AP EAPCET 2022 5th July Morning Shift Question Paper with Solutions

$𝑓\left(𝑥\right)=\log (\left(\frac{2{𝑥}^2-3}{𝑥}\right)+\sqrt{\surd \frac{4{𝑥}^4-11{𝑥}^2+9}{\left|𝑥\right|}})$

$\begin{array}{cc}& 𝑓(-𝑥)=\log ⎛⎝\frac{2(-𝑥{)}^2-3}{(-𝑥)}+\sqrt{\surd \frac{4(-𝑥{)}^4-11(-𝑥{)}^2+9}{|-𝑥|}}⎞⎠\\ & 𝑓(-𝑥)=\log ⎡⎣\frac{2{𝑥}^2-3}{-𝑥}+\sqrt{\surd \frac{4{𝑥}^4-11{𝑥}^2+9}{\left|𝑥\right|}}⎤⎦\end{array}$

We know that, if function $𝑓\left(𝑥\right)$ is an odd function. Then,

$\begin{array}{cc}& 𝑓\left(𝑥\right)+𝑓(-𝑥)=0\\ & \log ⎡⎣\left(\frac{2{𝑥}^2-3}{𝑥}\right)+\sqrt{\surd \frac{4{𝑥}^4-11{𝑥}^2+9}{\left|𝑥\right|}}⎤⎦\\ & +\log ⎡⎣\sqrt{\surd \frac{4{𝑥}^4-11{𝑥}^2+9}{\left|𝑥\right|}}-\frac{(2{𝑥}^2-3)}{𝑥}⎤⎦\\ & =\log [\frac{4{𝑥}^4-11{𝑥}^2+9}{{𝑥}^2}-\frac{{(2{𝑥}^2-3)}^2}{{𝑥}^2}]\\ & =\log \left[\frac{4{𝑥}^4-11{𝑥}^2+9-4{𝑥}^4-9+12{𝑥}^2}{{𝑥}^2}\right]\\ & =\log 1=0\end{array}$

$𝑓\left(𝑥\right)=\frac{𝑥-2}{2𝑥+1}$

$𝑓\left(𝑓\right(𝑥\left)\right)=-𝑥$ [Given]

$\begin{array}{cc}& \Rightarrow \frac{𝑓\left(𝑥\right)-2}{2\left(𝑓\right(𝑥\left)\right)+1}=-𝑥\\ & \Rightarrow \frac{\frac{𝑥-2}{2𝑥+1}-2}{\frac{2𝑥-4}{2𝑥+1}+1}=-𝑥\\ & \Rightarrow \frac{𝑥-2-4𝑥-2}{2𝑥-4+2𝑥+1}=-𝑥\\ & \Rightarrow \frac{-3𝑥-4}{4𝑥-3}=-𝑥\Rightarrow \frac{3𝑥+4}{4𝑥-3}=𝑥\\ & \Rightarrow 3𝑥+4=4{𝑥}^2-3𝑥\\ & \Rightarrow 4{𝑥}^2-6𝑥-4=0\\ & \Rightarrow 2{𝑥}^2-3𝑥-2=0\end{array}$

Now, $𝛼+𝛽=\frac{3}{2},𝛼𝛽=-1$

Now, $(𝛼+𝛽{)}^2={𝛼}^2+{𝛽}^2+2𝛼𝛽$

$\Rightarrow \frac{9}{4}={𝛼}^2+{𝛽}^2-2[\because 𝛼𝛽=-1]$

$\begin{array}{cc}& \Rightarrow {𝛼}^2+{𝛽}^2=\frac{17}{4}\\ & \Rightarrow 4({𝛼}^2+{𝛽}^2)=17\end{array}$

Here,

$\begin{array}{cc}& 𝐴=⎡⎣⎢\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}⎤⎦⎥\\ & {𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}3& 2& 0\\ -3& -3& -1\\ 4& 4& 1\end{array}⎤⎦⎥\\ & 𝐴{𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}3& 2& 0\\ -3& -3& -1\\ 4& 4& 1\end{array}⎤⎦⎥\\ & 𝐴{𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}9+9+16& 6+9+16& 0+3+4\\ 6+9+16& 4+9+16& 0+3+4\\ 0+3+4& 0+3+4& 0+1+1\end{array}⎤⎦⎥\\ & 𝐴{𝐴}^{𝑇}=⎡⎣⎢\begin{array}{ccc}34& 31& 7\\ 31& 29& 7\\ 7& 7& 2\end{array}⎤⎦⎥\\ & {\left(𝐴{𝐴}^{𝑇}\right)}^{𝑇}=⎡⎣⎢\begin{array}{ccc}34& 31& 7\\ 31& 29& 7\\ 7& 7& 2\end{array}⎤⎦⎥\end{array}$

So, $𝐴{𝐴}^{𝑇}$ is a symmetric matrix.

The given system of equation is $5𝑥-𝑦+2𝑧=3,𝑥+𝑦+𝑧=6$ and $2𝑥+𝑦-𝑧=-5$ In matrix form, if may be represented as where

$𝐴=⎡⎣⎢\begin{array}{ccc}1& 1& 1\\ 5& -1& 2\\ 2& 1& -1\end{array}⎤⎦⎥,𝑥=⎡⎣⎢\begin{array}{c}𝑥\\ 𝑦\\ 𝑧\end{array}⎤⎦⎥\text{ and }𝐷=⎡⎣⎢\begin{array}{c}6\\ 3\\ -5\end{array}⎤⎦⎥$

Thus, $𝐴$ is a non-singular matrix, so ${𝐴}^{-1}$ exist.

$\begin{array}{cc}& {𝐶}_{11}=(-1{)}^{1+1}\mid \begin{array}{cc}-1& 2\\ 1& -1\end{array}\mid =-1\\ & {𝐶}_{12}=(-1{)}^{1+2}\mid \begin{array}{cc}5& 2\\ 2& -1\end{array}\mid =9\\ & {𝐶}_{13}=(-1{)}^{1+3}\mid \begin{array}{cc}5& -1\\ 2& 1\end{array}\mid =7\\ & {𝐶}_{21}=(-1{)}^{2+1}\mid \begin{array}{cc}1& 1\\ 1& -1\end{array}\mid =2\end{array}$

$\begin{array}{cc}& {𝐶}_{22}=(-1{)}^{2+2}\mid \begin{array}{cc}1& 1\\ 2& -1\end{array}\mid =-3\\ & {𝐶}_{23}=(-1{)}^{2+3}\mid \begin{array}{cc}1& 1\\ 2& 1\end{array}\mid =1\\ & {𝐶}_{31}=(-1{)}^{3+1}\mid \begin{array}{cc}1& 1\\ -1& 2\end{array}\mid =3\\ & {𝐶}_{32}=(-1{)}^{3+2}\mid \begin{array}{cc}1& 1\\ 5& 2\end{array}\mid =3\\ & {𝐶}_{33}=(-1{)}^{3+3}\mid \begin{array}{cc}1& 1\\ 5& -1\end{array}\mid =-6\end{array}$

$\begin{array}{cc}\text{ Then, }\mathrm{adj}\left(𝐴\right)& ={⎡⎣⎢\begin{array}{ccc}-1& 9& 7\\ 2& -3& 1\\ 3& 3& -6\end{array}⎤⎦⎥}^{𝑇}=⎡⎣⎢\begin{array}{ccc}-1& 2& 3\\ 9& -3& 3\\ 7& 1& -6\end{array}⎤⎦⎥\\ (\mathrm{adj}𝐴)\cdot 𝐷& =⎡⎣⎢\begin{array}{ccc}-1& 2& 3\\ 9& -3& 3\\ 7& 1& -6\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}6\\ 3\\ -5\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{c}-6+6-15\\ 54-9-15\\ 42+3+30\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}-15\\ 30\\ 75\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& 𝐴=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right],𝐵=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\\ & \Rightarrow {𝐴}^2=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]\\ & \Rightarrow {𝐴}^2=\left[\begin{array}{cc}1+0& 0+0\\ 2+2& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]\\ & \Rightarrow {𝐴}^3=\left[\begin{array}{cc}1& 0\\ 4& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right][{𝐴}^2\cdot 𝐴={𝐴}^3]\\ & \Rightarrow {𝐴}^3=\left[\begin{array}{cc}1+0& 0+0\\ 4+2& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 6& 1\end{array}\right]\\ & \text{ Similarly, }{𝐴}^6=\left[\begin{array}{cc}1& 0\\ 12& 1\end{array}\right][{𝐴}^{𝑛}={𝐴}_{21}=𝑛\times 2]\\ & \text{ Now, }𝐵=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\\ & \Rightarrow {𝐵}^2=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\\ & \Rightarrow {𝐵}^2=\left[\begin{array}{cc}1+0& 3+3\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]\\ & \Rightarrow {𝐵}^3=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 9\\ 0& 1\end{array}\right]\end{array}$

Similarly, ${𝐵}^6=\left[\begin{array}{cc}1& 18\\ 0& 1\end{array}\right][{𝐵}^{𝑛}={𝐵}_{12}=𝑛\times 3]$

$\begin{array}{cc}& \begin{array}{cc}& \therefore ({𝐴}^6+{𝐵}^6)=\left[\begin{array}{cc}1& 0\\ 12& 1\end{array}\right]+\left[\begin{array}{cc}1& 18\\ 0& 1\end{array}\right]\\ & {𝐴}^6+{𝐵}^6=\left[\begin{array}{cc}2& 18\\ 12& 2\end{array}\right]\\ & \end{array}\\ & \mathrm{Det}({𝐴}^6+{𝐵}^6)=4-216=-212\end{array}$

$\text{ Here, }𝐺\left(𝑥\right)=⎡⎣⎢\begin{array}{ccc}\cos 𝑥& -\sin 𝑥& 0\\ \sin 𝑥& \cos 𝑥& 0\\ 0& 0& 1\end{array}⎤⎦⎥$

$𝐺\left(𝑦\right)=⎡⎣⎢\begin{array}{ccc}\cos 𝑦& -\sin 𝑦& 0\\ \sin 𝑦& \cos 𝑦& 0\\ 0& 0& 1\end{array}⎤⎦⎥$

So, $𝑥+𝑦=0\Rightarrow 𝑦=(-𝑥)$

Now, $𝐺(-𝑥)=⎡⎣⎢\begin{array}{ccc}\cos (-𝑥)& -\sin (-𝑥)& 0\\ \sin (-𝑥)& \cos (-𝑥)& 0\\ 0& 0& 1\end{array}⎤⎦⎥$

$\begin{array}{cc}𝐺\left(𝑥\right)\cdot 𝐺\left(𝑦\right)& =⎡⎣⎢\begin{array}{ccc}\cos 𝑥& -\sin 𝑥& 0\\ \sin 𝑥& \cos 𝑥& 0\\ 0& 0& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}\cos 𝑥& \sin 𝑥& 0\\ -\sin 𝑥& \cos 𝑥& 0\\ 0& 0& 1\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥=𝐼\end{array}$

$\begin{array}{cc}& {𝑖}^{18}-3{𝑖}^7+{𝑖}^2(1+{𝑖}^4)(𝑖{)}^{22}\\ & ={𝑖}^{4\times 4+2}-3{𝑖}^{4+3}+-1(1+1){\left({𝑖}^2\right)}^{11}\\ & ={𝑖}^2-3{𝑖}^3+2[\because {𝑖}^2=-1\text{ and }{𝑖}^4=1]\\ & =-1+3𝑖+2=1+3𝑖\end{array}$

Conjugate of $\cos 𝑥-𝑖\sin 2𝑥$ is $\cos 𝑥+𝑖\sin 2𝑥$

So, $\sin 𝑥+𝑖\cos 2𝑥=\cos 𝑥+𝑖\sin 2𝑥$

Comparing real and imaginary parts,

$\sin 𝑥=\cos 𝑥\text{ or }\tan 𝑥=1$

and $\cos 2𝑥=\sin 2𝑥$ or $\tan 2𝑥=1$

But for same value of $𝑥$, both $\tan 𝑥$ and $\tan 2𝑥$ can not. Hence, no solution is possible.

$\begin{array}{cc}\text{ Let }𝑧& =𝑥+𝑖𝑦\\ \left|𝑧\right|& =\sqrt{\surd {𝑥}^2+{𝑦}^2}\end{array}$

Now, $|𝑧{|}^2=\mathrm{Re}(𝑧)$

$\begin{array}{cc}{𝑥}^2+{𝑦}^2& =𝑥\\ {𝑥}^2+{𝑦}^2-𝑥& =0\\ 𝑔=1/2,𝑓& =0\end{array}$

So, centre of circle $(\frac{1}{2},0)$.

$\begin{array}{cc}\text{Given,}& {\sin }^4𝜃{\cos }^2𝜃=\sum _{𝑛=0}^{\infty }{𝑎}_{2𝑛}\cos 2𝑛𝜃\\ \Rightarrow & {({\sin }^2𝜃)}^2{\cos }^2𝜃=\sum _{𝑛=0}^{\infty }{𝑎}_{2𝑛}\cos 2𝑛𝜃\\ \Rightarrow & {(1-{\cos }^2𝜃)}^2{\cos }^2𝜃=\sum _{𝑛=0}^{\infty }{𝑎}_{2𝑛}\cos 2𝑛𝜃\\ \Rightarrow & (1+{\cos }^4𝜃-2{\cos }^2𝜃){\cos }^2𝜃=\sum _{𝑛=0}^{\infty }{𝑎}_{2𝑛}\cos 2𝑛𝜃\\ \therefore & {\cos }^2𝜃+{\cos }^4𝜃{\cos }^2𝜃-2{\cos }^4𝜃\\ & =\sum _{𝑛=0}^{\infty }{𝑎}_{2𝑛}\cos 2𝑛𝜃\\ \Rightarrow & \left(\frac{\cos 2𝜃+1}{2}\right)+{\left(\frac{\cos 2𝜃+1}{2}\right)}^2\left(\frac{1+\cos 2𝜃}{2}\right)\\ & -2{\left(\frac{\cos 2𝜃+1}{2}\right)}^2=\sum _{𝑛=0}^{\infty }{𝑎}_{2𝑛}\cos 2𝑛𝜃\end{array}$

On comparing both sides, we get minimum value of n is 1.

Given, equation ${𝑥}^2-2(1+3𝑚)𝑥+7(3+2𝑚)=0$

Here, $𝑎>0$ and $𝐷={𝑏}^2-4𝑎𝑐>0$

Expression is always positive, if roots are distinct then $𝐷>0$,

$\begin{array}{cc}& [-2(1+3𝑚){]}^2-4\times 7(3+2𝑚)>0\\ \Rightarrow & 4[1+9{𝑚}^2+6𝑚]-84-56𝑚>0\\ \Rightarrow & 36{𝑚}^2-32𝑚-80>0\\ \Rightarrow & 9{𝑚}^2-8𝑚-20>0\\ \Rightarrow & (𝑚-2)(𝑚+\frac{10}{9})>0\\ \Rightarrow & 𝑚\in (-\infty ,\frac{-10}{9})\cup [2,\infty )\end{array}$

$\therefore$ Integral value of $𝑚$ are .......... $3𝑖,1,2,3,4,5,6$. So, the number of element in $𝑆$ is infinite.

Given,

$\begin{array}{cc}& 4^{𝑥}-3^{𝑥-\frac{1}{2}}=3^{𝑥+\frac{1}{2}}-2^{2𝑥-1}\\ \Rightarrow & (2{)}^{2𝑥}+2^{2𝑥-1}=3^{𝑥+\frac{1}{2}}+3^{𝑥-\frac{1}{2}}\\ \Rightarrow & 2^{2𝑥}+\frac{2^{2𝑥}}{2}=3^{𝑥}\cdot \sqrt{\surd 3}+\frac{3^{𝑥}}{\sqrt{\surd 3}}\\ \Rightarrow & 2^{2𝑥}[1+\frac{1}{2}]=3^{𝑥}[\sqrt{\surd 3}+\frac{1}{\sqrt{\surd 3}}]\\ \Rightarrow & 2^{2𝑥}\times \frac{3}{2}=3^{𝑥}\times \frac{4}{\sqrt{\surd 3}}\\ \Rightarrow & 2^{2𝑥}\left(3\sqrt{\surd 3}\right)=3^{𝑥}\left(8\right)\\ \Rightarrow & 2^{2𝑥}(3{)}^{\frac{3}{2}}=3^{𝑥}\times (2{)}^3\end{array}$

Comparing power of $2,2𝑥=3$

and comparing power of $3,𝑥=\frac{3}{2}$

Both equations given us $𝑥=\frac{3}{2}$.

Considering the question to be, to find the roots of the equation ${𝑥}^4-2{𝑥}^3+𝑥-380=0$

Now, by trial error method we find that $𝑥=5$ is a solution to the equation

$\begin{array}{cc}& \Rightarrow 5^4-2\times 5^3+5-380=0\\ & =625-250+5-380=0\\ & =380-380=0\end{array}$

Thus, $𝑥=5$ is a solution.

We also find that $𝑥=-4$ is a solution

$\begin{array}{cc}& (-4{)}^4-2\times (-4{)}^3-4-380=0\\ & =256+128-4-380\\ & =380-380=0\end{array}$

The real roots are 5 and $-$4.

Sum of real roots are $5+(-4)=1$

Let $𝑎,2𝑎$ and $𝑏$ be roots of the cubic equation ${𝑥}^3+36-7{𝑥}^2=0$

So, sum of roots $=-\frac{(-7)}{1}$

$\begin{array}{cccc}& \therefore & 𝑎+2𝑎+𝑏& =7\\ & \Rightarrow & 3𝑎+𝑏& =7\text{... (i)}\end{array}$

$\text{ Now, }𝑎\left(2𝑎\right)𝑏=36[\text{ product of roots }=\frac{𝑑}{𝑎}]$

$\begin{array}{cc}& 2{𝑎}^2𝑏=36\text{ and }𝑎𝑏+𝑎\left(2𝑎\right)+2𝑎\left(𝑏\right)=0\\ & \Rightarrow 2{𝑎}^2+3𝑎𝑏=0\Rightarrow 𝑎(2𝑎+3𝑏)=0\\ & \Rightarrow 𝑏=\frac{-2𝑎}{3}\end{array}$

$\begin{array}{ccc}\Rightarrow & 3𝑎-\frac{2𝑎}{3}=7& \text{ [from Eq. (i)] }\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{9𝑎-2𝑎}{3}=7\Rightarrow 9𝑎-2𝑎=21\\ & \Rightarrow 7𝑎=21\Rightarrow 𝑎=3\\ & \text{ and }𝑏=7-3𝑎\\ & =7-3\left(3\right)=-2\text{[from Eq. (i)]}\end{array}$

Roots are $3,6,-2$

$\therefore$ Number of negative root is 1.

${10}^{𝑛}{𝐶}_2=3^{𝑛+1}{𝐶}_3$

$\begin{array}{cc}& \frac{10𝑛!}{(𝑛-2)!2!}=\frac{3(𝑛+1)!}{(𝑛+1-3)!\times 3!}\\ & \Rightarrow \frac{10𝑛!}{2}=\frac{3(𝑛+1)𝑛!}{3\times 2}\\ & \Rightarrow 10=𝑛+1\\ & \Rightarrow 𝑛=9\end{array}$

Here's how to solve this problem:

First, let's understand the concepts involved:

• Collinear points: Points that lie on the same straight line.
• Triangle: A polygon with three sides.

To form a triangle, we need to choose 3 points out of the 10. We can do this using combinations:

Total number of ways to choose 3 points out of 10 = ${}^{10}{𝐶}_3=\frac{10!}{3!7!}=120$

However, this count includes triangles formed using the 6 collinear points. We need to subtract these invalid triangles.

Number of ways to choose 3 points out of 6 collinear points = ${}^6{𝐶}_3=\frac{6!}{3!3!}=20$

Therefore, the total number of valid triangles (those not formed by the collinear points) is:

$𝑁=120-20=100$

So the answer is Option C: 100.

Total number of students $=205$

Number of students passed in Math $=70$

Number of students passed in English = 105

and number of students pass in both $=30$

Now,

Number of students passed in Math or English $=$ Number of students passed in Math + Number of student passed in English $-$ Number of students passed in both $=70+105-30=145$

So, number of students fail in both subjects $=$ Total number of students $-$ Students passed in Math or English $=205-145=60$

Total marks $=$ Marks for first 3 papers + Marks for fourth paper $=3𝑛+2𝑛=5𝑛$

Candidate needs to get $3𝑛$ marks.

Let ${𝑥}_1,{𝑥}_2,{𝑥}_3,{𝑥}_4$ be the marks of candidate in I, II, III and IV paper, respectively.

Then, ${𝑥}_1+{𝑥}_2+{𝑥}_3+{𝑥}_4=3𝑛(0\le {𝑥}_1,{𝑥}_2,{𝑥}_3,\le 𝑛$ and $0\le {𝑥}_4\le 2𝑛)$

Using multinomial theorem,

$\begin{array}{cc}& \begin{array}{c}\text{ Number of ways }=\text{ Coefficient of }{𝑥}^{3𝑛}\text{ in }\\ \underset{\text{GP series }}{\underset{}{{(1+𝑥+\dots +{𝑥}^{𝑛})}^3}}\underset{\text{GP Series }}{\underset{}{(1+𝑥+\dots {𝑥}^{2𝑛})}}\end{array}\\ & \Rightarrow {\left(\frac{1-{𝑥}^{𝑛+1}}{1-𝑥}\right)}^3\left(\frac{1-{𝑥}^{2𝑛+1}}{1-𝑥}\right)\\ & \Rightarrow 1-{𝑥}^{3(𝑛+1)}-3{𝑥}^{𝑛+1}(1-{𝑥}^{𝑛+1})(1-{𝑥}^{2𝑛+1})(1-𝑥{)}^{-4}\\ & \Rightarrow (1-{𝑥}^{3(𝑛+1)}-3{𝑥}^{𝑛+1}+3{𝑥}^{2𝑛+2})(1-{𝑥}^{2𝑛+1})(1-𝑥{)}^{-4}\\ & \Rightarrow (1-{𝑥}^{3(𝑛+1)}-3{𝑥}^{𝑛+1}+3{𝑥}^{2𝑛+2}-{𝑥}^{2𝑛+1}+{𝑥}^{5𝑛+4}\\ & +3{𝑥}^{3𝑛+2}-3{𝑥}^{4𝑛+3}(1-𝑥{)}^{-4})\end{array}$

Finding the coefficient of ${𝑥}^{3𝑛}$ using binomial expansion, we get

$\begin{array}{c}{}^{3𝑛+3}{𝐶}_3-3^{2𝑛+2}{𝐶}_3+3^{𝑛+2}{𝐶}_3-{}^{𝑛+3}{𝐶}_3\\ =\frac{1}{6}(𝑛+1)(5{𝑛}^2+10𝑛+6)\end{array}$

Here, $\frac{2{𝑥}^2+1}{{𝑥}^3-1}=\frac{𝐴}{𝑥-1}+\frac{𝐵𝑥+𝐶}{{𝑥}^2+𝑥+1}$

Resolving into partial fractions,

$\frac{2{𝑥}^2+1}{{𝑥}^3-1}=\frac{𝐴({𝑥}^2+𝑥+1)+(𝐵𝑥+𝐶)(𝑥-1)}{(𝑥-1)({𝑥}^2+𝑥+1)}$

$\begin{array}{cc}& \Rightarrow 2{𝑥}^2+1=𝐴{𝑥}^2+𝐴𝑥+𝐴+𝐵{𝑥}^2-𝐵𝑥+𝐶𝑥-𝐶\\ & \Rightarrow 2{𝑥}^2+1={𝑥}^2(𝐴+𝐵)+𝑥(𝐴-𝐵+𝐶)+𝐴-𝐶\end{array}$

Comparing with coefficient, we get

$𝐴+𝐵=2,𝐴-𝐵+𝐶=0\text{ and }𝐴-𝐶=1$

Clearly, $𝐴=1,𝐵=1$ and $𝐶=0$

So, $7𝐴+2𝐵+𝐶=7\left(𝑙\right)+2\left(𝑙\right)+0=9$

$\begin{array}{cc}\sin 𝜃& =\frac{-3}{4}\\ \cos 𝜃& =\sqrt{\surd 1-{\sin }^2𝜃}\\ \cos 𝜃& =\sqrt{\surd 1-{\left(\frac{-3}{4}\right)}^2}\\ \cos 𝜃& =\sqrt{\surd \frac{16-9}{4}}=\sqrt{\surd \frac{7}{16}}\\ \therefore \sin 2𝜃& =2\sin 𝜃\cdot \cos 𝜃\\ & =2\times \left(\frac{-3}{4}\right)\left(\frac{\sqrt{\surd 7}}{4}\right)=\frac{-3\sqrt{\surd 7}}{8}\end{array}$

Here,

$\begin{array}{cc}& \frac{1}{\sin 1^{\circ }\sin 2^{\circ }}+\frac{1}{\sin 2^{\circ }\sin 3^{\circ }}+\dots +\frac{1}{\sin {89}^{\circ }\sin {90}^{\circ }}\\ & =\frac{1}{\sin 1^{\circ }}⎡⎣⎢\begin{array}{c}\frac{\sin (2^{\circ }-1^{\circ })}{\sin 1^{\circ }\sin 2^{\circ }}+\frac{\sin (3^{\circ }-2^{\circ })}{\sin 2^{\circ }\sin 3^{\circ }}+\dots ..\\ +\frac{\sin ({90}^{\circ }-{89}^{\circ })}{\sin {89}^{\circ }\sin {90}^{\circ }}\end{array}⎤⎦⎥\end{array}$

We know that

$\sin (𝐴-𝐵)=\sin 𝐴\cdot \cos 𝐵-\cos 𝐴\cdot \sin 𝐵$

$\begin{array}{c}=\frac{1}{\sin 1^{\circ }}⎡⎣⎢\begin{array}{c}\frac{\sin 2^{\circ }\cdot \cos 1^{\circ }-\cos 2^{\circ }\cdot \sin 1^{\circ }}{\sin 1^{\circ }\cdot \sin 2^{\circ }}\\ +\frac{\sin 3^{\circ }\cdot \cos 2^{\circ }-\cos 3^{\circ }\cdot \sin 2^{\circ }}{\sin 3^{\circ }\cdot \sin 2^{\circ }}+\dots .\\ \dots .+\frac{\sin {90}^{\circ }\cdot \cos {89}^{\circ }-\cos {.90}^{\circ }\cdot \sin {89}^{\circ }}{\sin {89}^{\circ }\cdot \sin {90}^{\circ }}\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& =\frac{1}{\sin 1^{\circ }}[\cot 1^{\circ }-\cot 2^{\circ }+\cot 2^{\circ }-\cot 3^{\circ }+\dots .\\ & ++\cot {89}^{\circ }-\cot {90}^{\circ }]\\ & =\frac{1}{\sin 1^{\circ }}[\cot 1^{\circ }-\cot {90}^{\circ }]=\frac{\cot 1^{\circ }}{\sin 1^{\circ }}=\frac{\cos 1^{\circ }}{{\sin }^2{1}^{\circ }}\end{array}$