TG EAPCET 2022 19th July Morning Shift Question Paper with Solutions

Given,

$\begin{array}{cc}& 𝑓\left(𝑥\right)={\log }_2(2^{𝑥}-2)+\sqrt{\surd 1-𝑥}\\ & 1-𝑥\ge 0\\ \Rightarrow & 𝑥\le 1...\left(𝑖\right)\end{array}$

Also, $2^{𝑥}-2>0$

$\begin{array}{cc}{2}^{𝑥}& >2\\ 𝑥& >1...\left(𝑖𝑖\right)\end{array}$

From Eqs. (i) and (ii), it can be concluded that $𝑥$ contains no real values.

$\begin{array}{cc}& \begin{array}{cc}𝐺𝑖𝑣𝑒𝑛,𝑓\left(𝑥\right)& =1-𝑥,𝑔\left(𝑥\right)=\frac{1}{1-𝑥}\\ ℎ\left(𝑥\right)& =1/𝑥\\ 𝑓\left(𝐹\right(ℎ\left(𝑥\right)\left)\right)& =𝑔\left(𝑥\right)\\ \Rightarrow 𝑓\left(𝐹\left(\frac{1}{𝑥}\right)\right)& =\frac{1}{1-𝑥}\end{array}\end{array}$

$\begin{array}{cccc}\Rightarrow & & 1-𝐹\left(\frac{1}{𝑥}\right)& =\frac{1}{1-𝑥}\\ \Rightarrow & & 𝐹\left(\frac{1}{𝑥}\right)& =1-\frac{1}{1-𝑥}\\ \Rightarrow & & 𝐹\left(\frac{1}{𝑥}\right)& =\frac{-𝑥}{1-𝑥}\end{array}$

Substitute $𝑥=\frac{1}{2022}$ to determine $𝐹\left(2022\right)$

$\begin{array}{cc}𝐹\left(2022\right)& =\frac{-\frac{1}{2022}}{1-\frac{1}{2022}}\\ & =-\frac{1}{2021}=\frac{1}{1-2022}\\ & =𝑔\left(2022\right)\end{array}$

Let $𝐴=⎡⎣⎢\begin{array}{ccc}0& 2& 𝑎\\ 𝑏& 0& 4\\ -3& 𝑐& 0\end{array}⎤⎦⎥$

If $𝐴$ is a skew-symmetric, then ${𝐴}^{𝑇}=-𝐴$

$\begin{array}{cc}⎡⎣⎢\begin{array}{ccc}0& 𝑏& -3\\ 2& 0& 𝑐\\ 𝑎& 4& 0\end{array}⎤⎦⎥& =⎡⎣⎢\begin{array}{ccc}0& -2& -𝑎\\ -𝑏& 0& -4\\ 3& -𝑐& 0\end{array}⎤⎦⎥\\ \Rightarrow 𝑏=-2,𝑎& =3\text{ and }𝑐=-4\\ \left[\begin{array}{cc}𝑎& 𝑏\\ 𝑏& 𝑎\end{array}\right]\left[\begin{array}{cc}𝑏& 𝑐\\ 𝑐& 𝑏\end{array}\right]& =\left[\begin{array}{cc}3& -2\\ -2& 3\end{array}\right]\left[\begin{array}{cc}-2& -4\\ -4& -2\end{array}\right]\\ & =\left[\begin{array}{cc}2& -8\\ -8& 2\end{array}\right]\end{array}$

$\text{Let }𝐴=⎡⎣⎢\begin{array}{ccc}-1& 2& 𝑏\\ 𝑎& 5& 6\\ 3& 𝑐& 7\end{array}⎤⎦⎥$

If $𝐴$ is a symmetric matrix, then ${𝐴}^{𝑇}=𝐴$

$⎡⎣⎢\begin{array}{ccc}-1& 𝑎& 3\\ 2& 5& 𝑐\\ 𝑏& 6& 7\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}-1& 2& 𝑏\\ 𝑎& 5& 6\\ 3& 𝑐& 7\end{array}⎤⎦⎥$

On comparing both the matrices, we get $𝑎=2,𝑏=3$ and $𝑐=6$

$\begin{array}{cc}\therefore & \mid \begin{array}{ccc}𝑎& 𝑏& 𝑐\\ 𝑏& 𝑐& 𝑎\\ 𝑐& 𝑎& 𝑏\end{array}\mid =\mid \begin{array}{ccc}2& 3& 6\\ 3& 6& 2\\ 6& 2& 3\end{array}\mid \\ & =2(18-4)-3(9-12)+6(6-36)\\ & =28+9-180=-143\end{array}$

${𝐴}^2-4𝐴-5𝐼=0$

Post multiplying by ${𝐴}^{-1}$ on both the sides, we get

$\begin{array}{cc}& 𝐴-4𝐼-5{𝐴}^{-1}=0\\ \Rightarrow & 5{𝐴}^{-1}=𝐴-4𝐼\\ \Rightarrow & {𝐴}^{-1}=\frac{1}{5}\{𝐴-4𝐼\}\\ \therefore & {𝐴}^{-1}=\frac{1}{5}⎧⎩⎨⎪⎪⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}⎤⎦⎥-⎡⎣⎢\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& 4\end{array}⎤⎦⎥⎫⎭⎬⎪⎪\\ \because & 𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}⎤⎦⎥=\frac{1}{5}⎡⎣⎢\begin{array}{ccc}-3& 2& 2\\ 2& -3& 2\\ 2& 2& -3\end{array}⎤⎦⎥\end{array}$

$\because 𝐵$ is the unique solution of $𝐴𝑌=𝑄$ and $𝐴$ is an invertible matrix

$\begin{array}{cccc}\Rightarrow & & 𝐴𝐵& =𝑄\\ \Rightarrow & & 𝐵& ={𝐴}^{-1}𝑄\end{array}$

Now, $𝐴𝑋=𝐵$

$\begin{array}{cc}\Rightarrow 𝑋& ={𝐴}^{-1}𝐵\\ & ={𝐴}^{-1}\left({𝐴}^{-1}𝑄\right)={\left({𝐴}^{-1}\right)}^2𝑄\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & (2𝑥-𝑦+1)+𝑖(𝑥-2𝑦-1)=2-3𝑖\end{array}$

$\begin{array}{cccc}\Rightarrow & & 2𝑥-𝑦+1& =2\\ \text{ and }& & 𝑥-2𝑦-1& =-3\\ \Rightarrow & & 2𝑥-𝑦& =1\\ \text{ and }& & 𝑥& =2𝑦-2\end{array}$

Now, substitute value of $𝑥=2𝑦-2$ in equation $2𝑥-𝑦=1$, we get

$\begin{array}{cccc}& & 2(2𝑦-2)-𝑦& =1\\ \Rightarrow & & 4𝑦-4-𝑦& =1\end{array}$

$\begin{array}{cc}& 3𝑦=5\Rightarrow 𝑦=\frac{5}{3}\\ & \text{ Now, substitute }𝑦=\frac{5}{3}\text{ in }𝑥=2𝑦-2\\ & 𝑥=\frac{10}{3}-2=\frac{4}{3}\end{array}$

$\begin{array}{cc}& \text{ Multiplicative inverse of }(𝑥-𝑖𝑦)=\frac{1}{𝑥-𝑖𝑦}\\ & \begin{array}{cc}& =\frac{1}{\frac{4}{3}-𝑖\frac{5}{3}}=\frac{3}{4-5𝑖}\times \frac{(4+5𝑖)}{(4+5𝑖)}\\ & =\frac{12}{4^2+5^2}+\frac{15}{4^2+5^2}𝑖=\frac{12}{41}+\frac{15}{41}𝑖\end{array}\end{array}$

$\because 𝛼$ and $𝛽$ are the roots of the equation

$\begin{array}{cc}& {𝑥}^2-2\sqrt{\surd 3}𝑥+4=0\\ & \therefore 𝛼+𝛽=2\sqrt{\surd 3}\text{ and }𝛼𝛽=4\\ & \text{ Now, }{𝛼}^2+{𝛽}^2=(𝛼+𝛽{)}^2-2𝛼𝛽\\ & =12-8=4\\ & \begin{array}{cc}\therefore {𝛼}^6+{𝛽}^6& ={\left({𝛼}^2\right)}^3+{\left({𝛽}^2\right)}^3\\ & ={({𝛼}^2+{𝛽}^2)}^3-3{𝛼}^2{𝛽}^2({𝛼}^2+{𝛽}^2)\\ & =(4{)}^3-3(4{)}^2\left(4\right)\\ [\because & {𝑎}^3+{𝑏}^3=(𝑎+𝑏{)}^3-3𝑎𝑏(𝑎+𝑏\left)\right]\\ & =4^3(1-3)=64(-2)=-128\end{array}\end{array}$

$\begin{array}{cc}& (-1)=\cos 𝜋+𝑖\sin 𝜋=\mathrm{cis}𝜋\\ & \begin{array}{cc}(-1{)}^{1/4}& =(\cos 𝜋+𝑖\sin 𝜋{)}^{1/4}\\ & =\cos \frac{3𝜋}{4}+𝑖\sin \frac{3𝜋}{4}\\ & [\because \cos \frac{𝜋}{4}\text{ is not }-(\mathrm{ve}\left)\right]\end{array}\end{array}$

$\begin{array}{cc}& =\mathrm{cis}\frac{3𝜋}{4}(-𝑖{)}^{1/2}={(\cos \frac{3𝜋}{2}+𝑖\sin \frac{3𝜋}{2})}^{1/2}\\ & =(\cos \frac{3𝜋}{4}+𝑖\sin \frac{3𝜋}{4})=\mathrm{cis}\frac{3𝜋}{4}\\ \therefore & 𝛼=\frac{3𝜋}{4}\Rightarrow \tan 𝛼=\tan \frac{3𝜋}{4}=-1\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& 𝑐=-15𝑥(-2)=30\\ & \text{ Now, }{𝑥}^2+13𝑥+30=0\\ & {𝑥}^2+10𝑥+3𝑥+30=0\\ & 𝑥(𝑥+10)+3(𝑥+10)=0\\ & \begin{array}{cc}\Rightarrow & 𝑥=-10,-3\\ \therefore & |𝛼-𝛽|=|-3+10|=7\end{array}\end{array}\end{array}$

$\begin{array}{cc}& \text{(A) }2{𝑥}^2+4𝑥+5\\ & 2{𝑥}^2+4𝑥+2+3\Rightarrow 2(𝑥+1{)}^2+3\end{array}$

∴ The minimum value of $2(𝑥+1{)}^2+3𝑖𝑠3.$

$\begin{array}{cc}\left(𝐵\right)𝐿𝑒𝑡𝑓\left(𝑥\right)& =\frac{{𝑥}^2+4𝑥+1}{{𝑥}^2+𝑥+1}\\ & =\frac{{𝑥}^2+𝑥+1}{{𝑥}^2+𝑥+1}+\frac{3𝑥}{{𝑥}^2+𝑥+1}\\ & =1+\frac{3𝑥}{{𝑥}^2+𝑥+1}\end{array}$

The maximum value of $\frac{3𝑥}{{𝑥}^2+𝑥+1}$ can be 1.

$\frac{3𝑥}{{𝑥}^2+𝑥+1}=\frac{3}{𝑥+\frac{1}{𝑥}+1}\text{ and }𝑥+\frac{1}{𝑥}\ge 2$

Therefore, maximum value of $𝑓\left(𝑥\right)$ is 2 .

(C) Given, $1\le \frac{3{𝑥}^2-5𝑥+6}{{𝑥}^2+1}\le 2$

$\begin{array}{cc}& \Rightarrow 1\le \frac{3{𝑥}^2-5𝑥+6}{{𝑥}^2+1}\text{ and }\frac{3{𝑥}^2-5𝑥+6}{{𝑥}^2+1}\le 2\\ & \Rightarrow {𝑥}^2+1\le 3{𝑥}^2-5𝑥+5\\ & \text{ and }3{𝑥}^2-5𝑥+6\le 2{𝑥}^2+2\\ & \Rightarrow 2{𝑥}^2-5𝑥+4\ge 0\text{ and }{𝑥}^2-5𝑥+4\le 0\end{array}$

The value of $𝑥$ for which

$2{𝑥}^2-5𝑥+4\ge 0$ are not real.

$\because 2{𝑥}^2-5𝑥+4\ge 0$ is possible for all real value

$\begin{array}{cc}& \Rightarrow 𝑥\in 𝑅\\ & \text{ and }{𝑥}^2-5𝑥+4\le 0\\ & \Rightarrow (𝑥-4)(𝑥-1)\le 0\Rightarrow 𝑥\in [1,4]\end{array}$

Now common of $𝑥\in 𝑅$ and $𝑥\in [1,4]$

$\begin{array}{cc}\Rightarrow & 𝑥\in [1,4]\\ \therefore & 𝑏=4\end{array}$

(D) $𝑎=1$

Correct order is = IV, III, V, II

$\because 𝛼,𝛽$ and $𝛾$ are the roots of the equation $5{𝑥}^3-2𝑥-4=0$

$\therefore 𝛼+𝛽+𝛾=0,𝛼𝛽+𝛽𝛼+𝛾𝛼=-\frac{2}{5}$

and $𝛼𝛽𝛾=\frac{4}{5}$

If $𝛼+𝛽+𝛾=0$, then

${𝛼}^3+{𝛽}^3+{𝛾}^3=3𝛼𝛽𝛾=3\times \frac{4}{5}=\frac{12}{5}$

Let ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4$ and ${𝑎}_5$ are the roots of

${𝑥}^5-𝑎{𝑥}^4+𝑏{𝑥}^3-𝑐{𝑥}^2+𝑑𝑥-1=0$

If AM and GM of ${𝑎}_1,{𝑎}_2,{𝑎}_3,{𝑎}_4$ and ${𝑎}_5$ are equal, then

${𝑎}_1={𝑎}_2={𝑎}_3={𝑎}_4={𝑎}_5=1$

Sum of roots $={}^5{𝐶}_1=5$

$\because 𝑎=5$

Sum of roots taken 2 at a time

$={}^5{𝐶}_2=10$

$\therefore 𝑏=10$

Sum of roots taken 3 at a time

$={}^5{𝐶}_3=10$

$\therefore 𝑐=10$

Sum of roots taken 4 at a time $={}^5{𝐶}_4=5$

$\therefore 𝑑=5$

Hence, $𝑎+𝑏+𝑐+𝑑=5+10+10+5$

$=30$

The roots of the given equations are $\sqrt{\surd 3}+\sqrt{\surd 2}𝑖$ and $\sqrt{\surd 3}-\sqrt{\surd 2}$.

$\begin{array}{cccc}\therefore & & 𝑥& =\sqrt{\surd 3}+\sqrt{\surd 2}𝑖\\ \Rightarrow & & 𝑥-\sqrt{\surd 3}& =\sqrt{\surd 2}𝑖\\ \Rightarrow & & (𝑥-\sqrt{\surd 3}{)}^2& =(\sqrt{\surd 2}𝑖{)}^2\end{array}$

$\begin{array}{cc}& \Rightarrow {𝑥}^2+3-2\sqrt{\surd 3}𝑥=-2\Rightarrow {𝑥}^2+5=2\sqrt{\surd 3}𝑥\\ & \Rightarrow {({𝑥}^2+5)}^2=(2\sqrt{\surd 3}𝑥{)}^2\\ & \Rightarrow {𝑥}^4+25+10{𝑥}^2=12{𝑥}^2\\ & \Rightarrow {𝑥}^4-2{𝑥}^2+25=0\end{array}$

$\begin{array}{cc}\text{ Also, }& 𝑥=\sqrt{\surd 3}-\sqrt{\surd 2}\\ \Rightarrow & {𝑥}^2=(\sqrt{\surd 3}-\sqrt{\surd 2}{)}^2\\ \Rightarrow & {𝑥}^2=3+2-2\sqrt{\surd 6}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow & {𝑥}^2-5=-2\sqrt{\surd 6}\\ \Rightarrow & {({𝑥}^2-5)}^2=(-2\sqrt{\surd 6}{)}^2\\ \Rightarrow & {𝑥}^4+25-10{𝑥}^2=24\\ \Rightarrow & {𝑥}^4-10{𝑥}^2+1=0\end{array}\\ & \text{ ∴ The required equation is }\\ & ({𝑥}^4-2{𝑥}^2+25)({𝑥}^4-10{𝑥}^2+1)=0\end{array}$

$\begin{array}{cc}𝑃\left(𝑥\right):{𝑥}^{10}-3{𝑥}^8+5{𝑥}^6& -5{𝑥}^4+3{𝑥}^2-1=0\end{array}$

$\begin{array}{cc}& \because 𝑃\left(1\right)\text{ and }𝑃(-1)\text{ are zero }\\ & \Rightarrow (𝑥-1)\text{ and }(𝑥+1)\text{ are factors. }\\ & 𝑃\left(𝑥\right)=({𝑥}^2-1)({𝑥}^8-2{𝑥}^6+3{𝑥}^4-2{𝑥}^2+1)\\ & \text{ Let }𝑔\left(𝑥\right)={𝑥}^8-2{𝑥}^6+3{𝑥}^4-2{𝑥}^2+1\\ & ={𝑥}^2({𝑥}^6-2{𝑥}^4+3{𝑥}^2-2)+1\\ & ={𝑥}^2({𝑥}^2-1)({𝑥}^4-{𝑥}^2+2)+1\\ & ={𝑥}^2({𝑥}^2-1)\{{({𝑥}^2-\frac{1}{2})}^2+\frac{7}{4}\}+1\end{array}$

$𝑔\left(𝑥\right)\ne 0$ for any real value.

∴ Total number of non-real roots

$=10-2=8$

Case I When $𝑥=1$, then $𝑦+𝑧=11$

There are 4 triplets in this case.

i.e. $(1,2,9),(1,3,8),(1,4,7),(1,5,6)$

Case II When $𝑥=2$, then $𝑦+𝑧=10$

There are 2 triplets in this case i.e. $(2,3,7)$ and $(2,4,6)$.

Case III When $𝑥=3$, then $𝑦+𝑧=9$

There is only one triplet in this case i.e. $(3,4,5)$.

∴ Total number of triplets $=4+2+1=7$

There are total 12 questions.

A candidate can choose 8 questions from 12 in the following pattern

$\overline{)\begin{array}{ccc}\text{ Ist }& \text{ Ilnd }& \text{ Illrd }\\ 2& 2& 4\\ 2& 3& 3\\ 2& 4& 2\\ 3& 2& 3\\ 3& 3& 2\\ 4& 2& 2\end{array}}$

$\begin{array}{cc}& \text{ ∴ Total number of ways }\\ & \begin{array}{cc}=& ({}^4{𝐶}_2\times {}^4{𝐶}_2\times {}^4{𝐶}_4)+({}^4{𝐶}_2\times {}^4{𝐶}_3\times {}^4{𝐶}_3)\\ & +({}^4{𝐶}_2\times {}^4{𝐶}_4\times {}^4{𝐶}_2)+({}^4{𝐶}_3\times {}^4{𝐶}_2\times {}^4{𝐶}_3)\\ & +({}^4{𝐶}_3\times {}^4{𝐶}_3\times {}^4{𝐶}_2)+({}^4{𝐶}_4\times {}^4{𝐶}_2\times {}^4{𝐶}_2)\\ =& 6\times 6\times 1+6\times 4\times 4+6\times 1\times 6+4\times \\ & 6\times 4+4\times 4\times 6+1\times 6\times 6\\ =& 36+96+36+96+96+36=396\end{array}\end{array}$

$\begin{array}{cc}{𝑇}_{𝑟+1}& ={}^{𝑛}{𝐶}_{𝑟}{\left(2{𝑥}^2\right)}^{𝑛-𝑟}{\left(\frac{-1}{3{𝑥}^3}\right)}^{𝑟}\\ {𝑇}_{𝑟+1}& ={}^5{𝐶}_{𝑟}(2{)}^{5-𝑟}{\left(\frac{-1}{3}\right)}^{𝑟}{𝑥}^{10-2𝑟-3𝑟}\\ & ={}^5{𝐶}_{𝑟}(2{)}^{5-𝑟}{\left(\frac{-1}{3}\right)}^{𝑟}{𝑥}^{10-5𝑟}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}10-5𝑟& =5\Rightarrow 5𝑟=5\Rightarrow 𝑟=1\\ {𝑇}_2& ={}^5{𝐶}_1(2{)}^4\left(\frac{-1}{3}\right){𝑥}^5\end{array}\\ & \begin{array}{cc}& \text{ Coefficient of }{𝑥}^5=5(2{)}^4\left(\frac{-1}{3}\right)\\ & \therefore 𝑘=\frac{-16\times 5}{3}\Rightarrow \frac{3𝑘}{2}=-40\end{array}\end{array}$

$\begin{array}{cc}& \text{Let }\frac{42-13𝑥}{{𝑥}^2+𝑥-6}=\frac{𝑀}{𝑥+3}+\frac{𝑁}{𝑥-2}\\ & 42-13𝑥=𝑀𝑥-2𝑀+𝑁𝑥+3𝑁\\ & \Rightarrow 𝑀+𝑁=-13\text{ and }-2𝑀+3𝑁=42\\ & -2(-13-𝑁)+3𝑁=42\\ & 26+2𝑁+3𝑁=42\\ & 5𝑁=16\Rightarrow 𝑁=16/5\\ & 𝑀=-13-\frac{16}{5}=\frac{-65-16}{5}=-\frac{81}{5}\\ & \frac{42-3𝑥}{{𝑥}^2+𝑥-16}=\frac{-81}{5(𝑥+3)}+\frac{16}{5(𝑥-2)}\\ & =\frac{-81}{5𝑥+15}+\frac{16}{5𝑥-10}\end{array}$

Here,$𝐴=-81,𝐵=16,𝑙=𝑝=5$

$\begin{array}{cc}𝑚& =15,𝑞=-10\\ \frac{𝐴𝑙𝑝}{𝐵𝑚𝑞}& =\frac{-81\times 5\times 5}{16\times 15\times (-10)}=\frac{27}{32}\end{array}$

$\begin{array}{cc}& \frac{3𝑥+5}{(𝑥+1)(2{𝑥}^2+3)}=\frac{𝐴}{𝑥+1}+\frac{𝐵𝑥+𝐶}{2{𝑥}^2+3}\\ & 3𝑥+5=𝐴(2{𝑥}^2+3)+(𝐵𝑥+𝐶)(𝑥+1)\\ & 3𝑥+5=2𝐴{𝑥}^2+3𝐴+𝐵{𝑥}^2+𝐵𝑥+𝐶𝑥+𝐶\end{array}$

Therefore,  $2𝐴+𝐵=0,𝐵+𝐶=3$

$\begin{array}{c}3𝐴+𝐶=5\\ 𝐵=-2𝐴,-2𝐴+𝐶=3,3𝐴+𝐶=5\\ \Rightarrow 𝐴=2/5,𝐶=19/5\Rightarrow 𝐵=-4/5\\ 𝑓\left(𝑥\right)=\frac{2}{5}{𝑥}^3-\frac{4}{5}{𝑥}^2+7𝑥+\frac{19}{5}\\ {𝑓}^{\prime }\left(𝑥\right)=\frac{6}{5}{𝑥}^2-\frac{8}{5}𝑥+7\\ {𝑓}^{\prime }(-2)=\frac{6}{5}(-2{)}^2-\frac{8}{5}(-2)+7\\ =\frac{24}{5}+\frac{16}{5}+7\end{array}$

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\mid \begin{array}{ccc}-\sin 𝑥& 2\sin 2𝑥& 4{\cos }^2𝑥\\ \cos 𝑥& 4{\sin }^2𝑥& 2\sin 2𝑥\\ 0& -\cos 𝑥& \sin 𝑥\end{array}\mid \\ & \begin{array}{c}=-\sin 𝑥\mid \begin{array}{cc}4{\sin }^2𝑥& 2\sin 2𝑥\\ -\cos 𝑥& \sin 𝑥\end{array}\mid \\ -\cos 𝑥\mid \begin{array}{cc}2\sin 2𝑥& 4{\cos }^2𝑥\\ -\cos 𝑥& \sin 𝑥\end{array}\mid \\ =-\sin 𝑥[4{\sin }^3𝑥+4{\cos }^2𝑥\sin 𝑥]\\ -\cos 𝑥[4{\sin }^2𝑥\cos 𝑥+4{\cos }^3𝑥]\\ [\because \sin 2𝑥=2\sin 𝑥\cos 𝑥]\\ =4[-{\sin }^4𝑥-{\cos }^2𝑥{\sin }^2𝑥\\ -{\sin }^2𝑥{\cos }^2𝑥-{\cos }^4𝑥]\\ =-4[{\sin }^4𝑥+{\cos }^4𝑥+2{\sin }^2𝑥{\cos }^2𝑥]\\ =-4{({\sin }^2𝑥+{\cos }^2𝑥)}^2=-4\\ \therefore {𝑓}^{\prime }\left(𝑥\right)=0\\ 𝑓\left(\frac{5𝜋}{4}\right)+{𝑓}^{\prime }\left(\frac{5𝜋}{4}\right)=-4+0=-4\end{array}\end{array}$