MHT CET 2024 16th May Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& g\left(𝑥\right)=\left\{f\right[2f\left(𝑥\right)+2]{\}}^2\\ & \therefore g^{\prime }\left(𝑥\right)=2f\left[2f\right(𝑥)+2]\cdot f^{\prime }\left[2f\right(𝑥)+2]\cdot 2f^{\prime }\left(𝑥\right)\\ & \therefore {𝑔}^{\prime }\left(0\right)=2𝑓\left[2𝑓\right(0)+2]\cdot {𝑓}^{\prime }\left[2𝑓\right(0)+2]\cdot 2{𝑓}^{\prime }\left(0\right)\\ & =2f\left[2\right(-1)+2]\cdot f^{\prime }\left(2\right(-1)+2)\cdot 2\left(1\right)\\ & \Rightarrow g^{\prime }\left(0\right)=4f\left(0\right)\cdot f^{\prime }\left(0\right)\\ & =4(-1)\left(1\right)\\ & =-4\end{array}$

$\begin{array}{cc}& {\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)={\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}\right)\left(\frac{2}{9}\right)}\right)\\ & ={\tan }^{-1}\left(\frac{9+8}{36-2}\right)\\ & ={\tan }^{-1}\left(\frac{17}{34}\right)\\ & ={\tan }^{-1}\left(\frac{1}{2}\right)\\ & =\frac{1}{2}{\cos }^{-1}\left[\frac{1-{\left(\frac{1}{2}\right)}^2}{1+{\left(\frac{1}{2}\right)}^2}\right]\dots [\because 2{\tan }^{-1}𝑥={\cos }^{-1}\left(\frac{1-{𝑥}^2}{1+{𝑥}^2}\right)]\end{array}$

$\begin{array}{cc}& =\frac{1}{2}{\cos }^{-1}\left(\frac{3}{5}\right)\\ \therefore & 𝑥=\frac{3}{5}\end{array}$

The corner points of the feasible region are $O(0,0),A(6,0),B(6,4),C(3,7)$ and $D(0,5)$.

$𝑧=4𝑥+3𝑦$

At $O(0,0),z=4\left(0\right)+3\left(0\right)=0$

At $A(6,0),z=4\left(6\right)+3\left(0\right)=24$

At $B(6,4),z=4\left(6\right)+3\left(4\right)=36$

At $C(3,7),z=4\left(3\right)+3\left(7\right)=33$

At $D(0,5),z=4\left(0\right)+3\left(5\right)=15$

$\therefore$ Maximum value of $𝑧$ is 36.

$\begin{array}{cc}\frac{{𝑧}_1+{𝑧}_2}{{𝑧}_1-{𝑧}_2}& =\frac{8-𝑖}{2-3𝑖}\\ & =\frac{8-𝑖}{2-3𝑖}\times \frac{2+3𝑖}{2+3𝑖}\\ & =\frac{19+22𝑖}{13}=\frac{19}{13}+\frac{22𝑖}{13}\end{array}$

$\therefore \arg \left(\frac{z_1+z_2}{z_1-z_2}\right)={\tan }^{-1}\left(\frac{22}{19}\right)$

Let $\frac{𝑥+3}{5}=\frac{𝑦+1}{2}=\frac{𝑧+4}{3}=𝜆$

Any point on the line is

$𝑃\equiv (5𝜆-3,2𝜆-1,3𝜆-4)$

Given point is $A(0,2,3)$

$\therefore$ The d.r.s. of AP are $5𝜆-3,2𝜆-3,3𝜆-7$

Since the line AP is perpendicular to the given line.

$\begin{array}{cc}\therefore & 5(5𝜆-3)+2(2𝜆-3)+3(3𝜆-7)=0\\ & \Rightarrow 38𝜆-42=0\\ & \Rightarrow 𝜆=\frac{21}{19}\\ \therefore & 𝑃\equiv (\frac{48}{19},\frac{23}{19},\frac{-13}{19})\end{array}$

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}=\frac{𝑦}{𝑥}-{\cos }^2\frac{𝑦}{𝑥}\text{..... (i)}\\ & \text{ Put }𝑦=v𝑥\text{..... (ii)}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=v+𝑥\frac{\mathrm{dv}}{d𝑥}\text{..... (iii)}\end{array}$

Substituting (ii) and (iii) in (i), we get<?p>

$v+𝑥\frac{\mathrm{dv}}{d𝑥}=v-{\cos }^{2}v\Rightarrow 𝑥\frac{\mathrm{dv}}{d𝑥}=-{\cos }^{2}v$

$\begin{array}{cc}& \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \int {\sec }^2𝑣𝑑𝑣=-\int \frac{𝑑𝑥}{𝑥}+𝑐\\ & \Rightarrow \tan 𝑣=-\log 𝑥+𝑐\\ & \Rightarrow \tan \frac{𝑦}{𝑥}=-\log 𝑥+𝑐\text{.... (iv)}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Since the required curve passes through }(1,\frac{𝜋}{4})\text{, }\\ & \begin{array}{cc}& \tan \frac{𝜋}{4}=-\log 1+c\Rightarrow c=1\\ \therefore & \tan \frac{𝑦}{𝑥}=-\log 𝑥+1\dots \left[\text{ From (iv) }\right]\\ & \Rightarrow \tan \frac{𝑦}{𝑥}=-\log 𝑥+\log e\\ & \Rightarrow 𝑦=𝑥{\tan }^{-1}[\log \left(\frac{e}{𝑥}\right)]\end{array}\end{array}$

$\begin{array}{cc}& {\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{12}{13}\right)\\ & ={\sin }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\sqrt{\surd 1-{\left(\frac{12}{13}\right)}^2}\\ & ={\sin }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\\ & ={\sin }^{-1}⎡⎣⎢\frac{3}{5}\sqrt{\surd 1-{\left(\frac{5}{13}\right)}^2}+\frac{5}{13}\sqrt{\surd 1-{\left(\frac{3}{5}\right)}^2}⎤⎦⎥\\ & ={\sin }^{-1}(\frac{3}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{4}{5})\\ & ={\sin }^{-1}\left(\frac{56}{65}\right)\\ \therefore & 𝑃=\frac{56}{65}\end{array}$

Volume of sphere $\left(𝑉\right)=\frac{4}{3}𝜋r^3$

Surface area of sphere $\left(𝐴\right)=4𝜋r^2$

$\begin{array}{cc}& \therefore \frac{𝑑𝑉}{𝑑𝑟}=4𝜋{𝑟}^2\text{ and }\frac{𝑑𝐴}{𝑑𝑟}=8𝜋𝑟\\ & \therefore \frac{𝑑𝑉}{𝑑𝐴}=\frac{\frac{𝑑𝑉}{𝑑𝑟}}{\frac{𝑑𝐴}{𝑑𝑟}}=\frac{4𝜋{𝑟}^2}{8𝜋𝑟}=\frac{𝑟}{2}\\ & \therefore {\left(\frac{𝑑𝑉}{𝑑𝐴}\right)}_{𝑟=2}=\frac{2}{2}=1{\mathrm{cm}}^3/{\mathrm{cm}}^2\end{array}$

$\begin{array}{cc}\text{ Let }𝐼& =\int \frac{{𝑥}^2-4}{{𝑥}^4+9{𝑥}^2+16}𝑑𝑥\\ & =\int \frac{1-\frac{4}{{𝑥}^2}}{{𝑥}^2+\frac{16}{{𝑥}^2}+9}𝑑𝑥\\ & =\int \frac{1-\frac{4}{{𝑥}^2}}{{(𝑥+\frac{4}{𝑥})}^2+1}𝑑𝑥\end{array}$

$\begin{array}{cc}& \text{ Put }𝑥+\frac{4}{𝑥}=t\Rightarrow (1-\frac{4}{{𝑥}^2})d𝑥=\mathrm{dt}\\ & \begin{array}{cc}\therefore & =\int \frac{\mathrm{dt}}{t^2+1}\\ & ={\tan }^{-1}t+c\\ & ={\tan }^{-1}(𝑥+\frac{4}{𝑥})+c\\ \therefore & f\left(𝑥\right)=𝑥+\frac{4}{𝑥}\\ & \Rightarrow f\left(2\right)=2+2=4\end{array}\end{array}$

$\begin{array}{cc}3\overline{𝑎}+5\overline{𝑏}& =3(\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘})+5(2\widehat{𝑖}+3\widehat{𝑗}-\widehat{𝑘})\\ & =13\widehat{𝑖}+9\widehat{𝑗}+4\widehat{𝑘}\\ 5\overline{𝑎}+3\overline{𝑏}& =5(\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘})+3(2\widehat{𝑖}+3\widehat{𝑗}-\widehat{𝑘})\\ & =11\widehat{𝑖}-\widehat{𝑗}+12\widehat{𝑘}\\ \therefore \cos 𝜃& =\frac{(13\widehat{𝑖}+9\widehat{𝑗}+4\widehat{𝑘})\cdot (11\widehat{𝑖}-\widehat{𝑗}+12\widehat{𝑘})}{\sqrt{\surd 169+81+16}\sqrt{\surd 121+1+144}}\\ & =\frac{13\left(11\right)+9(-1)+1\left(12\right)}{\sqrt{\surd 266}\sqrt{\surd 266}}\\ & =\frac{143-9+48}{266}\\ & =\frac{182}{266}\\ & =\frac{13}{19}\\ \therefore & 𝜃={\cos }^{-1}\left(\frac{13}{19}\right)\end{array}$

$\begin{array}{cc}& \left|𝐴\right|=\mid \begin{array}{cc}1& 2\\ -1& 4\end{array}\mid =4+2=6\ne 0\\ & \text{ If }𝐴=\left[\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right]\text{ and }\mathrm{ad}-\mathrm{bc}\ne 0\text{, then }\\ & \begin{array}{cc}& {𝐴}^{-1}=\frac{1}{𝑎𝑑-𝑏𝑐}\left[\begin{array}{cc}𝑑& -𝑏\\ -𝑐& 𝑎\end{array}\right]\\ \therefore {𝐴}^{-1}& =\frac{1}{6}\left[\begin{array}{cc}4& -2\\ 1& 1\end{array}\right]\end{array}\end{array}$

$\begin{array}{cc}& {𝐴}^{-1}=𝛼𝐼+𝛽𝐴\\ & \Rightarrow \frac{1}{6}\left[\begin{array}{cc}4& -2\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}𝛼& 0\\ 0& 𝛼\end{array}\right]+\left[\begin{array}{cc}𝛽& 2𝛽\\ -𝛽& 4𝛽\end{array}\right]\\ & \Rightarrow \left[\begin{array}{cc}\frac{2}{3}& -\frac{1}{3}\\ \frac{1}{6}& \frac{1}{6}\end{array}\right]=\left[\begin{array}{cc}𝛼+𝛽& 2𝛽\\ -𝛽& 𝛼+4𝛽\end{array}\right]\end{array}$

$\begin{array}{cc}& \therefore \text{ By the equality of matrices, }\\ & \begin{array}{cc}& \frac{1}{6}=-𝛽\text{ and }𝛼+𝛽=\frac{2}{3}\\ & \Rightarrow 𝛽=\frac{-1}{6}\text{ and }𝛼-\frac{1}{6}=\frac{2}{3}\\ & \Rightarrow 𝛽=\frac{-1}{6}\text{ and }𝛼=\frac{5}{6}\\ \therefore & 4(𝛼-𝛽)=4\end{array}\end{array}$

$\begin{array}{cc}& \text{ By rationalising, we get }\\ & \begin{array}{cc}& \underset{𝑦\to 0}{lim}\frac{\sqrt{\surd 1+\sqrt{\surd 1+{𝑦}^4}-\sqrt{\surd 2}}}{{𝑦}^4}\\ & =\underset{𝑦\to 0}{lim}\frac{1+\sqrt{\surd 1+{𝑦}^4}-2}{{𝑦}^4(\sqrt{\surd 1+\sqrt{\surd 1+{𝑦}^4}}+\sqrt{\surd 2})}\\ & =\underset{𝑦\to 0}{lim}\frac{\sqrt{\surd 1+{𝑦}^4}-1}{{𝑦}^4(\sqrt{\surd 1+\sqrt{\surd 1+{𝑦}^4}}+\sqrt{\surd 2})}\\ & =\underset{𝑦\to 0}{lim}\frac{\sqrt{\surd 1+{𝑦}^4}-1}{{𝑦}^4(\sqrt{\surd 1+\sqrt{\surd 1+{𝑦}^4}}+\sqrt{\surd 2})}\times \frac{\sqrt{\surd 1+{𝑦}^4}+1}{\sqrt{\surd 1+{𝑦}^4}+1}\\ & =\underset{𝑦\to 0}{lim}\frac{{𝑦}^4}{{𝑦}^4(\sqrt{\surd 1+\sqrt{\surd 1+{𝑦}^4}}+\sqrt{\surd 2})(\sqrt{\surd 1+{𝑦}^4}+1)}\\ & =\frac{1}{(\sqrt{\surd 1+\sqrt{\surd 1+0}}+\sqrt{\surd 2})(\sqrt{\surd 1+0}+1)}=\frac{1}{4\sqrt{\surd 2}}\end{array}\end{array}$

Let $\hat{n}$ be the unit vector perpendicular to $\bar{b}$ and $\overline{𝑐}$.

$\begin{array}{cc}\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]& =\bar{a}\cdot (\bar{b}\times \bar{c})\\ & =\bar{a}\cdot \left(\right|\bar{b}\left|\right|\overrightarrow{c}|\sin 𝜃\hat{n})\\ & =\bar{a}\cdot (3\times 4\sin \frac{𝜋}{3}\cdot \hat{n})\\ & =\bar{a}\cdot (12\times \frac{\sqrt{\surd 3}}{2}\hat{n})\\ & =6\sqrt{\surd 3}\left|\bar{a}\right|\left|\hat{n}\right|\cos 0\\ & =6\sqrt{\surd 3}\times 2\times 1\\ & =12\sqrt{\surd 3}\end{array}$

$\begin{array}{cc}& \text{ Let the unknown numbers be }𝑥\text{ and }𝑦\text{. }\\ & \begin{array}{cc}& \text{ Mean }=8\\ & \Rightarrow \frac{2+4+10+12+14+𝑥+𝑦}{7}=8\\ & \Rightarrow 𝑥+𝑦=14\text{.... (i)}\\ & \text{ Variance }=16\\ & \Rightarrow \frac{2^2+4^2+{10}^2+{12}^2+{14}^2+{𝑥}^2+{𝑦}^2}{7}-(\text{ mean }{)}^2=16\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow 460+{𝑥}^2+{𝑦}^2=7[16+(8{)}^2]\\ & \Rightarrow 460+{𝑥}^2+{𝑦}^2=560\\ & \Rightarrow {𝑥}^2+{𝑦}^2=100\text{.... (ii)}\end{array}$

Solving (i) and (ii), we get

$𝑥=6,𝑦=8\text{ or }𝑥=8,𝑦=6$

$\therefore$ Product $=48$

$\begin{array}{cc}& 𝑦={\tan }^{-1}\left(\frac{2+3𝑥}{3-2𝑥}\right)+{\tan }^{-1}\left(\frac{4𝑥}{1+5{𝑥}^2}\right)\\ & ={\tan }^{-1}\left(\frac{\frac{2}{3}+𝑥}{1-\frac{2}{3}𝑥}\right)+{\tan }^{-1}\left(\frac{5𝑥-𝑥}{1+5{𝑥}^2}\right)\\ & ={\tan }^{-1}\frac{2}{3}+{\tan }^{-1}𝑥+{\tan }^{-1}5𝑥-{\tan }^{-1}𝑥\\ \therefore & 𝑦={\tan }^{-1}\frac{2}{3}+{\tan }^{-1}5𝑥\\ \therefore & \frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{1+(5𝑥{)}^2}\cdot 5=\frac{5}{1+25{𝑥}^2}\end{array}$

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{𝑎\sin 𝑥+𝑏\cos 𝑥}{𝑐\sin 𝑥+𝑑\cos 𝑥}\\ & f\left(𝑥\right)\text{ will be decreasing, if }{f}^{\prime }\left(𝑥\right)<0\end{array}$

$\begin{array}{c}\therefore \frac{1}{(𝑐\sin 𝑥+𝑑\cos 𝑥{)}^2}\left[\right(𝑐\sin 𝑥+𝑑\cos 𝑥\left)\right(𝑎\cos 𝑥-𝑏\sin 𝑥)\\ -(𝑎\sin 𝑥+𝑏\cos 𝑥)(𝑐\cos 𝑥-𝑑\sin 𝑥)]<0\end{array}$

$\begin{array}{c}\Rightarrow \mathrm{ac}\sin 𝑥\cos 𝑥-\mathrm{bc}{\sin }^2𝑥+\mathrm{ad}{\cos }^2𝑥\\ -\mathrm{bd}\sin 𝑥\cos 𝑥-\mathrm{acsin}𝑥\cos 𝑥+\mathrm{ad}{\sin }^2𝑥\\ -\mathrm{bc}{\cos }^2𝑥+\mathrm{bd}\sin 𝑥\cos 𝑥<0\end{array}$

$\begin{array}{cc}& \Rightarrow \mathrm{ad}({\sin }^2𝑥+{\cos }^2𝑥)-\mathrm{bc}({\sin }^2𝑥+{\cos }^2𝑥)<0\\ & \Rightarrow \mathrm{ad}-\mathrm{bc}<0\end{array}$

$\begin{array}{cc}& \text{ We have, }\sum _{𝑥=0}^{\infty }P(X=𝑥)=1\\ & \Rightarrow k\sum _{𝑥=0}^{\infty }(𝑥+1){\left(\frac{1}{5}\right)}^{𝑥}=1\\ & \Rightarrow k[1+2\left(\frac{1}{5}\right)+3{\left(\frac{1}{5}\right)}^2+4{\left(\frac{1}{5}\right)}^3+\dots ]=1\end{array}$

$\Rightarrow k[\frac{1}{1-\frac{1}{5}}+\frac{1\times \frac{1}{5}}{{(1-\frac{1}{5})}^2}]=1$

$\cdots \left[\begin{array}{c}\because 𝑎+(𝑎+𝑑)𝑟+(𝑎+2𝑑){𝑟}^2+\dots .\\ =\frac{𝑎}{1-𝑟}+\frac{𝑑𝑟}{(1-𝑟{)}^2}\end{array}\right]$

$\begin{array}{cc}& \Rightarrow 𝑘(\frac{5}{4}+\frac{5}{16})=1\\ & \Rightarrow \frac{25k}{16}=1\\ & \Rightarrow 𝑘=\frac{16}{25}\end{array}$

$\begin{array}{cc}& 3\overline{𝑎}+\overline{𝑏}-2\overline{𝑐}\\ & =3(2\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘})+(\widehat{𝑖}+\widehat{𝑗}-2\widehat{𝑘})-2(4\widehat{𝑖}-2\widehat{𝑗}+\widehat{𝑘})\\ & =-\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}\end{array}$

$\begin{array}{cc}& \therefore \text{ The unit vector in the direction of }3\overline{𝑎}+\overline{𝑏}-2\overline{𝑐}\text{ is }\\ & \frac{-\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}}{|-\widehat{𝑖}+2\widehat{𝑗}-\widehat{𝑘}|}\end{array}$

$\begin{array}{cc}& =\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{\surd (-1{)}^2+2^2+(-1{)}^2}}\\ & =\frac{1}{\sqrt{\surd 6}}(-\hat{i}+2\hat{j}-\hat{k})\end{array}$

Event E: Product of the two digits is 18.

$\begin{array}{cc}\therefore & 𝐸=\{29,36,63,92\}\\ \therefore & 𝑝=\frac{4}{90}\\ & \Rightarrow 𝑞=1-\frac{4}{90}=\frac{86}{90}\end{array}$

Here, $𝑛=4$

$\begin{array}{cc}\therefore P(X\ge 3)& =P(X=3)+P(X=4)\\ & ={}^4{C}_3{\left(\frac{4}{90}\right)}^3{\left(\frac{86}{90}\right)}^1+{}^4{C}_4{\left(\frac{4}{90}\right)}^4{\left(\frac{86}{90}\right)}^0\\ & =4{\left(\frac{4}{90}\right)}^3\left(\frac{86}{90}\right)+1{\left(\frac{4}{90}\right)}^4\\ & ={\left(\frac{4}{90}\right)}^4(86+1)\\ & =87{\left(\frac{4}{90}\right)}^4\end{array}$

$\begin{array}{cc}𝑥& =\sin 𝑡\\ \therefore \frac{d𝑥}{\mathrm{dt}}& =\cos t\\ 𝑦& =𝑎{𝑒}^{t\sqrt{\surd 2}}+𝑏{𝑒}^{𝑡\sqrt{\surd 2}}\\ & =e^{t\sqrt{\surd 2}}(𝑎+𝑏)\end{array}$

$\begin{array}{cc}\therefore & \frac{𝑑𝑦}{𝑑𝑡}=(𝑎+𝑏)\cdot {𝑒}^{𝑡\sqrt{\surd 2}}\cdot \sqrt{\surd 2}\\ \therefore & \frac{𝑑𝑦}{𝑑𝑥}=\frac{\frac{𝑑𝑦}{𝑑𝑡}}{\frac{𝑑𝑥}{𝑑𝑡}}=\frac{\sqrt{\surd 2}(𝑎+𝑏){𝑒}^{𝑡\sqrt{\surd 2}}}{\cos 𝑡}\\ \therefore & \frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{𝑑}{𝑑𝑡}\left(\frac{\sqrt{\surd 2}(𝑎+𝑏){𝑒}^{𝑡\sqrt{\surd 2}}}{\cos 𝑡}\right)\cdot \frac{𝑑𝑡}{𝑑𝑥}\end{array}$

$\begin{array}{cc}& =\sqrt{\surd 2}(𝑎+𝑏)⎡⎣⎢\frac{\cos 𝑡({𝑒}^{𝑡\sqrt{\surd 2}}\cdot \sqrt{\surd 2})-{𝑒}^{𝑖\sqrt{\surd 2}}(-\sin 𝑡)}{(\cos 𝑡{)}^2}⎤⎦⎥\cdot \frac{1}{\cos 𝑡}\\ & =\sqrt{\surd 2}(𝑎+𝑏)\left[\frac{{𝑒}^{𝑖\sqrt{\surd 2}}(\sqrt{\surd 2}\cos 𝑡+\sin 𝑡)}{{\cos }^3𝑡}\right]\end{array}$

$\begin{array}{cc}& (1-{𝑥}^2){𝑦}^{\prime \prime }-𝑥{𝑦}^{\prime }\\ & =(1-{\sin }^2𝑡)\sqrt{\surd 2}(𝑎+𝑏)\left[\frac{{𝑒}^{𝑡\sqrt{\surd 2}}(\sqrt{\surd 2}\cos 𝑡+\sin 𝑡)}{{\cos }^3𝑡}\right]-\sin 𝑡\left[\frac{\sqrt{\surd 2}(𝑎+𝑏){𝑒}^{𝑡\sqrt{\surd 2}}}{\cos 𝑡}\right]\end{array}$

$\begin{array}{c}=({\cos }^2𝑡)\sqrt{\surd 2}(𝑎+𝑏)\left[\frac{{𝑒}^{𝑡\sqrt{\surd 2}}(\sqrt{\surd 2}\cos 𝑡+\sin 𝑡)}{{\cos }^3𝑡}\right]-\sin 𝑡\left[\frac{\sqrt{\surd 2}(𝑎+𝑏){𝑒}^{𝑡\sqrt{\surd 2}}}{\cos 𝑡}\right]\end{array}$

$\begin{array}{cc}& =\frac{\sqrt{\surd 2}(𝑎+𝑏){𝑒}^{𝑡\sqrt{\surd 2}}(\sqrt{\surd 2}\cos 𝑡+\sin 𝑡-\sin 𝑡)}{\cos 𝑡}\\ & =\frac{\sqrt{\surd 2}(𝑎+𝑏){𝑒}^{𝑡\sqrt{\surd 2}}(\sqrt{\surd 2}\cos 𝑡)}{\cos 𝑡}\\ & =2(𝑎{𝑒}^{𝑡\sqrt{\surd 2}}+𝑏{𝑒}^{𝑡\sqrt{\surd 2}})\\ & =2𝑦\\ & \therefore 𝑘=2\end{array}$

$\begin{array}{cc}& \cos {20}^{\circ }+2{\sin }^2{55}^{\circ }-\sqrt{\surd 2}\sin {65}^{\circ }\\ & =\cos {20}^{\circ }+1-\cos 2\left({55}^{\circ }\right)-\sqrt{\surd 2}\sin {65}^{\circ }\\ & \dots [2{\sin }^2𝜃=1-\cos 2𝜃]\\ & =\cos {20}^{\circ }-\cos {110}^{\circ }-\sqrt{\surd 2}\sin {65}^{\circ }+1\\ & =2\sin {65}^{\circ }\sin {45}^{\circ }-\sqrt{\surd 2}\sin {65}^{\circ }+1\\ & =2\sin {65}^{\circ }\left(\frac{1}{\sqrt{\surd 2}}\right)-\sqrt{\surd 2}\sin {65}^{\circ }+1\\ & =\sqrt{\surd 2}\sin {65}^{\circ }-\sqrt{\surd 2}\sin {65}^{\circ }+1\\ & =1\end{array}$