MHT CET 2023 13th May Morning Shift (Mathematics) Question Paper with Solutions

\begin{aligned} & \text { Since } \sum_\limits{x=0}^2 \mathrm{P}(\mathrm{X}=x)=1 \\ & 4 \mathrm{k}-10 \mathrm{k}^2+5 \mathrm{k}-1+3 \mathrm{k}^3=1 \\ & \Rightarrow 3 \mathrm{k}^3-10 \mathrm{k}^2+9 \mathrm{k}-2=0 \\ & \Rightarrow(\mathrm{k}-1)(\mathrm{k}-2)(3 \mathrm{k}-1)=0 \\ & \Rightarrow \mathrm{k}=1 \text { or } \mathrm{k}=2 \text { or } \mathrm{k}=\frac{1}{3} \end{aligned}

For $\mathrm{k}=1$ or $\mathrm{k}=2$,

$\mathrm{P}(\mathrm{X}=0)<0$, which is not possible

$\therefore \quad \mathrm{k}=\frac{1}{3}$

$$\begin{aligned} \text{Now,} \quad & \mathrm{P}(\mathrm{X}<2)=P(X=0)+P(X=1) \\ & =4 k-10 k^2+5 k-1 \\ & =9 k-10 k^2-1 \\ & =9\left(\frac{1}{3}\right)-10\left(\frac{1}{9}\right)-1 \\ & =\frac{8}{9} \end{aligned}$$

$$\begin{aligned} & \text { Let } \mathrm{I}=\int x^5 \mathrm{e}^{-4 x^3} \mathrm{~d} x \\ & \text { Put } x^3=\mathrm{t} \Rightarrow 3 x^2 \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad I=\frac{1}{3} \int t e^{-4 t} d t=\frac{1}{3}\left(t \cdot \frac{e^{-4 t}}{-4}-\int 1 \cdot \frac{e^{-4 t}}{-4} d t\right) \\ & =\frac{1}{3}\left(\frac{-\mathrm{te}^{-4 \mathrm{t}}}{4}+\frac{1}{4} \cdot \frac{\mathrm{e}^{-4 \mathrm{t}}}{-4}\right)+\mathrm{c} \\ & =\frac{1}{48} \mathrm{e}^{-4 \mathrm{t}}(-4 \mathrm{t}-1)+\mathrm{c} \\ & =\frac{1}{48} e^{-4 x^3}\left(-4 x^3-1\right)+c \\ & \therefore \quad \mathrm{f}(x)=-4 x^3-1 \\ & \end{aligned}$$

Given equation of pair of lines is

$x^2-3 x y+\lambda y^2+3 x-5 y+2=0$

Here, $\mathrm{a}=1, \mathrm{~b}=\lambda, \mathrm{c}=2, \mathrm{f}=\frac{-5}{2}, \mathrm{~g}=\frac{3}{2}, \mathrm{~h}=\frac{-3}{2}$

$$\begin{aligned} & \theta=\tan ^{-1}\left(\frac{1}{3}\right) \Rightarrow \tan \theta=\frac{1}{3} \\ & \text { Since } \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \\ & \Rightarrow \frac{1}{3}=\left|\frac{2 \sqrt{\left(\frac{-3}{2}\right)^2-\lambda}}{\lambda+1}\right| \\ & \Rightarrow(\lambda+1)^2=9(9-4 \lambda) \Rightarrow \lambda^2+38 \lambda-80=0 \\ & \Rightarrow(\lambda+40)(\lambda-2)=0 \Rightarrow \lambda=2 \quad \ldots[\because \lambda \geq 0] \end{aligned}$$

$$\begin{aligned} & (p \wedge \sim q) \vee q \vee(\sim p \wedge q) \\ & \equiv[(p \vee q) \wedge(\sim q \vee q)] \vee(\sim p \wedge q) \quad \text{...[Distributive law]} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & \equiv[(p \vee q) \wedge T] \vee(\sim p \wedge q) \quad \ldots[\text { Complement law }] \\ & \equiv(p \vee q) \vee(\sim p \wedge q) \end{aligned}\\ & \text { ....[Identity law] } \end{aligned}$$

$\equiv(p \vee q \vee \sim p) \wedge(p \vee q \vee q)$ .... [Distributive law]

$\equiv(T \vee q) \wedge(p \vee q)$ ...[Complement law and Idempotent law]

$$\begin{aligned} & \equiv T \wedge(p \vee q) \quad ...[\text{Identity law}]\\ & \equiv p \vee q \quad ...[\text{Identity law}] \end{aligned}$$

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

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

By sine rule,

$$\begin{aligned} & \frac{\sin 120^{\circ}}{\mathrm{a}}=\frac{\sin 30^{\circ}}{\mathrm{b}}=\frac{\sin 30^{\circ}}{\mathrm{c}} \\ & \therefore \quad \mathrm{a}:(\mathrm{a}+\mathrm{b}+\mathrm{c}) \\ &=\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\ &=\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}: \sqrt{3}+2 \end{aligned}$$

Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the direction ratios of the required line. Since the line is perpendicular to the lines with d.r.s. $1,2,3$ and $-3,2,5$.

$\therefore \quad \mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}=0$ .... (i)

and $-3 a+2 b+5 c=0$ .... (ii)

$\Rightarrow \frac{\mathrm{a}}{2}=\frac{\mathrm{b}}{-7}=\frac{\mathrm{c}}{4}$ ...[From (i) and (ii)]

$\therefore \quad$ Equation of the required line is

$\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$

Here, $n(S)=52$

Let event A : A black card is drawn.

event B : A face card is drawn.

$\therefore \quad \mathrm{n}(\mathrm{A})=26, \mathrm{n}(\mathrm{B})=12, \mathrm{n}(\mathrm{A} \cap \mathrm{B})=6$

$\therefore \quad$ Required probability

$$\begin{aligned} & =P(A \cup B) \\ & =P(A)+P(B)-P(A \cap B) \\ & =\frac{26}{52}+\frac{12}{52}-\frac{6}{52} \\ & =\frac{32}{52}=\frac{8}{13} \end{aligned}$$

$$\begin{array}{ll} \mathrm{f}(x)=\sin x+\cos x+6 \\ \therefore \quad \mathrm{f}^{\prime}(x)=\cos x-\sin x \\ \text { Now, } \mathrm{f}^{\prime}(\mathrm{c})=0 \\ \Rightarrow \cos \mathrm{c}-\sin \mathrm{c}=0 \\ \Rightarrow \cos \mathrm{c}=\sin \mathrm{c} \\ \Rightarrow \tan \mathrm{c}=1 \\ \Rightarrow \mathrm{c}=\frac{\pi}{4}, \frac{5 \pi}{4} \quad \ldots[\because x \in[0,2 \pi]] \end{array}$$

\begin{aligned} \text { Required area } & =\int_\limits0^3[4 x+1-(3 x+1)] \mathrm{d} x \\ & =\int_\limits0^3 x \mathrm{~d} x \\ & =\left[\frac{x^2}{2}\right]_0^3 \\ & =\frac{9}{2} \text { sq. units } \end{aligned}

$$\begin{aligned} \mathrm{I} & =\int \frac{2 x-7}{\sqrt{3 x-2}} \mathrm{~d} x \\ & =\int \frac{\frac{2}{3}(3 x-2)-\frac{17}{3}}{\sqrt{3 x-2}} \mathrm{~d} x \\ & =\frac{2}{3} \int(3 x-2)^{\frac{1}{2}} \mathrm{~d} x-\frac{17}{3} \int(3 x-2)^{\frac{-1}{2}} \mathrm{~d} x \\ & =\frac{2}{3} \times \frac{(3 x-2)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{3}-\frac{17}{3} \times \frac{(3 x-2)^{\frac{1}{2}}}{\frac{1}{2}} \times \frac{1}{3}+\mathrm{c} \\ & =\frac{4}{27}(3 x-2)^{\frac{3}{2}}-\frac{34}{9}(3 x-2)^{\frac{1}{2}}+\mathrm{c} \end{aligned}$$

$$\begin{aligned} & \mathrm{f}(x)=x^3+x^2 \mathrm{f}^{\prime}(1)+x \mathrm{f}^{\prime \prime}(2)+6 \\ \therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+2 x \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \quad \text{... (i)}\\ \therefore \quad & \mathrm{f}^{\prime \prime}(x)=6 x+2 \mathrm{f}^{\prime}(1) \quad \text{... (ii)} \end{aligned}$$

Substituting $x=1$ in (i), we get

$$\begin{aligned} & \mathrm{f}^{\prime}(1)=3(1)^2+2(1) \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2) \\ & \Rightarrow \mathrm{f}^{\prime}(1)+\mathrm{f}^{\prime \prime}(2)=-3 \quad \text{... (iii)} \end{aligned}$$

Substituting $x=2$ in (ii), we get

$$\begin{aligned} & f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1) \\ & \Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1) \quad \text{... (iv)} \end{aligned}$$

From (iii) and (iv), we get

$$\begin{aligned} & \mathrm{f}^{\prime}(1)+12+2 \mathrm{f}^{\prime}(1)=-3 \\ & \Rightarrow 3 \mathrm{f}^{\prime}(1)=-15 \\ & \Rightarrow \mathrm{f}^{\prime}(1)=-5 \\ & \text { From (iii), }-5+f^{\prime \prime}(2)=-3 \\ & \Rightarrow \mathrm{f}^{\prime \prime}(2)=2 \\ & \therefore \quad \mathrm{f}(2)=2^3+2^2(-5)+2(2)+6 \\ & =8-20+4+6=-2 \\ \end{aligned}$$

$$\begin{aligned} & 3 \mathrm{f}(x)-\mathrm{f}\left(\frac{1}{x}\right)=8 \log _2 x^3 \quad \text{... (i)}\\ & \Rightarrow 3 \mathrm{f}\left(\frac{1}{x}\right)-\mathrm{f}(x)=8 \log _2\left(\frac{1}{x}\right)^3 \quad \text{... (ii)} \end{aligned}$$

From (i) and (ii), we get

$$\begin{aligned} & 8 \mathrm{f}(x)=24 \log _2 x^3+8 \log _2\left(\frac{1}{x}\right)^3 \\ & \Rightarrow 8 \mathrm{f}(x)=72 \log _2 x-24 \log _2 x \\ & \Rightarrow 8 \mathrm{f}(x)=48 \log _2 x \\ & \Rightarrow \mathrm{f}(x)=6 \log _2 x \end{aligned}$$

$$\begin{aligned} \therefore & \quad f(2)=6 \log _2 2=6 \\ & f(4)=6 \log _2 4=12 \\ & f(8)=6 \log _2 8=18 \end{aligned}$$

$\therefore \quad f(2), f(4), f(8)$ are in A.P.

Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$

$\therefore \quad$ The co-ordinates of any point on the line are

$P \equiv(3 \lambda+2,4 \lambda-1,12 \lambda+2)$

This point lies on the plane

$$\begin{array}{ll} & x-y+z=5 \\ \therefore \quad & 3 \lambda+2-(4 \lambda-1)+12 \lambda+2=5 \\ & \Rightarrow 11 \lambda=0 \\ & \Rightarrow \lambda=0 \\ \therefore \quad & \mathrm{P} \equiv(2,-1,2) \\ & \text { Let } \mathrm{Q} \equiv(-1,-5,-10) \\ \therefore \quad &\mathrm{PQ}=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2} \\ & =\sqrt{9+16+144} \\ & =13 \text { units } \end{array}$$

$$\begin{aligned} \bar{x} & =\frac{x_1+x_2+\ldots+x_{\mathrm{n}}}{\mathrm{n}} \\ & =\frac{\frac{\mathrm{n}}{2}\left[2 x_1+(\mathrm{n}-1) \mathrm{d}\right]}{\mathrm{n}} \\ & =\frac{(\mathrm{n}-1) \mathrm{d}}{2} \quad ... [\because x_1=0]\\ \sum x_{\mathrm{i}}^2 & =x_1^2+x_2^2+\ldots+x_{\mathrm{n}}^2 \\ & =0+\mathrm{d}^2+(2 \mathrm{~d})^2+\ldots+[(\mathrm{n}-1) \mathrm{d}]^2\\ & =\mathrm{d}^2\left[1+2^2+3^2+\ldots+(\mathrm{n}-1)^2\right] \\ & =\mathrm{d}^2\left[\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}\right] \\ \therefore \quad \sigma^2 & =\frac{1}{\mathrm{n}} \sum x_{\mathrm{i}}^2-(\bar{x})^2 \\ & =\frac{\mathrm{d}^2(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}-\left[\frac{(\mathrm{n}-1) \mathrm{d}}{2}\right]^2 \\ & =\frac{\mathrm{d}^2(\mathrm{n}-1)}{2}\left(\frac{2 \mathrm{n}-1}{3}-\frac{\mathrm{n}-1}{2}\right) \\ & =\frac{\mathrm{d}^2(\mathrm{n}-1)}{2}\left(\frac{\mathrm{n}+1}{6}\right)=\frac{\left(\mathrm{n}^2-1\right) \mathrm{d}^2}{12} \end{aligned}$$

$$\begin{aligned} & x^2+y^2+a x+b y=0 \\ & \Rightarrow\left(x+\frac{a}{2}\right)^2+\left(y+\frac{b}{2}\right)^2=\frac{a^2+b^2}{4} \end{aligned}$$

Comparing with $(x-\mathrm{h})^2+(y-\mathrm{k})^2=\mathrm{r}^2$, we get

$\mathrm{h}=-\frac{\mathrm{a}}{2}, \mathrm{k}=-\frac{\mathrm{b}}{2}, \mathrm{r}=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}}$

$\therefore\quad$ The parametric equations are

$x=-\frac{a}{2}+\sqrt{\frac{\mathrm{a}^2+b^2}{4}} \cos \theta, y=-\frac{b}{2}+\sqrt{\frac{a^2+b^2}{4}} \sin \theta$

$|(\bar{a}-\bar{b}) \times(\bar{a}+\bar{b})|^2+4(\bar{a}. \bar{b})^2$

$= {\left| {\left( {\overline a \times \overline a } \right) + \left( {\overline a \times \overline b } \right) - \left( {\overline b \times \overline a } \right) - \left( {\overline b \times \overline b } \right)} \right|^2} + 4{\left( {\overline a .\,\overline b } \right)^2}$

$=|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-(\overline{\mathrm{b}} \times \overline{\mathrm{a}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2$

$=|2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})|^2+4(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})^2 \ldots[(\overline{\mathrm{a}} \times \overline{\mathrm{b}})=-(\overline{\mathrm{b}} \times \overline{\mathrm{a}})]$

$= 4{\left| {\left( {\overline a \times \overline b } \right)} \right|^2} + 4{\left( {\overline a \,.\,\overline b } \right)^2}$

$= 4\left[ {{{\left| {\overline a \times \overline b } \right|}^2} + {{\left( {\overline a \,.\,\overline b } \right)}^2}} \right]$

$=4|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2$

$=4(4)^2(3)^2$

$=576$

$f(t)=\frac{1}{t^2+t-2}=\frac{1}{(t+2)(t-1)}$

$f(t)$ is not defined at $t=-2$ and $t=1$.

$$\begin{aligned} & t=-2 \\ & \Rightarrow \frac{1}{x-1}=-2 \\ & \Rightarrow x=\frac{1}{2} \\ & t=1 \\ & \Rightarrow \frac{1}{x-1}=1 \\ & \Rightarrow x=2 \end{aligned}$$

$\therefore \quad$ The function $\mathrm{f}(\mathrm{t})$ is discontinuous at $x=\frac{1}{2}$ and $x=2$.

$$\begin{aligned} & 3 \sec ^2 \theta=2 \operatorname{cosec} \theta \\ & \Rightarrow \frac{3}{\cos ^2 \theta}=\frac{2}{\sin \theta} \\ & \Rightarrow \frac{3}{1-\sin ^2 \theta}=\frac{2}{\sin \theta} \\ & \Rightarrow 2 \sin ^2 \theta+3 \sin \theta-2=0 \\ & \Rightarrow(2 \sin \theta-1)(\sin \theta+2)=0 \\ & \Rightarrow \sin \theta=\frac{1}{2} \end{aligned}$$

or $\sin \theta=-2$, which is not possible

$$\begin{aligned} \therefore \quad \sin \theta & =\frac{1}{2}=\sin \frac{\pi}{6} \\ \Rightarrow \theta & =\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{6}, \mathrm{n} \in \mathrm{Z} \end{aligned}$$

For the conical vessel, $\mathrm{h}=9 \mathrm{~cm}, \mathrm{r}=5 \mathrm{~cm}$

$\therefore \quad$ Full volume of the vessel,

$$\begin{aligned} \mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\ & =\frac{1}{3} \pi \times 25 \times 9 \\ & =75 \pi \mathrm{~cm}^3 \end{aligned}$$

Now, $\frac{\mathrm{h}}{\mathrm{r}}=\frac{9}{5}$

$$\begin{aligned} & \therefore \quad r=\frac{5 h}{9} \\ & \therefore \quad A=\pi r^2=\pi \frac{25 h^2}{81} \end{aligned}$$

According to the given condition,

$$\begin{aligned} \frac{\mathrm{dh}}{\mathrm{dt}} & =\frac{\pi}{\mathrm{A}}=\pi \frac{81}{\pi 25 \mathrm{~h}^2}=\frac{81}{25 \mathrm{~h}^2} \\ \therefore \quad \mathrm{h}^2 \mathrm{dh} & =\frac{81}{25} \mathrm{dt} \end{aligned}$$

Integrating on both sides, we get

$$\begin{aligned} & \frac{h^3}{3}=\frac{81}{25} t+c_1 \\ \therefore \quad h^3 & =\frac{243}{25} t+c, \text { where } c=3 c_1 \end{aligned}$$

Naturally, $\mathrm{h}=0$, when $\mathrm{t}=0$ and hence, $\mathrm{c}=0$

$$\begin{aligned} \therefore \quad \mathrm{h}^3 & =\frac{243}{25} \mathrm{t} \quad \text{.... (i)}\\ \therefore \quad \mathrm{V} & =\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h} \\ & =\frac{1}{3} \pi \frac{25 \mathrm{~h}^2}{81} \mathrm{~h} \\ & =\frac{25}{243} \pi \mathrm{h}^3 \\ & =\frac{25}{243} \pi \frac{243}{25} \mathrm{t} \quad \text{...[From (i)]}\\ \therefore \quad \mathrm{V} & =\pi \mathrm{t} \end{aligned}$$

But volume of vessel, $\mathrm{V}=75 \pi$

$$\begin{array}{ll} \therefore & \pi \mathrm{t}=75 \pi \\ \therefore & \mathrm{t}=75 \text { seconds. } \end{array}$$

Shaded region lies on origin side of $4 x+7 y=28$ and above the line $y=1$, and on non-origin side of $3 x+4 y=12$.

$\therefore \quad 3 x+4 y \geq 12,4 x+7 y \leq 28, y \geq 1, x \geq 0, y \geq 0$

If $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$, then the triangle is equilateral.

$$\begin{aligned} \therefore \quad \mathrm{A}(\triangle \mathrm{ABC}) & =\frac{\sqrt{3}}{4} \mathrm{a}^2 \\ & =\frac{\sqrt{3}}{4}\left(\frac{1}{\sqrt{6}}\right)^2 \\ & =\frac{\sqrt{3}}{24}=\frac{1}{8 \sqrt{3}} \text { sq. units } \end{aligned}$$

$$\begin{aligned} & \frac{i^{248}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{249}+i^{247}+i^{245}+i^{243}+i^{241}} \\ & =\frac{i^{240}\left(i^8+i^6+i^4+i^2+1\right)}{i^{241}\left(i^8+i^6+i^4+i^2+1\right)} \\ & =\frac{i^{240}}{i^{241}} \\ & =\frac{1}{i} \\ & =\frac{i}{i^2}=-i \end{aligned}$$