MHT CET 2021 24th September Morning Shift (Mathematics) Question Paper with Solutions

$$\begin{aligned} & a \sin \theta=b \cos \theta \Rightarrow \tan \theta=\frac{b}{a} \\ & a \cos 2 \theta+b \sin 2 \theta \\ & =a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \\ & =a\left[\frac{1-\left(\frac{b^2}{a^2}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]+b\left[\frac{2\left(\frac{b}{a}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]=a\left(\frac{a^2-b^2}{a^2+b^2}\right)+b\left(\frac{2 b}{a} \times \frac{a^2}{a^2+b^2}\right) \\ & =\frac{a\left(a^2-b^2\right)}{a^2+b^2}+\frac{b(2 a b)}{a^2+b^2}=\frac{a^3-a b^2+2 a b^2}{a^2+b^2}=\frac{a^3+a b^2}{a^2+b^2}=\frac{a\left(a^2+b^2\right)}{a^2+b^2}=a \end{aligned}$$

\text { Let } \quad I=\int_\limits0^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x \quad \text{.... (1)}

=\int_\limits0^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x=\int_0^{\frac{\pi}{2}} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x\quad \text{.... (2)}

Eq. (1) + (2) gives,

2 I=\log \left[\frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin x}\right] d x \quad=\int_\limits0^{\frac{\pi}{2}}(\log 1) d x=0

$$\begin{aligned} & y=\frac{\log _e x}{\log _e 10}+\frac{\log _e 10}{\log _e x}+1+1 \\ & \begin{aligned} \therefore \frac{d y}{d x} & =\left(\frac{1}{\log _e 10}\right)\left(\frac{1}{x}\right)+\left(\log _e 10\right)\left[\frac{-\frac{1}{x}}{\left(\log _e x\right)^2}\right] \\ & =\frac{1}{x \log _e 10}-\frac{\log _e 10}{x\left(\log _e x\right)^2} \end{aligned} \end{aligned}$$

Refer figure

$$\begin{aligned} & B=(r \cos \theta, r \sin \theta) \\ & \therefore A B=2 r \cos \theta \text { and } B C=2 r \sin \theta \\ & A(\square A B C D)=A B \times B C \\ & \therefore \quad f(\theta) \quad=(2 r \cos \theta)(2 r \sin \theta)=2 r^2 \sin 2 \theta \\ & f^{\prime}(\theta)=4 r^2 \cos 2 \theta \text { and when } f^{\prime}(\theta)=0 \text {, we write } \\ & \quad \cos 2 \theta=0 \Rightarrow \cos 2 \theta=\cos \frac{\pi}{2} \quad \Rightarrow \theta=\frac{\pi}{4} \\ & \quad f^{\prime \prime}(\theta)=-8 r^2 \sin 2 \theta \text { and } f^{\prime \prime}(\theta)_{\left[\theta=\frac{\pi}{4}\right]}=-8 r^2<0 \end{aligned}$$

Thus area of required rectangle is maximum when $\theta=\frac{\pi}{4}$.

$$\begin{aligned} & \therefore A B=2 r \cos \frac{\pi}{4}=2 r\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2} r \text { and } B C=2 r \sin \frac{\pi}{4}=2 r\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2} r \\ & \therefore A(\square A B C D)=(\sqrt{2} r)(\sqrt{2} r)=2 r^2 \end{aligned}$$

Required plane passes through $(1,2,3)$ and is parallel to plane $2 x+3 y-4 z=0$. Hence equation of required plane is

$2(x-1)+3(y-2)-4(z-3)=0 \quad \Rightarrow 2 x+3 y-4 z+4=0$

To find $(f \circ g)(x)$, we need to compose the functions $f(x)$ and $g(x)$. This means we will first apply $g(x)$, and then apply $f$ to the result of $g(x)$.

Given:

$f(x)=\frac{x}{2x+1}$

and

$g(x)=\frac{x}{x+1}$

First, compute $g(x)$ for a general $x$:

$g(x) = \frac{x}{x+1}$

Now, use this result as the input for $f$. Therefore, $f(g(x))$ is given by substituting $g(x)$ into $f(x)$:

$f(g(x)) = f\left(\frac{x}{x+1}\right)$

According to the definition of $f(x)$:

$f(t) = \frac{t}{2t+1}$

Here, $t = \frac{x}{x+1}$. Substitute $t$ into the equation for $f(t)$:

$f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{2 \left(\frac{x}{x+1}\right) + 1}$

Simplify the inside of the fraction:

$2 \left(\frac{x}{x+1}\right) + 1 = \frac{2x}{x+1} + 1 = \frac{2x + (x+1)}{x+1} = \frac{3x+1}{x+1}$

Therefore,

$f(g(x)) = \frac{\frac{x}{x+1}}{\frac{3x+1}{x+1}}$

When you divide fractions, you multiply by the reciprocal:

$f(g(x)) = \frac{x}{x+1} \cdot \frac{x+1}{3x+1} = \frac{x}{3x+1}$

Hence,

$(f \circ g)(x) = \frac{x}{3x+1}$

The correct answer is Option B:

$\frac{x}{3x+1}$

Let the initial population be $\mathrm{P}$ and rate of increase is $8 \%$ per year.

$$\begin{aligned} & \therefore \frac{d P}{d t}=\frac{8}{100} \mathrm{P} \\ & \therefore \int \frac{d P}{P}=\int 0.08 d t \\ & \therefore \log |P|=0.08 t+c \end{aligned}$$

When $t=0$, we get $c=\log P$

$\therefore \log \mathrm{P}=0.08 \mathrm{t}+\log \mathrm{P}$

When '$\mathrm{P}$' doubles, we write

$$\begin{aligned} & \log 2 \mathrm{p}=0.08 \mathrm{t}+\log \mathrm{P} \\ & \therefore \log \left(\frac{2 \mathrm{P}}{\mathrm{P}}\right)=\log 2=0.6912=0.08 \mathrm{t} \\ & \therefore \mathrm{t}=\frac{0.6912}{0.08}=8.64 \text { years } \end{aligned}$$

$\sin ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-\pi}{3}-\frac{\pi}{6}=\frac{-\pi}{2}$

$\tan \alpha$ and $\tan \beta$ are roots of the $a x^2-b x y-y^2=0$

$\therefore \tan \alpha+\tan \beta=\frac{-(-b)}{-1}=-b$ and $\tan \alpha \tan \beta=\frac{a}{(-1)}=-\mathrm{a}$

$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{-b}{1-(-a)}=\frac{-b}{1+a}$

$$\begin{aligned} & u=\cos ^3 x \text { and } v=\sin ^3 x \\ & \therefore \frac{d u}{d x}=3 \cos ^2 x(-\sin x) \quad \text { and } \quad \frac{d v}{d x}=3 \sin ^2 x(\cos x) \\ & \therefore \frac{d v}{d u}=\frac{3 \sin ^2 x \cos x}{-3 \sin x \cos ^2 x}=-\tan x \\ & \therefore\left(\frac{d v}{d u}\right)_{x=\frac{\pi}{4}}=-\tan \left(\frac{\pi}{4}\right)=-1 \end{aligned}$$

$$\begin{aligned} & (\bar{a}+k \bar{b}) \cdot(\bar{a}-k \bar{b})=0 \\ & \therefore|a|^2-k^2|\bar{b}|^2=0 \\ & \therefore(4)^2-k^2(5)^2=0 \Rightarrow k= \pm \frac{4}{5} \end{aligned}$$

Let

$$\begin{aligned} I & =\int \frac{\cos x-\sin x}{8-\sin 2 x} d x \\ & =\int \frac{\cos x-\sin x}{9-1-\sin 2 x} d x=\int \frac{\cos x-\sin x}{9-(1+\sin 2 x)} d x \\ & =\int \frac{\cos x-\sin x}{(3)^2-(\sin x+\cos x)^2} d x \end{aligned}$$

Put $\sin \mathrm{x}+\cos \mathrm{x}=\mathrm{t} \Rightarrow(\cos \mathrm{x}-\sin \mathrm{x}) d \mathrm{dx}=\mathrm{dt}$

$$\begin{aligned} \therefore I & =\int \frac{d t}{(3)^2-\left(t^2\right)}=\frac{1}{3(2)} \log \left|\frac{3+t}{3-t}\right|+c \\ & =\frac{1}{6} \log \left|\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right|+c \Rightarrow p=6 \end{aligned}$$

From 1 to 6, we have 1 and 4 as perfect squares.

Probability of getting perfect square in one throw of a die $=\frac{2}{6}=\frac{1}{3}$

$\therefore$ Probability of not getting perfect square in 4 throws of a die $=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{16}{81}$

$\therefore$ Required probability $=1-\frac{16}{81}=\frac{65}{81}$

We have $$|\mathrm{A}(\operatorname{adj} \mathrm{A})|=\left|\begin{array}{ccc}-10 & 0 & 0 \\ 0 & -10 & 2 \\ 0 & 0 & -10\end{array}\right|$$

$\therefore|A||\operatorname{adj} A|=(-10)(100)=-1000$

$\therefore|A||A|^{3-1}=-1000 \Rightarrow|A|^3=-1000 \Rightarrow|A|=-10$

$$\begin{aligned} & \frac{d y}{d x}=\frac{x+y+1}{x+y-1} \\ & \text { Put } x+y=v \Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x} \\ & \therefore \frac{d v}{d x}-1=\frac{v+1}{v-1} \Rightarrow \frac{d v}{d x}=\frac{v+1}{v-1}+1=\frac{2 v}{v-1} \\ & \therefore \int \frac{(v-1) d v}{2 v}=\int d x \\ & \therefore \int \frac{1}{2} d v-\int \frac{1}{2 v} d v=\int d x \Rightarrow \frac{v}{2}-\frac{1}{2} \log |v|=x+c \\ & \therefore \frac{x+y}{2}-\frac{1}{2} \log |x+y|=x+c \end{aligned}$$

We have $\mathrm{x}=\frac{2}{3}, \mathrm{y}=\frac{1}{3}$, we get

$$\begin{aligned} & \therefore \frac{1}{2}-\frac{1}{2} \log |1|=\frac{2}{3}+c \quad \Rightarrow c=\frac{1}{2}-\frac{2}{3}=\frac{-1}{6} \\ & \therefore \frac{x+y}{2}-\frac{1}{2} \log |x+y|=x-\frac{1}{6} \end{aligned}$$

Multiplying both sides by 2 , we get

$\therefore(x+y)-\log |x+y|=2 x-\frac{2}{6} \Rightarrow \cdot y-x+\frac{1}{3}=\log |x+y|$

$$\begin{aligned} & \left(B^{-1} A^{-1}\right)^{-1}=\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1}=A B \\ & \therefore\left(B^{-1} A^{-1}\right)^{-1}=\left[\begin{array}{ll} 2 & -2 \\ 2 & -3 \end{array}\right]\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} -2 & -2 \\ -3 & -2 \end{array}\right] \end{aligned}$$

Let $P \equiv(2,-1,5)$ and co-ordinates of any point on the given line be

$Q \equiv(10 \lambda+11,-4 \lambda-2,-11 \lambda-8)$

d.r. of $P Q$ are $(10 \lambda+9,-4 \lambda-1,-11 \lambda-13)$

d.r. of given line are $(10,-4,-11)$.

$$\begin{aligned} & \therefore(10 \lambda+9)(10)+(-4 \lambda-1)(-4)+(-11 \lambda-13)(-11)=0 \\ & \therefore 100 \lambda+90+16 \lambda+4+121 \lambda+143=0 \Rightarrow \lambda=-1 \end{aligned}$$

$\therefore \mathrm{Q} \equiv(1,2,3)$ and $\mathrm{d}(\mathrm{PQ})=\sqrt{1^2+3^2+2^2}=\sqrt{14}$ units

$$\begin{aligned} \text { Let } \quad\mathrm{I} & =\int \cos ^3 x \cdot e^{\log (\sin x)} d x \\ & =\int \cos ^3 x \cdot \sin x d x \end{aligned}$$

Put $\cos x=t \Rightarrow-\sin x d x=d t$

$\therefore I=\int-t^3 d t=\frac{-t^4}{4}+t=\frac{-\cos ^4 x}{4}+c$

Let $\sqrt{-5-12 \mathrm{i}}=\mathrm{a}+\mathrm{ib}\quad$ $\ldots[\mathrm{a}, \mathrm{b} \in \mathrm{R}]$

$$\begin{aligned} & \therefore(a+i b)^2=-5-12 i \\ & \therefore\left(a^2-b^2\right)+i(2 a b)=-5-i(12) \end{aligned}$$

$\therefore \mathrm{a}^2-\mathrm{b}^2=-5 \quad \text { and } \quad 2 \mathrm{ab}=-12 \Rightarrow \mathrm{ab}=-6 \quad \Rightarrow \mathrm{b}=\frac{-6}{\mathrm{a}}$

$$\begin{aligned} & \therefore a^2-\left(\frac{-6}{a}\right)^2=-5 \\ & \therefore a^4-36=-5 a^2 \Rightarrow a^4+5 a^2-36=0 \\ & \therefore\left(a^2+9\right)\left(a^2-4\right)=0 \quad \Rightarrow a^2=4 \\ & \therefore a= \pm 2 \quad \Rightarrow b=\frac{-6}{ \pm 2}=\mp 3 \\ & \therefore \sqrt{-5-12 i}= \pm(2-3 i) \end{aligned}$$

We have $\mathrm{K}+2 \mathrm{~K}+3 \mathrm{~K}+4 \mathrm{~K}+5 \mathrm{~K}+6 \mathrm{~K}=1 \Rightarrow \mathrm{K}=\frac{1}{21}$

$$\begin{aligned} \therefore \mathrm{P}(2<\mathrm{x}<6) & =\mathrm{P}(\mathrm{x}=3)+\mathrm{P}(\mathrm{x}=4)+\mathrm{P}(\mathrm{x}=5) \\ & =\frac{3}{21}+\frac{4}{21}+\frac{5}{21}=\frac{12}{21}=\frac{4}{7} \end{aligned}$$

$$\begin{aligned} & f(x)=\log |\sin x| \text {, where } x \in(0, \pi) \\ & \therefore f^{\prime}(x)=\frac{1}{\sin x} \times \cos x \quad=\cot x \end{aligned}$$

When $f^{\prime}(x)>0$, we say $\frac{\cos x}{\sin x}>0$

Here $\sin \mathrm{x}>0\quad$ ... $[x \in(0, \pi)]$

Thus for the function to be strictly increasing, $\cos x>0 \Rightarrow x \in\left(0, \frac{\pi}{2}\right)$ only.

$$\begin{aligned} & \text { Let } \quad \lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=L \quad \text{....(say)}\\ & \therefore L=\frac{a b-a b}{1-1}=\frac{0}{0} \text { form } \Rightarrow L=\lim _{x \rightarrow 1} \frac{\left(a b^x \log b\right)-\left(a^x \log a \cdot b\right)}{2 x} \quad \ldots\left[L^{\prime}\right. \text { Hospital Rule] } \\ &=\frac{a b \log b-a b \log a}{2}=\frac{a b}{2} \log \left(\frac{b}{a}\right) \end{aligned}$$

Refer figure,

$$|A|=\left|\begin{array}{lll} 1 & 2 & 3 \\ 1 & 1 & a \\ 2 & 4 & 7 \end{array}\right|=(7-4 a)-2(7-2 a)+3(2)=-1$$

Matrix B is inverse of matrix A. 'b' is element $(1 \times 3)$ in matrix B Here element $(3 \times 1)$ in matrix $\mathrm{A}$ is 2.

Cofactor of $$2=(-1)^{3+1}\left|\begin{array}{ll}2 & 3 \\ 1 & \mathrm{a}\end{array}\right|=2 \mathrm{a}-3$$

$\text { Now } \frac{2 a-3}{-1}=b \quad \Rightarrow \quad 2 a+b=3 \quad \text{.... (1)}$

For element $(1 \times 2)$ in matrix A i.e. 2.

Now cofactor of 2 is $$(-1)^{1+2}\left|\begin{array}{ll}1 & a \\ 2 & 7\end{array}\right|=-(7-2 a)$$ and $\frac{-(7-2 a)}{-1}=7-2 a$

Here element $(2 \times 1)$ in matrix B is $-$3

$\therefore 7-2a=-3$ .... (2)

Solving eq. (1) and (2), we get a = 5, b = $-$7

$$\begin{aligned} y & =\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right] \\ & =\tan ^{-1}\left[\frac{\log e-\log x^2}{\log e+\log x^2}\right]+\tan ^{-1} 3+\tan ^{-1}(2 \log x) \\ & =\tan ^{-1}\left[\frac{1-\log x^2}{1+\log x^2}\right]+\tan ^{-1} 3+\tan ^{-1}\left(\log x^2\right) \\ & =\tan ^{-1}(1)-\tan ^{-1}\left(\log x^2\right)+\tan ^{-1} 3+\tan ^{-1}\left(\log x^2\right) \\ & =\tan ^{-1}(1)+\tan ^{-1}(3) \\ \therefore & \frac{d y}{d x}=0 \Rightarrow \frac{d^2 y}{d x^2}=0 \end{aligned}$$