MHT CET 2021 21st September Evening Shift (Mathematics) Question Paper with Solutions

$p \rightarrow(\sim p \vee q) \equiv F$

We know that $\mathrm{T} \rightarrow \mathrm{F} \equiv \mathrm{F}$

$\therefore \mathrm{p} \equiv \mathrm{T} \text { and }(\sim \mathrm{p} \vee \mathrm{q}) \equiv \mathrm{F} \text {. }$

We know that $F \vee F \equiv F$

$\therefore \sim p \equiv F \text { and } q \equiv F$

Thus $\mathrm{p}, \mathrm{q}$ are $\mathrm{T}, \mathrm{F}$ respectively.

$$\begin{aligned} \bar{a}+\lambda \bar{b} & =(\hat{i}+2 \hat{j}-3 \hat{k})+\lambda(3 \hat{i}-\hat{j}+2 \hat{k}) \\ & =(1+3 \lambda) \hat{i}+(2-\lambda) \hat{j}+(-3+2 \lambda) \hat{k} \end{aligned}$$

Since $\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$ is $\perp$ er to $\overline{\mathrm{c}}$, we write

$$\begin{aligned} & (1)(1+3 \lambda)+(3)(2-\lambda)+(1)(-3+2 \lambda)=0 \\ & \therefore 1+3 \lambda+6-3 \lambda-3+2 \lambda=0 \quad \Rightarrow \quad \lambda=-2 \end{aligned}$$

\begin{aligned} & \lim _\limits{x \rightarrow 5} \frac{x^k-5^k}{x-5}=500 \\ & \therefore(k)(5)^{k-1}=500 \\ & \quad=4(125)=4(5)^3=4(5)^{4-1} \\ & \therefore k=4\end{aligned}

Refer figure

Required area is shaded.

Point of intersection of $y^2=x$ and $y=4$ is $A \equiv(16,4)$.

\begin{aligned} \therefore A & =\int_\limits0^4 y^2 d y \\ & =\left[\frac{y^3}{3}\right]_0^4=\frac{64}{3} \text { sq. units } \end{aligned}

$$\begin{aligned} & \text { We have }\left(\frac{d^2 y}{d x^2}\right)^5+\frac{4\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d^3 y}{d x^3}\right)}+\left(\frac{d^3 y}{d x^3}\right)=x^2-1 \\ & \therefore\left(\frac{d^3 y}{d x^3}\right)\left(\frac{d^2 y}{d x^2}\right)^5+4\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d^3 y}{d x^3}\right)^2=\left(x^2-1\right)\left(\frac{d^3 y}{d x^3}\right) \\ & \therefore \text { order }=3, \text { degree }=2\end{aligned}$$

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

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

$2\left(x^2-1\right)=0 \quad \Rightarrow \quad x= \pm 1$

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

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

We have lines $x^2-4 x y+y^2=0$ and slopes of the lines are $\tan \alpha$ and $\tan \beta$.

$\therefore \tan \alpha+\tan \beta=4$ and $\tan \alpha \cdot \tan \beta=1$

$\therefore \tan \alpha=\frac{1}{\tan \beta}=\cot \beta$.

$\therefore \cot \beta+\tan \beta=4$ and squaring, we get $\cot ^2 \beta+\tan ^2 \beta+2=16 \Rightarrow \tan ^2 \beta+\cot ^2 \beta=14$

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

Half life period $=5$ years.

Initial quantity of substance $=64$ grams

$\therefore$ Quantity left after 5 years $=\frac{64}{2}=32$ grams

$\therefore$ Quantity left after 10 years $=\frac{32}{2}=16$ grams

$\therefore$ Quantity left after 15 years $=\frac{16}{2}=8$ grams

We have probability of cases getting solved $=25 \%=\frac{1}{4}$

$\therefore \mathrm{p}=\frac{1}{4} \Rightarrow \mathrm{q}=\frac{3}{4}$ and we have $\mathrm{n}=6, \mathrm{x}=5,6$

Hence required probability

$$\begin{aligned} & =\left[{ }^6 \mathrm{C}_5\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^1\right]+\left[{ }^6 \mathrm{C}_6\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^0\right] \\ & =\frac{(6)(3)}{(4)^6}+\frac{1}{(4)^6}=\frac{19}{(4)^6}=\frac{19}{4096} \end{aligned}$$

Since circles touch $\mathrm{Y}$ axis at origin, the centres of the circles lie on $\mathrm{X}$ axis.

Let centre be $(h, 0)$ and radius $=h$.

$\therefore(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-0)^2=\mathrm{h}^2 \Rightarrow \mathrm{x}^2-2 \mathrm{hx}+\mathrm{y}^2=0$ ..... (1)

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

$2 x-2 h+2 y \frac{d y}{d x}=0 \quad \Rightarrow \quad x+y \frac{d y}{d x}=h$

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

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

We have $y=x^3-3 x^2-9 x+5$

$\therefore \frac{d y}{d x}=3 x^2-6 x-9$ and since tangent is parallel to $X$ axis, we write

$3 x^2-6 x-9=0 \quad \Rightarrow \quad x^2-2 x-3=0 \text { i.e. }(x-3)(x+1)=0 \Rightarrow \quad x=-1,3$

We have $\mathrm{A}=\{10,11,12,14,26\}$ and $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{N}$, where $\mathrm{f}(\mathrm{a})=$ Highest prime factor of $\mathrm{a}$, when $\mathrm{a} \in \mathrm{A}$.

Now $10=2 \times 5 \Rightarrow$ Highest prime factor is 5 $11=1 \times 11 \Rightarrow$ Highest prime factor is 11 $12=2 \times 2 \times 3 \Rightarrow$ Highest prime factor is 3 $14=2 \times 7 \quad \Rightarrow$ Highest prime factor is 7 $26=2 \times 13 \Rightarrow$ Highest prime factor is 13 $\therefore \mathrm{f}=\{3,5,7,11,13\}$

$$\begin{aligned} & {\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right] A^{-1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \\ & R_1 \rightarrow R_1+R_2, R_3 \rightarrow R_3-3 R_2 \\ & {\left[\begin{array}{ccc} 1 & 3 & 5 \\ 1 & 2 & 3 \\ 0 & -5 & -8 \end{array}\right] A^{-1}=\left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & -3 & 1 \end{array}\right]} \\ & R_2 \rightarrow R_2-R_1 \\ & {\left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & -1 & -2 \\ 0 & -5 & -8 \end{array}\right] A^{-1}=\left[\begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right]} \\ & R_1 \rightarrow R_1+3 R_2 \text { and } R_3 \rightarrow R_3-5 R_2 \\ & {\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & -1 & -2 \\ 0 & 0 & 2 \end{array}\right] A^{-1}=\left[\begin{array}{ccc} -2 & 1 & 0 \\ -1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right]} \\ & R_2 \rightarrow R_2+R_3 \text { and } R_1 \rightarrow 2 R_1+R_3 \\ & {\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{array}\right] A^{-1}=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 4 & -3 & 1 \\ 5 & -3 & 1 \end{array}\right]} \\ & R_1 \rightarrow \frac{1}{2} R_1, R_2 \rightarrow-R_2, R_3 \rightarrow \frac{1}{2} R_3 \\ & {\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A^{-1}=\left[\begin{array}{ccc} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{array}\right]} \end{aligned}$$

$$\begin{aligned} & \mathrm{y}=2 \sin \mathrm{x}+3 \cos \mathrm{x} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \cos \mathrm{x}-3 \sin \mathrm{x} \\ & \therefore \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=-2 \sin \mathrm{x}-3 \cos \mathrm{x}=-(2 \sin \mathrm{x}+3 \cos \mathrm{x})=-\mathrm{y} \\ & \therefore \mathrm{y}+\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=0 \\ & \text { We have } \mathrm{y}+\mathrm{A} \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=B \Rightarrow \mathrm{A}=1, \mathrm{~B}=0\end{aligned}$$

We will check all 3 equations.

I $: \frac{d y}{d x}=\frac{y+x}{x} \quad \Rightarrow$ All terms have same degree equal to 1 .

II $: \frac{d y}{d x}=\frac{x^2+y}{y^3} \quad \Rightarrow$ Here all terms have different degrees.

III $: \frac{d y}{d x}=\frac{2 x y}{y^2-x^2} \Rightarrow$ All terms have same degree equal to 2 .

$$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) & =\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x} \\ & =\lim _{x \rightarrow 0^{-1}} \frac{[(\sqrt{1+p x})-(\sqrt{1-p x})][(\sqrt{1+p x})+(\sqrt{1-p x})]}{x[(\sqrt{1+p x})+(\sqrt{1-p x})]} \\ & =\lim _{x \rightarrow 0^{-}} \frac{[(1+p x)-(1-p x)]}{x[\sqrt{1+p x}+\sqrt{1-p x}]}=\lim _{x \rightarrow 0^{-}} \frac{2 p}{\sqrt{1+p x}+\sqrt{1-p x}} \\ & =\frac{2 p}{2}=p \\ \lim _{x \rightarrow 0^{+}} f(x) & =\lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-2}=\frac{1}{-2} \end{aligned}$$

Since $f(x)$ is continuous at $x=0$, we get $p=\frac{-1}{2}$

\begin{aligned} & \text { Let } I=\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x \quad \text{.....(1)}\\ & =\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)}{1+e^{\left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)}} d x=\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos (-x)}{1+e^{-x}} d x=\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\left(\frac{1}{e^x}\right)} d x \\ & =\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{e}^{\mathrm{x}} \cos \mathrm{x}}{1+\mathrm{e}^{\mathrm{x}}} \mathrm{dx} \quad \text{.....(2)}\\ & \text { Eq. (1) }+(2) \text { gives } \\ & 2 I=\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x\left(1+e^x\right)}{\left(1+e^x\right)} d x=\int_\limits{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x d x=2 \int_\limits0^{\frac{\pi}{2}} \cos x d x \\ & \therefore I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {\cos xdx = [\sin x]_0^{{\pi \over 2}} = 1} \\ & \end{aligned}

Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the direction cosines of the required plane.

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

$$\begin{aligned} & \therefore \mathrm{a}(\mathrm{x}+1)+\mathrm{b}(\mathrm{y}-3)+\mathrm{c}(\mathrm{z}+2)=0 \quad \text{......(1)}\\ & \therefore \mathrm{a}(0+1)+\mathrm{b}(7-3)+\mathrm{c}(-7+2)=0 \quad \Rightarrow \mathrm{a}+4 \mathrm{~b}-5 \mathrm{c}=0 \quad \text{......(2)} \end{aligned}$$

Also $-3 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=0$ ..... (3)

From (2) and (3), we write

$$\begin{aligned} & \frac{a}{\left|\begin{array}{cc} 4 & -5 \\ 2 & 1 \end{array}\right|}=\frac{-b}{\left|\begin{array}{cc} 1 & -5 \\ -3 & 1 \end{array}\right|}=\frac{c}{\left|\begin{array}{cc} 1 & 4 \\ -3 & 2 \end{array}\right|} \\ & \therefore \frac{a}{14}=\frac{b}{14}=\frac{c}{14} \quad \Rightarrow \quad a=b=c=1 \end{aligned}$$

Hence eq. (1) becomes

$x+1+y-3+z+2=0 \Rightarrow x+y+z=0$

Required probability

$$\begin{aligned} & =\frac{{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^9 \mathrm{C}_3} \\ & =\frac{3 \times 4 \times 2}{\left(\frac{9 !}{3 ! 6 !}\right)}=\frac{24 \times 6}{9 \times 8 \times 7}=\frac{2}{7} \end{aligned}$$

We have p = 0.6 and np = 6 $\Rightarrow$ n = 10

$\therefore$ Var (X) = npq = (10) (0.6) (0.4) = 2.4

Let $\quad I=\int[\sin |\log x|+\cos |\log x|] d x$

Put $\log x=t \Rightarrow \frac{1}{x} d x=d t \Rightarrow d x=e^t d t$

$$\begin{aligned} \therefore I & =\int(\sin +\cos t) e^t d t \\ & =\int e^t(\sin t+\cos t) d t=e^t \sin t+c=x \sin |\log x|+c \end{aligned}$$

We have $2 \pi r=10 \pi \Rightarrow r=5$ and centre of circle is $(2,-3)$

Hence equation of circle is $(x-2)^2+(y+3)^2=(5)^2$. i.e. $x^2+y^2-4 x+6 y-12=0$

We have $\mathrm{O} \equiv(0,0,0)$ and $\mathrm{P} \equiv(1,2,3)$

Hence direction ratios of $\overline{\mathrm{OP}}$ are

$\frac{1-0}{\sqrt{1^2+2^2+3^2}}, \frac{2-0}{\sqrt{1^2+2^2+3^2}}, \frac{3-0}{\sqrt{1^2+2^2+3^2}}=\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

$$\begin{aligned} & \text { Let } \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ & \therefore \sin A=\frac{a}{k}, \sin B=\frac{b}{k}, \sin C=\frac{c}{k} \end{aligned}$$

With usual notations, from the given data, we write

$$\begin{aligned} & a+b+c=6\left[\frac{\left(\frac{a}{k}+\frac{b}{k}+\frac{c}{k}\right)}{3}\right] \\ & \therefore(a+b+c)=\frac{2(a+b+c)}{k} \Rightarrow k=2 \\ & \therefore \sin A=\frac{a}{k}=\frac{1}{2} \Rightarrow A=\frac{\pi^c}{6} \end{aligned}$$

$$\begin{aligned} & \text { } \\ & \begin{aligned} & f(x)=\log (1+x)-\frac{2 x}{2+x} \Rightarrow x \neq-2 \\ & \therefore f^{\prime}(x)=\frac{1}{(1+x)}-\left[\frac{(2+x)(2)-(2 x)(1)}{(2+x)^2}\right] \\ &=\frac{1}{1+x}-\left[\frac{4}{(2+x)^2}\right]=\frac{(x+2)^2-4(x+1)}{(x+2)^2(1+x)} \end{aligned} \end{aligned}$$

When $f^{\prime}(x)>0$, we write

$\frac{x^2}{(x+2)^2(1+x)}>0$

Since $x^2>0$ and $(x+2)^2>0$, we write $(1+x)>0 \Rightarrow x>-1$

\begin{array}{left}|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=25 & \\ \therefore|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| \cdot \sin \theta=(5)(13) \sin \theta =25 & \\ \therefore \sin \theta =\frac{5}{13} \Rightarrow \cos \theta=\frac{-12}{13} & \ldots\left[\because \frac{\pi}{2}<\theta \leq \pi\right] \\ \therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} =|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| \cdot \cos \theta & \\ =(5)(13)\left(\frac{-12}{13}\right)=-60 & \end{array}

Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the direction ratios of the required line.

$\therefore 2 \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0$ ....... (1) and $\mathrm{a}-2 \mathrm{~b}+2 \mathrm{c}=0$ ..... (2)

From (1) and (2), we write

$$\begin{aligned} & \frac{\mathrm{a}}{\left|\begin{array}{ll} -2 & 1 \\ -2 & 2 \end{array}\right|}=\frac{-\mathrm{b}}{\left|\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{ll} 2 & -2 \\ 1 & -2 \end{array}\right|} \Rightarrow \frac{\mathrm{a}}{-2}=\frac{\mathrm{b}}{-3}=\frac{\mathrm{c}}{-2} \\ & \therefore(\mathrm{a}, \mathrm{b}, \mathrm{c})=(2,3,2) \end{aligned}$$

So equation of required line is

$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$