MHT CET 2024 11th May Evening Shift (Mathematics) Question Paper with Solutions

There are 6 letters.

M repeats 2 times,

A repeats 3 times.

We first arrange all letters except 2 M's in $\frac{4!}{3!}$ $=4$ ways

These 4 letters create 5 gaps, where we can arrange $2M^{\prime }$ s in $\frac{{}^5{p}_2}{2!}=10$ ways

$\therefore$ Required number of arrangements $=4\times 10=40$

$\begin{array}{cc}P\left(E\right)& =P\left(X\text{ is a prime number }\right)\\ & =P(X=2\text{ or }X=3\text{ or }X=5\text{ or }X=7)\\ & =P(X=2)+P(X=3)+P(X=5)+P(X=7)\\ & =0.23+0.10+0.20+0.07\\ & =0.6\end{array}$

$\begin{array}{cc}& \begin{array}{cc}P\left(F\right)& =P(X<4)\\ & =P(X=1)+P(X=2)+P(X=3)\\ & =0.15+0.23+0.10\\ & =0.48\end{array}\\ & \begin{array}{cc}P(E\cap F)& =P\left(X\text{ is a prime number less than }4\right)\\ & =P(X=2)+P(X=3)\\ & =0.23+0.10\\ & =0.33\end{array}\\ & \begin{array}{cc}𝑃(𝐸\cup 𝐹)& =𝑃\left(𝐸\right)+𝑃\left(𝐹\right)-𝑃(𝐸\cap 𝐹)\\ & =0.6+0.48-0.33\\ & =0.75\end{array}\end{array}$

$\begin{array}{cc}\text{ Let }I& =\int \frac{𝑥+1}{𝑥{(1+𝑥e^{𝑥})}^2}d𝑥\\ & =\int \frac{(𝑥+1)e^{𝑥}}{𝑥e^{𝑥}{(1+𝑥e^{𝑥})}^2}d𝑥\end{array}$

$\begin{array}{cc}& \text{ Let }𝑥e^{𝑥}=t\\ & \therefore (e^{𝑥}+𝑥e^{𝑥})d𝑥=\mathrm{dt}\\ & \therefore 𝐼=\int \frac{\mathrm{dt}}{t(1+t{)}^2}\\ & =\int \frac{(1+t)-t}{t(1+t{)}^2}\mathrm{dt}\\ & =\int \frac{1}{𝑡}\mathrm{dt}-\int \frac{1}{(1-t)}\mathrm{dt}-\int \frac{1}{(1+t{)}^2}\mathrm{dt}\\ & =\log \left|t\right|-\log |1+t|+\frac{1}{(1+t)}+c\\ & =\log \mid \frac{𝑥e^{𝑥}}{1+𝑥e^{𝑥}}\mid +\frac{1}{(1+𝑥e^{𝑥})}+c\end{array}$

Given that half of the ice melts in 15 minutes.

$\therefore$ In 30 minutes, ${\left(\frac{1}{4}\right)}^h$ of the ice will melt.

$\therefore k=\frac{1}{4}$

Let p : two numbers are not equal.

$q:$ squares of two numbers are not equal.

Given statement is $p\to q$.

$\therefore$ Contrapositive is $\sim 𝑞<text>\to </text><text>\sim </text>𝑝$.

i.e., If squares of two numbers are equal, then the numbers are equal.

$\begin{array}{cc}& f\left(𝑥\right)=𝑥+\frac{1}{𝑥}\\ & \text{ let }𝑦=𝑥+\frac{1}{𝑥}\\ \therefore & 𝑥𝑦={𝑥}^2+1\\ \therefore & {𝑥}^2-𝑥𝑦+1=0\end{array}$

$\begin{array}{cc}\therefore & 𝑥=\frac{𝑦\pm \sqrt{\surd {𝑦}^2-4}}{2}\\ \therefore & 𝑥=\frac{𝑦+\sqrt{\surd {𝑦}^2-4}}{2}\dots [\because 𝑥\in [1,\infty \left)\right]\\ \therefore & f^{-1}\left(𝑥\right)=\frac{𝑥+\sqrt{\surd {𝑥}^2-4}}{2}\end{array}$

Let the angles of the triangle be $5𝑥,𝑥,6𝑥$

$\begin{array}{cc}\therefore & 5𝑥+𝑥+6𝑥={180}^{\circ }\\ \therefore & 12𝑥={180}^{\circ }\\ \therefore & 𝑥={15}^{\circ }\end{array}$

$\therefore$ Three angles are ${75}^{\circ },{15}^{\circ },{90}^{\circ }$.

$\begin{array}{cc}& \frac{\sin {75}^{\circ }}{a}=\frac{\sin {15}^{\circ }}{b}=\frac{\sin {90}^{\circ }}{c}=k\\ \therefore & \frac{b}{c}=\frac{\frac{\sin {15}^{\circ }}{k}}{\frac{\sin {90}^{\circ }}{k}}=\frac{\frac{\sqrt{\surd 3}-1}{2\sqrt{\surd 2}}}{1}\end{array}$

$\therefore \text{ Required ratio }=b:c=\sqrt{\surd 3}-1:2\sqrt{\surd 2}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑥\frac{d𝑦}{d𝑥}+𝑦=𝑥\log 𝑥\\ \therefore & \frac{d𝑦}{d𝑥}+\frac{1}{𝑥}𝑦=\log 𝑥\\ & \text{ Here, }P\left(𝑥\right)=\frac{1}{𝑥},Q\left(𝑥\right)=\log 𝑥\\ \therefore & \text{ I.F. }=e^{\int P\left(𝑥\right)\mathrm{dx}}=e^{\int \frac{1}{𝑥}\mathrm{dx}}=e^{\log 𝑥}=𝑥\\ \therefore & 𝑦\left(\text{ I.F. }\right)=\int Q\left(\text{ I.F. }\right)d𝑥+c\\ \therefore & 𝑥𝑦=\int 𝑥\log 𝑥d𝑥+c\\ \therefore & 𝑥𝑦=\frac{{𝑥}^2}{2}\log 𝑥-(\int \frac{{𝑥}^2}{2}\times \frac{1}{𝑥}d𝑥)+c\\ \therefore & 𝑥𝑦=\frac{{𝑥}^2}{2}\log 𝑥-\frac{1}{2}\int 𝑥d𝑥+c\\ \therefore & 𝑥𝑦=\frac{{𝑥}^2}{2}\log 𝑥-\frac{1}{4}{𝑥}^2+c\text{.... (i)}\end{array}\\ & \text{ Given that }2\left(𝑦\right(2\left)\right)=\log 4-1\\ & \therefore 𝑦\left(2\right)=\log 2-\frac{1}{2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\therefore & \text{ From equation (i), we get }\\ & 2(\log 2-\frac{1}{2})=\frac{(2{)}^2}{2}\log 2-\frac{1}{4}(2{)}^2+𝑐\\ \therefore & 2\log 2-1=2\log 2-1+𝑐\\ \therefore & 𝑐=0\text{... (ii)}\end{array}\\ & \therefore \text{ From (i) and (ii), we get }\\ & 𝑦\left(e\right)=\frac{e}{2}-\frac{e}{4}=\frac{e}{4}\end{array}$

$\begin{array}{cc}& 𝑦-𝑥=2\text{ and }{𝑥}^2=𝑦\\ \therefore & {𝑥}^2-𝑥-2=0\\ \therefore & (𝑥-2)(𝑥+1)=0\\ \therefore & 𝑥=2\text{ or }𝑥=-1\\ \therefore & 𝑦=4,𝑦=1\end{array}$

$\therefore$ Points of intersection of the two curves are $(2,4)$ and $(-1,1)$

$\begin{array}{cc}\text{ Required area }& =\underset{-1}{\overset{2}{\int }}(2+𝑥)-{𝑥}^{2}d𝑥\\ & =2[𝑥{]}_2^{-1}+\frac{1}{2}{\left[{𝑥}^2\right]}_2^{-1}-\frac{1}{3}{\left[{𝑥}^3\right]}_2^{-1}\\ & =6+\frac{3}{2}-3=\frac{9}{2}\text{ sq. units }\end{array}$

$\begin{array}{cc}& \bar{a}\times (\bar{b}\times \bar{c})=\frac{\sqrt{\surd 3}}{2}(\bar{b}+\bar{c})\\ & \Rightarrow (\bar{a}\cdot \bar{c})\bar{b}-(\bar{a}\cdot \bar{b})\bar{c}=\left(\frac{\sqrt{\surd 3}}{2}\right)\bar{b}+\left(\frac{\sqrt{\surd 3}}{2}\right)\bar{c}\\ & \Rightarrow \bar{a}\cdot \bar{c}=\frac{\sqrt{\surd 3}}{2}\text{ and }\bar{a}\cdot \bar{b}=\frac{-\sqrt{\surd 3}}{2}\\ & \Rightarrow \left|\bar{a}\right|\left|\bar{b}\right|\cos 𝜃=\frac{-\sqrt{\surd 3}}{2}\\ & \Rightarrow \cos 𝜃=\frac{-\sqrt{\surd 3}}{2}=\cos \frac{5𝜋}{6}\\ & \Rightarrow 𝜃=\frac{5𝜋}{6}\end{array}$

$\begin{array}{cc}& \underset{𝑥\to 0}{lim}\frac{𝑥\tan 2𝑥-2𝑥\tan 𝑥}{(1-\cos 2𝑥{)}^2}\\ & =\underset{𝑥\to 0}{lim}\frac{𝑥(\tan 2𝑥-2\tan 𝑥)}{{(2{\sin }^2𝑥)}^2}\end{array}$

$=\underset{𝑥\to 0}{lim}\frac{𝑥(\tan 2𝑥-2\tan 𝑥)}{4{\sin }^4𝑥}$

$=\frac{1}{4}\underset{𝑥\to 0}{lim}\frac{𝑥⎧⎩⎨⎪⎪\begin{array}{c}(2𝑥+\frac{1}{3}(2𝑥{)}^3+\frac{2}{15}(2𝑥{)}^5+\dots )\\ -2(𝑥+\frac{{𝑥}^3}{3}+\frac{2}{15}{𝑥}^5+\dots )\end{array}⎫⎭⎬⎪⎪}{{𝑥}^4{(1-\frac{{𝑥}^2}{3!}+\frac{{𝑥}^4}{5!}-\dots )}^4}$

$=\frac{1}{4}(\frac{8}{3}-\frac{2}{3})=\frac{2}{4}=\frac{1}{2}$

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}=\frac{a}{𝑥}+2b𝑥+1\Rightarrow {\left(\frac{d𝑦}{d𝑥}\right)}_{𝑥=-1}=-a-2b+1=0\\ & \Rightarrow a=-2b+1\\ & \text{ and }{\left(\frac{d𝑦}{d𝑥}\right)}_{𝑥=2}=\frac{a}{2}+4b+1=0\\ & \Rightarrow \frac{-2b+1}{2}+4b+1=0\\ & \Rightarrow -b+4b+\frac{3}{2}=0\\ & \Rightarrow 3b=\frac{-3}{2}\Rightarrow b=\frac{-1}{2}\text{ and }a=2\\ & \Rightarrow (\frac{a}{b}+\frac{b}{a})=\frac{-17}{4}\end{array}$

Here, $\widehat{𝑎}\cdot \widehat{𝑏}=0$

$\therefore \hat{a}$ and $\hat{b}$ are perpendicular unit vectors.

Now, $(2\widehat{𝑎}-\widehat{𝑏})\cdot \left\{\right(\widehat{𝑎}\times \widehat{𝑏})\times (\widehat{𝑎}+2\widehat{𝑏}\left)\right\}$

$\begin{array}{cc}& =\left[\begin{array}{c}2\widehat{𝑎}-\widehat{𝑏}\widehat{𝑎}\times \widehat{𝑏}\widehat{𝑎}+2\widehat{𝑏}\end{array}\right]\\ & =-[\begin{array}{cc}\widehat{𝑎}\times \widehat{𝑏}& 2\widehat{𝑎}-\widehat{𝑏}\widehat{𝑎}+2\widehat{𝑏}]\end{array}\\ & =-(\widehat{𝑎}\times \widehat{𝑏})\cdot \left\{\right(2\widehat{𝑎}-\widehat{𝑏})\times (\widehat{𝑎}+2\widehat{𝑏}\left)\right\}\\ & =-(\widehat{𝑎}\times \widehat{𝑏})\cdot 5(\widehat{𝑎}\times \widehat{𝑏})\\ & =-5|\widehat{𝑎}\times \widehat{𝑏}|=-5\left|\widehat{𝑎}{|}^2\right|\widehat{𝑏}{|}^2\dots [\because \widehat{𝑎}⟂\widehat{𝑏}]\\ & =-5\cdots [\because |\widehat{𝑎}|=|\widehat{𝑏}|=1]\end{array}$

Let $𝑦={\log }_{{𝑥}^2}(\log 𝑥)$

$\therefore {𝑥}^{2𝑦}=\log 𝑥$

Differentiating w.r.t. $𝑥$, we get

$\begin{array}{cc}& {𝑥}^{2𝑦}(\log 𝑥)\times 2\times \frac{d𝑦}{d𝑥}=\frac{1}{𝑥}\\ \therefore & (\log 𝑥{)}^2\times \frac{d𝑦}{d𝑥}=\frac{1}{2𝑥}\dots [\because {𝑥}^{2𝑦}=\log 𝑥]\end{array}$

At $𝑥=𝑒,f^{\prime }\left(𝑥\right)=\frac{d𝑦}{d𝑥}=\frac{1}{2e}$

Let $𝐼={\int }_{\frac{𝜋}{4}}^0\log (1+\tan 𝜃)𝑑𝜃$

$\begin{array}{cc}& ={\int }_{\frac{𝜋}{4}}^0\log [1+\tan (\frac{𝜋}{4}-𝜃)]𝑑𝜃\\ & \dots \cdot [\because {\int }_{𝑎}^0𝑓(𝑥)𝑑𝑥={\int }_{𝑎}^0𝑓(𝑎-𝑥\left)𝑑𝑥\right]\\ & ={\int }_{\frac{𝜋}{4}}^0\log (1+\frac{1-\tan 𝜃}{1+\tan 𝜃})𝑑𝜃\\ & ={\int }_{\frac{𝜋}{4}}^0\log 2𝑑𝜃-{\int }_{\frac{𝜋}{4}}^0\log (1+\tan 𝜃)𝑑𝜃\\ & \therefore 2𝐼={\int }_{\frac{𝜋}{4}}^0\log 2d𝜃\Rightarrow 𝐼=\frac{\log 2}{2}[𝜃{]}_{𝜋/4}^0=\frac{𝜋}{8}\log 2\end{array}$

Given equation can be written as

$\begin{array}{cc}& ({𝑥}^2-\mathrm{ax}+\frac{a^2}{4})+({𝑦}^2-b𝑦+\frac{b^2}{4})=\frac{a^2}{4}+\frac{b^2}{4}\\ & \Rightarrow {(𝑥-\frac{a}{2})}^2+{(𝑦-\frac{b}{2})}^2={\left(\sqrt{\surd \frac{a^2+b^2}{4}}\right)}^2\\ & \therefore h=\frac{a}{2},k=\frac{b}{2}\text{ and }r=\sqrt{\surd \frac{a^2+b^2}{4}}\end{array}$

$\therefore$ Parametric equations of circle are

$𝑥=\frac{𝑎}{2}+\frac{\sqrt{\surd {𝑎}^2+{𝑏}^2}}{2}\cos 𝜃$ and

$𝑦=\frac{b}{2}+\frac{\sqrt{\surd a^2+b^2}}{2}\sin 𝜃$

$\begin{array}{cc}& 𝑦=𝑎{𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+5\text{ touches X-axis at }(-2,0)\\ \\ & \Rightarrow 0=-8𝑎+4𝑏-2𝑐+5\\ \\ & \Rightarrow 8𝑎-4𝑏+2𝑐=5\text{.... (i)}\end{array}$

Also, it cuts Y -axis at a point Q

$\therefore$ Put $𝑥=0$ in the equation of curve, we get

$𝑦=5$

$\therefore Q\equiv (0,5)$

$𝑦=𝑎{𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+5$

$\therefore \frac{d𝑦}{d𝑥}=3a{𝑥}^2+2b𝑥+c$

$\begin{array}{cc}& {\left(\frac{d𝑦}{d𝑥}\right)}_{Q\left(0.5\right)}=3\\ \\ & \Rightarrow 3a(0{)}^2+2b(0)+c=3\\ \\ & \Rightarrow c=3\end{array}$

Equation (i) becomes,

$\begin{array}{cc}& 8𝑎-4𝑏+6=5\\ \\ & \Rightarrow 8𝑎-4𝑏+1=0\text{.... (ii)}\end{array}$

Option (C) satisfies equation (ii)

$\therefore$ Option (C) is correct

Let the plane be $\frac{𝑥}{𝑎}+\frac{𝑦}{𝑏}+\frac{𝑧}{𝑐}=1$

If passes through $(3,2,1)$

$\therefore \frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\text{.... (i)}$

Now, coordinates of points A, B, C are (a, 0, 0), $(0,𝑏,0)$ and $(0,0,𝑐)$ respectively.

$\therefore$ Equations of the planes passing through $A,B,C$ are $𝑥=a,𝑦=b$ and $z=c$ respectively.

$\therefore$ From equation (i), we get

Required locus is $\frac{3}{𝑥}+\frac{2}{𝑦}+\frac{1}{𝑧}=1$

$\begin{array}{cc}& \sin ({\cos }^{-1}(-\frac{1}{3})-{\sin }^{-1}\left(\frac{1}{3}\right))\\ & =\sin ({\cos }^{-1}(-\frac{1}{3})-[\frac{𝜋}{2}-{\cos }^{-1}\left(\frac{1}{3}\right)])\\ & =\sin ([𝜋-{\cos }^{-1}\left(\frac{1}{3}\right)]-\frac{𝜋}{2}+{\cos }^{-1}\left(\frac{1}{3}\right))\\ & =\sin \left(\frac{𝜋}{2}\right)\\ & =1\end{array}$

$\bar{a}+2\bar{b}$ is collinear with $\bar{c}$

$\therefore \bar{a}+2\bar{b}=\mathrm{nc}\text{.... (i)}$

Similarly $\bar{b}+3\bar{c}=\mathrm{ma}\text{.... (ii)}$

m and n are non-zero scalars.

$\therefore$ (i) $\Rightarrow \overline{𝑎}+2\overline{𝑏}+6\overline{𝑐}=(𝑛+6)\overline{𝑐}$

(ii) $\Rightarrow \bar{a}+2\bar{b}+6\bar{c}=(2m+1)\bar{a}$

$\Rightarrow n+6=0$ and $2m+1=0$

$\Rightarrow n=-6$ and $m=\frac{-1}{2}$

$\therefore$ (i) $\Rightarrow \overline{𝑎}+2\overline{𝑏}+6\overline{𝑐}=0$

Given that $A,B$ and C are pairwise independent.

$\begin{array}{cc}\therefore & 𝑃(𝐴\cap 𝐵\cap 𝐶)=0\\ & \Rightarrow 𝑃\left(𝐴\right)\cdot 𝑃\left(𝐵\right)\cdot 𝑃\left(𝐶\right)=0\\ & \Rightarrow 𝑃\left(𝐴\right)\cdot 𝑃\left(𝐵\right)=0\text{.... (i)}\\ & 𝑃\left(\right(\overline{𝐴}\cap \overline{𝐵})/𝐶)=\frac{𝑃(\overline{𝐴}\cap \overline{𝐵}\cap 𝐶)}{𝑃\left(𝐶\right)}\\ & =\frac{𝑃\left(\overline{𝐴}\right)\cdot 𝑃\left(\overline{𝐵}\right)\cdot 𝑃\left(\overline{𝐶}\right)}{𝑃\left(\overline{𝐶}\right)}\\ & =[1-𝑃(𝐴\left)\right]\cdot [1-𝑃(𝐵\left)\right]\\ & =1-𝑃\left(𝐴\right)-𝑃\left(𝐵\right)+𝑃\left(𝐴\right)𝑃\left(𝐵\right)\\ & =𝑃\left(\overline{𝐴}\right)-𝑃\left(𝐵\right)\text{...[From (i)]}\end{array}$

Let $𝑥$ and $𝑦$ denote the quantity of the basic ingredients of copper and brass respectively.

$\therefore$ cost, $𝑧=8𝑥+5𝑦$

Constraints are,

$\begin{array}{cc}& 𝑥+𝑦\ge 5\\ & 𝑥\ge 2\\ & 𝑦\le 4\\ & 𝑥\ge 0,𝑦\ge 0\end{array}$

$\therefore$ Feasible region is as shown in the figure.

$\begin{array}{cc}& \therefore \text{ Corner points of the feasible region are }\\ & \begin{array}{cc}& A(2,4),B(2,3)\text{ and }C(5,0)\\ \therefore & z\text{ at }A=36,\\ & z\text{ at }B=31,\\ & z\text{ at }C=40\end{array}\end{array}$

$\therefore$ Minimum cost is Rs.31.

Equation of the line can be written as $\frac{𝑥-1}{1}=\frac{𝑦+5}{1}=\frac{𝑧-9}{1}=𝜆$

$\therefore$ Co-ordinates of any point on the line are given as $𝑥=𝜆+1,𝑦=𝜆-5,𝑧=𝜆+9$

Substituting in the equation of the plane, we get

$\begin{array}{cc}& (𝜆+1)-(𝜆-5)+(𝜆+9)=5\\ \therefore & 𝜆+15=5\\ \therefore & 𝜆=-10\\ \therefore & \text{ Point on the plane is }(-9,-15,-1)\\ \therefore & \text{ Required distance }\\ & =\text{ Distance between }(1,-5,9)\text{ and }(-9,-15,-1)\\ & =\sqrt{\surd (-9-1{)}^2+(-15+5{)}^2+(-1-9{)}^2}\\ & =10\sqrt{\surd 3}\end{array}$

$\begin{array}{cc}& \int g\left(f\right(𝑥\left)\right)d𝑥\\ & =\int 1\times \log (1+𝑥)d𝑥\end{array}$

$\begin{array}{cc}& =𝑥\log (1+𝑥)-\int 𝑥\times \frac{1}{(1+𝑥)}d𝑥+c\\ & =𝑥\log (1+𝑥)-[\int \frac{1+𝑥}{1+𝑥}d𝑥-\int \frac{1}{1+𝑥}d𝑥]+c\\ & =𝑥\log (1+𝑥)-𝑥+\log (1+𝑥)+c\\ & =(1+𝑥)\log (1+𝑥)-𝑥+c\end{array}$