AP EAPCET 2021 20th August Morning Shift Question Paper with Solutions

$\begin{array}{cc}𝑓\left(𝑥\right)& =\sin 𝑥+\cos 𝑥\\ 𝑔\left(𝑥\right)& ={𝑥}^2-1\\ 𝑔\left[𝑓\right(𝑥\left)\right]& =(\sin 𝑥+\cos 𝑥{)}^2-1\\ & =1+\sin 2𝑥-1=\sin 2𝑥\end{array}$

Among the given options, $\sin 2𝑥$ is monotonous (here strictly increasing) in $-\frac{𝜋}{4}\le 2\le \frac{𝜋}{4}$.

$\begin{array}{cc}𝑓\left(𝑥\right)& ={𝑥}^9-11{𝑥}^8-2{𝑥}^7+22{𝑥}^6+{𝑥}^4\\ & -12{𝑥}^3+11{𝑥}^2+𝑥-3,\forall 𝑥\in 𝑍\\ & ={𝑥}^8(𝑥-11)-2{𝑥}^6(𝑥-11)+{𝑥}^3(𝑥-11)\\ & -{𝑥}^2(𝑥-11)+(𝑥-11)+8\\ & =(𝑥-11)[{𝑥}^8-2{𝑥}^6+{𝑥}^3-{𝑥}^2+1]+8\end{array}$

So, $𝑓\left(11\right)=0+8=8$

$𝑓\left(𝑥\right)={𝑥}^3$ and $𝑔\left(𝑥\right)=3^{𝑥}$

$\begin{array}{cc}& \Rightarrow 𝑓\left[𝑔\right(𝑥\left)\right]=𝑔\left[𝑓\right(𝑥\left)\right]\\ & \Rightarrow {\left(3^{𝑥}\right)}^3=3^{{𝑥}^3}\Rightarrow 3^{3𝑥}=3^{{𝑥}^3}\\ & \Rightarrow 3𝑥={𝑥}^3\Rightarrow 𝑥({𝑥}^2-3)=0\end{array}$

As, $𝑥\ne 0$, So ${𝑥}^2-3=0$

Trace of A = Sum of diagonal elements of A

Given, $𝐴=⎡⎣⎢\begin{array}{ccc}1& -5& 7\\ 0& 7& 9\\ 11& 8& 9\end{array}⎤⎦⎥$

$=1+7+9=17$

$\begin{array}{cc}& -𝑥+𝑦\cos 𝐶+𝑧\cos 𝐵=0\\ & 𝑥\cos 𝐶-𝑦+𝑧\cos 𝐴=0\\ & 𝑥\cos 𝐵+𝑦\cos 𝐴-𝑍=0\\ & \Delta =\mid \begin{array}{ccc}-1& \cos 𝐶& \cos 𝐵\\ \cos 𝐶& -1& \cos 𝐴\\ \cos 𝐵& \cos 𝐴& -1\end{array}\mid \\ & =-1(1-{\cos }^2𝐴)+\cos 𝐶(\cos 𝐴\cos 𝐵+\cos 𝐶)\\ & +\cos 𝐵(\cos 𝐴\cos 𝐶+\cos 𝐵)\\ & \end{array}$

$\begin{array}{cc}& =-1+{\cos }^2𝐴+2\cos 𝐴\cos 𝐵\cos 𝐶\\ & +{\cos }^2𝐶+{\cos }^2𝐵\\ & =-{\sin }^2𝐴+{\cos }^2𝐵+{\cos }^2𝐶+2\cos 𝐴\cos 𝐵\cos 𝐶\\ & \text{ Clearly, }\Delta >0.\\ & \Rightarrow \text{ System of equations has a non-zero solution. }\end{array}$

$\begin{array}{cc}\text{ Let }𝐴& =\left[\begin{array}{cc}1& \tan 𝜃\\ -\tan 𝜃& 1\end{array}\right]\\ {𝐴}^{-1}& =\frac{\mathrm{adj}\left(𝐴\right)}{\left|𝐴\right|}\\ & =\left(\frac{1}{{\sec }^2𝜃}\right)\left[\begin{array}{cc}1& -\tan 𝜃\\ \tan 𝜃& 1\end{array}\right]\end{array}$

So, $\frac{1}{{\sec }^2𝜃}\left[\begin{array}{cc}1& -\tan 𝜃\\ \tan 𝜃& 1\end{array}\right]\left[\begin{array}{cc}1& -\tan 𝜃\\ \tan 𝜃& 1\end{array}\right]$

$\begin{array}{cc}& =\frac{1}{{\sec }^2𝜃}\left[\begin{array}{cc}1-{\tan }^2𝜃& -2\tan 𝜃\\ 2\tan 𝜃& -{\tan }^2𝜃+1\end{array}\right]\\ & \left[\begin{array}{cc}\cos 2𝜃& -\sin 2𝜃\\ \sin 2𝜃& \cos 2𝜃\end{array}\right]=\left[\begin{array}{cc}𝑎& -𝑏\\ 𝑏& 𝑎\end{array}\right]\\ & \Rightarrow 𝑎=\cos 2𝜃,𝑏=\sin 2𝜃\end{array}$

$\begin{array}{cc}& {𝑧}_1=2+3𝑖,{𝑧}_2=3+2𝑖\\ & \Rightarrow \left[\begin{array}{cc}{𝑧}_1& {𝑧}_2\\ -{\overline{𝑧}}_2& {\overline{𝑧}}_1\end{array}\right]\left[\begin{array}{cc}{\overline{𝑧}}_1& -{𝑧}_2\\ {\overline{𝑧}}_2& {𝑧}_1\end{array}\right]\\ & =\left[\begin{array}{cc}{\left|{𝑧}_1\right|}^2+{\left|{𝑧}_2\right|}^2& 0\\ 0& {\left|{𝑧}_1\right|}^2+{\left|{𝑧}_2\right|}^2\end{array}\right]\\ & \Rightarrow {\left|{𝑧}_1\right|}^2=4+9=13\\ & \Rightarrow {\left|{𝑧}_2\right|}^2=9+4=13=\left[\begin{array}{cc}26& 0\\ 0& 26\end{array}\right]=26𝐼\\ & \end{array}$

$\begin{array}{cc}& ⎡⎣⎢\begin{array}{ccc}𝑎& 𝑏& 𝑐\\ 𝑎-𝑏& 𝑏-𝑐& 𝑐-𝑎\\ 𝑏+𝑐& 𝑐+𝑎& 𝑎+𝑏\end{array}⎤⎦⎥[{𝑅}_1\to {𝑅}_1+{𝑅}_3]\\ & =(𝑎+𝑏+𝑐)⎡⎣⎢\begin{array}{ccc}1& 1& 1\\ 𝑎-𝑏& 𝑏-𝑐& 𝑐-𝑎\\ 𝑏+𝑐& 𝑐+𝑎& 𝑎+𝑏\end{array}⎤⎦⎥\\ & =(𝑎+𝑏+𝑐)⎡⎣⎢\begin{array}{ccc}1& 1& 1\\ 𝑎+𝑐& 𝑏+𝑎& 𝑏+𝑐\\ 𝑏+𝑐& 𝑐+𝑎& 𝑎+𝑏\end{array}⎤⎦⎥[{𝑅}_2\to {𝑅}_2+{𝑅}_3]\\ & \end{array}$

$\begin{array}{cc}{𝐶}_2& \to {𝐶}_2-{𝐶}_1\text{ and }{𝐶}_3\to {𝐶}_3-{𝐶}_1\\ & =(𝑎+𝑏+𝑐)⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 𝑎+𝑐& 𝑏-𝑐& 𝑏-𝑎\\ 𝑏+𝑐& 𝑎-𝑏& 𝑎-𝑐\end{array}⎤⎦⎥\\ & =(𝑎+𝑏+𝑐)\left[\right(𝑏-𝑐\left)\right(𝑎-𝑐)-(𝑏-𝑎\left)\right(𝑎-𝑏\left)\right]\\ & =(𝑎+𝑏+𝑐)(𝑎𝑏-𝑏𝑐-𝑐𝑎+{𝑐}^2-𝑏𝑎+{𝑏}^2+{𝑎}^2-𝑎𝑏)\\ & =(𝑎+𝑏+𝑐)({𝑎}^2+{𝑏}^2+{𝑐}^2-𝑎𝑏-𝑏𝑐-𝑐𝑎)\\ & ={𝑎}^3+{𝑏}^3+{𝑐}^3-3𝑎𝑏𝑐\end{array}$

$\begin{array}{cccc}& & (1+𝑖)(1+3𝑖)(1+7𝑖)& =𝑥+𝑖𝑦\\ \Rightarrow & & (-2+4𝑖)(1+7𝑖)& =𝑥+𝑖𝑦\\ \Rightarrow & & -30-10𝑖& =𝑥+𝑖𝑦\\ \Rightarrow & & |𝑥+𝑖𝑦|& =\sqrt{\surd {30}^2+{10}^2}\\ & & =\sqrt{\surd 1000}=10\sqrt{\surd 10}\end{array}$

$\because 1,{𝛼}_1,{𝛼}_2,{𝛼}_3,{𝛼}_4$ are the roots of ${𝑧}^5-1=0$

$\begin{array}{c}\Rightarrow (𝑧-1)(𝑧-{𝛼}_1)(𝑧-{𝛼}_2)(𝑧-{𝛼}_3)(𝑥-{𝛼}_4)={𝑧}^5-1\\ \Rightarrow (𝜔-1)(𝜔-{𝛼}_1)(𝜔-{𝛼}_2)(𝜔-{𝑑}_3)(𝜔-{𝛼}_4)={𝜔}^5-1\\ (𝜔-1)(𝜔-{𝛼}_1)(𝜔-{𝛼}_2)(𝜔-{𝛼}_3)(𝜔-{𝛼}_4)+𝜔\\ =({𝜔}^5-1)+𝜔={𝜔}^2-1+𝜔\\ =-1-1=-2\end{array}$

$\begin{array}{cc}& 𝑎>0,𝑧=𝑥+𝑖𝑦\\ & {\log }_{{\cos }^2{}_{𝜃}}|𝑧-𝑎|>{\log }_{{\cos }^2{}_{𝜃}}|𝑧-𝑎𝑖|\\ & 0<{\cos }^2𝜃<1\\ & \text{ So, }|𝑧-𝑎|<|𝑧-𝑎𝑖|\\ & \Rightarrow (𝑥-𝑎{)}^2+{𝑦}^2<{𝑥}^2+(𝑦-𝑎{)}^2\\ & \Rightarrow {𝑥}^2+{𝑎}^2-2𝑎𝑥+{𝑦}^2<{𝑥}^2+{𝑦}^2+{𝑎}^2-2𝑎𝑦\\ & \Rightarrow -2𝑎𝑥<-2𝑎𝑦\\ & \Rightarrow 𝑥>𝑦\end{array}$

$𝑖{𝑥}^2-2(𝑖+1)𝑥+(2-𝑖)=0$ ..... (i)

Let the other roots be $𝛼$.

$\begin{array}{cc}& [𝑥-(2-𝑖\left)\right][𝑥-𝛼]=0\\ & \Rightarrow {𝑥}^2-(2-𝑖+𝛼)𝑥-\left(𝛼\right(2-𝑖\left)\right)=0\\ & \text{ {comparing with Eq. (i) }\\ & \Rightarrow 𝑖{𝑥}^2-(2𝑖+1+𝛼𝑖)𝑥-\left[𝛼\right(2𝑖+1\left)\right]=0\\ & \Rightarrow 2𝑖+1+𝛼𝑖=2𝑖+2\\ & \Rightarrow \\ & 𝛼𝑖=1\\ & \Rightarrow 𝛼=\frac{1}{𝑖}=-𝑖\\ & \end{array}$

${𝑥}^2+𝑥+1=0$

Roots : $𝛼=𝜔$ and $𝛽={𝜔}^2$

$\begin{array}{cc}\text{ where, }𝜔& =\frac{-1+\sqrt{\surd 3}𝑖}{2}\text{ and }{𝜔}^3=1\\ {𝛼}^{2021}& ={𝜔}^{2021}={𝜔}^2\\ \text{ and }{𝛽}^{2021}& ={𝜔}^{4042}=𝜔\end{array}$

$\therefore$ Equation whose roots are ${𝛼}^{2021}$ and ${𝛽}^{2021}$ is

$(𝑥-𝜔)(𝑥-{𝜔}^2)={𝑥}^2+𝑥+1$

This is the graph of $𝑓\left(𝑥\right)={𝑥}^3-4{𝑥}^2+5𝑥-2$

And the graph of $𝑓(𝑥+\frac{1}{3})$ will shift towards left by $1/3$ units, so the new roots will also shift by $1/3$ units towards left.

$\therefore \text{ Roots }=(1-\frac{1}{3},1-\frac{1}{3},2-\frac{1}{3})=(\frac{2}{3},\frac{2}{3},\frac{5}{3})$

$𝑓\left(𝑥\right)=2{𝑥}^3+𝑚{𝑥}^2-13𝑥+𝑛$

Let $𝛼,𝛽,𝛾$ are the roots of $𝑓\left(𝑥\right)=0$ where $𝛼=2,𝛽=3$ and let $𝛾=𝑘$

$\begin{array}{cc}\Rightarrow & 𝛼+𝛽+𝛾=-\frac{𝑚}{2}=2+3+𝑘\text{......(i)}\\ \Rightarrow & 𝛼𝛽+𝛽𝛾+𝛾𝛼=-\frac{13}{2}=6+5𝑘\text{......(ii)}\\ \Rightarrow & 𝛼𝛽𝛾=-\frac{𝑛}{2}=6𝑘\text{......(iii)}\end{array}$

From Eq. (ii), $-13=12+10𝑘$

$\Rightarrow \begin{array}{cc}𝑘& =\frac{-5}{2}\\ 𝑛& =-12𝑘=30\\ 𝑚& =-10-2𝑘=-5\end{array}$

$\begin{array}{cc}& {}^6{𝑃}_4+4\cdot {}^6{𝑃}_3\\ & (\frac{6!}{2!}+4\cdot \frac{6!}{3!})=360+480[\because {}^{𝑛}{𝑃}_{𝑟}=\frac{𝑛!}{(𝑛-𝑟)!}]\\ & =840\end{array}$

$\underset{–}{{𝐵}_1}{𝐺}_1\underset{–}{{𝐵}_2}{𝐺}_2\underset{–}{{𝐵}_3}$

${𝐺}_1$ and ${𝐺}_2$ can be arranged in 2! ways. ${𝐵}_1,{𝐵}_2$ and ${𝐵}_3$ can be arranged in 3! ways.

$\therefore$ Required number of arrangements

$=2!\times 3!$

$=12$

5 balls $\to$ 4 Tins

Each ball can be placed in 4 ways.

So, number of ways $=4\times 4\times 4\times 4\times 4=4^5$

$\begin{array}{cc}& \frac{3𝑥-2}{(𝑥+1{)}^2(𝑥+3)}=\frac{𝐴}{𝑥+1}+\frac{𝐵}{(𝑥+1{)}^2}+\frac{𝐶}{(𝑥+3)}\\ & \Rightarrow 3𝑥-2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1{)}^2\\ & \text{ Putting }𝑥=-1\\ & -5=2𝐵\\ & \Rightarrow 𝐵=\frac{-5}{2}\\ & \text{ Putting }𝑥=-3\\ & -11=4𝐶\Rightarrow 𝐶=\frac{-11}{4}\end{array}$

Putting $𝑥=-3$

$-11=4𝐶\Rightarrow 𝐶=\frac{-11}{4}$

Putting $𝑥=0$

$\begin{array}{cc}-2& =3𝐴+3𝐵+𝐶\Rightarrow -2=3𝐴-\frac{15}{2}-\frac{11}{4}\\ \Rightarrow \frac{33}{4}& =3𝐴\Rightarrow 𝐴=\frac{11}{4}\\ \Rightarrow 4𝐴+2𝐵& +4𝐶=11-5-11=-5\end{array}$

$\begin{array}{cc}\cos \left(\frac{𝜋}{8}\right)& =\sqrt{\surd \frac{1+\cos \frac{𝜋}{4}}{2}}\\ & =\sqrt{\surd \frac{1+\frac{1}{\sqrt{\surd 2}}}{2}}=\sqrt{\surd \frac{\sqrt{\surd 2}+1}{2\sqrt{\surd 2}}}\end{array}$

$\begin{array}{cc}& \cos 𝜃=-\sqrt{\surd \frac{3}{2}}\\ & \sin 𝛼=-\frac{3}{5}\\ & \text{ As, }-1\le \cos 𝜃\le 1\end{array}$

In the question given, $\cos 𝜃=-\sqrt{\surd \frac{3}{2}}$. So, this is a wrong question.

$\begin{array}{cc}& \tan 𝛽=\frac{\tan 𝛼+\tan 𝛾}{1+\tan 𝛼\tan 𝛾}\\ & \tan 𝛽=\frac{\sin 𝛼\cos 𝛾+\sin 𝛾\cos 𝛼}{\cos 𝛼\cos 𝛾+\sin 𝛼\sin 𝛾}\\ & \tan 𝛽=\frac{\sin (𝛼+𝛾)}{\cos (𝛼-𝛾)}\\ & \frac{\sin 2𝛼+\sin 2𝛾}{1+\sin 2𝛼\sin 2𝛾}\\ & =\frac{2\sin (𝛼+𝛾)\cos (𝛼-𝛾)}{1+(2\sin 𝛼\cos 𝛾)(2\sin 𝛾\cos 𝛼)}\\ & =\frac{2\sin (𝛼+𝛾)\cos (𝛼-𝛾)}{1+[\sin (𝛼+𝛾)+\sin (𝛼-𝛾\left)\right][\sin (𝛼+𝛾)-\sin (𝛼-𝛾\left)\right]}\end{array}$

$\begin{array}{cc}& =\frac{2\sin (𝛼+𝛾)\cos (𝛼-𝛾)}{1+{\sin }^2(𝛼+𝛾)-{\sin }^2(𝛼-𝛾)}\\ & =\frac{2\sin (𝛼+𝛾)\cos (𝛼-𝛾)}{{\sin }^2(𝛼+𝛾)+{\cos }^2(𝛼-𝛾)}=\frac{2\tan 𝛽}{1+{\tan }^2𝛽}=\sin 2𝛽\end{array}$

$\begin{array}{cc}& \sin (\frac{𝜋}{4}\cot 𝜃)=\cos (\frac{𝜋}{4}\tan 𝜃)\\ & \sin 𝐴=\cos 𝐵\\ & \text{ If }𝐴+𝐵=\frac{𝜋}{2}\\ & \Rightarrow \sin 𝐴=\cos 𝐵\\ & \Rightarrow \frac{𝜋}{4}(\tan 𝜃+\cot 𝜃)=\frac{𝜋}{2}\\ & \Rightarrow \tan 𝜃+\frac{1}{\tan 𝜃}=2\Rightarrow {\tan }^2𝜃\cdot 2\tan 𝜃+1=0\\ & \Rightarrow (\tan 𝜃-1{)}^2=0\Rightarrow {\tan }^2𝜃-1=0\\ & \Rightarrow \tan 𝜃=1\\ & 𝜃=𝑛𝜋+\frac{𝜋}{4}\end{array}$

$\begin{array}{cc}& 𝑥=\sin (2{\tan }^{-1}2)=\frac{2\cdot 3}{1+3^2}=\frac{3}{5}\\ & 𝑦=\cos (2{\tan }^{-1}3)=\frac{(1-3^2)}{(1+3^2)}=\frac{-4}{5}\\ & 𝑧=\sec (3{\tan }^{-1}4)={\tan }^{-1}4=𝜃\\ & \cos 3𝜃=4{\cos }^3𝜃-3\cos 𝜃\\ & =4\cdot \left(\frac{1}{17\sqrt{\surd 17}}\right)-3\left(\frac{1}{\sqrt{\surd 17}}\right)\\ & \end{array}$

$=\frac{-47}{17\sqrt{\surd 17}}$

$\sec 3𝜃=\frac{-17\sqrt{\surd 17}}{47}$

$\therefore 𝑥>𝑦>𝑧$.

Given : $𝐴𝐷=4;\angle 𝐷𝐴𝐵=\frac{𝜋}{6};\angle 𝐴𝐵𝐸=\frac{𝜋}{3}$

$\begin{array}{cc}𝐴𝑂=\frac{8}{3}\text{ and }𝑂𝐷=& \frac{4}{3}\\ \\ & [\because 𝑂\text{ divides }𝐴𝐷\text{ in }2:1]\end{array}$

Area of $△𝐴𝐵𝐶=2\left(\text{Area of}△𝐴𝐵𝐸\right)$

$\begin{array}{cc}& =2\left[\frac{3}{2}\right(\text{ Area of }△𝐴𝑂𝐵\left)\right]\\ \\ & =3\times \text{ area of }△𝐴𝑂𝐵\\ \\ & =3\times \frac{1}{2}\times 𝐵𝑂\times 𝐴𝑂\\ \\ & =3\cdot \frac{1}{2}\cdot \frac{8}{3\sqrt{\surd 3}}\cdot \frac{8}{3}\\ \\ & \{\because \tan \frac{𝜋}{6}=\frac{𝐵𝑂}{𝐴𝑂}\Rightarrow 𝐵𝑂=\frac{1}{\sqrt{\surd 3}}\times \frac{8}{3}\}\\ \\ & =\frac{32}{3\sqrt{\surd 3}}\end{array}$

$\begin{array}{cc}2{\Delta }^2({𝑎}^2+{𝑏}^2+{𝑐}^2)& ={𝑎}^2{𝑏}^2{𝑐}^2\\ {𝑎}^2+{𝑏}^2+{𝑐}^2& ={\left(\frac{𝑎𝑏𝑐}{\Delta }\right)}^2\cdot \frac{1}{2}=8{𝑅}^2\end{array}$

$[\because 𝑎𝑏𝑐/\Delta =4𝑅$ where $𝑅$-circumradius

$\begin{array}{cc}& \because 𝑎=2𝑅\sin 𝐴,𝑏=2𝑅\sin 𝐵,𝑐=2𝑅\sin 𝐶]\\ & \Rightarrow 4{𝑅}^2({\sin }^2𝐴+{\sin }^2𝐵+{\sin }^2𝐶)=8{𝑅}^2\\ & \Rightarrow {\sin }^2𝐴+{\sin }^2𝐵+{\sin }^2𝐶=2\\ & \Rightarrow (\cos 2𝐴+\cos 2𝐵+\cos 2𝐶)=-1\\ & \Rightarrow -1-4\cos 𝐴\cos 𝐵\cos 𝐶=-1\\ & \cos 𝐴\cos 𝐵\cos 𝐶=0\\ & \end{array}$

So, any one among $𝐴,𝐵$ and $𝐶$ has to be $\frac{𝜋}{2}$.

So, right angled triangle.

$\begin{array}{cc}& \cos (𝛼+\frac{𝜋}{2})=-\sin 𝛼\\ & \cos (𝛽+\frac{𝜋}{2})=-\sin 𝛽\\ & \cos (𝛾+\frac{𝜋}{2})=-\sin 𝛾\\ & \because 𝐴𝑀\ge 𝐺𝑀\\ & \frac{-(\sin 𝛼+\sin 𝛽+\sin 𝛾)}{3}\ge (-\sin 𝛼\sin 𝛽\sin 𝛾{)}^{1/3}\end{array}$

Equating holds, when $𝛼=𝛽=𝛾$

$\begin{array}{c}𝛼+𝛽+𝛾=360𝛾\\ 𝛼=𝛽=𝛾=120𝛾\\ \Rightarrow 𝐴𝑀=\frac{1}{3}[(-\frac{\sqrt{\surd 3}}{2})+(-\frac{\sqrt{\surd 3}}{2})+(-\frac{\sqrt{\surd 3}}{2})]\\ =-\frac{\sqrt{\surd 3}}{2}\end{array}$