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

Let $𝑥$ and $𝑦$ be any two elements in the domain (Z), such that

$𝑓(𝑥+𝑦)=𝑓\left(𝑥\right)+𝑓\left(𝑦\right)...\left(𝑖\right)$

Differentiating above expression w.r.t. $𝑦$, keeping $𝑥$ constant, we get

${𝑓}^{\prime }(𝑥+𝑦)={𝑓}^{\prime }\left(𝑦\right)$

Let $𝑦=0\Rightarrow {𝑓}^{\prime }(𝑥+0)={𝑓}^{\prime }\left(0\right)$

and Assume ${𝑓}^{\prime }\left(0\right)=𝐾$

$\therefore {𝑓}^{\prime }\left(𝑥\right)=𝐾$

Integrating on both sides, we get

$\begin{array}{cc}& 𝑓\left(𝑥\right)=𝐾𝑥+𝐶\\ & \{𝐶\to \text{ integration constant }\}...\left(𝑖𝑖\right)\end{array}$

Now putting $𝑥=𝑦=0$ in Eq. (i), we get $𝑓(0+0)=𝑓\left(0\right)+𝑓\left(0\right)\Rightarrow 𝑓\left(0\right)=2𝑓\left(0\right)$ or $𝑓\left(0\right)=0$

Let $𝑥=0$, from Eq. (ii)

$\begin{array}{cc}𝑓\left(0\right)& =𝐾\times 0+𝐶\text{ or }0=0+𝐶\text{ or }𝐶=0\\ \therefore 𝑓\left(𝑥\right)& =𝑘𝑥\end{array}$

Case (i) If $𝑘>0$, then $𝑓\left(𝑥\right)$ is strictly increasing.

Case (ii) If $𝑘<0$, then $𝑓\left(𝑥\right)$ is strictly decreasing

So, function is injective

Also $𝑓\left(𝑥\right)$ where every element in the codomain is a valid output of the function i.e., range is equal to codomain.

So, function is surjective also. Therefore, there are two bijective functions.

Given that, ${𝐴}_{𝑛}=\left\{\right(𝑛+1)𝑘\mid 𝑘\in 𝐍\}$

If $𝑛=1,{𝐴}_1=\{2𝑘\mid 𝑘\in 𝐍\}$

for, $𝑘=1,2,3\dots {𝐴}_1=\{2,4,6,8,\dots \}$

If $𝑛=2,{𝐴}_2=\{3𝑘\mid 𝑘\in 𝐍\}$

for $𝑘=1,2,3,\dots {𝐴}_2=\{3,6,9,\dots \}$

If $𝑛=3,{𝐴}_3=\{4𝑘\mid 𝑘\in 𝐍\}$

for $𝑘=1,2,3,\dots {𝐴}_3=\{4,8,12,\dots \}$

Now, $𝑋=\bigcup _{𝑛\in 𝐍}{𝐴}_{𝑛}$

or $𝑋={𝐴}_1\cup {𝐴}_2\cup {𝐴}_3\dots$

or $𝑋=\{2,3,4,5,6,\dots \}$

Here, $𝑓:𝑋\to 𝑁$ and $𝑓\left(𝑥\right)=𝑥\forall 𝑥\in 𝐗$

$\because 𝑓\left(𝑥\right)$ is linear function i.e., it is one-one function but there is no value of $𝑥$ in domain where $𝑓\left(𝑥\right)=1$

$\therefore 𝑓\left(𝑥\right)$ is not onto function.

Given that,

$\begin{array}{c}\begin{array}{cc}& (𝑥-𝑛)({({𝑥}^2-2𝑛𝑥)}^2+(2{𝑛}^2-5)({𝑥}^2-2𝑛𝑥)\\ & +({𝑛}^4-5{𝑛}^2+4))=0\\ & \Rightarrow (𝑥-𝑛)({𝑥}^4-4𝑛{𝑥}^3+4{𝑛}^2{𝑥}^2+2{𝑛}^2{𝑥}^2\\ & -4{𝑛}^3𝑥-5{𝑥}^2+10𝑛𝑥+{𝑛}^4-5{𝑛}^2+4)=0\\ & \Rightarrow (𝑥-𝑛)({𝑥}^4-4𝑛{𝑥}^3+6{𝑛}^2{𝑥}^2-4{𝑛}^3𝑥\\ & -5{𝑥}^2+10𝑛𝑥+{𝑛}^4-5{𝑛}^2+4)=0\\ & \Rightarrow {𝑥}^5-4𝑛{𝑥}^4+6{𝑛}^2{𝑥}^3-4{𝑛}^3{𝑥}^2\\ & -5{𝑥}^3+10𝑛{𝑥}^2+{𝑛}^4𝑥-5{𝑛}^2𝑥+4𝑥-{𝑥}^4𝑛\\ & +4{𝑛}^2{𝑥}^3-6{𝑛}^3{𝑥}^2+4{𝑛}^4𝑥\\ & +5{𝑥}^2𝑛-10{𝑛}^2𝑥-{𝑛}^5+5{𝑛}^3-4𝑛=0\\ & \Rightarrow {𝑥}^5-5𝑛{𝑥}^4+5(2{𝑛}^2-1){𝑥}^3\\ & +5𝑛(3-2{𝑛}^2){𝑥}^2\end{array}\\ +(5{𝑛}^4-15{𝑛}^2+4)𝑥-𝑛({𝑛}^4-5{𝑛}^2+4)=0\end{array}$

Now, product of roots,

$\begin{array}{cc}𝑃& =-\frac{\text{ Constant term }}{\text{ Coefficient of }{𝑥}^5}\\ & =-\frac{-𝑛({𝑛}^4-5{𝑛}^2+4)}{1}\end{array}$

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

Given that, $𝑛>2$

So, putting $𝑛=3$,

$\begin{array}{cc}𝑃& =3(3-1)(3+1)(3-1)(3+2)\\ & =3\left(2\right)\left(4\right)\left(2\right)\left(5\right)\\ 𝑃& =120\end{array}$

Putting $𝑛=4$,

$\begin{array}{cc}𝑃& =4(4-1)(4+1)(4-1)(4+2)\\ & =4\left(3\right)\left(5\right)\left(3\right)\left(6\right)=1080\end{array}$

Therefore, products of roots is always divisible by 120 .

Given that,

${𝑎}_{𝑖𝑗}=\{\begin{array}{cc}1,& \text{ if }𝑖+𝑗=7\\ 0,& \text{ if }𝑖+𝑗\ne 7\end{array}$

Now, $\left(𝐴\right(\mathrm{adj}𝐴\left){𝐴}^{-1}\right){𝐴}^2$

$\begin{array}{cc}=\left(𝐴\right(\mathrm{Adj}𝐴\left){𝐴}^{-1}\right){𝐴}^2& \\ =\left(\right(\left|𝐴\right|𝐼\left){𝐴}^{-1}\right){𝐴}^2& [\because 𝐴\mathrm{adj}𝐴=|𝐴\left|𝐼\right]\\ =-1\left({𝐴}^{-1}𝐴\right)𝐴& [\because |𝐴|=-1]\\ =-𝐼𝐴& [\because {𝐴}^{-1}𝐴=𝐼]\\ =-𝐴& \end{array}$

Given system of equations has a non-trivial solution

$\begin{array}{cc}& {\Pi }_1\equiv 𝑥+𝑎𝑦+𝑎𝑧=0\\ & {\Pi }_2\equiv 𝑏𝑥+𝑦+𝑏𝑧=0\\ & {\Pi }_3\equiv 𝑐𝑥+𝑐𝑦+𝑧=0\end{array}$

It can be written as $𝐴𝑋=0$

Where, $𝐴=⎡⎣⎢\begin{array}{ccc}1& 𝑎& 𝑎\\ 𝑏& 1& 𝑏\\ 𝑐& 𝑐& 1\end{array}⎤⎦⎥$

So, the system has non-trivial solutions hence it has infinitely many solutions or unique solutions.

$\therefore \left|𝐴\right|=0$

Now, let the system has infinitely many solutions

So, let $𝑦=𝑡;𝑡\in 𝐑$

Then from Eqs. (i) and (ii), we get

$𝑥+𝑎𝑧=-𝑎𝑡\Rightarrow 𝑏𝑥+𝑏𝑧=-𝑡$

Solving these, we get

$𝑧=\frac{𝑡(1-𝑎𝑏)}{𝑏(𝑎-1)},𝑥=\frac{𝑎}{𝑏}𝑡\frac{(1-𝑏)}{(1-𝑎)}$

Hence, if $𝑎\ne 𝑏\ne 𝑐$, then

$\begin{array}{cc}& 𝑥=\frac{𝑎}{𝑏}𝑡\frac{(1-𝑏)}{(1-𝑎)},𝑦=𝑡\\ & 𝑧=\frac{𝑡}{𝑏}\frac{(1-𝑎𝑏)}{(𝑎-1)},𝑡\in 𝐑\end{array}$

If $𝑎=𝑏=𝑐$

Let $𝑦={𝑡}_1,𝑧={𝑡}_2,{𝑡}_1,{𝑡}_2\in 𝐑$

Then by Eq. (i)

$𝑥=-𝑎({𝑡}_1+{𝑡}_2)$

So, the system of equations has unique solution.

Given that, Rank of matrix $𝐴=$ rank of matrix $𝐵$

$𝑓\left(𝐴\right)=𝑓\left(𝐵\right)\text{ and }𝑓\left(𝐵\right)=3$

∴ Order of matrix $𝐵\ge$ Rank of matrix $𝐵$.

Given $\mathrm{Re}\left(\frac{𝑧-2}{𝑧-4𝑖}\right)=0$ and

$\begin{array}{cc}& 𝑚\left(\frac{𝑧-2}{𝑧-4𝑖}\right)=1\\ & \Rightarrow \frac{𝑧-2}{𝑧-4𝑖}=𝑖\Rightarrow 𝑧-2=𝑖(𝑧-4𝑖)\\ & \Rightarrow 𝑧-2=𝑖𝑧+4\Rightarrow 𝑧(1-𝑖)=6\\ & \Rightarrow 𝑧=\frac{6}{1-𝑖}\Rightarrow 𝑧=3(1+𝑖)\end{array}$

∴ Only one $𝑧$ in Argand plane.

$\begin{array}{cc}& \text{ Circle }𝑧\overline{𝑧}+𝑎𝑧+\overline{𝑎}\overline{𝑧}-4=0\\ & [\because 𝑧\overline{𝑧}=|𝑧{|}^2]\\ & \Rightarrow |𝑧{|}^2+𝑎𝑧+(\overline{𝑎𝑧})-4=0\\ & [\because {\overline{𝑧}}_1{\overline{𝑧}}_2=\overline{{𝑧}_1{𝑧}_2}]\\ & \Rightarrow |𝑧{|}^2+2\cdot \mathrm{Re}(𝑎𝑧)-4=0...(𝑖)\\ & [𝑧+\overline{𝑧}=2\mathrm{Re}(𝑧\left)\right]\end{array}$

$\begin{array}{cc}& \text{ Now, given }𝑧=𝑥+𝑖𝑦\Rightarrow |𝑧{|}^2={𝑥}^2+𝑦\\ & \text{ and }𝑎=1+𝑖\text{, then }𝑎𝑧=(1+𝑖)(𝑥+𝑖𝑦\\ & \therefore 𝑎𝑧=(𝑥-𝑦)+𝑖(𝑥+𝑦)\\ & \Rightarrow \mathrm{Re}\left(𝑎𝑧\right)=𝑥-𝑦\\ & \therefore \text{ From Eq. (i), put }\mathrm{Re}\left(𝑎𝑧\right)=𝑥-𝑦\\ & \text{ and }|𝑧{|}^2={𝑥}^2+{𝑦}^2\\ & \Rightarrow {𝑥}^2+{𝑦}^2+2(𝑥-𝑦)-4=0\\ & {𝑥}^2+{𝑦}^2+2𝑥-2𝑦-4=0...\left(𝑖𝑖\right)\end{array}$

is a circle and given line

$\begin{array}{cc}& 𝐿:𝑧+\overline{𝑧}-𝑖(𝑧-\overline{𝑧})+2=0\\ \Rightarrow & 𝐿:2\mathrm{Re}\left(𝑧\right)-𝑖2\mathrm{Im}\left(𝑧\right)+2=0\\ \Rightarrow & 𝐿:2𝑥-𝑖\left(2𝑦𝑖\right)+2=0\\ \Rightarrow & 𝐿:𝑥+𝑦+1=0...\left(𝑖𝑖𝑖\right)\end{array}$

Now, equation of circle passing through point of intersection of circle $𝑆$ and line $𝐿$ is

$\begin{array}{cc}𝑆+𝜆𝐿=0({𝑥}^2& +{𝑦}^2+2𝑥-2𝑦-4)\\ & +𝜆(𝑥+𝑦+1)=0...\left(𝑖𝑣\right)\end{array}$

Pass through $(0,0)$

∴ Put in Eq. (iv), we get

$-4+𝜆=0\Rightarrow 𝜆=4$

Now, required circle is

$\begin{array}{cc}\Rightarrow & {𝑥}^2+{𝑦}^2+2𝑥-2𝑦-4\\ & +4(𝑥+𝑦+1)=0\\ \Rightarrow & {𝑥}^2+{𝑦}^2+6𝑥+2𝑦=0\end{array}$

Let $𝑧=\sqrt{\surd 3}+𝑖\Rightarrow \left|𝑧\right|=\sqrt{\surd 3+1}=2$ and $\arg \left(𝑧\right)={\tan }^{-1}\frac{1}{\sqrt{\surd 3}}=\frac{𝜋}{6}$

∴ Polar form of $𝑧=2(\cos \frac{𝜋}{6}+𝑖\sin \frac{𝜋}{6})$

$\begin{array}{cc}& \Rightarrow {𝑧}^{10}=2^{10}{(\cos \frac{𝜋}{6}+𝑖\sin \frac{𝜋}{6})}^{10}\\ & \Rightarrow {𝑧}^{10}=2^{10}(\cos \frac{10𝜋}{6}+𝑖\sin \frac{10𝜋}{6})\\ & \Rightarrow (\sqrt{\surd 3}+𝑖{)}^{10}=2^{10}(\frac{1}{2}-𝑖\frac{\sqrt{\surd 3}}{2})\\ & \Rightarrow (\sqrt{\surd 3}+𝑖{)}^{10}=512-𝑖512\sqrt{\surd 3}=𝑎+𝑖𝑏\\ & \Rightarrow 𝑎=512\text{ and }𝑏=-512\sqrt{\surd 3}\end{array}$

$\begin{array}{cc}& \text{ Given, }{𝑧}^2+𝑧+1=0\\ & \begin{array}{cc}& \therefore 𝑧=\frac{-1\pm \sqrt{\surd 1-4}}{2}=\frac{-1\pm \sqrt{\surd 3}𝑖}{2}\\ & \therefore 𝑧=𝜔,{𝜔}^2\\ & \text{ Now }𝑧+\frac{1}{𝑍}=𝜔+\frac{1}{𝜔}=𝜔+{𝜔}^2=-1\\ & {𝑧}^2+\frac{1}{{𝑧}^2}={𝜔}^2+\frac{1}{{𝜔}^2}={𝜔}^2+𝜔=-1\\ & \therefore {(𝑧+\frac{1}{𝑧})}^3+{({𝑧}^2+\frac{1}{{𝑧}^2})}^3+\\ & \dots +{({𝑧}^{2020}+\frac{1}{{𝑧}^{2020}})}^3\\ & \Rightarrow (-1{)}^3+(-1{)}^3+\dots +(-1{)}^3=-2020\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }𝛼,𝛽\text{ are roots of }\\ & \begin{array}{cc}& 𝑎{𝑥}^2+𝑏𝑥+𝑐=0\\ & \Rightarrow 𝑎{𝛼}^2+𝑏𝛼+𝑐=0\\ & \text{ and }𝑎{𝛽}^2+𝑏𝛽+𝑐=0\\ & 𝛼(𝑎𝛼+𝑏)=-𝑐\text{ and }\\ & 𝛽(𝑎𝛽+𝑏)=-𝑐\\ & 𝑎𝛼+𝑏=-\frac{𝑐}{𝛼}\Rightarrow 𝑎𝛽+𝑏=-\frac{𝑐}{𝛽}\end{array}\\ & \begin{array}{cc}& \mathrm{Now}{\left(\frac{𝛼}{𝑎𝛽+𝑏}\right)}^3-{\left(\frac{𝛽}{𝑎𝛼+𝑏}\right)}^3\\ & ={\left(\frac{𝛼}{-𝑐/𝛽}\right)}^3-{\left(\frac{𝛽}{-𝑐/𝛼}\right)}^3\\ & =-{\left(\frac{𝛼𝛽}{𝑐}\right)}^3+{\left(\frac{𝛼𝛽}{𝑐}\right)}^3=0\end{array}\end{array}$

Given $\{𝑥\in 𝐑:\sqrt{\surd 𝑥+2}>\sqrt{\surd 8-{𝑥}^2}\}$

∴ Defining the function

$\begin{array}{cc}𝑥+2\ge 0& \Rightarrow 𝑥\ge -2...\left(𝑖\right)\\ \text{ and }8-{𝑥}^2\ge 0& \Rightarrow 𝑥\in [-2\sqrt{\surd 2},2\sqrt{\surd 2}]....\left(𝑖𝑖\right)\end{array}$

Now, $\sqrt{\surd 𝑥+2}>\sqrt{\surd 8-{𝑥}^2}$

∴ Squaring on both sides,

$\begin{array}{cc}& 𝑥+2>8-{𝑥}^2\\ \Rightarrow & {𝑥}^2+𝑥-6>0\\ \Rightarrow & (𝑥+3)(𝑥-2)>0\\ \Rightarrow & 𝑥\in (-\infty ,-3)\cup (2,\infty )...\left(𝑖𝑖𝑖\right)\end{array}$

∴ According to Eqs. (i), (ii) and (iii)

$\Rightarrow 𝑥\in [2,2\sqrt{\surd 2}]$

∴ Maximum value of $𝑥=2\sqrt{\surd 2}$.

Given ${𝑛}_1,{𝑛}_2,{𝑛}_3$ are roots of the equation

$\begin{array}{c}{𝐸}_1:{𝑥}^3+{𝑥}^2+𝑙𝑥+𝑛=0...\left(𝑖\right)\\ \therefore \text{ Sum of roots }{𝑥}_1+{𝑥}_2+{𝑥}_3=-1...\left(𝑖𝑖\right)\end{array}$

Now, ${𝐸}_2:{𝑥}^3+𝑎{𝑥}^2+𝑏𝑥+𝑐=0$ is reciprocal equation of class one $\Rightarrow 𝑐=1$ and $𝑎=𝑏$

$\therefore {𝐸}_2:{𝑥}^3+𝑎{𝑥}^2+𝑎𝑥+1=0...\left(𝑖𝑖𝑖\right)$

Given, Eq. (iii) have roots

$\begin{array}{cc}& \frac{{𝑥}_1-1}{2},\frac{{𝑥}_2-1}{2},\frac{{𝑥}_3-1}{2}\\ \Rightarrow & \frac{{𝑥}_1-1}{2}+\frac{{𝑥}_2-1}{2}+\frac{{𝑥}_3-1}{2}=-𝑎\\ \Rightarrow & \frac{{𝑥}_1+{𝑥}_2+{𝑥}_3}{2}-\frac{3}{2}=-𝑎\end{array}$

From Eq. (ii),

$\begin{array}{cc}& {𝑥}_1+{𝑥}_2+{𝑥}_3=-1\\ & \therefore \frac{-1}{2}-\frac{3}{2}=-𝑎\Rightarrow 𝑎=2\\ & \therefore \text{ Eq. }\left(\text{ iii }\right),{𝑥}^3+2{𝑥}^2+2𝑥+1=0\\ & (𝑥+1)({𝑥}^2+𝑥+1)=0\therefore 𝑥=-1,\\ & {𝑥}^2+𝑥+1=0\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑥=\frac{-1\pm \sqrt{\surd 3}𝑖}{2}\\ & \therefore \text{ Given, }\frac{{𝑥}_1-1}{2}=-1\Rightarrow {𝑥}_1=-1\\ & \frac{{𝑥}_2-1}{2}=\frac{-1+\sqrt{\surd 3}𝑖}{2}\\ & \Rightarrow {𝑥}_2=\sqrt{\surd 3}𝑖\\ & \frac{{𝑥}_3-1}{2}=\frac{-1-\sqrt{\surd 3}𝑖}{2}\\ & \Rightarrow {𝑥}_3=-\sqrt{\surd 3}𝑖\end{array}$

∴ Roots of these two equation excluding common roots are

$\sqrt{\surd 3}𝑖,-\sqrt{\surd 3}𝑖,\frac{-1+\sqrt{\surd 3}𝑖}{2},\frac{-1-\sqrt{\surd 3}𝑖}{2}$

Given $𝛼,𝛽,𝛾,𝛿$ are roots of

$\begin{array}{cc}& {𝑥}^4+{𝑥}^2+1=0\\ & \therefore {𝑥}^2=\frac{-1\pm \sqrt{\surd 1-4}}{2}=\frac{-1\pm \sqrt{\surd 3}𝑖}{2}\\ & \therefore 𝑥=\pm \sqrt{\surd \frac{1+\sqrt{\surd 3}𝑖}{2}},𝑥=\pm \sqrt{\surd \frac{-1-\sqrt{\surd 3}𝑖}{2}}\\ & \text{ Let }𝛼=\sqrt{\surd \frac{1+\sqrt{\surd 3}𝑖}{2}},𝛽=-\sqrt{\surd \frac{1+\sqrt{\surd 3}𝑖}{2}},\\ & 𝛾=\sqrt{\surd \frac{-1-\sqrt{\surd 3}𝑖}{2}}\\ & 𝛿=-\sqrt{\surd \frac{-1-\sqrt{\surd 3}𝑖}{2}}\\ & \Rightarrow {𝛼}^3+{𝛽}^3+{𝛾}^3+{𝛿}^3=0\end{array}$

∴ Given expression,

$\frac{{𝛼}^3+{𝛽}^3+{𝛾}^3+{𝛿}^3}{{𝛼}^6+{𝛽}^6+{𝛾}^6+{𝛿}^6}=0$

Given,

$\begin{array}{cc}𝐴& =\{{𝑎}_{𝑛}/{𝑎}_{𝑛}=\left[\begin{array}{cc}(-1{)}^{𝑛/2}\left(\frac{𝑛}{2}\right),& 𝑛\text{ is even }\\ (-1{)}^{\frac{𝑛-1}{2}}\left(\frac{𝑛-1}{2}\right),& 𝑛\text{ is odd }\end{array}\right\}\\ & \therefore 𝐴=\{0,-1,2,-3,4,-5,\dots -25\}\\ & \therefore 𝐵=\{0,-1,2,-3,4,-5,\dots ,24,-25\}\end{array}$

Here given integer $=12$, total integer $=26$

∴ Required permutation $=\frac{26!}{12!}$

Given all men are together

∴ Linear arrangement of $𝑝$ men and $𝑞$ women $=𝑝!(𝑞+1)$ ! and circular arrangement of $𝑝$ men and $𝑞$ women $=𝑝!𝑞!$

$\Rightarrow \frac{𝛼}{𝛽}=\frac{𝑝!(𝑞+1)!}{𝑝!𝑞\mid !}=(𝑞+1):1$

Given $𝑛=10$, then $\sum _{10}^{𝑟=1}{}^{𝑛}{𝐶}_{𝑟}𝑟(𝑟-4)$

$\begin{array}{cc}& =\sum _{𝑟=1}^{10}{}^{𝑛}{𝐶}_{𝑟}𝑟\left(\right(𝑟-1)-3)\\ & =\sum _{𝑟=1}^{10}\left(𝑟\right(𝑟-1){}^{𝑛}{𝐶}_{𝑟}-3𝑟{}^{𝑛}{𝐶}_{𝑟})\\ & =\sum _{𝑟=1}^{10}\left(𝑟\right(𝑟-1)\frac{𝑛(𝑛-1)}{𝑟(𝑟-1)}{}^{𝑛-2}{𝐶}_{𝑟-2}-3\cdot 𝑟\frac{𝑛}{𝑟}\cdot {}^{𝑛-1}{𝐶}_{𝑟-1})\\ & =\sum _{𝑟=1}^{10}\left(𝑛\right(𝑛-1){}^{𝑛-2}{𝐶}_{𝑟-2}-3𝑛\cdot {}^{𝑛-1}{𝐶}_{𝑟-1})\\ & =𝑛(𝑛-1)\sum _{𝑟=1}^{10}{}^{𝑛-2}{𝐶}_{𝑟-2}-3𝑛\sum _{𝑟=1}^{10}{}^{𝑛-1}{𝐶}_{𝑟-1}\end{array}$

Given, $𝑛=10$ and $\sum _{10}^{𝑟=1}{}^{𝑛}{𝐶}_{𝑟}=2^{𝑛}$

$\begin{array}{cc}& \therefore 10(10-1)\cdot 2^{10-2}-3\times 10\cdot 2^{10-1}\\ & =10\times 9\times 2^8-30\times 2^9\\ & =(90-60)\cdot 2^8\\ & =30\times 256=7680\end{array}$

Given sum of coefficient in expansion of ${(1+3𝑥-2{𝑥}^2)}^{𝑛}$ is 128

∴ For sum of coefficient, putting $𝑛=1$ in the expression ${(1+3𝑥-2{𝑥}^2)}^{𝑛}$

$\begin{array}{cc}\Rightarrow & (1+3\cdot 1-2\cdot 1{)}^{𝑛}=128\\ \Rightarrow & 2^{𝑛}=2^7\\ \Rightarrow & 𝑛=7\end{array}$

Now $\sum _{2𝑛}^{𝑟=1}\frac{𝑟\cdot {}^{2𝑛}{𝐶}_{𝑟}}{{}^{2𝑛}{𝐶}_{𝑟-1}}=\sum _{2𝑛}^{𝑟=1}\frac{𝑟\cdot \frac{2𝑛!}{𝑟!(2𝑛-𝑟)!}}{\frac{2𝑛!}{(𝑟-1)!(2𝑛-𝑟+1)!}}$

$\begin{array}{cc}& =\sum _{𝑟=1}^{2𝑛}\frac{𝑟\cdot (2𝑛-𝑟+1)}{𝑟}=\sum _{𝑟=1}^{2𝑛}(2𝑛-𝑟+1)[\because 𝑛=7]\\ & =(2𝑛+1)\left(2𝑛\right)-\frac{2𝑛(2𝑛+1)}{2}\\ & =(14+1)\left(14\right)-7(14+1)\\ & =15\times 7=105\end{array}$