MHT CET 2024 9th May Evening Shift (Mathematics) Question Paper with Solutions

The feasible region lies on the origin side of $𝑥+𝑦=5$ and non-origin side of $3𝑥+4𝑦=12$, in the first quadrant.

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

$𝑧=4𝑥+2𝑦$

At $A(4,0),z=16$

At $𝐵(5,0),𝑧=20$

At $𝐶(0,5),𝑧=10$

At $D(0,3),z=6$

$\therefore$ Maximum value of $𝑧$ is 20.

Let $I=\int \frac{𝑥+1}{𝑥{(1+𝑥e^{𝑥})}^2}\mathrm{dx}=\int \frac{e^{𝑥}(𝑥+1)}{e^{𝑥}\cdot 𝑥{(1+𝑥e^{𝑥})}^2}\mathrm{dx}$

Put $𝑥\cdot e^{𝑥}=t$

$\Rightarrow (𝑥+1){𝑒}^{𝑥}d𝑥=\mathrm{dt}$

$\begin{array}{cc}\therefore 𝐼& =\int \frac{𝑑𝑡}{𝑡(1+𝑡{)}^2}\\ & =\int \frac{1+𝑡-𝑡}{𝑡(1+𝑡{)}^2}𝑑𝑡\\ & =\int \frac{1𝑑𝑡}{𝑡(1+𝑡)}-\int \frac{1}{(1+𝑡{)}^2}𝑑𝑡\\ & =\int \frac{1+𝑡-𝑡}{𝑡(1+𝑡)}-\int \frac{1}{(1+𝑡{)}^2}𝑑𝑡\\ & =\int \frac{1}{𝑡}𝑑𝑡-\int \frac{1}{(𝑡+1)}𝑑𝑡-\int \frac{1}{(1+𝑡{)}^2}𝑑𝑡\\ & =\log 𝑡-\log (1+𝑡)+\frac{1}{(1+𝑡)}+𝑐\\ & =\log 𝑥e^{𝑥}-\log (1+𝑥e^{𝑥})+\frac{1}{(1+𝑥e^{𝑥})}+𝑐\\ \therefore 𝐼& =\log \mid \frac{𝑥e^{𝑥}}{1+𝑥e^{𝑥}}\mid +\frac{1}{1+𝑥e^{𝑥}}+c\end{array}$

Consider option (C)

$\begin{array}{cc}& p\to (\sim p\vee \sim 𝑞)\\ & \equiv T\to (\sim T\vee \sim T)\\ & \equiv T\to (F\vee F)\\ & \equiv T\to F\\ & \equiv F\end{array}$

$\therefore$ Option (C) is the correct answer.

$𝛼$ and $𝛽$ are the roots of ${𝑥}^2-p𝑥+r=0$

$\therefore$ sum of roots $=𝛼+𝛽$

$\Rightarrow 𝛼+𝛽=𝑝\text{.... (i)}$

$\frac{𝛼}{2},2𝛽$ are the roots of ${𝑥}^2-𝑞𝑥+𝑟=0$

$\therefore$ sum of roots $=q$

$\begin{array}{cc}& \Rightarrow \frac{𝛼}{2}+2𝛽=𝑞\\ & \Rightarrow 𝛼+4𝛽=2𝑞\text{.... (ii)}\end{array}$

Subtracting (i) from (ii), we get

$\begin{array}{cc}& 3𝛽=2𝑞-𝑝\\ & \Rightarrow 𝛽=\frac{2𝑞-𝑝}{3}\end{array}$

From (i),

$\begin{array}{cc}& 𝛼+\frac{2𝑞-𝑝}{3}=𝑝\\ & \Rightarrow 𝛼=\frac{2(2𝑝-𝑞)}{3}\end{array}$

Product of roots $=𝛼𝛽$

$\begin{array}{cc}\Rightarrow 𝑟=𝛼𝛽& =\frac{2(2𝑝-𝑞)}{3}\times \frac{2𝑞-𝑝}{3}\\ & =\frac{2}{9}(2𝑝-𝑞)(2𝑞-𝑝)\end{array}$

Projection of $\bar{a}$ on $\overline{𝑐}$ is $\frac{10}{3}$

$\begin{array}{cc}& \therefore \frac{\overline{𝑎}\cdot \bar{c}}{\mid \overline{𝑐}}=\frac{10}{3}\\ & \Rightarrow \frac{(𝛼\hat{i}+3\hat{j}-\hat{k})\cdot (\hat{i}+2\hat{j}-2\hat{k})}{\sqrt{\surd 1+4+4}}=\frac{10}{3}\\ & \Rightarrow \frac{𝛼+3\left(2\right)-1(-2)}{\sqrt{\surd 9}}=\frac{10}{3}\\ & \Rightarrow \frac{𝛼+8}{3}=\frac{10}{3}\\ & \Rightarrow 𝛼=2\\ & \bar{b}\times \bar{c}=-6\hat{i}+10\hat{j}+7\hat{k}\\ & \Rightarrow \mid \begin{array}{cc}\hat{i}& \hat{j}\\ 3& -𝛽\\ 1& 4\\ 1& -2\end{array}\mid =-6\hat{i}+10\hat{j}+7\hat{k}\\ & \Rightarrow (2𝛽-8)\hat{i}+10\hat{j}+(6+𝛽)\hat{k}=-6\hat{i}+10\hat{j}+7\hat{k}\\ & \Rightarrow 6+𝛽=7\Rightarrow 𝛽=1\\ & {𝛼}^2+{𝛽}^2-𝛼𝛽=4+1-2=3\end{array}$

To determine where the function $𝑓\left(𝑥\right)=2{𝑥}^3-6𝑥+5$ is increasing, we first need to find its derivative:

${𝑓}^{\prime }\left(𝑥\right)=\frac{𝑑}{𝑑𝑥}(2{𝑥}^3-6𝑥+5)=6{𝑥}^2-6.$

For the function to be increasing, its derivative must be positive:

$6{𝑥}^2-6>0.$

Simplifying, we have:

$6({𝑥}^2-1)>0.$

Divide both sides by 6:

${𝑥}^2-1>0.$

This can be factored as:

$(𝑥-1)(𝑥+1)>0.$

The critical points are $𝑥=1$ and $𝑥=-1$. To determine the intervals where the inequality holds, we test intervals around the critical points:

For $𝑥<-1$, choose $𝑥=-2$: $(-2-1)(-2+1)=(-3)(-1)=3>0$. The inequality holds.

Between $-1<𝑥<1$, choose $𝑥=0$: $(0-1)(0+1)=(-1)\left(1\right)=-1<0$. The inequality does not hold.

For $𝑥>1$, choose $𝑥=2$: $(2-1)(2+1)=\left(1\right)\left(3\right)=3>0$. The inequality holds.

So, the function is increasing on the intervals $𝑥<-1$ and $𝑥>1$. Thus, the correct interval where the function is increasing is:

Option C: $𝑥<-1$ or $𝑥>1$.

$\begin{array}{cc}\therefore & 2{\sin }^2𝑥+3\sin 𝑥-2>0\\ & (2\sin 𝑥-1)(\sin 𝑥+2)>0\\ & \Rightarrow 2\sin 𝑥-1>0\dots [\because \sin 𝑥+2>0\text{ and }𝑥\in R]\\ & \Rightarrow \sin 𝑥>\frac{1}{2}\\ & \Rightarrow 𝑥\in (\frac{𝜋}{6},\frac{5𝜋}{6})\end{array}$

$\begin{array}{cc}& \text{ Also, }{𝑥}^2-𝑥-2<0\\ & \Rightarrow (𝑥-2)(𝑥+1)<0\\ & \Rightarrow -1<𝑥<2\end{array}$

Since $2<\frac{5𝜋}{6}$

$\therefore 𝑥$ must lie in $(\frac{𝜋}{6},2)$

$\begin{array}{cc}& \bar{a}⟂(\bar{b}+\bar{c}),\bar{b}⟂(\bar{c}+\bar{a})\text{ and }\bar{c}⟂(\bar{a}+\bar{b})\\ & \Rightarrow \bar{a}\cdot \bar{b}+\bar{a}\cdot \bar{c}=0,\bar{b}\cdot \bar{c}+\bar{b}\cdot \bar{a}=0,\bar{c}\cdot \bar{a}+\bar{c}\cdot \bar{b}=0\\ & \Rightarrow \bar{a}\cdot \bar{b}+\bar{b}\cdot \bar{c}+\bar{c}\cdot \bar{a}=0\end{array}$

$\begin{array}{cc}|\overline{𝑎}+\overline{𝑏}+\overline{𝑐}{|}^2& =|\overline{𝑎}{|}^2+|\overline{𝑏}{|}^2+|\overline{𝑐}{|}^2+2(\overline{𝑎}\cdot \overline{𝑏}+\overline{𝑏}\cdot \overline{𝑐}+\overline{𝑐}\cdot \overline{𝑎})\\ & =2^2+3^2+4^2+2\left(0\right)\\ & =4+9+16\\ \therefore |\overline{𝑎}+\overline{𝑏}+\overline{𝑐}{|}^2& =29\\ \Rightarrow |\overline{𝑎}+\overline{𝑏}+\overline{𝑐}|& =\sqrt{\surd 29}\end{array}$

The side length of a square is denoted by $𝑠$, and the area $𝐴$ of the square is given by:

$𝐴={𝑠}^2$

Given that the area of the square is contracting at a rate of $\frac{𝑑𝐴}{𝑑𝑡}=-3{\text{cm}}^2/\text{sec}$.

Using the chain rule, the rate of change of the area with respect to time is related to the rate of change of the side length as follows:

$\frac{𝑑𝐴}{𝑑𝑡}=2𝑠\frac{𝑑𝑠}{𝑑𝑡}$

Substitute the given values into this equation:

$-3=2\left(15\right)\frac{𝑑𝑠}{𝑑𝑡}$

Solving for $\frac{𝑑𝑠}{𝑑𝑡}$:

$-3=30\frac{𝑑𝑠}{𝑑𝑡}$

$\frac{𝑑𝑠}{𝑑𝑡}=-\frac{1}{10}\text{cm/sec}$

The perimeter $𝑃$ of the square is given by:

$𝑃=4𝑠$

Differentiating with respect to time $𝑡$:

$\frac{𝑑𝑃}{𝑑𝑡}=4\frac{𝑑𝑠}{𝑑𝑡}$

Substituting the value of $\frac{𝑑𝑠}{𝑑𝑡}$:

$\frac{𝑑𝑃}{𝑑𝑡}=4(-\frac{1}{10})$

$\frac{𝑑𝑃}{𝑑𝑡}=-\frac{4}{10}\text{cm/sec}$

$\frac{𝑑𝑃}{𝑑𝑡}=-\frac{2}{5}\text{cm/sec}$

Thus, the rate at which the perimeter is decreasing is $\frac{2}{5}\text{cm/sec}$. The correct answer is Option B:

$\frac{2}{5}\mathrm{cm}/\sec$

$\begin{array}{cc}\text{ Required area }& =\underset{0}{\overset{2}{\int }}[({𝑥}^2+2)-(𝑥+1\left)\right]d𝑥\\ & =\underset{0}{\overset{2}{\int }}({𝑥}^2-𝑥+1)d𝑥\\ & ={[\frac{{𝑥}^3}{3}-\frac{{𝑥}^2}{2}+𝑥]}_2^0\\ & =\frac{8}{3}-2+2-0\\ & =\frac{8}{3}\text{ sq. units }\end{array}$

Since $\bar{a}$ bisects the angle between $\bar{b}$ and $\bar{c}$.

$\therefore$ The equation of bisector of $\bar{b}$ and $\bar{c}$ is

$\begin{array}{cc}& \bar{a}=𝜆(\hat{b}+\hat{c})\\ & \Rightarrow 𝛼\hat{i}+2\hat{j}+𝛽\hat{k}=𝜆(\frac{\hat{i}+\hat{j}}{\sqrt{\surd 2}}+\frac{\hat{j}+\hat{k}}{\sqrt{\surd 2}})\end{array}$

$\begin{array}{cc}& \Rightarrow 𝛼\hat{i}+2\hat{j}+𝛽\hat{k}=\frac{𝜆}{\sqrt{\surd 2}}(\hat{i}+2\hat{j}+\hat{k})\\ & \text{ On comparing, we get }\\ & \begin{array}{cc}& 𝛼=\frac{𝜆}{\sqrt{\surd 2}},2=\sqrt{\surd 2𝜆}\text{ and }𝛽=\frac{𝜆}{\sqrt{\surd 2}}\\ & \Rightarrow 𝛼=1,𝛽=1\end{array}\end{array}$

$\begin{array}{cc}& 2𝜋-[({\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13})+{\sin }^{-1}\frac{16}{65}]\\ & =2𝜋-[({\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{5}{12})+{\sin }^{-1}\frac{16}{65}]\\ & =2𝜋-[{\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times \frac{5}{12}}\right)+{\sin }^{-1}\frac{16}{65}]\\ & =2𝜋-[{\tan }^{-1}\left(\frac{63}{16}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)]\\ & =2𝜋-[{\cos }^{-1}\left(\frac{16}{65}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)]\\ & =2𝜋-\frac{𝜋}{2}\\ & =\frac{3𝜋}{2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& A_{11}=(-1{)}^{1+1}\mid \begin{array}{cc}1& 2\\ 1& 2\end{array}\mid =1(0)=0\\ & A_{12}=(-1{)}^{1+2}\mid \begin{array}{cc}3& 2\\ -1& 2\end{array}\mid =(-1\left)\right(8)=-8\\ & A_{13}=(-1{)}^{1+3}\mid \begin{array}{cc}3& 1\\ -1& 1\end{array}\mid =(1\left)\right(4)=4\\ & A_{21}=(-1{)}^{2+1}\mid \begin{array}{cc}0& -1\\ 1& 2\end{array}\mid =(-1\left)\right(1)=-1\\ & A_{22}=(-1{)}^{2+2}\mid \begin{array}{cc}2& -1\\ -1& 2\end{array}\mid =(1\left)\right(3)=3\\ & A_{23}=(-1{)}^{2+3}\mid \begin{array}{cc}2& 0\\ -1& 1\end{array}\mid =(-1\left)\right(2)=-2\\ & A_{31}=(-1{)}^{3+1}\mid \begin{array}{cc}0& -1\\ 1& 2\end{array}\mid =(1\left)\right(1)=1\\ & A_{32}=(-1{)}^{3+2}\mid \begin{array}{cc}2& -1\\ 3& 2\end{array}\mid =(-1\left)\right(7)=-7\\ & A_{33}=(-1{)}^{3+3}\mid \begin{array}{cc}2& 0\\ 3& 1\end{array}\mid =(1\left)\right(2)=2\end{array}\end{array}$

$\begin{array}{cc}& \therefore \text{ The matrix of the cofactors is }\\ & ⎡⎣⎢\begin{array}{ccc}0& -8& 4\\ -1& 3& -2\\ 1& -7& 2\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& 𝑦={\sin }^{-1}(\frac{3𝑥}{2}-\frac{{𝑥}^3}{2})\\ & ={\sin }^{-1}(\frac{3𝑥}{2}-4{\left(\frac{𝑥}{2}\right)}^3)\\ & \text{ Put }\frac{𝑥}{2}=\sin 𝜃\Rightarrow 𝜃={\sin }^{-1}\left(\frac{𝑥}{2}\right)\\ & \therefore 𝑦={\sin }^{-1}(3\sin 𝜃-4{\sin }^3𝜃)\\ & ={\sin }^{-1}(\sin 3𝜃)\\ & =3𝜃\\ & \therefore 𝑦=3{\sin }^{-1}\left(\frac{𝑥}{2}\right)\\ & \therefore \frac{𝑑𝑦}{𝑑𝑥}=3\cdot \frac{1}{\sqrt{\surd 1-{\left(\frac{𝑥}{2}\right)}^2}}\cdot \frac{1}{2}=\frac{3}{\sqrt{\surd 4-{𝑥}^2}}\end{array}$

Let $\overline{𝑎}=\widehat{𝑖}+\widehat{𝑗}-\widehat{𝑘},\overline{𝑏}=\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘},\overline{𝑐}=2\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘}$.

Then, required unit vectors are given by

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

Now, $\bar{a}\times (\bar{b}\times \bar{c})=(\bar{a}\cdot \bar{c})\bar{b}-(\bar{a}\cdot \bar{b})\bar{c}$

$=2(\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘})-1(2\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘})=\widehat{𝑗}+\widehat{𝑘}$

$\therefore |\bar{a}\times (\bar{b}\times \bar{c}\left)\right|=\sqrt{\surd 1+1}=\sqrt{\surd 2}$

Hence, required unit vectors are

$\overline{𝛼}=\pm \frac{\hat{j}+\hat{k}}{\sqrt{\surd 2}}$

Let p : Horses have wings

q : Crows have tails

Given statement is $p\leftrightarrow q$

$\sim (𝑝\leftrightarrow 𝑞)\equiv (𝑝\wedge \sim 𝑞)\vee (𝑞\wedge \sim 𝑝)$

The equation of a plane passing through $(1,2,1)$ is $a(𝑥-1)+b(𝑦-2)+c(z-1)=0\text{.... (i)}$

Plane (i) is perpendicular to planes

$2𝑥-2𝑦+𝑧=0\text{ and }𝑥-𝑦+2𝑧=4$

$\begin{array}{cc}& \therefore 2𝑎-2𝑏+𝑐=0\text{.... (ii)}\\ & 𝑎-𝑏+2𝑐=0\text{.... (iii)}\end{array}$

Solving (ii) and (iii), we get

$a=-3,b=-3,c=0$

Substituting the values of $𝑎,𝑏,𝑐$ in equation (i), we get

$𝑥+𝑦-3=0$

$\therefore$ The distance of this plane from $(2,3,4)$ is $𝑙=\mid \frac{2+3-3}{\sqrt{\surd 1+1}}\mid =\sqrt{\surd 2}$ units

$\begin{array}{cc}& \log (𝑥+𝑦)=\sin (𝑥+𝑦)\end{array}$

Differentiating both sides w.r.t. $𝑥$, we get

$\begin{array}{cc}& \frac{1}{𝑥+𝑦}(1+\frac{d𝑦}{d𝑥})=\cos (𝑥+𝑦)[1+\frac{d𝑦}{d𝑥}]\\ & \Rightarrow \frac{1}{𝑥+𝑦}+\frac{1}{𝑥+𝑦}\cdot \frac{d𝑦}{d𝑥}=\cos (𝑥+𝑦)+\cos (𝑥+𝑦)\frac{d𝑦}{d𝑥}\\ & \Rightarrow \frac{d𝑦}{d𝑥}(\frac{1}{𝑥+𝑦}-\cos (𝑥+𝑦\left)\right)=\cos (𝑥+𝑦)-\frac{1}{𝑥+𝑦}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=-1\end{array}$

Let $I=\int \sqrt{\surd e^{𝑥}-1}d𝑥$

Put $e^{𝑥}-1=t^2$

$\Rightarrow e^{𝑥}\mathrm{dx}=2\mathrm{tdt}$

$\Rightarrow \mathrm{dx}=\frac{2t}{t^2+1}\mathrm{dt}$

$\therefore 𝐼=\int t\cdot \frac{2t}{t^2+1}\mathrm{dt}=\int \frac{2t^2}{t^2+1}\mathrm{dt}$

$=\int \frac{2(t^2+1)-2}{t^2+1}\mathrm{dt}$

$=2\int \mathrm{dt}-2\int \frac{\mathrm{dt}}{t^2+1}$

$=2t-2{\tan }^{-1}t+c$

$=2\sqrt{\surd e^{𝑥}-1}-2{\tan }^{-1}\sqrt{\surd e^{𝑥}-1}+c$

When each item of a data is multiplied by $𝜆$, variance is multiplied by ${𝜆}^2$.

$\therefore$ New variance $=3^2\times 5=9\times 5=45$

Since 8 is added to each number, the variance remains the same i.e. 45

$\begin{array}{cc}& \begin{array}{cc}& 2𝑥-4𝑦+𝑧=7\\ \therefore & 2\left(4\right)-4\left(2\right)-𝑚=7\\ & \Rightarrow 𝑚=-7\end{array}\\ & \text{ Point }(4,2,-m)\text{ should lie in the plane }\end{array}$

Let event $A:A$ can hit the target

event $B:B$ can hit the target

event $C:C$ can hit the target

event $D:D$ can hit the target

$\begin{array}{cc}& P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3},P\left(C\right)=\frac{1}{4},P\left(D\right)=\frac{1}{5}\\ & P\text{ (Target is not hit by any one of them) }\\ & \begin{array}{cc}& =P(A^{\prime }\cap B^{\prime }\cap C^{\prime }\cap D^{\prime })\\ & =P\left(A^{\prime }\right)\cdot P\left(B^{\prime }\right)\cdot P\left(C^{\prime }\right)\cdot P^{\prime }\left(D^{\prime }\right)=\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right)\\ & =\frac{1}{5}\end{array}\end{array}$

$\begin{array}{cc}& 𝑃\left(\text{ Target is hit }\right)\\ & =1-𝑃\left(\text{ Target is not hit by any one of them }\right)\\ & =1-\frac{1}{5}\\ & =\frac{4}{5}\end{array}$

Let height and breadth of the sheet be ' $𝑦$ ' m and ' $𝑥$ ' m respectively.

$\begin{array}{cc}\therefore & 𝑥𝑦=180000{\mathrm{cm}}^2\\ \therefore & 𝑦=\frac{180000}{𝑥}\end{array}$

$\therefore$ The area available for printing is

$\begin{array}{cc}𝐴& =(𝑦-150)(𝑥-100)\\ & =(\frac{180000}{𝑥}-150)(𝑥-100)\end{array}$

$\begin{array}{cc}& =180000-\frac{18000000}{𝑥}-150𝑥-15000\\ & =165000-150𝑥-\frac{18000000}{𝑥}\\ \therefore & \frac{\mathrm{dA}}{d𝑥}=0-150+\frac{18000000}{{𝑥}^2}\\ \therefore & \frac{\mathrm{dA}}{d𝑥}=0\Rightarrow {𝑥}^2=\frac{18000000}{150}=120000\\ \Rightarrow 𝑥& =200\sqrt{\surd 3}\mathrm{cm}\\ \Rightarrow 𝑦& =\frac{180000}{200\sqrt{\surd 3}}=300\sqrt{\surd 3}\mathrm{cm}\end{array}$