AP EAPCET 2023 15th May Evening Shift Question Paper with Solutions

Given, $𝑓\left(𝑎\right)=\log \mid \frac{1-𝑎}{1+𝑎}\mid$ for

$𝑖\ne \{-1,1\}$

Then, $𝑓\left(\frac{2𝑎}{1+{𝑎}^2}\right)=\log \mid \frac{1-\frac{2𝑎}{1+{𝑎}^2}}{1+\frac{2𝑎}{1+{𝑎}^2}}\mid$

$=\log \mid \frac{{𝑎}^2+1-2𝑎}{{𝑎}^2+1+2𝑎}\mid =\log {\left(\frac{𝑎-1}{𝑎+1}\right)}^2$

For, $𝑓\left(\frac{2𝑎}{1+{𝑎}^2}\right)>0$ or, $\log {\left(\frac{𝑎-1}{𝑎+1}\right)}^2>0$

$={\left(\frac{𝑎-1}{𝑎+1}\right)}^2>1\Rightarrow (𝑎-1{)}^2>(𝑎+1{)}^2$

$\Rightarrow -2𝑎>2𝑎\Rightarrow 4𝑎<0\Rightarrow 𝑎<0$

$\because 𝛼\ne \{-1,1\}$

∴ Set of values of all ' $𝑎$ ', for which

$\int \left(\frac{2𝛼}{1+{𝑎}^2}\right)>0$ is $𝑎\in (-\infty ,0)-\{-1\}$

$𝑓\left(𝑥\right)=\frac{𝑎𝑥+\sqrt{\surd {𝑎}^2-{𝑥}^2}}{𝑏𝑥}=\frac{𝑎}{𝑏}+\frac{\sqrt{\surd {𝑎}^2-{𝑥}^2}}{𝑏𝑥}$, $𝑥\in [-𝑎,𝑎]$ domain and $𝑥\ne 𝑏$

$\begin{array}{cc}& {𝑓}^{\prime }\left(𝑥\right)=0+\frac{1}{𝑏}⎡⎣⎢\frac{𝑥\left(\frac{-𝑥}{\sqrt{\surd {𝑎}^2-{𝑥}^2}}\right)-\sqrt{\surd {𝑎}^2-{𝑥}^2}}{{𝑥}^2}⎤⎦⎥\\ & {𝑓}^{\prime }\left(𝑥\right)=\frac{-{𝑥}^2-({𝑎}^2-{𝑥}^2)}{𝑏{𝑥}^2\sqrt{\surd {𝑎}^2-{𝑥}^2}}=\frac{-{𝑎}^2}{𝑏{𝑥}^2\sqrt{\surd {𝑎}^2-{𝑥}^2}}\\ & {𝑓}^{\prime }\left(𝑥\right)<0,\forall 𝑥\in (-𝑎,𝑎)\end{array}$

[Assuming $𝑏$ is +ve ]

$𝑓$ is decreasing on $(-𝑎,𝑎)$

and $𝑓\left(𝑎\right)=𝑓(-𝑎)=\frac{𝑎}{𝑏}$

and $𝑥\to 0^{+},𝑓\left(𝑥\right)\to \infty$ and $𝑥\to 0^{-}$

$𝑓\left(𝑥\right)\to -\infty \Rightarrow$ Range $=𝑅=$ co-domain

$\therefore 𝑓$ is not one-one but $𝑓$ is onto.

The given matrix is $⎡⎣⎢\begin{array}{ccc}1& 2& 3\\ 4& -1& 7\\ 2& 4& 6\end{array}⎤⎦⎥$

Co-factors of $3=+1(16+2)=18=𝑎$

Co-factors of $7=-1(4-4)=0=𝑏$

and co-factors of $6=+1(-1-8)=-9=𝑐$

$\begin{array}{cc}& \text{ Then, }\left[\begin{array}{ccc}𝑎& 𝑏& 𝑐\end{array}\right]⎡⎣⎢\begin{array}{c}1\\ 4\\ 2\end{array}⎤⎦⎥+\left[\begin{array}{ccc}𝑎& 𝑏& 𝑐\end{array}\right]⎡⎣⎢\begin{array}{c}3\\ 7\\ 6\end{array}⎤⎦⎥\\ & =\left[\begin{array}{ccc}18& 0& -9\end{array}\right]⎡⎣⎢\begin{array}{c}1\\ 4\\ 2\end{array}⎤⎦⎥+\left[\begin{array}{ccc}18& 0& -9\end{array}\right]⎡⎣⎢\begin{array}{c}3\\ 7\\ 6\end{array}⎤⎦⎥\\ & =[18+0-18]+[54+0-54]\\ & =\left[0\right]+\left[0\right]=0\end{array}$

Given, $𝐵=⎡⎣⎢\begin{array}{ccc}2& 6& 4\\ 1& 0& 1\\ -1& 1& -1\end{array}⎤⎦⎥$,

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

and $𝐵𝐴𝐶=𝐼$

$\because \left|𝐵\right|\ne 0\Rightarrow {𝐵}^{-1}$ exist.

So, $𝐵𝐴𝐶=𝐼\Rightarrow 𝐴𝐶={𝐵}^{-1}$

$\Rightarrow 𝐶={𝐴}^{-1}{𝐵}^{-1}\Rightarrow 𝐶𝐵={𝐴}^{-1}$

Now, $𝐶𝐵=⎡⎣⎢\begin{array}{ccc}-1& 0& 1\\ 1& 1& 3\\ 2& 0& 2\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}2& 6& 4\\ 1& 0& 1\\ -1& 1& -1\end{array}⎤⎦⎥$

$\begin{array}{cc}& =⎡⎣⎢\begin{array}{ccc}-2+0-1& -6+0+1& -4+0-1\\ 2+1-3& 6+0+3& 4+1-3\\ 4+0-2& 12+0+2& 8+0-2\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}-3& -5& -5\\ 0& 9& 2\\ 2& 14& 6\end{array}⎤⎦⎥\\ & \therefore {𝐴}^{-1}=𝐶𝐵=⎡⎣⎢\begin{array}{ccc}-3& -5& -5\\ 0& 9& 2\\ 2& 14& 6\end{array}⎤⎦⎥\end{array}$

We know that,

$𝐴\mathrm{adj}𝐴=\left|𝐴\right|{𝐼}_{𝑛}$

$\Rightarrow \mid 𝐴\mathrm{adj}𝐴]|=\left|\right|𝐴\left|{𝐼}_{𝑛}\right|$

$\Rightarrow \left|𝐴\right||\mathrm{adj}𝐴|=|𝐴{|}^{𝑛}\left|{𝐼}_{𝑛}\right|$

$[\because |𝐴𝐵|=|𝐴\left|\right|𝐵\left|\right|𝐾𝐴|={𝐾}^{𝑛}|𝐴\left|\right]$

$\therefore \left|𝐴\right||\mathrm{adj}𝐴|=|𝐴{|}^{𝑛}\left|{𝐼}_{𝑛}\right|$

On dividing by $\left|𝐴\right|$, we get

$|\mathrm{adj}𝐴|=|𝐴{|}^{𝑛-1}(\because \left|{𝐼}_{𝑛}\right|=1$ and since $𝐴$ is non-singular matrix i.e. $\left|𝐴\right|\ne 0$ )

So, if $𝐴$ is a square matrix of order 3, then $|\mathrm{adj}𝐴|=|𝐴{|}^2$

Given that $𝑆=\{\frac{𝑧\in 𝐶}{|𝑧-1+𝑖|}=1\}$

$\begin{array}{cc}& \Rightarrow |𝑧-1+𝑖|=1\\ & \Rightarrow |𝑧-(1-𝑖\left)\right|=1\end{array}$

It is of the form $|𝑧-{𝑧}_0|=𝑟$ and so it represents a circle, whose centre and radius are $(1,-1)$ and 1 respectively.

Given, $\mid 𝑧-\frac{2}{𝑧}\mid =2$

By triangle in equality, we know that

$\begin{array}{cc}& \mid 𝑧-\frac{2}{𝑧}\mid \ge \left|𝑧\right|-\mid \frac{2}{𝑧}\mid \\ & 2\ge \left|𝑧\right|-\mid \frac{2}{𝑧}\mid \end{array}$

Let $\left|𝑧\right|=𝑡$

$\Rightarrow 𝑡-\frac{2}{𝑡}\le 2\Rightarrow {𝑡}^2-2𝑡-2\le 0$

The roots of ${𝑡}^2-2𝑡-2=0$ are

$𝑡=\frac{2\pm \sqrt{\surd 4+8}}{2}=1\pm \sqrt{\surd 3}$

Therefore for ${𝑡}^2-2𝑡-2\le 0$, we get

$\begin{array}{cc}& 1-\sqrt{\surd 3}\le 𝑡\le 1+\sqrt{\surd 3}\\ & 1-\sqrt{\surd 3}\le \left|𝑧\right|\le 1+\sqrt{\surd 3}\end{array}$

But we know that $\left|𝑧\right|\ge 0$ and since

$\begin{array}{cc}& 1-\sqrt{\surd 3}<0\\ & \therefore 0\le \left|𝑧\right|\le 1+\sqrt{\surd 3}\end{array}$

Hence, greatest value of $\left|𝑧\right|=1+\sqrt{\surd 3}$

15 th roots of -1 are roots of the equation ${𝑥}^{15}=-1$

or    $𝑥=(-1{)}^{\frac{1}{15}}$

$\therefore 𝑥=(\cos (2𝑘+1)𝜋+𝑖\sin (2𝑘+1)𝜋{)}^{1/15}$

where, $𝑘=0,1,2,\dots ,14$.

By De-Moivre's Theorem,

$𝑥=\cos \left(\frac{2𝑘+1}{15}\right)𝜋+𝑖\sin \left(\frac{2𝑘+1}{15}\right)𝜋$

Hence, from the given option only (c) is $\frac{13𝜋}{15}$ is correct.

Given expression is $(1+𝑖\sqrt{\surd 3}{)}^{\frac{3}{4}}$.

Let $1+𝑖\sqrt{\surd 3}=𝑧$

$\Rightarrow \left|𝑧\right|=\sqrt{\surd 1^2+(\sqrt{\surd 3}{)}^2}=2$

and $\tan 𝜃=\frac{\sqrt{\surd 3}}{1}\Rightarrow 𝜃=\frac{𝜋}{3}$

$\begin{array}{cc}\text{ and }& (1+𝑖\sqrt{\surd 3}{)}^{\frac{3}{4}}={𝑧}^{\frac{3}{4}}={\left(2{𝑒}^{\frac{𝜋𝑖}{3}}\right)}^{\frac{3}{4}}=2^{\frac{3}{4}}\times {𝑒}^{𝑖\frac{𝜋}{4}}\\ & \begin{array}{cc}\therefore {\left({𝑧}^{\frac{3}{4}}\right)}^4& ={\left(2^{\frac{3}{4}}\right)}^4\times {\left({𝑒}^{𝑖\frac{𝜋}{4}}\right)}^4=2^3\times {𝑒}^{𝜋𝑖}\\ & =2^3\times (-1)=-8\end{array}\end{array}$

We have,

$𝑎{𝑥}^3+𝑏𝑥+𝑐=0$

and if one root $=𝛽$

then other root $=2𝛽$        [given]

Let third root $=𝑦$

Then, $𝛽+2𝛽+𝑦=0/𝑎=0$

$[\because \text{ Sum of roots in cubic }=\frac{-𝑏}{𝑎}]$

$𝑦=-3𝛽$

and $𝛽\times 2𝛽\times 𝑦=-\frac{𝑐}{𝑎}$

[As product of roots in cubic $=\frac{-𝑑}{𝑎}$ ]

$\begin{array}{c}\Rightarrow 2{𝛽}^2\times (-3𝛽)=\frac{-𝑐}{𝑎}\\ 6{𝛽}^3=\frac{𝑐}{𝑎}\Rightarrow {𝛽}^3=\frac{𝑐}{6𝑎}& ...\left(𝑖\right)\end{array}$

$\begin{array}{c}\Rightarrow 𝛽\times 2𝛽+2𝛽(-3𝛽)+(-3𝛽)\times 𝛽=\frac{𝑏}{𝑎}\\ [\text{ As }𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{𝑐}{𝑎}]\\ \Rightarrow 2{𝛽}^2-6{𝛽}^2-3{𝛽}^2=\frac{𝑏}{𝑎}\\ {𝛽}^2=\frac{-𝑏}{7𝑎}...\left(𝑖𝑖\right)\end{array}$

Divide Eq. (i) by Eq. (ii),

$𝛽=\frac{-7𝑐}{6𝑏}$

Put this value of $𝛽$ in Eq. (i),

$\begin{array}{cc}\Rightarrow {\left(\frac{-7𝑐}{6𝑏}\right)}^3& =\frac{𝑐}{6𝑎}\Rightarrow -343{𝑐}^2=\frac{36{𝑏}^3}{𝑎}\\ \Rightarrow -343𝑎{𝑐}^2& =36{𝑏}^3\\ \Rightarrow 36{𝑏}^3+343𝑎{𝑐}^2& =0\end{array}$

We have,

${𝑥}^2-𝑎𝑥+𝑏=0$

Let $𝛼$, and $𝛽$ be the roots of above equation.

So, $𝛼+𝛽=𝑎$ and $𝛼𝛽=𝑏$

$\begin{array}{cc}& {𝛼}^2+{𝛽}^2=(𝛼+𝛽{)}^2-2𝛼𝛽\\ \Rightarrow & {𝛼}^2+{𝛽}^2={𝑎}^2-2𝑏...\left(𝑖\right)\end{array}$

$\begin{array}{cc}𝑎𝑛𝑑{𝛼}^3+{𝛽}^3& =(𝛼+𝛽{)}^3-3𝛼𝛽(𝛼+𝛽)\\ & ={𝑎}^3-3𝑎𝑏...\left(𝑖𝑖\right)\end{array}$

We know that,

if two roots ${𝛼}^2+{𝛽}^2$ and ${𝛼}^3+{𝛽}^3$ of a quadratic equation are given, then the equation can be written as

$[𝑥-({𝛼}^2+{𝛽}^2)][𝑥-({𝛼}^3+{𝛽}^3)]=0$

$\begin{array}{cc}\Rightarrow {𝑥}^2-[({𝛼}^2+& {𝛽}^2)+({𝛼}^3+{𝛽}^3)]𝑥\\ & +({𝛼}^2+{𝛽}^2)({𝛼}^3+{𝛽}^3)=0\end{array}$

On comparing it with given equation,

$\begin{array}{cc}& 𝐴{𝑥}^2+𝐵𝑥+𝐶=0,\text{ we get }\\ & 𝐶=({𝛼}^2+{𝛽}^2)({𝛼}^3+{𝛽}^3)=0\\ & =({𝑎}^2-2𝑏)({𝑎}^3-3𝑎𝑏)\end{array}$

[from Eqs. (i) and (ii)]

              $={𝑎}^5-5{𝑎}^3𝑏+6𝑎{𝑏}^2$

Given that,

$𝑎{𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+1=0$

The two roots of given equation are -1 and -1 . Let third root be equal to $𝛼$.

$\begin{array}{cc}\text{ So, }& -1-1+𝛼=\frac{-𝑏}{𝑎}\\ & (-1)(-1)\left(𝛼\right)=\frac{-1}{𝑎}\Rightarrow 𝛼=\frac{-1}{𝑎}\\ & (-1)(-1)+(-1)𝛼+(-1)𝛼=\frac{𝑐}{𝑎}\\ & 1-2𝛼=\frac{𝑐}{𝑎}\\ & 1-2\left(\frac{-1}{𝑎}\right)=\frac{𝑐}{𝑎}(\because 𝛼=\frac{-1}{𝑎})\\ & \frac{𝑎+2}{𝑎}=\frac{𝑐}{𝑎}\Rightarrow 𝑎+2=𝑐\end{array}$

and from Eq. (i),

$\begin{array}{cccc}& & -2+𝛼& =-\frac{𝑏}{𝑎}\\ & \Rightarrow & -2-\frac{1}{𝑎}& =-\frac{𝑏}{𝑎}(\because 𝛼=-\frac{1}{𝑎})\\ & \Rightarrow & \frac{2𝑎+1}{𝑎}& =\frac{𝑏}{𝑎}\Rightarrow 2𝑎+1=𝑏\end{array}$

Let the minimum value of

$\frac{{𝑥}^2-2𝑥+3}{{𝑥}^2-4𝑥+7}=𝑚$

So, ${𝑥}^2-2𝑥+3=𝑚{𝑥}^2-4𝑚𝑥+7𝑚$

${𝑥}^2(1-𝑚)+𝑥(-2+4𝑚)+3-7𝑚=0$

Then, $𝐷=(4𝑚-2{)}^2-4(1-𝑚\left)\right(3-7𝑚)$

$\begin{array}{cc}& =16{𝑚}^2+4-16𝑚-4(3-7𝑚-3𝑚+7{𝑚}^2)\\ & =16{𝑚}^2+4-16𝑚-12+40𝑚-28{𝑚}^2\\ & =-12{𝑚}^2+24𝑚-8\\ & =-4(3{𝑚}^2-6𝑚+2)\end{array}$

For real values of $𝑚$,

$\begin{array}{cccc}& & 𝐷& \ge 0\\ \Rightarrow & & -4(3{𝑚}^2-6𝑚+2)& \ge 0\\ \Rightarrow & & 3{𝑚}^2-6𝑚+2& \le 0\end{array}$

Let ${𝑚}_1$ and ${𝑚}_2$ be the roots of $3{𝑚}^2-6𝑚+2=0$.

Then,

$\begin{array}{cc}𝑆𝑜,𝑚& =\frac{6\pm \sqrt{\surd (-6{)}^2-4(3\left)\right(2)}}{2\times 3}\\ & =\frac{6}{6}\pm \frac{\sqrt{\surd 12}}{6}=1\pm \frac{1}{\sqrt{\surd 3}}\end{array}$

Therefore, $1-\frac{1}{\sqrt{\surd 3}}\le 𝑚\le 1+\frac{1}{\sqrt{\surd 3}}$

Hence, minimum value of $𝑚=1-\frac{1}{\sqrt{\surd 3}}$

$=\frac{\sqrt{\surd 3}-1}{\sqrt{\surd 3}}\times \frac{\sqrt{\surd 3}}{\sqrt{\surd 3}}=\frac{3-\sqrt{\surd 3}}{3}$

i.e., The minimum value of $𝑓\left(𝑥\right)=\frac{3-\sqrt{\surd 3}}{3}$

$\begin{array}{cc}& \\ & \begin{array}{c}\because (1+𝑥{)}^{𝑛}={𝐶}_0+{𝐶}_1𝑥+{𝐶}_2{𝑥}^2+\dots +{𝐶}_{𝑛}{𝑥}^{𝑛}\\ \text{ and }{(1+\frac{1}{𝑥})}^{𝑛}={𝐶}_0+{𝐶}_1\left(\frac{1}{𝑥}\right)+{𝐶}_2{\left(\frac{1}{𝑥}\right)}^2+\dots \\ +{𝐶}_{𝑛}{\left(\frac{1}{𝑥}\right)}^{𝑛}\end{array}\end{array}$

$\begin{array}{cc}& \text{ So, coefficient of }{𝑥}^{𝑇}\text{ in }\\ & \begin{array}{cc}& \begin{array}{cc}(1+𝑥{)}^{𝑛}{(1+\frac{1}{𝑥})}^{𝑛}={𝐶}_0{𝐶}_{𝑟}& +{𝐶}_1{𝐶}_{𝑟+1}\\ & +\dots +{𝐶}_{𝑛-𝑟}+{𝐶}_{𝑛}\end{array}\\ & =\text{ coefficient of }{𝑥}^{𝑟}\text{ in }{\left[\right(1+𝑥\left)\left(\frac{𝑥+1}{𝑥}\right)\right]}^{𝑛}\\ & =\text{ coefficient of }{𝑥}^{𝑟}\text{ in }\frac{(1+𝑥{)}^{2𝑛}}{{𝑥}^{𝑛}}\\ & =\text{ coefficient of }{𝑥}^{𝑛-𝑟}\text{ in }(1+𝑥{)}^{2𝑛}={}^{2𝑛}{𝐶}_{𝑛+𝑟}\\ & =\frac{2𝑛!}{(𝑛+𝑟)!(2𝑛-𝑛-𝑟)!}=\frac{2𝑛!}{(𝑛+𝑟)!(𝑛-𝑟)!}\end{array}\end{array}$

Expansion of ${(𝑥+\sqrt{\surd {𝑥}^2-1})}^8$

$\begin{array}{cc}& ={}^8{𝐶}_0(𝑥{)}^8+{}^8{𝐶}_1(𝑥{)}^7\left(\sqrt{\surd {𝑥}^2-1}\right)+{}^8{𝐶}_2(𝑥{)}^6\\ & {\left(\sqrt{\surd {𝑥}^2-1}\right)}^2+\dots +{}^8{𝐶}_8{\left(\sqrt{\surd {𝑥}^2-1}\right)}^8\end{array}$

Every term has ${𝑥}^8$ and that is the highest power.

Similarly,

$\begin{array}{cc}& {(𝑥-\sqrt{\surd {𝑥}^2+1})}^8={}^8{𝐶}_0{𝑥}^8-{}^8{𝐶}_1{𝑥}^7\left(\sqrt{\surd {𝑥}^2-1}\right)\\ & +\dots +{}^8{𝐶}_8{\left(\sqrt{\surd {𝑥}^2-1}\right)}^8\text{ then, }{2}^{\text{nd }},4^{\text{th }},6^{\text{th }},\dots ,\end{array}$

terms will cancel out.

So, coefficient of highest power $𝑥$

$\begin{array}{cc}& ={}^8{𝐶}_0+{}^8{𝐶}_0+{}^8{𝐶}_2+{}^8{𝐶}_2+\dots +{}^8{𝐶}_8+{}^8{𝐶}_8\\ & =2({}^8{𝐶}_0+{}^8{𝐶}_2+{}^8{𝐶}_4+{}^8{𝐶}_6+{}^8{𝐶}_8)\\ & =2(1+\frac{8\cdot 7}{2}+\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}+\frac{8\cdot 7}{2}+1)\\ & =2(2+56+70)=2\times 128=256\end{array}$

Given that,

${𝑥}_1+{𝑥}_2+{𝑥}_3+{𝑥}_4=10$

and we know that,

Number of non-negative integral Solution of ${𝑥}_1+{𝑥}_2+{𝑥}_3+\dots +{𝑥}_{𝑟}=𝑛$ is ${}^{𝑛+𝑟-1}{𝐶}_{𝑟-1}$

Here, $𝑟=4,𝑛=10$

∴ Number of non-negative integral solution ${=}^{10+4-1}{𝐶}_{4-1}={}^{13}{𝐶}_3=286$

The given digits are 2, 3, 4, 0, 5, 6, 7, 8.

Let 6 digits natural number be $\underset{–}{𝑎}\underset{–}{𝑏}\underset{–}{𝑐}\underset{–}{𝑑}\underset{–}{𝑒}\underset{–}{𝑓}$ → For the values of $𝑎$ we have 7 choice $(:𝑎\ne 0$ as if we take $𝑎=0$, it make the desired number to a 5 digit number).

→ For the values of $𝑏,𝑐,𝑑,𝑒,𝑓$ we have 8 choice for each.

Hence, total number of six digit natural numbers that can be formed with the digits $2,3,4,0,5,6,7,8$

$\begin{array}{cc}& =7\times 8\times 8\times 8\times 8\times 8\\ & =7\times (8{)}^5=7\times {\left(2^3\right)}^5=7\times 2^{15}\end{array}$

$𝐴=\left[\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right]:𝑎,𝑏,𝑐,𝑑\in \{-1,1\}$

If $𝐴$ is a singular matrix then, $\left|𝐴\right|=0$

${𝑆}_{0,}\left|𝐴\right|=𝑎𝑑-𝑏𝑐=0\Rightarrow 𝑎𝑑=𝑏𝑐$

Case I ⇒ If $𝑎𝑑=-1$, i.e. $𝑎=1,𝑑=-1$

then $𝑏𝑐=-1$, when $𝑏=1,𝑐=-1$

or $𝑏=1,𝑐=1$

There will be a possible ways.

Similarly, when $𝑎=-1,𝑑=1$

then, there will be 2 more possible ways.

Case II If $𝑎𝑑=1$, i.e. $(𝑎,𝑑)=(1,1)$ or

$(𝑎,𝑑)=(-1,-1)$

Then, $𝑏𝑐=1$ when $(𝑏,𝑐)=(1,1)$ or $(𝑎,𝑑)$

$=(-1,-1)$

Then, $𝑏𝑐=1$ when $(𝑏,𝑐)=(1-1)$ or $(-1,-1)$

So, for this total possible ways $=2\times 2=4$

Hence, the number of singular matrices

in $𝐴=4+4=8$

Given that, $\cot 𝑥\cot 𝑦=𝑎$ and $𝑥+𝑦=\frac{𝜋}{6}$ if $\cot 𝑥$ and $\cot 𝑦$ are the roots of an equation then the equation will be

$\begin{array}{cc}& {𝑡}^2-(\cot 𝑥+\cot 𝑦)𝑡+\cot 𝑥\cot 𝑦=0\\ & {𝑡}^2-\left[\frac{\cot 𝑥\cot 𝑦-1}{\cot (𝑥+𝑦)}\right]𝑡+\cot 𝑥\cot 𝑦=0\\ & [\because \cot (𝑥+𝑦)=\frac{\cot 𝑥\cot 𝑦-1}{\cot 𝑥+\cot 𝑦}]\\ & {𝑡}^2-\left(\frac{𝑎-1}{\cot \frac{𝜋}{6}}\right)𝑡+𝑎=0\\ & \{\cot 𝑥\cot 𝑦=𝑎\text{ and }𝑥+𝑦=\frac{𝜋}{6}\}\\ & \Rightarrow {𝑡}^2-\left(\frac{𝑎-1}{\sqrt{\surd 3}}\right)𝑡+𝑎=0\\ & \Rightarrow \sqrt{\surd 3}{𝑡}^2-(𝑎-1)𝑡+\sqrt{\surd 3}𝑎=0\\ & \sqrt{\surd 3}{𝑡}^2+(1-𝑎)𝑡+\sqrt{\surd 3}𝑎=0\end{array}$

We have,

$\tan 𝐴+\tan 𝐵=𝑥...\left(𝑖\right)$

and $\cot 𝐴+\cot 𝐵=𝑌$

$\begin{array}{cc}& \Rightarrow 𝑦=\frac{1}{\tan 𝐴}+\frac{1}{\tan 𝐵}\\ & \Rightarrow 𝑦=\frac{\tan 𝐴+\tan 𝐵}{\tan 𝐴\cdot \tan 𝐵}\\ & \Rightarrow \tan 𝐴\cdot \tan 𝐵=\frac{𝑥}{𝑦}\text{ [from (Eq. (i)] }\end{array}$

We know that,

$\begin{array}{cc}\tan (𝐴+𝐵)& =\frac{\tan 𝐴+\tan 𝐵}{1-\tan 𝐴\cdot \tan 𝐵}=\frac{𝑥}{1-\frac{𝑥}{𝑦}}\\ & =\frac{𝑥𝑦}{𝑦-𝑥}\end{array}$

Hence, $\tan (𝐴+𝐵)=\frac{𝑥𝑦}{𝑦-𝑥}$

We have,

$\begin{array}{cc}& \frac{1}{\cos {290}^{\circ }}+\frac{1}{\sqrt{\surd 3}\sin {250}^{\circ }}\\ =& \frac{1}{\cos ({270}^{\circ }+{20}^{\circ })}+\frac{1}{\sqrt{\surd 3}\sin ({270}^{\circ }-{20}^{\circ })}\\ =& \frac{1}{\sin {20}^{\circ }}+\frac{1}{\sqrt{\surd 3}(-\cos {20}^{\circ })}\\ & [\cos ({270}^{\circ }+𝜃)=\sin 𝜃,\sin ({270}^{\circ }-𝜃)=-\cos 𝜃]\\ =& \frac{\sqrt{\surd 3}\cos {20}^{\circ }-\sin {20}^{\circ }}{\sqrt{\surd 3}\sin {20}^{\circ }\cos {20}^{\circ }}\end{array}$

On multiply by 2 in both numerator and denominator, we get

$\begin{array}{cc}& \frac{2(\frac{\sqrt{\surd 3}}{2}\cos {20}^{\circ }-\frac{1}{2}\sin {20}^{\circ })}{\frac{\sqrt{\surd 3}}{2}(2\sin {20}^{\circ }\cos {20}^{\circ })}\\ & =\frac{2(\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ })}{\frac{\sqrt{\surd 3}}{2}\sin {40}^{\circ }}\\ & [\text{ as }2\sin 𝜃\cos 𝜃=\sin 2𝜃]\end{array}$

$\begin{array}{cc}& =\frac{4}{\sqrt{\surd 3}}\frac{\sin ({60}^{\circ }-{20}^{\circ })}{\sin {40}^{\circ }}\\ & [\sin 𝐴\cos 𝐵-\cos 𝐴\sin 𝐵=\sin (𝐴-𝐵\left)\right]\\ & =\frac{4}{\sqrt{\surd 3}}\end{array}$