AP EAPCET 2024 22nd May Evening Shift Question Paper with Solutions

We need to determine the range of the function:

$𝑓\left(𝑥\right)=\frac{15}{3\sin 𝑥+4\cos 𝑥+10}$

Start with the expression inside the denominator:

$3\sin 𝑥+4\cos 𝑥$

The range of this expression is given by:

$-\sqrt{\surd 3^2+4^2}\le 3\sin 𝑥+4\cos 𝑥\le \sqrt{\surd 3^2+4^2}$

Calculating the bounds:

$-\sqrt{\surd 9+16}\le 3\sin 𝑥+4\cos 𝑥\le \sqrt{\surd 9+16}$

$-5\le 3\sin 𝑥+4\cos 𝑥\le 5$

Adding 10 to the entire inequality, we have:

$5\le 3\sin 𝑥+4\cos 𝑥+10\le 15$

The function $𝑓\left(𝑥\right)$ can then be rewritten as:

$𝑓\left(𝑥\right)=\frac{15}{3\sin 𝑥+4\cos 𝑥+10}$

Therefore, the minimum value of the denominator is 5, and the maximum is 15.

The range of $𝑓\left(𝑥\right)$ can be calculated as follows:

Minimum value of $𝑓\left(𝑥\right)$:

$𝑓(𝑥{)}_{min}=\frac{15}{15}=1$

Maximum value of $𝑓\left(𝑥\right)$:

$𝑓(𝑥{)}_{max}=\frac{15}{5}=3$

Hence, the range of the function $𝑓\left(𝑥\right)$ is $[1,3]$.

Given the functions:

$𝑓\left(𝑥\right)={𝑥}^2-1$

$𝑔\left(𝑥\right)=\sqrt{\surd {𝑥}^2+1}$

$ℎ\left(𝑥\right)=\{\begin{array}{cc}0,& \text{if }𝑥\le 0\\ 𝑥,& \text{if }𝑥\ge 0\end{array}$

Let's analyze the statements:

Function Composition:

$𝑓\circ 𝑔=𝑓\left(𝑔\right(𝑥\left)\right)=𝑓\left(\sqrt{\surd {𝑥}^2+1}\right)=(\sqrt{\surd {𝑥}^2+1}{)}^2-1={𝑥}^2+1-1={𝑥}^2$.

Since $𝑓\circ 𝑔={𝑥}^2$, it is clear that this function is not one-to-one because different values of $𝑥$ (such as $𝑥$ and $-𝑥$) yield the same ${𝑥}^2$. Therefore, $𝑓\circ 𝑔$ is not invertible.

Composition involving $ℎ$:

$ℎ\circ 𝑓\circ 𝑔=ℎ\left(𝑓\right(𝑔\left(𝑥\right)\left)\right)=ℎ\left({𝑥}^2\right)$.

The function $ℎ\left({𝑥}^2\right)$ has the rule:

$ℎ\left({𝑥}^2\right)=\{\begin{array}{cc}0,& \text{if }𝑥\le 0\\ {𝑥}^2,& \text{if }𝑥\ge 0\end{array}$

Since ${𝑥}^2$ is always non-negative and $ℎ\left({𝑥}^2\right)={𝑥}^2$ for $𝑥\ge 0$, this simplifies to $ℎ\left({𝑥}^2\right)={𝑥}^2$ across non-negative ranges, reinforcing that $ℎ\left({𝑥}^2\right)$ is equivalent to ${𝑥}^2$.

Considering these observations, statements (III) and (IV) are true:

Statement (III): $𝑓\circ 𝑔$ is not invertible.

Statement (IV): $ℎ\circ 𝑓\circ 𝑔={𝑥}^2$.

Thus, statements (III) and (IV) are correct.

Given that

$𝑃$ is the greatest divisor of ${49}^{𝑛}+16𝑛-1$, for $\forall$ all $𝑛\in 𝑁$.

$\begin{array}{cc}& {49}^{𝑛}+16𝑛-1=(1+48{)}^{𝑛}+16𝑛-1\\ =& 1+𝑛\cdot 48+\frac{𝑛(𝑛-1)}{2}\cdot (48{)}^2+\dots ..16𝑛-1\\ =& 64𝑛+{}^{𝑛}{𝐶}_2(48{)}^2+{}^{𝑛}{𝐶}_3(48{)}^3+\dots (48{)}^{𝑛}\\ =& 64(𝑛+{}^{𝑛}{𝐶}_2{6}^2+{}^{𝑛}{𝐶}_3{6}^3\cdot 8+\dots 6^{𝑛}\cdot 8^{𝑛-2})\end{array}$

So, clearly it is divisible by 64 .

Now, number factor of 64 is

$64=2^6=(6+1)=7$

Given,

$𝐴=⎡⎣⎢\begin{array}{ccc}0& 1& 2\\ 2& 3& 0\\ 4& 0& 3\end{array}⎤⎦⎥$

and $𝐵$ is a matrix such that $𝐴𝐵=𝐵𝐴$. And $𝐴𝐵$ is not an identity matrix.

Let

$𝐵=⎡⎣⎢\begin{array}{ccc}{𝑎}_1& {𝑎}_2& {𝑎}_3\\ {𝑏}_1& {𝑏}_2& {𝑏}_3\\ {𝑐}_1& {𝑐}_2& {𝑐}_3\end{array}⎤⎦⎥$

Now, $𝐴𝐵=𝐵𝐴$

$\begin{array}{cc}& ⎡⎣⎢\begin{array}{ccc}0& 1& 2\\ 2& 3& 0\\ 4& 0& 3\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}{𝑎}_1& {𝑎}_2& {𝑎}_3\\ {𝑏}_1& {𝑏}_2& {𝑏}_3\\ {𝑐}_1& {𝑐}_2& {𝑐}_3\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}{𝑎}_1& {𝑎}_2& {𝑎}_3\\ {𝑏}_1& {𝑏}_2& {𝑏}_3\\ {𝑐}_1& {𝑐}_2& {𝑐}_3\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}0& 1& 2\\ 2& 3& 0\\ 4& 0& 3\end{array}⎤⎦⎥\\ & ⎡⎣⎢\begin{array}{ccc}{𝑏}_1+2{𝑐}_1& {𝑏}_2+2{𝑐}_2& {𝑏}_3+2{𝑐}_3\\ 2{𝑎}_1+3{𝑏}_1& 2{𝑎}_2+3{𝑏}_2& 2{𝑎}_3+3{𝑏}_3\\ 4{𝑎}_1+3{𝑐}_1& 4{𝑎}_2+3{𝑐}_2& 4{𝑎}_3+3{𝑐}_3\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}2{𝑎}_2+4{𝑎}_3& {𝑎}_1+3{𝑎}_2& 2{𝑎}_1+3{𝑎}_3\\ 2{𝑏}_2+4{𝑏}_3& {𝑏}_1+3{𝑏}_2& 2{𝑏}_1+3{𝑏}_3\\ 2{𝑐}_2+4{𝑐}_3& {𝑐}_1+3{𝑐}_2& 2{𝑐}_1+3{𝑐}_3\end{array}⎤⎦⎥\end{array}$

On comparing two matrix and solve for ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑏}_1,{𝑏}_2,{𝑏}_3,{𝑐}_1,{𝑐}_2$ and ${𝑐}_3$, we get

$\begin{array}{cc}& {𝑎}_1=9,{𝑎}_2=-3\text{ and }{𝑎}_3=-6\\ & {𝑏}_1=-6,{𝑏}_2=-8\text{ and }{𝑏}_3=4\\ & {𝑐}_1=-12{𝑐}_2=4\text{ and }{𝑐}_3=-2\end{array}$

So, $𝐵=⎡⎣⎢\begin{array}{ccc}9& -3& -6\\ -6& -8& 4\\ -12& 4& -2\end{array}⎤⎦⎥$

Let $𝐴=⎡⎣⎢\begin{array}{ccc}𝑥-2& 0& 1\\ 1& 𝑥+3& 2\\ 2& 0& 2𝑥-1\end{array}⎤⎦⎥$ is a singular matrix.

So, $\left|𝐴\right|=0$

$\mid \begin{array}{ccc}𝑥-2& 0& 1\\ 1& 𝑥+3& 2\\ 2& 0& 2𝑥-1\end{array}\mid =0$

$\begin{array}{cc}& \Rightarrow (𝑥-2)\left[\right(𝑥+3\left)\right(2𝑥-1)-0]-0\\ & \Rightarrow (𝑥-2)(𝑥+3)(2𝑥-1)-2(𝑥+3)]=0\\ & \Rightarrow (𝑥+3)\left[\right(𝑥-2\left)\right(2𝑥-1)-2]=0\\ & \Rightarrow (𝑥+3)[2{𝑥}^2-𝑥-4𝑥+2-2]=0\\ & \Rightarrow (𝑥+3)[2{𝑥}^2-5𝑥]=0\\ & \Rightarrow 𝑥(𝑥+3)(2𝑥-5)=0\\ & \Rightarrow 𝑥=-3,0,5/2\end{array}$

So, $𝛼=-3,𝛽=0$ and $𝛾=5/2$

$\text{ [given }𝛼<𝛽<𝛾\text{ ] }$

Now,

$\begin{array}{cc}2𝛼+3𝛽+4𝛾& =2\times (-3)+3\times 0+4\times \frac{5}{2}\\ & =-6+10=4\end{array}$

To determine the nature of the solutions for the system of linear equations:

$\begin{array}{cc}𝑥+2𝑦+𝑧& =-3,\\ 3𝑥+3𝑦-2𝑧& =-1,\\ 2𝑥+7𝑦+7𝑧& =-4,\end{array}$

we begin by calculating the determinant $\Delta$ of the coefficient matrix:

$\Delta =\mid \begin{array}{ccc}1& 2& 1\\ 3& 3& -2\\ 2& 7& 7\end{array}\mid$

Calculating this determinant:

$\Delta =1(21+14)-2(21+4)+1(21-6)$

$=1\times 35-2\times 25+1\times 15$

$=35-50+15=0$

Since $\Delta =0$, the system of equations is either inconsistent (no solution) or dependent (infinitely many solutions). To resolve this, we check ${\Delta }_{𝑥}$, the determinant of the matrix when the first column is replaced by the constants from the right side of the equations:

${\Delta }_{𝑥}=\mid \begin{array}{ccc}-3& 2& 1\\ -1& 3& -2\\ -4& 7& 7\end{array}\mid$

Calculating ${\Delta }_{𝑥}$:

${\Delta }_{𝑥}=(-3)\left(35\right)-2(-15)+1\left(5\right)$

$=-105+30+5=-70$

Since $\Delta =0$ and ${\Delta }_{𝑥}\ne 0$, the system has no solution. Hence, the equations are inconsistent.

We have, $\arg \left[\frac{(1+𝑖\sqrt{\surd 3})(-\sqrt{\surd 3}-𝑖)}{(1-𝑖)(-𝑖)}\right]$

$\begin{array}{cc}𝐿𝑒𝑡𝑧& =\frac{(1+𝑖\sqrt{\surd 3})(-\sqrt{\surd 3}-𝑖)}{(1-𝑖)(-𝑖)}\\ & =\frac{(𝑖+{𝑖}^2\sqrt{\surd 3})(-\sqrt{\surd 3}-𝑖)}{(1-𝑖)(-{𝑖}^2)}\\ & =\frac{(-\sqrt{\surd 3}+𝑖)(-\sqrt{\surd 3}-𝑖)}{(1-𝑖)}\\ & =\frac{(-\sqrt{\surd 3}{)}^2-{𝑖}^2}{1-𝑖}=\frac{3+1}{1-𝑖}=\left(\frac{4}{1-𝑖}\right)\\ & \frac{4(1+𝑖)}{(1-𝑖)(1+𝑖)}=4\times \frac{1+𝑖}{2}\end{array}$

$\begin{array}{cc}\Rightarrow 𝑧& =2(1+𝑖)\\ \therefore \arg \left(𝑧\right)=\arg (2+2𝑖)& ={\tan }^{-1}\left(1\right)\\ & ={\tan }^{-1}(\tan \frac{𝜋}{4})\\ \Rightarrow \arg \left(𝑧\right)& =\frac{𝜋}{4}\end{array}$

Given, arg $\left(\frac{𝑧-3𝑖}{𝑧+4}\right)=\frac{𝜋}{2}$ and $𝑧=𝑥+𝑖𝑦$

$\begin{array}{cc}& \text{ Now, let }𝜔=\frac{𝑧-3𝑖}{𝑧+4}=\frac{𝑥+𝑖𝑦-3𝑖}{𝑥+𝑖𝑦+4}\\ & =\frac{𝑥+(𝑦-3)𝑖}{(𝑥+4)+𝑖𝑦}\times \frac{(𝑥+4)-𝑖𝑦}{(𝑥+4)-𝑖𝑦}\\ & =\frac{𝑥(𝑥+4)-𝑖𝑥𝑦+(𝑦-3)(𝑥+4)𝑖+𝑦(𝑦-3)}{(𝑥+4{)}^2+{𝑦}^2}\\ & \Rightarrow \frac{𝑥(𝑥+4)+𝑦(𝑦-3)}{(𝑥+4{)}^2+{𝑦}^2}+𝑖\frac{(𝑦-3)(𝑥+4)-𝑥𝑦}{(𝑥+4{)}^2+{𝑦}^2}\end{array}$

Now,

$\begin{array}{cc}& \arg \left(𝜔\right)={\tan }^{-1}\left(\frac{(𝑦-3)(𝑥+4)-𝑥𝑦)}{𝑥(𝑥+4)+𝑦(𝑦-3)}\right)=\frac{𝜋}{2}\\ & \Rightarrow \frac{(𝑦-3)(𝑥+4)-𝑥𝑦}{𝑥(𝑥+4)+𝑦(𝑦-3)}=\tan \frac{𝜋}{2}\\ & \Rightarrow \frac{4𝑦-3𝑥-12}{{𝑥}^2+{𝑦}^2+4𝑥-3𝑦}=\infty \end{array}$

So, clearly

${𝑥}^2+{𝑦}^2+4𝑥-3𝑦=0$

and $3𝑥-4𝑦<0$

We start with the given equations:

$\begin{array}{cc}\cos 𝛼+4\cos 𝛽+9\cos 𝛾& =0...\left(𝑖\right)\\ \sin 𝛼+4\sin 𝛽+9\sin 𝛾& =0...\left(𝑖𝑖\right)\end{array}$

Combine these equations using the complex form of the equations. Multiply both expressions:

$\cos 𝜃+𝑖\sin 𝜃={𝑒}^{𝑖𝜃}$

Thus, combining both equations, we have:

$\begin{array}{cc}& (\cos 𝛼+𝑖\sin 𝛼)+4(\cos 𝛽+𝑖\sin 𝛽)+9(\cos 𝛾+𝑖\sin 𝛾)=0\\ & {𝑒}^{𝑖𝛼}+4{𝑒}^{𝑖𝛽}+9{𝑒}^{𝑖𝛾}=0\\ & \Rightarrow 1+4{𝑒}^{𝑖(𝛽-𝛼)}=-9{𝑒}^{𝑖(𝛾-𝛼)}\end{array}$

Substitute using Euler's formula:

$\begin{array}{c}1+[4\cos (𝛽-𝛼)+4𝑖\sin (𝛽-𝛼\left)\right]=-[9\cos (𝛾-𝛼)+9𝑖\sin (𝛾-𝛼\left)\right]\end{array}$

Next, square both sides and compare the real parts:

$\begin{array}{cc}& [1+4\cos (𝛽-𝛼){]}^2-16{\sin }^2(𝛽-𝛼)\\ & =[9\cos (𝛾-𝛼){]}^2-[9\sin (𝛾-𝛼){]}^2\\ & =1+8\cos (𝛽-𝛼)+16[{\cos }^2(𝛽-𝛼)-{\sin }^2(𝛽-𝛼\left)\right]\\ & =81[{\cos }^2(𝛾-𝛼)-{\sin }^2(𝛾-𝛼\left)\right]\end{array}$

Use the identities for double angles:

$\begin{array}{cc}& \Rightarrow 81\cos 2(𝛾-𝛼)-16\cos 2(𝛽-𝛼)\\ & =1+8\cos (𝛽-𝛼)\end{array}$

Therefore, the expression $81\cos (2𝛾-2𝛼)-16\cos (2𝛽-2𝛼)$ simplifies to:

$1+8\cos (𝛽-𝛼)$

Given,

${𝑥}^2-3𝑎𝑥+{𝑎}^2-2𝑎-4=0$

where, $𝑎$ is a rational number

Then,

discriminant, $𝐷={𝐵}^2-4𝐴𝐶$

$\begin{array}{cc}& =(-3𝑎{)}^2-4\times 1\times ({𝑎}^2-2𝑎-4)\\ & =9{𝑎}^2-4{𝑎}^2+8𝑎+16\\ & =5{𝑎}^2+8𝑎+16\\ 𝐷& =\underset{+\mathrm{ve}}{\underset{}{5{𝑎}^2+16+8𝑎}}\end{array}$

So, $𝐷>0$

So, roots of equation will be different real number.

$\begin{array}{cc}& \text{ Given, }-1<\frac{2{𝑥}^2+𝑎𝑥+2}{{𝑥}^2+𝑥+1}<3\\ & \begin{array}{cc}& \text{ Now, }-1<\frac{2{𝑥}^2+𝑎𝑥+2}{{𝑥}^2+𝑥+1}\\ & \Rightarrow -{𝑥}^2-𝑥-1<2{𝑥}^2+𝑎𝑥+2\\ & \Rightarrow 0<3{𝑥}^2+(𝑎+1)𝑥+3\\ & \Rightarrow 𝐷<0\\ & \Rightarrow (𝑎+1{)}^2-4\times 3\times 3<0\\ & \Rightarrow (𝑎+1{)}^2-36<0\Rightarrow (𝑎+1{)}^2-6^2<0\\ & \Rightarrow (𝑎+1-6)(𝑎+1+6)<0\\ & \Rightarrow (𝑎-5)(𝑎+7)<0\\ & \text{ and }\frac{2{𝑥}^2+𝑎𝑥+2}{{𝑥}^2+𝑥+1}<3...\left(𝑖\right)\end{array}\end{array}$

$\begin{array}{cc}\Rightarrow & 2{𝑥}^2+𝑎𝑥+2<3{𝑥}^2+3𝑥+3\\ \Rightarrow & 0<{𝑥}^2+(3-𝑎)𝑥+1\Rightarrow 𝐷<0\\ \Rightarrow & (3-𝑎{)}^2-4\times 1\times 1<0\\ \Rightarrow & (3-𝑎{)}^2-2^2<0\\ \Rightarrow & (3-𝑎-2)(3-𝑎+2)<0\\ \Rightarrow & (-𝑎+1)(5-𝑎)<0\end{array}$

$\begin{array}{cc}& {𝑥}^2+𝑥-3)\\ & \begin{array}{c}3{𝑥}^5-4{𝑥}^4+5{𝑥}^3-3{𝑥}^2+6𝑥-8\\ (3{𝑥}^3-7{𝑥}^2+21𝑥-45\end{array}\end{array}$

Given,

${𝛼}_1,{𝛼}_2,{𝛼}_3,{𝛼}_4$ and ${𝛼}_5$ are roots of the equation

$\begin{array}{cc}& {𝑥}^5-5{𝑥}^4+9{𝑥}^3-9{𝑥}^2+5𝑥-1=0\\ \Rightarrow & {𝑥}^5-{𝑥}^4-4{𝑥}^4+4{𝑥}^3+5{𝑥}^3-5{𝑥}^2-4{𝑥}^2+4𝑥+𝑥-1=0\\ \Rightarrow & {𝑥}^4(𝑥-1)-4{𝑥}^3(𝑥-1)+5{𝑥}^2(𝑥-1)-4𝑥(𝑥-1)+1(𝑥-1)=0\\ \Rightarrow & (𝑥-1)[{𝑥}^4-4{𝑥}^3+5{𝑥}^2-4𝑥+1]=0\\ \Rightarrow & (𝑥-1)({𝑥}^2-3𝑥+1)({𝑥}^2-𝑥+1)=0\\ \Rightarrow & 𝑥=1,\frac{3+\sqrt{\surd 5}}{2},\frac{3-\sqrt{\surd 5}}{2},\frac{1+\sqrt{\surd 3}𝑖}{2}\end{array}$

$\begin{array}{cc}& \text{ and }\frac{1-\sqrt{\surd 3}𝑖}{2}\\ & {𝛼}_1=1\Rightarrow {𝛼}_2^1=1-\frac{1}{{𝛼}_2^1}=1\\ & {𝛼}_2=\frac{3+\sqrt{\surd 5}}{2}\Rightarrow {𝛼}_2^2=\frac{(3+\sqrt{\surd 5}{)}^2}{4}\\ & \Rightarrow \frac{1}{{𝛼}_2^2}=\frac{4}{(3+\sqrt{\surd 5}{)}^2}\\ & \Rightarrow \begin{array}{c}{𝛼}_3=\frac{3-\sqrt{\surd 5}}{2}\Rightarrow {𝛼}_2^3=\frac{(3-\sqrt{\surd 5}{)}^2}{4}\\ \Rightarrow \frac{1}{{𝛼}_2^3}=\frac{4}{(3-\sqrt{\surd 5}{)}^2}\end{array}\end{array}$

$\begin{array}{cc}{𝛼}_4& =\frac{1+\sqrt{\surd 3}𝑖}{2}\\ & \Rightarrow {𝛼}_2^4=\frac{(1+\sqrt{\surd 3}𝑖{)}^2}{4}\\ \Rightarrow & \frac{1}{{𝛼}_2^4}=\frac{4}{(1+\sqrt{\surd 3}𝑖{)}^2}\\ {𝛼}_5& =\frac{1-\sqrt{\surd 3}𝑖}{2}\Rightarrow {𝛼}_2^5=\frac{(1-\sqrt{\surd 3}𝑖{)}^2}{4}\\ & \Rightarrow \frac{1}{{𝛼}_2^5}=\frac{4}{(1-\sqrt{\surd 3}𝑖{)}^2}\end{array}$

Let there be $𝑛$ men participents, then the number of games that the men play between themselves is $2\times {}^{𝑛}{𝐶}_2$ and the number of games that the men played with the women is $2\times 2𝑛$

$\therefore 2\times {}^{𝑛}{𝐶}_2-2\times 2𝑛=66\left[𝑔𝑖𝑣𝑒𝑛\right]$

or ${𝑛}^2-5𝑛-66=0\Rightarrow 𝑛=11$

Hence, number of participents men and women $=11$ men +2 women $=13$

The number of ways arranging a men around a circular table $=8!$

We have, 9 places to sit of 5 women $={}^9{𝑃}_5$

$\therefore$ Number of ways $=8!\cdot {}^9{𝑃}_5$

Since the fruits are alike, you can choose any number of fruits from 1 up to 6. This gives $6$ possible selections (choosing 1, 2, 3, 4, 5, or 6 fruits).

For the vegetables, you can choose any number from 1 up to 7, which gives $7$ possible selections.

For the biscuits, you can choose any number from 1 up to 8, providing $8$ possible selections.

Since these choices are made independently (one choice does not affect the others), you multiply the number of ways in each category:

$\text{Total number of ways}=6\times 7\times 8.$

Calculating the product:

$6\times 7=42,$

$42\times 8=336.$

Thus, the number of ways of selecting at least one fruit, one vegetable, and one biscuit is $\overline{)336}$.

The correct answer, therefore, is Option B.

Given that coefficient of $𝑟$ th: $(𝑟+1)$ th : $(𝑟+2)$ th $=4:15:42$ in expansion of $(1+𝑥{)}^{𝑛}$.

$\begin{array}{cc}& \Rightarrow {}^{𝑛}{𝐶}_{𝑟-1}:{}^{𝑛}{𝐶}_{𝑟}:{}^{𝑛}{𝐶}_{𝑟+1}=4:15:42=𝐾\\ & \Rightarrow \frac{{}^{𝑛}{𝐶}_{𝑟-1}}{{}^{𝑛}{𝐶}_{𝑟}}=\frac{4𝐾}{15𝐾}\Rightarrow \frac{𝑟}{𝑛-𝑟+1}=\frac{4}{15}\cdots \left(𝑖\right)\\ & \text{ and }\frac{{}^{𝑛}{𝐶}_{𝑟}}{{}^{𝑛}{𝐶}_{𝑟+1}}=\frac{15𝐾}{42𝐾}\Rightarrow 𝑛-𝑟=\frac{14(𝑟+1)}{5}\dots \left(𝑖𝑖\right)\end{array}$

From Eqs. (i) and (ii), we get

$\begin{array}{cc}& \frac{5𝑟}{14𝑟+19}=\frac{4}{15}\\ \Rightarrow & 75𝑟=56𝑟+76\Rightarrow 19𝑟=76\\ \Rightarrow & 𝑟=4\\ \Rightarrow & 𝑛-𝑟=\frac{14\times (4+1)}{5}=14\text{ [from Eq. (i) }\end{array}$

Given that coefficients of $(2𝑟+6)$ th and $(𝑟-1)$ th terms in the expansion of $(1+𝑥{)}^{21}$ are equal.

$\begin{array}{cc}{𝑇}_{2𝑟+5+1}& ={}^{21}{𝐶}_{2𝑟+5}(1{)}^{21-2𝑟-5}{𝑥}^{2𝑟+5}\\ {𝑇}_{𝑟-2+1}& ={}^{21}{𝐶}_{𝑟-2}(𝑙{)}^{21-𝑟+2}{𝑥}^{𝑟-2}\\ \because {}^{21}{𝐶}_{2𝑟+5}& ={}^{21}{𝐶}_{𝑟-2}\\ \Rightarrow 2𝑟+5& =𝑟-2\\ 𝑟& =3\left(𝑛𝑜𝑡𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒\right)\end{array}$

So, $2𝑟+5=21-(𝑟-2)$

$\begin{array}{ccc}& & [\because {}^{𝑛}{𝐶}_{𝑟}={}^{𝑛}{𝐶}_{𝑛-𝑟}]\\ \Rightarrow & & 2𝑟+5=21-𝑟+2\\ \Rightarrow & & 3𝑟=18\\ \Rightarrow & & 𝑟=6\end{array}$

Given,

$\begin{array}{cc}& \frac{13𝑥+43}{2{𝑥}^2+17𝑥+30}=\frac{𝐴}{2𝑥+5}+\frac{𝐵}{𝑥+6}\\ & \Rightarrow \frac{13𝑥+43}{2{𝑥}^2+17𝑥+30}=\frac{𝐴(𝑥+6)+𝐵(2𝑥+5)}{(2𝑥+5)(𝑥+6)}\\ & \Rightarrow 13𝑥+43=𝑥(𝐴+2𝐵)+6𝐴+5𝐵\end{array}$

On compare both sides, we get

$\begin{array}{cc}& 𝐴+2𝐵=13\\ \text{ and }& 6𝐴+5𝐵=43\\ \Rightarrow & 6𝐴+12𝐵=78\\ & 6𝐴+5𝐵=43\\ & ---\\ & 7𝐵=35\\ & 𝐵=5\Rightarrow 𝐴=3\end{array}$

So, $𝐴+𝐵=3+5=8$