MHT CET 2023 9th May Morning Shift (Mathematics) Question Paper with Solutions

$x^2+y^2=t+\frac{1}{t}$

Squaring on both sides, we get

$x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2$

$$\begin{array}{ll} \therefore \quad \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 \quad ... [\because x^4+y^4=t^2+\frac{1}{t^2},\text{given}]\\ \\ 2 x^2 y^2=2 \\ x^2 y^2=1 \\ \text { differentiating w.r.t. } x, \text { we get } \\ x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y^2=0 \\ x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x y^2 \\ \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 x y^2}{2 x^2 y}=\frac{-y}{x} \end{array}$$

$$\begin{aligned} \text { Required area } & =2 \times \text { Area of } \triangle \mathrm{PBO} \\ & =2 \times \frac{1}{2} \times 2 \times 2 \sqrt{3} \\ & =4 \sqrt{3} \text { sq. units } \end{aligned}$$

The shaded region lies:

on non-origin side of line $3 x+4 y=12$ i.e., $3 x+4 y \geq 12$,

on the side of the line $y-x=0$, where $y \geq x$ i.e., $y-x \geq 0$,

on origin side of line $y=3$ i.e., $y \leq 3$,

and in first quadrant i.e., $x \geq 0, y \geq 0$.

\begin{aligned} & \text { Since } \sum_\limits{x=0}^n P(X=n)=1 \\ & \frac{{ }^n C_0+{ }^n C_1+{ }^n C_2 \ldots+\ldots+{ }^n C_n}{32}=1 \\ & 2^n=32 \\ & \therefore \quad n=5 \\ & P(X \leq 2)=P(X=0)+P(X=1)+P(X=2) \\ & =\frac{{ }^5 C_0}{32}+\frac{{ }^5 C_1}{32}+\frac{{ }^5 C_2}{32}=\frac{1}{2} \end{aligned}

$$\begin{aligned} \text { Let } f(x)=\log _{10} x=\frac{\log _e x}{\log _e 10} & =\left(\log _{10} e\right)\left(\log _e x\right) \\ & =0.4343\left(\log _e x\right) \end{aligned}$$

On differentiating w.r.t. $x$, we get

$$\begin{aligned} & \mathrm{f}^{\prime}(x)=\frac{0.4343}{x} \\ & \text { Let } x=998 \\ & =1000-2=\mathrm{a}+\mathrm{h} \\ & \therefore \quad \mathrm{a}=1000, \mathrm{~h}=-2 \\ & f(a)=f(1000) \\ & =\log _{10}(1000) \\ & =3 \log _{10} 10 \\ & \therefore \quad \mathrm{f}(\mathrm{a})=3 \\ & \text { Also, } \mathrm{f}^{\prime}(\mathrm{a})=\mathrm{f}^{\prime}(1000)=\frac{0.4343}{1000}=0.0004343 \\ & \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \\ & \therefore \quad \log _{10}(998) \approx 3-2(0.0004343) \\ & \approx 2.9991314 \\ \end{aligned}$$

$$\begin{array}{ll} & \left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0 \\ \therefore \quad & \left(1+y^2\right) \mathrm{d} x=x y \mathrm{~d} y \\ \therefore \quad & \frac{1}{x} \mathrm{~d} x=\frac{y \mathrm{~d} y}{1+y^2} \end{array}$$

Integrating both sides, we get

$$\begin{array}{ll} & \int \frac{1}{x} \mathrm{~d} x=\int \frac{y}{1+y^2} \mathrm{~d} y \\ & \log x=\frac{1}{2} \log \left(1+y^2\right)+\mathrm{c} \\ & \text { At } x=1, y=0 \quad \text{.... [Given]}\\ \therefore \quad & 0=\frac{1}{2} \log (1+0)+\mathrm{c} \\ \therefore \quad & \mathrm{c}=0 \\ \therefore \quad & \log x=\frac{1}{2} \log \left(1+y^2\right) \\ \therefore \quad & x^2=1+y^2 \\ \therefore \quad & x^2-y^2=1, \end{array}$$

Which is a rectangular hyperbola.

$$\begin{array}{ll} & \sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \pi-\cos ^{-1} x-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \cos ^{-1} x+\sin ^{-1} y=\pi-\frac{3 \pi}{10}=\frac{7 \pi}{10} \end{array}$$

Odd in favor of $\mathrm{A}$ is $4: 6$.

$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{4}{10}$

Odd against $\mathrm{B}$ is $7: 3$

$\therefore \quad \mathrm{P}(\mathrm{B})=\frac{3}{10}$

Since only one of the events A, B and C can happen, we get

$$\begin{array}{ll} & \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=1 \\ \therefore & \frac{4}{10}+\frac{3}{10}+\mathrm{P}(\mathrm{C})=1 \\ \therefore \quad & \mathrm{P}(\mathrm{C})=\frac{3}{10} \\ \therefore \quad & \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{7}{10} \end{array}$$

$\therefore \quad$ odds against the event $\mathrm{C}$ are $\mathrm{P}\left(\mathrm{C}^{\prime}\right): \mathrm{P}(\mathrm{C})$

$$\begin{aligned} & =\frac{7}{10}: \frac{3}{10} \\ & =7: 3 \end{aligned}$$

Volume of parallelopiped is $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]$

$$\begin{aligned} \therefore \quad V & =\left|\begin{array}{lll} 1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1 \end{array}\right| \\ & =1-\alpha\left(-\alpha^2\right)-\alpha \\ & =1+\alpha^3-\alpha \end{aligned}$$

Differentiating w.r.t. $\alpha$, we get

$$\begin{array}{ll} & \frac{d V}{d \alpha}=3 \alpha^2-1 \\ \therefore \quad & \frac{d^2 V}{d \alpha^2}=6 \alpha \\ & \text { Let } \frac{d V}{d \alpha}=0 \\ \therefore \quad & 3 \alpha^2-1=0 \\ \therefore \quad & \alpha= \pm \frac{1}{\sqrt{3}} \\ & \text { at } \alpha=\frac{-1}{\sqrt{3}}, \\ & \frac{d^2 V}{d \alpha^2}=\frac{-6}{\sqrt{3}}<0 \end{array}$$

$\therefore \quad \mathrm{V}$ is maximum at $\alpha=\frac{-1}{\sqrt{3}}$

Let $a=\sqrt{3}+1, b=\sqrt{3}-1, C=60^{\circ}$

Using cosine Rule,

$$\begin{aligned} & c^2=a^2+b^2-2 a b \cos C \\ & c^2=(\sqrt{3}+1)^2+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}-1) \frac{1}{2} \\ & c^2=6 \end{aligned}$$

$$\begin{aligned} \therefore \quad c^2 & =6 \\ c & =\sqrt{6} \end{aligned}$$

Using sine Rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Consider

$$\begin{array}{ll} & \frac{b}{\sin B}=\frac{c}{\sin C} \\ \therefore & \frac{\sqrt{3}-1}{\sin B}=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}} \\ \therefore & \sin B=\frac{\sqrt{3}-1}{2 \sqrt{2}} \\ \therefore & \angle B=15^{\circ} \\ \therefore & \angle A=105^{\circ} \end{array}$$

.... [Remaining angle of a Triangle]

$\therefore \quad \text { Difference }=90^{\circ}$

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

$$\begin{aligned} & \text { Put } \cos x=\mathrm{t} \\ & -\sin x \mathrm{~d} x=\mathrm{dt} \\ & I=\int \frac{t^2-2}{2 t^2-1} d t \\ & =\frac{1}{2} \int \frac{2 t^2-4}{2 t^2-1} d t \\ & =\frac{1}{2}\left[\int \frac{2 t^2-1}{2 t^2-1} d t-\int \frac{3}{2 t^2-1} d t\right] \\ & I=\frac{1}{2} t-\frac{1}{2} \times \frac{3}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} t-1}{\sqrt{2} t+1}\right|+c \\ & =\frac{1}{2} \cos x-\frac{3}{4 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+c \\ & \therefore \quad P=\frac{1}{2} \text { and } Q=\frac{-3}{4 \sqrt{2}} \\ \end{aligned}$$

Given Cartesian equation of the line is

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

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

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

$\therefore \quad$ Vector equation is

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

Three fair coins numbered 1,0 are tossed.̌/p>

$$\begin{array}{ll} \therefore \quad & \text { Sample space }=\{111,110,101,011,100,010, \\ & 001,000\} \\ \therefore \quad & n(S)=8 \end{array}$$

$\mathrm{X}$ represents the sum of numbers on upper most face

$$\begin{array}{r} \mathrm{P}(\mathrm{X}=0)=\frac{1}{8} \\ \mathrm{P}(\mathrm{X}=1)=\frac{3}{8} \\ \mathrm{P}(\mathrm{X}=2)=\frac{3}{8}, \\ \mathrm{P}(\mathrm{X}=3)=\frac{1}{8} \end{array}$$

$\therefore \quad$ Probability distribution of $\mathrm{X}$ is

$$\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum_{x=0}^3 x_{\mathrm{i}} \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & \therefore \quad \mathrm{E}(\mathrm{X})= 0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8} \\ &= \frac{12}{8}=\frac{3}{2} \\ & \mathrm{E}\left(\mathrm{X}^2\right)=\sum_{x=0}^3 x_{\mathrm{i}}^2 \mathrm{P}\left(x_{\mathrm{i}}\right)=3 \\ & \text { Variance of } \mathrm{X}=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ &=3-\left(\frac{3}{2}\right)^2 \\ &=3-\frac{9}{4} \\ &=\frac{3}{4}=0.75 \end{aligned}$$

$$\begin{aligned} & \text { Let } \mathrm{I}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\sin x(\pi-2 x)^3} \\ & \text { Put } x=\frac{\pi}{2}-\mathrm{h} \\ & \therefore \quad \pi-2 x=2 \mathrm{~h} \\ & \text { As } x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0 \end{aligned}$$

$$\begin{aligned} \therefore \quad I & =\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}-h\right)\left(1-\sin \left(\frac{\pi}{2}-h\right)\right)}{\sin \left(\frac{\pi}{2}-h\right)(2 h)^3} \\ & =\lim _{h \rightarrow 0} \frac{\sinh \cdot 2 \sin ^2\left(\frac{h}{2}\right)}{\cosh \cdot 8 h^3} \\ & =\frac{2}{8} \lim _{h \rightarrow 0} \frac{\sinh }{h} \times \lim _{h \rightarrow 0} \frac{\sin ^2 \frac{h}{2}}{\frac{h^2}{4} \cdot 4} \\ & =\frac{2}{8}(1) \times(1) \times \frac{1}{4} \\ & =\frac{1}{16} \end{aligned}$$

$$\begin{array}{ll} \quad y=\sqrt{x^2+16} \\ \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 x}{2 \sqrt{x^2+16}} \\ \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{\sqrt{x^2+16}} \quad \text{..... (i)}\\ \text { Let } \mathrm{z}=\frac{x}{x-1} \\ \therefore \quad \frac{\mathrm{dz}}{\mathrm{d} x}=\frac{(x-1)-x}{(x-1)^2} \\ \therefore \quad \frac{\mathrm{dz}}{\mathrm{d} x}=\frac{-1}{(x-1)^2} \quad \text{..... (ii)}\\ \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{dz}}\right)_{x=5}=\frac{\frac{x}{\sqrt{x^2+16}}}{\frac{-1}{(x-1)^2}} \\ \quad=\frac{-5}{\sqrt{25+16}} \times 16=\frac{-80}{\sqrt{41}} \end{array}$$

$$\begin{aligned} & x+2 y=8 \\ \therefore \quad & 2 y=8-x \\ \therefore \quad & y=\frac{8-x}{2} \end{aligned}$$

Let $\mathrm{f}(x)=x y$

$\therefore \quad \mathrm{f}(x)=x \cdot \frac{(8-x)}{2}$

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

$$\begin{aligned} & f^{\prime}(x)=\frac{(8-x)-x}{2} \\ & f^{\prime}(x)=4-x \end{aligned}$$

To find critical points,

$$\begin{aligned} & \mathrm{f}^{\prime}(x)=0 \\ \therefore \quad & 4-x=0 \\ \therefore \quad & x=4 \end{aligned}$$

critical point at $x=4$

$\therefore \quad \mathrm{f}(4)=\frac{4(8-4)}{2}=8$

$\therefore \quad$ Maximum value of the given function is 8.

$$\begin{aligned} & \mathrm{f}(x)=\sqrt{25-x^2} \\ & \therefore \quad \mathrm{f}^{\prime}(x=\mathrm{c})=\frac{-2 \mathrm{c}}{2 \sqrt{25-\mathrm{c}^2}} \\ &=\frac{-\mathrm{c}}{\sqrt{25-\mathrm{c}^2}} \end{aligned}$$

Applying Lagrange's mean value theorem, we get

$$\begin{array}{ll} & \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(1)-\mathrm{f}(5)}{1-5} \\ \therefore \quad & \frac{\mathrm{c}}{\sqrt{25-c^2}}=\frac{\sqrt{25-1}-\sqrt{25-5^2}}{1-5} \\ \therefore \quad & \frac{-c}{\sqrt{25-c^2}}=\frac{-\sqrt{24}}{4} \end{array}$$

$$\begin{array}{ll} \therefore & 4 c=\sqrt{24} \cdot \sqrt{25-c^2} \\ \therefore & 16 c^2=24\left(25-c^2\right) \\ \therefore & c^2=15 \\ \therefore & c= \pm \sqrt{15} \\ & \text { Since } c=-\sqrt{15} \text { does not lie in }[1,5] \\ & c=\sqrt{15} \end{array}$$

Let $\overrightarrow{\mathrm{n}}$ be inclined at angles $\alpha, \beta, \gamma$ to $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ axes respectively.

$$\begin{array}{ll} & \alpha=45^{\circ}, \beta=60^{\circ}, \gamma=? \\ & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ \therefore \quad & \frac{1}{2}+\frac{1}{4}+\cos ^2 \gamma=1 \\ \therefore \quad & \cos ^2 \gamma=\frac{1}{4} \\ \therefore \quad & \gamma=60^{\circ} \\ & \vec{n}=\cos \alpha \hat{i}+\cos \beta \hat{j}+\cos \gamma \hat{k} \\ \therefore \quad & \vec{n}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k} \end{array}$$

$\therefore \quad$ Equation of the required plane is

$\frac{1}{\sqrt{2}}(x+\sqrt{2})+\frac{1}{2}(y-1)+\frac{1}{2}(z-1)=0$

i.e., $\sqrt{2} x+y+z=0$

$$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=\int \frac{\mathrm{e}^{\prime}(x+1)}{\mathrm{e}^x \cdot x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x \\ & \text { Let } x \cdot \mathrm{e}^x=\mathrm{t} \\ & \Rightarrow(x+1) \mathrm{e}^x \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad I=\int \frac{d t}{t(1+t)^2} \\ & =\int \frac{1+t-t}{t(1+t)^2} d t \\ & =\int \frac{1 d t}{t(1+t)}-\int \frac{1}{(1+t)^2} d t \\ & =\int \frac{1+t-t}{t(1+t)}-\int \frac{1}{(1+t)^2} d t \\ & =\int \frac{1}{t} d t-\int \frac{1}{(t+1)} d t-\int \frac{1}{(1+t)^2} d t \\ & =\log \mathrm{t}-\int \frac{\mathrm{dt}}{(\mathrm{t}+1)}-\int \frac{1}{(1+\mathrm{t})^2} d \mathrm{dt}+\mathrm{c} \\ & \text { Let } y=1+\mathrm{t} \\ & \Rightarrow \mathrm{d} y=\mathrm{dt} \\ & \therefore \quad \mathrm{I}=\log \mathrm{t}-\int \frac{1}{y} \mathrm{~d} y-\int \frac{1}{y^2} \mathrm{~d} y+\mathrm{c} \\ & =\log \mathrm{t}-\log y+\frac{1}{y}+\mathrm{c} \\ & =\log t-\log (1+t)+\frac{1}{(1+t)}+c \\ & =\log x \mathrm{e}^x-\log \left(1+x \mathrm{e}^x\right)+\frac{1}{\left(1+x \mathrm{e}^x\right)}+c \\ & \therefore \quad \mathrm{I}=\log \left|\frac{x \mathrm{e}^x}{1+x \mathrm{e}^x}\right|+\frac{1}{1+x \mathrm{e}^x}+\mathrm{c} \\ \end{aligned}$$

As $\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$

$\frac{\mathrm{f}^{\prime}(x)}{\mathrm{f}(x)}=1$

Integrating on both sides, we get

$\log \mathrm{f}(x)=x+\mathrm{c}$ ..... (i)

As $\mathrm{f}(0)=1$

$$\begin{array}{ll} \therefore & \text { (i) } \Rightarrow \mathrm{c}=0 \\ \therefore & \log \mathrm{f}(x)=x \\ \therefore & \mathrm{f}(x)=\mathrm{e}^x \end{array}$$

As $\mathrm{f}(x)+\mathrm{g}(x)=x^2$

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

\begin{aligned} \therefore \quad \mathrm{f}(x) \mathrm{g}(x) & =\mathrm{e}^x\left(x^2-\mathrm{e}^x\right) \\ & =\int_\limits0^1\left(\mathrm{e}^x x^2-\mathrm{e}^{2 x}\right) \mathrm{d} x \\ & =\left[\left(x^2-2 x+2\right) \mathrm{e}^x\right]_0^1-\frac{1}{2} \mathrm{e}^2+\frac{1}{2} \\ & =\mathrm{e}-\frac{1}{2} \mathrm{e}^2-\frac{3}{2} \end{aligned}

Substitute $x=4, y=-2, \mathrm{z}=5$ in all options.

$\therefore \quad$ For option $\mathrm{A}$

$$\begin{aligned} 4 x-2 y+5 z & =4(4)-2(-2)+5(5) \\ & =16+4+25 \\ & =45 \end{aligned}$$