MHT CET 2023 11th May Evening Shift (Mathematics) Question Paper with Solutions

Here, $\bar{c}=2 \hat{i}-\hat{j}+4 \hat{k}$

And given that $\bar{c} \cdot \bar{d}=15$

We verify given options one by one to satisfy the above condition.

Consider option (B),

For $\overline{\mathrm{d}}=90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}$

$$\begin{aligned} \bar{c} \cdot \bar{d} & =(2)(90)+(-1)(-3)+(4)(-42) \\ & =180+3-168=15 \end{aligned}$$

$\therefore \quad$ Option (B) is correct.

$$\begin{aligned} & f(x)=\frac{1-x+x^2}{1+x+x^2} \\ \therefore \quad & f^{\prime}(x)=\frac{\left(1+x+x^2\right)(-1+2 x)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)^2} \\ = & \frac{\left(-1+2 x-x+2 x^2-x^2+2 x^3\right)-\left(-1+2 x-x-2 x^2+x^2+2 x^3\right)}{\left(1+x+x^2\right)^2} \\ = & \frac{-2+2 x^2}{\left(1+x+x^2\right)^2} \end{aligned}$$

If $\mathrm{f}^{\prime}(x)=0$, then $\frac{-2+2 x^2}{\left(1+x+x^2\right)^2}=0 \Rightarrow x^2=1$ $\Rightarrow x= \pm 1$

$\therefore \quad \mathrm{f}(x)$ at $x=1$ is $\frac{1}{3}$ and $\mathrm{f}(x)$ at $x=-1$ is 1.

$\therefore \quad$ Minimum value of $\mathrm{f}(x)$ is $\frac{1}{3}$.

Area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|$

$\therefore \quad|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=16$

$\therefore \quad$ Area of the required parallelogram

$$\begin{aligned} & =|(3 \overline{\mathrm{a}}+2 \overline{\mathrm{b}}) \times(\overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\ & =|3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+(\overline{\mathrm{b}} \times \overline{\mathrm{b}})| \\ & =0+9|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|-2|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|+0 \\ & =7|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \\ & =7 \times 16 \\ & =112 \end{aligned}$$

Let

$$\begin{aligned} \mathrm{I} & =\int(1-\cos x) \cdot \operatorname{cosec}^2 \mathrm{~d} x \\ & =\int \frac{2 \sin ^2 \frac{x}{2}}{[\sin x]^2} \mathrm{~d} x \quad \ldots\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right] \\ & =\int \frac{2 \sin ^2 \frac{x}{2}}{\left[2 \sin \frac{x}{2} \cos \frac{x}{2}\right]^2} \mathrm{~d} x \quad \ldots\left[\because \sin\theta=2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\right]\\ & =\frac{1}{2} \int \sec ^2 \frac{x}{2} \mathrm{~d} x \\ & =\tan \frac{x}{2}+\mathrm{c} \end{aligned}$$

Given that $\mathrm{n}=50, \bar{x}=16$ and $\sigma_x^2=256$

\begin{array}{ll} \therefore & \sigma_x^2=\frac{1}{n}\left(\sum_\limits{i=1}^{50} x_i^2\right)-(\bar{x})^2 \\ \therefore & 256=\frac{1}{50}\left(\sum_\limits{i=1}^{50} x_i^2\right)-256 \\ \therefore & \frac{1}{50}\left(\sum_\limits{i=1}^{50} x_i^2\right)=512 \\ \therefore & \sum_\limits{i=1}^{50} x_i^2=25600 \quad \text{..... (i)} \end{array}

\begin{aligned} & \text { Now } \sum_\limits{i=1}^{50}\left(x_i-5\right)^2=\sum_\limits{i=1}^{50} x_i^2+25 \times 50-10 \sum_\limits{i=1}^{50} x_i \\ & =25600+1250-8000 \quad \text {...[From (i) and (ii)] } \\ & =18850 \\ & \therefore \quad \text { Required Mean }=\frac{\sum_\limits{\mathrm{i}=1}^{50}\left(x_{\mathrm{i}}-5\right)^2}{50}=\frac{18850}{50}=377 \\ \end{aligned}

$$\begin{aligned} & \frac{\mathrm{d} x}{\mathrm{~d} y}+\frac{x}{y}=x^2 \\ & \frac{1}{x^2} \frac{\mathrm{d} x}{\mathrm{~d} y}+\frac{1}{x y}=1 \quad \text{.... (i)} \end{aligned}$$

Let $\frac{1}{x}=\mathrm{t}$

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

$$\begin{array}{ll} & \frac{-1}{x^2} \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{\mathrm{dt}}{\mathrm{d} y} \Rightarrow \frac{1}{x^2} \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{-\mathrm{dt}}{\mathrm{d} y} \\ \therefore & \text { (i) } \Rightarrow \frac{-\mathrm{dt}}{\mathrm{d} y}+\frac{\mathrm{t}}{y}=1 \\ \therefore \quad & \frac{\mathrm{dt}}{\mathrm{d} y}-\frac{\mathrm{t}}{y}=-1 \\ \therefore & \text { I.F. }=\mathrm{e}^{\int \frac{-1}{y} \mathrm{~d} y}=\mathrm{e}^{\log \left(\frac{1}{y}\right)}=\frac{1}{y} \end{array}$$

$\therefore \quad$ The solution of the given equation is

$$\begin{aligned} & \mathrm{t}(\text { I.F })=\int(-1)(\text { I.F. }) \mathrm{d} y+\mathrm{c} \\ & \mathrm{t}\left(\frac{1}{y}\right)=\int \frac{-1}{y} \mathrm{~d} y+\mathrm{c} \end{aligned}$$

$$\begin{aligned} & \therefore \quad \frac{t}{y}=-\log y+c \\ & \therefore \quad \frac{1}{x y}=-\log y+c \\ & \therefore \quad \frac{1}{x}=c y-y \log y \end{aligned}$$

$$\begin{aligned} & {[\sim(\sim p \vee q) \vee(p \wedge r)] \wedge(\sim q \wedge r)} \\ & \equiv[(p \wedge \sim q) \vee(p \wedge r)] \wedge(\sim q \wedge r) \quad \text{...[De Morgan's law]} \end{aligned}$$

$$\begin{aligned} & \equiv p \wedge(\sim q \vee r) \wedge(\sim q \wedge r) \quad \ldots[\text { Distributive law] } \\ & \equiv p \wedge[(\sim q \vee r) \wedge \sim q] \wedge r \quad \ldots \text { A Associative law] } \\ & \equiv p \wedge(\sim q) \wedge r \quad \ldots[\text { Absorption law] } \\ & \equiv(p \wedge r) \wedge \sim q \quad \text{...[Commutative and Associative law]} \end{aligned}$$

$$\begin{aligned} & \pi+\left[\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =\pi+\left[\left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =\pi+\left[\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}\right)+\sin ^{-1} \frac{16}{65}\right] \\ & =\pi+\left[\tan ^{-1}\left(\frac{63}{16}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =\pi+\left[\cos ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{16}{65}\right)\right] \\ & =\pi+\frac{\pi}{2} \\ & =\frac{3 \pi}{2} \end{aligned}$$

$$\begin{aligned} \mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right) & =\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)} \\ & =\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{1-\mathrm{P}(\mathrm{B})} \\ & =\frac{1-\frac{1}{3}}{1-\frac{1}{5}}=\frac{\frac{2}{3}}{\frac{4}{5}}=\frac{5}{6} \\ \mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right) & =\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)} \\ & =\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{1-\mathrm{P}(\mathrm{A})} \\ & =\frac{\frac{2}{3}}{1-\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{2}{3}}=1 \\ \therefore \quad \mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right) & +\mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right)=\frac{5}{6}+1=\frac{11}{6} \end{aligned}$$

Probability of getting ten $=\frac{4}{52}=\frac{1}{13}$

$\therefore \quad$ Probability of getting a card without ten $=\frac{12}{13}$

Let random variable $\mathrm{X}$ denotes the number of tens.

$\therefore \quad$ Possible values of $\mathrm{X}$ are $0,1,2$

Consider following probability distribution table.

$$\begin{aligned} \therefore \quad & \text { Required mean } \\ & =0+1 \times\left(\frac{12}{13 \times 13}+\frac{12}{13 \times 13}\right)+2 \times\left(\frac{1}{13} \times \frac{1}{13}\right) \\ & =\frac{24}{169}+\frac{2}{169} \\ & =\frac{26}{169} \\ & =\frac{2}{13} \end{aligned}$$

$$\begin{aligned} & \text { Let } \mathrm{f}(x)=\frac{x}{|x|+x^2} \\ & \qquad \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{-x+x^2}=\lim _{x \rightarrow 0} \frac{1}{-1+x}=-1 \\ & \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{x+x^2}=\lim _{x \rightarrow 0} \frac{1}{1+x}=1 \end{aligned}$$

Here, \lim _\limits{x \rightarrow 0^{-}} \mathrm{f}(x) \neq \lim _\limits{x \rightarrow 0^{+}} \mathrm{f}(x)

$\therefore$ Value of $l$ is non-existant

$$\begin{aligned} & A=\left[\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}\right] \\ & \therefore \quad A^2=\left[\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}\right] \times\left[\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 0 & i \\ i & 1 \end{array}\right] \\ & \therefore \quad A^3=\left[\begin{array}{ll} 0 & i \\ i & 1 \end{array}\right]\left[\begin{array}{ll} i & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right] \\ & \therefore \quad \mathrm{A}^6=\mathrm{A}^3 \times \mathrm{A}^3 \\ & =\left[\begin{array}{ll} \mathrm{i} & 0 \\ 0 & \mathrm{i} \end{array}\right] \times\left[\begin{array}{ll} \mathrm{i} & 0 \\ 0 & \mathrm{i} \end{array}\right] \\ & =\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]=(-1) I_2 \\ & \text { Now, } B=A^{2029}=A^{(6 \times 338+1)} \\ & \therefore \quad \mathrm{B}=\left(\mathrm{A}^6\right)^{338} \times \mathrm{A} \\ & =\left((-1) I_2\right)^{338} \times A \\ & =\mathrm{I}_2 \times \mathrm{A} \\ & \therefore \quad \mathrm{B}=\mathrm{A} \\ & \text { Now, }|A|=-1 \\ & \therefore \quad \mathrm{B}^{-1}=\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A} \\ & \therefore \quad B^{-1}=-\operatorname{adj} \mathrm{A} \\ \end{aligned}$$

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

$$\begin{aligned} \therefore \quad & 4 x+x+x=180^{\circ} \Rightarrow 6 x=180^{\circ} \Rightarrow x=30^{\circ} \\ & \frac{\sin 120^{\circ}}{a}=\frac{\sin 30^{\circ}}{b}=\frac{\sin 30^{\circ}}{c} \\ \therefore \quad & a:(a+b+c) \\ & =\left(\sin 120^{\circ}\right):\left(\sin 120^{\circ}+\sin 30^{\circ}+\sin 30^{\circ}\right) \\ & =\frac{\sqrt{3}}{2}: \frac{\sqrt{3}+2}{2}=\sqrt{3}:(\sqrt{3}+2) \end{aligned}$$

Objective function $\mathrm{z}=3 x+5 y$

The corner points of the feasible region are $\mathrm{O}(0,0), \mathrm{A}(4,0), \mathrm{B}(4,3), \mathrm{C}(2,6)$ and $\mathrm{D}(0,6)$

$\therefore \quad \mathrm{Z}$ at $\mathrm{A}(4,0)=12$

$\qquad\mathrm{Z}$ at $\mathrm{B}(4,3)=27$

$\qquad\mathrm{Z}$ at $\mathrm{C}(2,6)=36$

$\qquad\mathrm{Z}$ at $\mathrm{D}(0,6)=30$

$\therefore \quad$ Maximum value of $\mathrm{Z}$ is 36.

Note that only for option (A), i.e., for $\alpha=1$ and $\beta=1,|\vec{c}|=\sqrt{3}$ holds true.

$\therefore \quad$ Option (A) is correct.

$f(x)=x^3-3(a-2) x^2+3 a x+7$

As $\mathrm{f}(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$, we get that $\mathrm{f}(x)$ has critical point at $x=1$

$$\begin{aligned} & \Rightarrow \mathrm{f}^{\prime}(1)=0 \\ & \mathrm{f}^{\prime}(x)=3 x^2-6(\mathrm{a}-2) x+3 \mathrm{a} \\ & \therefore \quad 3(1)^2-6(\mathrm{a}-2)+3 \mathrm{a}=0 \\ & \therefore \quad \mathrm{a}=5 \\ & \therefore \quad \frac{\mathrm{f}(x)-14}{(x-1)^2}=\frac{x^3-9 x^2+15 x-7}{(x-1)^2} \\ &=\frac{(x-1)^2(x-7)}{(x-1)^2} \\ &=x-7 \end{aligned}$$

$\therefore \quad$ The required root is 7.

Joint equation of the liens passing through the point $\left(x_1, y_1\right)$ and perpendicular to the lines $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0$ is:

$\mathrm{b}\left(x-x_1\right)^2-2 \mathrm{~h}\left(x-x_1\right)\left(y-y_1\right)+\mathrm{a}\left(y-y_1\right)^2=0$

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

$$\begin{aligned} -3(x-3)^2-2(x-3)(y+2)+5(y+2)^2=0 \\ \therefore \quad-3\left(x^2-6 x+9\right)-2(x y+2 x-3 y-6)+5\left(y^2+4 y+4\right)=0 \end{aligned}$$

$$\begin{aligned} & \therefore \quad-3 x^2+18 x-27-2 x y-4 x+6 y+12+5 y^2 \\ & \quad+20 y+20=0 \\ & \therefore \quad 3 x^2+2 x y-5 y^2-14 x-26 y-5=0 \end{aligned}$$

Let $\mathrm{I}=\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} d x$

$=\int x \sqrt{\frac{2 \sin \left(x^2+1\right)-2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}{2 \sin x\left(x^2+1\right)+2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}} \mathrm{d} x$

$=\int x \sqrt{\frac{1-\cos \left(x^2+1\right)}{1+\cos \left(x^2+1\right)}} \mathrm{d} x$

$=\int x \sqrt{\frac{2 \sin ^2\left(\frac{x^2+1}{2}\right)}{2 \cos ^2\left(\frac{x^2+1}{2}\right)}} d x$

$=\int x \tan \left(\frac{x^2+1}{2}\right) d x$

Let $\left(\frac{x^2+1}{2}\right)=\mathrm{t} \Rightarrow x \mathrm{~d} x=\mathrm{dt}$

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

Required line is perpendicular to the lines $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$ and $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$

$\therefore \quad$ Required line is parallel to vector

$$\bar{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 2 & -3 & 1 \end{array}\right|=-4 \hat{i}-7 \hat{j}-13 \hat{k}$$

$\therefore \quad$

The equation of the required line is $(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-4 \hat{i}-7 \hat{j}-13 \hat{k})$

$$\begin{aligned} & \text { let } \mathrm{f}(\theta)=\mathrm{a} \sec \theta-\mathrm{b} \tan \theta \\ & \therefore \quad \mathrm{f}^{\prime}(\theta)=\mathrm{a} \sec \theta \tan \theta-\mathrm{b} \sec ^2 \theta \\ & =\sec \theta(a \tan \theta-b \sec \theta) \\ & \therefore \quad \mathrm{f}^{\prime}(\theta)=0 \Rightarrow \sec \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta)=0 \\ & \Rightarrow \mathrm{a} \tan \theta-\mathrm{b} \sec \theta=0 \\ & \ldots\left[\text { As } 0<\pi<\frac{\pi}{2}, \sec \theta \neq 0\right] \\ & \Rightarrow \mathrm{a} \sin \theta-\mathrm{b}=0 \\ & \ldots\left[\text { As } 0<\pi<\frac{\pi}{2}, \cos \theta \neq 0\right] \\ \end{aligned}$$

$$\begin{aligned} & \Rightarrow \sin \theta=\frac{b}{a} \\ & \Rightarrow \sec \theta=\frac{a}{\sqrt{a^2-b^2}} \text { and } \tan \theta=\frac{b}{\sqrt{a^2-b^2}} \quad \text{... (i)} \end{aligned}$$

Now,

$$\begin{aligned} & f^{\prime \prime}(\theta)=\sec \theta \tan \theta(a \tan \theta-b \sec \theta) \\ & \quad+\sec \theta\left(a \sec ^2 \theta-b \sec \theta \tan \theta\right) \\ & =a \tan ^2 \theta \sec \theta-b \sec ^2 \theta \tan \theta \\ & \quad+a \sec ^3 \theta-b \sec ^2 \theta \tan \theta \\ & =a \sec \theta\left(\tan ^2 \theta+\sec ^2 \theta\right) \\ & =a \sec \theta\left(1+2 \tan ^2 \theta\right) \\ & >0 \quad \ldots\left[\because \text { a is positive and } 0<\theta<\frac{\pi}{2}\right] \end{aligned}$$

$\therefore \mathrm{f}(\theta)$ is minimum when $\sin \theta=\frac{\mathrm{b}}{\mathrm{a}}$.

$\therefore \quad$ Minimum value of $f(\theta)$

$$\begin{aligned} & =a\left(\frac{a}{\sqrt{a^2-b^2}}\right)-b\left(\frac{b}{\sqrt{a^2-b^2}}\right) \quad \ldots[\text { From (i) }] \\ & =\frac{a^2-b^2}{\sqrt{a^2-b^2}} \\ & =\sqrt{a^2-b^2} \end{aligned}$$