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

We have to choose 3 consonants and 2 vowels from 7 consonants and 4 vowels.

$$\begin{aligned} \therefore \text { Number of words } & =\left({ }^7 C_3\right) \times\left({ }^4 C_2\right) \times 5! \\ & =\frac{7!}{3!4!} \times \frac{4!}{2!2!} \times 5!=25200 \end{aligned}$$

Let $\mathrm{p}: 3+6>8$ and $\mathrm{q}: 2+3<6$

The logical form of given statement is $\mathrm{p} \wedge \mathrm{q}$.

$\therefore \sim(p \wedge q) \equiv \sim p \vee \sim \text { q i.e. } 3+6 \leq 8 \text { or } 2+3 \geq 6$

$f(x)=e^{\frac{1}{x}}$

$f^{\prime}(x)=e^{\frac{1}{x}}(-1)\left(\frac{-1}{x^2}\right)=\frac{1}{x^2 e^{\frac{1}{x}}}$

When $f^{\prime}(x)>0, x^2 e^{\frac{1}{x}}>0$ and $x \neq 0$.

Now $x^2>0$ and $\mathrm{e}^{\frac{1}{x}}>0$ for $\forall x \in R$.

Hence $f(x)$ is an increasing function for $\forall x$, except $x=0$.

We have $\mathrm{n}=16, \Sigma \mathrm{x}_{\mathrm{i}}=528$, and $\frac{528}{16}=33 \Rightarrow \overline{\mathrm{x}}=33$ and $\Sigma\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2=9158$

Variance $=\frac{1}{n} \Sigma\left(x_i-\bar{x}\right)^2=\frac{9158}{16}=572.375$

We can use the triple angle formula for sine: $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$. Substituting this into the given equation, we get:

$3 \sin \theta = 2 (3 \sin \theta - 4 \sin^3 \theta)$

Simplifying, we get:

$3 \sin \theta = 6 \sin \theta - 8 \sin^3 \theta$

$8 \sin^3 \theta - 3 \sin \theta = 0$

$\sin \theta (8 \sin^2 \theta - 3) = 0$

This gives us two possibilities:

$\sin \theta = 0$ or

$8 \sin^2 \theta - 3 = 0$

If $\sin \theta = 0$, then $\theta = 0$ or $\theta = \pi$, but these values are outside the given interval $0 < \theta < \pi$.

So we must have $8 \sin^2 \theta - 3 = 0$. Solving for $\sin \theta$, we get:

$\sin^2 \theta = \frac{3}{8}$

$\sin \theta = \pm \sqrt{\frac{3}{8}} = \pm \frac{\sqrt{3}}{2 \sqrt{2}}$

Since $0 < \theta < \pi$, $\sin \theta$ is positive. Therefore, $\sin \theta = \boxed{\frac{\sqrt{3}}{2 \sqrt{2}}}$.

$f(x)=\frac{(1+\cos x)-(\sin x)}{(1+\cos x)+(\sin x)}$, is continuous at $x=\pi$.

\begin{aligned} \lim _\limits{x \rightarrow \pi} f(x) & =\lim _\limits{x \rightarrow \pi} \frac{\left(2 \cos ^2 \frac{x}{2}\right)-\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{\left(2 \cos ^2 \frac{x}{2}\right)+\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}=\lim _\limits{x \rightarrow \pi} \frac{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)} \\ & =\lim _\limits{x \rightarrow \pi} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\lim _\limits{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan \left(\frac{\pi}{4}-\frac{\pi}{2}\right)=-1 \end{aligned}

$$\begin{aligned} & \text { Let } \quad I=\int e^{e^x+x} d x=\int e^{e^x} \cdot e^x d x \\ & \text { Put } \quad e^x=t \Rightarrow e^x d x=d t \\ & \therefore I=\int e^t d t=e^t+c=e^{e^x}+c \end{aligned}$$

$$\begin{aligned} & \tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]=\tan \left[\tan ^{-1}\left(\frac{\left(\frac{3}{4}\right)+\left(\frac{2}{3}\right)}{1-\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{17}{6}\right)\right]=\frac{17}{6} \end{aligned}$$

$$\begin{aligned} & {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]} \\ & R_3 \rightarrow R_3-3 R_1 \text { and } R_2 \rightarrow R_2-2 R_1 \\ & {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 6 & -7 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 1 \\ -1 \\ -1 \end{array}\right]} \\ & \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2 \\ & {\left[\begin{array}{ccc} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 5 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 1 \\ -1 \\ 5 \end{array}\right]} \\ & \therefore \mathrm{x}_1-\mathrm{x}_2+\mathrm{x}_3=1 \\ & \mathrm{x}_2-2 \mathrm{x}_3=-1 \\ & 5 x_3=5 \\ & \therefore \mathrm{x}_3=1, \mathrm{x}_2=1, \mathrm{x}_1=1 \\ \end{aligned}$$

We have $\tan \alpha=\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]$

$$\begin{aligned} & \text { Put } \mathrm{x}^2=\cos \theta \\ & \therefore \tan \alpha=\frac{\sqrt{2 \cos ^2 \frac{\theta}{2}}-\sqrt{2 \sin ^2 \frac{\theta}{2}}}{\sqrt{2 \cos ^2 \frac{\theta}{2}}+\sqrt{2 \sin ^2 \frac{\theta}{2}}} \\ & \therefore \tan \alpha=\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}=\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right) \\ & \therefore \alpha=\frac{\pi}{4}-\frac{\theta}{2} \Rightarrow 2 \alpha=\frac{\pi}{2}-\theta \\ & \therefore \sin 2 \alpha=\sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta=x^2 \end{aligned}$$

$$\begin{aligned} & \lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\log x} \\ & =\lim _{x \rightarrow 1} \frac{\frac{1}{2 \sqrt{x}}}{\left(\frac{1}{x}\right)} \quad \text{... [L' Hospital Rule]}\\ & =\frac{1}{2} \end{aligned}$$

The equations of two required lines are $y=\sqrt{3} x$ and $y=\frac{1}{\sqrt{3}} x$

Their combined equation is $(\sqrt{3} x-y)(x-\sqrt{3} y)=0$

i.e. $\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$

By Newton's law of cooling, we write

$$\begin{aligned} & \frac{\mathrm{d} \theta}{\mathrm{dt}} \propto\left(\theta-\theta_0\right) \\ & \therefore \frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta_0\right) \quad \Rightarrow \int\left(\frac{\mathrm{d} \theta}{\theta-\theta_0}\right)=\int \mathrm{kdt} \end{aligned}$$

$$\begin{aligned} & \therefore \log \left|\theta-\theta_0\right| \quad=k t+c \quad \text{... (1)}\\ & \text { When } t=0, \theta=80 \text { and } \theta_0=25 \\ & \therefore \log |80-25|=0+c \quad \Rightarrow \quad c=\log |55| \quad \text{... (2)}\\ & \text { When } t=30, \theta=50 \\ & \therefore \log |50-25|=30 \mathrm{k}+\log |55| \end{aligned}$$

$\therefore \mathrm{k}=\frac{1}{30} \log \left|\frac{5}{11}\right| \quad \text{... (3)}$

From (1), (2), (3) we write

$\log \left|\theta-\theta_0\right|=\frac{1}{30} \log \left|\frac{5}{11}\right| t+\log |55|$

When $t=60$, we get

$$\begin{aligned} & \text { When } t=60 \text {, we get } \\ & \log \left|\theta-\theta_0\right|=\frac{60}{30} \log \left|\frac{5}{11}\right|+\log |55|=\log \left(\left|\frac{5}{11}\right|\right)^2+\log |55|=\log \left|\frac{25}{121} \times 55\right|=\log \left|\frac{125}{11}\right| \\ & \therefore \theta-25=\frac{125}{11} \Rightarrow \theta=36.36^{\circ} \mathrm{C} \end{aligned}$$

Let $\mathrm{p}:-7$ is an integer.

$\mathrm{q}: \sqrt{-7}$ is a complex number.

Logical form of $S_1 \equiv p \rightarrow q$

Logical form of $S_2 \equiv \sim p \vee q$

We know that $\mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \vee \mathrm{q} . \quad \therefore \mathrm{S}_1$ and $\mathrm{S}_2$ are equivalent statements.

Let $\quad I=\int_0^{\frac{\pi}{2}} \frac{\cos x}{3 \cos x+\sin x} d x$

Put $\cos x=A(3 \cos x+\sin x)+B \frac{d}{d x}(3 \cos x+\sin x)$

$=A(3 \cos x+\sin x)+B(-3 \sin x+\cos x)$

$=\cos x(3 A+B)+\sin x(A-3 B)$

Thus $3 \mathrm{~A}+\mathrm{B}=1$ and $\mathrm{A}-3 \mathrm{~B}=0$

Solving, we get $B=\frac{1}{10}, A=\frac{3}{10}$

\therefore I=\int_\limits0^{\frac{\pi}{2}} \frac{\frac{3}{10}(3 \cos x+\sin x)+\frac{1}{10}(-3 \sin x+\cos x)}{3 \cos x+\sin x} d x

=\frac{3}{10} \int_\limits0^{\frac{\pi}{2}} d x+\frac{1}{10} \int_\limits0^{\frac{\pi}{2}} \frac{\frac{d}{d x}(3 \cos x+\sin x)}{3 \cos x+\sin x} d x

$$\begin{aligned} & =\frac{3}{10}[x]_0^{\frac{\pi}{2}}+\frac{1}{10}[\log |3 \cos x+\sin x|]_0^{\frac{\pi}{2}}=\frac{3}{10}\left(\frac{\pi}{2}\right)+\frac{1}{10}[\log |1|-\log |3|] \\ & =\frac{3 \pi}{20}-\frac{1}{10} \log 3 \end{aligned}$$

To find the general solution of the differential equation

$\frac{d x}{d t}=\frac{x \log x}{t}$,

we will follow these steps:

First, separate the variables. Rewrite the given differential equation as:

$\frac{1}{x \log x} \frac{d x}{d t} = \frac{1}{t}$.

Now, separate the variables:

$\frac{1}{x \log x} \, dx = \frac{1}{t} \, dt$.

Integrate both sides:

$\int \frac{1}{x \log x} \, dx = \int \frac{1}{t} \, dt$.

For the left-hand side, let $u = \log x$. Therefore, $du = \frac{1}{x} dx$. Substitute $u$ and $du$ into the integral:

$\int \frac{1}{x \log x} \, dx = \int \frac{1}{u} \, du = \log u + C_1 = \log (\log x) + C_1$.

For the right-hand side, the integral is straightforward:

$\int \frac{1}{t} \, dt = \log t + C_2$.

Now, combine the results:

$\log (\log x) + C_1 = \log t + C_2$.

To simplify the constants, we can combine them into a single arbitrary constant, say $C$:

$\log (\log x) = \log t + C$.

Exponentiate both sides to eliminate the logarithms:

$\log x = t e^C = c t$.

We obtain the general solution:

$\log x = c t.$

Therefore, exponentiating both sides yields:

$x = e^{c t}.$

So the correct option is:

Option D

We have $\bar{a}=3 \hat{j}, \bar{b}=4 \hat{k}$ and $\bar{c}=3 \hat{j}+4 \hat{k}$

Angle bisector $\mathrm{AD}$ divides $\mathrm{BC}$ in the ratio $\mathrm{AB}$ : $\mathrm{AC}$.

$\mathrm{A B}=\sqrt{9+16}=5 \quad$ and $\quad \mathrm{A C}=\sqrt{0+16}=4$

Thus D divides $\mathrm{BC}$ in the ratio $5: 4$

$$\begin{aligned} \therefore \mathrm{D} & =\left[0, \frac{5(3)}{5+4}, \frac{5(4)+4(4)}{5+4}\right] \\ & =\left(0, \frac{15}{9}, \frac{20+16}{9}\right)=\left(0, \frac{5}{3}, 4\right) \Rightarrow \frac{5}{3} \hat{j}+4 \hat{k} \end{aligned}$$

Let the two parts of 10 be $x$ and $(10-x)$.

$f(x)=2(10-x)+x^2=x^2-2 x+20$

$f^{\prime}(x)=2 x-2$ and when $f^{\prime}(x)=0$, we get $x=1$

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

$\therefore f(x)$ is minimum at $x=1$.

Thus the parts are 1, 9.

Let $z=x+i y$ and $|z+1|=1$

$$\begin{aligned} & \therefore|(x+1)+i y|=1 \\ & \therefore \sqrt{(x+1)^2+y^2}=1 \Rightarrow(x+1)^2+y^2=1 \end{aligned}$$

This is an equation of a circle with centre $(-1,0)$ and radius 1.

Refer figure.

Required area is shaded.

Point of intersection of $x=a$ and $y^2=4 a x$, is $y^2=4 a^2 \quad \Rightarrow y= \pm 2 a$ and $x=a \Rightarrow(a, \pm 2 a)$

\therefore A=2 \int_\limits\limits0^a(2 \sqrt{a} \sqrt{x}) d x

=4 \sqrt{a} \int_\limits0^a x^{\frac{1}{2}} d x=4 \sqrt{a}\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_0^a=(4 \sqrt{a})\left(\frac{2}{3}\right)(a \sqrt{a})=\frac{8}{3} a^2 \text { sq. units }

$x=a\left(t-\frac{1}{t}\right) \quad \text { and } \quad y=b\left(t+\frac{1}{t}\right) \quad \text{.... (1)}$

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

$\text { Now } \mathrm{x}=\mathrm{a}\left(\frac{\mathrm{t}^2-1}{\mathrm{t}}\right) \text { and } \quad \mathrm{y}=\mathrm{b}\left(\frac{\mathrm{t}^2+1}{\mathrm{t}}\right) \quad \ldots \text { [From (1)] }$

$\therefore\left(t^2-1\right)=\left(\frac{x}{a}\right) t$ and $\left(t^2+1\right)=\left(\frac{y}{b}\right)(t)$

$\therefore$ Eq. (2) becomes

$\frac{d y}{d x}=\left(\frac{b}{a}\right)\left(\frac{x}{a}\right)(t) \times \frac{1}{\left(\frac{y}{b}\right) \times t}=\frac{b^2 x}{a^2 y}$

To find the coordinates of the center of the circle passing through the points $(0,0), (x, 0)$, and $(0, y)$, we can follow these steps.

First, let's understand that the general equation of a circle with center $(h, k)$ and radius $r$ can be written as:

$(x - h)^2 + (y - k)^2 = r^2$

Since the circle passes through the origin $(0,0)$, substituting $(0,0)$ in the above equation, we get:

$h^2 + k^2 = r^2 \quad \text{(1)}$

Next, since the circle passes through $(x,0)$, substituting $(x,0)$ in the circle equation, we get:

$(x - h)^2 + k^2 = r^2 \quad \text{(2)}$

Substituting $r^2$ from equation (1) into equation (2) gives:

$(x - h)^2 + k^2 = h^2 + k^2$

Canceling $k^2$ from both sides, we get:

$(x - h)^2 = h^2$

Solving for $h$, we get two possible solutions:

$x - h = h \quad \text{or} \quad x - h = -h$

$x = 2h \quad \text{or} \quad x = 0$

Since $x \neq 0$ in our case, we have:

$h = \frac{x}{2}$

Similarly, since the circle passes through $(0, y)$, substituting $(0, y)$ into the circle equation, we get:

$h^2 + (y - k)^2 = r^2 \quad \text{(3)}$

Substituting $r^2$ from equation (1) into equation (3) gives:

$h^2 + (y - k)^2 = h^2 + k^2$

Canceling $h^2$ from both sides, we get:

$(y - k)^2 = k^2$

Solving for $k$, we get two possible solutions:

$y - k = k \quad \text{or} \quad y - k = -k$

$y = 2k \quad \text{or} \quad y = 0$

Since $y \neq 0$ in our case, we have:

$k = \frac{y}{2}$

Therefore, the coordinates of the center of the circle are:

$\left(\frac{x}{2}, \frac{y}{2}\right)$

So, the correct answer is:

$\left(\frac{x}{2}, \frac{y}{2}\right)$

d.r. of line through $P Q$ are $[(2-1),(3-2),(4-3)]$ i.e. $(1,1,1)$ Hence required equation of line is $(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$

Let $\bar{a}_1=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{a}}_2=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

$\therefore \bar{a}_2-\bar{a}_1=-\hat{i}-6 \hat{j}-4 \hat{k}$ and let $\bar{b}=3 \hat{i}+5 \hat{j}+2 \hat{k}$

$$\begin{aligned} & \quad \overline{\mathrm{b}} \times\left|\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right|=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 5 & 2 \\ -1 & -6 & -4 \end{array}\right|=\hat{\mathrm{i}}(-20+12)-\hat{\mathrm{j}}(-12+2)+\hat{\mathrm{k}}(-18+5)=-8 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \\ & \therefore\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|=\sqrt{64+100+169}=\sqrt{333} \\ & \text { Also }|\overline{\mathrm{b}}|=\sqrt{9+25+4}=\sqrt{38} \end{aligned}$$

$\therefore$ Distance between given lines $=\sqrt{\frac{333}{38}}$ units.

$2 x+y-2 z=18$

Dividing both sides by $\sqrt{(2)^2+(1)^2+(-2)^2}=3$, we get

$\left(\frac{2}{3}\right) x+\left(\frac{1}{3}\right) y-\left(\frac{2}{3}\right) z=6$

Hence length of perpendicular from origin to the plane is 6.

Therefore coordinates of foot of perpendicular are $\left[\left(\frac{2}{3}\right)(6),\left(\frac{1}{3}\right)(6),\left(\frac{-2}{3}\right)(6)\right]$ i.e. $(4,2,-4)$