TG EAPCET 2023 13th May Morning Shift Question Paper with Solutions

$𝑓:𝑅\to 𝑅$ defined by

$𝑓\left(𝑥\right)=⎧⎩⎨⎪⎪\begin{array}{cc}2𝑥-3,& 𝑥<-2\\ {𝑥}^2-1,& -2\le 𝑥\le 2\\ 3𝑥+2,& 𝑥>2\end{array}$

For $𝑥=2,𝑓\left(2\right)=2^2-1=3$

For $𝑥=-2,𝑓(-2)=(-2{)}^2-1=3$

$\therefore 𝑓$ is not one-one (not injective).

Now, for $𝑥<-2\Rightarrow 2𝑥-3<-7$

For $-2\le 𝑥\le 2$

$\Rightarrow 0\le {𝑥}^2\le 4\Rightarrow -1\le {𝑥}^2-1\le 3$

For   $𝑥>2$

$3𝑥+2>8$

$\therefore 𝑓\left(𝑥\right)\in (-\infty ,-7)\cup [-1,3]\cup (8,\infty )$

⇒ Range of $𝑓\subset$ co-domain of $𝑓$.

$\therefore 𝑓$ is not onto and not surjective.

Hence, $𝑓\left(𝑥\right)$ is neither injection nor surjection.

$𝑓\left(𝑥\right)=\frac{\sqrt{\surd {\log }_{10}\left(\frac{𝑥}{𝑥-2}\right)}}{\sqrt{\surd [𝑥{]}^2-5[𝑥]+6}}$

Function is defined, when

$\begin{array}{cc}& [𝑥{]}^2-5[𝑥]+6>0\\ & \Rightarrow \left(\right[𝑥]-2)\left(\right[𝑥]-3)>0\\ & \Rightarrow \left[𝑥\right]\in (-\infty ,2)\cup (3,\infty )\\ & \Rightarrow 𝑥\in (-\infty ,2)\cup [4,\infty )\end{array}$

and ${\log }_{10}\left(\frac{𝑥}{𝑥-2}\right)\ge 0$

$\begin{array}{cc}\Rightarrow & \frac{𝑥}{𝑥-2}\ge 1\Rightarrow \frac{2}{𝑥-2}\ge 0\\ \Rightarrow & 𝑥-2\ge 0\Rightarrow 𝑥\ge 2\\ \Rightarrow & 𝑥\in [2,\infty )\end{array}$

∴ Domain of $𝑓\left(𝑥\right)$ is $[4,\infty )$.

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{1}{𝑥-\left|𝑥\right|}\\ & \because \left|𝑥\right|>𝑥[\because 𝑥\ne |𝑥\left|\right]\\ & \Rightarrow 𝑥-\left|𝑥\right|<0\Rightarrow \frac{1}{𝑥-\left|𝑥\right|}<0\\ & \Rightarrow 𝑓\left(𝑥\right)\in (-\infty ,0)\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}1& 2& -1\\ -1& 0& 2\\ 1& 2& 0\end{array}⎤⎦⎥,𝐵=⎡⎣⎢\begin{array}{ccc}-3& -2& 4\\ 2& 2& -1\\ -2& 0& 3\end{array}⎤⎦⎥\\ {𝐴}^2& =⎡⎣⎢\begin{array}{ccc}1& 2& -1\\ -1& 0& 2\\ 1& 2& 0\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}1& 2& -1\\ -1& 0& 2\\ 1& 2& 0\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}-2& 0& 3\\ 1& 2& 1\\ -1& 2& 3\end{array}⎤⎦⎥=𝐴+𝐵\end{array}\end{array}$

Let ${\Delta }_1=\mid \begin{array}{ccc}2& 3& 5\\ 3& 5& 2\\ 5& 2& 3\end{array}\mid =-70$

$\begin{array}{cc}𝑎𝑛𝑑{\Delta }_2& =\mid \begin{array}{ccc}1& 1& 1\\ 7& 11& 13\\ 49& 121& 169\end{array}\mid \\ & {𝐶}_2\to {𝐶}_2-{𝐶}_1\\ & {𝐶}_3\to {𝐶}_3-{𝐶}_1\\ & =\mid \begin{array}{ccc}1& 0& 0\\ 7& 4& 6\\ 49& 72& 120\end{array}\mid =48\end{array}$

Now, ${\Delta }_1+{\Delta }_2=-70+48=-22$

$\det 𝐴=190$

$\begin{array}{cc}& ⎡⎣⎢\begin{array}{ccc}𝑘& 5& 2\\ 2& -𝑘& 5\\ 5& 2& -𝑘\end{array}⎤⎦⎥=190\\ \Rightarrow & (𝑘-3)({𝑘}^2+3𝑘+19)=0\Rightarrow 𝑘=3\end{array}$

$\mathrm{adj}𝐴=$ Transpose of co-factor matrix of $𝐴$

$=⎡⎣⎢\begin{array}{ccc}-1& 19& 31\\ 31& -19& -11\\ 19& 19& -19\end{array}⎤⎦⎥$

System of equations

$\begin{array}{cc}3𝑥-2𝑦+𝑧& =5𝑘\\ 2𝑥+3𝑦-2𝑧& =-5𝑘\\ 𝑥+4𝑦+3𝑧& =𝑘\end{array}$

It has unique solution.

If $\Delta \ne 0$ and $𝑥=\frac{{\Delta }_1}{\Delta },𝑦=\frac{{\Delta }_2}{\Delta },𝑧=\frac{{\Delta }_3}{\Delta }$

$\begin{array}{cc}\text{ Now, }& \Delta =\mid \begin{array}{ccc}3& -2& 1\\ 2& 3& -2\\ 1& 4& 3\end{array}\mid =72\ne 0\\ \text{ and }& {\Delta }_3=\mid \begin{array}{ccc}3& -2& 5𝑘\\ 2& 3& -5𝑘\\ 1& 4& 𝑘\end{array}\mid \\ & =𝑘[69+14+25]=108𝑘\\ \therefore & 𝑧=3\left(𝑔𝑖𝑣𝑒𝑛\right)\\ \Rightarrow & \frac{{\Delta }_3}{\Delta }=3\Rightarrow \frac{108𝑘}{72}=3\\ \Rightarrow & 𝑘=2\end{array}$

$\sqrt{\surd -5-12𝑖}=\pm (\frac{\sqrt{\surd \frac{\sqrt{\surd (-5{)}^2+(-12{)}^2}+(-5)}{2}}-𝑖}{\sqrt{\surd \frac{\sqrt{\surd (-5{)}^2+(-12{)}^2}-(-5)}{2}}})$

$=\pm (\sqrt{\surd 4}-𝑖\sqrt{\surd 9})=\pm (2-3𝑖)$

$\begin{array}{cc}\sqrt{\surd 7+24𝑖}=\pm ⎛⎝⎜\sqrt{\surd \frac{\sqrt{\surd 7^2+{24}^2}+7}{2}}& +𝑖\sqrt{\surd \frac{\sqrt{\surd 7^2+{24}^2}-7}{2}}⎞⎠⎟\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& =\pm (\sqrt{\surd 16}+𝑖\sqrt{\surd 9})=\pm (4+3𝑖)\\ \because & \sqrt{\surd -5-12𝑖}+\sqrt{\surd 7+24𝑖}=𝑘\end{array}\\ & \text{ (a negative real number) }\\ & \begin{array}{ccc}\Rightarrow & -(2-3𝑖)-(4+3𝑖)& =𝑘\\ \Rightarrow & 𝑘& =-6\end{array}\end{array}$

$\mathrm{amp}\left(\frac{𝑧-3}{𝑧+2𝑖}\right)=\frac{𝜋}{2}$

$\begin{array}{cc}& \Rightarrow \mathrm{amp}(\frac{(𝑥-3)+𝑖𝑦}{𝑥+𝑖(𝑦+2)}\times \frac{𝑥-𝑖(𝑦+2)}{𝑥-𝑖(𝑦+2)})=\frac{𝜋}{2}\\ & \Rightarrow \mathrm{amp}\\ & \begin{array}{c}\left(\frac{𝑥(𝑥-3)+𝑦(𝑦+2)+𝑖(𝑥𝑦-(𝑥-3\left)\right(𝑦+2)}{{𝑥}^2+(𝑦+2{)}^2}\right)=\frac{𝜋}{2}\end{array}\end{array}$

Here, $\mathrm{amp}\left(𝑧\right)=\frac{𝜋}{2}$ means $𝑧$ is purely imaginary

$\begin{array}{cc}\Rightarrow & \text{ Real part }=0\\ \therefore & 𝑥(𝑥-3)+𝑦(𝑦+2)=0\\ & {𝑥}^2-3𝑥+{𝑦}^2+2𝑦=0\\ \Rightarrow & {𝑥}^2+{𝑦}^2-3𝑥+2𝑦=0\end{array}$

which represents a circle containing the origin.

$\begin{array}{c}\frac{𝑧-(2+𝑖)}{𝑧+(1-2𝑖)}=\frac{(𝑥-2)+𝑖(𝑦-1)}{(𝑥+1)+𝑖(𝑦-2)}\times \frac{(𝑥+1)-𝑖(𝑦-2)}{(𝑥+1)-𝑖(𝑦-2)}\end{array}$

$=\frac{\begin{array}{c}(𝑥-2)(𝑥+1)+(𝑦-1)(𝑦-2)\\ +𝑖\left[\right(𝑥+1\left)\right(𝑦-1)-(𝑥-2\left)\right(𝑦-2\left)\right]\end{array}}{(𝑥+1{)}^2+(𝑦-2{)}^2}$

$\begin{array}{cc}& \because \text{ This is purely real. }\\ & \Rightarrow \text{ Imaginary point }=0\\ & \Rightarrow (𝑥+1)(𝑦-1)-(𝑥-2)(𝑦-2)=0\\ & [𝑥\ne -1,𝑦\ne 2]\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑥𝑦-𝑥+𝑦-1=𝑥𝑦-2𝑥-2𝑦+4\\ & \Rightarrow -𝑥+2𝑥+𝑦+2𝑦-1-4=0\\ & \Rightarrow 𝑥+3𝑦-5=0,[𝑥\ne -1,𝑦\ne 2]\end{array}$

We know that $1,𝜔$ and ${𝜔}^2$ are cube roots of unit, where

$\begin{array}{cc}& 𝜔=\frac{-1+𝑖\sqrt{\surd 3}}{2}\\ & {𝜔}^2=\frac{-1-𝑖\sqrt{\surd 3}}{2}\end{array}$

and ${𝜔}^3=1$

$\begin{array}{cc}\therefore -2𝜔& =1-𝑖\sqrt{\surd 3}\\ \text{ and }-2{𝜔}^2& =1+𝑖\sqrt{\surd 3}\end{array}$

Now, $(1+𝑖\sqrt{\surd 3}{)}^{2023}+(1-\sqrt{\surd 3}𝑖{)}^{2023}$

$\begin{array}{cc}& ={(-2{𝜔}^2)}^{2023}+(-2𝜔{)}^{2023}\\ & =(-2{)}^{2023}({𝜔}^{4046}+{𝜔}^{2023})\\ & =(-2{)}^{2023}({𝜔}^2+𝜔)[\because {𝜔}^{3𝑛}=1]\\ & =(-2{)}^{2023}(-1)[𝜔+{𝜔}^2=-1]\\ & =2^{2023}\end{array}$

Let $𝑧=\sqrt{\surd 3}-𝑖$

$\begin{array}{cc}𝑟& =\left|𝑧\right|=\sqrt{\surd 3+1}=2\\ 𝜃& =\arg \left(𝑧\right)=-{\tan }^{-1}\frac{1}{\sqrt{\surd 3}}=-\frac{𝜋}{6}\\ \therefore \sqrt{\surd 3}-𝑖& =2\mathrm{cis}(-\frac{𝜋}{6})\\ \sqrt{\surd 3}-𝑖& =2\mathrm{cis}(2𝑛𝜋-\frac{𝜋}{6})\\ (\sqrt{\surd 3}-𝑖{)}^{1/6}& =2^{1/6}\mathrm{cis}(\frac{2𝑛𝜋}{6}-\frac{𝜋}{36})\\ 𝑛& =0,1,2,3,4,5\\ & =2^{1/6}\mathrm{cis}\frac{(12𝑛-1)𝜋}{36}\end{array}$

For $𝑛=5$ one of the value is $2^{1/6}\mathrm{cis}\left(\frac{59𝜋}{36}\right)$.

Let roots of ${𝑥}^2+3𝑥-2𝑘=0$ be $𝛼$ and $𝛽$ and roots of ${𝑥}^2-2𝑥-7𝑘=0$ be $𝛼$ and $𝛾$,

$\begin{array}{cc}𝛼+𝛽& =-3...\left(𝑖\right)\\ 𝛼+𝛾& =2...\left(𝑖𝑖\right)\\ 𝛼𝛽& =-2𝑘...\left(𝑖𝑖𝑖\right)\\ 𝛼𝛾& =-7𝑘...\left(𝑖𝑣\right)\end{array}$

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

$𝛽-𝛾=-5$

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

$\frac{𝛽}{𝛾}=\frac{2}{7}\Rightarrow 𝛽=\frac{2}{7}𝛾$

$\therefore \frac{2}{7}𝛾-𝛾=-5\Rightarrow \frac{-5}{7}𝛾=-5$

$\begin{array}{cc}& \Rightarrow 𝛾=7\\ & \because 𝛾\text{ is a root of }{𝑥}^2-2𝑥-7𝑘=0\\ & \therefore 7^2-2\left(7\right)-7𝑘=0\\ & \Rightarrow 𝑘=5\end{array}$

Now, roots of $5{𝑥}^2+7𝑥-6=0$

$𝑥=\frac{3}{5},𝑥=-2$

Positive root is $𝑥=\frac{3}{5}$.

Let $𝑦=\frac{{𝑥}^2-2𝑥+1}{{𝑥}^2+𝑥-1}$

$\Rightarrow (𝑦-1){𝑥}^2+(𝑦+2)𝑥-𝑦-1=0$

value does not lie in the given interval, if discriminant $<0$.

$\begin{array}{cc}\Rightarrow & (𝑦+2{)}^2+4(𝑦-1\left)\right(𝑦+1)<0\\ \Rightarrow & {𝑦}^2+4𝑦+4+4{𝑦}^2-4<0\\ \Rightarrow & 5{𝑦}^2+4𝑦<0\Rightarrow 𝑦(5𝑦+4)<0\\ \Rightarrow & -\frac{4}{5}<𝑦<0\\ \therefore & 𝑦\text{ does not lie in }(-\frac{4}{5},0)\end{array}$

$𝛼,𝛽$ and $𝛾$ are roots of the equation

${𝑥}^3-3{𝑥}^2+3𝑥+1=0$

Transforming the equation whose roots are ${𝛼}^2,{𝛽}^2$ and ${𝛾}^2$.

For this replacing $𝑥$ with ${𝑥}^{1/2}$

$\begin{array}{cc}& {𝑥}^{3/2}-3𝑥+3{𝑥}^{1/2}+1=0\\ \Rightarrow & {({𝑥}^{3/2}+3{𝑥}^{1/2})}^2=(3𝑥-1{)}^2\\ \Rightarrow & {𝑥}^3+9𝑥+6{𝑥}^2=9{𝑥}^2+1-6𝑥\\ \Rightarrow & {𝑥}^3-3{𝑥}^2+15𝑥-1=0\end{array}$

Then, sum of product of roots taken two at a time is ${𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2=15$

Given equation

${𝑥}^4-4{𝑥}^3-7{𝑥}^2+22𝑥+24=0$

Let $𝑎,𝑏,𝑐$ and $𝑑$ are roots of the equation, such that

$𝑎+𝑏=𝑐+𝑑\text{, then by relation }$

between roots and coefficients

$\begin{array}{cc}& 𝑎+𝑏+𝑐+𝑑=4...\left(𝑖\right)\\ \Rightarrow & 𝑎+𝑏=2,𝑐+𝑑=2...\left(𝑖𝑖\right)\end{array}$

and $𝑎𝑏+𝑎𝑐+𝑎𝑑+𝑏𝑐+𝑏𝑑+𝑐𝑑=-7$

$\begin{array}{c}𝑎𝑏+𝑎(𝑐+𝑑)+𝑏(𝑐+𝑑)+𝑐𝑑=-7\\ 𝑎𝑏+𝑐𝑑+(𝑎+𝑏)(𝑐+𝑑)=-7\\ 𝑎𝑏+𝑐𝑑+4=-7\\ 𝑎𝑏+𝑐𝑑=-11...\left(𝑖𝑖𝑖\right)\end{array}$

Now, $(𝑎+𝑏+𝑐+𝑑{)}^3=(𝑎+𝑏{)}^3+(𝑐+𝑑{)}^3$

$\begin{array}{cc}+3(𝑎+𝑏{)}^2(𝑐+𝑑)+3(𝑎+𝑏\left)\right(𝑐+𝑑{)}^2& \\ (𝑎+𝑏+𝑐+𝑑{)}^3=& {𝑎}^3+{𝑏}^3+{𝑐}^3+{𝑑}^3\\ +& 3𝑎𝑏(𝑎+𝑏)+3𝑐𝑑(𝑐+𝑑)\\ +& 3(𝑎+𝑏)(𝑐+𝑑)(𝑎+𝑏+𝑐+𝑑)\end{array}$

$\begin{array}{cc}& \text{ Using Eqs. (i), (ii) and (iii), we get }\\ & \begin{array}{cc}& 4^3={𝑎}^3+{𝑏}^3+{𝑐}^3+{𝑑}^3+6𝑎𝑏+6𝑐𝑑+3\left(2\right)\left(2\right)\left(4\right)\\ & 64={𝑎}^3+{𝑏}^3+{𝑐}^3+{𝑑}^3-66+48\\ & \Rightarrow {𝑎}^3+{𝑏}^3+{𝑐}^3+{𝑑}^3=82\end{array}\end{array}$

We know that

${}^{𝑛-1}{𝑃}_{𝑟}+𝑟\cdot {}^{𝑛-1}{𝑃}_{𝑟-1}={}^{𝑛}{𝑃}_{𝑟}$

Then,

$\begin{array}{c}{}^{𝑛}{𝑃}_{𝑟+1}+{𝑟}^2{}^{𝑛-1}{𝑃}_{𝑟-1}+(𝑟+1){}^{𝑛-1}{𝑃}_{𝑟}+𝑟\cdot {}^{𝑛-1}{𝑃}_{𝑟-1}\\ ={}^{𝑛}{𝑃}_{𝑟+1}+𝑟(𝑟{}^{𝑛-1}{𝑃}_{𝑟-1}+{}^{𝑛-1}{𝑃}_{𝑟})+({}^{𝑛-1}{𝑃}_{𝑟}+𝑟{}^{𝑛-1}{𝑃}_{𝑟-1})\\ ={}^{𝑛}{𝑃}_{𝑟+1}+𝑟\cdot {}^{𝑛}{𝑃}_{𝑟}+{}^{𝑛}{𝑃}_{𝑟}\\ ={}^{𝑛}{𝑃}_{𝑟+1}+(𝑟+1){}^{𝑛}{𝑃}_{𝑟}={}^{𝑛+1}{𝑃}_{𝑟+1}\end{array}$

The number of ways of arranging $𝑛$ boys and $𝑛$ girls in a row $=𝑛!\times 𝑛!\times 2!$ Also, the number of ways $𝑛$ boys and $𝑛$ girls can be arranged such that no two girls are together and no two boys are together $=𝑛!\times 𝑛!\times 2!$

$12600=2^3\times 3^2\times 5^2\times 7$

For multiples of 3

$\begin{array}{cc}12600& =3[2^3\times 3\times 5^2\times 7]\\ {𝑛}_1& =(3+1)(1+1)(2+1)(1+1)\\ & =4\times 2\times 3\times 2\\ & =48\end{array}$

For multiples of 14

$\begin{array}{cc}12600& =14[2^2\times 3^2\times 5^2]\\ {𝑛}_2& =(2+1)(2+1)(2+1)\\ & =3\times 3\times 3=27\\ \therefore {𝑛}_1+{𝑛}_2& =48+27=75\end{array}$

Number of terms in expansion of

$\begin{array}{cc}(𝑥-2𝑦+3𝑧{)}^5& ={}^{5+3-1}{𝐶}_{3-1}\\ 𝑃& ={}^7{𝐶}_2\Rightarrow 𝑃=21\end{array}$

Coefficient of ${𝑥}^{𝑝}{𝑦}^{𝑞}{𝑧}^{𝑟}$ in $(𝑎𝑥+𝑏𝑦+𝑐𝑧{)}^{𝑛}$ is

$\frac{𝑛!}{𝑝!𝑞!𝑟!}({𝑎}^{𝑝}\cdot {𝑏}^{𝑞}\cdot {𝑐}^{𝑟})$

$𝑞=$ Coefficient of ${𝑥}^2𝑦{𝑧}^2$ in $(𝑥-2𝑦+3𝑧{)}^5$

$=\frac{5!}{2!\times 1!\times 2!}\{1^2\cdot (-2{)}^1\left(3{)}^2\right\}=-540$

Now, $\frac{𝑞}{𝑝}=-\frac{540}{21}=-\frac{180}{7}$

$\begin{array}{cc}& \begin{array}{cc}{𝑆}_{𝑛+1}=5\cdot {𝐶}_0+8\cdot & {𝐶}_1+11\cdot {𝐶}_2\\ & +\dots +(𝑛+1)\text{ terms }\\ =& \sum _{𝑟=0}^{𝑛}(3𝑟+5)\cdot {𝐶}_{𝑟}\\ =& \sum _{𝑟=0}^{𝑛}(3\cdot 𝑟\cdot {𝐶}_{𝑟})+\sum _{𝑟=0}^{𝑛}5\cdot {𝐶}_{𝑟}\\ =& 3\sum _{𝑟=0}^{𝑛}𝑟{𝐶}_{𝑟}+5\sum _{𝑟=0}^{𝑛}{𝐶}_{𝑟}\\ {𝑆}_{𝑛+1}=& 3\left(𝑛2^{𝑛-1}\right)+5\cdot 2^{𝑛}\end{array}\end{array}$

On putting $𝑛=10$

${𝑆}_{11}=3(10\cdot 2^9)+5\cdot 2^{10}=20480$

$\frac{1}{\sqrt{\surd 4-𝑥}(2+𝑥{)}^3}$

Using binomial expansion for negative index

$\begin{array}{cc}& \frac{1}{16}{(1-\frac{𝑥}{4})}^{-1/2}{(1+\frac{𝑥}{2})}^{-3}\\ =& \frac{1}{16}(1+\frac{𝑥}{8}+\frac{3}{128}{𝑥}^2)(1-\frac{3𝑥}{2}+\frac{3}{2}{𝑥}^2)\\ =& \frac{1}{16}(1-\frac{3𝑥}{2}+\frac{3}{2}{𝑥}^2+\frac{𝑥}{8}-\frac{3{𝑥}^2}{16}+\frac{3{𝑥}^2}{128})\\ =& \frac{1}{16}(1-\frac{11𝑥}{8}+\frac{171}{128}{𝑥}^2)\end{array}$

$\begin{array}{cc}& \frac{6{𝑥}^4+13{𝑥}^3+2{𝑥}^2-𝑥+3}{2{𝑥}^2+3𝑥-2}\\ & =3{𝑥}^2+2𝑥+1+\frac{5}{2{𝑥}^2+3𝑥-2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝑓\left(𝑥\right)& =\frac{𝐴}{𝑎𝑥-1}+\frac{𝐵}{𝑥+𝑏}\\ & =3{𝑥}^2+2𝑥+1+\frac{5}{(2𝑥-1)(𝑥+2)}\\ & =3{𝑥}^2+2𝑥+1+\frac{2}{2𝑥-1}+\frac{(-1)}{𝑥+2}\end{array}\\ & \text{ On comparing both sides, we get }\\ & \begin{array}{c}𝑓\left(𝑥\right)=3{𝑥}^2+2𝑥+1\\ 𝐴=2,𝐵=-1\\ 𝑎=2,𝑏=2\end{array}\\ & 𝑓\left(1\right)+𝑎\cdot 𝐵+𝑏\cdot 𝐴=6-2+4=8\end{array}$