TG EAPCET 2022 18th July Morning Shift Question Paper with Solutions

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{\sqrt{\surd 6{𝑥}^2+5𝑥-6}}{\sqrt{\surd 4-𝑥}-\sqrt{\surd 𝑥+4}}\\ & 6{𝑥}^2+5𝑥-6\ge 0\\ & \Rightarrow 6{𝑥}^2+9𝑥-4𝑥-6\ge 0\\ & \Rightarrow 3𝑥(2𝑥+3)-2(2𝑥+3)\ge 0\\ & \Rightarrow (3𝑥-2)(2𝑥+3)\ge 0\\ & \end{array}$

$\begin{array}{cc}& 𝑥\in (-\infty ,\frac{-3}{2}]\cup [\frac{2}{3},\infty )...\left(𝑖\right)\\ \Rightarrow & 4-𝑥\ge 0\Rightarrow 𝑥\le 4\\ \Rightarrow & 𝑥\in (-\infty ,4]...\left(𝑖𝑖\right)\\ \Rightarrow & 𝑥+4\ge 0\Rightarrow 𝑥\ge -4\end{array}$

$𝑥\in [-4,\infty )...\left(𝑖𝑖𝑖\right)$

$\therefore 𝑥\in [-4,\frac{-3}{2}]\cup [\frac{2}{3},4]$

$\begin{array}{cc}& \text{ Given, }𝑓\left(𝑥\right)=\frac{1}{\sqrt{\surd [𝑥{]}^2+[𝑥]-2}}\\ & =\frac{1}{\sqrt{\surd {\left(\right[𝑥]+\frac{1}{2})}^2-\frac{9}{4}}}\end{array}$

Maximum value of ${\left(\right[𝑥]+\frac{1}{2})}^2$ is $\infty$.

So, $𝑓\left(𝑥\right)=\frac{1}{\sqrt{\surd \infty -\frac{9}{4}}}\to 0$

${\left(\right[𝑥]+\frac{1}{2})}_2^{min}-\frac{9}{4}$ to be positive, $\left[𝑥\right]$ must be 2 .

$\begin{array}{cc}& \Rightarrow {(2+\frac{1}{2})}^2-\frac{9}{4}=\frac{25}{4}-\frac{9}{4}=\frac{16}{4}=4\\ & \Rightarrow 𝑓\left(𝑥\right)=\frac{1}{\sqrt{\surd 4}}=\frac{1}{2}\\ & \text{ Range : }(0,\frac{1}{2}]\end{array}$

$\begin{array}{cc}& \text{ Given, trace of }𝐴=-4\\ & \begin{array}{cc}& \therefore 𝑎+(-1)+(-4)=-4\\ & \Rightarrow 𝑎=1\\ & \text{ Now, }𝐴𝐵=⎡⎣⎢\begin{array}{ccc}-1& 0& 17\\ -3& 10& 25\\ 28& -8& 3\end{array}⎤⎦⎥\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}⎡⎣⎢\begin{array}{ccc}1& 3& 5\\ 5& -1& 3\\ 2& 3& -4\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}𝑏& 1& 4\\ 4& 𝑐& 1\\ -3& 1& 𝑑\end{array}⎤⎦⎥& =⎡⎣⎢\begin{array}{ccc}-1& 0& 17\\ -3& 10& 25\\ 28& -8& 3\end{array}⎤⎦⎥\\ \Rightarrow ⎡⎣⎢\begin{array}{ccc}𝑏-3& 3𝑐+6& 7+5𝑑\\ 5𝑏-13& 8-𝑐& 19+3𝑑\\ 2𝑏+24& 3𝑐-2& 11-4𝑑\end{array}⎤⎦⎥& =⎡⎣⎢\begin{array}{ccc}-1& 0& 17\\ -3& 10& 25\\ 28& -8& 3\end{array}⎤⎦⎥\end{array}\\ & \text{ On comparing, }\\ & \begin{array}{cc}𝑏-3=-1\Rightarrow 𝑏& =2\\ 3𝑐+6=0\Rightarrow 𝑐& =-2\\ 7+5𝑑=17\Rightarrow 𝑑& =2\\ \therefore 𝑎+𝑏+𝑐+𝑑=1+2-2+2& =3\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }\mid \begin{array}{ccc}1& 1& 1\\ {𝑎}^2& {𝑏}^2& {𝑐}^2\\ {𝑎}^3& {𝑏}^3& {𝑐}^3\end{array}\mid \\ & \begin{array}{c}=({𝑏}^2{𝑐}^3-{𝑏}^3{𝑐}^2)-({𝑎}^2{𝑐}^3-{𝑎}^3{𝑐}^2)+({𝑎}^2{𝑏}^3-{𝑎}^3{𝑏}^2)\\ =({𝑎}^2{𝑏}^3-{𝑎}^2{𝑐}^3)+({𝑏}^2{𝑐}^3-{𝑎}^3{𝑏}^2)+({𝑎}^3{𝑐}^2-{𝑏}^3{𝑐}^2)\\ ={𝑎}^2({𝑏}^3-{𝑐}^3)+{𝑏}^2({𝑐}^3-{𝑎}^3)+{𝑐}^2({𝑎}^3-{𝑏}^3)\end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }𝛼,𝛽,𝛾\text{ be real numbers, then }\\ & \begin{array}{cc}& 𝐴=⎡⎣⎢\begin{array}{ccc}7& 3& 𝛼\\ 𝛽& 1& -11\\ -5& 𝛾& 19\end{array}⎤⎦⎥\\ & \Rightarrow 𝐴⎡⎣⎢\begin{array}{c}5\\ -13\\ 11\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}-290\\ -119\\ 210\end{array}⎤⎦⎥\\ & \Rightarrow ⎡⎣⎢\begin{array}{ccc}7& 3& 𝛼\\ 𝛽& 1& -11\\ -5& 𝛾& 19\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}5\\ -13\\ 11\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}-290\\ -119\\ 210\end{array}⎤⎦⎥\\ & 11𝛼-4=-290\Rightarrow 11𝛼=-286\\ & \Rightarrow 𝛼=-265𝛽-134=-119\\ & \Rightarrow 5𝛽=15\Rightarrow 𝛽=3-13𝛾+184=210\\ & \Rightarrow -13𝛾=26\Rightarrow 𝛾=-2\\ & 𝐴=⎡⎣⎢\begin{array}{ccc}7& 3& -26\\ 3& 1& -11\\ -5& -2& 19\end{array}⎤⎦⎥\end{array}\end{array}$

$\begin{array}{cc}\left|𝐴\right|& =7(-3)-3\left(2\right)-26(-1)\\ & =-27+26=-1\\ {𝐴}^{-1}& =\frac{\mathrm{adj}\left(𝐴\right)}{\left|𝐴\right|}\\ \Rightarrow \mathrm{adj}\left(𝐴\right)& ={𝐴}^{-1}\left|𝐴\right|\end{array}$

$\begin{array}{cc}(\mathrm{adj}𝐴{)}^{-1}+\mathrm{adj}{𝐴}^{-1}& =(\mathrm{adj}𝐴{)}^{-1}+(\mathrm{adj}𝐴{)}^{-1}\\ & =2(\mathrm{adj}𝐴{)}^{-1}\\ =2{\left({𝐴}^{-1}\right|𝐴\left|\right)}^{-1}& =2𝐴\cdot |𝐴{|}^{-1}\\ =\frac{2𝐴}{\left|𝐴\right|}=\frac{2𝐴}{-1}& =-2𝐴\end{array}$

Given, $[𝛼,𝛽𝛾]⎡⎣⎢\begin{array}{ccc}1& 2& 3\\ 2& 3& -5\\ 1& 2& 5\end{array}⎤⎦⎥=\left[\begin{array}{ccc}3& 5& 2\end{array}\right]$

$\begin{array}{cc}& \text{ Let }𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 3\\ 2& 3& -5\\ 1& 2& 5\end{array}⎤⎦⎥\\ & \left[\begin{array}{ccc}𝛼& 𝛽& 𝛾\end{array}\right]\cdot 𝐴=\left[\begin{array}{ccc}3& 5& 2\end{array}\right]\\ & \left[\begin{array}{ccc}𝛼& 𝛽& 𝛾\end{array}\right]=\left[\begin{array}{ccc}3& 5& 2\end{array}\right]{𝐴}^{-1}\\ & \left|𝐴\right|=1(15+10)-2(10+5)+3(4-3)\\ & \left|𝐴\right|=25-30+3=-2\\ & \mathrm{adj}\left(𝐴\right)=⎡⎣⎢\begin{array}{ccc}25& -4& -19\\ -15& 2& 11\\ 1& 0& -1\end{array}⎤⎦⎥\end{array}$

Now, $\left[\begin{array}{ccc}𝛼& 𝛽& 𝛾\end{array}\right]=\left[\begin{array}{ccc}3& 5& 2\end{array}\right]\frac{\mathrm{adj}𝐴}{\left|𝐴\right|}$

$\left[\begin{array}{ccc}𝛼& 𝛽& 𝛾\end{array}\right]=\frac{1}{-2}\left[\begin{array}{ccc}3& 5& 2\end{array}\right]⎡⎣⎢\begin{array}{ccc}25& -4& -19\\ -15& 2& 11\\ 1& 0& -1\end{array}⎤⎦⎥$

$\left[\begin{array}{ccc}𝛼& 𝛽& 𝛾\end{array}\right]=-\frac{1}{2}\left[\begin{array}{ccc}2& -2& -4\end{array}\right]=\left[\begin{array}{ccc}-1& 1& 2\end{array}\right]$

On comparing $𝛼=-1,𝛽=1$ and $𝛾=2$

$\therefore {𝛼}^3+{𝛽}^3+{𝛾}^3=-1+1+8=8$

Given, $\sqrt{\surd (-3+4𝑖)(8+6𝑖)}$

$=\sqrt{\surd -48+14𝑖}=\sqrt{\surd 2}\sqrt{\surd -24+7𝑖}$

We know that,

$\begin{array}{cc}& \sqrt{\surd 𝑧}=\pm (\sqrt{\surd \frac{\left|𝑧\right|+\mathrm{Re}\left(𝑧\right)}{2}}+𝑖\sqrt{\surd \frac{\left|𝑧\right|-\mathrm{Re}\left(𝑧\right)}{2}})\text{ for }\\ & \mathrm{Im}\left(𝑧\right)>0\\ & \therefore \sqrt{\surd -24+7𝑖}\\ & =\pm (\sqrt{\surd \frac{25-24}{2}}+𝑖\sqrt{\surd \frac{25+24}{2}})\\ & =\pm \frac{1}{\sqrt{\surd 2}}(1+7𝑖)\\ & \therefore \sqrt{\surd (-3+4𝑖)(8+6𝑖)}=\pm (1+7𝑖)\end{array}$

$\begin{array}{cc}& \text{ We know that, }\\ & \begin{array}{cc}& 𝜔=\frac{-1}{2}+𝑖\frac{\sqrt{\surd 3}}{2}\text{ and }{𝜔}^2=\frac{-1}{2}-𝑖\frac{\sqrt{\surd 3}}{2}\\ \Rightarrow & -2𝑖𝜔=\sqrt{\surd 3}+𝑖\text{ and }2𝑖{𝜔}^2=\sqrt{\surd 3}-𝑖\\ & {\left(\frac{\sqrt{\surd 3}+𝑖}{\sqrt{\surd 3}-𝑖}\right)}^{𝑚}=1\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow {\left(\frac{-2𝑖𝜔}{2𝑖{𝜔}^2}\right)}^{𝑚}=1\Rightarrow (-𝜔{)}^{𝑚}=1\\ & \Rightarrow (-1{)}^{𝑚}{𝜔}^{𝑚}=1\end{array}$

$\Rightarrow 𝑚$ is even number and multiple of 3 . Clearly, 2028 is even and multiple of 3.

Here, 1, $𝜔$ and ${𝜔}^2$ are the cube root of unity, $𝑛\in 𝑁,𝑛>2$

$\begin{array}{cc}\therefore & 1+𝜔+{𝜔}^2=0,{𝜔}^3=1\\ & {𝑥}^{𝑛}-𝑥=0\to 1+𝜔\text{ is a root. }\end{array}$

$\begin{array}{cc}& \text{ So, }(1+𝜔{)}^{𝑛}-(1+𝜔)=0\\ & \Rightarrow (1+𝜔{)}^{𝑛}=(1+𝜔)\\ & (1+𝜔{)}^{𝑛-1}=1\Rightarrow {(-{𝜔}^2)}^{𝑛-1}=1\end{array}$

Least value of $𝑛$ must be 7 .

$\begin{array}{cc}& 𝐴=\{𝑥\in 𝑅/\sqrt{\surd {𝑥}^2-8𝑥+15}\in 𝑅\}\\ & \begin{array}{cc}\Rightarrow & {𝑥}^2-8𝑥+15\ge 0\\ \Rightarrow & (𝑥-5)(𝑥-3)\ge 0\\ \Rightarrow & 𝑥\in (-\infty ,3]\cup [5,\infty ]\\ \therefore & 𝐴=(-\infty ,3]\cup [5,\infty ]\\ & 𝐵=\{𝑥\in 𝑅/\frac{𝑥-3}{2𝑥-5}<\frac{𝑥-6}{2𝑥-11}\}\\ \Rightarrow & \frac{𝑥-3}{2𝑥-5}-\frac{𝑥-6}{2𝑥-11}<0\\ \Rightarrow & \frac{-10𝑥+3}{(2𝑥-5)(2𝑥-11)}<0\\ \Rightarrow & \frac{10𝑥-3}{(2𝑥-5)(2𝑥-11)}>0\end{array}\end{array}$

$\begin{array}{cc}\Rightarrow & 𝑥\in (\frac{3}{10},\frac{5}{2})\cup (\frac{11}{2},\infty )\\ \therefore & 𝐵=(\frac{3}{10},\frac{5}{2})\cup (\frac{11}{2},\infty )\\ & 𝐴\cap 𝐵=(\frac{5}{2},3]\cup [5,\frac{11}{2})\end{array}$

Let $𝑓\left(𝑥\right)=3𝑥-2{𝑥}^2+1$

Given, extreme value of $𝑓\left(𝑥\right)$ is $𝑘$.

$\begin{array}{cc}& \Rightarrow {𝑓}^{\prime }\left(𝑘\right)=03-2\left(2𝑘\right)=0𝑘=3/4\\ & \because 𝑘{𝑥}^2+2𝑥+1>0\\ & \Rightarrow \frac{3}{4}{𝑥}^2+2𝑥+1>0\Rightarrow 3{𝑥}^2+8𝑥+4>0\\ & \Rightarrow (𝑥+\frac{2}{3})(𝑥+2)>0\\ & \Rightarrow 𝑥\in (-\infty ,\frac{-2}{3})\cup (-2,\infty )\end{array}$

Let $𝛼,𝛽,𝛾$ be the roots of the equation

$\begin{array}{c}{𝑥}^3-5{𝑥}^2-2𝑥+24=0\\ 𝛼+𝛽+𝛾=5...\left(𝑖\right)\end{array}$

$\begin{array}{c}𝛼𝛽+𝛽𝛾+𝛾𝛼=-2...\left(𝑖𝑖\right)\\ 𝛼𝛽𝛾=-24...\left(𝑖𝑖𝑖\right)\end{array}$

$\begin{array}{cc}& \therefore \frac{𝛽𝛾}{𝛼}+\frac{𝛾𝛼}{𝛽}+\frac{𝛼𝛽}{𝛾}\\ & =\frac{𝛽𝛾}{-24}\times 𝛽𝛾+\frac{𝛾𝛼}{-24}\times 𝛾𝛼+\frac{𝛼𝛽}{-24}\times 𝛼𝛽\\ & \text{ using Eq. (iii) }]\\ & =\frac{1}{-24}\left[\right(𝛽𝛾{)}^2+(𝛾𝛼{)}^2+(𝛼𝛽{)}^2]\\ & =\frac{-1}{24}\left[\right(𝛼𝛽+𝛽𝛾+𝛾𝛼{)}^2-2(𝛼{𝛽}^2𝛾+𝛼𝛽{𝛾}^2+{𝛼}^2𝛽𝛾)]\\ & =\frac{-1}{24}\left[\right(-2{)}^2-2(-24(𝛼+𝛽+𝛾\left)\right)]\\ & =\frac{-1}{24}[4+2\times 24\times 5]=-\frac{61}{6}\end{array}$

Let $𝛼,𝛽,𝛾$ be the roots of the equation

$\begin{array}{c}3{𝑥}^3-26{𝑥}^2+52𝑥-24=0\\ 𝛼+𝛽+𝛾=\frac{26}{3}...\left(𝑖\right)\\ 𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{52}{3}...\left(𝑖\right)\\ 𝛼𝛽𝛾=8...\left(𝑖𝑖𝑖\right)\end{array}$

$𝛼,𝛽,𝛾$ are in GP $\Rightarrow {𝛽}^2=𝛼𝛾$

From Eq. (iii), ${𝛽}^3=8\Rightarrow 𝛽=2...\left(𝑖𝑣\right)$

∴ From Eqs. (i) and (iv),

$\begin{array}{cc}& 𝛼+𝛾=\frac{20}{3}𝛼𝛾=4\Rightarrow 𝛾=\frac{4}{𝛼}\\ & \therefore 𝛼+\frac{4}{𝛼}=\frac{20}{3}\Rightarrow 3{𝛼}^2-20𝛼+12=0\\ & \Rightarrow (𝛼-6)(3𝛼-2)=0𝛼=6,𝛼=\frac{2}{3}\end{array}$

For $𝛼=6,𝛾=\frac{2}{3}$ and for $𝛼=\frac{2}{3},𝛾=6$

$\begin{array}{cc}& \therefore 𝛼=6,𝛽=2\text{ and }𝛾=\frac{2}{3}\\ & \text{ or }𝛼=\frac{2}{3},𝛽=2\text{ and }𝛾=6\end{array}$

$\begin{array}{cc}𝑁𝑜𝑤,3𝛼+2𝛽+𝛾& =3\times \frac{2}{3}+2\times 2+6\\ & =2+4+6=12\end{array}$

$𝑝\left(𝑥\right)$ is a quadratic polynomial having only purely imaginary roots.

Let $𝑝\left(𝑥\right)={𝑥}^2+{𝑎}^2=0,𝑎\in 𝑅$

whose roots are $𝑥=\pm 𝑖𝑎$

$\begin{array}{cc}\text{ Then, }𝑝\left(𝑝\right(𝑥\left)\right)=& {({𝑥}^2+{𝑎}^2)}^2+{𝑎}^2\\ =& {𝑥}^4+2{𝑎}^2{𝑥}^2+{𝑎}^4+{𝑎}^2\\ \text{ Discriminant }=& 4{𝑎}^4-4({𝑎}^4+{𝑎}^2)\\ =& -4{𝑎}^2<0\end{array}$

∴ Roots of $𝑝\left(𝑝\right(𝑥)$ are non-real.

Given equation, $4{𝑥}^3+12{𝑥}^2-7𝑥+165=0$ has roots $𝛼,𝛽$ and $𝛾$, respectively.

$𝛼+𝛽+𝛾=-\frac{12}{4}𝛼𝛽𝛾=-\frac{165}{4}$

$𝛼𝛽+𝛽𝛾+𝛾𝛼=-\frac{7}{4}$

Now, $𝑎{𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+𝑑=0$ has roots $𝛼+5,𝛽+5$ and $𝛾+5$, respectively. Product of roots $=(𝛼+5)(𝛽+5)(𝛾+5)$

$\begin{array}{cc}& =𝛼𝛽𝛾+5(𝛼𝛽+𝛽𝛾+𝛾𝛼)+25(𝛼+𝛽+𝛾)+125\\ & =\frac{-165}{4}-\frac{35}{4}-\frac{300}{4}+125=0\end{array}$

$\text{ Taking }1\text{ at one’s place, }$

So, two numbers can be formed using the above combination.

So, two numbers can be formed using the above combination.

Total numbers that can be formed when 1 is at the unit's place

$=2+2+2+2=8$

Now, taking 3 at one's place,

Total numbers that can be formed when 3 is at the unit's place $=2+2+2+2=8$

Now, taking 5 at one's place,

Total numbers that can be formed when 5 is at the unit's place $=2+2+2+2=8$ ∴ Total number of 3 -digit odd numbers divisible by 3 that can be formed using the digits $1,2,3,4,5,6$ when repetition is not allowed $=8+8+8=24$

(A) The number of ways of not selecting $(𝑛-𝑟)$ things from $𝑛$ different things $={}^{𝑛}{𝐶}_{𝑛-𝑟}$

(B) $(𝑛-𝑟+1)\cdot {}^{𝑛}{𝐶}_{𝑟-1}$

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

(C) The number of ways of selecting atleast $(𝑛-𝑟)$ things from $𝑛$ different things

$\begin{array}{cc}& ={}^{𝑛}{𝐶}_{𝑛-𝑟}+{}^{𝑛}{𝐶}_{𝑛-𝑟+1}+{}^{𝑛}{𝐶}_{𝑛-𝑟+2}+\dots +{}^{𝑛}{𝐶}_{𝑛}\\ & ={}^{𝑛}{𝐶}_{𝑛}+{}^{𝑛}{𝐶}_{𝑛-1}+{}^{𝑛}{𝐶}_{𝑛-2}+\dots +{}^{𝑛}{𝐶}_{𝑛-𝑟+1}+{}^{𝑛}{𝐶}_{𝑛-𝑟}\end{array}$

$\begin{array}{cc}& =1+{}^{𝑛}{𝐶}_1+{}^{𝑛}{𝐶}_2+\dots +{}^{𝑛}{𝐶}_{𝑟}\\ & \text{ (D) }(𝑛-𝑟)({}^{𝑛-1}{𝐶}_{𝑟-1}+{}^{𝑛-1}{𝐶}_{𝑟})\\ & =(𝑛-𝑟)\left({}^{𝑛}{𝐶}_{𝑟}\right)\\ & [\because {}^{𝑛}{𝐶}_{𝑟}+{}^{𝑛}{𝐶}_{𝑟+1}={}^{𝑛+1}{𝐶}_{𝑟+1}]\\ & =(𝑛-𝑟)\times \frac{𝑛!}{(𝑛-𝑟)!𝑟!}=\frac{𝑛!}{(𝑛-𝑟-1)!𝑟!}\\ & =(𝑟+1)\frac{𝑛!}{(𝑟+1)!(𝑛-𝑟-1)!}=(𝑟+1){}^{𝑛}{𝐶}_{𝑟+1}\end{array}$

General term of ${(𝑎𝑥+\frac{𝑏}{{𝑥}^2})}^{11}$

${𝑇}_{𝑟+1}={}^{11}{𝐶}_{𝑟}{𝑎}^{𝑟}{𝑏}^{11-𝑟}{𝑥}^{3𝑟-22}$

For ${𝑥}^{-7}3𝑟-22=-7$

$\begin{array}{cc}\Rightarrow 𝑟& =5\\ 𝐿& =\text{ Coefficient of }{𝑥}^{-7}={}^{11}{𝐶}_5{𝑎}^5{𝑏}^6\end{array}$

General term of ${(𝑏{𝑥}^2+\frac{𝑎}{𝑥})}^{11}$

${𝑇}_{𝑟+1}={}^{11}{𝐶}_{𝑟}{𝑎}^{11-𝑟}{𝑏}^{𝑟}{𝑥}^{3𝑟-11}$

For ${𝑥}^7$

$\begin{array}{cc}& 3𝑟-11=7\Rightarrow 𝑟=6\\ & 𝑀=\text{ Coefficient of }{𝑥}^7={}^{11}{𝐶}_6{𝑎}^5{𝑏}^6\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝐿+𝑀& =({}^{11}{𝐶}_5+{}^{11}{𝐶}_6){𝑎}^5{𝑏}^6\\ & ={}^{12}{𝐶}_6{𝑎}^5{𝑏}^6\end{array}\\ & \text{ Now, coefficient of }{𝑥}^6\text{ in }\\ & \begin{array}{cc}& {(𝑎{𝑥}^2+\frac{𝑏}{𝑥})}^{12}={}^{12}{𝐶}_0{𝑎}^6{𝑏}^6\\ & \therefore 𝐿+𝑀=\frac{1}{𝑎}\\ & \left[\text{ coefficient of }{𝑥}^{-6}\text{ in }{(𝑎{𝑥}^2+\frac{𝑏}{𝑥})}^{12}\right]\end{array}\end{array}$