AP EAPCET 2025 23rd May Morning Shift Question Paper with Solutions

Given, $𝑓\left(𝑥\right)={\cos }^{-1}\left(\frac{3}{\sqrt{\surd 9{𝑥}^2-12𝑥+22}}\right)$

Since, $9{𝑥}^2-12𝑥+22=9$

$\begin{array}{cc}& \left[\underset{\ge 2}{\underset{}{{(𝑥-\frac{2}{3})}^2+2}}\right]\ge 18\\ \Rightarrow & \sqrt{\surd 9{(𝑥-\frac{3}{2})}^2+18}\ge \sqrt{\surd 18}\\ \Rightarrow & 0<\frac{1}{\sqrt{\surd 9{𝑥}^2-12𝑥+22}}\le \frac{1}{\sqrt{\surd 18}}\\ \Rightarrow & 0<\frac{3}{\sqrt{\surd 9{𝑥}^2-12𝑥+22}}\le \frac{3}{\sqrt{\surd 18}}\\ \Rightarrow & 0<\frac{3}{\sqrt{\surd 9{𝑥}^2-12𝑥+22}}\le \frac{1}{\sqrt{\surd 2}}\end{array}$

$\begin{array}{c}\Rightarrow {\cos }^{-1}0>{\cos }^{-1}\left(\frac{3}{\sqrt{\surd 9{𝑥}^2-12𝑥+22}}\right)\\ \ge {\cos }^{-1}\left(\frac{1}{\sqrt{\surd 2}}\right)\\ \Rightarrow \frac{𝜋}{4}\le 𝑓\left(𝑥\right)<\frac{𝜋}{2}\end{array}$

It is obvious. Since, $𝑓$ is said to be one-one if $𝑓\left(𝑥\right)=𝑓\left(𝑦\right)\Rightarrow 𝑥=𝑦$ or $𝑓\left(𝑥\right)\ne 𝑓\left(𝑦\right)\Rightarrow 𝑥\ne 𝑦$. Also, in a function, one image may have more than one pre-image.

Given, ${𝑡}_{𝑛}=\frac{1}{4}(𝑛+2)(𝑛+3),𝑛\in 𝑁\dots$ (i)

Also, $\frac{1}{{𝑡}_1}+\frac{1}{{𝑡}_2}+\dots +\frac{1}{{𝑡}_{2003}}=\frac{2003}{3009}...\left(𝑖𝑖\right)$

From Eq. (i), we get

$\begin{array}{cc}\frac{1}{{𝑡}_{𝑛}}& =\frac{4}{(𝑛+2)(𝑛+3)}\\ & =\frac{4}{𝑛+2}-\frac{4}{𝑛+3}\end{array}$

(By partial fraction)

$\begin{array}{cc}\Rightarrow & \sum _{𝑛=1}^{2003}\frac{1}{{𝑡}_{𝑛}}=\sum _{𝑛=1}^{2003}(\frac{4}{𝑛+2}-\frac{4}{𝑛+3})\\ & =4(\frac{1}{3}-\frac{1}{2006})=4\left(\frac{2006-3}{3\times 2006}\right)\\ & =4\times \frac{2003}{3\times 2006}\\ & =\frac{8012}{6018}=\frac{4006}{3009}\end{array}$

So, Assertion is false.

Also, by putting $𝑛=2003$ in Reason then we observe that it is wrong.

Given, $𝐴=⎡⎣⎢\begin{array}{ccc}0& 𝑘& 𝑘\\ 𝑘& -4& -6\\ 𝑘& -3& -5\end{array}⎤⎦⎥$

$\because 𝐴$ is an singular matrix.

$\begin{array}{cc}& \therefore \left|𝐴\right|=0\\ & \Rightarrow \mid \begin{array}{ccc}0& 𝑘& 𝑘\\ 𝑘& -4& -6\\ 𝑘& -3& -5\end{array}\mid =0\\ & \Rightarrow 0-𝑘(-5𝑘+6𝑘)+𝑘(-3𝑘+4𝑘)=0\end{array}$

$\Rightarrow -{𝑘}^2+{𝑘}^2=0$ which is true for all real values of $𝑘$.

$\therefore 𝑘\in 𝑅$

Given, matrix $𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 𝑥\\ 4& -1& 7\\ 2& 4& -6\end{array}⎤⎦⎥$ and the rank of $𝐴=2$

$\begin{array}{cc}& =1(6-28)-2(-24-14)+𝑥(16+2)\\ & \left|𝐴\right|=54+18𝑥\end{array}$

Since, the rank of $𝐴$ is 2 then the determinant of $𝐴$ must be zero.

So, we set the determinant equal to zero and solve for $𝑥$.

i.e. $54+18𝑥=0\Rightarrow 𝑥=-3$

$\begin{array}{cc}& \mid \begin{array}{cc}2& 1\\ 3& 1\end{array}\mid +\mid \begin{array}{cc}1& \frac{1}{3}\\ 3& 1\end{array}\mid +\mid \begin{array}{cc}\frac{1}{2}& \frac{1}{9}\\ 3& 1\end{array}\mid +\mid \begin{array}{cc}\frac{1}{4}& \frac{1}{27}\\ 3& 1\end{array}\mid +\dots +\infty \\ & =(2-3)+(1-3\times \frac{1}{3})+(\frac{1}{2}-3\times \frac{1}{9})+(\frac{1}{4}-3\times \frac{1}{27})+(\frac{1}{8}-3\times \frac{1}{81})+\dots +\infty \\ & =-1+0+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{2^3}-\frac{1}{3^3})+\dots +\infty \end{array}$

$\begin{array}{cc}& \begin{array}{cc}& =-1+[(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots \infty )-(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\dots +\infty )]\\ & =-1+\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)-\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)\end{array}\\ & \text{ (Using sum of infinite GP formula) }\\ & =-1+\frac{\frac{1}{2}}{\frac{1}{2}}-\frac{\frac{1}{3}}{\frac{2}{3}}=-1+1-\frac{1}{2}=-\frac{1}{2}\end{array}$

$\begin{array}{cc}& \text{ Given }{|{𝑧}_1+{𝑧}_2|}^2={\left|{𝑧}_1\right|}^2+{\left|{𝑧}_2\right|}^2\\ & \Rightarrow {\left|{𝑧}_1\right|}^2+{\left|{𝑧}_2\right|}^2+{𝑧}_1{\overline{𝑧}}_2+{\overline{𝑧}}_1{𝑧}_2={\left|{𝑧}_1\right|}^2+{\left|{𝑧}_2\right|}^2\\ & \Rightarrow {𝑧}_1{\overline{𝑧}}_2+{\overline{𝑧}}_1{𝑧}_2=0\Rightarrow \frac{{𝑧}_1}{{𝑧}_2}+\frac{{\overline{𝑧}}_1}{{\overline{𝑧}}_2}=0\\ & \text{ (on dividing by }{𝑧}_2\cdot {\overline{𝑧}}_2)\\ & \Rightarrow \frac{{𝑧}_1}{{𝑧}_2}+\left(\frac{{\overline{𝑧}}_1}{{𝑧}_2}\right)=0\end{array}$

$\therefore \frac{{𝑧}_1}{{𝑧}_2}$ is purely imaginary i.e. its real pant

$=0$

Given, $1,𝜔,{𝜔}^2$ are cube roots of unity.

Also, we can write the general term $\left({𝑡}_{𝑛}\right)$ of given expression

$\begin{array}{cc}{𝑡}_{𝑛}& =(𝑛-1)(𝑛+\frac{1}{𝜔})(𝑛+\frac{1}{{𝜔}^2})\\ & =(𝑛-1)(𝑛+{𝜔}^2)(𝑛+𝜔)\\ & =(𝑛-1)({𝑛}^2+𝑛(𝜔+{𝜔}^2)+{𝜔}^3)\\ & =(𝑛-1)({𝑛}^2-𝑛+1)[\because 1+𝜔+{𝜔}^2=0]\\ & =({𝑛}^3-2{𝑛}^2+2𝑛-1)...\left(𝑖\right)\end{array}$

Now, $\sum _{𝑛}^{𝑟=1}𝑟=\frac{𝑛(𝑛+1)}{2}$,

$\begin{array}{cc}& \sum _{𝑟=1}^{𝑛}{𝑟}^2=\frac{𝑛(𝑛+1)(2𝑛+1)}{6}\\ & \sum _{𝑟=1}^{𝑛}{𝑟}^3={\left(\frac{𝑛(𝑛+1)}{2}\right)}^2\end{array}$

So, we have;

$\begin{array}{cc}& \begin{array}{cc}\sum _{𝑛=2}^{11}𝑛& =(\sum _{𝑛=2}^{11}𝑛-1)\\ & =\frac{11\left(12\right)}{2}-1=66-1=65\end{array}\\ & \sum _{𝑛=2}^{11}{𝑛}^2-\left(\sum _{𝑛=1}^{11}{𝑛}^2\right)-1\end{array}$

$\begin{array}{cc}& =\frac{11\left(12\right)\left(23\right)}{6}-1=506-1=505\\ & \begin{array}{cc}\sum _{𝑛=2}^{11}{𝑛}^3& =\left(\sum _{𝑛=1}^{11}{𝑛}^3\right)-1\\ & ={\left(\frac{11\left(12\right)}{2}\right)}^2-1=(66{)}^2-1\\ & =4356-=4355\\ 𝑎𝑛𝑑\sum _{𝑛=2}^{11}1& =10\end{array}\end{array}$

So, final answer

$\begin{array}{cc}& =4355-2\left(505\right)+2\left(65\right)-10\\ & =4355-1010+130-10=3465\end{array}$

$\begin{array}{cc}& (1+\sqrt{\surd 3}𝑖{)}^6-(\sqrt{\surd 3}+𝑖{)}^6\\ & =2^6\cdot {(\frac{1}{2}+\frac{\sqrt{\surd 3}}{2}𝑖)}^6-{(\frac{\sqrt{\surd 3}}{2}+\frac{1}{2}𝑖)}^6{2}^6\\ & =2^6[{(\cos \frac{𝜋}{3}+\sin \frac{𝜋}{3}𝑖)}^6-(\cos \frac{𝜋}{6}+\sin \frac{𝜋}{6}𝑖)]\end{array}$

(Using De-Moivre's theorem)

$\begin{array}{cc}& =2^6\left[\right(\cos 2𝜋+\sin 2𝜋𝑖)-(\cos 𝜋+\sin 𝜋𝑖\left)\right]\\ & =64\left[\right(1+0)-(-1+0\left)\right]\\ & =64(1+1)=128\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \because 𝛼+𝛽=5\Rightarrow \frac{-𝑏}{1}=5\Rightarrow 𝑏=-5\dots \text{ (i) }\\ & \text{ and }{𝛼}^3+{𝛽}^3=60\\ & \Rightarrow (𝛼+𝛽{)}^3-3𝛼𝛽(𝛼+𝛽)=60\\ & \Rightarrow (5{)}^3-3\times 𝑐(5)=60\\ & \Rightarrow 125-15𝑐=60\\ & \Rightarrow 15𝑐=125-60=65\\ & \Rightarrow 𝑐=\frac{65}{15}=\frac{13}{3}\\ & \therefore 3𝑐+2=3\times \frac{13}{3}+2=13+2=15\\ & =(-3)(-5)=-3𝑏\text{ (from Eq. }\left(i\right))\end{array}\end{array}$

Let $𝑦=\frac{{𝑥}^2+𝑥+𝑎}{{𝑥}^2+2𝑥+𝑎}=\frac{𝑥+\frac{𝑎}{𝑥}+1}{𝑥+\frac{𝑎}{𝑥}+2}$

Let $𝑔\left(𝑥\right)=𝑥+\frac{𝑎}{𝑥}\Rightarrow {𝑔}^{\prime }\left(𝑥\right)=1-\frac{𝑎}{{𝑥}^2}$

For maximum or minimum put ${𝑔}^{\prime }\left(𝑥\right)=0$

$\Rightarrow 1-\frac{𝑎}{{𝑥}^2}=𝑥\Rightarrow 𝑥=\pm \sqrt{\surd 𝑎}$

Minimum value at $𝑥=\sqrt{\surd 𝑎}$ is $=\frac{2\sqrt{\surd 𝑎}+1}{2\sqrt{\surd 𝑎}+2}$

Maximum value at $𝑥=-\sqrt{\surd 𝑎}$ is

$\begin{array}{cc}& =\frac{2\sqrt{\surd 𝑎}-1}{2\sqrt{\surd 𝑎}-2}\\ & \text{ Clearly, }\frac{2\sqrt{\surd 𝑎}-1}{2\sqrt{\surd 𝑎}-2}=2\\ & \Rightarrow 2\sqrt{\surd 𝑎}-1=4\sqrt{\surd 𝑎}-4\\ & \Rightarrow 2\sqrt{\surd 𝑎}=3\\ & \Rightarrow \sqrt{\surd 𝑎}=\frac{3}{2}\Rightarrow 𝑎=\frac{9}{4}\end{array}$

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

$\begin{array}{cc}& {𝑥}^3+𝑎{𝑥}^2+𝑏𝑥+𝑐=0\\ & 𝛼+𝛽+𝛾=-𝑎\\ & 𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑏\\ & 𝛼𝛽𝛾=-𝑐\end{array}$

Now, $(𝛼+𝛽-2𝑦)(𝛽+𝛾-2𝛼)(𝛾+𝛼-2𝛽)$

$\begin{array}{cc}& =(𝛼+𝛽+𝛾-3𝛾)(𝛼+𝛽+𝛾-3𝛼)(𝛼+𝛽+𝛾-3𝛽)\\ & =-(𝛼+3𝛼)(𝑎+3𝛽)(𝑎+3𝛾)\end{array}$

$\begin{array}{cc}& =-({𝑎}^3+3(𝛼+𝛽+𝛾){𝑎}^2+9(𝛼𝛽+𝛽𝛾+𝛾𝛼)𝑎+27\left(𝛼𝛽𝛾\right)\\ & =-({𝑎}^3-3{𝑎}^3+9𝑎𝑏-27𝑐)\\ & =2{𝑎}^3-9𝑎𝑏+27𝑐\end{array}$

Given,

$\begin{array}{cc}& {𝑥}^4+2{𝑥}^3-7{𝑥}^2-8𝑥+12=0\\ \Rightarrow & (𝑥-1)(𝑥-3)(𝑥-2)(𝑥+2)=0\\ \Rightarrow & 𝑥=1,3,2-2\end{array}$

Since sum of two roots of Eq. (i) is zero.

So, $-2+2=0$

Hence, sum of the squares of the other two roots

$(1{)}^2+(3{)}^2=10$

The total number of ways that all sisters and brothers seated around a circle is 10 !

The number of ways that the three sisters seated together along with the brothers is $8!\times 3$ !

So, required number of ways

$=10!-8!3!=8!(10\times 9-6)=8!\times 84$

Given, 8 students in a class room 4 of them are chosen and arrange in ground table and 4 in row

$\begin{array}{cc}& \therefore \text{ Total arrangement }={}^8{𝐶}_4(3!+4!)\\ & =\frac{8!}{4!4!}(6+24)=70\times 30\\ & =2100\end{array}$

The integers divisible by 5 from 1 to 100 are $5,10,15,20,25,\dots ,100$. Which are in A.P. with first term (a) $=5$ and common difference $\left(𝑑\right)=5$ ) last term $\left(𝑙\right)=100$

So, total number of terms,

$\begin{array}{cc}\left(𝑛\right)& =\frac{𝑙-𝑎}{𝑑}+1=\frac{100-5}{5}+1=\frac{95}{5}+1\\ & =19+1=20\\ \therefore {𝑆}_{𝑛}& =\frac{𝑛}{2}(𝑎+𝑙)\\ & =10(5+100)=1050\end{array}$

Similarly, the integers divisible by 13 from 1 to 100 are $13,26,39,52,\dots ,91$ which are in AP with to $𝑎=13,𝑑=13$, $𝑙=91$.

$\begin{array}{cc}& \begin{array}{c}𝑆𝑜,𝑛=\frac{𝑙-𝑎}{𝑑}+1=\frac{91-13}{13}+1=\frac{78}{13}+1\\ =6+1=7\end{array}\\ & \therefore {𝑆}_{𝑛}=\frac{7}{2}(13+91)=7\times 52=364\end{array}$

Now, the integer divisible by both 5 and 13 from 1 to 100 is 65 .

Hence, the sum of integer from 1 to 100 that are divisible by 5 or 13 is $=1050+364-65=1414-65=1349$

As we know that, coefficient of ${𝑥}^{11}$ in the expansion of $(1+𝑥{)}^{11}$ is ${}^{11}{𝐶}_{11}$.

Now, we have to find coefficient of ${𝑥}^{11}$ in the expansion of $(1+𝛼𝑥+𝛽{𝑥}^2)(1+𝑥{)}^{11}$. When we choosing 1 from $(1+𝛼𝑥+𝛽{𝑥}^2)$ then we need ${𝑥}^{11}$ from $(1+𝑥{)}^{11}$.

So, coefficient $={}^{11}{𝐶}_{11}=1$

Similarly, we choosing $𝛼𝑥$ from $(1+𝛼𝑥+𝛽{𝑥}^2)$ then we need ${𝑥}^{10}$ from $(1+𝑥{)}^{11}=$ Coefficient $=𝛼\times {}^{11}{𝐶}_{10}=11𝛼$ and when we choosing $𝛽{𝑥}^2$ from $(1+𝛼𝑥+𝛽{𝑥}^2)$ then we need ${𝑥}^9$ from $(1+𝑥)=$ Coefficient $=𝛽\times {}^{11}{𝐶}_9=55\cdot 𝛽$

$\therefore$ Coefficient of ${𝑥}^{11}$ in the expansion of $(1+𝛼𝑥+𝛽{𝑥}^2)$

$(1+𝑥{)}^{11}=1+11𝛼+55𝛽=144$ ( given )          ...(i)

Just, in similar way, we have to find coefficient of ${𝑥}^{10}$ such that

$11+55𝛼+165𝛽=396...\left(𝑖𝑖\right)$

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

$\begin{array}{c}𝛼=-2\text{ and }𝛽=3\\ \therefore {𝛼}^2+{𝛽}^2=(-2{)}^2+(3{)}^2=13\end{array}$

Given, $\frac{-2}{3}<𝑥<\frac{2}{3}$

$\Rightarrow -2<3𝑥<2\Rightarrow -1<\frac{3𝑥}{2}<1$

Now, $\frac{1}{3\sqrt{\surd 2-3𝑥}}=\frac{1}{(2-3𝑥{)}^{1/3}}$

$=\frac{1}{2^{1/3}}{(1-\frac{3𝑥}{2})}^{-1/3}$

$=\frac{1}{3\sqrt{\surd 2}}[1+\left(\frac{1}{3}\right)\left(\frac{3𝑥}{2}\right)+\frac{\left(\frac{1}{3}\right)(\frac{1}{3}+1)}{2\times 1}{\left(\frac{3𝑥}{2}\right)}^2+\frac{\left(\frac{1}{3}\right)(\frac{1}{3}+1)(\frac{1}{3}+2)}{3\times 2\times 1}{\left(\frac{3𝑥}{2}\right)}^3+\frac{\frac{1}{3}(\frac{1}{3}+1)(\frac{1}{3}+2)(\frac{1}{3}+3)}{4\times 3\times 2\times 1}{\left(\frac{3𝑥}{2}\right)}^4+\dots ]$

$\begin{array}{cc}& \therefore 5\text{ th term when }𝑥=\frac{1}{2}\text{ is given by }\\ & \begin{array}{c}\frac{1}{\text{exception 25:}}[\frac{1}{3}\times \frac{4}{3}\times \frac{7}{3}\times \frac{10}{3}\times \frac{1}{24}\times \frac{81}{16}\times \frac{1}{16}]\\ =\frac{1}{3\sqrt{\surd 2}}\left[\frac{7\times 5}{24\times 2\times 6}\right]=\frac{35}{768\left(\text{exception 25:}\right)}\end{array}\end{array}$

Given, $𝑥>\sqrt{\surd 3}\Rightarrow {𝑥}^2>3\Rightarrow 0<\frac{1}{{𝑥}^2}<\frac{1}{3}$ Also, $\frac{{𝑥}^2+1}{({𝑥}^2+2)({𝑥}^2+3)}=\frac{-1}{{𝑥}^2+2}+\frac{2}{{𝑥}^2+3}=(-1){({𝑥}^2+2)}^{-1}+2{({𝑥}^2+3)}^{-1}$

$\begin{array}{cc}& \begin{array}{cc}& =(-1){\left({𝑥}^2\right)}^{-1}{(1+\frac{2}{{𝑥}^2})}^{-1}+2{\left({𝑥}^2\right)}^{-1}{(1+\frac{3}{{𝑥}^2})}^{-1}\\ & =\frac{-1}{{𝑥}^2}(1-\frac{2}{{𝑥}^2}+\frac{4}{{𝑥}^4}-\frac{8}{{𝑥}^6}+\frac{16}{{𝑥}^8}-\dots \infty )\\ & +\frac{2}{{𝑥}^2}(1-\frac{3}{{𝑥}^2}+\frac{9}{{𝑥}^4}-\frac{27}{{𝑥}^6}+\dots \infty )\end{array}\\ & \text{ So, the coefficient of }\\ & \begin{array}{cc}{𝑥}^{-8}& =(-1)(-8)+\left(2\right)(-27)\\ & =8-54=-46\end{array}\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \begin{array}{cc}𝐿𝑒𝑡𝑓\left(𝑥\right)& ={\cos }^2\left(\frac{𝑥}{4}\right)+\sin \left(\frac{𝑥}{4}\right)\\ & =1-{\sin }^2\left(\frac{𝑥}{4}\right)+\sin \left(\frac{𝑥}{4}\right)\\ & =-{\sin }^2\left(\frac{𝑥}{4}\right)+\sin \left(\frac{𝑥}{4}\right)+1\\ & =\frac{5}{4}-{(\sin \left(\frac{𝑥}{4}\right)-\frac{1}{2})}^2\end{array}\\ & {\therefore 𝑓\left(𝑥\right)|}_{max}=𝛼=\frac{5}{4}\text{ when }\sin \frac{𝑥}{4}=\frac{1}{2}\\ & \text{ and }{𝑓\left(𝑥\right)|}_{min}=𝛽=-1\text{ when }\sin \frac{𝑥}{4}=-1\\ & \therefore 𝛼-𝛽=\frac{5}{4}-(-1)=\frac{9}{4}\end{array}\end{array}$

$\begin{array}{cc}& 3{\cos }^2𝐴+2{\cos }^2𝐵=4\\ & \Rightarrow 3(1-{\sin }^2𝐴)+2(1-{\sin }^2𝐵)=4\end{array}$

Now, $3\sin 𝐴\cdot \cos 𝐴=2\sin 𝐵\cdot \cos 𝐵$ (given)

$\Rightarrow 3\sin 2𝐴=2\sin 2𝐵...\left(𝑖\right)$

$\begin{array}{cc}& \cos (𝐴+2𝐵)=\cos 𝐴\cos 2𝐵-\sin 𝐴\sin 2𝐵\\ & =\cos 𝐴\cdot 3{\sin }^2𝐴-\sin 𝐴\cdot \frac{3}{2}\cdot 2\sin 𝐴\cos 𝐴(𝑓𝑟𝑜𝑚𝐸𝑞.(𝑖\left)\right)\end{array}$

$=3\cos 𝐴{\sin }^2𝐴-3\cos 𝐴{\sin }^2𝐴=0$

$\therefore 𝐴+2𝐵={90}^{\circ }$

$\begin{array}{cc}& \text{ Given, }\sin 𝑥-\sin 𝑦=27/65\text{ and }\\ & \begin{array}{cc}& \cos 𝑥-\cos 𝑦=\frac{-21}{65}\\ & 2\sin \left(\frac{𝑥-𝑦}{2}\right)\cos \left(\frac{𝑥+𝑦}{2}\right)=\frac{27}{65}\text{ and }\\ & 2\sin \left(\frac{𝑥+𝑦}{2}\right)\sin \left(\frac{𝑥-𝑦}{2}\right)=\frac{21}{65}\\ & \therefore \tan \left(\frac{𝑥+𝑦}{2}\right)=\frac{21}{27}=\frac{7}{9}\\ & \sin (𝑥+𝑦)=\frac{2\tan \left(\frac{𝑥+𝑦}{2}\right)}{1+{\tan }^2\frac{𝑥+𝑦}{2}}=\frac{2\times \frac{7}{9}}{1+\frac{49}{81}}\\ & \sin (𝑥+𝑦)=\frac{63}{65}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, trigonometric equation is }\\ & \begin{array}{cc}& \sec 𝑥\cdot \cos 5𝑥+1=0\\ & \Rightarrow \frac{\cos 5𝑥+\cos 𝑥}{\cos 𝑥}=0\\ & \Rightarrow 2\cdot \cos 3𝑥\cdot \cos 2𝑥=0\end{array}\end{array}$

$\begin{array}{c}\Rightarrow \cos 3𝑥=0\text{ or }\cos 2𝑥=0\\ \Rightarrow 3𝑥=\frac{𝜋}{2},\frac{3𝜋}{2},\frac{5𝜋}{2},\frac{7𝜋}{2},\frac{9𝜋}{2},\frac{11𝜋}{2}\\ (\frac{\because 0\le 𝑥\le 2𝜋}{\therefore 0\le 3𝑥\le 6𝜋})\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow 𝑥=\frac{𝜋}{6},\frac{3𝜋}{6},\frac{5𝜋}{6},\frac{7𝜋}{6},\frac{9𝜋}{6},\frac{11𝜋}{6}\\ & \text{ Also, }2𝑥=\frac{𝜋}{2},\frac{3𝜋}{2},\frac{5𝜋}{2},\frac{7𝜋}{2}(\frac{\because 0\le 𝑥\le 2𝜋}{\therefore 0\le 2𝑥\le 4𝜋})\\ & \Rightarrow 𝑥=\frac{𝜋}{4},\frac{3𝜋}{4},\frac{5𝜋}{4},\frac{7𝜋}{4}\end{array}\\ & \therefore \text{ Total number of solutions }=10\end{array}$