AP EAPCET 2021 20th August Evening Shift Question Paper with Solutions

Let $𝑓\left(𝑥\right)=𝑦$, then $𝑥={𝑓}^{-1}\left(𝑦\right)$, here

$𝑦=(𝑥+2{)}^2-2$

or

$\begin{array}{cc}& 𝑦+2=(𝑥+2{)}^2\\ & 𝑥+2=\sqrt{\surd 𝑦+2}\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑥=\sqrt{\surd 𝑦+2}-2\\ & \therefore {𝑓}^{-1}\left(𝑦\right)=\sqrt{\surd 𝑦+2}-2\\ & \therefore {𝑓}^{-1}\left(𝑥\right)=\sqrt{\surd 𝑥+2}-2\end{array}$

$𝑓\left(𝑥\right)=\left[𝑥\right],𝑔\left(𝑥\right)=\left|𝑥\right|$

Now, $𝑓(-\frac{5}{3})=[-\frac{5}{3}]$

$=[-1.667]=-2$

Then, $𝑔\left(𝑓(-\frac{5}{3})\right)=𝑔(-2)$

$(𝑔\circ 𝑓)(-\frac{5}{3})=|-2|=2$

$\therefore (𝑔\circ 𝑓)(-\frac{5}{3})=2$

Given, $𝑓\left(𝑥\right)=𝑎𝑥+𝑏$ and $𝑔\left(𝑥\right)=𝑐{𝑥}^3+𝑑$

$\begin{array}{cc}𝑓\left(𝑔\right(𝑥\left)\right)& =𝑓(𝑐{𝑥}^3+𝑑)\\ & =𝑎(𝑐{𝑥}^3+𝑑)+𝑏\\ (𝑓\circ 𝑔)\left(𝑥\right)& =𝑎𝑐{𝑥}^3+𝑎𝑑+𝑏.....\left(\text{i}\right)\end{array}$

Let $(𝑓\circ 𝑔)\left(𝑥\right)=𝑦$, then

$𝑥=(𝑓\circ 𝑔{)}^{-1}𝑦$

From Eq. (i), we get

$𝑦=𝑎𝑐{𝑥}^3+𝑎𝑑+𝑏$

i.e.

$\begin{array}{cc}𝑎𝑐{𝑥}^3& =𝑦-𝑎𝑑-𝑏\\ {𝑥}^3& =(𝑦-𝑎𝑑-𝑏)/𝑎𝑐\\ 𝑥& ={\left[\frac{(𝑦-𝑎𝑑-𝑏)}{𝑎𝑐}\right]}^{1/3}\end{array}$

$\begin{array}{cc}\therefore & (𝑓\circ 𝑔{)}^{-1}(𝑦)={\left[\frac{(𝑦-𝑎𝑑-𝑏)}{𝑎𝑐}\right]}^{1/3}\\ \text{ or }& (𝑓\circ 𝑔{)}^{-1}(𝑥)={\left[\frac{(𝑥-𝑎𝑑-𝑏)}{𝑎𝑐}\right]}^{1/3}\end{array}$

Given, ${𝑎}_0=1,{𝑎}_{𝑛+1}=3{𝑛}^2+𝑛+{𝑎}_{𝑛}$ .... (i)

From Eq.(i), we get

$\begin{array}{cc}& {𝑎}_1=3(0{)}^2+0+{𝑎}_0=1\\ \Rightarrow & {𝑎}_1=1\\ & {𝑎}_2=3(1{)}^2+1+{𝑎}_1=3+1+1=5\\ \Rightarrow & {𝑎}_2=5\end{array}$

$\begin{array}{cc}& {𝑎}_3=3(2{)}^2+2+{𝑎}_2=12+2+5=19\\ & \Rightarrow {𝑎}_3=19\end{array}$

Now, check the option for $𝑛=3$ which expression satisfy by ${𝑎}_3=19$

Option (a), ${𝑎}_{𝑛}={𝑛}^3+{𝑛}^2+1$

Put $𝑛=3$, then

$\begin{array}{cc}{𝑎}_3& =(3{)}^3+(3{)}^2+1\\ & =27+9+1=37\\ \because {𝑎}_3& \ne 37\\ \therefore {𝑎}_{𝑛}& \ne {𝑛}^3+{𝑛}^2+1\\ & \text{ Option (b), }{𝑎}_{𝑛}={𝑛}^3-{𝑛}^2+1\end{array}$

Option (b), ${𝑎}_{𝑛}={𝑛}^3-{𝑛}^2+1$

Put $𝑛=3$, then

$\begin{array}{cc}{𝑎}_3& =(3{)}^3-(3{)}^2+1\\ & =27-9+1=19\\ \therefore & {𝑎}_{𝑛}={𝑛}^3-{𝑛}^2+1\end{array}$

$\det 𝐴=\mid \begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}\mid =𝑘$ (given)

$\begin{array}{cc}\text{ Then, }\det 𝐵& =\mid \begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2+2{𝑎}_1& {𝑏}_2+2{𝑏}_1& {𝑐}_2+2{𝑐}_1\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}\mid \\ & =\mid \begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}\mid +\mid \begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ 2{𝑎}_1& 2{𝑏}_1& 2{𝑐}_1\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}\mid \\ & =𝑘+0\\ \Rightarrow \det 𝐵& =𝑘\end{array}$

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{cccc}\sqrt{\surd 2020}& \sqrt{\surd 2022}& \sqrt{\surd 2022}& \sqrt{\surd 2023}\\ \sqrt{\surd 4040}& \sqrt{\surd 4042}& \sqrt{\surd 4044}& \sqrt{\surd 4046}\\ \sqrt{\surd 6060}& \sqrt{\surd 6063}& \sqrt{\surd 6066}& \sqrt{\surd 6069}\\ \sqrt{\surd 8080}& \sqrt{\surd 8084}& \sqrt{\surd 8088}& \sqrt{\surd 8092}\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{cccc}\sqrt{\surd 2020}& \sqrt{\surd 2021}& \sqrt{\surd 2022}& \sqrt{\surd 2023}\\ \sqrt{\surd 2}\cdot \sqrt{\surd 2020}& \sqrt{\surd 2}\sqrt{\surd 2021}& \sqrt{\surd 4044}& \sqrt{\surd 4046}\\ \sqrt{\surd 3}\sqrt{\surd 2020}& \sqrt{\surd 3}\sqrt{\surd 2021}& \sqrt{\surd 6066}& \sqrt{\surd 6069}\\ \sqrt{\surd 4}\sqrt{\surd 2020}& \sqrt{\surd 4}\sqrt{\surd 2021}& \sqrt{\surd 8088}& \sqrt{\surd 8092}\end{array}⎤⎦⎥\\ {𝑅}_2& \to {𝑅}_2-\sqrt{\surd 2}{𝑅}_1,{𝑅}_3\to {𝑅}_3-\sqrt{\surd 3}{𝑅}_1,{𝑅}_4\to {𝑅}_4-\sqrt{\surd 4}{𝑅}_1\\ 𝐴& \sim ⎡⎣⎢\begin{array}{cccc}\sqrt{\surd 2020}& \sqrt{\surd 2021}& \sqrt{\surd 2022}& \sqrt{\surd 2023}\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}⎤⎦⎥\end{array}$

$\because$ A has one non-zero row

$\therefore$ Rank (A) = 1

Option (a) is correct.

The given equation is:

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

We can rewrite the third column as (1) + (the cube term), so this is the same as:

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

Let’s call the first determinant as $𝐴$ and the second as $𝐵$.

Step 1: Find the value of $\left|𝐴\right|$

Let: $𝐴=\mid \begin{array}{ccc}𝑥& {𝑥}^2& 1\\ 𝑦& {𝑦}^2& 1\\ 𝑧& {𝑧}^2& 1\end{array}\mid$

Subtract first row from second and third, ${𝑅}_2\to {𝑅}_2-{𝑅}_1,{𝑅}_3\to {𝑅}_3-{𝑅}_1$ so, $\mid \begin{array}{ccc}𝑥& {𝑥}^2& 1\\ 𝑦-𝑥& {𝑦}^2-{𝑥}^2& 0\\ 𝑧-𝑥& {𝑧}^2-{𝑥}^2& 0\end{array}\mid$

Now, since last column for ${𝑅}_2$ and ${𝑅}_3$ is 0, expand along third column:

$\left|𝐴\right|=(𝑦-𝑥)({𝑧}^2-{𝑥}^2)-(𝑧-𝑥)({𝑦}^2-{𝑥}^2)$

$=(𝑦-𝑥)(𝑧-𝑥)(𝑧+𝑥)-(𝑧-𝑥)(𝑦-𝑥)(𝑦+𝑥)$

$=(𝑦-𝑥)(𝑧-𝑥)(𝑧+𝑥-𝑦-𝑥)$

$=(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)$

Step 2: Find the value of $\left|𝐵\right|$

$𝐵=\mid \begin{array}{ccc}𝑥& {𝑥}^2& {𝑥}^3\\ 𝑦& {𝑦}^2& {𝑦}^3\\ 𝑧& {𝑧}^2& {𝑧}^3\end{array}\mid$

Factor $𝑥,𝑦,𝑧$ from each row: $=𝑥𝑦𝑧\mid \begin{array}{ccc}1& 𝑥& {𝑥}^2\\ 1& 𝑦& {𝑦}^2\\ 1& 𝑧& {𝑧}^2\end{array}\mid$

Now, do the same row operation as before: ${𝑅}_2\to {𝑅}_2-{𝑅}_1,{𝑅}_3\to {𝑅}_3-{𝑅}_1$

So determinant becomes: $𝑥𝑦𝑧\mid \begin{array}{ccc}1& 𝑥& {𝑥}^2\\ 0& 𝑦-𝑥& {𝑦}^2-{𝑥}^2\\ 0& 𝑧-𝑥& {𝑧}^2-{𝑥}^2\end{array}\mid$

Expand this along the first column:

$\left|𝐵\right|=𝑥𝑦𝑧\left\{\right(𝑦-𝑥\left)\right({𝑧}^2-{𝑥}^2)-(𝑧-𝑥\left)\right({𝑦}^2-{𝑥}^2\left)\right\}$

$=𝑥𝑦𝑧(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)$

Step 3: Use the original equation

We have: $\left|𝐴\right|+\left|𝐵\right|=0$ So, $(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)+𝑥𝑦𝑧(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)=0$

Factor out $(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)$: $(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)(1+𝑥𝑦𝑧)=0$

Because $𝑥,𝑦,𝑧$ are all different, $(𝑦-𝑥)(𝑧-𝑥)(𝑧-𝑦)\ne 0$. So, $1+𝑥𝑦𝑧=0$

Therefore, $𝑥𝑦𝑧=-1$

Let $𝑛=2$ (say) and $𝐴=\left[\begin{array}{cc}𝑎& 𝑏\\ 0& 𝑐\end{array}\right]$, Then $\mathrm{adj}𝐴=\left[\begin{array}{cc}𝑐& -𝑏\\ 0& 𝑎\end{array}\right]$ which is a upper triangular matrix. Then, for any $𝑛\times 𝑛$ matrix $𝐴$, $\mathrm{adj}𝐴$ is also upper triangular matrix.

$\text{begin{array}{left} left|begin{array}{aligned} a \& b \& c b \& c \& a c \& a \& b end{array}right|=aleft(b c-a^2right)-bleft(b^2-a cright)+cleft(a b-c^2right) Rightarrow a b c-a^3-b^3+a b c+a b c-c^3=0 Rightarrow 3 a b c-a^3-b^3-c^3=0 Rightarrow a^3+b^3+c^3-3 a b c=0 Rightarrow (a+b+c)=0 end{array}}$

Since, $𝑎=\left|{𝑍}_1\right|,𝑏=\left|{𝑍}_2\right|,𝑐=\left|{𝑍}_3\right|$

$\therefore \left|{𝑍}_1\right|+\left|{𝑍}_2\right|+\left|{𝑍}_3\right|=0$

${|{𝑧}_1+{𝑧}_2|}^2={\left|{𝑧}_1\right|}^2+{\left|{𝑧}_2\right|}^2$ $$ (given) ..... (i)

Let ${𝑧}_1=𝑥+𝑖𝑦,{𝑧}_2=𝑎+𝑖𝑏$.

Then

$\begin{array}{cc}& \frac{{𝑧}_1}{{𝑧}_2}=\frac{𝑥+𝑖𝑦}{𝑎+𝑖𝑏}=\frac{𝑥+𝑖𝑦}{𝑎+𝑖𝑏}\times \frac{𝑎-𝑖𝑏}{𝑎-𝑖𝑏}\\ & \frac{{𝑧}_1}{{𝑧}_2}=\frac{(𝑎𝑥+𝑦𝑏)+𝑖(𝑎𝑦-𝑥𝑏)}{{𝑎}^2+{𝑏}^2}\text{...... (ii)}\end{array}$

From Eq. (i), we get

$\begin{array}{cc}& \text{ LHS }={|{𝑧}_1+{𝑧}_2|}^2=|𝑥+𝑎+𝑖(𝑦+𝑏){|}^2\\ & =(𝑥+𝑎{)}^2+(𝑦+𝑏{)}^2\\ & ={𝑥}^2+{𝑎}^2+2𝑎𝑥+{𝑦}^2+{𝑏}^2+2𝑏𝑦\\ & \text{ RHS }=|𝑥+𝑖𝑦{|}^2+|𝑎+𝑖𝑏{|}^2\\ & ={𝑥}^2+{𝑦}^2+{𝑎}^2+{𝑏}^2\\ & \because \text{ LHS }=\text{ RHS }\\ & \Rightarrow {𝑥}^2+{𝑦}^2+{𝑎}^2+{𝑏}^2+2(𝑎𝑥+𝑏𝑦)\\ & ={𝑥}^2+{𝑦}^2+{𝑎}^2+{𝑏}^2\\ & \Rightarrow 2(𝑎𝑥+𝑏𝑦)=0\Rightarrow 𝑎𝑥+𝑏𝑦=0\end{array}$

Put $𝑎𝑥+𝑏𝑦=0$ in Eq. (ii),

$\frac{{𝑧}_1}{{𝑧}_2}=\frac{𝑖(𝑎𝑦-𝑥𝑏)}{{𝑎}^2+{𝑏}^2}$, which is purely imaginary.

$\begin{array}{cc}& \text{ LHS }=\frac{3-4𝑖𝑥}{3+4𝑖𝑥}\times \frac{3-4𝑖𝑥}{3-4𝑖𝑥}\\ & =\frac{9+16{𝑖}^2{𝑥}^2-24𝑖𝑥}{9-\left(16{𝑖}^2{𝑥}^2\right)}=\frac{9-16{𝑥}^2-24𝑖𝑥}{9+16{𝑥}^2}\\ & =\frac{9-16{𝑥}^2}{9+16{𝑥}^2}-\frac{24𝑖𝑥}{9+16{𝑥}^2}=𝛼-𝑖𝛽=\text{ RHS }\\ & \therefore 𝛼=\frac{9-16{𝑥}^2}{9+16{𝑥}^2}\\ & 𝛽=\frac{24𝑥}{9+16{𝑥}^2}\\ & \Rightarrow {𝑎}^2+{𝛽}^2=\frac{{(9-16{𝑥}^2)}^2+(24𝑥{)}^2}{{(9+16{𝑥}^2)}^2}\\ & =\frac{{(9+16{𝑥}^2)}^2}{{(9+16{𝑥}^2)}^2}=1\\ & \therefore {𝛼}^2+{𝛽}^2=1\\ & \end{array}$

$\begin{array}{cc}(1-𝑖\sqrt{\surd 3}{)}^9& =2^9{(\frac{1}{2}-𝑖\frac{\sqrt{\surd 3}}{2})}^9\\ & =2^9{(\cos \frac{𝜋}{3}-𝑖\sin \frac{𝜋}{3})}^9=2^9{\left({𝑒}^{-𝑖𝜋/3}\right)}^9\\ & =2^9\left({𝑒}^{-𝑖3𝜋}\right)=2^9(\cos 3𝜋-𝑖\sin 3𝜋)\\ & =2^9\left(\right(-1{)}^3-0)=-2^9\end{array}$

${\left(\sqrt{\surd 2}\frac{(\sqrt{\surd 3}-1)}{2.2}+𝑖\sqrt{\surd 2}\frac{(\sqrt{\surd 3}+1)}{2.2}\right)}^{2020}$

$={\left(\frac{\left(\frac{\sqrt{\surd 3}}{2}-\frac{1}{2}\right)}{\sqrt{\surd 2}}+𝑖\frac{\left(\frac{\sqrt{\surd 3}}{2}+\frac{1}{2}\right)}{\sqrt{\surd 2}}\right)}^{2020}$

$={\left(\frac{\frac{1}{\sqrt{\surd 2}}\left(\frac{\sqrt{\surd 3}}{2}-\frac{1}{2}\right)}{1}+\frac{𝑖\frac{1}{\sqrt{\surd 2}}\left(\frac{\sqrt{\surd 3}}{2}+\frac{1}{2}\right)}{1}\right)}^{2020}$

$={\left(\left(\frac{1}{\sqrt{\surd 2}}.\frac{\sqrt{\surd 3}}{2}-\frac{1}{\sqrt{\surd 2}}.\frac{1}{2}\right)+𝑖\left(\frac{1}{\sqrt{\surd 2}}.\frac{\sqrt{\surd 3}}{2}+\left(\frac{1}{\sqrt{\surd 2}}.\frac{1}{2}\right)\right)\right)}^{2020}$

$=(\cos {45}^{\circ }\cos {30}^{\circ }-\sin {45}^{\circ }\sin {30}^{\circ })+𝑖(\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ }){)}^{2020}$

$=(\cos {75}^{\circ }+𝑖\sin {75}^{\circ }{)}^{2020}$

$={\left(\cos \frac{5𝜋}{12}+𝑖\sin \frac{5𝜋}{12}\right)}^{2020}$

$=\left(\cos \frac{5𝜋}{12}\times 2020+𝑖\sin \frac{5𝜋}{12}\times 2020\right)$ [De-moivre]

$=\left(\cos \frac{2525𝜋}{3}+𝑖\sin \frac{2525𝜋}{3}\right)$

$=(\cos (2525\times 60{)}^{\circ }+𝑖\sin (2525\times 60{)}^{\circ }$

$=(\cos (𝑛𝜋-60{)}^{\circ }+𝑖\sin (𝑛𝜋-60{)}^{\circ })$, where n = 842

$=((-1{)}^{𝑛}\cos {60}^{\circ }+𝑖(-\sin {60}^{\circ }\left)\right)$

$=(\cos {60}^{\circ }-𝑖\sin {60}^{\circ })$ [$\because$ n = even]

$=\frac{1}{2}-𝑖\frac{\sqrt{\surd 3}}{2}$

$𝑓(10-𝑥)=3{𝑥}^2+4𝑥-5$ and $𝑓\left(𝑥\right)=𝑝{𝑥}^2+𝑞𝑥+𝑟$

Let $10-𝑥=𝑦$, then $𝑥=10-𝑦$

$\begin{array}{cc}𝑓\left(𝑦\right)& =3(10-𝑦{)}^2+4(10-𝑦)-5\\ & =3(100+{𝑦}^2-20𝑦)+40-4𝑦-5\\ & =340+3{𝑦}^2-60𝑦-4𝑦-5\\ & =3{𝑦}^2-64𝑦+335\end{array}$

Replace $𝑦$ by $𝑥\Rightarrow 𝑓\left(𝑥\right)=3{𝑥}^2-64𝑥+335$

compare with $𝑓\left(𝑥\right)=𝑝{𝑥}^2+𝑞𝑥+𝑟$, we obtain

$\begin{array}{cc}& 𝑝=3,𝑞=-64,𝑟=335\\ \Rightarrow & 𝑝+𝑞+𝑟=274\end{array}$

Given, equations are

$\begin{array}{cc}& {𝑥}^2+𝑎𝑥+𝑏=0.....\left(\text{i}\right)\\ & {𝑥}^2+𝑏𝑥+𝑎=0.....\left(\text{ii}\right)\end{array}$

Let $𝛼$ be common root of Eqs. (i) and (ii), then

$\begin{array}{cc}& {𝛼}^2+𝑎𝛼+𝑏=0.....\left(\text{iii}\right)\\ & {𝛼}^2+𝑏𝛼+𝑎=0.....\left(\text{iv}\right)\end{array}$

Compare Eqs. (iii) and (iv), we get

$\begin{array}{cc}& {𝛼}^2+𝑎𝛼+𝑏={𝛼}^2+𝑏𝛼+𝑎\\ & \Rightarrow 𝛼(𝑎-𝑏)=(𝑎-𝑏)\\ & \Rightarrow 𝛼=1\\ & \text{ Common root }=1\\ & \text{ From Eq. (iii), we get }\\ & 1+𝑎+𝑏=0\\ & \Rightarrow 𝑎+𝑏=-1\end{array}$

Given equation is

$9{𝑥}^3+112{𝑥}^2-120𝑥+𝑎=0$

Let $𝛼,𝛽,𝛾$ be roots of Eq.(i), then

Product of roots $=𝛼𝛽𝛾=-\frac{𝑎}{9}$

Given, $-\frac{𝑎}{9}=12$

$\begin{array}{cc}& \Rightarrow 𝑎=-12\times 9=-108\\ & \therefore 𝑎=-108\end{array}$

Roots of cubic equations are 1 and $2+\sqrt{\surd 5}$.

Since, $2+\sqrt{\surd 5}$ is root, then its conjugate also become roots, i.e. $2-\sqrt{\surd 5}$ is root of cubic equation.

$\therefore \text{ Roots }=1,2+\sqrt{\surd 5},2-\sqrt{\surd 5}\text{ i.e. }𝛼,𝛽,𝛾$

Let cubic equation be

${𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+𝑑=0$

Then,

$\begin{array}{cc}& \text{ Sum of Roots }=-𝑏=1+2+\sqrt{\surd 5}+2-\sqrt{\surd 5}=5\\ & \Rightarrow 𝑏=-5\\ & \text{ Product of Roots }=-𝑑=\left(1\right)(2+\sqrt{\surd 5})(2-\sqrt{\surd 5})\\ & =4-5=-1\\ & \Rightarrow 𝑑=1\\ & \therefore 𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑐\\ & \text{ (l) }(2+\sqrt{\surd 5})+(2+\sqrt{\surd 5})(2-\sqrt{\surd 5})+(2-\sqrt{\surd 5})\left(l\right)=𝑐\\ & 2+\sqrt{\surd 5}+2-\sqrt{\surd 5}+4-5=𝑐\\ & \Rightarrow \\ & 3=𝑐\end{array}$

$\therefore$ Cubic equation is

${𝑥}^3-5{𝑥}^2+3𝑥+1=0$

Here, n = 11 and subset have atmost 5 elements

$\therefore$ Total subset = (Subset with 0 element) + (Subset with 1 element) + (Subset with 2 element) + (Subset with 3 element) + (Subset with 4 element) + (Subset with 5 element)

$\begin{array}{cc}& =({}^{11}{𝐶}_0+{}^{11}{𝐶}_1)+({}^{11}{𝐶}_2+{}^{11}{𝐶}_3)+({}^{11}{𝐶}_4+{}^{11}{𝐶}_5)\\ & ={}^{12}{𝐶}_1+{}^{12}{𝐶}_3+{}^{12}{𝐶}_5[\because {}^{𝑛}{𝐶}_{𝑟}+{}^{𝑛}{𝐶}_{𝑟-1}={}^{𝑛+1}{𝐶}_{𝑟}]\end{array}$

$\frac{2{𝑥}^4-{𝑥}^3+3{𝑥}^2-𝑥+4}{{𝑥}^2-3𝑥+2}$

$\because$ Degree of numerator is greater than the degree of denominator.

$\therefore$ First we divide,

$\begin{array}{cc}\text{ or }& =\frac{2{𝑥}^4-{𝑥}^3+3{𝑥}^2-𝑥+4}{{𝑥}^2-3𝑥+2}\\ & =(2{𝑥}^2+5𝑥+14)+\frac{31𝑥-24}{(𝑥-2)(𝑥-1)}\text{.....(i)}\end{array}$

Let $\frac{31𝑥-24}{(𝑥-2)(𝑥-1)}=\frac{𝐴}{𝑥-1}+\frac{𝐵}{𝑥-2}$

$\Rightarrow 31-24=-𝐴(𝑥-2)+𝐵(𝑥-1)$ ..... (ii)

Put $𝑥=1$

$\Rightarrow 31-24=-𝐴\Rightarrow 𝐴=-7$

Put $𝑥=2$ in Eq. (ii)

$\begin{array}{cc}& 31\times 2-24=0+𝐵\Rightarrow 𝐵=38\\ & \therefore \text{ From Eq. (i) }𝑓\left(𝑥\right)=2{𝑥}^2+5𝑥+14\\ & \text{ and }𝐴+𝐵=-7+38=31\end{array}$

$\tan 𝜃=\frac{-5}{12}$

Then, $|\tan 𝜃|=5/12$

and $-\tan 𝜃=-\left(\frac{-5}{12}\right)=\frac{5}{12}$

$\therefore$ $|\tan 𝜃|=-\tan 𝜃$

$\begin{array}{cc}& \cos \frac{𝜋}{4}\cos \frac{𝜋}{8}\cos \frac{𝜋}{16}\cos \frac{𝜋}{32}\\ & =\cos \frac{𝜋}{2^2}\cdot \cos \frac{𝜋}{2^3}\cdot \cos \frac{𝜋}{2^4}\cdot \cos \frac{𝜋}{2^5}\\ & =\frac{1}{2\sin \left(\frac{𝜋}{2^5}\right)}\\ & [2\sin \left(\frac{𝜋}{2^5}\right)\cdot \cos \left(\frac{𝜋}{2^5}\right)\cdot \cos \frac{𝜋}{2^2}\cdot \cos \frac{𝜋}{2^3}\cdot \cos \frac{𝜋}{2^4}]\\ & =\frac{1}{2}\mathrm{cosec}\left(\frac{𝜋}{2^5}\right)[\sin \frac{𝜋}{2^4}\cdot \cos \frac{𝜋}{2^4}\cdot \cos \frac{𝜋}{2^2}\cdot \cos \frac{𝜋}{2^3}]\\ & =\frac{1}{4}\mathrm{cosec}\left(\frac{𝜋}{2^5}\right)[\sin \frac{𝜋}{2^3}\cdot \cos \frac{𝜋}{2^3}\cdot \cos \frac{𝜋}{2^2}]\\ & =\frac{1}{8}\mathrm{cosec}\left(\frac{𝜋}{2^5}\right)[\sin \frac{𝜋}{2^2}\cdot \cos \frac{𝜋}{2^2}]\\ & =\frac{1}{16}\mathrm{cosec}\frac{𝜋}{2^5}\sin \frac{𝜋}{2}\\ & =2^{-4}\mathrm{cosec}\frac{𝜋}{2^5}[\because \sin \frac{𝜋}{2}=1]\\ & \therefore 𝑚=-4\text{ and }𝑛=2^5\text{, then }\\ & 𝑚+𝑛=-4+32=28\\ & \end{array}$