MHT CET 2021 20th September Evening Shift (Mathematics) Question Paper with Solutions

We have $\frac{\mathrm{dA}}{\mathrm{dt}}=2$ and $\mathrm{A}=4 \pi \mathrm{r}^2$

$$\begin{aligned} & \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=8 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow 2=8 \pi(6) \frac{\mathrm{dr}}{\mathrm{dt}} \\ & \therefore \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{24 \pi}\end{aligned}$$

Here $V=\frac{4}{3} \pi r^3$

$$\begin{aligned} \therefore \frac{\mathrm{dV}}{\mathrm{dt}} & =4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \\ & =4 \pi(6)^2\left(\frac{1}{24 \pi}\right)=6 \mathrm{~cm}^3 / \mathrm{sec} \end{aligned}$$

$$\begin{aligned} & P(X=x)=\frac{1}{32}{ }^5 C_x \text {, where } x=0,1,2,3,4,5 \\ & =0 \text {, otherwise } \\ & \therefore \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ & =\frac{1}{32}\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2\right]=\frac{1}{32}(1+5+10)=\frac{16}{32}=\frac{1}{2} \\ & \end{aligned}$$

$\mathrm{P(X\ge3)=P(X=3)+P(X=4)+P(X=5)}$

$= {1 \over {32}}\left[ {{}^5{C_3} + {}^5{C_4} + {}^5{C_5}} \right] = {1 \over {31}}[10 + 5 + 1] = {{16} \over {32}} = {1 \over 2}$

$$\begin{aligned} & \sin ^{-1}\left[\sin (-600)^{\circ}\right]+\cot ^{-1}(-\sqrt{3}) \\ & =\sin ^{-1}\left[-\sin \left(360^{\circ}+180^{\circ}+60^{\circ}\right)\right]+\left[-\cot ^{-1}(\sqrt{3})\right] \\ & =\sin ^{-1}\left[-\sin \left(3 \pi+60^{\circ}\right)\right]+\left[-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right] \\ & =\sin ^{-1}\left[\sin 60^{\circ}\right]-\frac{\pi}{6}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\end{aligned}$$

$$\begin{aligned} & y=\frac{d p}{d x}+\sqrt{a^2 p^2-b^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)+\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2} \\ & \therefore y-\frac{d^2 y}{d x^2}=\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2} \end{aligned}$$

Squaring both sides, we write

$$\begin{aligned} & y^2+\left(\frac{d^2 y}{d x}\right)^2-2 y \frac{d^2 y}{d x^2}=a^2\left(\frac{d y}{d x}\right)^2-b^2 \\ & \therefore \text { order }=2 \text { and degree }=2 \end{aligned}$$

$$\begin{aligned} & x=a(t+\sin t) \quad \text { and } y=a(1-\cos t) \\ & \therefore \frac{d x}{d t}=a(1+\cos t) \quad \text { and } \quad \frac{d y}{d t}=a(\sin t) \\ & \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^2 \frac{t}{2}}=\tan \frac{t}{2}\end{aligned}$$

Let $\triangle \mathrm{OAB}$ be the required triangle. Since $\triangle \mathrm{OAB}$ is an equilateral triangle.

Slope of line $\mathrm{OA}=\tan 60^{\circ}=\sqrt{3}$ and Slope of line $\mathrm{OB}=\tan 120^{\circ}=-\sqrt{3}$

$\therefore$ Equation of $\mathrm{OA}$ is $y=\sqrt{3} x$ i.e. $\sqrt{3} x-y=0$ and equation of OB is $y=-\sqrt{3} x$ i.e. $\sqrt{3} x+y=0$

Hence required joint equation is

$(\sqrt{3} x-y)(\sqrt{3} x+y)=0 \text { i.e. } 3 x^2-y^2=0$

We have $\frac{\mathrm{dP}}{\mathrm{dt}} \propto \mathrm{P} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP}$

$\therefore \int \frac{\mathrm{dP}}{\mathrm{P}}=\int \mathrm{kdt} \Rightarrow \log \mathrm{P}=\mathrm{kt}+\mathrm{c}$

From given data, we write

$$\begin{aligned} & \log 20=\mathrm{k}(0)+\mathrm{c} \Rightarrow \mathrm{c}=\log 20 \\ & \therefore \log \mathrm{P}=\mathrm{kt}+\log 20 \end{aligned}$$

Also $\log 40=30 \mathrm{k}+\log 20$

$$\begin{aligned} & \therefore \log 40-\log 20=30 \mathrm{k} \Rightarrow \mathrm{k}=\frac{1}{30} \log 2 \\ & \therefore \log \mathrm{P}=\left(\frac{\log 2}{30}\right) \mathrm{t}+\log 20 \end{aligned}$$

When $t=30+15=45$,

$$\begin{aligned} \therefore \log P & =\left(\frac{\log 2}{30}\right)(45)+\log 20=(\log 2)\left(\frac{3}{2}\right)+\log 20 \\ & =\log (2)^{\frac{3}{2}}+\log 20=\log (2 \sqrt{2} \times 20) \\ \therefore \quad P & =40 \sqrt{2} \text { lakhs } \end{aligned}$$

$$\begin{aligned} & \text { Let } I=\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x \\ & =\int \frac{d x}{\sin (x-a) \sin x} \\ & =\int \frac{\sin a}{\sin a \sin (x-a) \sin x} d x \\ & =\frac{1}{\sin a} \int \frac{\sin (a+x-x)}{\sin (x-a) \sin x} d x \\ & =\frac{1}{\sin a} \int \frac{\sin -[x-a-x]}{\sin (x-a) \sin x} d x \\ & =\frac{1}{\sin a} \int \frac{\sin -[(x-a)-(x)]}{\sin (x-a) \sin x} d x \quad=\frac{-1}{\sin a} \int \frac{\sin [(x-a)-x]}{\sin (x-a) \sin x} d x \\ & =\frac{-1}{\sin a} \int \frac{\sin (x-a) \cos x-\cos (x-a) \sin x}{\sin (x-a) \sin x} d x \\ & =\frac{-1}{\sin a} \int[\cot x-\cot (x-a)] d x \\ & =\frac{-1}{\sin a}[\log |\sin x|-\log |\sin (x-a)|]+c \\ & =\frac{-1}{\sin a}\left[\int \cot x d x-\int \cot (x-a) d x\right] \\ & =\frac{1}{\sin a}[\log |\sin (x-a)|-\log |\sin x|]+c \\ & =(\operatorname{cosec} a)\left[\log \left|\frac{\sin (x-a)}{\sin x}\right|\right]+c \\ & =\operatorname{cosec} a \cdot \log |\sin (x-a) \cdot \operatorname{cosec} x|+c \\ & \end{aligned}$$

The equation of the required plane is $(x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0$ i.e. ...... (1)

$(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0$

Since (1) is perpendicular to the plane $5 x+3 y-6 z+8=0$, we write

$$\begin{aligned} & (1+2 \lambda)(5)+(2+\lambda)(3)+(3-\lambda)(-6)=0 \\ & \therefore 5+10 \lambda+6+3 \lambda-18+6 \lambda=0 \end{aligned}$$

$19 \lambda=7 \Rightarrow \lambda=\frac{7}{19}$

Substituting value of $\lambda$ in eq. (1), we get

$$\begin{aligned} & \left(1+\frac{14}{19}\right) x+\left(2+\frac{7}{19}\right) y+\left(3-\frac{7}{19}\right) z+\left(-4+\frac{35}{19}\right)=0 \\ & \therefore 33 x+45 y+50 z-41=0 \end{aligned}$$

The required area is shaded.

The point of intersection of the curves are

\begin{aligned} \mathrm{x}^2=\mathrm{x} & \Rightarrow \mathrm{x}(\mathrm{x}-1)=0 \text { i.e. } \mathrm{O}(0,0) \text { and } \mathrm{P}(1,1) \\ \therefore \mathrm{A} & =\int_\limits0^1\left(\mathrm{x}-\mathrm{x}^2\right) \mathrm{dx} \\ & =\left[\frac{\mathrm{x}^2}{2}\right]_0^1-\left[\frac{\mathrm{x}^3}{3}\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \text { sq. units }\end{aligned}

Lines are $x+y=1,7x+9y=63,y=5,x=6$

The required area is shaded.

$$\begin{aligned} & \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right) \\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right) \\ & =\tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}\right] \\ & =\tan ^{-1}\left(\frac{36+20}{48-15}\right)=\tan ^{-1}\left(\frac{56}{33}\right)=\cos ^{-1}\left(\frac{33}{65}\right)\end{aligned}$$

We have $\bar{a} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})=0, \overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})=0$ and $\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0$

$\therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0$ ....... (1)

$\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=0$ ..... (2)

$\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0$ ..... (3)

From (1), (2) and (3), we get

$$\begin{aligned} & \quad 2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}})=0 \\ & \text { Now }|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & \therefore|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=(5)^2+(4)^2+(3)^2+2(0)=50 \end{aligned}$$

From the data given, symbolic form of the given statement is

$$\begin{aligned} & (\sim p \vee \sim q) \rightarrow(r \wedge s) \\ & \equiv \sim(p \wedge q) \rightarrow(r \wedge s) \end{aligned}$$

$y=\sqrt{e^{\sqrt{x}}} \Rightarrow y^2=e^{\sqrt{x}}$

Takig log on both sides,

$2 \log y=\sqrt{x} \log e \Rightarrow 2 \log y=\sqrt{x}$

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

$\frac{2}{y} \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{4 \sqrt{x}}\right]=\frac{\sqrt{e^{\sqrt{x}}}}{4 \sqrt{x}}$

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

Let $\mathrm{u}=\mathrm{x}^{\mathrm{x}} \Rightarrow \log \mathrm{u}=\mathrm{x} \log \mathrm{x}$

$$\begin{aligned} & \therefore \quad \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{x}}+\log \mathrm{x} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x}) \\ & \therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{1+\mathrm{x}}}\right]+\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x}) \\ & \therefore\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=1}=\left(\frac{-1}{\sqrt{2}}\right)\left(\frac{1}{2 \sqrt{2}}\right)+1 \quad=\frac{-1}{4}+1=\frac{3}{4} \end{aligned}$$

$$\begin{aligned} & (1+\omega)^7=A+B \omega \\ & \begin{aligned} & \therefore A+B \omega=\left(-\omega^2\right)^7 \quad ...[\because 1+\omega+\omega^2=0]\\ &=(-1) \omega^{14}=-\omega^{12} \omega^2=-\omega^2=\left(1+\omega\right) \\ & \therefore A=1, B=1\end{aligned}\end{aligned}$$

$$\begin{aligned} & \text { Let } A^{-1} A=I \\ & \therefore\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5\end{array}\right] \mathrm{A}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ & R_2 \rightarrow 3 R_2-R_1 \text { and } R_3 \rightarrow 3 R_3-2 R_1 \\ & {\left[\begin{array}{ccc}3 & 2 & 6 \\ 0 & 1 & 0 \\ 0 & 11 & 3\end{array}\right] A=\left[\begin{array}{ccc}1 & 0 & 0 \\ -1 & 3 & 0 \\ -2 & 0 & 3\end{array}\right]} \\ & \mathrm{R}_1 \rightarrow \mathrm{R}_1-2 \mathrm{R}_2 \text { and } \mathrm{R}_3 \rightarrow \mathrm{R}_3-11 \mathrm{R}_2 \\ & {\left[\begin{array}{lll}3 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{array}\right] A^{-}=\left[\begin{array}{ccc}3 & -6 & 0 \\ -1 & 3 & 0 \\ 9 & -33 & 3\end{array}\right]} \\ & \mathrm{R}_1 \rightarrow \mathrm{R}_1-2 \mathrm{R}_3 \\ & {\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{array}\right] \mathrm{A}=\left[\begin{array}{ccc}-15 & 60 & -6 \\ -1 & 3 & 0 \\ 9 & -33 & 3\end{array}\right]} \\ & \mathrm{R}_1 \rightarrow \frac{1}{3} \mathrm{R}_1 \text { and } \mathrm{R}_3 \rightarrow \frac{1}{3} \mathrm{R}_3 \\ & {\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A=\left[\begin{array}{ccc}-5 & 20 & -2 \\ -1 & 3 & 0 \\ 3 & -11 & 1\end{array}\right]} \\ & \end{aligned}$$

We have $|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=20$ and we have to find value of $|(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})|$

$$\begin{aligned} & (3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}) \\ & =6(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{b}} \times \overline{\mathrm{b}}) \\ & =0+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0=7(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \end{aligned}$$

Hence area of the required parallelogram $=7 \times 20=140$

To find the x-intercept of a line, set $y = 0$ and solve for $x$.

First, let's determine the equation of the line passing through the points $\left(\frac{-1}{2}, 1\right)$ and $(1, 2)$.

Using the point-slope formula, the slope $m$ of the line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is : $m = \frac{y_2 - y_1}{x_2 - x_1}$

Given $x_1 = \frac{-1}{2}, y_1 = 1, x_2 = 1$, and $y_2 = 2$ :

$m = \frac{2 - 1}{1 - (-1/2)}$

$m = \frac{1}{1.5}$

$m = \frac{2}{3}$

Now, using the point-slope form :

$y - y_1 = m(x - x_1)$

Substitute the point $\left(\frac{-1}{2}, 1\right)$ and the slope $m = \frac{2}{3}$ into the equation :

$y - 1 = \frac{2}{3}(x + \frac{1}{2})$

$y = \frac{2}{3}x + \frac{1}{3} + 1$

$y = \frac{2}{3}x + \frac{4}{3}$

To find the x-intercept, set $y = 0$ :

$0 = \frac{2}{3}x + \frac{4}{3}$

$-\frac{4}{3} = \frac{2}{3}x$

$x = -2$

Therefore, the x-intercept of the line is $-2$. The correct answer is Option B.

We find that in R$_4$, f(a) = 1 and f(a) = 2. Hence R$_4$ is not a function.

Let $I=\int \sec ^4 x \tan ^4 x d x$

Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$

$\therefore$ $I=\int \sec ^2 x\left(\sec ^2 x\right)\left(\tan ^4 x\right) d x$

$=\int\left(1+t^2\right)(t)^4 d t=\int\left(t^4+t^6\right) d t=\frac{t^5}{5}+\frac{t^7}{7}+c=\frac{\tan ^5 x}{5}+\frac{\tan ^7 x}{7}+c$

$$\begin{aligned} & f(x)=2 x^3-15 x^2-144 x-7 \\ & f^{\prime}(x)=6 x^2-30 x-144 \\ & \text { When } f^{\prime}(x) < 0, \text { we get } x^2-5 x-24 < 0 \\ & \therefore(x-8)(x+3) < 0 \Rightarrow-3 < x <8\end{aligned}$$