MHT CET 2024 4th May Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \text{ Given that }(𝐴-3𝐼)(𝐴-5𝐼)=0\\ \therefore & {𝐴}^2-3𝐴-5𝐴+15𝐼=0\\ \therefore & {𝐴}^2-8𝐴+15𝐼=0\\ \therefore & {𝐴}^2+15𝐼=8𝐴\end{array}$

Multiplying entire equation by $\frac{A^{-1}}{2}$, we get

$\frac{1}{2}𝐴+\frac{15}{2}{𝐴}^{-1}=4𝐼$

Comparing with $𝛼A+𝛽A^{-1}=4I$, we get

$𝛼=\frac{1}{2}$ and $𝛽=\frac{15}{2}$

$\therefore 𝛼+𝛽=\frac{1}{2}+\frac{15}{2}=8$

$\begin{array}{cc}& \text{ If the given lines are co-planar, we get }\\ & \begin{array}{cc}& \mid \begin{array}{ccc}2-1& 3-4& 4-5\\ 1& 1& -k\\ k& 2& 1\end{array}\mid =0\\ & \therefore \mid \begin{array}{ccc}1& -1& -1\\ 1& 1& -𝑘\\ 𝑘& 2& 1\end{array}\mid =0\\ & \therefore 1(1+2k)+1(1+k^2)-1(2-k)=0\\ & \therefore 1+2k+1+k^2-2+k=0\\ & \therefore k^2+3k=0\\ & \therefore k=0\text{ or }k=-3\end{array}\end{array}$

$\begin{array}{cc}& \text{ Variance }=\frac{\sum {𝑥}_2^{𝑖}}{𝑛}-(\overline{𝑥}{)}^2\\ & =\frac{(2^2+4^2+\dots +{100}^2)}{50}-{\left(\frac{2+4+\dots .+100}{50}\right)}^2\\ & =\frac{4(1^2+2^2+\dots .+{50}^2)}{50}-(51{)}^2\\ & =4\left(\frac{50\times 51\times 101}{50\times 6}\right)-(51{)}^2\\ & =3434-2601\\ & =833\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& [𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee [\sim 𝑟\wedge \sim 𝑞\wedge 𝑝]\\ & \equiv [𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee [𝑝\wedge (\sim 𝑟\wedge \sim 𝑞\left)\right]\\ & \cdots \left[\text{ Commutativity and associativity }\right]\\ & \equiv [𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee [𝑝\wedge \sim (𝑟\vee 𝑞)\end{array}\\ & \text{ ...[DeMorgan’s Law] }\end{array}$

$\begin{array}{cc}& \equiv 𝑝\wedge \left[\right(𝑞\vee 𝑟)\vee \sim (𝑞\vee 𝑟\left)\right]\\ & \dots \dots [\text{ Distributive and commutative Law] }\\ & \equiv 𝑝\wedge 𝑇\\ & \dots [\text{ Absorption Law] }\\ & \equiv 𝑝\text{... [Identity Law]}\end{array}$

$8𝑓\left(𝑥\right)+6𝑓\left(\frac{1}{𝑥}\right)=𝑥+5\text{.... (i)}$

Differentiating w.r.t. $𝑥$, we get

$8{𝑓}^{\prime }\left(𝑥\right)+6{𝑓}^{\prime }\left(\frac{1}{𝑥}\right)\left(\frac{-1}{{𝑥}^2}\right)=1\text{.... (ii)}$

Substituting $𝑥=-1$ in (i), we get

$8f(-1)+6f(-1)=4$

$\therefore 14f(-1)=4$

$\therefore 𝑓(-1)=\frac{4}{14}=\frac{2}{7}\text{.... (iii)}$

Substituting $𝑥=-1$ in (ii), we get

$\begin{array}{cc}& f^{\prime }(-1)=\frac{1}{2}\text{.... (iv)}\\ & 𝑦={𝑥}^{2}f\left(𝑥\right)\\ \therefore & \frac{d𝑦}{d𝑥}=2𝑥f\left(𝑥\right)+{𝑥}^2{f}^{\prime }\left(𝑥\right)\end{array}$

$\begin{array}{cc}{\therefore \frac{d𝑦}{d𝑥}\mid }_{𝑥=-1}& =2(-1)f(-1)+(-1{)}^2{f}^{\prime }(-1)\\ & =\frac{-4}{7}+\frac{1}{2}\dots \left[\text{ From (iii) and (iv) }\right]\\ & =-\frac{1}{14}\end{array}$

First, we arrange 4 boys and 2 girls (excluding ${𝐵}_1$ and ${𝐺}_1$ ) around the table, which can be done in 5 ! ways.

In any such arrangement, ${𝐵}_1$ and ${𝐺}_1$ can be arranged in 6 available gaps in ${}^6{P}_2=6\times 5$ ways.

$\begin{array}{cc}\therefore \text{ Total number of arrangements }& =5!\times 6\times 5\\ & =6!\times 5\end{array}$

$\begin{array}{cc}& \cot (\mathrm{cosec}\frac{}{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3})\\ & =\cot ({\sin }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{2}{3})\\ & =\cot ⎛⎝⎜{\tan }^{-1}\frac{\frac{3}{5}}{\sqrt{\surd 1-{\left(\frac{3}{5}\right)}^2}}+{\tan }^{-1}\frac{2}{3}⎞⎠⎟\\ & =\cot ({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3})\\ & =\cot [{\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)]\end{array}$

$\begin{array}{cc}& =\cot [{\tan }^{-1}\left(\frac{17}{6}\right)]\\ & =\cot [{\cot }^{-1}\left(\frac{6}{17}\right)]=\frac{6}{17}\end{array}$

Note that $\int \left[𝑥f^{\prime }\right(𝑥)+f(𝑥\left)\right]d𝑥=𝑥f\left(𝑥\right)+c$

$\begin{array}{cc}\therefore & \int (1+𝑥-\frac{1}{𝑥}){𝑒}^{𝑥+\frac{1}{𝑥}}𝑑𝑥\\ & =\int [𝑥{𝑒}^{𝑥+\frac{1}{𝑥}}(1-\frac{1}{{𝑥}^2})+{𝑒}^{𝑥+\frac{1}{𝑥}}]𝑑𝑥\\ & =𝑥{𝑒}^{𝑥+\frac{1}{𝑥}}+𝑐\end{array}$

Let $f\left(𝑥\right)=\frac{\ln (𝜋+𝑥)}{\ln (e+𝑥)}$

$\begin{array}{cc}{f}^{\prime }\left(𝑥\right)& =\frac{\ln (e+𝑥)\times \frac{1}{𝜋+𝑥}-\ln (𝜋+𝑥)\times \frac{1}{e+𝑥}}{[\ln (e+𝑥){]}^2}\\ & =\frac{(e+𝑥)\ln (e+𝑥)-(𝜋+𝑥)\ln (𝜋+𝑥)}{[\ln (e+𝑥){]}^2\times (e+𝑥\left)\right(𝜋+𝑥)}\\ \Rightarrow f^{\prime }\left(𝑥\right)& <0\text{ for all }𝑥>0\dots .[\because 𝜋>e]\end{array}$

$\therefore f\left(𝑥\right)$ is decreasing on $(0,\infty )$.

$\begin{array}{cc}& \sin 𝑥\frac{d𝑦}{d𝑥}+𝑦\cos 𝑥=4𝑥\\ \therefore & \frac{d𝑦}{d𝑥}+𝑦\cot 𝑥=\frac{4𝑥}{\sin 𝑥}\\ & \text{ Here, }P\left(𝑥\right)=\cot 𝑥,Q\left(𝑥\right)=\frac{4𝑥}{\sin 𝑥}\\ & \text{ Integrating factor (I.F. })=e^{\int P\left(𝑥\right)d𝑥}\\ & =e^{\int \mathrm{out}𝑥\mathrm{dx}}=e^{\log |\sin 𝑥|}=\sin 𝑥\\ \therefore & 𝑦\left(\text{ I.F. }\right)=\int Q\left(\text{ I.F. }\right)d𝑥+c\\ \therefore & 𝑦(\sin 𝑥)=\int \frac{4𝑥}{\sin 𝑥}\times \sin 𝑥d𝑥=C\\ \therefore & 𝑦\sin 𝑥=4\int 𝑥d𝑥+C\end{array}$

$\begin{array}{cc}\therefore & 𝑦=\frac{2{𝑥}^2+𝐶}{\sin 𝑥}\text{.... (i)}\\ & 𝑦\left(\frac{𝜋}{2}\right)=0\\ & \Rightarrow \frac{2{\left(\frac{𝜋}{2}\right)}^2+𝐶}{\sin \left(\frac{𝜋}{2}\right)}=0\\ & \Rightarrow \frac{{𝜋}^2}{2}+𝐶=0\\ & \Rightarrow 𝐶=\frac{-{𝜋}^2}{2}\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑦=\frac{2{𝑥}^2-\frac{{𝜋}^2}{2}}{\sin 𝑥}\\ & \text{ is the solution of the given differential equation. }\\ & \begin{array}{cc}\therefore 𝑦\left(\frac{𝜋}{6}\right)& =\frac{2{\left(\frac{𝜋}{6}\right)}^2-\frac{{𝜋}^2}{2}}{\sin \left(\frac{𝜋}{6}\right)}\\ & =\frac{\frac{2{𝜋}^2}{36}-\frac{{𝜋}^2}{2}}{\frac{1}{2}}=\frac{-8}{9}{𝜋}^2\end{array}\end{array}$

$\begin{array}{cc}& I=\int \frac{{𝑥}^2}{(a+b𝑥{)}^2}d𝑥\\ & \text{ Let }a+b𝑥=t\Rightarrow 𝑥=\frac{t-a}{b}\\ \therefore & \mathrm{bd}d𝑥=\mathrm{dt}\\ \therefore & d𝑥=\frac{\mathrm{dt}}{b}\end{array}$

$\begin{array}{cc}\therefore 𝐼& =\int \frac{{\left(\frac{𝑡-𝑎}{𝑏}\right)}^2}{{𝑡}^2}\times \frac{𝑑𝑡}{𝑏}\\ & =\frac{1}{{𝑏}^3}\int \frac{(𝑡-𝑎{)}^2}{{𝑡}^2}𝑑𝑡\\ & =\frac{1}{{𝑏}^3}\int \frac{{𝑡}^2-2𝑎𝑡+{𝑎}^2}{{𝑡}^2}𝑑𝑡\\ & =\frac{1}{{𝑏}^3}[\int 1𝑑𝑡-2𝑎\int \frac{1}{𝑡}+{𝑎}^2\int \frac{1}{{𝑡}^2}𝑑𝑡]\\ & =\frac{1}{{𝑏}^3}[𝑡-2𝑎\log |𝑡|-\frac{{𝑎}^2}{𝑡}]+𝑐\\ & =\frac{1}{{𝑏}^3}[𝑎+𝑏𝑥-2𝑎\log |𝑎+𝑏𝑥|-\frac{{𝑎}^2}{𝑎+𝑏𝑥}]+𝑐\end{array}$

Projection of $\bar{a}$ on $\bar{c}=\frac{\bar{a}\cdot \bar{c}}{\left|\bar{c}\right|}$

$\begin{array}{cc}& \therefore \frac{𝛼+6+2}{\sqrt{\surd 1+4+4}}=\frac{10}{3}\\ & \therefore 𝛼=2\\ & \bar{b}\times \bar{c}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -𝛽& 4\\ 1& 2& -2\end{array}\mid \\ & =(2𝛽-8)\widehat{𝑖}-(-6-4)\widehat{𝑗}+(6+𝛽)\widehat{𝑘}\\ & =(2𝛽-8)\widehat{𝑖}+10\widehat{𝑗}+(6+𝛽)\widehat{𝑘}\end{array}$

Comparing with $-6\widehat{𝑖}+10\widehat{𝑗}+7\widehat{𝑘}$, we get

$\begin{array}{cc}& 2𝛽-8=-6\\ \therefore & 2𝛽=2\\ \therefore & 𝛽=1\\ \therefore & 2𝛼+𝛽=5\end{array}$

Here, ${𝑚}_1+{𝑚}_2=\frac{-2ℎ}{𝑏}\text{.... (i)}$

and $m_1{m}_2=\frac{a}{b}$

$\begin{array}{cc}\therefore {({𝑚}_1-{𝑚}_2)}^2& ={({𝑚}_1+{𝑚}_2)}^2-4{𝑚}_1{𝑚}_2\\ & =\frac{4{ℎ}^2-4𝑎𝑏}{{𝑏}^2}\\ & =\frac{4{ℎ}^2-3{ℎ}^2}{{𝑏}^2}\dots [\because 4𝑎𝑏=3{ℎ}^2\text{ (given) }]\\ & =\frac{{ℎ}^2}{{𝑏}^2}\end{array}$

$\therefore m_1-m_2=\frac{h}{b}\text{.... (ii)}$

On solving (i) and (ii), we get

$\begin{array}{cc}& m_1=\frac{-h}{2b}\text{ and }{m}_2=\frac{-3h}{2b}\\ & \therefore m_1:m_2=1:3\end{array}$

$\begin{array}{cc}& \sin ({\cot }^{-1}(𝑥+1\left)\right)=\cos ({\tan }^{-1}𝑥)\\ & \sin [{\sin }^{-1}\frac{1}{\sqrt{\surd {𝑥}^2+2𝑥+2}}]=\cos [{\cos }^{-1}\frac{1}{\sqrt{\surd 1+{𝑥}^2}}]\\ & \frac{1}{\sqrt{\surd {𝑥}^2+2𝑥+2}}=\frac{1}{\sqrt{\surd 1+{𝑥}^2}}\\ & \Rightarrow \sqrt{\surd 1+{𝑥}^2}=\sqrt{\surd {𝑥}^2+2𝑥+2}\\ & \Rightarrow 1+{𝑥}^2={𝑥}^2+2𝑥+2\\ & \Rightarrow 2𝑥+1=0\\ & \Rightarrow 𝑥=\frac{-1}{2}\end{array}$

$\begin{array}{cc}& \text{ Here, }\\ & 0+2𝑝+2𝑝+3𝑝+{𝑝}^2+2{𝑝}^2+7{𝑝}^2+2𝑝=1\\ \therefore & 9𝑝+10{𝑝}^2=1\\ \therefore & 10{𝑝}^2+9𝑝-1=0\\ \therefore & 10{𝑝}^2+10𝑝-𝑝-1=0\\ \therefore & 10𝑝(𝑝+1)-1(𝑝+1)=0\\ \therefore & 𝑝=\frac{1}{10}\text{ or }𝑝=-1\\ & \text{ But }0\le 𝑝\le 1\\ \therefore & 𝑝=\frac{1}{10}\end{array}$

Given that, $P[(A\cap B^{\prime })\cup (A^{\prime }\cap B)]=\frac{2}{5}\text{.... (i)}$

and $𝑃(𝐴\cup 𝐵)=\frac{1}{2}\text{.... (ii)}$

From (i), we get

$𝑃\left(𝐴\right)+𝑃\left(𝐵\right)-2𝑃(𝐴\cap 𝐵)=\frac{2}{5}$

$\begin{array}{cc}\therefore & 𝑃(𝐴\cup 𝐵)-𝑃(𝐴\cap 𝐵)=\frac{2}{5}\text{.... [From (i)]}\\ \therefore & \frac{1}{2}-𝑃(𝐴\cap 𝐵)=\frac{2}{5}\\ \therefore & 𝑃(𝐴\cap 𝐵)=\frac{1}{2}-\frac{2}{5}=\frac{1}{10}=0.1\end{array}$

$\begin{array}{cc}& {(-2-\frac{1}{3}i)}^3=\frac{𝑥+i𝑦}{27}\\ & {(-2-\frac{1}{3}i)}^3=\frac{1}{27}(-6-i{)}^3\end{array}$

Consider, $(-6-i{)}^3$

$\begin{array}{cc}& =(-6{)}^3+3(-6{)}^2(-𝑖)+3(-6)(-𝑖{)}^2+(-𝑖{)}^3\\ & =-216-108i+18+i\\ & =-198-107i\end{array}$

$\therefore {(-2-\frac{1}{3}i)}^3=\frac{-198-107i}{27}$

Comparing with $\frac{𝑥+i𝑦}{27}$, we get

$\begin{array}{cc}& 𝑥=-198,𝑦=-107\\ & 𝑦-𝑥=-107+198=91\end{array}$

The line passing through $(3,0)$ and $(0,5)$ has the equation

$\frac{𝑥}{3}+\frac{𝑦}{5}=1⟹5𝑥+3𝑦=15$

Since the shaded region does not contain the origin $(0,0)$, the corresponding inequality is $5𝑥+3𝑦\ge 15$.

The line passing through $(6,0)$ and $(0,3)$ has the equation

$\frac{𝑥}{6}+\frac{𝑦}{3}=1⟹𝑥+2𝑦=6$

Since the shaded region does not contain the origin $(0,0)$, the corresponding inequality is $𝑥+2𝑦\ge 6$.

The vertical boundary line is $𝑥=7$, and the shaded region lies to its left, which gives $𝑥\le 7$.

The horizontal boundary line is $𝑦=6$, and the shaded region lies below it, which gives $𝑦\le 6$.

Since the region is located in the first quadrant, $𝑥\ge 0$ and $𝑦\ge 0$.

Therefore, the system of inequations defining the shaded region is $𝑥+2𝑦\ge 6$, $5𝑥+3𝑦\ge 15$, $𝑥\le 7$, $𝑦\le 6$, and $𝑥,𝑦\ge 0$.

Hence, the correct option is (B).

Let $\overline{𝑟}$ and $\overline{𝑠}$ be the position vectors of points $𝑅$ and $𝑆$ respectively.

$\therefore \bar{r}=\frac{3p+2q}{3+2}$ and $\bar{s}=\frac{3p-2q}{3-2}$

As OS and OR are perpendicular, we get

$\begin{array}{cc}& \overline{𝑟}\cdot s=0\\ \therefore & \left(\frac{3𝑝+2𝑞}{5}\right)\left(\frac{3𝑝-2𝑞}{1}\right)=0\\ \therefore & 9{𝑝}^2-6𝑝𝑞+6𝑝𝑞-4{𝑞}^2=0\\ \therefore & 9{𝑝}^2=4{𝑞}^2\end{array}$

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

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

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

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

$\therefore 𝑎=\frac{1}{\sqrt{\surd 3}}$.

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

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

Let $f\left(𝑥\right)={\log }_0𝑥$

$\therefore {𝑓}^{\prime }\left(𝑥\right)=\frac{1}{𝑥}$

Here, $𝑎=1000,h=2$

$\therefore f(a+h)\approx f\left(a\right)+{\mathrm{hf}}^{\prime }\left(a\right)$

$\approx 𝑓\left(1000\right)+2f^{\prime }\left(1000\right)$

$\approx {\log }_{𝑒}{10}^3+2\times \frac{1}{1000}$

$\approx 3\left(0.4343\right)+0.002$

$\approx 1.3029+0.002$

$\approx 1.3049$

$\therefore {\log }_{𝑒}1002\approx 1.3049$

Now, ${\log }_{10}1002\approx \frac{{\log }_{8}1002}{{\log }_{𝑒}10}$

$\approx \frac{1.3049}{0.4343}\approx 3.0009$

$\sqrt{\surd 𝑥}+\sqrt{\surd 𝑦}=\sqrt{\surd 𝑎}$

Differentiating w.r.t. $𝑥$, we get

$\frac{1}{2\sqrt{\surd 𝑥}}+\frac{1}{2\sqrt{\surd 𝑦}}\frac{𝑑𝑦}{𝑑𝑥}=0$

$\therefore$ Slope of the tangent is $\frac{d𝑦}{\mathrm{dx}}=-\sqrt{\surd \frac{𝑦}{𝑥}}$

$\therefore$ Equation of the tangent at $({𝑥}_1,{𝑦}_1)$ is

$\begin{array}{cc}\therefore & (𝑦-{𝑦}_1)=-\sqrt{\surd \frac{{𝑦}_1}{{𝑥}_1}}(𝑥-{𝑥}_1)\\ & \sqrt{\surd {𝑥}_1}𝑦+\sqrt{\surd {𝑦}_1}𝑥=\sqrt{\surd {𝑥}_1}{𝑦}_1+\sqrt{\surd {𝑦}_1}{𝑥}_1\\ \therefore & \sqrt{\surd {𝑦}_1}𝑥+\sqrt{\surd {𝑥}_1}𝑦=\sqrt{\surd {𝑥}_1{𝑦}_1}(\sqrt{\surd {𝑥}_1}+\sqrt{\surd {𝑦}_1})\\ \therefore & \sqrt{\surd {𝑦}_1}𝑥+\sqrt{\surd {𝑥}_1}𝑦=\sqrt{\surd {𝑥}_1{𝑦}_{1}a}\\ \therefore & \frac{𝑥}{\sqrt{\surd {𝑥}_1𝑎}}+\frac{𝑦}{\sqrt{\surd {𝑦}_{1}a}}=1\\ \therefore & 𝑥\text{-intercept }+𝑦\text{-intercept }=\\ & =\sqrt{\surd a}(\sqrt{\surd {𝑥}_1}+{𝑦}_1)\\ & =\sqrt{\surd a}\times \sqrt{\surd a}=a\end{array}$

Let the angles of the triangle be $4𝑥,𝑥$ and $𝑥$.

$\therefore 4𝑥+𝑥+𝑥={180}^{\circ }\Rightarrow 6𝑥={180}^{\circ }\Rightarrow 𝑥={30}^{\circ }$

$\frac{\sin {120}^{\circ }}{a}=\frac{\sin {30}^{\circ }}{b}=\frac{\sin {30}^{\circ }}{c}$

$\begin{array}{cc}\therefore 𝑎& :(𝑎+𝑏+𝑐)\\ & =(\sin {120}^{\circ }):(\sin {120}^{\circ }+\sin {30}^{\circ }+\sin {30}^{\circ })\\ & =\frac{\sqrt{\surd 3}}{2}:\frac{\sqrt{\surd 3}+2}{2}=\sqrt{\surd 3}:\sqrt{\surd 3}+2\end{array}$

$\begin{array}{cc}& {\tan }^{-1}\left(2𝑥\right)+{\tan }^{-1}\left(3𝑥\right)=\frac{𝜋}{4}\\ \therefore & {\tan }^{-1}\left(\frac{2𝑥+3𝑥}{1-\left(2𝑥\right)\left(3𝑥\right)}\right)=\frac{𝜋}{4}\end{array}$

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