-\frac{1}{25}]\\ & =-5{(𝑥+\frac{1}{5})}^2+1+\frac{1}{5}\\ & =-5{(𝑥+\frac{1}{5})}^2+\frac{6}{5}\\ & \therefore 𝑓\left(𝑥\right)\le \frac{6}{5};\therefore 𝛼=\frac{6}{5}\\ & \text{ Again, }𝑔\left(𝑥\right)={𝑥}^2-2𝑥+𝑟\\ & =(𝑥-1{)}^2+𝑟-1\\ & \therefore 𝑔\left(𝑥\right)\ge (𝑟-1)\\ & \therefore 𝛽=𝑟-1\end{array}$

So, $5𝛼{𝑥}^2+𝛽𝑥+6=6{𝑥}^2+(𝑟-1)𝑥+6$

Since, $5𝛼{𝑥}^2+𝛽𝑥+6>0$ and coefficient of ${𝑥}^2>0$

$\begin{array}{cc}\therefore & 𝐷<0\\ \Rightarrow & (𝑟-1{)}^2-4\times 6\times 6<0\\ \Rightarrow & (𝑟+11)(𝑟-13)<0\\ \Rightarrow & 𝑟\in (-11,13)\end{array}$

$\begin{array}{cc}& \text{ Let }𝑝\left(𝑥\right)={𝑥}^4+{𝑥}^3-4{𝑥}^2+𝑥-1\\ & \begin{array}{c}=(𝑥-1)({𝑥}^3+2{𝑥}^2-2𝑥-1)\\ =(𝑥-1)(𝑥-1)({𝑥}^2+3𝑥+1)\\ =(𝑥-1)(𝑥-1)[𝑥-\left(\frac{-3+\sqrt{\surd 5}}{2}\right)]\\ [𝑥-\left(\frac{-3-\sqrt{\surd 5}}{2}\right)]\end{array}\end{array}$

∴ Roots of the equation, $𝑝\left(𝑥\right)=0$ are

$𝑥=1,1,\frac{-3+\sqrt{\surd 5}}{2},\frac{-3-\sqrt{\surd 5}}{2}$

∴ Required ratio

$\begin{array}{cc}& =\frac{(1{)}^2+(1{)}^2+{\left(\frac{-3+\sqrt{\surd 5}}{2}\right)}^2+{\left(\frac{-3-\sqrt{\surd 5}}{2}\right)}^2}{1\times \left(\frac{-3+\sqrt{\surd 5}}{2}\right)\left(\frac{-3-\sqrt{\surd 5}}{2}\right)}\\ & =\frac{1+1+\frac{5+9-6\sqrt{\surd 5}}{4}+\frac{5+9+6\sqrt{\surd 5}}{4}}{\frac{9-5}{4}}\\ & =\frac{8+5+9-6\sqrt{\surd 5}+5+9+6\sqrt{\surd 5}}{4}=\frac{36}{4}=9\\ & \therefore \text{ Ratio }=9:1\end{array}$

Given,

${𝛼}_1,{𝛽}_1,{𝛾}_1,{𝛿}_1$ are the roots of equation

$𝑎{𝑥}^4+𝑏{𝑥}^3+𝑐{𝑥}^2+𝑑𝑥+𝑒=0\text{ and }$

${𝛼}_2,{𝛽}_2,{𝛾}_2,{𝛿}_2$ are the roots of equation

$𝑒{𝑥}^4+𝑑{𝑥}^3+𝑐{𝑥}^2+𝑏𝑥+𝑎=0$

∴ Clearly,

${𝛼}_1,{𝛽}_1,{𝛾}_1,{𝛿}_1$ are the reciprocal of roots of

$𝑎{𝑥}^4+𝑏{𝑥}^3+𝑐{𝑥}^2+𝑑𝑥+𝑒=0$

Also, ${𝛿}_1>{𝛾}_1>{𝛽}_1>{𝛼}_1>0$

and ${𝛿}_2>{𝛾}_2>{𝛽}_2>{𝛼}_2>0$

$\therefore {𝛿}_1$ and ${𝛼}_2$ are reciprocal

Similarly ${𝛾}_1$ and ${𝛽}_2$,

${𝛾}_2$ and ${𝛽}_1$ and ${𝛼}_1$ and ${𝛿}_2$ are reciprocal

$\begin{array}{cc}& \therefore {𝛼}_1{𝛿}_2=1,{𝛽}_1{𝛾}_2=1,{𝛽}_2{𝛾}_1=1,{𝛼}_2{𝛿}_1=1\end{array}$

$\begin{array}{cc}& {𝛼}_1-{𝛿}_2=2\\ & \therefore {𝛼}_1-\frac{1}{{𝛼}_1}=2\Rightarrow {𝛼}_2^1-2{𝛼}_1-1=0\\ & \text{ and }{𝛿}_1-{𝛼}_2=4\\ & \\ & {𝛿}_1-\frac{1}{{𝛿}_1}=4={𝛿}_2^1-4{𝛿}_1-1=0\end{array}$

$\therefore {𝛼}_1$ and ${𝛿}_1$ are roots of

$\begin{array}{c}𝑎{𝑥}^4+𝑏{𝑥}^3+𝑐{𝑥}^2+𝑑𝑥+𝑒=0\\ \therefore ({𝑥}^2-2𝑥-1)({𝑥}^2-4𝑥-1)\\ =𝑎{𝑥}^4+𝑏{𝑥}^3+𝑐{𝑥}^2+𝑑𝑥+𝑒\end{array}$

Put $𝑥=1$,

$\begin{array}{c}(1-2-1)(1-4-1)=𝑎+𝑏+𝑐+𝑑+𝑐\\ 8=𝑎+𝑏+𝑐+𝑑+𝑒\end{array}$

Total number of three digit numbers $=5\times 5\times 4=100$

Total number of five digit numbers

$=5\times 5\times 4\times 3\times 2=600$

∴ Total number of required numbers

$=100+600=700$

$\begin{array}{cc}& \text{ Number of ways of selections }\\ & ={}^{12}{𝐶}_1+{}^{12}{𝐶}_2+{}^{12}{𝐶}_3+{}^{12}{𝐶}_4\end{array}$

$\begin{array}{cc}& =12+\frac{12\times 11}{2\times 1}+\frac{12\times 11\times 10}{3\times 2\times 1}+\frac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}\\ & =12+66+220+495=793\end{array}$

We have,

$(1-𝑥{)}^3=1-3𝑥+3{𝑥}^2$ [neglecting

higher term]

$(2+𝑥{)}^6={}^6{𝐶}_0{2}^6{𝑥}^0+{}^6{𝐶}_1{2}^5{𝑥}^1+{}^6{𝐶}_2{2}^4{𝑥}^2$

[neglecting higher term]

$=64+192𝑥+240{𝑥}^2$

Now, $(1-𝑥{)}^3(2+𝑥{)}^6=(1-3𝑥+3{𝑥}^2)$

$\begin{array}{c}(64+192𝑥+240{𝑥}^2)\\ =64+192𝑥+240{𝑥}^2-192𝑥\\ -576{𝑥}^2+192{𝑥}^2\end{array}$

(neglecting higher term)

$\begin{array}{cc}& =-144{𝑥}^2+64\\ \therefore & 𝑎=64,𝑏=0,𝑐=-144\\ \therefore & 𝑎+𝑏+𝑐=64+0-144=-80\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ Given, }\\ & {(1+\frac{1}{𝑥})}^{1/2}\\ & =1+\frac{1}{2}\left(\frac{1}{𝑥}\right)+\frac{\left(\frac{1}{2}\right)(\frac{1}{2}-1)}{2!}{\left(\frac{1}{𝑥}\right)}^2\\ & +\frac{\left(\frac{1}{2}\right)(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}{\left(\frac{1}{𝑥}\right)}^3\\ & +\frac{\left(\frac{1}{2}\right)(\frac{1}{2}-1)(\frac{1}{2}-2)(\frac{1}{2}-3)}{4!}{\left(\frac{1}{𝑥}\right)}^4+\dots \\ & =1+\frac{1}{2𝑥}-\frac{1}{2!}{\left(\frac{1}{2𝑥}\right)}^2+\frac{1\cdot 3}{3!}{\left(\frac{1}{2𝑥}\right)}^3\\ & -\frac{1\cdot 3\cdot 5}{4!}{\left(\frac{1}{2𝑥}\right)}^4+\dots .\infty \end{array}\end{array}$

We have,

$\begin{array}{cc}& \frac{4{𝑥}^2+5{𝑥}^4+7}{({𝑥}^2+1)({𝑥}^4+{𝑥}^2+1)}=\frac{𝐴𝑥+𝐵}{{𝑥}^2+1}\\ & +\frac{𝐶{𝑥}^3+𝐷{𝑥}^2+𝐸𝑥+𝐹}{{𝑥}^4+{𝑥}^2+1}\\ & \Rightarrow 4{𝑥}^2+5{𝑥}^4+7=(𝐴𝑥+𝐵)({𝑥}^4+{𝑥}^2+1)\\ & +(𝐶{𝑥}^3+𝐷{𝑥}^2+𝐸𝑥+𝐹)({𝑥}^2+1)\\ & \Rightarrow 5{𝑥}^4+4{𝑥}^2+7=𝐴{𝑥}^5+𝐴{𝑥}^3+𝐴𝑥\\ & +𝐵{𝑥}^4+𝐵{𝑥}^2+𝐵+𝐶{𝑥}^5+𝐶{𝑥}^3+𝐷{𝑥}^4\\ & +𝐷{𝑥}^2+𝐸{𝑥}^3+𝐸𝑥+𝐹{𝑥}^2+𝐹\\ & \Rightarrow 5{𝑥}^4+4{𝑥}^2+7\\ & =(𝐴+𝐶){𝑥}^5+(𝐵+𝐷){𝑥}^4+(𝐴+𝐶+𝐸){𝑥}^3\\ & +(𝐵+𝐷+𝐹){𝜒}^2+(𝐴+𝐸)𝜒+(𝐵+𝐹)\end{array}$