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

Let the equation of the line be

$\frac{x}{a}+\frac{y}{b}=1$

According to the given condition,

$$\begin{aligned} & p=\left|\frac{a b}{\sqrt{a^2+b^2}}\right| \\ & \Rightarrow \frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2} \\ & \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2} \end{aligned}$$

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

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

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

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

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

Greatest integer function is discontinuous on integer values.

$\therefore \quad \mathrm{f}(x)$ is discontinuous at all integer points.

The equation of plane passing through $(1,-1,2)$ is

$\mathrm{a}(x-1)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}-2)=0$ ..... (i)

Since plane (i) is perpendicular to the planes $x+2 y-2 z=4$ and $3 x+2 y+z=6$

$$\begin{aligned} & \therefore \quad a+2 b-2 c=0 \\ & \text { and } 3 a+2 b+c=0 \\ & \Rightarrow \frac{a}{6}=\frac{b}{-7}=\frac{c}{-4} \end{aligned}$$

$\therefore \quad$ The equation of the required plane is

$$\begin{aligned} & 6(x-1)-7(y+1)-4(z-2)=0 \\ & \Rightarrow 6 x-7 y-4 z-5=0 \end{aligned}$$

$$\begin{aligned} & \bar{a}+2 \bar{b}+2 \bar{c}=\overline{0} \\ & \Rightarrow a+2 \bar{c}=-2 \bar{b} \end{aligned}$$

Squaring on both sides, we get

$$\begin{aligned} & |\overline{\mathrm{a}}|^2+4 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+4|\overline{\mathrm{c}}|^2=4|\overline{\mathrm{b}}|^2 \\ & \Rightarrow 1+4|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{c}}| \cos \theta+4=4 \\ & \Rightarrow \cos \theta=-\frac{1}{4} \\ & \Rightarrow \sin \theta=\frac{\sqrt{15}}{4} \\ & |\overline{\mathrm{a}} \times \overline{\mathrm{c}}|=|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \sin \theta \\ & =(1)(1)\left(\frac{\sqrt{15}}{4}\right)=\frac{\sqrt{15}}{4} \\ \end{aligned}$$

\text { Let } I=\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x

\begin{aligned} & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\mathrm{d} x}{\cos ^{\frac{2}{3}} x \cdot \sin ^{\frac{4}{3}} x} \\ & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\mathrm{d} x}{\frac{\sin ^{\frac{4}{3}} x}{\cos ^{\frac{4}{3}} x} \cdot \cos ^2 x} \\ & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec ^2 x}{\tan^{\frac{4}{3}}x} d x \end{aligned}

Put $\tan x=\mathrm{t}$

$\Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt}$

\therefore \quad I=\int_\limits{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d t}{t^{\frac{4}{3}}}

$$\begin{aligned} & =\left[-3 \mathrm{t}^{-\frac{1}{3}}\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \\ & =-3\left[(\sqrt{3})^{\frac{-1}{3}}-\left(\frac{1}{\sqrt{3}}\right)^{\frac{-1}{3}}\right] \\ & =-3\left(3^{\frac{1}{6}}-3^{\frac{1}{6}}\right) \\ & =3^{\frac{7}{6}}-3^{\frac{5}{6}} \end{aligned}$$

$$\begin{aligned} & \text { Put } \sin ^2 x+\cos ^3 x=\mathrm{t} \\ & \Rightarrow\left(2 \sin x \cos x-3 \cos ^2 x \sin x\right) \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow\left(\sin 2 x-\frac{3}{2} \sin 2 x \cos x\right) \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow \sin 2 x\left(1-\frac{3}{2} \cos x\right) \mathrm{d} x=\mathrm{dt} \\ \therefore \quad & \int \frac{\sin 2 x\left(1-\frac{3}{2} \cos x\right)}{\mathrm{e}^{\sin ^2 x+\cos ^3 x}} \mathrm{~d} x \\ & =\int \frac{1}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt} \\ & =\int \mathrm{e}^{-t} \mathrm{dt} \\ & =-\mathrm{e}^{-t}+\mathrm{c} \\ & =-\mathrm{e}^{-\left(-\sin ^2 x+\cos ^3 x\right)}+\mathrm{c} \end{aligned}$$

$$\begin{aligned} \text { Let } I & =\int \frac{\sin x+\sin ^3 x}{\cos 2 x} d x \\ & =\int \frac{\sin x\left(1+\sin ^2 x\right)}{\cos 2 x} d x \\ & =\int \frac{\sin x\left(1+1-\cos ^2 x\right)}{2 \cos ^2 x-1} d x \\ & =\int \frac{\sin x\left(2-\cos ^2 x\right)}{2 \cos ^2 x-1} d x \end{aligned}$$

Put $\cos x=\mathrm{t} \Rightarrow \sin x \mathrm{~d} x=-\mathrm{dt}$

$$\begin{aligned} \therefore \quad I & =-\int \frac{2-t^2}{2 t^2-1} d t \\ & =\int \frac{t^2-2}{2 t^2-1} d t \\ & =\frac{1}{2} \int \frac{2 t^2-4}{2 t^2-1} d t \\ & =\frac{1}{2} \int\left(1-\frac{3}{2 t^2-1}\right) d t \\ & =\frac{1}{2} \int d t-\frac{3}{2} \int \frac{d t}{(\sqrt{2} t)^2-1^2} \\ & =\frac{1}{2} t-\frac{3}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \log \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}\right|+c \\ & =\frac{1}{2} \cos x-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+c \\ \therefore \quad A & =\frac{1}{2}, B=\frac{-3}{4 \sqrt{2}}, f(x)=\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1} \end{aligned}$$

$$\begin{aligned} & \lim _{x \rightarrow \infty} x^3\left(\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right) \\ & =\lim _{x \rightarrow \infty} \frac{x^3\left(x^2+\sqrt{1+x^4}-2 x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}} \\ & =\lim _{x \rightarrow \infty} \frac{x^3\left(\sqrt{1+x^4}-x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}} \\ & =\mathop {\lim }\limits_{x \to \infty } {{{x^3}(1 + {x^4} - {x^4})} \over {\left( {\sqrt {{x^2} + \sqrt {1 + {x^4}} } + x\sqrt 2 } \right)\left( {\sqrt {1 + {x^4}} + {x^2}} \right)}} \end{aligned}$$

=\lim _\limits{x \rightarrow \infty} \frac{x^3}{x^3\left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}+\sqrt{2}}\right)\left(\sqrt{\frac{1}{x^4}+1}+1\right)}

=\lim _\limits{x \rightarrow \infty} \frac{1}{\left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}+\sqrt{2}}\right)\left(\sqrt{\frac{1}{x^4}+1}+1\right)}

$$\begin{aligned} & =\frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} \\ & =\frac{1}{4 \sqrt{2}} \end{aligned}$$

The probability that the first critic favours the book is $\mathrm{P}(\mathrm{A})=\frac{2}{2+5}=\frac{2}{7}$

$\therefore \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\frac{2}{7}=\frac{5}{7}$

The probability that the second critic favours the book is $P(B)=\frac{3}{3+4}=\frac{3}{7}$

$\therefore \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{3}{7}=\frac{4}{7}$

The probability that the third critic favours the book is $\mathrm{P}(\mathrm{C})=\frac{4}{4+3}=\frac{4}{7}$

$\therefore \quad P\left(C^{\prime}\right)=1-\frac{4}{7}=\frac{3}{7}$

$\therefore$ Majority will be in favour of the book if at least two critics favour the book. Hence, the probability is

$$\begin{aligned} & \mathrm{P}\left(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}^{\prime}\right)+\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}\right) \\ & \quad+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B} \cap \mathrm{C}\right)+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) \\ & =\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}(\mathrm{C}) \\ & \quad+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C}) \end{aligned}$$

$$\begin{aligned} & =\frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7} \\ & =\frac{18}{343}+\frac{32}{343}+\frac{60}{343}+\frac{24}{343} \\ & =\frac{134}{343} \end{aligned}$$

$$\begin{aligned} & y=[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)]^{\mathrm{n}} \\ & \Rightarrow \log y=\operatorname{nlog}[(x+1)(2 x+1)(3 x+1) \ldots(\mathrm{n} x+1)] \end{aligned}$$

$\Rightarrow \log y=\mathrm{n}[\log (x+1)+\log (2 x+1) +\log (3 x+1)+\ldots+\log (n x+1)]$

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

$$\begin{aligned} & \frac{1}{y} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{n}\left(\frac{1}{x+1}+\frac{2}{2 x+1}+\frac{3}{3 x+1}+\ldots+\frac{\mathrm{n}}{\mathrm{n} x+1}\right) \\ & \Rightarrow \frac{1}{1} \cdot\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\mathrm{n}(1+2+3+\ldots+\mathrm{n}) \quad ...[\text{At} ~x=0,y=1]\\ & \Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\mathrm{n}\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]=\frac{\mathrm{n}^2(\mathrm{n}+1)}{2} \end{aligned}$$

P\left(\frac{1}{4}< x<\frac{1}{3}\right)=\int_\limits{\frac{1}{4}}^{\frac{1}{3}} \mathrm{f}(x) \mathrm{d} x=\int_\limits{\frac{1}{4}}^{\frac{1}{3}} 3\left(1-2 x^2\right) \mathrm{d} x

$$\begin{aligned} & =\left[3 x-2 x^3\right]_{\frac{1}{4}}^{\frac{1}{3}} \\ & =\left(1-\frac{2}{27}\right)-\left(\frac{3}{4}-\frac{1}{32}\right) \\ & =\frac{1}{4}+\frac{1}{32}-\frac{2}{27} \\ & =\frac{179}{864} \end{aligned}$$

$\therefore \quad[(p \rightarrow q) \wedge \sim q] \rightarrow r$ is a tautology when all the entries in the last column are $\mathrm{T}$, which is only possible when $r \equiv \sim q$

Lines $L_1$ and $L_2$ are parallel to the vectors $\bar{b}_1=3 \hat{i}+\hat{j}+2 \hat{k} \text { and } \bar{b}_2=\hat{i}+2 \hat{j}+3 \hat{k}$ respectively.

$\therefore \quad$ The unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is $\hat{\mathrm{n}}=\frac{\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2}{\left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|}$

Now, $$\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 1 & 2 \\ 1 & 2 & 3\end{array}\right|=-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$$

$\therefore \quad \hat{\mathrm{n}}=\frac{1}{5 \sqrt{3}}(-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$

$\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in H.P.

$\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}$ will be in A.P.

$$\begin{aligned} & \therefore \quad \frac{2}{\sin \theta}=\frac{1}{\sin (\theta-\alpha)}+\frac{1}{\sin (\theta+\alpha)} \\ & \Rightarrow \frac{2}{\sin \theta}=\frac{\sin (\theta+\alpha)+\sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)} \\ & \Rightarrow \frac{2}{\sin \theta}=\frac{2 \sin \theta \cos \alpha}{\sin ^2 \theta-\sin ^2 \alpha} \\ & \Rightarrow \sin ^2 \theta-\sin ^2 \alpha=\sin ^2 \theta \cos \alpha \\ & \Rightarrow \sin ^2 \theta(1-\cos \alpha)=\sin ^2 \alpha \\ & \Rightarrow \sin ^2 \theta\left(2 \sin ^2 \frac{\alpha}{2}\right)=4 \sin ^2 \frac{\alpha}{2} \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow 1-\cos ^2 \theta=2 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow \cos ^2 \theta=1-2 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow 2 \cos ^2 \theta-1=1-4 \cos ^2 \frac{\alpha}{2} \\ & \Rightarrow \cos ^2 \theta=1-4 \cos ^2 \frac{\alpha}{2} \end{aligned}$$

$$\begin{aligned} & \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=10 \\ & \Rightarrow|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \frac{\pi}{3}=10 \\ & \Rightarrow(\mathrm{s})|\overline{\mathrm{c}}|\left(\frac{1}{2}\right)=10 \\ & \Rightarrow|\overline{\mathrm{c}}|=4 \end{aligned}$$

$$\begin{aligned} & |\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})|=|\overline{\mathrm{a}}||\overline{\mathrm{b}} \times \overline{\mathrm{c}}| \sin \frac{\pi}{2} \\ & =|\overline{\mathrm{a}}||\overline{\mathrm{b}} \times \overline{\mathrm{c}}| \\ & =|\overline{\mathrm{a}}||\overline{\mathrm{b}}||\overline{\mathrm{c}}| \sin \frac{\pi}{3} \\ & =(\sqrt{3})(5)(4)\left(\frac{\sqrt{3}}{2}\right) \\ & =30 \end{aligned}$$

$$\begin{aligned} & f(x)=\int \frac{x^2-3 x+2}{x^4+1} \mathrm{~d} x \\ & \Rightarrow \mathrm{f}^{\prime}(x)=\frac{x^2-3 x+2}{x^4+1} \end{aligned}$$

For $\mathrm{f}(x)$ to be decreasing,

$$\begin{aligned} & \mathrm{f}^{\prime}(x)<0 \\ & \Rightarrow \frac{x^2-3 x+2}{x^4+1}<0 \\ & \Rightarrow \frac{(x-1)(x-2)}{x^4+1}<0 \\ & \Rightarrow(x-1)(x-2)<0 \\ & \Rightarrow x \in(1,2) \end{aligned}$$

A, B, C are in A.P.

$$\begin{array}{ll} \therefore \quad & \mathrm{A}+\mathrm{C}=2 \mathrm{~B} \\ & \mathrm{Also}, \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \\ \therefore \quad & \quad \mathrm{B}=60^{\circ} \end{array}$$

By sine rule,

$$\begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k \\ \therefore \quad & \sin A=a k, \sin B=b k, \sin C=c k \\ \therefore \quad & \frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A \end{aligned}$$

$$\begin{aligned} \therefore \quad & \frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A \\ & =\frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\ & =\frac{a}{c}(2 c k \cos C)+\frac{c}{a}(2 a k \cos A) \\ & =2 k a \cos C+2 k c \cos A \\ & =2 k(a \cos C+c \cos A) \end{aligned}$$

$=2 \mathrm{~kb} \quad \ldots[\because \mathrm{b}=\mathrm{a} \cos \mathrm{C}+\mathrm{c} \cos \mathrm{A}]$

$$\begin{aligned} & =2 \sin \mathrm{B} \\ & =2 \times \frac{\sqrt{3}}{2} \quad \ldots .\left[\because \angle \mathrm{B}=60^{\circ}\right] \\ & =\sqrt{3} \end{aligned}$$

Let's denote the original set of observations as $X = \{x_1, x_2, ..., x_{20}\}$. The variance of these observations ($Var(X)$) is given as 5. Variance is a measure of how spread out the values in a dataset are, and it is calculated as the average of the squared differences from the mean: