MHT CET 2021 22nd September Evening Shift (Mathematics) Question Paper with Solutions

Refer figure

Required area is shaded.

It is unbounded and non-origin side.

We have \int_\limits0^a \sqrt{\frac{a-x}{x}} d x=\frac{k}{2}

Put $x=a \sin ^2 \theta \Rightarrow d x=a(2 \sin \theta) \cos \theta d \theta=2 a \sin \theta \cos \theta d \theta$

When $\mathrm{x}=0, \theta=0$ and when $\mathrm{x}=\mathrm{a}, \theta=\frac{\pi}{2}$

\begin{aligned} & \therefore \int_\limits0^{\frac{\pi}{2}} \sqrt{\frac{a-a \sin ^2 \theta}{a \sin ^2 \theta}}(2 a \sin \theta \cos \theta) d \theta=\frac{k}{2} \\ & \int_\limits0^{\frac{\pi}{2}} \sqrt{\frac{\cos ^2 \theta}{\sin ^2 \theta}} 2(a \sin \theta \cos \theta) d \theta=\frac{k}{2} \Rightarrow \int_\limits0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta}(2 a \sin \theta \cos \theta) d \theta=\frac{k}{2} \\ & \int_\limits0^{\frac{\pi}{2}} 2 a \cos \theta^2 d \theta=\frac{k}{2} \Rightarrow 2 a \int_\limits0^{\frac{\pi}{2}} \frac{1+\cos 2 \theta}{2} d \theta=\frac{k}{2} \\ & a \int_\limits0^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta=\frac{k}{2} \Rightarrow a\left[\theta+\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{2}}=\frac{k}{2} \\ & a\left[\left(\frac{\pi}{2}+0\right)-(0+0)\right]=\frac{k}{2} \Rightarrow a \frac{\pi}{2}=\frac{k}{2} \Rightarrow k=\pi a \end{aligned}

We know that Coefficient of Variation $= \mathrm{{{S\tan dard\,Deviation} \over {Mean}}}$

$\therefore$ ${(C.V.)_{Physics}} = {3 \over {20}} = 0.15$ and ${(C.V.)_{Chemistry}} = {2 \over {25}} = 0.08$

${(C.V.)_{maths}} = {4 \over {23}} = 0.174$ and ${(C.V.)_{Bio\log y}} = {5 \over {27}} = 0.185$

$$\begin{aligned} & \frac{d y}{d x} \quad=\frac{y+1}{x^2-x} \\ & \therefore \int \frac{d y}{y+1}=\int \frac{d x}{x(x-1)} \\ & \therefore \int \frac{d y}{y+1}=\int\left[\frac{1}{x-1}-\frac{1}{x}\right] d x \Rightarrow \log |y+1|=\log |x-1|-\log |x|+\log c \end{aligned}$$

We have $x=2$ and $y=1$

$$\begin{aligned} & \therefore \log |2|=\log |1|-\log |1|+\log c \Rightarrow c=2 \\ & \therefore \log |y+1|=\log |x-1|-\log |x|+\log |2| \\ & \log |y+1|=\log \left|\frac{2(x-1)}{x}\right| \\ & \therefore y+1=\frac{2(x-1)}{x} \Rightarrow x y+x=2 x-2 \Rightarrow x y=x-2 \\ \end{aligned}$$

We have $y^2=8 a x+8 a^2\quad \text{..... (1)}$

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

$2 y \frac{d y}{d x}=8 a \Rightarrow a=\left(\frac{y}{4}\right) \frac{d y}{d x}$

Substituting value of '$a$' in eq. (1), we get

$$\begin{aligned} & y^2=8\left(\frac{y}{4}\right) \frac{d y}{d x}(x)+8\left[\left(\frac{y}{4}\right)\left(\frac{d y}{d x}\right)\right]^2=2 x y \frac{d y}{d x}+\left(\frac{y^2}{2}\right)\left(\frac{d y}{d x}\right)^2 \\ & \therefore 2 y^2=4 x y\left(\frac{d y}{d x}\right)+y^2\left(\frac{d y}{d x}\right)^2 \end{aligned}$$

Hence order $=1$, degree $=2$

We have vertices $A(a, 31) ; B(4,5, b) ; C(6 ; c, 5)$ and centroid $G(4,3,3)$ of $\triangle A B C$.

$\therefore \frac{a+4+6}{3}=4, \frac{3+5+c}{3}=3, \frac{1+b+5}{3}=3 \Rightarrow a=2, c=1, b=3$

We have,

$$\begin{aligned} & \therefore P(x=1)=0.2, P(x=2)=0.17, P(x=3)=0.11 \\ & P(x=4)=0.14, P(x=5)=0.23, P(x=6)=0.15 \\ & \text { Thus } P(x=4)+P(x=5)=0.14+0.23=0.37\end{aligned}$$

We know that $\frac{a}{\sin A}=\frac{b}{\sin B}$

$\therefore a \sin B=b \sin A\quad \text{.... (1)}$ and we have

$a \cos B=b \cos A\quad \text{.... (2)}$

From (1) and (2), we write

$\frac{a}{b}=\frac{\sin A}{\sin B}=\frac{\cos A}{\cos B} \Rightarrow \tan A=\tan B \Rightarrow A=B$

Thus $\Delta$ is an isosceles triangle.

The greatest integer function, denoted by $[x]$, returns the greatest integer less than or equal to $x$. To find $f(\pi)$ where $\pi$ is approximately equal to 3.14 (though more accurately $\pi \approx 3.14159$), we substitute $\pi$ into the given function $f(x)=[8x]-3$.

Given that the approximation for $\pi$ is 3.14, when calculating $f(\pi)$, we first compute the product inside the greatest integer function:

$8\pi = 8 \times 3.14 = 25.12$

The greatest integer less than or equal to 25.12 is 25 because the greatest integer function drops all decimal places and takes the largest integer not greater than the given number. So, $[8\pi] = [25.12] = 25$.

Substituting this result back into our function:

$f(\pi) = [8\pi] - 3 = 25 - 3 = 22$

Therefore, the correct answer is Option D, 22.

We have $n p=4$ and $n p q=2$

$\therefore \quad q = {2 \over 4} = {1 \over 2} \Rightarrow p = 1 - {1 \over 2} = {1 \over 2}$ and $n\left( {{1 \over 2}} \right) = 4 \Rightarrow n = 8$

$\therefore\quad P(x - 2) = {}^8{C_2}{\left( {{1 \over 2}} \right)^2}{\left( {{1 \over 2}} \right)^6} = {{8!} \over {2!6!}}{\left( {{1 \over 2}} \right)^8} = {{8 \times 7} \over {2 \times {2^3} \times {2^5}}} = {7 \over {64}} = {{28} \over {256}}$

Refer figure

$$\begin{aligned} & \text { Slope of AB }=\frac{1-0}{3-2}=1 \\ & \therefore \tan \theta=1 \Rightarrow \theta=45^{\circ} \end{aligned}$$

Line AB is rotated through $15^\circ$ in anticlockwise direction about A. Therefore in a new position, slope of line $=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}$ and it passes through A.

Required equation of line is $$\begin{aligned} (y-0)=\sqrt{3}(x-2) & \Rightarrow \sqrt{3} x-2 \sqrt{3}=y \end{aligned}$$

For three vectors to be coplanar, the scalar triple product of the vectors must equal zero. The scalar triple product of vectors $\vec{a}, \vec{b},$ and $\vec{c}$ can be found by calculating the determinant of a matrix composed of their components. If the vectors are $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$, and $\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}$, we can set up the following determinant:

To solve the determinant, we expand it as follows:

For the vectors to be coplanar, this determinant must equal zero, so we set the expression equal to zero:

Solving for $\lambda$:

Therefore, $\lambda = -8$, which corresponds to Option A.

$$\begin{aligned} & \left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2 \\ & =\left(3+3 \omega+3 \omega^2+2 \omega\right)^2+\left(3+3 \omega+3 \omega^2+2 \omega^2\right)^2 \\ & =\left[3\left(1+\omega+\omega^2\right)+2 \omega\right]^2+\left[3\left(1+\omega+\omega^2\right)+2 \omega^2\right]^2 \\ & =(3(0)+2 \omega)^2+\left[3(0)+2 \omega^2\right]^2=4 \omega^2+4 \omega^4=4 \omega^2\left(1+\omega^2\right)=4 \omega^2(-\omega) \\ & =-4 \omega^3 \quad=-4 \end{aligned}$$

Let $\quad$ I=\int_\limits0^2|2 x-3| d x

When $\mathrm{x}=\frac{3}{2}, 2 \mathrm{x}-3=0$

\begin{aligned} \therefore I & =\int_\limits0^{3 / 2}(3-2 x) d x+\int_\limits{\frac{3}{2}}^2(2 x-3) d x=[3 x]_0^{3 / 2}-\frac{2}{2}\left[x^2\right]_0^{3 / 2}+\frac{2}{2}\left[x^2\right]_{3 / 2}^2-[3 x]_{3 / 2}^2 \\ & =\left(\frac{9}{2}\right)-\left(\frac{9}{4}\right)+\left(4-\frac{9}{4}\right)-3\left(2-\frac{3}{2}\right)=\frac{5}{2} \end{aligned}

$$\begin{aligned} & \text { We have } 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) \\ & \therefore \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\ & \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow \sin x \cos x=\sin ^2 x \\ & \therefore \sin x(\sin x-\cos x)=0 \Rightarrow \sin x=0 \text { or } \tan x=1 \\ & \therefore x=0 \text { or } x=\frac{\pi}{4} \end{aligned}$$

Refer figure

$\mathrm{A}(\triangle \mathrm{ABC}) \quad=\frac{1}{2}|\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}|$

Here

$$\begin{aligned} & \overline{\mathrm{BA}}=2 \hat{\mathrm{j}}-a \hat{\mathrm{k}} \\ & \overline{\mathrm{BC}}=-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+(1-\mathrm{a}) \hat{\mathrm{k}} \end{aligned}$$

Now

$$\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & 2 & -\mathrm{a} \\ -1 & 3 & 1-\mathrm{a} \end{array}\right|$$

$=\hat{i}(2-2 a+3 a)-\hat{j}(-a)+\hat{k}(2)=(a+2) \hat{i}+(a) \hat{j}+2 \hat{k}$

$\therefore|\overline{\mathrm{BA}} \times \overline{\mathrm{BC}}|=\sqrt{(\mathrm{a}+2)^2+(\mathrm{a})^2+(2)^2}$

From given data, we write

$$\begin{aligned} & \sqrt{6}=\frac{1}{2} \sqrt{2 a^2+4 a+8} \\ & \therefore 4(6)=2 a^2+4 a+8 \quad \Rightarrow a^2+2 a-8=0 \Rightarrow(a+4)(a-2)=0 \Rightarrow a=-4,2 \end{aligned}$$

The point of intersection of $y^2=x$ and $x+y=2$ is,

$(2-x)^2=x \quad \Rightarrow \quad x^2-5 x+4=0 \quad \Rightarrow \quad(x-4)(x-1)=0$

Let $A=(1,1)$ in first quadrant and $B=(4,-2)$ in fourth quadrant

The line $x+y=2$ cuts $X$ axis at $P(2,0)$

Refer figure

Required area is shaded.

$\therefore A=\int_0^1 \sqrt{x} d x+\int_1^2(2-x) d x$

$=\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_0^1+[2 x]_1^2-\left[\frac{x^2}{2}\right]_1^2$

$=\left[\left(\frac{2}{3}\right)(1)\right]+[2(2-1)]-\left[\left(\frac{4-1}{2}\right)\right]=\frac{2}{3}+2-\frac{3}{2}=\frac{7}{6}$ sq. units

$f(x)=x^2+a x+b$

$\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+\mathrm{a}$ and when $\mathrm{f}^{\prime}(\mathrm{x})=0$, we get $\mathrm{x}=\frac{-\mathrm{a}}{2}$

Now $\mathrm{f}^{\prime \prime}(\mathrm{x})=2$ and $2>0$

$\therefore \mathrm{f}(\mathrm{x})$ has minima at $\mathrm{x}=\frac{-\mathrm{a}}{2}=3\quad$ ...[as per given data]

$\therefore a=-6$

Since minimum value of $\mathrm{f}(\mathrm{x})$ is 5 at $\mathrm{x}=3$, we write

$5=(3)^2+(-6)(3)+b \Rightarrow b=14$

$f(x)=\frac{\lambda \sin x+6 \cos x}{2 \sin x+3 \cos x}$

$\therefore f^{\prime}(x)=\frac{\{[(2 \sin x+3 \cos x)(\lambda \cos x-6 \sin x)]-[(\lambda \sin x+6 \cos x)(2 \cos x-3 \sin x)]\}}{(2 \sin x+3 \cos x)^2}$

When $\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$, we get

$\left[\left(2 \lambda \sin x \cos x+3 \lambda \cos ^2 x-12 \sin ^2 x-18 \sin x \cos x\right)-\left(2 \lambda \sin x \cos x+12 \cos ^2 x-3 \lambda \sin ^2 x\right.\right.-18 \sin x \cos x)] \geq 0$

$\therefore 3 \lambda\left(\sin ^2 x+\cos ^2 x\right)-12\left(\sin ^2 x+\cos ^2 x\right) \geq 0$

$\therefore 3 \lambda-12 \geq 0 \Rightarrow \lambda \geq 4$

We have $\mathrm{a x^2+8 x y+5 y^2=0}$

Here $\mathrm{m_1+m_2=\frac{-8}{5}, m_1 m_2=\frac{a}{5}}$

As per condition given, we write $\left(-\frac{8}{5}\right)=2\left(\frac{a}{5}\right) \Rightarrow a=-4$

To solve this problem, let's start by using the given equation $2 \cos \theta=x+\frac{1}{x}$ and then find $2 \cos 3 \theta$ using trigonometric identities.

First, remember the triple angle identity for cosine:

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$

Given $2 \cos \theta = x + \frac{1}{x}$, we also have:

$\cos \theta = \frac{x+\frac{1}{x}}{2}$

Substitute this into the triple angle formula:

$\cos 3\theta = 4\left(\frac{x+\frac{1}{x}}{2}\right)^3 - 3\left(\frac{x+\frac{1}{x}}{2}\right)$

By simplifying, we arrive at:

$\cos 3\theta = 4\left(\frac{x^3+3x+\frac{3}{x}+\frac{1}{x^3}}{8}\right) - \frac{3(x+\frac{1}{x})}{2}$

Further simplification gives us:

$\cos 3\theta = \frac{x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}}{2} - \frac{3x + \frac{3}{x}}{2}$

Upon combining like terms, the terms $3x$ and $\frac{3}{x}$ will cancel out and we get:

$\cos 3\theta = \frac{x^3 + \frac{1}{x^3}}{2}$

However, to find $2 \cos 3\theta$, we need to multiply the entire expression by 2, yielding:

$2 \cos 3\theta = x^3+\frac{1}{x^3}$

Therefore, the correct answer is Option D: $x^3+\frac{1}{x^3}$.

We have $$A(\operatorname{adj} A)=\left[\begin{array}{cc}20 & 0 \\ 0 & 20\end{array}\right]$$

$$\therefore|A||\operatorname{adj} A|=\left|\begin{array}{cc} 20 & 0 \\ 0 & 20 \end{array}\right| \Rightarrow|A|\left(|A|^{2-1}\right)=400 \Rightarrow(|A|)^2=(20)^2 \Rightarrow|A|=20$$

Let $\quad I=\int \frac{1}{\cos x+\sqrt{3} \sin x} d x$

Dividing numerator and denominator by 2 , we get

$=\frac{1}{2} \int \frac{d x}{\left(\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x\right)}=\frac{1}{2} \int \frac{d x}{\sin \left(x+\frac{\pi}{6}\right)}=\frac{1}{2} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right|+C$

To find $\frac{d y}{d x}$ given that $x^y \cdot y^x = 16$, we first need to start with the given expression:

$x^y \cdot y^x = 16$

Since we seek $\frac{d y}{d x}$, we'll differentiate both sides of the equation with respect to $x$, employing the product rule $(uv)' = u'v + uv'$ where necessary, and remembering that when differentiating $x$ with respect to $x$, we simply get 1, but differentiating $y$ with respect to $x$ gives us $y'$ or $\frac{d y}{d x}$. In cases involving variables raised to variables, we apply the logarithmic differentiation method. However, because this equation involves a product where each term has a different base and exponent, we have to treat each part carefully, keeping in mind that the differentiation of $x^y$ and $y^x$ with respect to $x$ necessitates implicit differentiation.

Start by applying logarithm on both sides to simplify the differentiation process:

$\log(x^y \cdot y^x) = \log(16)$

Using properties of logarithms, this simplifies to:

$y\log(x) + x\log(y) = \log(16)$

Next, differentiate both sides with respect to $x$:

$\frac{d}{dx}[y\log(x) + x\log(y)] = \frac{d}{dx}[\log(16)]$

Noting that the right side is a constant, its derivative is 0. On the left side, apply the product rule and remember $\frac{d}{dx} \log(x) = \frac{1}{x}$:

$\frac{d}{dx}[y\log(x)] + \frac{d}{dx}[x\log(y)] = 0$

It becomes:

$y'\log(x) + \frac{y}{x} + \log(y) + xy'\frac{1}{y} = 0$

Upon simplification, this yields:

$y'\log(x) + y' + \frac{y}{x} + \log(y) = 0$

Then, we can solve for $y'$ or $\frac{dy}{dx}$:

$y'(\log(x) + 1) = -\frac{y}{x} - \log(y)$

So, $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{-\frac{y}{x} - \log(y)}{\log(x) + 1}$

To find $\frac{d y}{d x}$ at $(x,y) = (2,2)$, substitute $x = 2$ and $y = 2$ into the derived expression:

$\frac{dy}{dx} \bigg|_{(2,2)} = \frac{-\frac{2}{2} - \log(2)}{\log(2) + 1}$

Since $\log(2)$ is the natural logarithm of $2$, this further simplifies to:

$\frac{dy}{dx} \bigg|_{(2,2)} = \frac{-1 - \log(2)}{\log(2) + 1}$

To simplify, note that both the numerator and denominator have $\log(2)$, so by factoring it out, you can see that the value simplifies neatly:

$\frac{dy}{dx} \bigg|_{(2,2)} = -1$

Therefore, the correct answer is:

Option A: $-1$

Equation of the plane passing through the line of intersection of the given planes, is

$$\begin{aligned} & (x+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0 \\ & \therefore(1+4 \lambda) x+(2+3 \lambda) y+(3+2 \lambda) z+(-4-\lambda)=0 \quad \text{.... (1)} \end{aligned}$$

Since the plane (1) passes through origin,

We get $-4-\lambda=0 \Rightarrow \lambda=-4$

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

$$\begin{aligned} & -15 x-10 y-5 z=0 \quad \Rightarrow \quad 3 x+2 y+z=0 \\ & \therefore \text { d.r.s. are }(3,2,1) \text {. } \end{aligned}$$

$$\begin{aligned} & \lim _{x \rightarrow 2}(x-1)^{ \frac{1}{3 x-6}} \\ & =\lim _{x \rightarrow 2}(x-2+1) ^{\frac{1}{3(x-2)}}=\lim _{x \rightarrow 2}\left\{[1+(x-2)]^{\frac{1}{(x-2)}}\right\}^{\frac{1}{3}}=e^{\frac{1}{3}} \end{aligned}$$