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

$$\begin{aligned} & P=\operatorname{adj} A \quad\text{.... [Given]}\\ \therefore \quad & |P|=|A|^2 \quad.... [\because |\mathrm{adjA|=|A|^{n-1}}]\\ & \Rightarrow\left|\begin{array}{lll} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}\right|=4^2 \\ & \Rightarrow 2 \alpha-6=16 \\ & \Rightarrow \alpha=11 \end{aligned}$$

$$\begin{aligned} & \text { 9. } p(X=4)=p(X=2) \text { and } n=6 \\ & \Rightarrow 9 \times{ }^6 C_4 p^4 q^2={ }^6 C_2 p^2 q^4 \\ & \Rightarrow 9 p^2=q^2 \\ & \Rightarrow 3 p=q \quad ...[\because 0 < p, q < 1]\\ & \Rightarrow 3 p=1-p \\ & \Rightarrow p=\frac{1}{4} \end{aligned}$$

$$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2} \\ & \Rightarrow \frac{\mathrm{d} y}{1+y^2}=\frac{\mathrm{d} x}{1+x^2} \end{aligned}$$

Integrating on both sides, we get

$$\begin{aligned} & \int \frac{1}{1+y^2} d y-\int \frac{1}{1+x^2} d x=\tan ^{-1} c \\ & \Rightarrow \tan ^{-1} y-\tan ^{-1} x=\tan ^{-1} c \\ & \Rightarrow \tan ^{-1}\left(\frac{y-x}{1+x y}\right)=\tan ^{-1} c \\ & \Rightarrow \frac{y-x}{1+x y}=c \\ & \Rightarrow y-x=c(1+x y) \end{aligned}$$

Let $I=\int\limits_0^\pi \frac{d x}{4+3 \cos x}$

Put $\tan \frac{x}{2}=\mathrm{t}$

$\therefore \quad \mathrm{d} x=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2}$ and $\cos x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}$

\begin{aligned} \therefore \quad \mathrm{I}=\int_\limits0^\pi \frac{\mathrm{d} x}{4+3 \cos x} & =\int_\limits0^{\infty} \frac{2 \mathrm{dt}}{7+\mathrm{t}^2} \\ & =\left[\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}\right)\right]_0^{\infty} \\ & =\frac{2}{\sqrt{7}}\left[\tan ^{-1} \infty-0\right] \\ & =\frac{2}{\sqrt{7}} \cdot \frac{\pi}{2}=\frac{\pi}{\sqrt{7}} \end{aligned}

$$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{x \sqrt{1-x^3}} \\ & \qquad \begin{aligned} & =\int \frac{x^2 \mathrm{~d} x}{x^3 \sqrt{1-x^3}} \mathrm{~d} x \end{aligned} \\ & \begin{array}{l} \text { Put } 1-x^3=\mathrm{t}^2 \\ \Rightarrow-3 x^2 \mathrm{~d} x=2 \mathrm{tdt} \\ \Rightarrow x^2 \mathrm{~d} x=\frac{-2 \mathrm{tdt}}{3} \end{array} \end{aligned}$$

$$\begin{aligned} \therefore \quad \mathrm{I} & =\int \frac{-\left(\frac{2 \mathrm{tdt}}{3}\right)}{\left(1-\mathrm{t}^2\right) \mathrm{t}} \\ & =-\frac{2}{3} \int \frac{1}{1-\mathrm{t}^2} \mathrm{dt} \\ & =\frac{2}{3} \int \frac{1}{\mathrm{t}^2-1} \mathrm{dt} \\ & =\frac{2}{3} \times \frac{1}{2} \cdot \log \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{c} \\ & =\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+\mathrm{c} \\ \therefore \quad \mathrm{k} & =\frac{1}{3} \end{aligned}$$

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

$$\begin{array}{ll} & x^2+y^2-6 x-14 y+48=0 \\ \therefore \quad & \mathrm{C}_1(3,7), \mathrm{r}_1=\sqrt{10} \\ & \text { Again } x^2+y^2-6 x=0 \\ \therefore \quad & \mathrm{C}_2(3,0), \mathrm{r}_2=3 \end{array}$$

Now $l\left(\mathrm{C}_1 \mathrm{C}_2\right)=$ distance between centres

$\therefore \quad l\left(\mathrm{C}_1 \mathrm{C}_2\right)=\sqrt{0^2+7^2}=7 \text { and }$

$\qquad\mathrm{r}_1+\mathrm{r}_2=\sqrt{10}+3 < l\left(\mathrm{C}_1 \mathrm{C}_2\right)$

$\Rightarrow$ The given circles are disjoint.

$\Rightarrow$ Number of common tangents is 4 .

$$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \quad \text{.... (i)}\\ & \text { Put } y=\mathrm{v} x \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x} \\ \therefore \quad & \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x}=\mathrm{v}+\sec \mathrm{v} \quad \text{... [From (i)]}\\ & \Rightarrow \cos \mathrm{v} d \mathrm{v}=\frac{1}{x} \mathrm{~d} x \end{aligned}$$

Integrating both sides, we get

$$\begin{aligned} & \sin \mathrm{v}=\log (x)+\mathrm{c} \\ & \Rightarrow \sin \left(\frac{y}{x}\right)=\log (x)+\mathrm{c} \quad \text{.... (ii)} \end{aligned}$$

The curve passes through $\left(1, \frac{\pi}{6}\right)$.

$\sin \left(\frac{\pi}{6}\right)=\log (1)+c \Rightarrow c=\frac{1}{2}$

$\therefore \quad$ Equation (ii) becomes

$\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$

$$\begin{aligned} & \text { Volume of parallelopiped }=\left|\begin{array}{ccc} 1 & 1 & n \\ 2 & 4 & -n \\ 1 & n & 3 \end{array}\right|=158 \\ & \Rightarrow 1\left(12+n^2\right)-1(6+n)+n(2 n-4)=158 \\ & \Rightarrow 3 n^2-5 n-152=0 \\ & \Rightarrow(3 n+19)(n-8)=0 \\ & \Rightarrow n=8 \quad \ldots[\because n \geq 0] \end{aligned}$$

$y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$

Taking $\log$ on both sides, we get

$\log y=\frac{1}{2}\left[\log \left(1-\sin ^{-1} x\right)-\log \left(1+\sin ^{-1} x\right)\right]$

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

$$\begin{aligned} & \frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2}\left[\frac{1}{1-\sin ^{-1} x} \cdot \frac{-1}{\sqrt{1-x^2}}-\frac{1}{1+\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}\right] \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-y}{2 \sqrt{1-x^2}}\left(\frac{1}{1-\sin ^{-1} x}+\frac{1}{1+\sin ^{-1} x}\right) \\ & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(0,1)}=\frac{-1}{2(1)}\left(\frac{1}{1-0}+\frac{1}{1+0}\right)=-1 \end{aligned}$$

Shaded region lies on non-origin side of $2 x+y=2$ and on origin side of the lines $x-y=1$ and $x+2 y=8$

$\therefore \quad 2 x+y \geq 2, x-y \leq 1, x+2 y \leq 8$

$$\begin{aligned} & f(x)=[x] \sin (\pi x) \\ & \begin{aligned} \text { LHD } & =\lim _{h \rightarrow 0} \frac{f(k-h)-f(k)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{[k-h] \sin \pi(k-h)-[k] \sin k \pi}{-h} \\ & =\lim _{h \rightarrow 0} \frac{(k-1) \sin (k \pi-\pi h)-k \sin k \pi}{-h} \quad \ldots[\because k \in I]\\ & =\lim _{h \rightarrow 0} \frac{(-1)^{k+1}(k-1) \sinh \pi-0}{-h} \quad \ldots[\because \sin k\pi=0,k \in I] \\ & =(-1)^k(k-1) \pi \end{aligned} \end{aligned}$$

The equation of line is

$$\begin{aligned} & 2 x+4=3 y+1=6 z-3 \\ & \Rightarrow 2(x+2)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{2}\right) \\ & \Rightarrow \frac{x+2}{\frac{1}{2}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-\frac{1}{2}}{\frac{1}{6}} \\ & \Rightarrow \frac{x+2}{3}=\frac{y+\frac{1}{3}}{2}=\frac{z-\frac{1}{2}}{1} \end{aligned}$$

$\therefore \quad$ The given line passes through $\left(-2,-\frac{1}{3}, \frac{1}{2}\right)$ and has direction ratios proportional to $3,2,1$.

$\therefore \quad$ Vector equation of the line is

$\bar{r}=\left(-2 \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{2} \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})$

Let $$A=\left[\begin{array}{ccc}1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1\end{array}\right]$$

For non-singular matrix $|A| \neq 0$

$$\Rightarrow\left|\begin{array}{ccc} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{array}\right| \neq 0$$

$$\begin{aligned} & \Rightarrow 1(1-\omega c)-a\left(\omega-\omega^2 c\right)+b(0) \neq 0 \\ & \Rightarrow 1(1-\omega c)-a \omega(1-\omega c) \neq 0 \\ & \Rightarrow(1-\omega c)(1-a \omega) \neq 0 \\ & \Rightarrow c \neq \frac{1}{\omega} \text { and } a \neq \frac{1}{\omega} \\ & \Rightarrow c \neq \omega^2 \text { and } a \neq \omega^2 \quad \ldots\left[\because \omega^3=1\right] \end{aligned}$$

So possible value of $a$ and $c$ is $\omega$ only and $b$ can take values $\omega$ or $\omega^2$.

$\therefore \quad$ The possible number of distinct matrices = 2.

Equation of the plane passing through $(1,1,1)$ is given as

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

As the plane is parallel to the lines having direction ratios $1,0,-1$ and $-1,1,0$, we get

$\mathrm{a}-\mathrm{c}=0 \text { and }-\mathrm{a}+\mathrm{b}=0$

$\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}$ .... (ii)

$\therefore \quad$ From (i) and (ii), we get

$$\begin{aligned} & x \quad x-1+y-1+z-1=0 \\ & \therefore \quad x+y+z=3 \quad \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1 \end{aligned}$$

$\therefore \quad$ Co-ordinates of A, B, C are $(3,0,0),(0,3,0)$ and $(0,0,3)$ respectively.

$$\begin{aligned} \therefore \quad \text { Volume of tetrahedron } \mathrm{OABC} & =\frac{1}{6}\left|\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right| \\ & =\frac{1}{6} \times 27 \\ & =\frac{9}{2} \text { cu. units } \end{aligned}$$

$$\begin{array}{ll} & \text { Let } \tan ^{-1} 2=\alpha \Rightarrow \tan \alpha=2 \\ & \text { And } \cot ^{-1} 3=\beta \Rightarrow \cot \beta=3 \\ \therefore \quad & \sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right) \\ & =\sec ^2 \alpha+\operatorname{cosec}^2 \beta \\ & =1+\tan ^2 \alpha+1+\cot ^2 \beta \\ & =2+(2)^2+(3)^2 \\ & =15 \end{array}$$

\begin{aligned} \text { Let } \quad \mathrm{I} & =\int_\limits{-1}^3\left(\cot ^{-1}\left(\frac{x}{x^2+1}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \\ = & \int_\limits{-1}^3\left(\tan ^{-1}\left(\frac{x^2+1}{x}\right)+\cot ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x \\ & \ldots\left[\because \cot ^{-1}(x)=\tan ^{-1}\left(\frac{1}{x}\right)\right] \end{aligned}

\begin{aligned} & =\int_\limits{-1}^3 \frac{\pi}{2} \mathrm{~d} x \quad \ldots\left[\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \\ & =\frac{\pi}{2}[x]_{-1}^3 \\ & =\frac{\pi}{2}(4) \\ \therefore \quad \mathrm{I} & =2 \pi \end{aligned}

The number of triangles that can be drawn using 6 vertices is given by

$\mathrm{n}(\mathrm{S})={ }^6 \mathrm{C}_3=20$

$\mathrm{A}$ : Event of selecting equilateral triangle. The equilateral triangle can be drawn if selected three vertices are alternate.

$$\begin{array}{ll} \therefore & \mathrm{n}(\mathrm{A})=2 \\ \therefore & \mathrm{P}(\mathrm{A})=\frac{2}{20}=\frac{1}{10} \end{array}$$

\begin{aligned} & \mathrm{I}_1=\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x \text { and } \quad ... \text{(i)}\\ & \mathrm{I}_2=\int_\limits{1-h}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x \quad ... \text{(ii)}\\ & \mathrm{I}_1=\int_\limits{1-h}^{\mathrm{h}}(1-x) \mathrm{f}[(1-x)(1-1+x)] \mathrm{d} x \\ & \ldots\left[\because \int_{\mathrm{a}}^b \mathrm{f}(x) \mathrm{d} x=\int_\limits{\mathrm{a}}^b \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d} x\right] \end{aligned}

\begin{aligned} \therefore \quad \mathrm{I}_1 & =\int_\limits{1-\mathrm{h}}^{\mathrm{h}}(1-x) \mathrm{f}(x(1-x)) \mathrm{d} x \\ & =\int_\limits{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x-\int_\limits{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x \\ & \Rightarrow \mathrm{I}_1=\mathrm{I}_2-\mathrm{I}_1 \quad \text{....[From (i) and (ii)]}\\ & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2 \\ & \Rightarrow \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{1}{2} \end{aligned}

Since $\mathrm{f}(x)$ is continuous in $[-1,1]$, it is continuous at $x=0$.

$$\begin{aligned} & \therefore \quad \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2} \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{(1+\mathrm{m} x-1+\mathrm{m} x)}{x(\sqrt{1+\mathrm{m} x}+\sqrt{1-\mathrm{m} x})}=\frac{2(0)+1}{0-2} \\ & \Rightarrow \frac{2 \mathrm{~m}}{1+1}=\frac{1}{-2} \\ & \Rightarrow \mathrm{m}=\frac{-1}{2} \end{aligned}$$

$\alpha=3 \sin ^{-1} \frac{6}{11} \text { and } \beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$

Since $\frac{6}{11}>\frac{6}{12}$

Taking $\sin ^{-1}$ on both sides, we get

$\sin ^{-1}\left(\frac{6}{11}\right)>\sin ^{-1}\left(\frac{6}{12}\right)\quad$ ..... $\left[\because \sin ^{-1} x\right.$ is an increasing function $]$

$$\begin{aligned} & \Rightarrow 3 \sin ^{-1}\left(\frac{6}{11}\right)>3 \sin ^{-1}\left(\frac{1}{2}\right) \\ & \Rightarrow \alpha>3\left(\frac{\pi}{6}\right) \\ & \Rightarrow \alpha>\frac{\pi}{2} \quad \text{.... (i)} \end{aligned}$$

Now, $\frac{4}{9}<\frac{4}{8}$

Taking $\cos ^{-1}$ on both sides, we get

$$\begin{aligned} & \cos ^{-1}\left(\frac{4}{9}\right)>\cos ^{-1}\left(\frac{4}{8}\right) \\ & \ldots\left[\because \cos ^{-1} x \text { is a decreasing function }\right]\\ & \Rightarrow 3 \cos ^{-1}\left(\frac{4}{9}\right)>3 \cos ^{-1}\left(\frac{1}{2}\right) \\ & \Rightarrow \beta>3\left(\frac{\pi}{3}\right) \\ & \Rightarrow \beta>\pi \quad \text{.... (ii)} \end{aligned}$$

From (i) and (ii), we have $\alpha$ lies in II$^{\text {nd }}$ quadrant and $\beta$ lies in III$^{\text {rd }}$ quadrant.

$$\begin{array}{ll} \therefore \quad & \cos \alpha<0, \cos \beta<0 \text { and } \sin \beta<0 \\ & \text { Also, } \alpha+\beta>\frac{\pi}{2}+\pi \quad \ldots[\text { From (i) and (ii) }] \\ \therefore \quad & \alpha+\beta>\frac{3 \pi}{2} \end{array}$$

Thus, $\alpha+\beta$ lies in IV${ }^{\text {th }}$ quadrant.

So, $\cos (\alpha+\beta)>0$

[Note: Options (B), (C) and (D) are correct.]

Since, $\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}$

Putting $\frac{\mathrm{A}}{2}=\frac{\pi}{8}$, we get

$$\begin{aligned} \tan \left(\frac{\pi}{8}\right) & =\frac{1-\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} \\ & =\frac{1-\frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)} \\ \therefore \quad \tan \left(\frac{\pi}{8}\right) & =\sqrt{2}-1 \end{aligned}$$

$\mathrm{M}$ is the midpoint.

$\therefore \quad \mathrm{M} \equiv\left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right)$

D.r.s of $A B$ are $\frac{-10}{3}, \frac{-10}{3}, \frac{-10}{3}$ i.e., $1,1,1$

Equation of plane is

$$\begin{aligned} & 1\left(x+\frac{2}{3}\right)+1\left(y-\frac{1}{3}\right)+1\left(z-\frac{4}{3}\right)=0 \\ & \Rightarrow x+y+z=1 \end{aligned}$$

Option (A) satisfies this equation of the plane.

$$\begin{aligned} & p \rightarrow \sim(p \wedge \sim q) \\ & \equiv \sim p \vee \sim(p \wedge \sim q) \quad\ldots[\because p \rightarrow q \equiv \sim p \vee q]\\ & \equiv \sim p \vee(\sim p \vee q) \quad \text{...[De Morgan's law]}\\ & \equiv(\sim p \vee \sim p) \vee q \quad \text{...[Associative law]}\\ & \equiv \sim p \vee q \qquad \text{...[Idempotent law]} \end{aligned}$$

Let $y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$ and $\mathrm{z}=\cos ^{-1}\left(x^2\right)$

Put $x^2=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2$

$$\begin{aligned} & \therefore \quad y=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right) \\ & \Rightarrow y=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right) \\ & \Rightarrow y=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\ & \Rightarrow y=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right) \\ & \Rightarrow y=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x^2 \\ & \Rightarrow y=\frac{\pi}{4}-\frac{1}{2} z \\ & \therefore \quad \frac{\mathrm{dy}}{\mathrm{dz}}=-\frac{1}{2} \end{aligned}$$