MHT CET 2021 23rd September Evening Shift (Mathematics) Question Paper with Solutions

Let \quad I=\int_\limits0^\pi \frac{1}{4+3 \cos x} d x

Put $\tan \frac{x}{2}=t \Rightarrow \cos x=\frac{1-t^2}{1+t^2} \quad$ and $\quad \sec ^2 \frac{x}{2}\left(\frac{1}{2}\right) d x=d t \Rightarrow d t=\frac{2}{1+t^2} d t$

When $\mathrm{x}=0, \mathrm{t}=0$ and when $\mathrm{x}=\pi, \mathrm{t}=\infty$

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

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

Let p : Two triangles are congruent.

$\mathrm{q}$ : Their areas are equal.

Logical form of given statement is $\mathrm{p} \rightarrow \mathrm{q}$.

Inverse of given statement is $\sim \mathrm{p} \rightarrow \sim \mathrm{q}$.

Contrapositive of inverse of given statement is $\sim(\sim \mathrm{q}) \rightarrow \sim(\sim \mathrm{p})$ i.e. $\mathrm{q} \rightarrow \mathrm{p}$ i.e.

If areas of two triangles are equal, then they are congruent.

Given $f$ is a the p.d.f. of a r.v.x.

\begin{aligned} & \therefore \quad \int_\limits0^1 f(x) d x=1 \Rightarrow \int_\limits0^1 K\left(x-x^2\right) d x=1 \\ & K\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=1 \Rightarrow K\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow \frac{K}{6}=1 \\ & K=6 \\ & \therefore P\left(X<\frac{1}{2}\right)=\int_\limits0^{\frac{1}{2}} 6\left(x-x^2\right) d x=\left[\frac{6 x^2}{2}\right]_0^1-\left[\frac{6 x^3}{3}\right]_0^1 \\ & =\left[3 x^2-2 x^3\right]_0^{\frac{1}{2}}=\frac{3}{4}-\frac{2}{8}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2} \end{aligned}

Projection $\bar{a}$ on $\bar{b}=\frac{\bar{a} \cdot \bar{b}}{|\bar{b}|}=\frac{(\mathrm{i}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{(2)^2+(-1)^2+(1)^2}}=\frac{2+2+1}{\sqrt{6}}=\frac{5}{\sqrt{6}}$

Since given vectors are coplanar, we write

$$\begin{aligned} &\left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{array}\right|=0 \\ & \therefore 2(10+3 \lambda)+1(5+9)+1(\lambda-6)=0 \\ & \therefore 20+6 \lambda+14+\lambda-6=0 \Rightarrow 7 \lambda+28=0 \quad \Rightarrow \lambda=-4 \end{aligned}$$

Put $x=-4$ in all options.

(A) $8-12=-4 \neq 6$

(B) $16-8=8 \neq 4$

(C) $16-12=4$

(D) $16-8=8 \neq 6$

$$\begin{aligned} & y=\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right) \\ & \frac{d y}{d x}=\sqrt{2} \cos \left(2 x+\frac{\pi}{4}\right)(2) \\ & \therefore\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=(2 \sqrt{2}) \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=2 \sqrt{2}\left(-\sin \frac{\pi}{4}\right)=-2 \\ & \text { When } x=\frac{\pi}{4}, y=\sqrt{2} \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right)=1 \end{aligned}$$

Hence equation of required tangent is

$(y-1)=-2\left(x-\frac{\pi}{4}\right) \Rightarrow 2 x+y-\frac{\pi}{2}-1 =0$

Let

\begin{aligned} & I =\int_\limits1^3\left[\tan ^1\left(\frac{x}{x^2-1}\right)+\tan ^{-1}\left(\frac{x^2-1}{x}\right)\right] d x \\ & =\int_\limits1^3\left[\tan ^{-1}\left(\frac{x}{x^2-1}\right)+\cot ^{-1}\left(\frac{x}{x^2-1}\right)\right] d x \\ & =\int_\limits1^3\left(\frac{\pi}{2}\right) d x=\frac{\pi}{2} \int_\limits1^3 d x=\frac{\pi}{2}[x]_1^3=\pi \end{aligned}

$$\begin{aligned} & \text { Let } \quad I=\int \frac{d x}{\cos x \sqrt{\cos 2 x}} \\ & =\int \frac{d x}{\cos x \cdot \cos x \sqrt{\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}}}=\int \frac{d x}{\cos ^2 x \sqrt{1-\tan ^2 x}} \\ & I=\int \frac{\sec ^2 x}{\sqrt{1-\tan ^2 x}} d x \\ & \text { Put } \tan x=t \Rightarrow \sec ^2 d x=d t \\ & \therefore I=\int \frac{d t}{\sqrt{1-t^2}}=\sin ^{-1}(t)+c=\sin ^{-1}(\tan x)+c \end{aligned}$$

Let $\mathrm{A}$ be the origin.

Then $\mathrm{B}=(3,0,4)$ and $\mathrm{C}=(5,-2,4)$

Mid point of $\mathrm{BC}=\left(\frac{3+5}{2}, \frac{0-2}{2}, \frac{4+4}{2}\right)$ i.e. $(4,-1,4)$

$\therefore$ Length of median $=\sqrt{(4)^2+(-1)^2+(4)^2}=\sqrt{33}$

We know that centroid divides a line joining orthocenter to circumcentre in the ratio $2: 1$.

$\therefore \overline{\mathrm{g}}=\frac{\overline{\mathrm{h}}+2 \overline{\mathrm{p}}}{1+2} \Rightarrow 2 \overline{\mathrm{p}}+\overline{\mathrm{h}}-3 \overline{\mathrm{g}}=0$

$\therefore x=2, y=1, z=-3$ as per data given.

$\mathrm{P} \%$ amount disappears in one year.

Let initial amount of radium $=\mathrm{x}_0$.

$\therefore$ Amount left after 1 year $=\mathrm{x}_0-\frac{\mathrm{P}}{100} \times \mathrm{x}_0=\mathrm{x}_0\left(1-\frac{\mathrm{P}}{100}\right)$.

Amount left after 2 years

$$\begin{aligned} & =x_0\left(1-\frac{P}{100}\right)-\frac{P}{100} \times x_0\left(1-\frac{P}{100}\right) \\ & =x_0\left(1-\frac{P}{100}\right)\left(1-\frac{P}{100}\right)=x_0\left(1-\frac{P}{100}\right)^2 \end{aligned}$$

$$\begin{aligned} & |A|=\left|\begin{array}{cc} 2 & 3 \\ 10 & 15 \end{array}\right|=30-30=0 \\ & |B|=\left|\begin{array}{ccc} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 1 & 2 & 3 \end{array}\right|=0 \quad \ldots\left[R_1 \text { and } R_3 \text { are identical }\right] \\ & |C|=\left|\begin{array}{lll} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 6 & 8 \end{array}\right|=1(2)-2(4)+3(2)=0 \\ & |D|=\left|\begin{array}{lll} 2 & 4 & 2 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{array}\right|=2(5)-4(5)+2(3)=-4 \neq 0 \end{aligned}$$

We have $\sim \mathrm{p} \rightarrow \mathrm{q}$

Inverse of given statement is

$\sim(\sim \mathrm{p}) \rightarrow \sim \mathrm{q} \text { i.e. } \mathrm{p} \rightarrow \sim \mathrm{q}$

Negation of inverse of given statement is $\sim(p \rightarrow \sim q)$

$\equiv \sim(\sim p \vee \sim q) \equiv p \wedge q$

Amplitude of $(z-2-3 i)$ is $\frac{3 \pi}{4}$ and we have $z=x+i y$

$\therefore \operatorname{Amp}[(x-2)+\mathrm{i}(\mathrm{y}-3)]$ is $\frac{3 \pi}{4}$

$\therefore \tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{3 \pi}{4} \Rightarrow \tan \left(\frac{3 \pi}{4}\right)=\frac{y-3}{x-2}$

$\therefore-1=\frac{y-3}{x-2} \Rightarrow-x+2=y-3 \quad \Rightarrow x+y=5$

$\frac{1}{1+\sqrt{2}}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$

Hence equations of lines passing through origin and having slopes $(\sqrt{2}+1)$ and $(\sqrt{2}-1)$ are $y=(\sqrt{2}+1) x$ and $y=(\sqrt{2}-1) x$.

Their joint equation is $[(\sqrt{2}+1) x-y][(\sqrt{2}-1) x-y]=0$

$\therefore x^2-2 \sqrt{2} x y+y^2=0$

Let's denote the following events:

• $A$: The event that the man reports a 6.
  


• $B$: The event that the die actually shows a 6.

We need to find the probability that the die actually shows a 6 given that the man reports a 6. This is represented as $P(B|A)$. To find this, we can use Bayes' theorem:

$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}$

We are given that the man tells the truth 3 out of 4 times, so:

$P(\text{Truth}) = \frac{3}{4}$

$P(\text{Lie}) = 1 - P(\text{Truth}) = \frac{1}{4}$

If the die actually shows a 6 (event $B$), the probability that the man reports a 6 (event $A$) is:

$P(A|B) = P(\text{Truth}) = \frac{3}{4}$

The probability that the die actually shows any particular number, including 6, is:

$P(B) = \frac{1}{6}$

Next, we need to determine $P(A)$: the total probability that the man reports a 6. This can be calculated using the law of total probability:

$P(A) = P(A|B) \cdot P(B) + P(A|B^c) \cdot P(B^c)$

Where $B^c$ is the complement event that the die does not show a 6. We have:

$P(A|B^c) = P(\text{Lie}) = \frac{1}{4}$

$P(B^c) = 1 - P(B) = 1 - \frac{1}{6} = \frac{5}{6}$

Plugging in the values, we get:

$P(A) = \left( \frac{3}{4} \cdot \frac{1}{6} \right) + \left( \frac{1}{4} \cdot \frac{5}{6} \right)$

$P(A) = \frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3}$

Now we can use Bayes' theorem:

$P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}$

$P(B|A) = \frac{\left( \frac{3}{4} \right) \cdot \left( \frac{1}{6} \right)}{\frac{1}{3}} = \frac{\frac{3}{24}}{\frac{1}{3}} = \frac{3}{24} \cdot 3 = \frac{3}{8}$

Therefore, the probability that the die actually shows a 6 given that the man reports a 6 is:

Option C

$\frac{3}{8}$

${ }^{\text {11}} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_7$

We know that ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$

$$\begin{aligned} & \therefore{ }^{14} \mathrm{C}_{\mathrm{r}}=\left({ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5\right)+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7 \\ & =\left({ }^{12} \mathrm{C}_5+{ }^{12} \mathrm{C}_6\right)+{ }^{13} \mathrm{C}_7 \\ & ={ }^{13} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_7 \\ & \therefore r=7 \\ \end{aligned}$$

We have $a x^2+2 h x y+b y^2=0$ and let $m_1$ and $m_2$ be the slopes of lines.

Now $\mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$ and $\mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}$

$$\begin{aligned} \left(m_1-m_2\right)^2 & =\left(m_1+m_2\right)^2-4 m_1 m_2 \\ & =\left(\frac{-2 h}{b}\right)^2-4\left(\frac{a}{b}\right)=\frac{4 h^2}{b^2}-\frac{4 a}{b}=\frac{4 h^2-4 a b}{b^2}=\frac{4 h^2-3 h^2}{b^2} \ldots \text { [From data given] } \\ & =\frac{h^2}{b^2} \\ \therefore m_1-m_1 & =\frac{h}{b} \end{aligned}$$

Thus we have $m_1+m_2=\frac{-2 h}{b}$ and $m_1-m_2=\frac{h}{b}$.

Solving, we get $m_1=\frac{-h}{2 b}$ and $m_2=\frac{-3 h}{2 b} \Rightarrow m_1: m_2=1: 3$

$$\begin{aligned} & \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)} \\ & \therefore \sin A(\sin B \cos C-\cos B \sin C)=\sin C(\sin A \cos B-\cos A \sin B) \\ & \therefore \sin A \sin B \cos C+\cos A \sin B \sin C \quad=2 \sin A \cos B \sin C \\ & \therefore \sin B(\sin A \cos C+\cos A \sin C)=2 \sin A \cos B \sin C \\ & \therefore \sin B[\sin (A+C)]=2 \sin A \cos B \sin C \\ & \therefore \sin B \sin B=2 \sin A \cos B \sin C \quad \cdots[\because A+B+C=\pi] \end{aligned}$$

$\therefore \frac{1}{2 \cos B}=\frac{\sin A}{\sin B} \times \frac{\sin C}{\sin B}$

By sine rule, we know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$$\begin{aligned} & \therefore \frac{1}{2\left(\frac{c^2+a^2-b^2}{2 a c}\right)}=\frac{a}{b} \times \frac{c}{b} \Rightarrow \frac{a c}{c^2+a^2-b^2}=\frac{a c}{b^2} \\ & \therefore c^2+a^2-b^2=b^2 \Rightarrow 2 b^2=a^2+c^2 \end{aligned}$$

$$\begin{aligned} & y(1+\log x) \frac{d x}{d y}-x \log x=0 \\ & \therefore \operatorname{y}(1+\log x) \frac{d x}{d y}=x \log x \Rightarrow \frac{d y}{d x}=\frac{y(1+\log x)}{x \log x} \\ & \therefore \int \frac{d y}{y}=\int \frac{1+\log x}{x \log x} d x \\ & \therefore \log |y|=\log |x \log x|+\log c \\ & \text { We haye } x=e, y=e^2 \\ & \therefore \log \left|e^2\right|=\log |e \log e|+\log c \\ & 2=1+\log c \quad \Rightarrow \log c=1=\log e \\ & \therefore \log |y|=\log |x \log x|+\log e \\ & y = e x \log x \end{aligned}$$

We have $f(x)=\frac{1-x+x^2}{1+x+x^2}$

$$\begin{aligned} \therefore f^{\prime}(x) & =\frac{\left(1+x+x^2\right)(2 x-1)-\left(1-x+x^2\right)(2 x+1)}{\left(1+x+x^2\right)^2} \\ & =\frac{\left(2 x+2 x^2+2 x^3-x-1-x^2\right)-\left(2 x-2 x^2+2 x^3+1-x+x^2\right)}{\left(1+x+x^2\right)^2} \\ & =\frac{\left(x+x^2+2 x^3-1\right)-\left(x-x^2+2 x^3+1\right)}{\left(1+x+x^2\right)^2} \\ & =\frac{2 x^2-2}{\left(1+x+x^2\right)^2} \text { and when } f^{\prime}(x)=0 \text {, we get } \\ 2\left(x^2-1\right) & =0 \quad x= \pm 1 \end{aligned}$$

When $x=1, f(x)=\frac{1}{3}$ and when $x=-1, f(x)=3$

Hence minimum value of $f(x)$ is $\frac{1}{3}$.

\lim _\limits{x \rightarrow-\frac{\pi^{-}}{2}} f(x)=\lim _\limits{x \rightarrow-\frac{\pi^{-}}{2}}-2 \sin x=-2 \sin \left(-\frac{\pi}{2}\right) \quad=2

\lim _\limits{x \rightarrow \frac{-\pi^{+}}{2}} f(x)=\lim _\limits{x \rightarrow \frac{-\pi^{+}}{2}} a \sin x+b=a \sin \left(\frac{-\pi}{2}\right)+b=-a+b

Since $f(x)$ is continuous at $x=\frac{-\pi}{2}$, we write

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

\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} a \sin x+b=a \sin \left(\frac{\pi}{2}\right)+b=a+b

\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} \cos x=\cos \left(\frac{\pi}{2}\right)=0

Since $f(x)$ is continuous at $x=\frac{\pi}{2}$, we write

$a+b=0 \quad \text{... (2)}$

From (1) and (2), we get $a=-1, b=1$

$\therefore(3 a+2 b)^3=(-3+2)^3=-1$

$$\begin{aligned} & x=a(\theta+\sin \theta) \text { and } y=a(1-\cos \theta) \\ & \therefore \frac{d x}{d \theta}=a(1+\cos \theta) \text { and } \frac{d y}{d \theta}=a(\sin \theta) \\ & \therefore \frac{d y}{d x}=\frac{a \sin \theta}{a(1+\cos \theta)}=\frac{\sin \theta}{1+\cos \theta} \\ & \therefore \frac{d^2 y}{d x^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right)\left(\frac{d \theta}{d x}\right)=\frac{d}{d \theta}\left(\frac{\sin \theta}{1+\cos \theta}\right) \times \frac{1}{\left(\frac{d x}{d \theta}\right)} \\ & =\frac{(1+\cos \theta)(\cos \theta)-\sin \theta(-\sin \theta)}{(1+\cos \theta)^2 a(1+\cos \theta)} \\ & =\frac{\cos \theta+\cos ^2 \theta+\sin ^2 \theta}{a(1+\cos \theta)^3}=\frac{1+\cos \theta}{a(1+\cos \theta)^3}=\frac{1}{a(1+\cos \theta)^2} \\ & \therefore\left(\frac{d^2 y}{d x^2}\right)_{\theta=\frac{\pi}{2}}=\frac{1}{a\left(1+\cos \frac{\pi}{2}\right)^2}=\frac{1}{a} \\ & \end{aligned}$$

$3 x+4 y=9 \Rightarrow 6 x+8 y=18$ and $6 x+8 y=15$ are the two parallel lines.

Distance between the lines $=\frac{|18-15|}{\sqrt{(6)^2+(8)^2}}=\frac{3}{10}=0.3$ units

Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the intercepts on the coordinate axes and we have $\mathrm{a}=\mathrm{b}=\mathrm{c}$.

$\therefore \frac{x}{a}+\frac{y}{a}+\frac{z}{a}=1$ is required equation of plane.

Since plane passes through point $(2,2,2)$, we write

$\frac{2+2+2}{a}=1 \quad \Rightarrow a=6$

$\therefore$ Equation of plane is $x+y+z=6$

To determine the principal solutions of the equation $\cot x = \sqrt{3}$, we start by recalling the definition and properties of the cotangent function. The cotangent function, $\cot x$, is the reciprocal of the tangent function, so we have:

$\cot x = \frac{1}{\tan x}$

Given that $\cot x = \sqrt{3}$, we can rewrite this as: