MHT CET 2023 12th May Evening Shift (Mathematics) Question Paper with Solutions

Feasible region lies on the origin side of lines $x+y=20, x=10$ and on non-origin side of $y=5$.

$\therefore \quad$ Corner points of the feasible region ant $\mathrm{A}(0,5), \mathrm{B}(10,5), \mathrm{C}(10,10)$ and $\mathrm{D}(0,20)$

$\mathrm{z}$ at $\mathrm{A}(0,5)=40$

$\mathrm{z}$ at $\mathrm{B}(10,5)=110$

$\mathrm{z}$ at $\mathrm{C(10,10)=150}$

$\mathrm{z}$ at $\mathrm{D}(0,20)=160$

$\therefore \quad$ Maximum value of z is 160.

\begin{aligned} & \int_\limits0^\pi\left|\sin x-\frac{2 x}{\pi}\right| d x \\ & =\int_\limits0^{\frac{\pi}{2}}\left(\sin x-\frac{2 x}{\pi}\right) d x+\int_\limits{\frac{\pi}{2}}^\pi\left(\frac{2 x}{\pi}-\sin x\right) d x \\ & =[-\cos x]_0^{\frac{\pi}{2}}-\left[\frac{x^2}{\pi}\right]_0^{\frac{\pi}{2}}+\left[\frac{x^2}{\pi}\right]_{\frac{\pi}{2}}^\pi+[\cos x]_{\frac{\pi}{2}}^\pi \\ & =(0+1)-\left(\frac{\pi}{4}-0\right)+\left(\pi-\frac{\pi}{4}\right)+(-1-0) \\ & =\frac{\pi}{2} \end{aligned}

Given lines are: $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{0}$

$\therefore \quad$ Required distance

$$=\left|\frac{\left|\begin{array}{lll} 3 & 0 & 4 \\ 3 & 1 & 2 \\ 1 & 2 & 0 \end{array}\right|}{\sqrt{(6-1)^2+(0-2)^2+(0-4)^2}}\right|$$

$$\begin{aligned} & =\left|\frac{3(0-4)+0+4(6-1)}{\sqrt{25+4+16}}\right| \\ & =\left|\frac{8}{\sqrt{45}}\right| \\ & =\frac{8}{3 \sqrt{5}} \end{aligned}$$

$\begin{gathered} \quad(x \cos \alpha+y \sin \alpha)^2=\left(x^2+y^2\right) \sin ^2 \alpha \\ \therefore \quad x^2 \cos ^2 \alpha+y^2 \sin ^2 \alpha+2 x y \sin \alpha \cos \alpha \\ \quad=x^2 \sin ^2 \alpha+y^2 \sin ^2 \alpha \\ \therefore \quad x^2\left(\cos ^2 \alpha-\sin ^2 \alpha\right)+2 x y \sin \alpha \cos \alpha=0 \end{gathered}$

This represents a pair of straight lines where $\mathrm{a}=\cos ^2 \alpha-\sin ^2 \alpha, \mathrm{h}=\sin \alpha \cos \alpha$ and $\mathrm{b}=0$

As lines are perpendicular, we get $a+b=0$

$\therefore \cos ^2 \alpha=\sin ^2 \alpha \Rightarrow \alpha=\frac{\pi}{4}$

Required equation of line is

$$\begin{aligned} & \frac{x-1}{\left|\begin{array}{cc} -1 & 2 \\ 1 & 1 \end{array}\right|}=\frac{y-2}{-\left|\begin{array}{cc} 1 & 2 \\ 3 & 1 \end{array}\right|}=\frac{z-3}{\left|\begin{array}{cc} 1 & -1 \\ 3 & 1 \end{array}\right|} \\ & \therefore \quad \frac{x-1}{-3}=\frac{y-2}{5}=\frac{z-3}{4} \\ \end{aligned}$$

$$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right) \\ \therefore \quad f^{\prime}(x)= & \frac{1}{\sqrt{1-\left(\frac{2 \log x}{1+(\log x)^2}\right)^2}} \end{aligned}$$

$$\begin{aligned} & =\frac{1+(\log x)^2}{\sqrt{1+(\log x)^4+2(\log x)^2-4(\log x)^2}} \times \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 \log x}{1+(\log x)^2}\right) \\ & =\frac{1+(\log x)^2}{\sqrt{1-2(\log x)^2+(\log x)^4}} \\ & =\frac{1}{1-(\log x)^2} \times \frac{2+2(\log x)^2-4(\log x)^2}{x\left[1+(\log x)^2\right]} \\ = & \frac{1}{\left.1-(\log x)^2\right] \times \frac{2}{x}-(2 \log x)\left(\frac{2 \log x}{x}\right)} \times \frac{2\left[1-(\log x)^2\right]}{x\left[1+(\log x)^2\right]} \\ = & \frac{2}{x\left[1+(\log x)^2\right]} \\ \therefore \quad \mathrm{f}^{\prime}(\mathrm{e}) & =\frac{1}{\mathrm{e}} \end{aligned}$$

$$\begin{aligned} & \mathrm{e}^{y-x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y(\sin x+\cos x)}{(1+y \log y)} \\ \therefore \quad & \mathrm{e}^y \frac{(1+y \log y)}{y} \mathrm{~d} y=\mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\ \therefore \quad & \int \mathrm{e}^y\left(\log y+\frac{1}{y}\right) \mathrm{d} y=\int \mathrm{e}^x(\sin x+\cos x) \mathrm{d} x \\ \Rightarrow & \mathrm{e}^y \log y=\mathrm{e}^x \sin x+\mathrm{c} \\ & \ldots\left[\because \int \mathrm{e}^x\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right] \end{aligned}$$

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cot ^2(2 x)} & =\lim _{x \rightarrow 0} \frac{x \tan ^2 2 x}{\sin ^2 x \tan 4 x} \\ & =\lim _{x \rightarrow 0} \frac{4\left(\frac{\tan 2 x}{2 x}\right)^2}{4\left(\frac{\sin x}{x}\right)^2\left(\frac{\tan 4 x}{4 x}\right)} \\ & =1 \end{aligned}$$

$$\begin{aligned} \text{Let} \quad I & =\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} d x \\ & =\frac{1}{\sqrt{2}} \int \frac{\sec ^3 x}{\sqrt{\sin x \cos x}} d \\ & =\frac{1}{\sqrt{2}} \int \frac{\sec ^4 x}{\sqrt{\tan x}} d x \end{aligned}$$

Let $\tan x=t \Rightarrow \sec ^2 x=d t$

$$\begin{aligned} \therefore \quad I & =\frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} d t \\ & =\frac{1}{\sqrt{2}} \int t^{\frac{1}{2}} d t+\frac{1}{\sqrt{2}} \int t^{\frac{3}{2}} d t \\ & =\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)+c \\ & =\sqrt{2}\left(\sqrt{\tan x}+\frac{1}{5}(\tan x)^{\frac{5}{2}}\right)+c \end{aligned}$$

$(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$

Taking '$\log _e$' on both sides, we get

$$\begin{aligned} & 2 y \log _{\mathrm{e}}(2 x)=\log _{\mathrm{e}} 4+(2 x-2 y) \log _{\mathrm{e}} \mathrm{e} \\ & y \log _{\mathrm{e}}(2 x)=\log _{\mathrm{e}} 2+x-y \quad \text{..... (i)} \end{aligned}$$

Differentiating w.r.t. $x$, we get

$$\begin{array}{ll} & {\left[\log _{\mathrm{e}} 2 x\right] \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 y}{2 x}=0+1-\frac{\mathrm{d} y}{\mathrm{~d} x}} \\ & {\left[1+\log _{\mathrm{e}} 2 x\right] \frac{\mathrm{d} y}{\mathrm{~d} x}=1-\frac{y}{x}} \quad \text{.... (ii)}\\ & \text { Now, (i) } \Rightarrow y=\frac{\log _{\mathrm{e}} 2+x}{1+\log _{\mathrm{e}} 2 x} \\ \therefore \quad & (\text { ii }) \Rightarrow\left(1+\log _{\mathrm{e}} 2 x\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x\left(1+\log _{\mathrm{e}} 2 x\right)} \\ \therefore \quad & \left(1+\log _{\mathrm{e}} 2 x\right)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x} \end{array}$$

$$\begin{aligned} & \text { Let } A=\cos ^2 48^{\circ}-\sin ^2 12^{\circ} \\ & =\cos ^2\left(30^{\circ}+18^{\circ}\right)-\sin ^2\left(30^{\circ}-18^{\circ}\right) \\ & =\left[\cos 30^{\circ} \cos 18^{\circ}-\sin 30^{\circ} \sin 18^{\circ}\right]^2 \\ & -\left[\sin 30^{\circ} \cos 18^{\circ}-\sin 18^{\circ} \cos 30^{\circ}\right]^2 \\ & =\left[\frac{\sqrt{3} \cos 18^{\circ}-\sin 18^{\circ}}{2}\right]^2-\left[\frac{\cos 18^{\circ}-\sqrt{3} \sin 18^{\circ}}{2}\right]^2 \\ & =\frac{3 \cos ^2 18^{\circ}+\sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \\ & -\frac{\cos ^2 18^{\circ}+3 \sin ^2 18-2 \sqrt{3} \sin 18^{\circ} \cos 18^{\circ}}{4} \\ & =\frac{\cos ^2 18^{\circ}-\sin ^2 18^{\circ}}{2} \\ & \text { Note that } \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \Rightarrow \cos ^2 18^{\circ}=\frac{5+\sqrt{5}}{8} \\ & \text { and } \sin ^2 18^{\circ}=\frac{3-\sqrt{5}}{8} \\ & \therefore \quad A=\frac{\frac{5+\sqrt{5}-3+\sqrt{5}}{8}}{2}=\frac{1+\sqrt{5}}{8} \\ \end{aligned}$$

$$\begin{aligned} & (\bar{a}-\bar{d}) \cdot(\bar{b}-\bar{c})=(\bar{b}-\bar{d})(\bar{c}-\bar{a})=0 \\ & \overline{A D} \cdot \overline{B C}=\overline{B D} \cdot \overline{C A}=0 \\ & \Rightarrow \overline{A D} \perp \overline{B C} \text { and } \overline{B D} \perp \overline{C A} \end{aligned}$$

$\Rightarrow D \text { is the orthocentre of } \triangle A B C \text {. }$

$$\begin{aligned} & \text { In } \triangle A B D \text {, } \\ & \frac{\sin (\angle B A D)}{B D}=\frac{\sin (\angle A B D)}{A D} \\ & \Rightarrow \frac{\sin (\angle \mathrm{BAD})}{x}=\frac{\frac{\sqrt{3}}{2}}{\mathrm{AD}} \\ & \Rightarrow \mathrm{AD}=\frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})} \quad \text{.... (i)}\\ & \text { In } \triangle \mathrm{ADC} \text {, } \\ & \frac{\sin (\angle C A D)}{D C}=\frac{\sin (\angle A C D)}{A D} \\ & \Rightarrow \frac{\sin (\angle C A D)}{3 x}=\frac{\frac{1}{\sqrt{2}}}{\mathrm{AD}} \\ & \therefore \quad \mathrm{AD}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})} \quad \text{.... (ii)}\\ \end{aligned}$$

From (i) and (ii), we get

$$\begin{aligned} & \frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})} \\ \therefore \quad & \frac{\sin (\angle \mathrm{BAD})}{\sin (\angle \mathrm{CAD})}=\frac{\sqrt{6}}{6}=\frac{1}{\sqrt{6}} \end{aligned}$$

Let $\mathrm{p}$ : A quadrilateral is a square

$\quad\mathrm{q}:$ All sides of quadrilateral are equal.

$\therefore \quad$ Statement 1 is $p \rightarrow q$,

$\quad$ Statement 2 is $q \rightarrow p$

$\therefore \quad$ Statement 2 is the converse of statement 1.

Substitute $x=\frac{\mathrm{a}}{\sqrt{2}}$ in $x^2+y^2=\mathrm{a}^2$, we get $\frac{\mathrm{a}^2}{2}+y^2=\mathrm{a}^2 \Rightarrow y= \pm \frac{\mathrm{a}}{\sqrt{2}}$

$\therefore \quad$ Required area

$$\begin{aligned} & = 2\int\limits_{{a \over {\sqrt 2 }}}^a {\sqrt {{a^2} - {x^2}} dx}\\ & =2\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]_{\frac{a}{\sqrt2}}^2 \\ & =2\left\{\left[0+\frac{a^2}{2} \times \frac{\pi}{2}\right]-\left[\frac{a}{2 \sqrt{2}} \sqrt{a^2-\frac{a^2}{2}}+\frac{a^2}{2} \times \frac{\pi}{4}\right]\right\} \\ & =2\left[\frac{a^2 \pi}{4}-\frac{a^2}{4}-\frac{a^2 \pi}{8}\right] \\ & =\frac{a^2}{2}\left|\pi-1-\frac{\pi}{2}\right| \\ & =\frac{a^2}{2}\left|\frac{\pi}{2}-1\right| \end{aligned}$$

$3 x^2-y^2=8$

Differentiating w.r.t. $x$, we get

$$\begin{aligned} & 6 x-2 y \frac{d y}{d x}=0 \\ \therefore \quad & \frac{d y}{d x}=\frac{3 x}{y} \end{aligned}$$

$\therefore \quad$ Slope of the tangent to the curve is $\frac{3 x}{y}$.

$\therefore \quad$ Slope of the normal is $\frac{-y}{3 x}$.

It is parallel to line $x+3 y=10 \Rightarrow$ slope $=-\frac{1}{3}$

$\therefore \quad \frac{-y}{3 x}=\frac{-1}{3} \Rightarrow x=y$

$\therefore \quad$ When $x=y$, equation of the curve becomes $3 x^2-x^2=8$

$\therefore \quad x^2=4$

$\therefore \quad x=2,-2 \Rightarrow y=2,-2$

$\therefore \quad(2,2)$ and $(-2,-2)$ are the points of contact of the normal and the curve.

$$\begin{aligned} \therefore \quad & \text { Equations are }(y-2)=\frac{-1}{3}(x-2) \text { or } \\ & (y+2)=\frac{-1}{3}(x+2) \\ & \text { i.e., } x+3 y-8=0 \text { or } x+3 y+8=0 \end{aligned}$$

Consider option (C)

$(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot\left(\frac{-8 \hat{i}+3 \hat{j}+3 \hat{k}}{\sqrt{82}}\right)=0$

This is valid for only option (C)

$\therefore$ Option (c) is correct.

The question cannot be solved due to insufficient data.

$$\begin{array}{ll} & \overline{\mathrm{a}}+2 \overline{\mathrm{b}} \text { is collinear with } \overline{\mathrm{c}} \\ \therefore \quad & \overline{\mathrm{a}}+2 \overline{\mathrm{b}}=\mathrm{nc} \quad \text{.... (i)}\\ & \text { Similarly } \overline{\mathrm{b}}+3 \overline{\mathrm{c}}=\mathrm{ma} \quad \text{.... (ii)}\\ & \mathrm{m} \text { and } \mathrm{n} \text { are non-zero scalars. } \\ \therefore \quad & \text { (i) } \Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}=(\mathrm{n}+6) \overline{\mathrm{c}} \\ & \text { (ii) } \Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}=(2 \mathrm{~m}+1) \overline{\mathrm{a}} \\ &\Rightarrow \mathrm{n}+6 =0 \text { and } 2 \mathrm{~m}+1=0 \\ & \Rightarrow \mathrm{n}=-6 \text { and } \mathrm{m}=\frac{-1}{2} \\ \therefore \quad & \text { (i) } \Rightarrow \vec{a}+2 \vec{~b}=-6 \overline{\mathrm{c}} \end{array}$$

Let $\theta$ be the angle between $\overline{\mathrm{AB}}$ and $\overline{\mathrm{AD}}$

$$\begin{aligned} \therefore \quad \cos \theta & =\frac{\overline{A B} \cdot \overline{A D}}{|\overline{A B}| \overline{A D} \mid} \\ & =\frac{(2 \hat{i}+10 \hat{j}+11 \hat{k}) \cdot(-\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{4+100+121} \sqrt{1+4+4}} \\ & =\frac{-2+20+22}{\sqrt{225} \sqrt{9}} \\ & =\frac{40}{45} \\ & =\frac{8}{9} \\ \therefore \quad \sin \theta & =\sqrt{1-\left(\frac{8}{9}\right)^2}=\frac{\sqrt{17}}{9} \end{aligned}$$

$\alpha$ is the angle of rotation of AD The angle between side $A B$ and $A D$

$$\begin{aligned} & =\alpha+\theta \\ & =90^{\circ} \quad \text{...[Given]} \end{aligned}$$

$\therefore \quad \cos (\alpha+\theta)=\cos \left(90^{\circ}\right)$

$\therefore \quad \cos \alpha \cos \theta-\sin \alpha \sin \theta=0$

$\therefore \quad 8 \cos \alpha=\sqrt{17} \sin \alpha$

$\therefore \quad 64 \cos ^2 \alpha=17\left(1-\cos ^2 \alpha\right)$

$\therefore \quad 81 \cos ^2 \alpha=17$

$\therefore \quad \cos \alpha=\frac{\sqrt{17}}{9}$

$$\begin{aligned} & \text { Let } \mathrm{I}=\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 \mathrm{~d} x \\ & \text { Let } \frac{1}{x}=\mathrm{t} \Rightarrow \frac{1}{x^2} \mathrm{~d} x=-\mathrm{dt} \\ & \therefore \quad \mathrm{l}=-\int \tan ^2 \mathrm{t} d \mathrm{t} \\ & =\int\left(1-\sec ^2 t\right) d t \\ & =\mathrm{t}-\tan \mathrm{t}+\mathrm{c} \\ & =\frac{1}{x}-\tan \left(\frac{1}{x}\right)+c \\ & \end{aligned}$$

Let $\mathrm{X}$ : Event that number on the ticket is perfect square.

$$\begin{array}{ll} \therefore \quad \mathrm{X}= \{1,4,9,16,25,36,49,64,81,100\} \\ \therefore \quad \mathrm{n}(\mathrm{X})=10 \end{array}$$

Also, $\mathrm{n}(\mathrm{S})=100$

$\therefore \quad$ Required probability $\mathrm{=\frac{n(X)}{n(S)}=\frac{10}{100}=\frac{1}{10}}$

Let $\mathrm{p}$ : The switch $\mathrm{S}_1$

$\quad \mathrm{q}$ : The switch $\mathrm{S}_2$