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

Given cartesian equation of line is

$\begin{array}{cc}& 4𝑥-3𝑧+5=0,𝑦=2\\ & \Rightarrow 4𝑥=3𝑧-5,𝑦=2\\ & \Rightarrow 4𝑥=3(𝑧-\frac{5}{3}),𝑦=2\\ & \Rightarrow \frac{𝑥}{3}=\frac{𝑧-\frac{5}{3}}{4},𝑦=2\end{array}$

$\therefore$ The given line passes $(0,2,\frac{5}{3})$ and the direction ratios are proportional to $3,0,4$.

$\therefore$ The vector equation is

$\overline{𝑟}=(2\widehat{𝑗}+\frac{5}{3}\widehat{𝑘})+𝜆(3\widehat{𝑖}+4\widehat{𝑘})$

There are 3 books on physics, 2 books on Chemistry and 4 books on Mathematics. All Physics books and Mathematics books are kept together.

$\therefore$ All Physics books will be considered as 1 unit and all Mathematics books are also considered as 1 unit.

$\therefore$ Physics, Mathematics and 2 Chemistry books can be arranged in 4 ! ways.

Also, 3 Physics and 4 Mathematics books can be arranged themselves in $3!\times 4$ ! ways

$\begin{array}{cc}\therefore \text{ Total number of arrangements }& =4!\times 3!\times 4!\\ & =3456\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ L.H.L. }=\underset{𝑥\to 0^{-}}{lim}\frac{𝑥}{\left|𝑥\right|+{𝑥}^2}\\ & =\underset{𝑥\to 0^{-}}{lim}\frac{𝑥}{-𝑥+{𝑥}^2}\\ & =\underset{𝑥\to 0^{-}}{lim}\frac{1}{𝑥-1}=-1\\ & \text{ R.H.L. }=\underset{𝑥\to 0^{+}}{lim}\frac{𝑥}{\left|𝑥\right|+{𝑥}^2}\\ & =\underset{𝑥\to 0^{+}}{lim}\frac{𝑥}{𝑥+{𝑥}^2}\\ & =\underset{𝑥\to 0^{+}}{lim}\frac{1}{1+𝑥}=1\\ & =\text{ L.H.L. }\ne \text{ R.H.L. }\end{array}\\ & \therefore \text{ Limit does not exist. }\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \bar{a}=2\hat{i}+\hat{j}-2\hat{k}\text{ and }\bar{b}=\hat{i}+\hat{j}\\ & \left|a\right|=\sqrt{\surd 4+1+4}=3\\ & \bar{a}\times \bar{b}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\mid =2\hat{i}-2\hat{j}+\hat{k}\\ \therefore & |\bar{a}\times \bar{b}|=\sqrt{\surd 4+4+1}=3\end{array}\\ & \text{ Angle between }\bar{c}\text{ and }\bar{a}\times \bar{b}\text{ is }\frac{𝜋}{6}\dots \text{ [Given] }\\ & \begin{array}{cc}\therefore & \sin \frac{𝜋}{6}=\frac{\left|\right(\bar{a}\times \bar{b})\times \bar{c}|}{|\bar{a}\times \bar{b}|\left|\bar{c}\right|}\\ & \Rightarrow \frac{1}{2}=\frac{3}{3\times \left|\bar{c}\right|}\\ & \Rightarrow \left|\bar{c}\right|=2\\ & \text{ Now, }|\bar{c}-\bar{a}|=3\\ & \Rightarrow |\bar{c}{|}^2+|\bar{a}{|}^2-2\bar{a}\cdot \bar{c}=9\\ & \Rightarrow 4+9-2\bar{a}\cdot \bar{c}=9\\ & \Rightarrow \bar{a}\cdot \bar{c}=2\end{array}\end{array}$

$\begin{array}{cc}& \cos (2{\cos }^{-1}𝑥+{\sin }^{-1}𝑥)\\ & =\cos [({\sin }^{-1}𝑥+{\cos }^{-1}𝑥)+{\cos }^{-1}𝑥]\\ & =\cos (\frac{𝜋}{2}+{\cos }^{-1}𝑥)\\ & =-\sin ({\cos }^{-1}𝑥)\\ & =-\sin ({\sin }^{-1}\sqrt{\surd (1-{𝑥}^2)})\\ & =-\sqrt{\surd 1-{𝑥}^2}\\ & =-\sqrt{\surd 1-{\left(\frac{1}{5}\right)}^2}\\ & =-\sqrt{\surd \frac{24}{25}}=-\frac{2\sqrt{\surd 6}}{5}\end{array}$

$\begin{array}{cc}& {\tan }^{-1}\left(2𝑥\right)+{\tan }^{-1}\left(3𝑥\right)=\frac{𝜋}{4}\\ & \Rightarrow {\tan }^{-1}\left[\frac{2𝑥+3𝑥}{1-\left(2𝑥\right)\left(3𝑥\right)}\right]=\frac{𝜋}{4}\\ & \Rightarrow \frac{5𝑥}{1-6{𝑥}^2}=\tan \frac{𝜋}{4}=1\\ & \Rightarrow 6{𝑥}^2+5𝑥-1=0\\ & \Rightarrow (𝑥+1)(6𝑥-1)=0\\ & \Rightarrow 𝑥=-1\text{ or }𝑥=\frac{1}{6}\\ & \text{ But, }𝑥\ge 0\\ & \therefore 𝑥=\frac{1}{6}\\ & \therefore A\text{ is a singleton set. }\end{array}$

Given equation of pair of lines is

$\begin{array}{cc}& {𝑥}^2+2ℎ𝑥𝑦+2{𝑦}^2=0\\ \therefore & 𝐴=1,𝐵=2,𝐻=ℎ\end{array}$

Let the slopes of the lines be $m_1,m_2$.

$\begin{array}{cc}\therefore & m_1+m_2=\frac{-2ℎ}{2},m_1{m}_2=\frac{1}{2}\\ & \text{ Given }\frac{m_1}{m_2}=\frac{1}{2}\\ & \Rightarrow m_1=\frac{1}{2}{m}_2\\ & \Rightarrow \frac{1}{2}{m}_2\times m_2=\frac{1}{2}\\ & \Rightarrow m_2^2=1\\ & \Rightarrow m_2=\pm 1\\ \therefore & m_1=\pm \frac{1}{2}.\end{array}$

Now, ${𝑚}_1+{𝑚}_2=\frac{-2ℎ}{2}$

$\therefore$ when $m_1=\frac{1}{2},m_2=1$

$\begin{array}{cc}& {𝑚}_1+{𝑚}_2=\frac{-2ℎ}{2}\\ \therefore & ℎ=\frac{-3}{2}\\ & \text{ When }{𝑚}_1=\frac{-1}{2},{𝑚}_2=-1\\ \therefore & {𝑚}_1+{𝑚}_2=\frac{-2ℎ}{2}\\ & \frac{-1}{2}-1=-\frac{2ℎ}{2}\\ \therefore & ℎ=\frac{3}{2}\end{array}$

$𝑦=e^{4𝑥}{\left(\frac{𝑥-4}{𝑥+3}\right)}^{\frac{3}{4}}$

Taking log on both sides,

$\begin{array}{cc}& \log 𝑦=\log \left[{𝑒}^{4𝑥}{\left(\frac{𝑥-4}{𝑥+3}\right)}^{\frac{3}{4}}\right]\\ & \log 𝑦=\log {𝑒}^{4𝑥}+\log {\left(\frac{𝑥-4}{𝑥+3}\right)}^{\frac{3}{4}}\\ & \log 𝑦=4𝑥\log 𝑒+\frac{3}{4}\log \left(\frac{𝑥-4}{𝑥+3}\right)\\ & \log 𝑦=4𝑥+\frac{3}{4}\log \left(\frac{𝑥-4}{𝑥+3}\right)\end{array}$

Differentiating w.r.to $𝑥$ on both sides,

$\begin{array}{cc}\frac{1}{𝑦}\frac{d𝑦}{d𝑥}& =4+\frac{3}{4}\times \frac{1}{\frac{(𝑥-4)}{(𝑥+3)}}\cdot \frac{d}{d𝑥}\left(\frac{𝑥-4}{𝑥+3}\right)\\ \frac{1}{𝑦}\frac{d𝑦}{d𝑥}& =4+\frac{3}{4}\left(\frac{𝑥+3}{𝑥-4}\right)\times \left(\frac{(𝑥+3)-(𝑥-4)}{(𝑥+3{)}^2}\right)\\ \frac{1}{𝑦}\frac{d𝑦}{d𝑥}& =4+\frac{3}{4}\left(\frac{1}{𝑥-4}\right)\times \left(\frac{𝑥+3-𝑥+4}{(𝑥+3)}\right)\\ \frac{1}{𝑦}\frac{d𝑦}{d𝑥}& =4+\frac{3}{4}\times \left(\frac{1}{𝑥-4}\right)\times \left(\frac{7}{𝑥+3}\right)\\ \therefore \frac{d𝑦}{d𝑥}& =𝑦[4+\frac{21}{4(𝑥-4)(𝑥+3)}]\end{array}$

$\begin{array}{cc}& \text{ Let }I=\int 3^{3^{𝑥}}\cdot 3^{𝑥}d𝑥\\ & \text{ Let }{3}^{𝑥}=t\\ & 3^{𝑥}\log 3d𝑥=\mathrm{dt}\\ & 3^{𝑥}d𝑥=\frac{1}{\log 3}\mathrm{dt}\\ & \begin{array}{cc}\therefore I& =\int 3^t\cdot \frac{1}{\log 3}\cdot \mathrm{dt}\\ & =\frac{1}{\log 3}\int 3^t\mathrm{dt}\\ & =\frac{1}{\log 3}\times \frac{3^t}{\log 3}+c\\ I& =\frac{3^{3^{𝑥}}}{(\log 3{)}^2}+c\end{array}\end{array}$

$\begin{array}{cc}𝑓\left(𝑥\right)& =\frac{1+\cos 𝜋𝑥}{𝜋(1-𝑥{)}^2}\\ \therefore 𝑓\left(1\right)& =\underset{𝑥\to 1}{lim}𝑓\left(𝑥\right)\\ =& \underset{𝑥\to 1}{lim}\frac{1+\cos 𝜋𝑥}{𝜋(1-𝑥{)}^2}\end{array}$

$\begin{array}{cc}& \text{ Put }1-𝑥=ℎ\\ & \Rightarrow 1-h=𝑥\\ & \text{ As }𝑥\to 1,h\to 0\\ & \begin{array}{cc}\therefore 𝑓\left(1\right)& =\underset{ℎ\to 0}{lim}\frac{1+\cos 𝜋(1-ℎ)}{𝜋{ℎ}^2}\\ & =\underset{ℎ\to 0}{lim}\frac{1+\cos (𝜋-𝜋ℎ)}{𝜋{ℎ}^2}\\ & =\underset{ℎ\to 0}{lim}\frac{1-\cos 𝜋ℎ}{𝜋{ℎ}^2}\\ & =\underset{ℎ\to 0}{lim}\frac{2{\sin }^2\frac{𝜋ℎ}{2}}{𝜋{ℎ}^2}\\ & =2\underset{ℎ\to 0}{lim}\frac{{\sin }^2\frac{𝜋ℎ}{2}}{\frac{{𝜋}^2{ℎ}^2}{4}\times 4}\cdot 𝜋\\ & =\frac{2𝜋}{4}\underset{ℎ\to 0}{lim}{\left(\frac{\sin \frac{𝜋ℎ}{2}}{\frac{𝜋ℎ}{2}}\right)}^2\\ \therefore f\left(1\right)=\frac{𝜋}{2}\end{array}\end{array}$

$\begin{array}{cc}& f\left(𝑥\right)=3\sin 𝑥-4{\sin }^3𝑥\\ & \Rightarrow f\left(𝑥\right)=\sin 3𝑥\\ & \Rightarrow f^{\prime }\left(𝑥\right)=3\cos 3𝑥\end{array}$

For $𝑓\left(𝑥\right)$ to be increasing,

$\begin{array}{cc}& f^{\prime }\left(𝑥\right)\ge 0\\ & \Rightarrow 3\cos 3𝑥\ge 0\\ & \Rightarrow \cos 3𝑥\ge 0\\ & \Rightarrow \frac{-𝜋}{2}\le 3𝑥\le \frac{𝜋}{2}\\ & \Rightarrow \frac{-𝜋}{6}\le 𝑥\le \frac{𝜋}{6}\end{array}$

$\therefore$ largest length in which $f\left(𝑥\right)$ is increasing

$=\frac{𝜋}{6}-\left(\frac{-𝜋}{6}\right)=\frac{𝜋}{3}$

$\begin{array}{cc}& \bar{a}\left[\right(\bar{b}\times \bar{c})\times (\bar{a}+\bar{b}+\bar{c}\left)\right]\\ & =\bar{a}\left[\right(\bar{b}\times \bar{a}+\bar{b}\times \bar{b}+\bar{b}\times \bar{c}+\bar{c}\times \bar{a}+\bar{c}\times \bar{b}+\bar{c}\times \bar{c}\left)\right]\\ & =\bar{a}(\bar{b}\times \bar{a})+\bar{a}(\bar{b}\times \bar{c})+\bar{a}(\bar{c}\times \bar{a})+\bar{a}(\bar{c}\times \bar{b})\\ & =\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]+\left[\begin{array}{ccc}\bar{a}& \bar{c}& \bar{b}\end{array}\right]\\ & =\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]-\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & =0\end{array}$

Volume of the parallelopiped formed by vectors is i.e., $𝑉=\mid \begin{array}{ccc}1& 𝑚& 1\\ 0& 1& 𝑚\\ 𝑚& 0& 1\end{array}\mid =1-𝑚+{𝑚}^3$

$\therefore \frac{𝑑𝑉}{𝑑𝑚}=-1+3{𝑚}^2,\frac{{𝑑}^2𝑉}{𝑑{𝑎}^2}=6𝑚$

For $max$. or $min$. of $𝑉,\frac{𝑑𝑉}{𝑑𝑚}=0$

$\therefore m^2=\frac{1}{3}$

$\therefore m=\frac{1}{\sqrt{\surd 3}}$

$\frac{d^2𝑉}{𝑑m^2}=6m>0$ for $𝑚=\frac{1}{\sqrt{\surd 3}}$

$\therefore V$ is minimum for $m=\frac{1}{\sqrt{\surd 3}}$

$\begin{array}{cc}& \overline{𝑎}=\widehat{𝑖}-\widehat{𝑗}+2\widehat{𝑘},\overline{𝑏}=2\widehat{𝑖}+4\widehat{𝑗}+\widehat{𝑘}\text{ and }\overline{𝑐}=𝑚\widehat{𝑖}+\widehat{𝑗}+𝑛\widehat{𝑘}\end{array}$

$\bar{a}$ and $\bar{c}$ are orthogonal

$\begin{array}{cc}\therefore & \overline{𝑎}\cdot \overline{𝑐}=0\\ & 𝑚\left(1\right)+\left(1\right)(-1)+𝑛\left(2\right)=0\\ & 𝑚-1+2𝑛=0\\ & 𝑚+2𝑛=1\text{... (i)}\end{array}$

Also, $\bar{b}$ and $\bar{c}$ are orthogonal

$\begin{array}{cc}\therefore & \bar{b}\cdot \bar{c}=0\\ & 2\left(m\right)+4\left(1\right)+n\left(1\right)=0\\ & 2m+n=-4\text{.... (ii)}\end{array}$

Solving (i) and (ii), we get

$\begin{array}{cc}& 𝑚=-3,𝑛=2\\ \therefore & (𝑚,𝑛)=(-3,2)\end{array}$

Let $I=\int \log (1+𝑥{)}^{1+𝑥}d𝑥$

$I=\int (1+𝑥)\cdot \log (1+𝑥)d𝑥$

Put $1+𝑥=t$

$\begin{array}{cc}\therefore d𝑥& =\mathrm{dt}\\ \therefore I& =\int t\cdot \log t\cdot \mathrm{dt}\\ & =\log t\int t\mathrm{dt}-\int (\frac{d}{\mathrm{dt}}\log t\cdot \int \mathrm{tdt})\mathrm{dt}\\ & =\log t\cdot \frac{t^2}{2}-\int (\frac{1}{t}\times \frac{t^2}{2})\mathrm{dt}\\ & =\log t\cdot \frac{t^2}{2}-\frac{1}{2}\int t\mathrm{dt}\\ & =\log \frac{t^2}{2}-\frac{t^2}{4}+c\\ & =t^2\frac{\log t}{2}-\frac{t^2}{4}+c\\ & =\frac{t^2}{2}[\log t-\frac{1}{2}]+c\\ I& =\frac{(1+𝑥{)}^2}{2}[\log (1+𝑥)-\frac{1}{2}]+c\end{array}$

$\begin{array}{cc}& 𝐼=\int {\left(\frac{𝑥+2}{𝑥+4}\right)}^2{e}^{𝑥}d𝑥\\ & =\int \left(\frac{{𝑥}^2+4𝑥+4}{(𝑥+4{)}^2}\right)e^{𝑥}d𝑥\\ & =\int e^{𝑥}[\frac{𝑥(𝑥+4)}{(𝑥+4{)}^2}+\frac{4}{(𝑥+4{)}^2}]d𝑥\\ & =\int e^{𝑥}[\frac{𝑥}{(𝑥+4)}+\frac{4}{(𝑥+4{)}^2}]d𝑥\\ & =e^{𝑥}\left(\frac{𝑥}{𝑥+4}\right)+c\\ & \cdots \cdot [\int \left(f\right(𝑥)+f^{\prime }(𝑥)e^{𝑥}d𝑥=e^{𝑥}f(𝑥)+c)]\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \frac{1}{𝑏+𝑐}+\frac{1}{𝑐+𝑎}=\frac{3}{𝑎+𝑏+𝑐}\\ & \frac{𝑎+𝑏+𝑐}{𝑏+𝑐}+\frac{𝑎+𝑏+𝑐}{𝑐+𝑎}=\frac{3(𝑎+𝑏+𝑐)}{𝑎+𝑏+𝑐}\\ & \frac{𝑎}{𝑏+𝑐}+1+\frac{𝑏}{𝑐+𝑎}+1=3\\ & \Rightarrow \frac{𝑎}{𝑏+𝑐}+\frac{𝑏}{𝑐+𝑎}=1\\ & \Rightarrow 𝑎(𝑐+𝑎)+𝑏(𝑏+𝑐)=(𝑏+𝑐)(𝑐+𝑎)\\ & \Rightarrow 𝑎𝑐+{𝑎}^2+{𝑏}^2+𝑏𝑐=𝑏𝑐+𝑎𝑏+{𝑐}^2+𝑎𝑐\\ & \Rightarrow {𝑎}^2+{𝑏}^2-{𝑐}^2=𝑎𝑏\text{.... (i)}\end{array}\\ & \therefore \text{ By cosine Rule, }\\ & \begin{array}{cc}& \cos 𝐶=\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}\\ & \cos 𝐶=\frac{𝑎𝑏}{2𝑎𝑏}\text{... [From (i)]}\\ & \Rightarrow \cos 𝐶=\frac{1}{2}\\ & \Rightarrow 𝐶=\frac{𝜋}{3}\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝑧& =\frac{(1+𝑖{)}^2}{𝑎-𝑖}\\ & =\frac{2𝑖}{𝑎-𝑖}\\ & =\frac{2𝑖(𝑎+𝑖)}{{𝑎}^2+1}\dots [i^2=-1]\\ & =\frac{-2+2𝑎𝑖}{{𝑎}^2+1}\\ \therefore \left|𝑧\right|& =\sqrt{\surd \frac{4+4{𝑎}^2}{{({𝑎}^2+1)}^2}}\\ & \Rightarrow \sqrt{\surd \frac{2}{5}}=\frac{2}{\sqrt{\surd {𝑎}^2+1}}\\ & \Rightarrow 𝑎=3\end{array}\\ & \end{array}$

$\begin{array}{cc}\therefore & 𝑧=\frac{-2+2\left(3\right)𝑖}{3^2+1}=\frac{-2+6𝑖}{10}\\ & \Rightarrow 𝑧=\frac{-1+3𝑖}{5}\\ & \Rightarrow \overline{𝑧}=\frac{-1}{5}-\frac{3𝑖}{5}\end{array}$

(i) Probability that the first ball is black and second is red.

Total number of black balls $=6$

Total number of red balls $=4$

Probability of getting black ball in first draw $=\frac{6}{10}$.

Now, number of black balls $=9$ and Total number of balls = 13

Probability of getting red ball in second draw $=\frac{4}{13}$.

ii. Probability that both the balls are red.

Probability of getting red ball in first draw $=\frac{4}{10}$.

Now for second draw,

Number of red balls $=7$ and

Total number of balls $=13$.

$\therefore$ Probability of getting red ball in second draw $=\frac{7}{13}$

$\therefore$ Total probability of drawing red ball

$\begin{array}{cc}& =(\frac{6}{10}\times \frac{4}{13})+(\frac{4}{10}\times \frac{7}{13})\\ & =\frac{24+28}{130}\\ & =\frac{26}{65}\end{array}$

$\begin{array}{cc}f\left(𝑥\right)& =\sin \left|𝑥\right|-\left|𝑥\right|+2(𝑥-𝜋)\cos \left|𝑥\right|\\ \therefore f\left(𝑥\right)& =\sin 𝑥-𝑥+2(𝑥-𝜋)\cos 𝑥,𝑥\ge 0\\ & =-\sin 𝑥+𝑥+2(𝑥-𝜋)\cos 𝑥,𝑥<0\end{array}$

Now,

$\begin{array}{cc}{𝑓}^{\prime }\left(𝑥\right)& =\cos 𝑥-1-2(𝑥-𝜋)\sin 𝑥+2\cos 𝑥,𝑥\ge 0\\ & =-\cos 𝑥+1-2(𝑥+𝜋)\sin 𝑥+2\cos 𝑥,𝑥<0\end{array}$

$\begin{array}{cc}& 𝑓\left(0^{+}\right)=\cos 0-1-2(0-𝜋)\sin 0+2\cos 0=2\\ & 𝑓\left(0^{-}\right)=-\cos 0+1-2(0-𝜋)\sin 0+2\cos 0=2\end{array}$

$\therefore f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)=2$

$\therefore f\left(𝑥\right)$ is differentiable at $𝑥=0$

$\therefore f\left(𝑥\right)$ is differentiable everywhere.

$\therefore k=𝜙$

Given,

$\begin{array}{cc}& 𝑓\left(𝑥\right)=𝑓(𝑎-𝑥)\\ & 𝑔\left(𝑥\right)+𝑔(𝑎-𝑥)=4\end{array}$

Let $I={\int }_a^{0}f\left(𝑥\right)g\left(𝑥\right)\mathrm{dx}$

$\begin{array}{cc}& ={\int }_{𝑎}^0𝑓(𝑎-𝑥)\cdot 𝑔(𝑎-𝑥)𝑑𝑥\\ & ={\int }_{𝑎}^0𝑓\left(𝑥\right)\cdot \left[\right(4-𝑔\left(𝑥\right)]𝑑𝑥\\ & =4{\int }_{𝑎}^0𝑓\left(𝑥\right)𝑑𝑥-{\int }_{𝑎}^0𝑓\left(𝑥\right)𝑔\left(𝑥\right)𝑑𝑥\end{array}$

$\begin{array}{cc}& 𝐼=4{\int }_{𝑎}^0𝑓\left(𝑥\right)𝑑𝑥-𝐼\\ & 2𝐼=4{\int }_2^0𝑓\left(𝑥\right)𝑑𝑥\\ \therefore & 𝐼=2{\int }_2^0𝑓\left(𝑥\right)𝑑𝑥\end{array}$