MHT CET 2024 15th May Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& 𝑥=\sin 𝜃\text{ and }𝑦={\sin }^3𝜃\\ & \therefore 𝑦={𝑥}^3\\ & \therefore \frac{d𝑦}{d𝑥}=3{𝑥}^2\\ & \therefore \frac{d^2𝑦}{d{𝑥}^2}=6𝑥\\ & \text{ At }𝜃=\frac{𝜋}{2},𝑥=\sin \frac{𝜋}{2}=1\\ & \therefore {\left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)}_{𝜃=\frac{𝜋}{2}}={\left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)}_{𝑥=1}=6\left(1\right)=6\end{array}$

$m_{\mathrm{PQ}}=\frac{1}{6},m_{\mathrm{SR}}=\frac{1}{6},m_{\mathrm{RQ}}=-3,m_{\mathrm{SP}}=-3$

$◻\mathrm{PQRS}$ is a parallelogram.

But neither $\mathrm{PR}=\mathrm{SQ}$ nor $\mathrm{PR}⟂\mathrm{SQ}$.

$\therefore$ Parallelogram, which is neither a rhombus nor a rectangle.

$\begin{array}{cc}& 𝑦=\frac{{𝑥}^{\frac{2}{3}}-{𝑥}^{-\frac{1}{3}}}{{𝑥}^{\frac{2}{3}}+{𝑥}^{-\frac{1}{3}}}\\ & =\frac{{𝑥}^{-\frac{1}{3}}(𝑥-1)}{{𝑥}^{-\frac{1}{3}}(𝑥+1)}\\ & \therefore 𝑦=\frac{𝑥-1}{𝑥+1}\\ & \therefore \frac{d𝑦}{d𝑥}=\frac{(𝑥+1)\cdot 1-(𝑥-1)\cdot 1}{(𝑥+1{)}^2}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{2}{(𝑥+1{)}^2}\\ & \Rightarrow (𝑥+1{)}^2\frac{d𝑦}{d𝑥}=2\end{array}$

$\begin{array}{cc}& f\left(𝑥\right)=\sin 2𝜋𝑥\\ & \therefore f^{\prime }\left(𝑥\right)=2𝜋\cos 2𝜋𝑥\\ & \text{ Now, }{f}^{\prime }\left(C\right)=0\\ & \Rightarrow 2𝜋\cos 2𝜋C=0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow \cos 2𝜋C=0\\ & \Rightarrow 2𝜋C=\frac{-3𝜋}{2},\frac{-𝜋}{2},\frac{𝜋}{2},\frac{3𝜋}{2}\\ & \dots [\because 𝑥\in (-1,1)\Rightarrow 2𝜋𝑥\in (-2𝜋,2𝜋\left)\right]\\ & \Rightarrow 𝐶=\frac{-3}{4},-\frac{1}{4},\frac{1}{4},\frac{3}{4}\end{array}\\ & \text{ Number of values of }C=4\end{array}$

Let $𝑥$ be the length of the side of square bottom, h be the height, V be the volume and A be the surface area of open tank.

Then,

$\begin{array}{cc}& V={𝑥}^{2}h=4000\text{.... (i)}\\ & A={𝑥}^2+4𝑥h\text{.... (ii)}\end{array}$

From (i),

$h=\frac{4000}{{𝑥}^2}$

Substituting the value of $ℎ$ in (ii), we get

$\begin{array}{cc}& A={𝑥}^2+\frac{16000}{𝑥}\\ \therefore & \frac{\mathrm{dA}}{\mathrm{dx}}=2𝑥-\frac{16000}{{𝑥}^2}\end{array}$

$𝐴$ is minimum, if $\frac{𝑑𝐴}{𝑑𝑥}=0$

$\begin{array}{cc}& \Rightarrow 2𝑥-\frac{16000}{{𝑥}^2}=0\\ & \Rightarrow {𝑥}^3=8000\\ & \Rightarrow 𝑥=20\end{array}$

Now, $\frac{d^{2}A}{{\mathrm{dx}}^2}=2+\frac{32000}{{𝑥}^3}$

$\Rightarrow {\left(\frac{d^{2}A}{\mathrm{dx}}\right)}_{𝑥=20}=6>0$

$\therefore$ A is minimum when $𝑥=20\mathrm{cm}$.

$h=\frac{4000}{{𝑥}^2}=\frac{4000}{400}=10\mathrm{cm}$

$\begin{array}{cc}\text{ Let }𝐼& =\int \mathrm{cosec}(𝑥-𝑎)\cdot \mathrm{cosec}𝑥𝑑𝑥\\ & =\int \frac{1}{\sin (𝑥-𝑎)\sin 𝑥}𝑑𝑥\\ & =\frac{1}{\sin 𝑎}\int \frac{\sin 𝑎}{\sin (𝑥-𝑎)\sin 𝑥}𝑑𝑥\\ & =\frac{1}{\sin 𝑎}\int \frac{\sin [𝑥-(𝑥-𝑎\left)\right]}{\sin (𝑥-𝑎)\sin 𝑥}𝑑𝑥\\ & =\frac{1}{\sin 𝑎}\int \frac{\sin 𝑥\cos (𝑥-𝑎)-\cos 𝑥\sin (𝑥-𝑎)}{\sin (𝑥-𝑎)\sin 𝑥}𝑑𝑥\\ & =\frac{1}{\sin 𝑎}\int [\cot (𝑥-𝑎)-\cot 𝑥]𝑑𝑥\\ & =\frac{1}{\sin 𝑎}[\log \sin (𝑥-𝑎)-\log (\sin 𝑥\left)\right]+𝑐\\ & \begin{array}{cc}& =\frac{1}{\sin 𝑎}\log \mid \frac{\sin (𝑥-𝑎)}{\sin 𝑥}\mid +𝑐\\ & =\frac{1}{\sin 𝑎}\log |\sin (𝑥-𝑎)\cdot \mathrm{cosec}𝑥|+𝑐\end{array}\end{array}$

Inverse of $𝑝\to (𝑝\to 𝑞)$ is

$\sim 𝑝<text>\to </text><text>\sim </text>(𝑝\to 𝑞)$

Contrapositive of inverse of $𝑝\to (𝑝\to 𝑞)$ is

$\begin{array}{cc}& \sim [\sim (𝑝\to 𝑞\left)\right]<text>\to </text><text>\sim </text>(\sim 𝑝)\\ & \equiv (𝑝\to 𝑞)\to 𝑝\\ & \equiv (\sim 𝑝\vee 𝑞)\to 𝑝\end{array}$

$\begin{array}{cc}& \text{ Putting }𝑥=\frac{{𝑦}^2}{4}\text{ in }𝑦=2𝑥-4\text{, we get }\\ & \begin{array}{cc}& 𝑦=2\left(\frac{{𝑦}^2}{4}\right)-4\\ & \Rightarrow {𝑦}^2-2𝑦-8=0\\ & \Rightarrow (𝑦-4)(𝑦+2)=0\\ & \Rightarrow 𝑦=4,-2\end{array}\end{array}$

$\begin{array}{cc}\therefore & \text{ Required area }=\underset{-2}{\overset{4}{\int }}(\frac{𝑦+4}{2}-\frac{{𝑦}^2}{4})d𝑦\\ & =\frac{1}{2}{[\frac{{𝑦}^2}{2}+4𝑦]}_4^{-2}-\frac{1}{4}{\left[\frac{{𝑦}^3}{3}\right]}_4^{-2}\\ & =\frac{1}{2}[8+16-(2-8\left)\right]-\frac{1}{12}[64-(-8\left)\right]\\ & =15-6\\ & =9\text{ sq. units }\end{array}$

Total number of ways

$\begin{array}{cc}& ={}^5{C}_1\times {}^5{C}_1\times {}^5{C}_3+{}^5{C}_1\times {}^5{C}_2\times {}^5{C}_2\\ & +{}^5{C}_1\times {}^5{C}_3\times C_1+{}^5{C}_2\times C_1\times {}^5{C}_2\\ & +{}^5{C}_2\times {}^5{C}_2\times C_1+{}^5{C}_3\times C_1\times {}^5{C}_1\\ & =5\times 5\times 10+5\times 10\times 10+5\times 10\times 5\\ & +10\times 5\times 10+10\times 10\times 5+10\times 5\times 5\\ & =250+500+250+500+500+250\\ & =2250\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \begin{array}{cc}𝑦& =(C_1+C_2)e^{𝑥}+C_3{e}^{𝑥+C_4}\\ & ={\mathrm{Ae}}^{𝑥}+{\mathrm{Be}}^{𝑥}\text{, where }A=C_1+C_2\text{ and }B=C_3{e}_4\\ & =(A+B)e^{𝑥}\end{array}\\ & ={\mathrm{De}}^x,\text{where }D=A+B\end{array}\\ & \text{ This equation consist of one arbitrary constant. }\end{array}$

The given equations of the sides are $𝑥=-2,𝑥=6,𝑦=-2$ and $𝑦=5$.

Here, the diagonals AC and BD of rectangle $𝐴𝐵𝐶𝐷$ are diameters of the circle passing through the vertices $A,B,C$ and $D$.

Considering diagonal AC with end points $A(-2,-2)$ and $C(6,5)$, we get

Equation of circle in diameter form as,

$\begin{array}{cc}& (𝑥-6)(𝑥+2)+(𝑦-5)(𝑦+2)=0\\ & \Rightarrow {𝑥}^2-4𝑥-12+{𝑦}^2-3𝑦-10=0\\ & \Rightarrow {𝑥}^2+{𝑦}^2-4𝑥-3𝑦-22=0\end{array}$

The given equation is defined for $𝑥\ne \frac{𝜋}{2},\frac{3𝜋}{2}$.

Now, $\tan 𝑥+\sec 𝑥=2\cos 𝑥$

$\Rightarrow \frac{\sin 𝑥}{\cos 𝑥}+\frac{1}{\cos 𝑥}=2\cos 𝑥$

$\Rightarrow (\sin 𝑥+1)=2{\cos }^2𝑥$

$\Rightarrow (\sin 𝑥+1)=2(1-{\sin }^2𝑥)$

$\Rightarrow (\sin 𝑥+1)=2(1-\sin 𝑥)(1+\sin 𝑥)$

$\Rightarrow (1+\sin 𝑥)\left[2\right(1-\sin 𝑥)-1]=0$

$\Rightarrow 2(1-\sin 𝑥)-1=0$

$\dots \left[\begin{array}{c}\because \sin 𝑥\ne -1\text{ otherwise }\cos 𝑥=0\text{ and }\\ \tan 𝑥,\sec 𝑥\text{ will be undefined }\end{array}\right]$

$\Rightarrow \sin 𝑥=\frac{1}{2}$

$\Rightarrow 𝑥=\frac{𝜋}{6},\frac{5𝜋}{6}$ in $(0,2𝜋)$

Given equation of pair of lines is

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0$

$\begin{array}{cc}& \Rightarrow a+2h\left(\frac{𝑦}{𝑥}\right)+b{\left(\frac{𝑦}{𝑥}\right)}^2=0\\ & \Rightarrow a+2\mathrm{hk}+{\mathrm{bk}}^2=0\text{.... (i)}\end{array}$

Now, slope of line $𝑚𝑥+𝑛𝑦=18$ is $\frac{-𝑚}{𝑛}$

$\therefore$ Slope of the line perpendicular to $m𝑥+n𝑦=18$ is $k=\frac{n}{m}$

Substituting the value of $𝑘$ in (i), we get

$\begin{array}{cc}& 𝑎+2ℎ\left(\frac{𝑛}{𝑚}\right)+𝑏{\left(\frac{𝑛}{𝑚}\right)}^2=0\\ & \Rightarrow 𝑎{𝑚}^2+2ℎ𝑚𝑛+𝑏{𝑛}^2=0\end{array}$

$\begin{array}{cc}{\sin }^{-1}\left(\frac{𝑥}{13}\right)& =\frac{𝜋}{2}-{\mathrm{cosec}}^{-1}\left(\frac{13}{12}\right)\\ & ={\sec }^{-1}\left(\frac{13}{12}\right)={\cos }^{-1}\left(\frac{12}{13}\right)\end{array}$

$\begin{array}{cc}& \therefore {\sin }^{-1}\left(\frac{𝑥}{13}\right)={\sin }^{-1}\left(\frac{5}{13}\right)\\ & \therefore 𝑥=5\end{array}$

Feasible region lies on origin side of $2𝑥+3𝑦=6$ and non-origin side of $𝑥+4𝑦=4$, in $1^{\text{st }}$ quadrant.

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

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

$\begin{array}{cc}& \begin{array}{cc}& A_{11}=(-1{)}^{1+1}(2)=2,A_{12}=(-1{)}^{1+2}\left(3\right)=-3,\\ & A_{21}=(-1{)}^{2+1}(-b)=b,A_{22}=(-1{)}^{2+2}\left(5a\right)=5a\\ \therefore & \text{ adj }A={\left[\begin{array}{cc}2& -3\\ b& 5a\end{array}\right]}^T=\left[\begin{array}{cc}2& b\\ -3& 5a\end{array}\right]\\ & \text{ Given, }A\text{ adj }A={\mathrm{AA}}^T\\ & \Rightarrow \left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}2& b\\ -3& 5a\end{array}\right]=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}5a& 3\\ -b& 2\end{array}\right]\\ & \Rightarrow \left[\begin{array}{cc}10a+3b& 0\\ 0& 3b+10a\end{array}\right]=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]\end{array}\end{array}$

$\begin{array}{cc}& \therefore \text{ by the equality of matrices, }\\ & \begin{array}{cc}& 15𝑎-2𝑏=0\text{ and }3𝑏+10𝑎=13\\ & \Rightarrow 𝑎=\frac{2}{5}\text{ and }𝑏=3\\ \therefore & 5𝑎+𝑏=5\left(\frac{2}{5}\right)+3=2+3=5\end{array}\end{array}$

To determine where the function is decreasing, we need to find the derivative of the function and analyze where the derivative is negative.

First, compute the derivative of the function:

$𝑓\left(𝑥\right)=2{𝑥}^3-9{𝑥}^2+12𝑥+2$

The derivative is:

${𝑓}^{\prime }\left(𝑥\right)=\frac{𝑑}{𝑑𝑥}(2{𝑥}^3-9{𝑥}^2+12𝑥+2)=6{𝑥}^2-18𝑥+12$

Next, solve the inequality ${𝑓}^{\prime }\left(𝑥\right)<0$ to determine where the function is decreasing.

Set the derivative equal to zero to find the critical points:

$6{𝑥}^2-18𝑥+12=0$

Divide by 6:

${𝑥}^2-3𝑥+2=0$

Factor the quadratic equation:

$(𝑥-1)(𝑥-2)=0$

Thus, the critical points are $𝑥=1$ and $𝑥=2$.

To determine where the derivative is negative, test the intervals created by these critical points: $𝑥<1$, $1<𝑥<2$, and $𝑥>2$.

For $𝑥<1$, select $𝑥=0$:

${𝑓}^{\prime }\left(0\right)=6(0{)}^2-18(0)+12=12>0$

For $1<𝑥<2$, select $𝑥=1.5$:

${𝑓}^{\prime }\left(1.5\right)=6(1.5{)}^2-18(1.5)+12=6\cdot 2.25-27+12=13.5-27+12=-1.5<0$

For $𝑥>2$, select $𝑥=3$:

${𝑓}^{\prime }\left(3\right)=6(3{)}^2-18(3)+12=54-54+12=12>0$

Therefore, the function is decreasing in the interval:

Option A: $1<𝑥<2$

$\begin{array}{cc}& \int (1+𝑥-{𝑥}^{-1})e^{𝑥+{𝑥}^{-1}}d𝑥\\ & =\int [𝑥e^{𝑥+{𝑥}^{-1}}(1-\frac{1}{{𝑥}^2})+e^{𝑥+{𝑥}^{-1}}]d𝑥\\ & =𝑥e^{𝑥+{𝑥}^{-1}}+c\dots .[\because \int \left[𝑥f^{\prime }\right(𝑥)+f(𝑥\left)\right]d𝑥=𝑥f(𝑥)+c]\end{array}$

Since $𝑓\left(𝑥\right)$ is continuous everywhere.

$\therefore f\left(𝑥\right)$ is continuous at $𝑥=-\frac{𝜋}{2}$ and $𝑥=\frac{𝜋}{2}$.

$\begin{array}{cc}\therefore & \underset{𝑥\to \frac{-{𝜋}^{-}}{2}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to \frac{-{𝜋}^{+}}{2}}{lim}𝑓\left(𝑥\right)\\ & \Rightarrow \underset{𝑥\to -\frac{-𝜋}{2}}{lim}(-2\sin 𝑥)=\underset{𝑥\to \frac{-𝜋}{2}}{lim}(𝐴\sin 𝑥+𝐵)\\ & \Rightarrow -2(-1)=𝐴(-1)+𝐵\\ & \Rightarrow -𝐴+𝐵=2\text{.... (i)}\end{array}$

Also, $\underset{𝑥\to \frac{{𝜋}^{-}}{2}}{lim}f\left(𝑥\right)=\underset{𝑥\to \frac{{𝜋}^{+}}{2}}{lim}f\left(𝑥\right)$

$\begin{array}{cc}& \Rightarrow \underset{𝑥\to \frac{𝜋}{2}}{lim}(A\sin 𝑥+B)=\underset{𝑥\to \frac{𝜋}{2}}{lim}(\cos 𝑥)\\ & \Rightarrow A\left(1\right)+B=0\\ & \Rightarrow A+B=0\text{.... (ii)}\end{array}$

From (i) and (ii), we get

$A=-1,B=1$

$\begin{array}{cc}\text{ Let }I& ={\int }_{𝜋/2}^{𝜋/3}\frac{\sqrt{\surd 1+\cos 𝑥}}{(1-\cos 𝑥{)}^{\frac{5}{2}}}𝑑𝑥\\ & ={\int }_{𝜋/2}^{𝜋/3}\frac{\sqrt{\surd 1+\cos 𝑥}}{(1-\cos 𝑥{)}^{\frac{5}{2}}}\times \frac{\sqrt{\surd 1-\cos 𝑥}}{\sqrt{\surd 1-\cos 𝑥}}𝑑𝑥\\ & ={\int }_{𝜋/2}^{𝜋/3}\frac{\sin 𝑥}{(1-\cos 𝑥{)}^3}𝑑𝑥\end{array}$

$\begin{array}{cc}& \text{ Put }1-\cos 𝑥=t\\ & \Rightarrow \sin 𝑥d𝑥=\mathrm{dt}\\ & \therefore I={\int }_1^{1/2}\frac{\mathrm{dt}}{t^3}={\left[\frac{t^{-2}}{-2}\right]}_1^{1/2}=\frac{3}{2}\end{array}$

$\begin{array}{cc}& |z+1|=1\\ & \Rightarrow |𝑥+i𝑦+1|=1\\ & \Rightarrow (𝑥+1{)}^2+{𝑦}^2=(1{)}^2\\ & \Rightarrow (𝑥+1{)}^2+{𝑦}^2=1\end{array}$

This is an equation of circle with centre $(-1,0)$ and radius 1 .

$\begin{array}{cc}& \begin{array}{cc}& \frac{d𝑦}{d𝑥}=\frac{3e^{2𝑥}+3e^{4𝑥}}{e^{𝑥}+e^{-𝑥}}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{3e^{2𝑥}(1+e^{2𝑥})}{\left(\frac{e^{2𝑥}+1}{e^{𝑥}}\right)}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=3e^{2𝑥}\cdot e^{𝑥}\\ & \Rightarrow d𝑦=3e^{3𝑥}d𝑥\end{array}\\ & \text{ Integrating on both sides, we get }\\ & 𝑦=e^{3𝑥}+c\end{array}$

Given equation is

${\sin }^{-1}\left(2𝑥\sqrt{\surd 1-{𝑥}^2}\right)+{\sec }^{-1}\left(\frac{1}{2{𝑥}^2-1}\right)=\frac{2𝜋}{3}$

Also, ${\cos }^{-1}𝑥=𝛼$

$\Rightarrow 𝑥=\cos 𝛼$

and $0<𝛼<\frac{𝜋}{2}\dots [\because 0<𝑥<1]$

Putting $𝑥=\cos 𝛼$ in the given equation, we get

$\begin{array}{cc}& {\sin }^{-1}(2\cos 𝛼\sqrt{\surd 1-{\cos }^2𝛼})+{\sec }^{-1}\left(\frac{1}{2{\cos }^2𝛼-1}\right)=\frac{2𝜋}{3}\\ & \Rightarrow {\sin }^{-1}(2\cos 𝛼\sin 𝛼)+{\sec }^{-1}\left(\frac{1}{\cos 2𝛼}\right)=\frac{2𝜋}{3}\\ & \Rightarrow {\sin }^{-1}(\sin 2𝛼)+{\cos }^{-1}(\cos 2𝛼)=\frac{2𝜋}{3}\\ & \Rightarrow 2𝛼+2𝛼=\frac{2𝜋}{3}\\ & \Rightarrow 𝛼=\frac{2𝜋}{12}=\frac{𝜋}{6}\end{array}$

$\begin{array}{cc}\overline{\mathrm{PQ}}=& -2\widehat{𝑖}+3\widehat{𝑗}+2\widehat{𝑘}-(\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘})\\ & =-3\widehat{𝑖}+5\widehat{𝑗}-\widehat{𝑘}\end{array}$

$\begin{array}{cc}\overline{𝑄𝑅}& =-8\widehat{𝑖}+13\widehat{𝑗}-(-2\widehat{𝑖}+3\widehat{𝑗}+2\widehat{𝑘})\\ & =-6\widehat{𝑖}+10\widehat{𝑗}-2\widehat{𝑘}\\ & =2(-3\widehat{𝑖}+5\widehat{𝑗}-\widehat{𝑘})=2\overline{𝑃𝑄}\end{array}$

$\therefore \overline{\mathrm{QR}}$ is a scalar multiple of $\overline{\mathrm{PQ}}$.

$\therefore \overline{\mathrm{QR}}$ and $\overline{\mathrm{PQ}}$ are parallel to each other with point Q in common.

$\therefore$ Points $P,Q$ and R are collinear and Q lies between P and R .