MHT CET 2023 10th May Evening Shift (Mathematics) Question Paper with Solutions

$$\begin{aligned} &\begin{array}{ll} & \text { Using }|\operatorname{adj} A|=|A|^{n-1} \\ & B u t \quad B=\operatorname{Adj}(A) \quad ....[\text{Given}] \\ \therefore \quad & |B|=|A|^2 \\ \Rightarrow & \left|\begin{array}{ccc} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right|=|A|^2 \\ \Rightarrow & 24-4 \alpha-4=4^2 \\ \Rightarrow & 20-4 \alpha=16 \\ \Rightarrow & \alpha=1 \end{array}\\ \end{aligned}$$

$$\begin{aligned} & \tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right) \\ & =\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \\ & =\frac{3 \pi}{4} \end{aligned}$$

Given differential equation is

$$\begin{aligned} & \quad \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{3 x^2}{1+x^3}\right) y=\frac{1}{x^3+1} \\ & \text { Here, } \mathrm{P}=\frac{3 x^2}{1+x^3}, \mathrm{Q}=\frac{1}{x^3+1} \\ & \therefore \quad \text { I.F. }=\mathrm{e}^{\int \frac{3 x^2}{1+x^3} \mathrm{~d} x}=\mathrm{e}^{\log \left(1+x^3\right)}=\left(1+x^3\right) \end{aligned}$$

$\therefore \quad$ Solution of the given equation is

$$\begin{aligned} & y\left(1+x^3\right)=\int \frac{1}{1+x^3} \cdot\left(1+x^3\right) \mathrm{d} x+\mathrm{c} \\ & \Rightarrow y\left(1+x^3\right)=x+\mathrm{c} \end{aligned}$$

$$\begin{array}{ll} (p \wedge q) \vee r & (p \vee s) \leftrightarrow(q \wedge r) \\ \equiv(T \wedge T) \vee F & \equiv(T \vee F) \leftrightarrow(T \wedge F) \\ \equiv T \vee F & \equiv T \leftrightarrow F \\ \equiv T & \equiv F \end{array}$$

Differentiability at $x=\pi$ :

L.h.lim

=\lim _\limits{h \rightarrow 0} \frac{|\pi-h-\pi|\left(e^{|\pi-h|}-1\right) \sin |\pi-h|-0}{-h}=0

R.h.lim

=\lim _\limits{h \rightarrow 0} \frac{|\pi+h-\pi|\left(e^{|\pi+h|}-1\right) \sin |\pi+h|-0}{h}=0

Differentiability at $\boldsymbol{x}=\mathbf{0}$ :

L.h.\lim =\lim _\limits{\mathrm{h} \rightarrow 0} \frac{|-\mathrm{h}-\pi|\left(\mathrm{e}^{|-\mathrm{h}|}-1\right) \sin |-\mathrm{h}|-0}{-\mathrm{h}}=0

R.h.lim =\lim _\limits{h \rightarrow 0} \frac{|h-\pi|\left(e^{|h|}-1\right) \sin |h|-0}{h}=0

The function $\mathrm{f}(x)$ is differentiable at $x=0, \pi$.

$\Rightarrow$ Set $\mathrm{S}$ is an empty set.

$$\begin{aligned} & \text { Let } \bar{a}=\hat{i}-6 \hat{j}+10 \hat{k} \text {, } \\ & \overline{\mathrm{b}}=-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \text {, } \\ & \overline{\mathrm{c}}=5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}} \text {, } \\ & \overline{\mathrm{d}}=7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & \therefore \quad \overline{\mathrm{AB}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \overline{\mathrm{AC}}=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+(\lambda-10) \hat{\mathrm{k}} \\ & \overline{\mathrm{AD}}=6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \therefore \quad \text { Volume of tetrahedron }=\frac{1}{6}\left[\begin{array}{lll} \overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}} \end{array}\right] \\ & \Rightarrow 11=\frac{1}{6}\left|\begin{array}{ccc} -2 & 3 & -3 \\ 4 & 5 & \lambda-10 \\ 6 & 2 & -3 \end{array}\right| \\ & \Rightarrow 11=\frac{1}{6}\{-2(-15-2 \lambda+20)-3(-12-6 \lambda+60) -3(8-30)\}\\ & \Rightarrow \lambda=7 \\ \end{aligned}$$

$$\begin{aligned} & x=3 \tan \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=3 \sec ^2 \mathrm{t} \\ & y=3 \sec \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{dt}}=3 \sec \mathrm{t} \tan \mathrm{t} \\ & \text { Now, } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{3 \operatorname{sect} \tan \mathrm{t}}{3 \sec ^2 \mathrm{t}}=\sin \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d}^2 y}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dt}}(\sin \mathrm{t}) \cdot \frac{\mathrm{dt}}{\mathrm{d} x} \\ & =\cos \mathrm{t} \times \frac{1}{3 \sec ^2 \mathrm{t}} \\ & \therefore \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\cos ^3 \mathrm{t}}{3} \\ & \therefore \quad\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)_{\left(1-\frac{\pi}{4}\right)}=\frac{\left(\cos \frac{\pi}{4}\right)^3}{3}=\frac{1}{6 \sqrt{2}} \\ \end{aligned}$$

$$\begin{aligned} & \lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} \\ & =\lim _{x \rightarrow a} \frac{(\sqrt{a+2 x}-\sqrt{3 x})(\sqrt{a+2 x}+\sqrt{3 x})(\sqrt{3 a+x}+2 \sqrt{x})}{(\sqrt{3 a+x}-2 \sqrt{x})(\sqrt{3 a+x}+2 \sqrt{x})(\sqrt{a+2 x}+\sqrt{3 x})} \\ & =\lim _{x \rightarrow a} \frac{(a+2 x-3 x)(\sqrt{3 a+x}+2 \sqrt{x})}{(3 a+x-4 x)(\sqrt{a+2 x}+\sqrt{3 x})} \\ & =\frac{(\sqrt{3 a+a}+2 \sqrt{a})}{3(\sqrt{a+2 a}+\sqrt{3 a})} \\ & =\frac{1}{3} \cdot \frac{(2 \sqrt{a}+2 \sqrt{a})}{(\sqrt{3 a}+\sqrt{3 a})} \\ & =\frac{4 \sqrt{a}}{6 \sqrt{3 a}} \\ & =\frac{2}{3 \sqrt{3}} \end{aligned}$$

$$\begin{aligned} & \mathrm{h}(x)=(f \circ g)(x) \\ & \Rightarrow \mathrm{h}(x)=\mathrm{f}\left(x^2-x\right) \\ & \mathrm{h}(x)=\mathrm{e}^{x^2-x}-x^2+x \\ \therefore \quad & \mathrm{h}^{\prime}(x)=\mathrm{e}^{x^2-x}(2 x-1)-2 x+1 \\ & \Rightarrow \mathrm{h}^{\prime}(x)=\left(\mathrm{e}^{x^2-x}-1\right)(2 x-1) \end{aligned}$$

For function $\mathrm{h}(x)$ to be increasing,

$$\begin{aligned} & \mathrm{h}^{\prime}(x) \geq 0 \\ & \Rightarrow\left(\mathrm{e}^{x^2-x}-1\right)(2 x-1) \geq 0 \\ & \Rightarrow x \in\left[0, \frac{1}{2}\right] \cup[1, \infty) \end{aligned}$$

$$\begin{aligned} & \text { Volume of parallelopiped }=\left|\begin{array}{lll} 1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1 \end{array}\right| \\ & \therefore \quad \mathrm{V}=1+\alpha^3-\alpha \\ \end{aligned}$$

For maxima or minima,

$$\begin{aligned} & \frac{\mathrm{dV}}{\mathrm{d} \alpha}=0 \\ & \Rightarrow 3 \alpha^2-1=0 \\ & \Rightarrow \alpha= \pm \frac{1}{\sqrt{3}} \end{aligned}$$

Now, $\frac{d^2 V}{d \alpha^2}=6 \alpha$

For $\alpha=\frac{1}{\sqrt{3}}$,

$\frac{d^2 \mathrm{~V}}{d x^2}>0$

$\therefore \quad V$ is minimum at $\alpha=\frac{1}{\sqrt{3}}$.

$$\begin{aligned} f(x) & =\frac{4}{x^4}\left[1-\cos \frac{x}{2}-\cos \frac{x}{4}+\cos \frac{x}{2} \cdot \cos \frac{x}{4}\right] \\ & =\frac{4}{x^4}\left[\left(1-\cos \frac{x}{2}\right)-\cos \frac{x}{4}\left(1-\cos \frac{x}{2}\right)\right] \\ & =\frac{4}{x^4}\left[\left(1-\cos \frac{x}{2}\right)\left(1-\cos \frac{x}{4}\right)\right] \end{aligned}$$

$\mathrm{f}(x)$ is continuous at $x=0$. .... [Given]

$$\begin{aligned} & \therefore \quad \mathrm{f}(0)=\lim _{x \rightarrow 0} \mathrm{f}(x) \\ & =\lim _{x \rightarrow 0} \frac{4}{x^4}\left(1-\cos \frac{x}{2}\right)\left(1-\cos \frac{x}{4}\right) \\ & =4 \lim _{x \rightarrow 0}\left(\frac{2 \sin ^2 \frac{x}{4}}{x^2}\right) \times\left(\frac{2 \sin ^2 \frac{x}{8}}{x^2}\right) \\ & =16 \lim _{x \rightarrow 0}\left(\frac{\sin ^2 \frac{x}{4}}{\frac{x^2}{16}}\right) \times \frac{1}{16} \times \lim _{x \rightarrow 0}\left(\frac{\sin ^2 \frac{x}{8}}{\frac{x^2}{64}}\right) \times \frac{1}{64} \\ & =16 \times \frac{1}{16} \times \frac{1}{64} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{4}}{\frac{x}{4}}\right)^2 \times \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{8}}{\frac{x}{8}}\right)^2 \\ & =\frac{1}{64} \times 1 \times 1 =\frac{1}{64}\\ \end{aligned}$$

$$\begin{aligned} & \quad y=\tan ^{-1}\left(\frac{\log \left(\frac{\mathrm{e}}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right) \\ & =\tan ^{-1}\left(\frac{\log e-\log x^2}{\log e+\log x^2}\right)+\tan ^{-1}(4)+\tan ^{-1}(2 \log x) \\ & =\tan ^{-1}\left(\frac{1-2 \log x}{1+2 \log x}\right)+\tan ^{-1}(4)+\tan ^{-1}(2 \log x) \\ & \quad=\tan ^{-1}(1)-\tan ^{-1}(2 \log x)+\tan ^{-1}(4)+\tan ^{-1}(2 \log x) \\ & \therefore \quad y=\tan ^{-1}(1)+\tan ^{-1}(4) \\ & \therefore \quad \frac{d y}{d x}=0 \end{aligned}$$

$$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{\left(x^2-1\right) \mathrm{d} x}{x^3 \sqrt{2 x^4-2 x^2+1}} \\ & =\int \frac{\left(x^2-1\right) \mathrm{d} x}{x^3 \cdot x^2 \sqrt{\left(2-\frac{2}{x^2}+\frac{1}{x^4}\right)}} \\ & =\int \frac{\left(\frac{x^2-1}{x^5}\right) \mathrm{d} x}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} \\ & \text { Put } 2-\frac{2}{x^2}+\frac{1}{x^4}=\mathrm{t} \\ & \Rightarrow\left(\frac{4}{x^3}-\frac{4}{x^5}\right) \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow \frac{x^2-1}{x^5} \mathrm{~d} x=\frac{\mathrm{dt}}{4} \\ & \therefore \quad I = \int {{{{{dt} \over 4}} \over {\sqrt t }}} \\ & =\frac{1}{4} \int \mathrm{t}^{\frac{-1}{2}} \mathrm{dt} \\ & =\frac{1}{4} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+c \\ \therefore \quad & \mathrm{I}=\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+\mathrm{c} \\ \end{aligned}$$

Given: $\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$ for all $x \in \mathrm{R}$

$\Rightarrow \frac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}=1$

Integrating on both sides, we get

$$\begin{aligned} & \log |\mathrm{f}(x)|=x+\mathrm{c} \\ & \Rightarrow \mathrm{f}(x)=\mathrm{e}^{x+\mathrm{c}} \\ & \Rightarrow \mathrm{f}(x)=\mathrm{e}^x \cdot \mathrm{e}^{\mathrm{c}} \\ & \Rightarrow \mathrm{f}(x)=\mathrm{e}^x \cdot \mathrm{c}_1 \quad \ldots \text { (i) }\left[\mathrm{e}^{\mathrm{c}}=\mathrm{c}_1\right] \end{aligned}$$

As $\mathrm{f}(1)=2$

$$\begin{aligned} & \therefore \quad c_1 \cdot e=2 \\ & \Rightarrow c_1=\frac{2}{e} \end{aligned}$$

Equation (i) becomes

$f(x)=\mathrm{e}^x \cdot \frac{2}{\mathrm{e}}$

Now, $\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))$

$$\begin{aligned} \therefore \quad & h^{\prime}(x)=f^{\prime}(f(x)) \times f^{\prime}(x) \\ \therefore \quad & h^{\prime}(1)=f^{\prime}(f(1)) \times f^{\prime}(1) \\ & \Rightarrow h^{\prime}(1)=f^{\prime}(2) \times f^{\prime}(1) \\ & \Rightarrow h^{\prime}(1)=e^2 \times \frac{2}{e} \times 2 \\ & \Rightarrow h^{\prime}(1)=4 e \end{aligned}$$

Let $\bar{r}=a \hat{i}+b \hat{j}+c \hat{k}$

As $\overline{\mathrm{r}}$ is perpendicular to $\overline{\mathrm{q}}$.

$$\begin{aligned} \therefore \quad & \bar{r} \cdot \bar{q}=0 \\ & \Rightarrow a-2 b+c=0 \quad \text{.... (i)} \end{aligned}$$

Also, $\bar{r}$ is coplanar with vectors $\bar{p}$ and $\bar{q}$

$$\begin{aligned} & \therefore \quad\left[\begin{array}{ccc} \bar{p} & \bar{q} & \bar{r} \end{array}\right]=0 \\ \Rightarrow & \quad\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ a & b & c \end{array}\right|=0 \\ & \Rightarrow 3 a-3 c=0 \\ & \Rightarrow a-c=0 \\ & \Rightarrow a=c \quad \text{..... (ii)} \end{aligned}$$

From (i) and (ii), we get

$$\begin{aligned} \quad \mathrm{b} & =\mathrm{c} \\ \therefore \quad \overline{\mathrm{r}} & =\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \end{aligned}$$

Now, the magnitude of required vector is $5 \sqrt{3}$ units.

$$\begin{aligned} \therefore \quad \text { Required vector } & =5 \sqrt{3} \times \frac{\overline{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|} \\ & =5 \sqrt{3} \times \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}=5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \end{aligned}$$

$$\begin{aligned} & \int \mathrm{e}^x\left(1-\cot x+\cot ^2 x\right) \mathrm{d} x \\ & =\int \mathrm{e}^x\left(1+\cot ^2 x-\cot x\right) \mathrm{d} x \\ & =\int \mathrm{e}^x\left(-\cot x+\operatorname{cosec}^2 x\right) \mathrm{d} x \\ & =\mathrm{e}^x(-\cot x)+\mathrm{c} \\ & \quad \ldots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right] \\ & =-\mathrm{e}^x \cdot \cot x+\mathrm{c} \end{aligned}$$

$$\begin{aligned} & \text {For X-axis, } \\ & y=0 \\ & \therefore \quad x(x-2)(x+1)=0 \\ & \Rightarrow x=0 \text { or } x=2 \text { or } x=-1 \end{aligned}$$

Required area $= \int\limits_{ - 1}^0 {y\,dx + \left| {\int\limits_0^2 {y\,dx} } \right|}$

\begin{aligned} & =\int_\limits{-1}^0 x(x-2)(x+1) \mathrm{d} x+\left|\int_\limits0^2 x(x-2)(x+1) \mathrm{d} x\right| \\ & =\int_\limits{-1}^0\left(x^3-x^2-2 x\right) \mathrm{d} x+\left|\int_\limits0^2\left(x^3-x^2-2 x\right) \mathrm{d} x\right| \\ & =\left[\frac{x^4}{4}-\frac{x^3}{3}-x^2\right]_{-1}^0+\left|\left[\frac{x^4}{4}-\frac{x^3}{3}-x^2\right]_0^2\right| \\ & =\frac{5}{12}+\left|-\frac{8}{3}\right| \\ & =\frac{37}{12} \text { sq. units } \end{aligned}

Greatest value of ${ }^{10} \mathrm{C}_{\mathrm{r}}$ is at $\mathrm{r}=5$

Greatest value of ${ }^{11} \mathrm{C}_{\mathrm{r}}$ is at $\mathrm{r}=5$

$\therefore \quad \frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{11} \mathrm{C}_{\mathrm{r}}}=\frac{{ }^{10} \mathrm{C}_5}{{ }^{11} \mathrm{C}_5}=\frac{\frac{10 !}{5 ! 5 !}}{\frac{11 !}{5 ! 6 !}}=\frac{6}{11}$

Let $\mathrm{h}(x)=[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)]$

$$\begin{aligned} \therefore \quad \mathrm{h}(-x) & =[\mathrm{f}(-x)+\mathrm{f}(x)][\mathrm{g}(-x)-\mathrm{g}(x)] \\ & =-[\mathrm{f}(x)+\mathrm{f}(-x)][\mathrm{g}(x)-\mathrm{g}(-x)] \\ & =-\mathrm{h}(x) \end{aligned}$$

$\therefore \quad \mathrm{h}(x)$ is an odd function.

\therefore \quad \int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \mathrm{~h}(x)=0

$\mathrm{X}$ can take values 0,1 and 2 .

$\mathrm{P}(\mathrm{X}=0)=$ Probability of not getting six $=\frac{25}{36}$

$P(X=1)=$ Probability of getting one six

$=\frac{10}{36}=\frac{5}{18}$

$P(X=2)=$ Probability of getting two sixes $=\frac{1}{36}$

The probability distribution of $\mathrm{X}$ is

Given, $S=5+\frac{48}{t}+t^3$

$\operatorname{velocity}(\mathrm{V})=\frac{\mathrm{dS}}{\mathrm{dt}}=0-\frac{48}{\mathrm{t}^2}+3 \mathrm{t}^2$

$V=\frac{-48}{t^2}+3 t^2$ ..... (i)

But $\mathrm{V}=0$ ...[Given]

$$\begin{aligned} & \Rightarrow \frac{-48}{t^2}+3 t^2=0 \\ & \Rightarrow t=2 \end{aligned}$$

Now,

$$\begin{aligned} A & =\frac{d V}{d t} \\ & =\frac{d}{d t}\left(\frac{-48}{t^2}+3 t^2\right) \\ & =\frac{96}{t^3}+6 t \end{aligned}$$