TG EAPCET 2025 3rd May Evening Shift Question Paper with Solutions

Given, $𝑓\left(𝑥\right)=5^{\left|𝑥\right|}+\mathrm{Sgn}\left(5^{𝑥}\right)$

$\begin{array}{cc}\because & 5^{𝑥}>0\forall 𝑥\in 𝑅\\ \therefore & \mathrm{sgn}\left(5^{𝑥}\right)=1\\ \text{ And }& \left|𝑥\right|\ge 0\Rightarrow -\left|𝑥\right|\le 0\\ \Rightarrow & 5^{𝑥}\le 5^0\Rightarrow 5^{𝑥}\le 1\\ & 5^{\left|𝑥\right|}\le 5^0\Rightarrow 5^{\left|𝑥\right|}\le 1\\ \therefore & 𝑓\left(𝑥\right)=5^{\left|𝑥\right|}+1\le 1+1\le 2\\ \because & 𝑓\left(1\right)=𝑓(-1)\end{array}$

$\therefore 𝑓\left(𝑥\right)$ is neither one-one nor onto.

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

$\text{ ∵ Range }=[\frac{1}{3},3]$

For range let $𝑓\left(𝑥\right)=𝑦$

$\begin{array}{cc}& 𝑦=\frac{{𝑥}^2+𝑥+𝑘}{{𝑥}^2-𝑥+𝑘}\\ & \Rightarrow {𝑥}^2(𝑦-1)-𝑥(𝑦+1)+𝑘𝑦-𝑘=0\\ & \therefore 𝐷\ge 0\\ & \Rightarrow (𝑦+1{)}^2-4𝑘(𝑦-1\left)\right(𝑦-1)\ge 0\\ & \Rightarrow (𝑦+1{)}^2-(2\sqrt{\surd 𝑘}(𝑦-1){)}^2\ge 0\\ & \Rightarrow (𝑦+1+2\sqrt{\surd 𝑘}(𝑦-1\left)\right)\\ & (𝑦+1-2\sqrt{\surd 𝑘}(𝑦-1)\ge 0\\ & \Rightarrow \left(𝑦\right(2\sqrt{\surd 𝑘}+1)+1-2\sqrt{\surd 𝑘})\\ & \left(𝑦\right(2\sqrt{\surd 𝑘}-1)-2\sqrt{\surd 𝑘}-1)\le 0...\left(𝑖\right)\\ \because & 𝑦\in [\frac{1}{3},3]\\ & (3𝑦-1)(𝑦-3)\le 0...\left(𝑖𝑖\right)\end{array}$

Comparing Eqs. (i) and (ii),

$2\sqrt{\surd 𝑘}+1=3\Rightarrow 2\sqrt{\surd 𝑘}=2\Rightarrow 𝑘=1$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \begin{array}{cc}{2}^{𝑛+4}& +12\ge 𝑘(𝑛+4)\\ 𝑘& \le \frac{2^{𝑛+4}+12}{𝑛+4}\\ \text{ Let }𝑛+4& =𝑡\\ \Rightarrow 𝑘& \le \frac{2^{𝑡}+12}{𝑡},𝑡\ge 5\\ 𝑡& =5,𝑘\le \frac{2^5+12}{5}=8.81\\ 𝑡& =6,𝑘\le \frac{2^6+12}{6}=1267\\ 𝑡& =7,𝑘\le \frac{2^7+12}{7}=20\\ \because 𝑡& =5\Rightarrow 𝑘\text{ will be minimum }=8.8\\ \therefore \left[𝑘\right]& =8\end{array}\end{array}\end{array}$

$\begin{array}{c}𝑥-2𝑦+𝑧=0\\ 2𝑥+3𝑦+𝑧=6\\ 𝑥+2𝑦+𝑝𝑧=𝑞\end{array}$

$\begin{array}{cc}& \text{ System has infinite solution, }\\ & \begin{array}{cc}\therefore & \Delta ={\Delta }_{𝑥}={\Delta }_{𝑦}={\Delta }_{𝑧}=0\\ & \Delta =\mid \begin{array}{ccc}1& -2& 1\\ 2& 3& 1\\ 1& 2& 𝑝\end{array}\mid =0\\ \Rightarrow & 1(3𝑝-2)+2(2𝑝-1)+1(4-3)=0\\ \Rightarrow & 3𝑝-2+4𝑝-2+1=0\\ \Rightarrow & 7𝑝=3\Rightarrow 𝑝=\frac{3}{7}\\ & {\Delta }_{𝑧}=\mid \begin{array}{ccc}1& -2& 0\\ 2& 3& 6\\ 1& 2& 𝑞\end{array}\mid =0\\ \Rightarrow & 1(3𝑞-12)+2(2𝑞-6)=0\\ \Rightarrow & 3𝑞-12+4𝑞-12=0\\ \Rightarrow & 7𝑞=24\Rightarrow 𝑞=\frac{24}{7}\\ \therefore & 𝑞-𝑝=\frac{24}{7}-\frac{3}{7}=3\end{array}\end{array}$

We have,

$\begin{array}{cc}& (\sin 𝜃)𝑥-𝑦+𝑧=0\\ & 𝑥-(\cos 𝜃)𝑦+𝑧=0\\ & 𝑥+𝑦+(\sin 𝜃)𝑧=0\end{array}$

System has non-trivial solution.

$\therefore \Delta =0$

$\begin{array}{cc}& \mid \begin{array}{ccc}\sin 𝜃& -1& 1\\ 1& -\cos 𝜃& 1\\ 1& 1& \sin 𝜃\end{array}\mid =0\\ & \Rightarrow \sin 𝜃(-\sin 𝜃\cos 𝜃-1)+1(\sin 𝜃-1)+1(1+\cos 𝜃)=0\\ & \Rightarrow -{\sin }^2𝜃\cos 𝜃-\sin 𝜃+\sin 𝜃-1+1+\cos 𝜃=0\\ & \Rightarrow \cos 𝜃(1-{\sin }^2𝜃)=0\\ & \Rightarrow {\cos }^3𝜃=0\Rightarrow \cos 𝜃=0\Rightarrow 𝜃=\frac{𝜋}{2}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}1& 2& 3\\ 2& 1& 1\\ 1& 3& 1\end{array}⎤⎦⎥\\ \left|𝐴\right|& =1(1-3)-2(2-1)+3(6-1)\\ & =-2-2+15=11\\ 𝐵& =⎡⎣⎢\begin{array}{ccc}2& 3& 4\\ 3& 2& 2\\ 2& 4& 2\end{array}⎤⎦⎥\\ \Rightarrow & \left|𝐵\right|=2(4-8)-3(6-4)+4(12-4)\\ & =-8-6+32=18\\ \therefore & \sqrt{\surd |\mathrm{adj}𝐴𝐵|}=\sqrt{\surd |\mathrm{adj}𝐴||\mathrm{adj}𝐵|}\\ & =\sqrt{\surd \left|𝐴{|}^{3-1}\right|𝐵{|}^{3-1}}\\ & =\left|𝐴\right|\left|𝐵\right|=11\left(18\right)=198\end{array}\end{array}$

Step 1: Find the determinant of $𝐴$

Matrix $𝐴$ is: $𝐴=⎡⎣⎢\begin{array}{ccc}1& 5& 2\\ 4& 1& 3\\ 2& 6& 3\end{array}⎤⎦⎥$

So, $\left|𝐴\right|=1(3-18)-5(12-6)+2(24-2)$

Calculate each part:

• $1\times (3-18)=1\times (-15)=-15$
• $-5\times (12-6)=-5\times 6=-30$
• $2\times (24-2)=2\times 22=44$

Add them up: $\left|𝐴\right|=-15-30+44=-1$

Step 2: Find the value of $\mid (\mathrm{adj}𝐴{)}^{-1}\mid$

$\mid (\mathrm{adj}𝐴{)}^{-1}\mid =\frac{1}{|\mathrm{adj}𝐴|}$

The order of matrix $𝐴$ is 3, so: $|\mathrm{adj}𝐴|=|𝐴{|}^{3-1}=|𝐴{|}^2$

So, $\mid (\mathrm{adj}𝐴{)}^{-1}\mid =\frac{1}{|𝐴{|}^2}$

Since $\left|𝐴\right|=-1$, $\mid (\mathrm{adj}𝐴{)}^{-1}\mid =\frac{1}{(-1{)}^2}=\frac{1}{1}=1$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& =\frac{(\sqrt{\surd 3}+𝑖)(1-\sqrt{\surd 3}𝑖)}{(-1+𝑖)(-1-𝑖)}\\ & =\frac{\sqrt{\surd 3}-3𝑖+𝑖+\sqrt{\surd 3}}{1+𝑖-𝑖+1}=\frac{2\sqrt{\surd 3}-2𝑖}{2}\\ & =\sqrt{\surd 3}-𝑖\\ \mathrm{Amp}\left(𝑧\right)& =-{\tan }^{-1}\mid \frac{1}{\sqrt{\surd 3}}\mid =\frac{-𝜋}{6}\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \mathrm{Im}\left(\frac{𝑧-3}{𝑧+3𝑖}\right)=0,𝑧+3𝑖\ne 0\\ & \therefore 𝑥\ne 0,𝑦\ne -3\\ & \Rightarrow \frac{𝑧-3}{𝑧+3𝑖}-\left(\frac{\overline{𝑧-3}}{𝑧+3𝑖}\right)=0\\ & \Rightarrow \frac{𝑧-3}{𝑧+3𝑖}-\frac{\overline{𝑧}-3}{\overline{𝑧}-3𝑖}=0\\ & \Rightarrow (𝑧-3)(\overline{𝑧}-3𝑖)-(𝑧+3𝑖)(\overline{𝑧}-3)=0\\ & \Rightarrow 𝑧\overline{𝑧}-3𝑖𝑧-3\overline{𝑧}+9𝑖-𝑧\overline{𝑧}+3𝑧-3𝑖\overline{𝑧}+9𝑖=0\\ & \Rightarrow -3𝑖(𝑧+\overline{𝑧})+3(𝑧-\overline{𝑧})+18𝑖=0\\ & \because 𝑧=𝑥+𝑖𝑦,\overline{𝑧}=𝑥-𝑖𝑦\\ & \Rightarrow -3𝑖\left(2𝑥\right)+3\left(2𝑖𝑦\right)+18𝑖=0\\ & \Rightarrow 𝑥-𝑦-3=0,𝑥,𝑦\ne (0,-3)\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& 𝑧=(\sqrt{\surd 3}+𝑖{)}^{10}+(\sqrt{\surd 3}-𝑖{)}^{10}\\ & ={\left(2{𝑒}^{\frac{𝜋}{6}}\right)}^{10}+{\left(2{𝑒}^{\frac{-𝜋𝑖}{6}}\right)}^{10}\\ & =2^{10}({𝑒}^{\frac{10𝜋𝑖}{6}}+{𝑒}^{\frac{-𝑖10𝜋}{6}})\\ & =2^{10}(\cos \frac{10𝜋}{6}+𝑖\sin \frac{10𝜋}{6}+\cos \frac{10𝜋}{6}-𝑖\sin \frac{10𝜋}{6})\\ & =2^{10}\cdot 2\cos \frac{5𝜋}{3}=2^{11}\cos (2𝜋-\frac{𝜋}{3})\\ & =2^{11}\times \frac{1}{2}=2^{10}=1024\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \begin{array}{cc}𝐿𝑒𝑡& (-1-\sqrt{\surd 3}𝑖{)}^{3/4}\\ & 𝑧=-1-\sqrt{\surd 3}𝑖\end{array}\\ & \begin{array}{cc}𝑧& =2{𝑒}^{𝑖\left(\frac{-2𝜋}{3}\right)}\\ {𝑧}^{3/4}& ={𝑒}^{𝑖\left(\frac{-2𝜋}{3}\right)}\\ & ={\left(2{𝑒}^{-𝑖\frac{2𝜋}{3}}\right)}^{3/4}\\ & =2^{3/4}{\left({𝑒}^{-𝑖𝜋}\right)}^{1/2}\\ \therefore \text{ At }𝑘& =0,{𝑧}_1=2^{3/4}{𝑒}^0=2^{3/4}\\ \text{ At }𝑘& =1,{𝑧}_2=2^{3/4}{𝑒}^{-𝑖𝜋}=-2^{3/4}\end{array}\\ & \text{ Number of real value }=2\end{array}\end{array}$

The equation $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$ has the roots $\tan 𝜃$ and $\cot 𝜃$.

For a quadratic equation $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$, if $𝑝$ and $𝑞$ are roots, then:

Sum of roots: $𝑝+𝑞=-\frac{𝑏}{𝑎}$
Product of roots: $𝑝𝑞=\frac{𝑐}{𝑎}$

Here, the roots are $\tan 𝜃$ and $\cot 𝜃$. So, $\tan 𝜃+\cot 𝜃=-\frac{𝑏}{𝑎}$ and $\tan 𝜃\cdot \cot 𝜃=\frac{𝑐}{𝑎}$.

$\tan 𝜃+\cot 𝜃=\frac{\sin 𝜃}{\cos 𝜃}+\frac{\cos 𝜃}{\sin 𝜃}$

This simplifies to: $\frac{{\sin }^2𝜃+{\cos }^2𝜃}{\sin 𝜃\cos 𝜃}=\frac{1}{\sin 𝜃\cos 𝜃}$,
because ${\sin }^2𝜃+{\cos }^2𝜃=1$

Now, $\tan 𝜃+\cot 𝜃=\frac{1}{\sin 𝜃\cos 𝜃}$
But from earlier, $\tan 𝜃+\cot 𝜃=-\frac{𝑏}{𝑎}$.
So, $\frac{1}{\sin 𝜃\cos 𝜃}=-\frac{𝑏}{𝑎}$.

For $\tan 𝜃\cdot \cot 𝜃=\frac{𝑐}{𝑎}$:
$\tan 𝜃\cdot \cot 𝜃=1$, so $\frac{𝑐}{𝑎}=1⟹𝑐=𝑎$.

Now, take the result $\frac{1}{\sin 𝜃\cos 𝜃}=-\frac{𝑏}{𝑎}$.
Flip both sides: $\sin 𝜃\cos 𝜃=-\frac{𝑎}{𝑏}$

$2\sin 𝜃\cos 𝜃=2\times (-\frac{𝑎}{𝑏})=-\frac{2𝑎}{𝑏}$

But $2\sin 𝜃\cos 𝜃=\sin 2𝜃$.
So, $\sin 2𝜃=-\frac{2𝑎}{𝑏}$.

Since $𝑎=𝑐$, we can also write $\sin 2𝜃=-\frac{2𝑐}{𝑏}$.

We have,

$\begin{array}{cc}& \left|\right|2𝑥-3|-4|=2\\ \Rightarrow & |2𝑥-3|-4=\pm 2\\ \Rightarrow & |2𝑥-3|=\pm 2+4\\ \Rightarrow & |2𝑥-3|=6\text{ or }|2𝑥-3|=2\\ \Rightarrow & 2𝑥-3=\pm 6\text{ or }2𝑥-3=\pm 2\\ \Rightarrow & 2𝑥=\pm 6+3\text{ or }2𝑥=\pm 2+3\\ \Rightarrow & 2𝑥=9,-3\text{ or }2𝑥=5,1\\ \Rightarrow & 𝑥=\frac{9}{2},\frac{-3}{2}\text{ or }𝑥=\frac{5}{2},\frac{1}{2}\end{array}$

∴ Sum of roots

$\begin{array}{cc}& =\frac{9}{2}-\frac{3}{2}+\frac{5}{2}+\frac{1}{2}\\ & =\frac{9-3+5+1}{2}=\frac{12}{2}=6\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& 3{𝑥}^5-6{𝑥}^4+2{𝑥}^3+4{𝑥}^2-5𝑥+8\\ \Rightarrow & 3{𝑥}^3({𝑥}^2-2𝑥+3)-7𝑥({𝑥}^2-2𝑥+3)-10({𝑥}^2-2𝑥+3)-4𝑥+38\\ \Rightarrow & ({𝑥}^2-2𝑥+3)(3{𝑥}^3-7𝑥-10)-4𝑥+38\\ \therefore & 𝑎=3,𝑏=0,𝑐=-7,𝑑=-10\\ & 𝑝=-4,𝑞=38\\ \therefore & 𝑎𝑏+𝑐𝑑=70\\ & 𝑝+2𝑞-2=-4+76-2=70\\ \therefore & 𝑎𝑏+𝑐𝑑=𝑝+2𝑞-2\end{array}\end{array}$

$\begin{array}{cc}& 𝛼𝛽=𝛼𝛿=1\\ & 12({𝑥}^2+\frac{1}{{𝑥}^2})-56(𝑥+\frac{1}{𝑥})+89=0\\ & 12{(𝑥+\frac{1}{𝑥})}^2-24-56(𝑥+\frac{1}{𝑥})+89=0\\ & 12{(𝑥+\frac{1}{𝑥})}^2-56(𝑥+\frac{1}{𝑥})+65=0\\ & 12{𝑡}^2-56𝑡+65=0,𝑥+\frac{1}{𝑥}=𝑡\\ & 𝑡=\frac{56\pm \sqrt{\surd (56{)}^2-4(12\left)\right(65)}}{2\times 12}\end{array}$

$\begin{array}{cc}& 𝑥+\frac{1}{𝑥}=\frac{5}{2},𝑥+\frac{1}{𝑥}=\frac{13}{6}\\ & 𝑥=2,\frac{1}{2},𝑥=\frac{2}{3},\frac{3}{2}\\ \therefore & \frac{𝛼+𝛽}{𝛾+𝛿}=\frac{\frac{5}{2}}{\frac{13}{6}}=\frac{15}{13}\end{array}$

$\text{ Arrangement }=\frac{4!}{2!}$

Selection of 3 gaps out of 5 for $3{𝐴}^{\prime }5={}^5{𝐶}_3$

∴ Required arrangements

$\begin{array}{cc}& =\frac{4!}{2!}\cdot {}^5{𝐶}_3\cdot 1\times \frac{5!}{2!}\\ & =\frac{4!}{2!}\times \frac{5\times 4}{2\times 1}\times \frac{5!}{2!}\\ & =12\times 5\times 120=7200\end{array}$

Selecting 1 vowel 2 consonants (arrangement in which vowel is at middle place)

$\begin{array}{cc}& =({}^3{𝐶}_1\cdot {}^4{𝐶}_2)2!+({}^3{𝐶}_1\cdot {}^1{𝐶}_1)\cdot 1\\ & =3\times 6\times 2+3=39\end{array}$

Similarly,

Selecting, 2 vowels 1 consonant

$={}^3{𝐶}_2\cdot {}^4{𝐶}_1(2!)\times 2=3\times 4\times 4=48$

Selecting 3 vowels $=3!=6$

Total words $=39+48+6=93$

6 Boys and 4 girls

Number of arrangements in a row such that between any 2 girls there must be exactly 2 boys is

$\begin{array}{cc}& =\stackrel{{𝐺}_1}{\to \leftarrow }\uparrow \stackrel{{𝐺}_2}{\to \leftarrow }\uparrow \stackrel{{𝐺}_3}{\to \leftarrow }\uparrow \stackrel{{𝐺}_4}{\to \leftarrow }\\ & =4!\left(\frac{6!}{(2!{)}^3\cdot 3!}\right)\cdot 3!(2!{)}^3\\ & =6!\times 24=144(5!)\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& =\sum _{𝑟=0}^5{𝑟}^3\cdot {}^5{𝐶}_{𝑟}=\sum _{𝑟=1}^5{𝑟}^3\cdot \frac{5}{𝑟}\cdot {}^4{𝐶}_{𝑟-1}\\ & =\sum _{𝑟=1}^5{𝑟}^2\cdot 5\cdot {}^4{𝐶}_{𝑟-1}=𝑡=𝑟-1\\ & =5\sum _{𝑡=0}^4(𝑡+1{)}^2\cdot {}^4{𝐶}_{𝑡}\\ & 5(\Sigma {𝑡}^2\cdot {}^4{𝐶}_{𝑡}+2\Sigma 𝑡\cdot {}^4{𝐶}_{𝑡}+{\Sigma }^4{𝐶}_{𝑡})\\ & =5(\Sigma {𝑡}^2\cdot \frac{4}{𝑡}\cdot {}^3{𝐶}_{𝑡-1}+2\Sigma 𝑡\cdot \frac{4}{𝑡}\cdot {}^3{𝐶}_{𝑡-1}+2^{𝑡})\\ & =5(4\Sigma 𝑡{}^3{𝐶}_{𝑡-1}+8\left(2^3\right)+2^4)\\ & 𝑝=𝑡-1\\ & =5(4\cdot \Sigma (𝑝+1){}^3{𝐶}_{𝑝}+64+16)\\ & =5(4\cdot \Sigma 𝑝\cdot {}^3{𝐶}_{𝑝}+4\left(2^3\right)+64+16)\\ & =5(4\cdot 3\cdot 2^2+32+64+16)\\ & =5(48+32+80)=5\left(160\right)=800\end{array}\end{array}$

We need to find the coefficient of ${𝑥}^{12}$ in the expansion of ${({𝑥}^2+2𝑥+2)}^8$.

Step 1: General Term Formula

Each term in the expansion is made by choosing ${𝑃}_1$ times ${𝑥}^2$, ${𝑃}_2$ times $2𝑥$, and ${𝑃}_3$ times $2$ from the $8$ factors, so ${𝑃}_1+{𝑃}_2+{𝑃}_3=8$. The general term is:
$\frac{8!}{{𝑃}_1!{𝑃}_2!{𝑃}_3!}\left({𝑥}^2{)}^{{𝑃}_1}\right(2𝑥{)}^{{𝑃}_2}(2{)}^{{𝑃}_3}$

Step 2: Power of $𝑥$

The power of $𝑥$ in each term is $2{𝑃}_1$ (from ${𝑥}^2$) plus ${𝑃}_2$ (from $2𝑥$), for a total of $2{𝑃}_1+{𝑃}_2$.

Step 3: Find Values for ${𝑥}^{12}$

We want $2{𝑃}_1+{𝑃}_2=12$ and ${𝑃}_1+{𝑃}_2+{𝑃}_3=8$.
Possible solutions are: $\begin{array}{ccc}{𝑃}_1& {𝑃}_2& {𝑃}_3\\ 6& 0& 2\\ 5& 2& 1\\ 4& 4& 0\end{array}$

Step 4: Write Each Term's Coefficient

For each solution, we substitute into the formula:

• When ${𝑃}_1=6,{𝑃}_2=0,{𝑃}_3=2$:
  Coefficient: $\frac{8!}{6!0!2!}\cdot 2^0\cdot 2^0\cdot 2^2$
• When ${𝑃}_1=5,{𝑃}_2=2,{𝑃}_3=1$:
  Coefficient: $\frac{8!}{5!2!1!}\cdot 2^2\cdot 2^1$
• When ${𝑃}_1=4,{𝑃}_2=4,{𝑃}_3=0$:
  Coefficient: $\frac{8!}{4!4!0!}\cdot 2^4$

Step 5: Calculate Coefficients

$\begin{array}{cc}& \text{Total Coefficient}\\ & =\frac{8!}{6!2!}\cdot 2^2+\frac{8!}{5!2!1!}\cdot 2^2\cdot 2^1+\frac{8!}{4!4!}\cdot 2^4\\ & =28\times 4+168\times 4\times 2+70\times 16\\ & =112+1344+1120\\ & =2576\end{array}$

The coefficient of ${𝑥}^{12}$ is $\overline{)2576}$.

$\begin{array}{cc}& \frac{{𝑥}^2+1}{({𝑥}^2+2)({𝑥}^2+3)}=\frac{𝐴𝑥+𝐵}{{𝑥}^2+2}+\frac{𝐶𝑥+𝐷}{{𝑥}^2+3}\\ & {𝑥}^2+1=(𝐴𝑥+𝐵)({𝑥}^2+3)+(𝐶𝑥+𝐷)({𝑥}^2+2)\end{array}$

Coefficient of ${𝑥}^3=0$, coefficient of $𝑥=0$

$\Rightarrow 𝐴+𝐶=0,3𝐴+2𝐶=0\Rightarrow 𝐴=𝐶=0$

Coefficient of ${𝑥}^2=1$, constant term $=1$

$𝐵+𝐷=1,1=3𝐵+2𝐷$

$\begin{array}{cc}& 3𝐵+2𝐷=1\\ & 2𝐵+2𝐷=2\\ & \underset{–}{---}\\ & 𝐵=-1,𝐷=2\end{array}$

$\therefore 𝐴+𝐵+𝐶+𝐷=1$

$\begin{array}{cc}& 2\sin 𝜃+3\cos 𝜃=2\\ & \Rightarrow (2\sin 𝜃+3\cos 𝜃{)}^2=4\\ & \Rightarrow 4{\sin }^2𝜃+9{\cos }^2𝜃+12\sin 𝜃\cos 𝜃=4\\ & \Rightarrow 5{\cos }^2𝜃+12\sin 𝜃\cos 𝜃=0\\ & \Rightarrow \cos 𝜃(5\cos 𝜃+12\sin 𝜃)=0\\ & \Rightarrow \tan 𝜃=\frac{-5}{12}\\ & \sin 𝜃=\frac{-5}{13},\cos 𝜃=\frac{12}{13}\\ & \therefore \sin 𝜃+\cos 𝜃=\frac{-5}{13}+\frac{12}{13}=\frac{7}{13}\end{array}$

$\sin 𝐴=\frac{-24}{25},\cos 𝐵=\frac{15}{17}$

According to the question,

$𝐴$ is 3rd quadrant, $𝐵$ in 4th quadrant.

Now, $\sin (𝐴+𝐵)=\sin 𝐴\cos 𝐵+\cos 𝐴\sin 𝐵$