AP EAPCET 2022 4th July Morning Shift Question Paper with Solutions

$𝑓\left(𝑥\right)=\sqrt{\surd \frac{{𝑥}^2+2𝑥+8}{{𝑥}^2+2𝑥+4}}$

Let $𝑦=\frac{{𝑥}^2+2𝑥+8}{{𝑥}^2+2𝑥+4}[\because 𝑓(𝑥)=\sqrt{\surd 𝑦}]$

$\begin{array}{cc}& \Rightarrow 𝑦{𝑥}^2+2𝑥𝑦+4𝑦={𝑥}^2+2𝑥+8\\ & \Rightarrow (𝑦-1){𝑥}^2+2𝑥(𝑦-1)+4𝑦-8=0\end{array}$

For $𝑥$ to be real, discriminant of equation must be greater than equals to zero.

$\begin{array}{ccc}& \left[2\right(𝑦-1){]}^2-4(𝑦-1\left)\right(4𝑦-8)& \ge 0\\ \Rightarrow & 4(𝑦-1{)}^2-4(𝑦-1\left)4\right(𝑦-2)& \ge 0\\ \Rightarrow & (𝑦-1{)}^2-(𝑦-1\left)4\right(𝑦-2)& \ge 0\\ \Rightarrow & (𝑦-1)[𝑦-1-4𝑦+8]& \ge 0\\ \Rightarrow & (𝑦-1)(-3𝑦+7)& \ge 0\\ & 𝑦=1,\frac{7}{3}\end{array}$

$\begin{array}{cc}& \therefore (𝑦-1)(-3𝑦+7)\ge 0\Rightarrow 𝑦\in (1,\frac{7}{3}]\\ & [\because \text{ At }𝑦=1,{𝑥}^2+2𝑥+4\ne {𝑥}^2+2𝑥+8]\\ & \because 𝑦\in (1,\frac{7}{3}]\\ & \therefore 𝑓\left(𝑥\right)\in (1,\sqrt{\surd \frac{7}{3}}]\\ & \end{array}$

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

$\begin{array}{c}2-{𝑥}^2\ge 0\\ \Rightarrow {𝑥}^2\le 2\\ \Rightarrow \left|𝑥\right|\le \sqrt{\surd 2}\\ \Rightarrow -\sqrt{\surd 2}\le 𝑥\le \sqrt{\surd 2}\end{array}$

Also, $1-𝑥>0\Rightarrow 𝑥<1$

Required domain : $-\sqrt{\surd 2}\le 𝑥\le \sqrt{\surd 2}\cap 𝑥<1$

$\Rightarrow -\sqrt{\surd 2}\le 𝑥<1$

$\begin{array}{cc}& \text{ Given, }{𝑎}_2^{𝑖}+{𝑏}_2^{𝑖}+{𝑐}_2^{𝑖}=1\\ & \Rightarrow {𝑎}_2^1+{𝑏}_2^1+{𝑐}_2^1=1\\ & {𝑎}_2^2+{𝑏}_2^2+{𝑐}_2^2=1\\ & {𝑎}_2^3+{𝑏}_2^3+{𝑐}_2^3=1\end{array}$

Given, ${𝑎}_{𝑖}{𝑎}_{𝑗}+{𝑏}_{𝑖}{𝑏}_{𝑗}+{𝑐}_{𝑖}{𝑐}_{𝑗}=0[𝑖,𝑗=1,2,3,𝑖\ne 𝑗]$

$\begin{array}{cc}\text{ Let }𝐩& ={𝑎}_1𝐢+{𝑏}_1𝐣+{𝑐}_1𝐤\\ 𝐪& ={𝑎}_2𝐢+{𝑏}_2𝐣+{𝑐}_2𝐤\\ 𝐫& ={𝑎}_3𝐢+{𝑏}_3𝐣+{𝑐}_3𝐤\\ \left|𝐩\right|& =\left|𝐪\right|=\left|𝐫\right|=1\\ 𝐩\cdot 𝐪& =({𝑎}_1𝐢+{𝑏}_1𝐣+{𝑐}_1𝐤)\cdot ({𝑎}_2𝐢+{𝑏}_2𝐣+{𝑐}_3𝐤)\\ & ={𝑎}_1{𝑎}_2+{𝑏}_1{𝑏}_2+{𝑐}_1{𝑐}_2=0\\ \Rightarrow 𝐩& ⟂𝐪\end{array}$

Similarly, $𝐪⟂𝐫,𝐫⟂𝐩$

$\begin{array}{cc}& \text{ Now, }𝐴=\mid \begin{array}{ccc}{𝑎}_1& {𝑎}_2& {𝑎}_3\\ {𝑏}_1& {𝑏}_2& {𝑏}_3\\ {𝑐}_1& {𝑐}_2& {𝑐}_3\end{array}\mid =\mid \begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}\mid =\left[\begin{array}{ccc}𝐩& 𝐪& 𝐫\end{array}\right]=1\\ & \Rightarrow \det \left(𝐴{𝐴}^{𝜏}\right)=1\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝐴=\frac{1}{7}⎡⎣⎢\begin{array}{ccc}3& -2& 6\\ -6& -3& 2\\ -2& 6& 3\end{array}⎤⎦⎥\\ & \left|𝐴\right|=\frac{1}{7^3}⎡⎣⎢\begin{array}{ccc}3& -2& 6\\ -6& -3& 2\\ -2& 6& 3\end{array}⎤⎦⎥[\because \det (𝑘𝐴)={𝑘}^{𝑛}\det 𝐴]\\ & =\frac{1}{7^3}\left[3\right(-9-12)+2(-18+4)+6(-36-6\left)\right]\\ & =\frac{1}{7^3}[-63-28-252]\\ & =\frac{-343}{7^3}=-1\ne -0\end{array}\end{array}$

$\mathrm{Adj}\left(𝐴\right)=\frac{1}{7^2}⎡⎣⎢\begin{array}{ccc}-21& 14& -42\\ 42& 21& -14\\ 14& -42& -21\end{array}⎤⎦⎥$

$[\because \mathrm{adj}(𝑘𝐴)={𝑘}^{𝑛-1}\mathrm{adj}𝐴]$

$=\frac{1}{7^2}⎡⎣⎢\begin{array}{ccc}-21& 42& 14\\ 14& 21& -42\\ -42& -14& -21\end{array}⎤⎦⎥$

Now, ${𝐴}^{-1}=\frac{\mathrm{adj}\left(𝐴\right)}{\left|𝐴\right|}=\frac{-1}{7^2}⎡⎣⎢\begin{array}{ccc}-21& 42& 14\\ 14& 21& -42\\ -42& -14& -21\end{array}⎤⎦⎥$

$\begin{array}{cc}& =\frac{1}{7}⎡⎣⎢\begin{array}{ccc}3& -6& -2\\ -2& -3& 6\\ 6& 2& 3\end{array}⎤⎦⎥\\ {𝐴}^{-1}& ={𝐴}^{𝑇}\end{array}$

$𝐴=\left[\begin{array}{cc}{𝛼}^2& 5\\ 5& -𝛼\end{array}\right]$ and $\mid {𝐴}^{10}\mid =1024$

$\begin{array}{cc}& \left|𝐴\right|=-{𝛼}^3-25\\ & \because \mid {𝐴}^{10}\mid =1024\\ & \Rightarrow |𝐴{|}^{10}=2^{10}\\ & \Rightarrow \left|𝐴\right|=\pm 2\end{array}$

Taking positive value, $-{𝛼}^3-25=2\Rightarrow -{𝛼}^3=27$

$\begin{array}{cc}& \Rightarrow {𝛼}^3=-27\\ & \Rightarrow 𝛼=-3\end{array}$

Taking negative value, ${𝛼}^3-25=-2\Rightarrow -{𝛼}^3=23$

$\Rightarrow 𝛼=(-23{)}^{\frac{1}{3}}$

$𝐴=⎡⎣⎢\begin{array}{ccc}5& {\sin }^2𝜃& {\cos }^2𝜃\\ -{\sin }^2𝜃& -5& 1\\ {\cos }^2𝜃& 1& 5\end{array}⎤⎦⎥$

$\begin{array}{cc}& \left|𝐴\right|=5(-25-1)-{\sin }^2𝜃(-5{\sin }^2𝜃-{\cos }^2𝜃)\\ & +{\cos }^2𝜃(-{\sin }^2𝜃+5{\cos }^2𝜃)\\ & =-130-{\sin }^2𝜃[-5(1-{\cos }^2𝜃)-{\cos }^2𝜃]\\ & +{\cos }^2𝜃[-{\sin }^2𝜃+5(1-{\sin }^2𝜃)]\\ & =-130-{\sin }^2𝜃(-5+4{\cos }^2𝜃)+{\cos }^2𝜃(-6{\sin }^2𝜃+5)\\ & =-130+5{\sin }^2𝜃-4{\sin }^2𝜃{\cos }^2𝜃\\ & -6{\sin }^2𝜃{\cos }^2𝜃+5{\cos }^2𝜃\\ & =-130+5-10{\sin }^2𝜃{\cos }^2𝜃\\ & \left|𝐴\right|=-125-10{\sin }^2𝜃{\cos }^2𝜃\\ & \therefore |𝐴{|}_{max}=-125\\ & \end{array}$

$\begin{array}{cc}& 𝑖{𝑧}^3+{𝑧}^2-𝑧+𝑖=0\\ \Rightarrow & 𝑖{𝑧}^3-𝑧+{𝑧}^2+𝑖=0\\ \Rightarrow & 𝑖{𝑧}^3+{𝑖}^2𝑧+{𝑧}^2+𝑖=0\\ \Rightarrow & 𝑖𝑧({𝑧}^2+𝑖)+({𝑧}^2+𝑖)=0\\ \Rightarrow & ({𝑧}^2+𝑖)(𝑖𝑧+1)=0\\ \Rightarrow & ({𝑧}^2+𝑖)(𝑖𝑧-{𝑖}^2)=0\\ \Rightarrow & ({𝑧}^2+𝑖)𝑖(𝑧-𝑖)=0\\ \Rightarrow & {𝑧}^2=-𝑖\text{ or }𝑧=𝑖\\ & {𝑧}^2=1(\cos \frac{3𝜋}{2}+𝑖\sin \frac{3𝜋}{2})\\ \text{ or }& 𝑧=1(\cos \frac{𝜋}{2}+𝑖\sin \frac{𝜋}{2})\end{array}$

We can see that all the roots have modulus as 1.

So, $\left|𝑧\right|=1$

$\begin{array}{cc}& \frac{𝑥-1}{3+𝑖}+\frac{𝑦-1}{3-𝑖}=𝑖\\ \Rightarrow & \frac{(𝑥-1)(3-𝑖)+(𝑦-1)(3+𝑖)}{(3+𝑖)(3-𝑖)}=𝑖\\ \Rightarrow & \frac{3𝑥-𝑖𝑥-3+𝑖+3𝑦+𝑖𝑦-3-𝑖}{9-{𝑖}^2}=𝑖\\ \Rightarrow & 3𝑥+3𝑦-𝑖𝑥+𝑖𝑦-6=𝑖[9-(-1\left)\right]\end{array}$

$\begin{array}{c}\Rightarrow 3𝑥+3𝑦-𝑖𝑥+𝑖𝑦-6=10𝑖\\ \Rightarrow 3(𝑥+𝑦)+𝑖(-𝑥+𝑦)=6+10𝑖\end{array}$

On comparing both the sides, we get

$3(𝑥+𝑦)=6\text{ and }-𝑥+𝑦=10$

$\begin{array}{ccc}\Rightarrow & 𝑥+𝑦& =2\text{... (i)}\\ \text{ and }& -𝑥+𝑦& =10\text{.... (ii)}\end{array}$

On solving Eqs. (i) and (ii), we get

$2𝑦=12\Rightarrow 𝑦=6\text{ and }𝑥=-4$

$\begin{array}{cc}& {\cos }^4𝜃=𝑎\cos 4𝜃+𝑏\cos 2𝜃+𝑐\\ & \Rightarrow {\cos }^4𝜃=𝑎[2{\cos }^{2}2𝜃-1]+𝑏[2{\cos }^2𝜃-1]+𝑐\\ & \Rightarrow {\cos }^4𝜃=𝑎\cdot 2{\cos }^{2}2𝜃-𝑎+2𝑏{\cos }^2𝜃-𝑏+𝑐\\ & \Rightarrow {\cos }^4𝜃=2𝑎(\cos 2𝜃{)}^2+2𝑏{\cos }^2𝜃-𝑎-𝑏+𝑐\\ & \Rightarrow {\cos }^4𝜃=2𝑎{(2{\cos }^2𝜃-1)}^2+2𝑏{\cos }^2𝜃-𝑎-𝑏+𝑐\\ & \Rightarrow {\cos }^4𝜃=2𝑎(4{\cos }^4𝜃+1-4{\cos }^2𝜃)\\ & +2𝑏{\cos }^2𝜃-𝑎-𝑏+𝑐\\ & \Rightarrow {\cos }^4𝜃=8𝑎{\cos }^4𝜃+2𝑎-8𝑎{\cos }^2𝜃\\ & +2𝑏{\cos }^2𝜃-𝑎-𝑏+𝑐\\ & \Rightarrow {\cos }^4𝜃=8𝑎{\cos }^4𝜃+{\cos }^2𝜃(-8𝑎+2𝑏)+𝑎-𝑏+𝑐\end{array}$

On comparing the corresponding coefficients, we get

$\begin{array}{cc}& 8𝑎=1,-8𝑎+2𝑏=0\text{ and }𝑎-𝑏+𝑐=0\\ & \Rightarrow 𝑎=\frac{1}{8}\\ & \therefore -8\left(\frac{1}{8}\right)+2𝑏=0\text{ gives }𝑏=\frac{1}{2}\\ & \text{ and }\frac{1}{8}-\frac{1}{2}+𝑐=0\text{ gives }𝑐=\frac{3}{8}\text{. }\end{array}$

$|1-𝑖{|}^{𝑥}=2^{𝑥}$

$\begin{array}{cc}\Rightarrow & {\left(\sqrt{\surd \Lambda {)}^2+(-1{)}^2}\right)}^{𝑥}=2^{𝑥}\\ \Rightarrow & (\sqrt{\surd 2}{)}^{𝑥}=2^{𝑥}\\ \Rightarrow & 2^{\frac{𝑥}{2}}=2^{𝑥}\\ \Rightarrow & \frac{𝑥}{2}=𝑥\\ \Rightarrow & 0=\frac{𝑥}{2}\\ \Rightarrow & 𝑥=0\end{array}$

Number of integer solution is 1.

Given: $𝑓\left(𝑥\right)=𝑎{𝑥}^2+𝑏𝑥+𝑐$ for some $𝑎,𝑏,𝑐\in ℝ$ with $𝑎+𝑏+𝑐=3$, and $𝑓(𝑥+𝑦)=𝑓\left(𝑥\right)+𝑓\left(𝑦\right)+𝑥𝑦$ for all $𝑥,𝑦\in ℝ$.

Let's analyze the function step by step:

1. Start with the given quadratic function:

$𝑓\left(𝑥\right)=𝑎{𝑥}^2+𝑏𝑥+𝑐$

1. Since $𝑎+𝑏+𝑐=3$, and given the functional equation $𝑓(𝑥+𝑦)=𝑓\left(𝑥\right)+𝑓\left(𝑦\right)+𝑥𝑦$, we need to verify and utilize this relationship:

$𝑓\left(1\right)=𝑎+𝑏+𝑐=3$

1. Next, consider $𝑓\left(2\right)$:

$𝑓\left(2\right)=𝑓(1+1)=𝑓\left(1\right)+𝑓\left(1\right)+1=2𝑓\left(1\right)+1=2\times 3+1=7$

1. Now, calculate $𝑓\left(3\right)$:

$𝑓\left(3\right)=𝑓(2+1)=𝑓\left(2\right)+𝑓\left(1\right)+2=7+3+2=12$

1. Then compute $𝑓\left(4\right)$:

$𝑓\left(4\right)=𝑓(3+1)=𝑓\left(3\right)+𝑓\left(1\right)+3=12+3+3=18$

1. For the sum $\sum _{10}^{𝑛=1}𝑓\left(𝑛\right)$:

$\sum _{10}^{𝑛=1}𝑓\left(𝑛\right)=𝑓\left(1\right)+𝑓\left(2\right)+𝑓\left(3\right)+\dots +𝑓\left(10\right)$

Let's express each term:

$𝑓\left(1\right)=3$

$𝑓\left(2\right)=2𝑓\left(1\right)+1=7$

$𝑓\left(3\right)=3𝑓\left(1\right)+3=12$

$𝑓\left(4\right)=4𝑓\left(1\right)+6=18$

$\dots$

$𝑓\left(10\right)=10𝑓\left(1\right)+45=75$

1. Summarize the formula:

$\sum _{10}^{𝑛=1}𝑓\left(𝑛\right)=𝑓\left(1\right)+𝑓\left(2\right)+𝑓\left(3\right)+\dots +𝑓\left(10\right)$

$=𝑓\left(1\right)[1+2+3+\dots +10]+[1+3+6+10+15+21+28+36+45]$

$=3\left(\frac{10(10+1)}{2}\right)+165$

$=3\times 55+165$

$=165+165$

$=330$

Therefore, the sum $\sum _{10}^{𝑛=1}𝑓\left(𝑛\right)=330$.

$\begin{array}{cc}& 3^{𝑥+1}+3^{-𝑥+1}=10\\ & \Rightarrow 3^{𝑥}\cdot 3^1+3^{-𝑥}\cdot 3^1=10\\ & \Rightarrow -3𝑦+\frac{3}{𝑦}=10[\text{Let}{3}^{𝑥}=𝑦]\\ & \Rightarrow 3{𝑦}^2-10𝑦+3=0\\ & \Rightarrow 3{𝑦}^2-9𝑦-𝑦+3=0\\ & \Rightarrow 3𝑦(𝑦-3-1(𝑦-3)=0\\ & \Rightarrow (𝑦-3)(3𝑦-1)=0\\ & \Rightarrow 𝑦=3\text{ and }𝑦=\frac{1}{3}\\ & \Rightarrow 3^{𝑥}=3^1\text{ and }{3}^{𝑥}=3^{-1}\\ & \Rightarrow 𝑥=1\text{ and }𝑥=-1\end{array}$

Number of positive real roots is 1.

$\sqrt{\surd \frac{𝑥}{1-𝑥}}+\sqrt{\surd \frac{1-𝑥}{𝑥}}=\frac{13}{6}$

$\begin{array}{cc}& \Rightarrow \frac{𝑥+(1-𝑥)}{\left(\sqrt{\surd 1-𝑥}\right)\left(\sqrt{\surd 𝑥}\right)}=\frac{13}{6}\Rightarrow \frac{1}{\sqrt{\surd 𝑥-{𝑥}^2}}=\frac{13}{6}\\ & \Rightarrow 6=13\sqrt{\surd 𝑥-{𝑥}^2}\\ & \Rightarrow 36=169(𝑥-{𝑥}^2)\\ & \Rightarrow 169{𝑥}^2-169𝑥+36=0\\ & 𝐷=(-169{)}^2-4(169\left)\right(36)=4225>0\\ & \because 𝐷>0\end{array}$

$\therefore$ Number of roots will be 2.

$\begin{array}{cc}& 4^{𝑥}-3^{𝑥-\frac{1}{2}}=3^{𝑥+\frac{1}{2}}-2^{2𝑥-1}\\ & \Rightarrow 4^{𝑥}-3^{𝑥}\cdot 3^{-\frac{1}{2}}=3^{𝑥}\cdot 3^{\frac{1}{2}}-2^{2𝑥}\cdot 2^{-1}\\ & \Rightarrow 4^{𝑥}-3^{𝑥}\cdot \frac{1}{\sqrt{\surd 3}}=\sqrt{\surd 3}\cdot 3^{𝑥}-\frac{4^{𝑥}}{2}\\ & \Rightarrow 4^{𝑥}+\frac{4^{𝑥}}{2}=\sqrt{\surd 3}\cdot 3^{𝑥}+\frac{1}{\sqrt{\surd 3}}\cdot 3^{𝑥}\\ & \Rightarrow 4^{𝑥}(1+\frac{1}{2})=3^{𝑥}(\sqrt{\surd 3}+\frac{1}{\sqrt{\surd 3}})\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{3}{2}\cdot 4^{𝑥}=\frac{4}{\sqrt{\surd 3}}\cdot 3^{𝑥}\\ & \Rightarrow \frac{4^{𝑥}}{3^{𝑥}}=\frac{2\cdot 4}{3\sqrt{\surd 3}}\\ & \Rightarrow {\left(\frac{4}{3}\right)}^{𝑥}=\frac{4^{\frac{1}{2}}\cdot 4^1}{3^1\frac{1}{3^2}}={\left(\frac{4}{3}\right)}^{\frac{1}{2}+1}={\left(\frac{4}{3}\right)}^{\frac{3}{2}}\\ & \Rightarrow 𝑥=\frac{3}{2}\end{array}$

Total number of things $=𝑛$

Repetition of things is allowed and atmost $𝑟$ can be taken at a time.

The number of ways of taking 1 thing at a time $=𝑛$

The number of ways of taking 2 things at a time $={𝑛}^2$

The number of ways of taking 3 things at a time $={𝑛}^3$ ..........

The number of ways of taking $𝑟$ things at a time $={𝑛}^3$

Total number of permutations of $𝑛$ different things taken not more than $𝑟$ at a time i.e. number of permutations of taking $1,2,\dots .,𝑟$ at a time.

$\begin{array}{cc}& =𝑛+{𝑛}^2+{𝑛}^3+\dots .+{𝑛}^{\prime }\\ & =\frac{𝑛({𝑛}^2-1)}{𝑛-1}[\because \text{sum of GP}]\end{array}$

There are 21 points on the circle, denoted as $𝑛=21$. A chord is formed by connecting any two points on the circle.

Therefore, the number of possible chords is given by the combination formula ${}^{𝑛}{𝐶}_2$:

${}^{21}{𝐶}_2=\frac{21\times 20}{2\times 1}=210$

Hence, 210 chords can be drawn through the 21 points on the circle.

$\begin{array}{cc}& \text{ Number of sides }=𝑛\\ & \text{ Number of diagonals }=560\\ & \because \text{ Number of diagonals }=\frac{𝑛(𝑛-3)}{2}\\ & \Rightarrow 560=\frac{𝑛(𝑛-3)}{2}\\ & \Rightarrow 1120={𝑛}^2-3𝑛\\ & \Rightarrow {𝑛}^2-3𝑛-1120=0\\ & \Rightarrow {𝑛}^2-35𝑛+32𝑛-1120=0\\ & \Rightarrow 𝑛(𝑛-35)+32(𝑛-35)=0\\ & \Rightarrow (𝑛-35)(𝑛+32)=0\\ & \Rightarrow 𝑛=35,-32\\ & 𝑛\ne -32\\ & \therefore 𝑛=35.\end{array}$

If the total number of letters is $𝑛$, then the number of ways in which $𝑟$ letters goes into wrong envelopes $={}^{𝑛}{𝐶}_{𝑛-𝑟}{𝐷}_{𝑟}$

The number of ways in which atleast two of the letters are in wrong envelopes

$\begin{array}{cc}& ={}^6{𝐶}_{6-2}{𝐷}_2+{}^6{𝐶}_{6-3}{𝐷}_3+{}^6{𝐶}_{6-4}{𝐷}_4\\ & +{}^6{𝐶}_{6-5}{𝐷}_5+{}^6{𝐶}_{6-6}{𝐷}_6[\because 𝑟=2,3,4,5,6]\\ & =\sum _{𝑟=2}^6{}^6{𝐶}_{6-},{𝐷}_{𝑟}[\text{ where, }{𝐷}^{𝑟}=𝑟!\left(\sum _{𝑖=0}^{𝑟}\frac{(-1{)}^{𝑟}}{𝑖!}\right)]\end{array}$

$\frac{{𝑥}^4+24{𝑥}^2+28}{{({𝑥}^2+1)}^3}=\frac{𝐴𝑥+𝐵}{{𝑥}^2+1}+\frac{𝐶𝑥+𝐷}{{({𝑥}^2+1)}^2}+\frac{𝐸𝑥+𝐹}{{({𝑥}^2+1)}^3}$

$\begin{array}{cc}\text{ Let }{𝑥}^2+1& =𝑘\\ \Rightarrow {𝑥}^2& =𝑘-1\end{array}$

Substituting ${𝑥}^2=𝑘-1$ in LHS, we get

$\begin{array}{cc}& \frac{(𝑘-1{)}^2+24(𝑘-1)+28}{{𝑘}^3}=\frac{{𝑘}^2+22𝑘+5}{{𝑘}^3}\\ & =\frac{1}{𝑘}+\frac{22}{{𝑘}^2}+\frac{5}{{𝑘}^3}\\ & =\frac{1}{({𝑥}^2+1)}+\frac{22}{{({𝑥}^2+1)}^2}+\frac{5}{{({𝑥}^2+1)}^3}\end{array}$

On comparing with RHS, we get

$\begin{array}{cc}& 𝐴𝑥+𝐵=1,𝐶𝑥+𝐷=22\text{ and }𝐸𝑥+𝐹=5\\ & \text{ At }𝑥=1,\\ & 𝐴+𝐵=1,𝐶+𝐷=22\text{ and }𝐸+𝐹=5\\ & \therefore 𝐴+𝐵+𝐶+𝐷+𝐸+𝐹=1+22+5=28\end{array}$

$\begin{array}{cc}& \frac{\sin 𝜃+\sin 3𝜃}{\cos 𝜃+\cos 3𝜃}\\ & =\frac{\sin 𝜃+3\sin 𝜃-4{\sin }^3𝜃}{\cos 𝜃+4{\cos }^3𝜃-3\cos 𝜃}\\ & =\frac{4\sin 𝜃-4{\sin }^3𝜃}{4{\cos }^3𝜃-2\cos 𝜃}\\ & =\frac{4\sin 𝜃\cdot (1-{\sin }^2𝜃)}{2\cos 𝜃(2{\cos }^2𝜃-1)}\\ & =\frac{4\sin 𝜃\cdot {\cos }^2𝜃}{2\cos 𝜃\cdot \cos 2𝜃}\\ & =\frac{2\sin 𝜃\cos 𝜃}{\cos 2𝜃}\\ & =\frac{\sin 2𝜃}{\cos 2𝜃}=\tan 2𝜃\end{array}$

$(1+\tan 1^{\circ })(1+\tan 2^{\circ })(1+\tan 3^{\circ })\dots (1+\tan {45}^{\circ })$

Considering the following two terms

$(1+\tan 𝑥)(1+\tan ({45}^{\circ }-𝑥))$

$\begin{array}{cc}& =1+\tan 𝑥+\tan ({45}^{\circ }-𝑥)+\tan 𝑥\cdot \tan ({45}^{\circ }-𝑥)\\ & =1+[\tan 𝑥+\tan ({45}^{\circ }-𝑥)]+\tan 𝑥\cdot \tan ({45}^{\circ }-𝑥)\\ & =1+\tan (𝑥+({45}^{\circ }-𝑥))\cdot [1-\tan 𝑥\cdot \tan ({45}^{\circ }-𝑥)]\\ & +\tan 𝑥\cdot \tan ({45}^{\circ }-𝑥)\\ & =1+[1-\tan 𝑥\cdot \tan ({45}^{\circ }-𝑥)]+\tan 𝑥\cdot \tan ({45}^{\circ }-𝑥)\\ & =1+1=2\end{array}$

There are 22 such terms in the given multiplication

$\begin{array}{cc}& =(2{)}^{22}\cdot (1+\tan {45}^{\circ })\\ & =2^{22}\cdot 2^1=2^{23}=2^{𝑛}\\ & \therefore 𝑛=23\end{array}$