MHT CET 2023 14th May Evening Shift (Mathematics) Question Paper with Solutions

$$\begin{aligned} & \text { Given, } \overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \\ & \Rightarrow \overline{\mathrm{b}} \times(\overline{\mathrm{c}}-\overline{\mathrm{a}})=\overline{0} \\ & \Rightarrow \overline{\mathrm{b}} \text { is parallel to }(\overline{\mathrm{c}}-\overline{\mathrm{a}}) . \\ & \Rightarrow \overline{\mathrm{c}}-\overline{\mathrm{a}}=\lambda \overline{\mathrm{b}} \text { for some scalar } \lambda \\ & \Rightarrow \overline{\mathrm{c}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}} \quad \text{.... (i)}\\ & \Rightarrow \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}) \\ & \Rightarrow 0=|\ldots|^2+\lambda(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}) \quad \ldots[\because \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0 \text { (given) }] \\ & \Rightarrow 0=6+4 \lambda \\ & \Rightarrow \lambda=-\frac{3}{2} \end{aligned}$$

Substituting the value of $\lambda$ in (i), we get

$$\begin{aligned} \bar{c} & =(\hat{i}+2 \hat{j}-\hat{k})-\frac{3}{2}(\hat{i}+\hat{j}-\hat{k}) \\ & =-\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \\ \therefore \quad \bar{c} \cdot \bar{b} & =-\frac{1}{2}(\hat{i}-\hat{j}-\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k}) \\ & =-\frac{1}{2}(1-1+1)=-\frac{1}{2} \end{aligned}$$

$$\begin{aligned} & \sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}=p \\ \Rightarrow & \sum_{r=1}^{50} \tan ^{-1}\left(\frac{2}{4 r^2}\right)=p \\ \Rightarrow & \sum_{r=1}^{50} \tan ^{-1}\left[\frac{(2 r+1)-(2 r-1)}{1+(2 r+1)(2 r-1)}\right]=p \\ \Rightarrow & \sum_{r=1}^{50}\left[\tan ^{-1}(2 r+1)-\tan ^{-1}(2 r-1)\right]=p \\ \Rightarrow & \tan ^{-1}(101)-\tan ^{-1}(1)=p \\ \Rightarrow & \tan ^{-1}\left(\frac{101-1}{1+101}\right)=p \\ \Rightarrow & \frac{100}{102}=\tan p \\ \Rightarrow & \tan p=\frac{50}{51} \end{aligned}$$

$\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p})$ is false

$\Rightarrow \mathrm{p} \equiv \mathrm{F}$ and $\mathrm{q} \equiv \mathrm{F}$

Consider option (B),

$$\begin{aligned} \mathrm{p} \rightarrow(\mathrm{p} \vee \sim \mathrm{q}) & \equiv \mathrm{F} \rightarrow(\mathrm{F} \vee \sim \mathrm{F}) \\ & \equiv \mathrm{F} \rightarrow(\mathrm{F} \vee \mathrm{T}) \\ & \equiv \mathrm{F} \rightarrow \mathrm{T} \\ & \equiv \mathrm{T} \end{aligned}$$

$$\begin{aligned} & f(x)=x^3-6 x^2+9 x+2 \\ \therefore \quad & f^{\prime}(x)=3 x^2-12 x+9 \end{aligned}$$

For maximum or minimum,

$$\begin{aligned} & \mathrm{f}^{\prime}(x)=0 \\ & \Rightarrow 3 x^2-12 x+9=0 \\ & \Rightarrow 3(x-1)(x-3)=0 \\ & \Rightarrow x=1,3 \end{aligned}$$

Now, $\mathrm{f}^{\prime \prime}(x)=6 x-12$

$\therefore \quad \mathrm{f}^{\prime \prime}(1)=-6<0$

$\therefore \quad \mathrm{f}(x)$ is maximum at $x=1$.

$$\begin{aligned} x y & =\sqrt{\mathrm{e}^{\sin ^{-1} t}} \cdot \sqrt{\mathrm{e}^{\cos ^{-1} t}} \\ & =\sqrt{\mathrm{e}^{\sin ^{-1} t+\cos ^{-1} t}} \\ \therefore \quad x y & =\sqrt{\mathrm{e}^{\frac{\pi}{2}}} \end{aligned}$$

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

$$\begin{aligned} & x \frac{\mathrm{d} y}{\mathrm{~d} x}+y \cdot 1=0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y}{x} \quad \text{.... (i)} \end{aligned}$$

$$\begin{aligned} \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =-\left(\frac{x \frac{\mathrm{d} y}{\mathrm{~d} x}-y \cdot 1}{x^2}\right) \\ \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =-\left(\frac{x\left(-\frac{y}{x}\right)-y}{x^2}\right) \quad \text{... [From (i)]}\\ & =\frac{2 y}{x^2} \end{aligned}$$

Projection of $\overline{\mathrm{v}}$ along $\overline{\mathrm{u}}=$ Projection of $\overline{\mathrm{w}}$ along $\bar{u}$

$$\begin{aligned} & \Rightarrow \frac{\overline{\mathrm{v}} \cdot \overline{\mathrm{u}}}{|\overrightarrow{\mathrm{u}}|}=\frac{\overline{\mathrm{w}} \cdot \overline{\mathrm{u}}}{|\overrightarrow{\mathrm{u}}|} \\ & \Rightarrow \overline{\mathrm{v}} \cdot \overline{\mathrm{u}}=\overline{\mathrm{w}} \cdot \overline{\mathrm{u}} \quad \text{.... (i)} \end{aligned}$$

Also, $\overline{\mathrm{v}}$ and $\overline{\mathrm{w}}$ are perpendicular to each other.

$\therefore \quad \overline{\mathrm{v}} \cdot \overline{\mathrm{w}}=0$ ..... (ii)

Now, $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2=|\overline{\mathrm{u}}|^2+|\overline{\mathrm{v}}|^2+|\overline{\mathrm{w}}|^2-2(\overline{\mathrm{u}} \cdot \overline{\mathrm{v}})-2(\bar{v} \cdot \bar{w})+2(\bar{u} \cdot \bar{w})$

$\Rightarrow|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|^2=1+4+9$ ....[From (i) and (ii)]

$\Rightarrow|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|=\sqrt{14}$

$S=\{1,2,3,4,5,6\}$

The values of $\mathrm{X}$ for the possible outcomes of the experiment are as follows:

$$\begin{aligned} & \text { experiment are as follows: } \\ & X(1)=1, X(2)=2, X(3)=2, X(4)=3 \\ & X(5)=2, X(6)=4 \\ & P(X=1)=P[\{1\}]=\frac{1}{6} \\ & P(X=2)=P[\{2,3,5\}]=\frac{3}{6}=\frac{1}{2} \\ & P(X=3)=P[\{4\}]=\frac{1}{6} \\ & P(X=4)=P[\{6\}]=\frac{1}{6} \end{aligned}$$

The probability distribution of X is

Number of diagonals in a polygon of n sides is ${ }^{\mathrm{n}} C_2-\mathrm{n}$.

$$\begin{aligned} \therefore \quad & { }^{\mathrm{n}} C_2-\mathrm{n}=54 \\ & \Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}=54 \\ & \Rightarrow \mathrm{n}^2-3 \mathrm{n}-108=0 \\ & \Rightarrow(n-12)(\mathrm{n}+9)=0 \end{aligned}$$

But, number of sides cannot be negative,

$\therefore$ n = 12

$$\begin{aligned} & \int \mathrm{e}^x\left[\frac{x^2+4 x+4}{(x+4)^2}\right] \mathrm{d} x \\ = & \int \mathrm{e}^x\left[\frac{x(x+4)+4}{(x+4)^2}\right] \mathrm{d} x \\ = & \int \mathrm{e}^x\left[\frac{x}{x+4}+\frac{4}{(x+4)^2}\right] \mathrm{d} x \\ = & \mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c} \end{aligned}$$

$\ldots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]$

$$\begin{aligned} \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} & =|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos 60^{\circ} \\ & =(2)(3)\left(\frac{1}{2}\right) \\ & =3 \\ \overline{\mathrm{b}} \cdot \overline{\mathrm{c}} & =|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \cos 60^{\circ} \\ & =(3)(5)\left(\frac{1}{2}\right)=\frac{15}{2} \end{aligned}$$

$$\begin{aligned} \overline{\mathrm{a}} \cdot \overline{\mathrm{c}} & =|\overrightarrow{\mathrm{a}}| \overrightarrow{\mathrm{c}} \cos 60^{\circ} \\ & =(2)(5)\left(\frac{1}{2}\right) \\ & =5 \\ \therefore \quad|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2 & =|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & =2^2+3^2+5^2+2\left(3+\frac{15}{2}+5\right) \\ & =4+9+25+31 \\ & =69 \\ \therefore \quad|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}| & =\sqrt{69} \end{aligned}$$

$$\begin{aligned} & |\bar{c}-\vec{a}|=2 \sqrt{2} \\ & \Rightarrow|\vec{c}|^2+|\vec{a}|^2-2(\bar{a} \cdot \bar{c})=8 \\ & \Rightarrow|\vec{c}|^2+9-2|\vec{c}|=8 \quad \ldots[\because \bar{a} \cdot \bar{c}=|\vec{c}| \text { (given) }] \\ & \Rightarrow(|\vec{c}|-1)^2=0 \\ & \Rightarrow|\vec{c}|=1 \end{aligned}$$

$$\begin{aligned} \text{Now,}\quad |(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}|= & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \sin \frac{2 \pi}{3} \\ & =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|(1)\left(\frac{\sqrt{3}}{2}\right) \\ & =\frac{3 \sqrt{3}}{2} \\ & \ldots[\because \overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}] \end{aligned}$$

The d.r.s. of the normal to the plane are $1,-1,2$.

$\therefore \quad$ The equation of line PM is

$$\begin{aligned} & \frac{x-2}{1}=\frac{y-4}{-1}=\frac{z+1}{2}=\lambda \text { (say) } \\ \Rightarrow & x=\lambda+2, y=-\lambda+4, z=2 \lambda-1 \end{aligned}$$

Let $M \equiv(\lambda+2,-\lambda+4,2 \lambda-1)$

$\therefore \quad$ Equation of plane becomes

$$\begin{array}{ll} \therefore \quad & \text { Equation of plane becomes } \\ & 1(\lambda+2)-1(-\lambda+4)+2(2 \lambda-1)-2=0 \\ & \Rightarrow \lambda=1 \\ \therefore \quad & M \equiv(3,3,1) \end{array}$$

Since $M$ is the mid-point of $P Q$.

$$\begin{aligned} \therefore & \quad\frac{2+a}{2}=3, \frac{4+b}{2}=3, \frac{-1+c}{2}=1 \\ & \Rightarrow a=4, b=2, c=3 \\ & \Rightarrow a+b+c=9 \end{aligned}$$

$$A+B=\left[\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right]+\left[\begin{array}{cc} 1 & 1 \\ 4 & -1 \end{array}\right]=\left[\begin{array}{cc} 2 & 0 \\ 6 & -2 \end{array}\right]$$

$$\therefore \quad|A+B|=\left|\begin{array}{cc} 2 & 0 \\ 6 & -2 \end{array}\right|=-4 \neq 0$$

$$\begin{aligned} & \text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \text { and } a d-b c \neq 0, \\ & \text { then } A^{-1}=\frac{1}{(a d-b c)}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \\ & \therefore \quad(A+B)^{-1}=\frac{1}{-4}\left[\begin{array}{cc} -2 & 0 \\ -6 & 2 \end{array}\right] \\ &=\left[\begin{array}{cc} \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} \end{array}\right] \end{aligned}$$

Shaded region lies on origin side of $3 x+6 y=18, x-3 y=3,-x+2 y=2$ and on non-origin side of $2 x+3 y=6$.

$$\begin{aligned} \therefore 2 x+3 y & \geq 6,3 x+6 y \leq 18, x-3 y \leq 3,-x+2 y \leq 2, x \geq 0, y \geq 0 \end{aligned}$$

$$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=-y+\mathrm{e}^{-x} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}+y=\mathrm{e}^{-x} \\ \therefore \quad & \text { I.F. }=\mathrm{e}^{\int \mathrm{dx}}=\mathrm{e}^x \end{aligned}$$

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

$$\begin{aligned} & y \mathrm{e}^x=\int \mathrm{e}^x \cdot \mathrm{e}^{-x} \mathrm{~d} x+c \\ & \Rightarrow y \mathrm{e}^x=\int \mathrm{d} x+c \\ & \Rightarrow y \mathrm{e}^x=x+\mathrm{c} \quad \text{.... (i)} \end{aligned}$$

Since the curve passes through $(0,0)$.

$$\begin{array}{ll} \therefore \quad & 0=0+\mathrm{c} \\ & \Rightarrow \mathrm{c}=0 \\ \therefore \quad & y \mathrm{e}^x=x \quad \text{... [From (i)]}\\ & \Rightarrow y \mathrm{e}^x-x=0 \end{array}$$

$$\begin{aligned} \mathrm{PC}_1 & =\sqrt{(2+1)^2+(3+1)^2} \\ & =\sqrt{25} \\ & =5 \end{aligned}$$

$\mathrm{P}$ divides $\mathrm{C}_1 \mathrm{C}_2$ externally in the ratio $\mathrm{r}_1: \mathrm{r}_2$ i.e. $5: 3$

$$\begin{aligned} & \therefore \quad-1=\frac{5(\mathrm{~h})-3(2)}{5-3} \text { and }-1=\frac{5(\mathrm{k})-3(3)}{5-3} \\ & \Rightarrow-2=5 \mathrm{~h}-6 \text { and }-2=5 \mathrm{k}-9 \\ & \Rightarrow \mathrm{h}=\frac{4}{5} \text { and } \mathrm{k}=\frac{7}{5} \end{aligned}$$

$$\begin{aligned} & \operatorname{Im}(\mathrm{z})=10 \\ & \text { Let } \mathrm{z}=x+10 \mathrm{i} \\ & \frac{2 \mathrm{z}-\mathrm{n}}{2 \mathrm{z}+\mathrm{n}}=2 \mathrm{i}-1 \\ & \Rightarrow \frac{2(x+10 \mathrm{i})-\mathrm{n}}{2(x+10 \mathrm{i})+\mathrm{n}}=2 \mathrm{i}-1 \\ & \Rightarrow(2 x-\mathrm{n})+20 \mathrm{i}=(2 \mathrm{i}-1)(2 x+20 \mathrm{i}+\mathrm{n}) \\ & \Rightarrow(2 x-\mathrm{n})+20 \mathrm{i}=(-2 x-\mathrm{n}-40)+(4 x+2 \mathrm{n}-20) \mathrm{i} \end{aligned}$$

Equating real and imaginary parts, we get

$$\begin{aligned} & 2 x-\mathrm{n}=-2 x-\mathrm{n}-40 \text { and } 20=4 x+2 \mathrm{n}-20 \\ & \Rightarrow x=-10 \text { and } 20=4(-10)+2 \mathrm{n}-20 \\ & \Rightarrow x=-10 \text { and } \mathrm{n}=40 \end{aligned}$$

$\mathrm{f}$ is continuous at $x=0$.

$$\begin{aligned} \therefore \quad \mathrm{f}(0) & =\lim _{x \rightarrow 0} \mathrm{f}(x) \\ \therefore \quad \mathrm{k} & =\lim _{x \rightarrow 0}\left(\frac{1}{x} \log (1+3 x)-\frac{1}{x} \log (1-2 x)\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{3 \log (1+3 x)}{3 x}+\frac{2 \log (1-2 x)}{-2 x}\right) \\ & =3+2=5 \end{aligned}$$

$$\begin{aligned} & \text { Mean }=\frac{\sum x_{\mathrm{i}}}{\mathrm{n}} \\ & \Rightarrow 16=\frac{\sum x_{\mathrm{i}}}{50} \\ & \Rightarrow \sum x_{\mathrm{i}}=800 \\ & \text { Variance }=\frac{\sum x_{\mathrm{i}}^2}{\mathrm{n}}-(\bar{x})^2 \\ & \Rightarrow 256=\frac{\sum x_{\mathrm{i}}^2}{50}-(16)^2 \\ & \begin{aligned} \Rightarrow \frac{\sum x_{\mathrm{i}}^2}{50}= & 512 \\ \Rightarrow \sum x_{\mathrm{i}}^2= & 25600 \\ \text { New mean } & =\frac{\sum\left(x_{\mathrm{i}}-5\right)^2}{\mathrm{n}} \\ & =\frac{\sum x_{\mathrm{i}}^2-10 \sum x_{\mathrm{i}}+\sum 25}{50} \\ & =\frac{25600-10(800)+25(50)}{50} \\ & =\frac{18850}{50} \\ & =377 \end{aligned} \end{aligned}$$

\begin{aligned} \int_\limits0^{\frac{\pi}{4}}\left(\tan ^{\mathrm{n}} x\right. & \left.+\tan ^{\mathrm{n}-2} x\right) \mathrm{d} x=\frac{1}{\mathrm{n}-1} \\ \therefore \quad \mathrm{I}_{12}+\mathrm{I}_{10} & =\int_\limits0^{\frac{\pi}{4}}\left(\tan ^{12} \theta+\tan ^{10} \theta\right) \mathrm{d} \theta \\ & =\frac{1}{12-1} \\ & =\frac{1}{11} \end{aligned}

Slope of $Q R=\frac{3 \sqrt{3}-0}{3-0}=\sqrt{3}$ i.e., $\theta=60^{\circ}$