AP EAPCET 2021 19th August Morning Shift Question Paper with Solutions

Given,

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

Then,

$\begin{array}{cc}𝑔\circ 𝑓\left(𝑥\right)& =𝑔\left[𝑓\right(𝑥\left)\right]\\ & =𝑔(2𝑥+1)=(2𝑥+1{)}^2-2\\ & =4{𝑥}^2+4𝑥+1-2=4{𝑥}^2+4𝑥-1\end{array}$

Given, $𝑓\left(𝑥\right)=\frac{{𝑎}^{𝑥}+{𝑎}^{-𝑥}}{2}$

Then,

$\begin{array}{cc}& 𝑓(𝑥+𝑦)+𝑓(𝑥-𝑦)=\frac{{𝑎}^{𝑥+𝑦}+{𝑎}^{-(𝑥+𝑦)}}{2}\\ & +\frac{{𝑎}^{𝑥-𝑦}+{𝑎}^{-(𝑥-𝑦)}}{2}\\ & =\frac{{𝑎}^{𝑥}\cdot {𝑎}^{𝑦}+{𝑎}^{-𝑥}\cdot {𝑎}^{-𝑦}+{𝑎}^{𝑥}\cdot {𝑎}^{-𝑦}+{𝑎}^{-𝑥}{𝑎}^{𝑦}}{2}\\ & =\frac{{𝑎}^{𝑥}({𝑎}^{𝑦}+{𝑎}^{-𝑦})+{𝑎}^{-𝑥}({𝑎}^{𝑦}+{𝑎}^{-𝑦})}{2}\end{array}$

$\begin{array}{cc}& =\frac{({𝑎}^{𝑥}+{𝑎}^{-𝑥})({𝑎}^{𝑦}+{𝑎}^{-𝑦})}{2}\\ & =2\cdot \frac{({𝑎}^{𝑥}+{𝑎}^{-𝑥})}{2}\cdot \frac{({𝑎}^{𝑦}+{𝑎}^{-𝑦})}{2}=2𝑓\left(𝑥\right)𝑓\left(𝑦\right)\end{array}$

$\because \left\{𝑥\right\}=𝑥-\left[𝑥\right]$

$𝑓:(0,1)\to (0,1)$

Defined by $𝑓\left(𝑥\right)=min\{𝑥-[𝑥],-𝑥-[𝑥\left]\right\}$ which is an identity function, as $𝑥\in (0,1)$ and $\left\{𝑥\right\}\in (0,1)$ Where, $\{\cdot \}$ is fractional part function.

$\Rightarrow (𝑓\circ 𝑓\circ 𝑓\circ 𝑓)\left(𝑥\right)=𝑥$

$\begin{array}{c}8𝑛+16\le 2^{𝑛},𝑛\in 𝑁\\ \Rightarrow 8(𝑛+2)\le 2^{𝑛}\\ \Rightarrow 𝑛+2\le 2^{𝑛-3}\end{array}$

Clearly, for $𝑛=6$

$6+2\le 2^{6-3}$ or $8\le 8$ is true.

$\begin{array}{c}\mid \begin{array}{ccc}1& -3& 1\\ 1& 6& 4\\ 1& 3𝑥& {𝑥}^2\end{array}\mid =0\\ \Rightarrow 1(6{𝑥}^2-12𝑥)+3({𝑥}^2-4)+1(3𝑥-6)=0\\ \Rightarrow 6{𝑥}^2-12𝑥+3{𝑥}^2-12+3𝑥-6=0\\ \Rightarrow 9{𝑥}^2-9𝑥-18=0\\ \Rightarrow {𝑥}^2-𝑥-2=0\end{array}$

$\begin{array}{cc}& 𝐴=\left[\begin{array}{cc}𝑎& -𝑏\\ 𝑏& 𝑎\end{array}\right]\\ & \text{ Then, }{𝐴}^{𝑇}=\left[\begin{array}{cc}𝑎& 𝑏\\ -𝑏& 𝑎\end{array}\right]\\ & \because 𝐴{𝐴}^{𝑇}=20𝐼\\ & \Rightarrow \left[\begin{array}{cc}𝑎& -𝑏\\ 𝑏& 𝑎\end{array}\right]\left[\begin{array}{cc}𝑎& 𝑏\\ -𝑏& 𝑎\end{array}\right]=20\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ & \Rightarrow \left[\begin{array}{cc}{𝑎}^2+{𝑏}^2& 0\\ 0& {𝑎}^2+{𝑏}^2\end{array}\right]=\left[\begin{array}{cc}20& 0\\ 0& 20\end{array}\right]\\ & \therefore {𝑎}^2+{𝑏}^2=20\\ & \Rightarrow (𝑎+𝑏{)}^2-2𝑎𝑏=20\\ & \Rightarrow (𝑎+𝑏{)}^2=20+2\times \frac{5}{2}[\because 𝑎𝑏=\frac{5}{2}]\\ & \text{ or }(𝑎+𝑏{)}^2=25\\ & 𝑜𝑟𝑎+𝑏=\pm 5\end{array}$

or ${𝑥}^2\pm 5𝑥+\frac{5}{2}=0$ or $2{𝑥}^2\pm 10𝑥+5=0$

$\begin{array}{cc}& \text{ (c) }𝐴=⎡⎣⎢\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}⎤⎦⎥\\ & \therefore \left|𝐴\right|=1(1+3)+1(2+3)+1(2-1)=10...\left(i\right)\\ & \text{ and }10𝐵=⎡⎣⎢\begin{array}{ccc}4& 2& 2\\ -5& 0& 𝛼\\ 1& -2& 3\end{array}⎤⎦⎥\\ & \therefore \left|10𝐵\right|=4\left(2𝛼\right)-2(-15-𝛼)+2\left(10\right)\\ & \Rightarrow {10}^3\left|𝐵\right|=10𝛼+50[\because |𝑘𝐴|={𝑘}^{𝑛}|𝐴\left|\right]\dots \left(\mathrm{ii}\right)\\ & \because 𝐵={𝐴}^{-1}\\ & \Rightarrow \left|𝐵\right|=\mid {𝐴}^{-1}\mid =|𝐴{|}^{-1}\\ & \Rightarrow \left|𝐵\right|=\frac{1}{\left|𝐴\right|}=\frac{1}{10}[\text{ [rom Eq. (i)}\end{array}$

[from Eq. (i)] From Eq. (ii), we get

$1000.\frac{1}{10}=10𝛼+50$

or $𝛼=\frac{50}{10}=5$

Let $𝐴=⎡⎣⎢\begin{array}{ccc}4& 2& 1-𝑥\\ 5& 𝑘& 1\\ 6& 3& 1+𝑥\end{array}⎤⎦⎥$

Given that, rank of matrix $𝐴$ is 1 .

$\Rightarrow$ All the minors of order 2 are zero.

$\therefore {𝑀}_{33}=0$

$\Rightarrow \mid \begin{array}{cc}4& 2\\ 5& 𝑘\end{array}\mid =0$

or $4𝑘-10=0$ or $𝑘=\frac{5}{2}$

and ${𝑀}_{11}=\mid \begin{array}{cc}𝑘& 1\\ 3& 1+𝑥\end{array}\mid =0$

or $\mid \begin{array}{cc}\frac{5}{2}& 1\\ 3& 1+𝑥\end{array}\mid =0$

or $\frac{5}{2}+\frac{5}{2}𝑥-3=0$

or $\frac{5}{2}𝑥=\frac{1}{2}$

or $𝑥=\frac{1}{5}$

${𝑎}_1,{𝑎}_2,\dots ,{𝑎}_9$, are in GP.

Then, $\frac{{𝑎}_2}{{𝑎}_1}=\frac{{𝑎}_3}{{𝑎}_2}=\dots \frac{{𝑎}_9}{{𝑎}_8}=𝑟$ [common ratio]

Now, $\mid \begin{array}{ccc}\log {𝑎}_1& \log {𝑎}_2& \log {𝑎}_3\\ \log {𝑎}_4& \log {𝑎}_5& \log {𝑎}_6\\ \log {𝑎}_7& \log {𝑎}_8& \log {𝑎}_9\end{array}\mid$

${𝐶}_3\to {𝐶}_3-{𝐶}_2,{𝐶}_2\to {𝐶}_2-{𝐶}_1$

$\mid \begin{array}{ccc}\log {𝑎}_1& \log \frac{{𝑎}_2}{{𝑎}_1}& \log \frac{{𝑎}_3}{{𝑎}_2}\\ \log {𝑎}_4& \log \frac{{𝑎}_5}{{𝑎}_4}& \log \frac{{𝑎}_6}{{𝑎}_5}\\ \log {𝑎}_7& \log \frac{{𝑎}_8}{{𝑎}_7}& \log \frac{{𝑎}_9}{{𝑎}_8}\end{array}\mid$

or $\mid \begin{array}{ccc}\log {𝑎}_1& \log 𝑟& \log 𝑟\\ \log {𝑎}_4& \log 𝑟& \log 𝑟\\ \log {𝑎}_7& \log 𝑟& \log 𝑟\end{array}\mid =0$

[$\because$ C${}_2$ and C${}_3$ are identical]

$(\sin 𝜃-𝑖\cos 𝜃{)}^3=(-𝑖{)}^3(\cos 𝜃+𝑖\sin 𝜃{)}^3$

$=(-𝑖{)}^3(\cos 3𝜃+𝑖\sin 3𝜃)$ [by De-Moivre theorem]

Real part of $(\cos 4+𝑖\sin 4+1{)}^{2020}=$ Real part of ${(2{\cos }^{2}2+𝑖2\sin 2\cos 2)}^{2020}$

$\begin{array}{cc}& =\text{ Real part of }{2}^{2020}{\cos }^{2020}\left(2\right)(\cos 2+𝑖\sin 2{)}^{2020}\\ & =\text{ Real part of }{2}^{2020}{\cos }^{2020}\left(2\right)(\cos 4040+𝑖\sin 4040)\\ & =2^{2020}{\cos }^{2020}2\cdot \cos 4040[𝑧+\overline{𝑧}=2\mathrm{Re}(𝑧\left)\right]\end{array}$

$\because$ $({𝑥}^2+5𝑥+5{)}^{𝑥+5}=1$

Case I $𝑥+5=0$

$\Rightarrow 𝑥=-5$

Case II ${𝑥}^2+5𝑥+5=1$

or, ${𝑥}^2+5𝑥+4=0$

or, $(𝑥+1)(𝑥+4)=0$

$\Rightarrow 𝑥=-1,-4$

$𝑥=-1,-4$ and $-5$ are the three integers satisfying given equation.

$𝑓\left(𝑥\right)={𝑥}^3+𝑎{𝑥}^2+𝑏𝑥+𝑐$

Let roots of $𝑓\left(𝑥\right)$ are $𝛼,𝛽,𝛾$ and since roots are in AP.

$\therefore 2𝛽=𝛼+𝛾$

Now, Sum of roots

$\begin{array}{cc}𝛼+𝛽+𝛾& =-𝑎\\ \text{ or }2𝛽+𝛽& =-𝑎\text{ or }𝛽=-\frac{𝑎}{3}\end{array}$

It is given that, roots are integers.

$\therefore 𝛽$ is an integer when $𝑎$ is multiple of 3 .

Option (a), (c) and (d) are multiple of 3.

$\Rightarrow 𝑎\ne 1214$

Option (b) is correct.

${𝑒}^{4𝑡}-10{𝑒}^{3𝑡}+29{𝑒}^{2𝑡}-22{𝑒}^{𝑡}+4=0$ ...... (i)

Let ${𝑒}^{𝑡}=𝑥$ and roots of (1) be ${𝑡}_1,{𝑡}_2,{𝑡}_3,{𝑡}_4$

$\therefore {𝑥}^4-10{𝑥}^3+29{𝑥}^2-22𝑥+4=0$

Then, roots are ${𝑥}_1,{𝑥}_2,{𝑥}_3,{𝑥}_4$

product of roots ${𝑥}_1\cdot {𝑥}_2\cdot {𝑥}_3\cdot {𝑥}_4=4$

$\begin{array}{cc}& \Rightarrow {𝑒}^{{𝑡}_1}\cdot {𝑒}^{{𝑡}_2}\cdot {𝑒}^{{𝑡}_3}\cdot {𝑒}^{{𝑡}_4}=4\Rightarrow {𝑒}^{{𝑡}_1+{𝑡}_2+{𝑡}_3+{𝑡}_4}=4\\ & \Rightarrow {𝑡}_1+{𝑡}_2+{𝑡}_3+{𝑡}_4={\log }_{𝑒}4={\log }_{𝑒}{2}^2=2{\log }_{𝑒}2\end{array}$

Sum of roots $=2{\log }_{𝑒}2$

Total number of coins = 3

The number of different sums

= ³C₁ + ³C₂ + ³C₃

= 3 + 3 + 1 = 7

Number of white balls = 7

Number of black balls = 3 $\times$ W $\times$ W $\times$ W $\times$ W $\times$ W $\times$ W $\times$ W $\times$

$\times$ are the places in row where black balls can be placed, which are 8 in number for 3 black balls.

Number of arrangements $={}^8{𝐶}_3=\frac{8\times 7\times 6}{1\times 2\times 3}=56$

Total number of ways of dividing n identical rings to r different girls, so that every girl gets at least one ring $={}^{𝑛-1}{𝐶}_{𝑟-1}$

Here, we have to distribute 8 identical rings to 3 different girls.

This can be done in ${}^{8-1}{𝐶}_{3-1}={}^7{𝐶}_2=\frac{7\times 6}{1\times 2}=21$ ways

$\frac{{𝑥}^4}{(𝑥-1)(𝑥-2)}=\frac{{𝑥}^4}{{𝑥}^2-3𝑥+2}$ is an improper fraction.

$\therefore \frac{{𝑥}^4}{(𝑥-1)(𝑥-2)}={𝑥}^3+3𝑥+7+\frac{15𝑥-14}{(𝑥-1)(𝑥-2)}$

$={𝑥}^2+3𝑥+7+\frac{𝐴}{𝑥-1}+\frac{𝐵}{𝑥-2}$

On comparing with given equation,

$\frac{{𝑥}^4}{(𝑥-1)(𝑥-2)}=𝑓\left(𝑥\right)+\frac{𝐴}{𝑥-1}+\frac{𝐵}{𝑥-2}$, we have $𝑓\left(𝑥\right)={𝑥}^2+3𝑥+7$

$\begin{array}{cc}& \because \tan (𝐴+𝐵+𝐶)\\ & =\frac{\tan 𝐴+\tan 𝐵+\tan 𝐶-\tan 𝐴\tan 𝐵\tan 𝐶}{1-\tan 𝐴\tan 𝐵-\tan 𝐵\tan 𝐶-\tan 𝐶\tan 𝐴}\end{array}$

If $𝐴+𝐵+𝐶=90{\rm Y}$, then

$1-\tan 𝐴\tan 𝐵-\tan 𝐵\tan 𝐶-\tan 𝐶\tan 𝐴=0$

$\text{ or }\tan 𝐴\tan 𝐵+\tan 𝐵\tan 𝐶+\tan 𝐶\tan 𝐴=1$

Here, $𝐴=2𝛼,𝐵=30{\rm Y}-𝛼,𝐶=60{\rm Y}-𝛼$, then

$\begin{array}{cc}& 𝐴+𝐵+𝐶=2𝛼+30𝑌-𝛼+60𝑌-𝛼={90}^{\circ }\\ & \therefore \tan 2𝛼\tan (30𝑌-𝛼)+\tan 2𝛼\cdot \tan (60𝑌-𝛼)\\ & +\tan (30𝑌-𝛼)\tan (60𝑌-𝛼)=1\end{array}$

$\sin 𝛼-\cos 𝛼=𝑚$ and $\sin 2𝛼=𝑛-{𝑚}^2$

$\therefore$ $(\sin 𝛼-\cos 𝛼{)}^2={𝑚}^2$

${\sin }^2𝛼+{\cos }^2𝛼-2\sin 𝛼\cos 𝛼={𝑚}^2$

or $1-\sin 2𝛼={𝑚}^2$

or $\sin 2𝛼=1-{𝑚}^2$

But $\sin 2𝛼=𝑛-{𝑚}^2$ (given)

$\Rightarrow 𝑛=1$

$3\mathrm{cosec}𝑥=4\sin 𝑥$

or ${\sin }^2𝑥=3/4$

or ${\sin }^2𝑥={\sin }^2\frac{𝜋}{3}\Rightarrow 𝑥=𝑛𝜋\pm \frac{𝜋}{3},𝑛\in 𝑍$

or $𝑥=\pm \frac{𝜋}{3}$

$\begin{array}{cc}& {\tan }^{-1}\left(\frac{1}{1+1\cdot 2}\right)+{\tan }^{-1}\left(\frac{1}{1+2\cdot 3}\right)+\dots \\ & +{\tan }^{-1}\left(\frac{1}{1+𝑛(𝑛+1)}\right)={\tan }^{-1}\left(𝑥\right)\\ & \Rightarrow {\tan }^{-1}\left(\frac{2-1}{1+1\cdot 2}\right)+{\tan }^{-1}\left(\frac{3-2}{1+2\cdot 3}\right)+\dots \\ & +{\tan }^{-1}\left(\frac{𝑛+1-𝑛}{1+𝑛(𝑛+1)}\right)={\tan }^{-1}\left(𝑥\right)\\ & \Rightarrow {\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)+\dots \\ & \Rightarrow {\tan }^{-1}(𝑛+1)-{\tan }^{-1}\left(𝑛\right)={\tan }^{-1}\left(𝑥\right)\\ & \Rightarrow {\tan }^{-1}\left(\frac{𝑛+1-1}{1+(𝑛+1)\cdot 1}\right)={\tan }^{-1}\left(𝑥\right)={\tan }^{-1}\left(𝑥\right)\\ & \Rightarrow {\tan }^{-1}\left(\frac{𝑛}{𝑛+2}\right)={\tan }^{-1}\left(𝑥\right)\Rightarrow 𝑥=\frac{𝑛}{𝑛+2}\end{array}$

$\begin{array}{cc}& \because \tan 𝜃=\sinh 𝑢\\ & \Rightarrow \tan 𝜃=\sqrt{\surd {\cosh }^2𝑢-1}[{\cosh }^2𝑥-{\sinh }^2𝑥=1]\\ & \Rightarrow \cosh 𝑢=\sqrt{\surd 1+{\tan }^2𝜃}\Rightarrow \cosh 𝑢=\sec 𝜃\end{array}$

In $\Delta 𝐴𝐵𝐶,𝑎=3,𝑏=4,\sin 𝐴=3/4$

$\therefore$ $\cos 𝐴=\sqrt{\surd 1-{\sin }^2𝐴}=\sqrt{\surd 1-\frac{9}{16}}=\frac{\sqrt{\surd 7}}{4}$

But $\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}$

$\Rightarrow \frac{\sqrt{\surd 7}}{4}=\frac{4^2+{𝑐}^2-3^3}{2.4.𝑐}$

$\begin{array}{c}\Rightarrow 2\sqrt{\surd 7}𝑐=7+{𝑐}^2\\ \Rightarrow {𝑐}^2-2\sqrt{\surd 7}𝑐+7=0\\ \Rightarrow (𝑐-\sqrt{\surd 7}{)}^2=0\\ \therefore 𝑐=\sqrt{\surd 7}\\ \text{ Now, }\cos 𝐵=\frac{{𝑐}^2+{𝑎}^2-{𝑏}^2}{2𝑎𝑐}\\ \Rightarrow \cos 𝐵=\frac{7+9-16}{2\cdot 3\cdot \sqrt{\surd 7}}=0\Rightarrow 𝐵=90{\rm Y}\\ \therefore \angle 𝐶𝐵𝐴=90{\rm Y}\end{array}$

$</mo><mo>,</mo><mi>𝑡</mi><mi>ℎ</mi><mi>𝑒</mi><mi>𝑛</mi><mo>$

$\because \frac{𝑎}{\sin 𝐴}=\frac{𝑏}{\sin 𝐵}=\frac{𝐶}{\sin 𝐶}=2𝑅$ where $𝑅$ is circumradius

$\Rightarrow 𝑎=2𝑅\sin 𝐴,𝑏=2𝑅\sin 𝐵,𝑐=2𝑅\sin 𝐶$

Then, $𝑏+𝑐\sqrt{\surd 2}=2𝑅(\sin 𝐵+\sqrt{\surd 2}\sin 𝐶)$

$\begin{array}{cc}& =2𝑅(\sin 45𝑌+\sqrt{\surd 2}\sin 60𝑌)\\ & =2𝑅(\frac{1}{\sqrt{\surd 2}}+\frac{\sqrt{\surd 3}}{\sqrt{\surd 2}})=2𝑅\left(\frac{\sqrt{\surd 3}+1}{\sqrt{\surd 2}}\right)\\ & =2\cdot 2𝑅\left(\frac{\sqrt{\surd 3}+1}{2\sqrt{\surd 2}}\right)\\ & =2\cdot (2𝑅\sin 75𝑌)\\ & =2(2𝑅\sin 𝐴)\\ & =2𝑎\end{array}$

In $△𝐴𝐵𝐶$

$\begin{array}{cc}& 𝑎=2𝑅\sin 𝐴=4𝑅\sin \frac{𝐴}{2}\cos \frac{𝐴}{2}\\ & 𝑏=2𝑅\sin 𝐵=4𝑅\sin \frac{𝐵}{2}\cos \frac{𝐵}{2}\\ & 𝑟=4𝑅\sin \frac{𝐴}{2}\sin \frac{𝐵}{2}\sin \frac{𝐶}{2}\\ & {𝑟}_1=4𝑅\sin \frac{𝐴}{2}\cos \frac{𝐵}{2}\cos \frac{𝐶}{2}\\ & {𝑟}_2=4𝑅\cos \frac{𝐴}{2}\sin \frac{𝐵}{2}\cos \frac{𝐶}{2}\\ & {𝑟}_3=4𝑅\cos \frac{𝐴}{2}\cos \frac{𝐵}{2}\sin \frac{𝐶}{2}\end{array}$

$\text{ Now, }\begin{array}{cc}& 𝑎𝑏-{𝑟}_1{𝑟}_2=16{𝑅}^2\sin \frac{𝐴}{2}\sin \frac{𝐵}{2}\cos \frac{𝐴}{2}\\ & \cos \frac{𝐵}{2}(1-{\cos }^2\frac{𝐶}{2})\\ & =16{𝑅}^2\sin \frac{𝐴}{2}\sin \frac{𝐵}{2}\cos \frac{𝐴}{2}\cos \frac{𝐵}{2}{\sin }^2\frac{𝐶}{2}\end{array}$

$\therefore \frac{𝑎𝑏-{𝑟}_1{𝑟}_2}{{𝑟}_3}=4𝑅\sin \frac{𝐴}{2}\sin \frac{𝐵}{2}\sin \frac{𝐶}{2}=𝑟$

$𝑙=\cos 𝛼$

$𝑚=\cos 𝛼$

$𝑛=\cos 90𝛾=0$

$\because {𝑙}^2+{𝑚}^2+{𝑛}^2=1$

$\Rightarrow 2{\cos }^2𝛼=1$

$\Rightarrow {\cos }^2𝛼=\frac{1}{2}$

$\Rightarrow \cos 𝛼=\pm \frac{1}{\sqrt{\surd 2}}$

$\Rightarrow 𝛼=45{\rm Y}$ or ${135}^{\circ }$

Let $𝐴(5,-4,-3)$

Direction ratios of $𝐎𝐀=5,-4,-3$

Direction ratios of $𝑋$-axis $=1,0,0$

$\begin{array}{cc}\text{ Angle, }\cos 𝜃& =\frac{5\cdot 1+(-4)\cdot 0+(-3)\cdot 0}{\sqrt{\surd 5^2+(-4{)}^2+(-3{)}^2}\sqrt{\surd 1^2+0^2+0^2}}\\ & =\frac{5}{\sqrt{\surd 50}}=\frac{1}{\sqrt{\surd 2}}\\ \therefore 𝜃& =\frac{𝜋}{4}\end{array}$