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

Given the word "ABRACADABRA":

1. Vowels : A, A, A, A, A (5 A's)
2. Consonants : B, R, C, D, B, R (2 B's, 2 R's, 1 C, and 1 D)

When considering arrangements where the vowels are together, we can treat all the vowels as one single entity.

Thus, the entities for our purpose are :

Entity of Vowels (AAAAA), B, R, C, D, B, R

Given the repetitions, the number of ways to arrange these 7 entities is : $\text{Arrangements} = \frac{7!}{2! \times 2!}$

Since all the vowels are identical, there's only 1 way to arrange them within their entity.

Therefore, the total number of arrangements where vowels are together is : $\text{Total} = \frac{7!}{2! \times 2!} \times 1 = 1260$

Thus, the correct answer is $1260$.

The plane passes through the point $(0,7,-7)$ and contains the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$.

$\therefore$ Required equation of the plane is

$$\begin{aligned} & \left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1-0 & 3-7 & -2+7 \\ -3 & 2 & 1 \end{array}\right| \quad=0 \quad \Rightarrow\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & -4 & 5 \\ -3 & 2 & 1 \end{array}\right|=0 \\ & \therefore \hat{\mathrm{i}}(-4-10)-\hat{\mathrm{j}}(-1+15)+\hat{\mathrm{k}}(-2-12)=0 \\ & \therefore-14 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}-14 \hat{\mathrm{k}}=0 \quad \Rightarrow \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}=0 \end{aligned}$$

Hence Cartesian equation of the plane is $x+y+z=0$

Let $$\begin{aligned} & =\int \frac{x+\sin x}{1+\cos x} d x=\int \frac{x+\sin x}{2 \cos ^2 \frac{x}{2}} d x \\ & \quad=\int \frac{x}{2 \cos ^2 \frac{x}{2}} d x+\int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x=\frac{1}{2} \int x \sec ^2 \frac{x}{2} d x+\int \tan \frac{x}{2} d x \\ & \quad=\frac{1}{2}\left[x \tan \frac{x}{2}(2)-\int 2 \tan \frac{x}{2} d x\right]-2 \log \left|\cos \frac{x}{2}\right|+c \\ & \quad=x \tan \frac{x}{2}+2 \log \left|\cos \frac{x}{2}\right|-2 \log \left|\cos \frac{x}{2}\right|+c \quad=x \tan \frac{x}{2}+c \end{aligned}$$

Let the angles be $\mathrm{x}, 2 \mathrm{x}, 3 \mathrm{x}$

$\therefore \mathrm{x}+2 \mathrm{x}+3 \mathrm{x}=180^{\circ} \Rightarrow 6 \mathrm{x}=180^{\circ} \Rightarrow \mathrm{x}=30^{\circ}$

Thus angles of the triangle are $30^{\circ}, 60^{\circ}, 90^{\circ}$.

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

$$\begin{aligned} & \therefore \frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin 90^{\circ}} \Rightarrow \frac{a}{\left(\frac{1}{2}\right)}=\frac{b}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{c}{(1)} \\ & \therefore 2 a=\frac{2 b}{\sqrt{3}}=c \Rightarrow a=\frac{c}{2} \text { and } b=\frac{\sqrt{3} c}{2} \\ & \therefore a: b: c=\frac{c}{2}: \frac{\sqrt{3} c}{2}: c=1: \sqrt{3}: 2 \end{aligned}$$

We have $f(x)=[x]$

Let $[x]=K$, an integer.

\therefore \quad \lim _\limits{x \rightarrow \mathrm{K}^{+}} f(x)=K \text { and } \lim _\limits{x \rightarrow K^{-}} f(x)=K-1

Thus given function is not continuous at all integral values in its domain.

$\because \mathrm{f}$ is discontinuous at $\mathrm{x}=0,1$.

$$\begin{aligned} \mathrm{p} \wedge(\sim \mathrm{p} \vee \sim \mathrm{q}) \wedge \mathrm{q} & =\mathrm{p} \wedge \mathrm{q} \wedge \sim(\mathrm{p} \wedge \mathrm{q}) \\ & =(\mathrm{p} \wedge \mathrm{q}) \wedge \sim(\mathrm{p} \wedge \mathrm{q}) \equiv \mathrm{F}\end{aligned}$$

$$\begin{aligned} & x=a \cos \theta, y=b \sin \theta \\ & \therefore \frac{d x}{d \theta}=-a \sin \theta, \frac{d y}{d \theta}=b \cos \theta \\ & \therefore \frac{d y}{d x}=\frac{b \cos \theta}{-a \sin \theta}=\left(\frac{-b}{a}\right) \cot \theta \\ & \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \cdot \frac{d \theta}{d x} \\ & =\frac{d}{d \theta}\left[\left(\frac{-b}{a}\right) \cot \theta\right] \times \frac{1}{\left(\frac{d x}{d \theta}\right)} \\ & =\frac{\left(\frac{-b}{a}\right)\left(-\operatorname{cosec}^2 \theta\right)}{-a \sin \theta}=\left(\frac{-b}{a}\right) \times \frac{1}{a} \times \frac{1}{\sin ^3 \theta} \\ & \therefore\left(\frac{d^2 y}{d x^2}\right)_{\theta=\frac{\pi}{4}}=\left(\frac{-b}{a^2}\right)\left[\frac{1}{\left(\sin \frac{\pi}{4}\right)^3}\right]=\left(\frac{-b}{a^2}\right)(\sqrt{2})^3=-2 \sqrt{2}\left(\frac{b}{a^2}\right) \\ &\end{aligned}$$

${a_{11}}{A_{11}} + {a_{12}}{A_{12}} + {a_{13}}{A_{13}} = |A|$

$\therefore$ |A| = \left| {\matrix{ 3 & 2 & 4 \cr 1 & 2 & 1 \cr 3 & 2 & 6 \cr } } \right|

$= 3(12 - 2) - 2(6 - 3) + 4(2 - 6)$

$= 30 - 6 - 16 = 8$

As per data given, we write

$24 = {1 \over 6}\left\{ {\left( {\overline a + \overline b } \right)\,.\,\left[ {\left( {\overline b + \overline c } \right) \times \left( {\overline c + \overline a } \right)} \right]} \right\}$

$= {1 \over 6}\left\{ {\left( {\overline a + \overline b } \right)\,.\,\left[ {\left( {\overline b + \overline c } \right) + \left( {\overline b + \overline a } \right) + \left( {\overline c + \overline a } \right)} \right]} \right\}$

... [$\because$ $\overline c \times \overline c = 0$]

$\therefore$ $144 = \left[ {\overline a \,.\,\left( {\overline b \times \overline c } \right)} \right] + \left[ {\overline b \,.\,\left( {\overline b \times \overline c } \right)} \right] + \left[ {\overline a \,.\,\left( {\overline b \times \overline a } \right)} \right] + \left[ {\overline b \,.\,\left( {\overline b \times \overline a } \right)} \right] + \left[ {\overline a \,.\,\left( {\overline c \times \overline a } \right)} \right] + \left[ {\overline b \,.\,\left( {\overline c \times \overline a } \right)} \right]$

$= \left[ {\overline a \,.\,\left( {\overline b \times \overline c } \right)} \right] + 0 + 0 + 0 + 0 + \left[ {\overline b \,.\,\left( {\overline c \times \overline a } \right)} \right]$

$= 2\left[ {\overline a \,.\,\left( {\overline b \times \overline c } \right)} \right]$ ...... [$\because$ $\overline b \,.\,\,\left( {\overline c \times \overline a } \right) = \overline a \,.\,\,\left( {\overline b \times \overline c } \right)$]

$\therefore$ $\overline a \,.\,\,\left( {\overline b \times \overline c } \right) = 72$, i.e. volume of parallelopiped with conterminous edges $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$.

As per data given, we write

$\tan \frac{\pi}{4}=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=1$

Squaring both sides, we get

$$\begin{aligned} & (a+b)^2=4\left(h^2-a b\right) \\ & \therefore 4 h^2=a^2+6 a b+b^2 \end{aligned}$$

S = at$^2$ + bt + 6

$\mathrm{V=\frac{dS}{dt}=2at+b}$ and $\mathrm{A=\frac{d^2S}{dt^2}=2a}$

When particle comes to rest,

S = 16, t = 4, V = 0

$\therefore$ 16 = a(4)$^2$ + b(4) + 6 $\Rightarrow$ 16a + 4b = 10 ..... (1)

Also 0 = 2a(4) + b $\Rightarrow$ b = $-$8a ..... (2)

From (1) and (2), we get

16a + 4(8a) = 10 $\Rightarrow$ $-$16a = 10 $\Rightarrow$ a = $\frac{-5}{8}$

We have acceleration $\mathrm{A=2a=2\left(\frac{-5}{8}\right)=\frac{-5}{4}}$ m/sec$^2$

Refer figure.

Angle made by line L with positive direction of X axis is (90$^\circ$ + 30$^\circ$) i.e. 120$^\circ$.

$\therefore$ Slope of line L = $\tan (120^\circ ) = \tan (\pi - 60^\circ ) = - \tan 60^\circ = - \sqrt 3$

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$

Since given lines intersect, we write

$2 \lambda+1=\mu+3$ ..... (1)

$3 \lambda-1=2 \mu+\mathrm{k}$ .... (2)

$4 \lambda+1=\mu$ ..... (3)

Substituting value of $\mu$ from (3) in (1), we get

$$\begin{aligned} & 2 \lambda+1=(4 \lambda+1)+3 \Rightarrow 2 \lambda=-3 \Rightarrow \lambda=\frac{-3}{2} \\ & \therefore \mu=4\left(\frac{-3}{2}\right)+1=-6+1=-5 \end{aligned}$$

Substituting values of $\lambda$ and $\mu$ in (2), we get

$3\left(\frac{-3}{2}\right)-1=2(-5)+k \quad \Rightarrow k=\frac{9}{2}$

Here equation of line OC is $y=x$ i.e. $x-y=0$ and equation of line AB is $x=5$ i.e. $x-5=0$

Equation of line BC is $y=3$ i.e. $y-3=0$

Hence constraints for the shaded region are $x, y \geq 0, x-5 \leq 0, x-y \geq 0, y-3 \leq 0$ i.e. $x, y \geq 0, x \leq 5, x-y \geq 0, y \leq 3$

We have to find variance of numbers $2,3,4, \ldots, 11$.

$$\begin{aligned} & \text { Here } \quad \bar{x}=\frac{2+3+4+\ldots+11}{10}=\frac{65}{10}=6.5 \\ & \text { Variance }=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n} \\ & =\frac{\left[(-4.5)^2+(-3.5)^2+(-2.5)^2+(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2+(2.5)^2+(3.5)^2+(4.5)^2\right]}{10} \\ & =\frac{2(20.25+12.25+6.25+2.25+0.25)}{10}=\frac{41.25}{5}=8.25 \end{aligned}$$

From given data, we write

$$\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{array}\right|=0 \\ & \therefore \hat{\mathrm{i}}(6 \mu-27 \lambda)-\hat{\mathrm{j}}(2 \mu-27)+\hat{\mathrm{k}}(2 \lambda-6)=0 \\ \end{aligned}$$

$\therefore$ 6$\mu$ $-$ 27$\lambda$ = 0 ..... (1)

27 $-$ 2$\mu$ = 0 ..... (2)

2$\lambda$ $-$ 6 = 0 ..... (3)

From (2) and (3), we get $\mu=\frac{27}{2}$ and $\lambda=3$.

These values of $\lambda$ & $\mu$ satisfy eq. (1)

Let (0, K) be the centre of the circle.

Since the circle passes through origin, we write

$$\begin{aligned} & (x-0)^2+(y-K)^2=K^2 \\ & \therefore x^2+y^2-2 K y+K^2=K^2 \\ & \therefore x^2+y^2-2 K y=0\quad ... (1) \end{aligned}$$

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

$2 x+2 y \frac{d y}{d x}-2 K \frac{d y}{d x}=0$

$\therefore K=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}$ and substituting value of K in (1), we get

$$\begin{aligned} & x^2+y^2-2\left(\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}\right) y=0 \\ \therefore & \left(x^2+y^2\right) \frac{d y}{d x}-2 x y-2 y^2 \frac{d y}{d x}=0 \quad \Rightarrow\left(x^2-y^2\right) \frac{d y}{d x}-2 x y \quad=0\end{aligned}$$

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

Here $x+|x| \geq 0$ and $\sqrt{x+|x|} \neq 0$

$\therefore x+|x|>0$

Now when $x>0, x+|x|=x+x \Rightarrow 2 x>0$.

When $\mathrm{x}<0, \mathrm{x}+|\mathrm{x}|=\mathrm{x}-\mathrm{x}=0$

$\therefore \mathrm{x}>0$ is the required domain.

$$\begin{aligned} \text { Let I } & =\tan ^{-1}(\sec x+\tan x) d x \\ & \left.=\int \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x=\int \tan ^{-1} x\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right]\right]_{d x} \\ & =\int \tan ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right) d x=\int \tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) d x \\ & =\int \tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right] d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x \\ & =\frac{\pi x}{4}+\frac{x^2}{4}+c\end{aligned}$$

$\left|\overline{\mathrm{e}}_1+\overline{\mathrm{e}}_2\right|^2=\left|\overline{\mathrm{e}}_1^2\right|+\left|\overline{\mathrm{e}}_2^2\right|+2 \overline{\mathrm{e}}_1 \overline{\mathrm{e}}_2 \cos \theta$

Here $\left|\overline{\mathrm{e}}_1\right|=1,\left|\overline{\mathrm{e}}_2\right|=1$ and $\left|\overline{\mathrm{e}}_1+\overline{\mathrm{e}}_2\right|=1$

$$\begin{aligned} & \therefore(1)=(1)^2+(1)^2+2(1)(1) \cos \theta \\ & \therefore \quad \frac{-1}{2}=\cos \theta \quad \Rightarrow \theta=120^{\circ} \end{aligned}$$

Let $x$ be the one part and $y$ be the other part.

We have $x+y=20 \Rightarrow y=20-x$

As per condition given, we write

$$\begin{aligned} & f(x)=(20-x) x^3 \\ & \quad=20 x^3-x^4 \\ & \therefore f^{\prime}(x)=60 x^2-4 x^3 \end{aligned}$$

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

$$\begin{aligned} & 4 x^2(15-x)=0 \Rightarrow x=0,15 \\ & f^{\prime \prime}(x)=120 x-12 x^2 \\ & {\left[f^{\prime \prime}(x)\right]_{x=15}=(120)(15)-(12)(15)^2=-900<0} \\ & \therefore f(x) \text { is maximum when } x=15 \end{aligned}$$

$\therefore \mathrm{f}(\mathrm{x})$ is maximum when $\mathrm{x}=15$.

$\therefore y=5 \Rightarrow x y=(15)(5)=75$

$$\begin{aligned} & \text { Here }\left|\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right|=0 \\ & \therefore(6-28)-2(-24-14)+x(16+2)=0 \\ & \therefore-22+76+18 x=0 \Rightarrow x=-3\end{aligned}$$

Given that the rate of melting ($\frac{dx}{dt}$) is proportional to the amount of ice ($x$) present at that moment, we can write the differential equation as :

$\frac{dx}{dt} = -kx$

(Here the negative sign indicates melting or decrease in quantity)

Let's solve this ordinary differential equation :

Separate the variables :

$\frac{dx}{x} = -k dt$

Now, integrate both sides :

$\int \frac{dx}{x} = -k \int dt$

This gives :

$\ln|x| = -kt + C$

Where $C$ is the constant of integration.

Given that half the quantity of ice melts in 20 minutes, we have :

$x = \frac{x_0}{2}$ when $t = 20$

Plug this into our equation to find $C$ :

$\ln\left(\frac{x_0}{2}\right) = -20k + C$

Now, using the initial condition that $x = x_0$ when $t = 0$ :

$\ln(x_0) = C$

From these two, we get :

$\ln\left(\frac{x_0}{2}\right) = \ln(x_0) - 20k$

Which gives :

$k = \frac{\ln 2}{20}$

After 40 minutes, let $x = Kx_0$, plug into the equation :

$\ln(Kx_0) = -\frac{\ln 2}{20} \times 40 + \ln(x_0)$

$\ln(Kx_0) = -2\ln 2 + \ln(x_0)$

$\ln(K) = -2\ln 2$

$K = e^{-2\ln 2}$

$K = e^{\ln(2^{-2})}$

$K = 2^{-2}$

$K = \frac{1}{4}$

So, the correct answer is :

Option C :

$K = \frac{1}{4}$

We have $n=4$ and $P(x=0)=\frac{16}{81}$

$$\begin{aligned} & \therefore \frac{16}{81}={ }^4 C_0(p)^4(q)^0 \\ & \therefore \frac{16}{81}=(p)^4 \quad \Rightarrow P=\frac{2}{3} \quad \Rightarrow q=\frac{1}{3} \\ & \therefore P(x=4)={ }^4 C_4(p)^0(q)^4=(1)(1)\left(\frac{1}{3}\right)^4=\frac{1}{81} \end{aligned}$$

The sum of numbers on the two dice can be $(2,3, \ldots, 12)$ and prime numbers in this list are 2, 3, 5, 7, 11.

$$\begin{aligned} & 2 \Rightarrow(1,1) \text { and } 3 \Rightarrow(1,2),(2,1) \quad \text { and } \quad 5 \Rightarrow(1,4),(2,3),(3,2),(4,1), \\ & 7 \Rightarrow(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \quad \text { and } 11 \Rightarrow(5,6),(6,5)\end{aligned}$$

Thus $\mathrm{n(s)=6 \times 6=36}$ and $\mathrm{n(E)=15}$

Hence required probability $=\frac{15}{36}=\frac{5}{12}$