MHT CET 2024 2nd May Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \text{ Length of tangent segment from }P(1,7)\text{ is }\\ & \begin{array}{cc}\sqrt{\surd 1^2+7^2-5^2}& =5\\ \text{ Required area }& =2\times \text{ Area of }△\mathrm{PQO}\\ & =2\times \frac{1}{2}\times 5\times 5\\ & =25\text{ sq. }\mathrm{units}\end{array}\end{array}$

$\begin{array}{cc}& \mathrm{AX}=B\\ \therefore & ⎡⎣⎢\begin{array}{ccc}1& -1& 2\\ 2& 0& 1\\ 3& 2& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}a\\ b\\ c\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}3\\ 1\\ 4\end{array}⎤⎦⎥\\ \therefore & a-b+2c=3\text{.... (i)}\\ & 2a+c=1\text{... (ii)}\\ & 3a+2b+c=4\text{.... (iii)}\end{array}$

Solving (i), (ii) and (iii), we get

$\begin{array}{cc}& \begin{array}{cc}𝑎=-1,𝑏=2,𝑐& =3\\ \therefore 2𝑎-3𝑏+4𝑐& =2(-1)-3\left(2\right)+4\left(3\right)\\ & =4\end{array}\end{array}$

Let p be the probability of success.

$\begin{array}{cc}& P(X=1)=0.4096\text{ and }P(X=2)=0.2048\\ & \Rightarrow {}^5{C}_1{p}^1{q}^4=0.4096\text{ and }{}^5{C}_2{p}^2{q}^3=0.2048\\ & \Rightarrow 5{\mathrm{pq}}^4=0.4096\text{ and }10p^2{q}^3=0.2048\\ & \Rightarrow \frac{10p^2{q}^3}{5{\mathrm{pq}}^4}=\frac{0.2048}{0.4096}\\ & \Rightarrow \frac{2𝑝}{𝑞}=\frac{1}{2}\\ & \Rightarrow 4p=q\\ & \Rightarrow 4p=1-p\\ & \Rightarrow 5p=1\\ & \Rightarrow p=\frac{1}{5}\\ & \Rightarrow q=1-\frac{1}{5}=\frac{4}{5}\\ & 𝑃(𝑋=3)={}^5{𝐶}_3{𝑝}^3{𝑞}^2\\ & =10{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^2\\ & =10\times \frac{1}{125}\times \frac{16}{25}=\frac{32}{625}\end{array}$

Let $\bar{a}=5\widehat{𝑖}+2\widehat{𝑗}+6\widehat{𝑘},\overline{𝑏}=2\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘}$ and $\overline{𝑐}=\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘}$

Then, required unit vectors are given by

$\overline{𝛼}=\pm \frac{\bar{a}\times (\bar{b}\times \bar{c})}{|\bar{a}\times (\bar{b}\times \bar{c}\left)\right|}$

$\begin{array}{cc}& \text{ Now, }\bar{a}\times (\bar{b}\times \bar{c})=(\bar{a}\cdot \bar{c})\bar{b}-(\bar{a}\cdot \bar{b})\bar{c}\\ & =9(2\hat{i}+\hat{j}+\hat{k})-18(\hat{i}-\hat{j}+\hat{k})\\ & =27\hat{j}-9\hat{k}\\ \therefore & |\bar{a}\times (\bar{b}\times \bar{c}\left)\right|=\sqrt{\surd 729+81}=\sqrt{\surd 810}=9\sqrt{\surd 10}\end{array}$

Hence, required unit vectors are

$\overline{𝛼}=\pm \frac{27\hat{j}-9\hat{k}}{9\sqrt{\surd 10}}=\pm \frac{3\hat{j}-\hat{k}}{\sqrt{\surd 10}}$

$\begin{array}{cc}E\left(X\right)& =\sum {𝑥}_i\cdot P\left({𝑥}_i\right)\\ & =0\left(0.4\right)+1\left(0.3\right)+2\left(0.1\right)+3\left(0.1\right)+4\left(0.1\right)\\ & =1.2\\ \mathrm{Var}\left(X\right)& =E\left(X^2\right)-\left[E\right(X){]}^2\\ & =0^2\left(0.4\right)+1^2\left(0.3\right)+2^2\left(0.1\right)+3^2\left(0.1\right)+4^2\left(0.1\right)-(1.2{)}^2\\ & =0.3+0.4+0.9+1.6-1.44\\ & =1.76\end{array}$

Let $I=\int \frac{𝑥+1}{\sqrt{\surd 2𝑥-1}}d𝑥$

Put $2𝑥-1=t^2\Rightarrow 𝑥+1=\frac{t^2+3}{2}$

$\Rightarrow \mathrm{dx}=\mathrm{tdt}$

$\begin{array}{cc}\therefore 𝐼& =\int \frac{\left(\frac{{𝑡}^2+3}{2}\right)\mathrm{tdt}}{t}\\ & =\frac{{𝑡}^3}{6}+\frac{3}{2}𝑡+𝑐\\ & =\frac{𝑡}{2}\left(\frac{{𝑡}^2+9}{3}\right)+𝑐\\ & =\frac{\sqrt{\surd 2𝑥-1}}{2}\left(\frac{2𝑥-1+9}{3}\right)+𝑐\\ & =\sqrt{\surd 2𝑥-1}\left(\frac{𝑥+4}{3}\right)+𝑐\end{array}$

Comparing with $f\left(𝑥\right)\sqrt{\surd 2𝑥-1}+c$, we get

$𝑓\left(𝑥\right)=\frac{𝑥+4}{3}$

$\begin{array}{cc}& {\sin }^2𝛼+{\sin }^2𝛽-{\sin }^2𝛾\\ & ={\sin }^2𝛼+\sin (𝛽+𝛾)\sin (𝛽-𝛾)\\ & ={\sin }^2𝛼+\sin (𝜋-𝛼)\sin (𝛽-𝛾)\\ & =\sin 𝛼[\sin 𝛼+\sin (𝛽-𝛾\left)\right]\\ & =\sin 𝛼[\sin (𝛽+𝛾)+\sin (𝛽-𝛾\left)\right]\\ & =2\sin 𝛼\sin 𝛽\cos 𝛾\end{array}$

$\begin{array}{cc}& \bar{R}\times \bar{B}=\bar{C}\times \bar{B}\\ & \Rightarrow \bar{A}\times (\bar{R}\times \bar{B})=\bar{A}\times (\bar{C}\times \bar{B})\\ & \Rightarrow (\bar{A}\cdot \bar{B})\bar{R}-(\bar{A}\cdot \bar{R})\bar{B}=(\bar{A}\cdot \bar{B})\bar{C}-(\bar{A}\cdot \bar{C})\bar{B}\\ & \Rightarrow (\bar{A}\cdot \bar{B})\bar{R}-0=(\bar{A}\cdot \bar{B})\bar{C}-(\bar{A}\cdot \bar{C})\bar{B}\\ & \Rightarrow \bar{R}=\bar{C}-\left(\frac{\bar{A}\cdot \bar{C}}{\bar{A}\cdot \bar{B}}\right)\bar{B}\\ & \bar{A}\cdot \bar{C}=(2\hat{i}+\hat{k})\cdot (4\hat{i}-3\hat{j}+7\hat{k})\\ & =2\left(4\right)+0(-3)+1\left(7\right)\\ & =15\\ & \overline{𝐴}\cdot \overline{𝐵}=(2\widehat{𝑖}+\widehat{𝑘})\cdot (\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘})\\ & =2\left(1\right)+0\left(1\right)+1\left(1\right)\\ & =3\\ & \therefore \bar{R}=4\hat{i}-3\hat{j}+7\hat{k}-\left(\frac{15}{3}\right)(\hat{i}+\hat{j}+\hat{k})\\ & =4\hat{i}-3\hat{j}+7\hat{k}-5(\hat{i}+\hat{j}+\hat{k})\\ & =-\widehat{𝑖}-8\widehat{𝑗}+2\widehat{𝑘}\end{array}$

The vector equations of the given lines are $\overline{𝑟}=\widehat{𝑗}+𝜆(3\widehat{𝑖}-2\widehat{𝑗}+\widehat{𝑘})$ and $\bar{r}=-4\hat{i}+3\hat{j}-2\hat{k}+𝜇(3\hat{i}-2\hat{j}+\hat{k})$

The distance between the parallel lines $\overline{𝑟}={\overline{𝑎}}_1+𝜆\overline{𝑏}$ and $\overline{𝑟}={\overline{𝑎}}_2+𝜇\overline{𝑏}$ is given by

$𝑑=\mid \frac{({\overline{𝑎}}_2-{\overline{𝑎}}_1)\times \overline{𝑏}}{\left|\overline{𝑏}\right|}\mid$

Here, ${\overline{𝑎}}_1=\widehat{𝑗},{\overline{𝑎}}_2=-4\widehat{𝑖}+3\widehat{𝑗}-2\widehat{𝑘},\overline{𝑏}=3\widehat{𝑖}-2\widehat{𝑗}+\widehat{𝑘}$

$\begin{array}{cc}& \therefore {\bar{a}}_2-{\bar{a}}_1=-4\hat{i}+2\hat{j}-2\hat{k}\\ & ({\overline{𝑎}}_2-{\overline{𝑎}}_1)\times \overline{𝑏}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ -4& 2& -2\\ 3& -2& 1\end{array}\mid \\ & =-2\widehat{𝑖}-2\widehat{𝑗}+2\widehat{𝑘}\\ & \left|\overline{𝑏}\right|=\sqrt{\surd 9+4+1}=\sqrt{\surd 14}\\ & \therefore d=\mid \frac{-2\hat{i}-2\hat{j}+2\hat{k}}{\sqrt{\surd 14}}\mid \\ & =\sqrt{\surd \frac{4+4+4}{14}}\\ & =\sqrt{\surd \frac{12}{14}}=\sqrt{\surd \frac{6}{7}}\text{ units }\end{array}$

$\begin{array}{cc}𝑓\left(𝑥\right)& ={\log }_{{𝑥}^2}({\log }_{𝑒}𝑥)\\ & =\frac{\log ({\log }_{𝑒}𝑥)}{\log {𝑥}^2}\\ & =\frac{\log ({\log }_e𝑥)}{2\log 𝑥}\end{array}$

$\begin{array}{cc}\therefore & f^{\prime }\left(𝑥\right)=\frac{1}{2}\left[\frac{\log 𝑥\cdot \frac{1}{{\log }_e𝑥}\cdot \frac{d}{d𝑥}(\log 𝑥)-\log ({\log }_e𝑥)\cdot \frac{1}{𝑥}}{(\log 𝑥{)}^2}\right]\\ \therefore & f^{\prime }\left(𝑥\right)=\frac{1}{2}\left[\frac{\frac{1}{𝑥}-\frac{\log ({\log }_e𝑥)}{𝑥}}{{\left({\log }_{𝑥}\right)}^2}\right]\\ \therefore & f^{\prime }\left(e\right)=\frac{1}{2}\left[\frac{\frac{1}{(-0}}{(1{)}^2}\right]=\frac{1}{2e}\end{array}$

$\begin{array}{cc}& \text{ Let }𝜃\text{ be the temperature of the body at any time }𝑡\text{. }\\ & \begin{array}{cc}& \therefore \frac{d𝜃}{\mathrm{dt}}\propto (𝜃-20)\\ & \Rightarrow \frac{d𝜃}{\mathrm{dt}}=-k(𝜃-20),k>0\end{array}\end{array}$

$\begin{array}{cc}& \text{ Integrating on both sides, we get }\\ & \log |𝜃-20|=-\mathrm{kt}+c\\ & \text{ When }t=0,𝜃={100}^{\circ }\\ \therefore & \log 80=-k\left(0\right)+c\\ & \Rightarrow c=\log 80\\ \therefore & \log |𝜃-20|=-\mathrm{kt}+\log 80\text{.... (i)}\\ & \text{ When }t=15,𝜃={60}^{\circ }\\ \therefore & \log 40=-15k+\log 80\\ & \Rightarrow k=\frac{-1}{15}\log \frac{1}{2}\end{array}.$

$\begin{array}{cc}\therefore & \log |𝜃-20|=\frac{t}{15}\log \frac{1}{2}+\log 80\dots \left[\text{ From (i) }\right]\\ & \text{ When }t=1\text{ hour }=60\text{ minutes, }\\ & \log |𝜃-20|=\frac{60}{15}\log \frac{1}{2}+\log 80\\ & \Rightarrow \log \left(\frac{𝜃-20}{80}\right)=4\log \frac{1}{2}\\ & \Rightarrow \frac{𝜃-20}{80}={\left(\frac{1}{2}\right)}^4\\ & \Rightarrow 𝜃=5+20={25}^{\circ }C\end{array}$

$\begin{array}{cc}& \text{ Since }f\left(𝑥\right)\text{ is continuous at }𝑥=\frac{𝜋}{2}\text{. }\\ & \begin{array}{cc}& \therefore f\left(\frac{𝜋}{2}\right)=\underset{𝑥\to \frac{𝜋}{2}}{lim}{\left(\frac{4}{5}\right)}^{\frac{\ln 4𝑥}{\mathrm{lan}5𝑥}}\\ & \Rightarrow 𝑘+\frac{2}{5}=\underset{𝑥\to \frac{𝜋}{2}}{lim}{\left(\frac{4}{5}\right)}^{\underset{𝑥\to \frac{𝜋}{2}}{lim}(\ln 4𝑥\cot 5𝑥)}\\ & \Rightarrow 𝑘+\frac{2}{5}={\left(\frac{4}{5}\right)}^0=1\\ & \Rightarrow 𝑘=\frac{3}{5}\end{array}\end{array}$

$\begin{array}{cc}& 𝑦=𝑥\log 𝑥\text{.... (i)}\\ \therefore & \frac{d𝑦}{d𝑥}=1+\log 𝑥\end{array}$

Slope of the normal $=-\frac{1}{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}=\frac{-1}{1+\log 𝑥}$

Slope of the given line is 1.

Since the normal is parallel to the given line.

$\begin{array}{cc}\therefore & \frac{-1}{1+\log 𝑥}=1\\ & \Rightarrow \log 𝑥=-2\\ & \Rightarrow 𝑥=e^{-2}\end{array}$

From (i), $𝑦=-2e^{-2}$

$\therefore$ Co-ordinates of the point are $(e^{-2},-2e^{-2})$.

Let $I=\underset{\sqrt{\surd \log 2}}{\overset{\sqrt{\surd \log 3}}{\int }}\frac{𝑥\sin {𝑥}^2}{\sin {𝑥}^2+\sin (\log 6-{𝑥}^2)}𝑑𝑥$

$\begin{array}{cc}& \text{ Put }{𝑥}^2=𝑡\Rightarrow 2𝑥𝑑𝑥=𝑑𝑡\\ \therefore & 𝐼=\frac{1}{2}\underset{\log 2}{\overset{\log 3}{\int }}\frac{\sin 𝑡}{\sin 𝑡+\sin (\log 6-𝑡)}𝑑𝑡\text{.... (i)}\\ \therefore & 𝐼=\frac{1}{2}\underset{\log 2}{\overset{\log 3}{\int }}\frac{\sin (\log 6-𝑡)}{\sin (\log 6-𝑡)+\sin 𝑡}𝑑𝑡\text{.... (ii)}\end{array}$

$\dots [\because \underset{𝑎}{\overset{𝑏}{\int }}𝑓(𝑥)𝑑𝑥=\underset{𝑎}{\overset{𝑏}{\int }}𝑓(𝑎+𝑏-𝑥\left)𝑑𝑥\right]$

$\begin{array}{cc}& \text{ Adding (i) and (ii), we get }\\ & \begin{array}{cc}& 2I=\frac{1}{2}\underset{\log 2}{\overset{\log 3}{\int }}\mathrm{dt}=\frac{1}{2}(\log 3-\log 2)=\frac{1}{2}\log \left(\frac{3}{2}\right)\\ & \Rightarrow 𝐼=\frac{1}{4}\log \left(\frac{3}{2}\right)\end{array}\end{array}$

Let $I=\int \frac{1}{{𝑥}^2{({𝑥}^4+1)}^{\frac{3}{4}}}\mathrm{dx}=\int \frac{d𝑥}{{𝑥}^5{(1+\frac{1}{{𝑥}^4})}^{\frac{3}{4}}}$

Put $1+\frac{1}{{𝑥}^4}=t\Rightarrow \frac{-4}{{𝑥}^5}d𝑥=\mathrm{dt}$

$\begin{array}{cc}\therefore 𝐼& =-\frac{1}{4}\int \frac{𝑑𝑡}{{𝑡}^{\frac{3}{4}}}=-\frac{1}{4}\times 4{𝑡}^{\frac{1}{4}}+𝑐=-{𝑡}^{\frac{1}{4}}+𝑐\\ & =-{(1+\frac{1}{{𝑥}^4})}^{\frac{1}{4}}+𝑐\end{array}$

$\begin{array}{cc}& \text{ Total number of ways }=3\times 3\times 3=27\\ & \text{ Favourable number of cases }=3\\ \therefore & \text{ Required probability }=\frac{3}{27}=\frac{1}{9}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}f\left(𝑥\right)& ={\log }_e𝑥\\ \therefore f\left(1\right)& ={\log }_{e}1=0\\ f\left(3\right)& ={\log }_{e}3\text{ and }{f}^{\prime }\left(𝑥\right)=\frac{1}{𝑥}\end{array}\\ & \text{ By Lagrange’s mean value theorem, }\\ & \begin{array}{cc}& f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}\\ & \Rightarrow \frac{1}{c}=\frac{{\log }_{c}3-0}{2}\Rightarrow c=\frac{2}{{\log }_{e}3}\Rightarrow c=2{\log }_{3}e\end{array}\end{array}$

Take a test point $(10,40)$ which lies within the feasible region.

$\begin{array}{cc}& \text{ Since }10+2\left(40\right)=90\ge 50\\ & 2\left(10\right)+40=60\le 100\\ & 2\left(10\right)-40=-20\le 0\end{array}$

$\therefore 𝑥+2𝑦\ge 50,2𝑥+𝑦\le 100,2𝑥-𝑦\le 0,𝑥,𝑦\ge 0$

Slope of $𝑥+3𝑦=4$ is $m_1=-\frac{1}{3}$

Slope of $6𝑥-2𝑦=7$ is ${𝑚}_2=3$

Here, ${𝑚}_1\cdot {𝑚}_2=-1$

$\therefore$ The diagonals are perpendicular to each other.

$\therefore$ Parallelogram ABCD is a rhombus.

$\begin{array}{cc}& 𝑥=\sec 𝜃-\cos 𝜃\text{ and }𝑦={\sec }^{10}𝜃-{\cos }^{10}𝜃\\ & \therefore \frac{d𝑥}{d𝜃}=\sec 𝜃\tan 𝜃+\sin 𝜃\text{ and }\\ & \frac{𝑑𝑦}{𝑑𝜃}=10{\sec }^9𝜃\cdot \sec 𝜃\tan 𝜃-10{\cos }^9𝜃\cdot (-\sin 𝜃)\\ & =10{\sec }^{10}𝜃\tan 𝜃+10{\cos }^9𝜃\sin 𝜃\\ & \therefore \frac{d𝑦}{d𝑥}=\frac{\frac{d𝑦}{d𝜃}}{\frac{d𝑥}{d𝜃}}=\frac{10{\sec }^{10}𝜃\tan 𝜃+10{\cos }^9𝜃\sin 𝜃}{\sec 𝜃\tan 𝜃+\sin 𝜃}\\ & =\frac{10({\sec }^{10}𝜃+{\cos }^{10}𝜃)}{\sec 𝜃+\cos 𝜃}\\ & \therefore {\left(\frac{d𝑦}{d𝑥}\right)}^2=\frac{{10}^2{({\sec }^{10}𝜃+{\cos }^{10}𝜃)}^2}{(\sec 𝜃+\cos 𝜃{)}^2}\\ & =\frac{100[{({\sec }^{10}𝜃-{\cos }^{10}𝜃)}^2+4{\sec }^{10}𝜃{\cos }^{10}𝜃]}{(\sec 𝜃-\cos 𝜃{)}^2+4\sec 𝜃\cos 𝜃}\\ & \therefore {\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^2=\frac{100({𝑦}^2+4)}{{𝑥}^2+4}\\ & \Rightarrow ({𝑥}^2+4){\left(\frac{d𝑦}{d𝑥}\right)}^2=100({𝑦}^2+4)\Rightarrow k=100\end{array}$

$\begin{array}{cc}& \left(\frac{5+e^{𝑥}}{2+𝑦}\right)\frac{d𝑦}{d𝑥}+e^{𝑥}=0\\ & \Rightarrow \frac{d𝑦}{2+𝑦}=\frac{-e^{𝑥}}{5+e^{𝑥}}d𝑥\end{array}$

Integrating on both sides, we get

$\log |2+𝑦|=-\log |5+e^{𝑥}|+\log \left|c\right|$

$\Rightarrow \log |2+𝑦|=\log \mid \frac{c}{5+e^{𝑥}}\mid$

Since $𝑦\left(0\right)=1$ i.e., $𝑦=1$ when $𝑥=0$

$\begin{array}{cc}\therefore \log 3& =\log \mid \frac{𝑐}{6}\mid \\ \Rightarrow 3& =\frac{𝑐}{6}\\ \Rightarrow 𝑐& =18\end{array}$

$\begin{array}{cc}& \therefore \log |2+𝑦|=\log \mid \frac{18}{5+e^{𝑥}}\mid \\ & \Rightarrow 2+𝑦=\frac{18}{5+e^{𝑥}}\text{...[From (i)]}\\ & \Rightarrow 𝑦=\frac{18}{5+e^{𝑥}}-2\end{array}$

$\begin{array}{cc}\Rightarrow 𝑦(\log 13)& =\frac{18}{5+e^{\log 13}}-2\\ & =\frac{18}{5+13}-2\\ & =-1\end{array}$

M is the midpoint

$M=(2,3,4)$

d.r.s of $𝑃𝑄:2,2,-2$

Equation of plane is

$\begin{array}{cc}& 2(𝑥-2)+2(𝑦-3)-2(𝑧-4)=0\\ & \Rightarrow 2𝑥+2𝑦-2𝑧-2=0\\ & \Rightarrow 𝑥+𝑦-𝑧-1=0\end{array}$