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

$𝑦=\left[\right(𝑥+1\left)\right(2𝑥+1\left)\right(3𝑥+1)\dots (𝑛𝑥+1){]}^2$

Taking ' log' on both sides, we get

$\begin{array}{cc}& \log 𝑦=2[\log (𝑥+1)+\log (2𝑥+1)+\log (3𝑥+1)\\ & +\dots +\log (𝑛𝑥+1)]\end{array}$

Differentiating w.r.t. $𝑥$, we get

$\begin{array}{cc}& \frac{1}{𝑦}\cdot \frac{d𝑦}{d𝑥}=2(\frac{1}{𝑥+1}+\frac{2}{2𝑥+1}+\frac{3}{3𝑥+1}+\dots +\frac{n}{n𝑥+1})\\ \therefore & \frac{d𝑦}{d𝑥}=2𝑦(\frac{1}{𝑥+1}+\frac{2}{2𝑥+1}+\frac{3}{3𝑥+1}+\dots +\frac{n}{n𝑥+1})\end{array}$

Now at $𝑥=0,𝑦=$ $[\underset{𝑛\text{ times }}{\underset{}{\left(1\right)\left(1\right)\left(1\right)\dots \left(1\right)}}{]}^2$ $=1$

$\begin{array}{cc}\therefore {\left(\frac{d𝑦}{d𝑥}\right)}_{𝑥=0}& =2\left(1\right)(\frac{1}{0+1}+\frac{2}{0+1}+\frac{3}{0+1}+\dots +\frac{n}{0+1})\\ & =2(1+2+3+\dots +n)\\ & =2\times \frac{n(n+1)}{2}=n(n+1)\end{array}$

The sum of all probabilities in a probability distribution is always unity.

$\begin{array}{cc}& \therefore k+0.3+0.15+0.15+0.1+2k=1\\ & \Rightarrow 3k+0.7=1\\ & \Rightarrow 3k=0.3\\ & \Rightarrow k=0.1\\ & \therefore E\left(X\right)=\sum {𝑥}_i\cdot P\left({𝑥}_i\right)\\ & =0\left(0.1\right)+1\left(0.3\right)+2\left(0.15\right)+3\left(0.15\right)\\ & +4\left(0.1\right)+5\left(0.2\right)\\ & =0.3+0.3+0.45+0.4+1\\ & =2.45\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\int \frac{1}{7+6𝑥-{𝑥}^2}d𝑥& =\int \frac{1}{7+6𝑥-{𝑥}^2-9+9}d𝑥\\ & =\int \frac{1}{16-({𝑥}^2-6𝑥+9)}d𝑥\\ & =\int \frac{1}{4^2-(𝑥-3{)}^2}d𝑥\\ & =\frac{1}{2\left(4\right)}\log \mid \frac{4+(𝑥-3)}{4-(𝑥-3)}\mid +c\\ & =\frac{1}{8}\log \mid \frac{1+𝑥}{7-𝑥}\mid +c\end{array}\end{array}$

$\begin{array}{cc}& \cos ({18}^{\circ }-A)\cos ({18}^{\circ }+A)-\cos ({72}^{\circ }-A)\cos ({72}^{\circ }+A)\\ \\ & =\cos ({18}^{\circ }-A)\cos [{90}^{\circ }-({72}^{\circ }-A)]-\cos ({72}^{\circ }-A)\cos [{90}^{\circ }-({18}^{\circ }-A)]\\ \\ & =\sin ({72}^{\circ }-A)\cos ({18}^{\circ }-A)-\cos ({72}^{\circ }-A)\sin ({18}^{\circ }-A)\\ \\ & =\sin [({72}^{\circ }-𝐴)-({18}^{\circ }-𝐴)]=\sin {54}^{\circ }\end{array}$

$\begin{array}{cc}\int \frac{𝑑𝑥}{1+3{\sin }^2𝑥}& =\int \frac{𝑑𝑥}{{\sin }^2𝑥+{\cos }^2𝑥+3{\sin }^2𝑥}\\ & =\int \frac{𝑑𝑥}{4{\sin }^2𝑥+{\cos }^2𝑥}\\ & =\int \frac{{\sec }^2𝑥𝑑𝑥}{4{\tan }^2𝑥+1}\\ & =\frac{1}{4}\int \frac{{\sec }^2𝑥𝑑𝑥}{{\tan }^2𝑥+\frac{1}{4}}\end{array}$

$\begin{array}{cc}& \text{ Put }t=\tan 𝑥\Rightarrow \mathrm{dt}={\sec }^2𝑥d𝑥\\ & \begin{array}{cc}\therefore \int \frac{d𝑥}{1+3{\sin }^2𝑥}& =\frac{1}{4}\int \frac{\mathrm{dt}}{t^2+{\left(\frac{1}{2}\right)}^2}=\frac{1}{4}\cdot 2{\tan }^{-1}\left(2t\right)+c\\ & =\frac{1}{2}{\tan }^{-1}\left(2t\right)+c\\ & =\frac{1}{2}{\tan }^{-1}(2\tan 𝑥)+c\end{array}\\ \therefore & 𝑓\left(𝑥\right)=2\tan 𝑥\end{array}$

$\begin{array}{cc}& \text{ Since }\sum _{𝑥=1}^5𝑃(𝑋=𝑥)=1\\ & {𝑘}^2+2𝑘+𝑘+2𝑘+5{𝑘}^2=1\\ & \Rightarrow 6{𝑘}^2+5𝑘-1=0\\ & \Rightarrow (𝑘+1)(6𝑘-1)=0\\ & \Rightarrow 𝑘=\frac{1}{6}\dots [\because 𝑘\ge 0]\end{array}$

$\begin{array}{cc}𝑃(𝑋\ge 2)& =𝑃(𝑋=2)+𝑃(𝑋=3)+𝑃(𝑋=4)+𝑃(𝑋=5)\\ & =\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{5}{36}\\ & =\frac{12+6+12+5}{36}\\ & =\frac{35}{36}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& f\left(𝑥\right)=⎧⎩⎨⎪⎪\begin{array}{cc}\left|𝑥\right|+\left[𝑥\right],& -1\le 𝑥<1\\ 𝑥+\left|𝑥\right|,& 1\le 𝑥<2\\ 𝑥+\left[𝑥\right],& 2\le 𝑥\le 3\end{array}\\ \therefore & f\left(𝑥\right)=⎧⎩⎨⎪⎪\begin{array}{cc}-(𝑥+1),& -1\le 𝑥<0\\ 𝑥,& 0\le 𝑥<1\\ 2𝑥,& 1\le 𝑥<2\\ 𝑥+2,& 2\le 𝑥<3\\ 𝑥+3,& 𝑥=3\end{array}\end{array}\\ & \therefore f\left(𝑥\right)\text{ is discontinuous at }𝑥=0,1,3.\end{array}$

$\overline{𝑎}=\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘},\overline{𝑏}=\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘}\text{ and }\overline{𝑐}=\widehat{𝑖}-\widehat{𝑗}-\widehat{𝑘}$

$\bar{v}$ is in the plane of $\bar{a}$ and $\bar{b}$.

$\begin{array}{cc}& \bar{v}=m\bar{a}+n\bar{b}.\\ & \Rightarrow \bar{v}=(𝑚+𝑛)\hat{i}+(𝑚-𝑛)\widehat{𝑗}+(𝑚+𝑛)\hat{k}\text{... (i)}\end{array}$

Projection of $\overline{𝑣}$ on $\overline{𝑐}=\frac{\overline{𝑣}\cdot \overline{𝑐}}{\left|\overline{𝑐}\right|}=\frac{1}{\sqrt{\surd 3}}$

$\begin{array}{cc}& \Rightarrow \frac{(𝑚+𝑛)\left(1\right)+(𝑚-𝑛)(-1)+(𝑚+𝑛)(-1)}{\sqrt{\surd 1+1+1}}=\frac{1}{\sqrt{\surd 3}}\\ & \Rightarrow -𝑚+𝑛=1\\ & \Rightarrow n=1+m\\ & \bar{v}=(2m+1)\hat{i}-\hat{j}+(2m+1)\hat{k}\dots \left[\text{ From (i) }\right]\end{array}$

When $m=1$ then $\bar{v}=3\hat{i}-\hat{j}+3\hat{k}$

$A{𝑥}^2+\mathrm{By}{𝑦}^2=1$

Differentiating w.r.t. $𝑥$, we get

$2A𝑥+2B𝑦\frac{d𝑦}{d𝑥}=0\text{.... (i)}$

Again, differentiating w.r.t. $𝑥$, we get

$2A+2B[{\left(\frac{d𝑦}{d𝑥}\right)}^2+𝑦\frac{d^2𝑦}{\mathrm{dx}}{}^2]=0\text{.... (ii)}$

$\begin{array}{cc}& \text{ Solving (i) and (ii), we get }\\ & 𝑥𝑦\frac{d^2𝑦}{d{𝑥}^2}+𝑥{\left(\frac{d𝑦}{d𝑥}\right)}^2-𝑦\frac{d𝑦}{d𝑥}=0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \left|z\right|+z=2+i\\ & \Rightarrow \sqrt{\surd {𝑥}^2+{𝑦}^2}+𝑥+i𝑦=2+i\end{array}\\ & \text{ Equating real and imaginary parts, we get }\\ & \begin{array}{cc}& \sqrt{\surd {𝑥}^2+{𝑦}^2}+𝑥=2\text{ and }𝑦=1\\ & \Rightarrow \sqrt{\surd {𝑥}^2+1}=2-𝑥\\ & \Rightarrow {𝑥}^2+1=4-4𝑥+{𝑥}^2\\ & \Rightarrow 4𝑥=3\\ & \Rightarrow 𝑥=\frac{3}{4}\\ \therefore & 𝑧=\frac{3}{4}+𝑖\\ & \Rightarrow \left|𝑧\right|=\sqrt{\surd {\left(\frac{3}{4}\right)}^2+1^2}=\frac{5}{4}\end{array}\end{array}$

Let $𝜃$ be the angle between $\bar{a}$ and $\bar{b}$.

Since $5\overline{𝑎}+4\overline{𝑏}$ and $\overline{𝑎}-2\overline{𝑏}$ are perpendicular to each other.

$\begin{array}{cc}\therefore & (5\overline{𝑎}+4\overline{𝑏})\cdot (\overline{𝑎}-2\overline{𝑏})=0\\ & \Rightarrow 5(\overline{𝑎}\cdot \overline{𝑎})-6(\overline{𝑎}\cdot \overline{𝑏})-8(\overline{𝑏}\cdot \overline{𝑏})=0\\ & \Rightarrow 5|\bar{a}{|}^2-6|\overline{𝑎}\left|\right|\overline{𝑏}|\cos 𝜃-8|\bar{b}{|}^2=0\\ & \Rightarrow 5-6\cos 𝜃-8=0\\ & \Rightarrow \cos 𝜃=\frac{1}{2}\\ & \Rightarrow 𝜃=\frac{2𝜋}{3}\end{array}$

$\begin{array}{cc}\text{ Let }𝐼& =\int \frac{\sec 𝑥\tan 𝑥}{9-16{\tan }^2𝑥}𝑑𝑥\\ \\ & =\int \frac{\sec 𝑥\tan 𝑥}{9-16({\sec }^2𝑥-1)}d𝑥\\ \\ & =\int \frac{\sec 𝑥\tan 𝑥}{25-16{\sec }^2𝑥}d𝑥\end{array}$

$\begin{array}{cc}& \text{ Put }\sec 𝑥=t\Rightarrow \sec 𝑥\tan 𝑥d𝑥=\mathrm{dt}\\ \\ & \begin{array}{cc}\therefore 𝐼& =\int \frac{𝑑𝑡}{5^2-(4𝑡{)}^2}\\ \\ & =\frac{1}{2\left(5\right)}\cdot \frac{1}{4}\log \mid \frac{5+4𝑡}{5-4𝑡}\mid \\ \\ & =\frac{1}{40}\log \mid \frac{5+4\sec 𝑥}{5-4\sec 𝑥}\mid +𝑐\end{array}\end{array}$

$\begin{array}{cc}& n\left(A\right)=4,n\left(B\right)=2\\ \therefore & n(A\times B)=4\times 2=8\\ & \text{ Required numbers }={}^8{C}_3+{}^8{C}_4+\dots +{}^8{C}_8\\ & =2^8-({}^8{C}_0+{}^8{C}_1+{}^8{C}_2)\\ & =256-37\\ & =219\end{array}$

Let $𝑀$ be the mass of substance at time $𝑡$. Then,

$\begin{array}{cc}& \frac{\mathrm{dM}}{\mathrm{dt}}=-\mathrm{kM}\text{, where }k>0\\ & \Rightarrow \frac{\mathrm{dM}}{M}=-\mathrm{kdt}\end{array}$

Integrating on both sides, we get

$\log 𝑀=-𝑘𝑡+𝑐$

When $t=0,M=m_0$

$\begin{array}{cc}\therefore & \log {𝑚}_0=0+𝑐\\ \therefore & \Rightarrow c=\log {𝑚}_0\\ & \log 𝑀=-𝑘𝑡+\log {𝑚}_0\\ & \Rightarrow \log \frac{𝑀}{{𝑚}_0}=-𝑘𝑡\end{array}$

When $t=h,M=\frac{1}{2}{m}_0$

$\begin{array}{cc}\therefore & \log \left(\frac{\frac{1}{2}{𝑚}_0}{{𝑚}_0}\right)=-𝑘ℎ\\ & \Rightarrow \log \frac{1}{2}=-𝑘ℎ\\ & \Rightarrow \log 2=𝑘ℎ\\ & \Rightarrow 𝑘=\frac{\log 2}{ℎ}\text{.... (i)}\end{array}$

Initial decay rate,

$\begin{array}{cc}\frac{\mathrm{dM}}{\mathrm{dt}}& =-{\mathrm{km}}_0\\ & =\frac{-m_0}{h}\log 2\text{...[From (i)]}\end{array}$

$\begin{array}{cc}& (𝑎+\sqrt{\surd 2}𝑏\cos 𝑥)(𝑎-\sqrt{\surd 2}𝑏\cos 𝑦)={𝑎}^2-{𝑏}^2\\ \\ & \text{ Differentiating both sides w.r.t. }𝑦\text{, we get }\\ \\ & \begin{array}{cc}& (𝑎+\sqrt{\surd 2}𝑏\cos 𝑥)(\sqrt{\surd 2}𝑏\sin 𝑦)\\ \\ & +(a-\sqrt{\surd 2}b\cos 𝑦)(-\sqrt{\surd 2}b\sin 𝑥\frac{d𝑥}{d𝑦})=0\\ \\ & \Rightarrow \frac{d𝑥}{d𝑦}=\frac{\sqrt{\surd 2}b\sin 𝑦(a+\sqrt{\surd 2}b\cos 𝑥)}{\sqrt{\surd 2}b\sin 𝑥(a-\sqrt{\surd 2}b\cos 𝑦)}\\ \\ & \Rightarrow {\left(\frac{d𝑥}{d𝑦}\right)}_{(\frac{𝜋}{4}\cdot \frac{𝜋}{4})}=\frac{b(a+b)}{b(a-b)}=\frac{a+b}{a-b}\end{array}\end{array}$

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

Let $P=A^2{B}^2-B^2{A}^2$

$\begin{array}{cc}\therefore P^{𝑇}& ={(A^2{B}^2-B^2{A}^2)}^T\\ & ={\left(A^2{B}^2\right)}^T-{\left(B^2{A}^2\right)}^T\\ & ={\left(B^2\right)}^T{\left(A^2\right)}^T-{\left(A^2\right)}^T{\left(B^2\right)}^T\\ & =B^2{A}^2-A^2{B}^2\cdots [\because A^T=A\text{ and }{B}^T=-B]\\ & =-(A^2{B}^2-B^2{A}^2)\\ & =-P\end{array}$

$\therefore 𝑃$ is a skew - symmetric matrix.

$\therefore \det \left(P\right)=0$

$\therefore$ The given system of equations has infinitely many solutions.

$\begin{array}{cc}& f\left(𝑥\right)={𝑥}^3-10{𝑥}^2+200𝑥-10\\ & \Rightarrow f^{\prime }\left(𝑥\right)=3{𝑥}^2-20𝑥+200\end{array}$

For $f\left(𝑥\right)$ to be increasing $f^{\prime }\left(𝑥\right)>0$

$\begin{array}{cc}& \Rightarrow 3{𝑥}^2-20𝑥+200>0\\ & \Rightarrow 3({𝑥}^2-\frac{20}{3}𝑥+\frac{200}{3}+\frac{100}{9}-\frac{100}{9})>0\\ & \Rightarrow 3[{(𝑥-\frac{10}{3})}^2+\frac{500}{9}]>0\\ & \Rightarrow 3{(𝑥-\frac{10}{3})}^2+\frac{500}{3}>0\end{array}$

Always increasing throughout real line.

$\begin{array}{cc}& {𝑥}^2+{𝑦}^2-𝑥=0\\ & C_1=(\frac{1}{2},0),r_1=\sqrt{\surd {\left(\frac{1}{2}\right)}^2+0^2-0}=\frac{1}{2}\\ & {𝑥}^2+{𝑦}^2+𝑥=0\\ & C_2=(-\frac{1}{2},0),r_2=\sqrt{\surd {\left(\frac{1}{2}\right)}^2+0^2-0}=\frac{1}{2}\\ & C_1{C}_2=\sqrt{\surd {(\frac{1}{2}+\frac{1}{2})}^2+0^2}=1\\ & r_1+r_2=\frac{1}{2}+\frac{1}{2}=1\\ \therefore & C_1{C}_2=r_1{r}_2\\ & \Rightarrow \text{ The given circles touch each other externally. }\\ & \Rightarrow \text{ Number of common tangents }=3\end{array}$

$\begin{array}{cc}& \bar{a}\times \bar{b}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 1& 1& 0\end{array}\mid \\ & =2\hat{i}-2\hat{j}+\hat{k}\\ \therefore & |\bar{a}\times \bar{b}|=\sqrt{\surd 4+4+1}=3\text{.... (i)}\\ & |\bar{c}-\bar{a}|=2\sqrt{\surd 2}\\ & \Rightarrow (\bar{c}-\bar{a}{)}^2=8\end{array}$

$\begin{array}{cc}& \Rightarrow |\bar{c}{|}^2+|\bar{a}{|}^2-2\bar{c}\cdot \bar{a}=8\\ & \Rightarrow |\bar{c}{|}^2+9-2|\bar{c}|=8\dots [\because \bar{a}\cdot \bar{c}=\left|\bar{c}\right|]\\ & \Rightarrow |\bar{c}{|}^2-2|\bar{c}|+1=0\\ & \Rightarrow \left(\right|\overrightarrow{c}|-\bar{c}\cdot \bar{c}=|\bar{c}\left|\right]\\ & <text>\Rightarrow </text><text>\mid </text>\bar{c}=1\text{.... (i)}\end{array}$

$\begin{array}{cc}& \text{ Now, }\left|\right(\bar{a}\times \bar{b})\times \bar{c}|\\ & =|\bar{a}\times \bar{b}|\cdot \left|\bar{c}\right|\sin {30}^{\circ }\\ & =\left(3\right)\left(1\right)\left(\frac{1}{2}\right)\text{.... [From (i) and (ii)]}\\ & =\frac{3}{2}\end{array}$

${𝑒}^{\sin 𝑥}-{𝑒}^{-\sin 𝑥}-4=0$

Let $e^{\sin 𝑥}=𝑦$, then the given equation can be written as ${𝑦}^2-4𝑦-1=0\Rightarrow 𝑦=2\pm \sqrt{\surd 5}$

But the value of $𝑦=e^{\sin 𝑥}$ is always positive, so

$\begin{array}{cc}𝑦=2+\sqrt{\surd 5}& \dots [\because 2<\sqrt{\surd 5}]\\ \Rightarrow {𝑒}^{\sin 𝑥}=2+\sqrt{\surd 5}& \end{array}$

$\begin{array}{cc}\Rightarrow e^{\sin 𝑥}>e& \dots [\because 2+\sqrt{\surd 5}>e]\\ \Rightarrow \sin 𝑥>1\end{array}$

which is not possible, since $\sin 𝑥$ cannot be greater than 1.

Hence, no real value of $𝑥$ satisfies the given equation.

$\begin{array}{cc}& \text{ Let }𝐼=\int \frac{𝑑𝑥}{5+4\sin 𝑥}\\ & \text{ Put }\tan \left(\frac{𝑥}{2}\right)=t\\ & \therefore 𝑥=2{\tan }^{-1}t\\ & \therefore d𝑥=\frac{2\mathrm{dt}}{1+t^2}\text{ and }\sin 𝑥=\frac{2t}{1+t^2}\\ & \therefore I=\int \frac{\frac{2\mathrm{dt}}{1+t^2}}{5+4\left(\frac{2t}{1+t^2}\right)}\\ & =2\int \frac{\mathrm{dt}}{5+5t^2+8t}\\ & =\frac{2}{5}\int \frac{\mathrm{dt}}{t^2+\frac{8t}{5}+1}\\ & =\frac{2}{5}\int \frac{𝑑𝑡}{{𝑡}^2+\frac{8}{5}𝑡+1-\frac{16}{25}+\frac{16}{25}}\\ & =\frac{2}{5}\int \frac{\mathrm{dt}}{{(𝑡+\frac{4}{5})}^2+{\left(\frac{3}{5}\right)}^2}\\ & \begin{array}{cc}& =\frac{2}{5}\cdot \frac{1}{\frac{3}{5}}{\tan }^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)+c\\ & =\frac{2}{3}{\tan }^{-1}\left[\frac{5\tan \left(\frac{𝑥}{2}\right)+4}{3}\right]+c\end{array}\end{array}$