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

$$\begin{aligned} & \left(3 x y+y^2\right) d x+\left(x^2+x y\right) d y=0 \\ & \therefore \frac{d y}{d x}=-\frac{\left(3 x y+y^2\right)}{\left(x^2+x y\right)} \end{aligned}$$

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

$$\begin{aligned} & \therefore v+x \frac{d v}{d x}=-\frac{\left(3 v x^2+v^2 x^2\right)}{\left(x^2+v^2\right)}=-\frac{3 v+v^2}{1+v} \\ & \therefore x \frac{d v}{d x}=-\frac{3 v+v^2}{1+v}-v=-\frac{\left(2 v^2+4 v\right)}{1+v} \\ & \therefore \int \frac{1+v}{\left(2 v^2+4 v\right)} d v=\int \frac{-1}{x} d x \\ & \therefore \frac{1}{2} \int \frac{v+1}{v^2+2 v} d v=-\int \frac{d x}{x} \Rightarrow \frac{1}{4} \int \frac{2(v+1)}{v^2+2 v} d v=-\log x+\log c_1 \\ & \therefore \frac{1}{4} \log \left(v^2+2 v\right)+\log x=\log c_1 \Rightarrow \frac{1}{4} \log \left(\frac{y^2}{x^2}+\frac{2 y}{x}\right)+\log x=\log c_1 \end{aligned}$$

$$\begin{aligned} & \therefore \frac{1}{4} \log \left(\frac{y^2+2 x y}{x^2}\right)+\log x=\log c_1 \Rightarrow \log \left(\frac{y^2+2 x y}{x^2}\right)+\log x^4=4 \log c_1 \\ & \therefore \log \left[\left(\frac{y^2+2 x y}{x^2}\right)\left(x^4\right)\right]=\log c_1^4 \Rightarrow \log \left[x^2\left(y^2+2 x y\right)\right]=\log c_1^4 \\ & \therefore x^2\left(y^2+2 x y\right)=c \end{aligned}$$

$$\begin{aligned} & \mathrm{y}(1+\log \mathrm{x})\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)-\mathrm{x} \log \mathrm{x}=0 \\ & \therefore \int \frac{(1+\log \mathrm{x})}{\mathrm{x} \log \mathrm{x}} \mathrm{dx}=\int \frac{\mathrm{dy}}{\mathrm{y}} \\ & \therefore \int \frac{\mathrm{dx}}{\mathrm{x} \log \mathrm{x}}+\int \frac{\mathrm{dx}}{\mathrm{x}}=\int \frac{\mathrm{dy}}{\mathrm{y}} \\ & \therefore \log (\log \mathrm{x})+\log \mathrm{x}=\log \mathrm{y}+\log \mathrm{c} \\ & \therefore \log [\mathrm{x} \log \mathrm{x}]=\log (\mathrm{yc}) \Rightarrow \mathrm{x} \log \mathrm{x}=\mathrm{yc} \end{aligned}$$

Refer figure.

$\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s}, \mathrm{m}, \mathrm{n}$ be the position vectors of points $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}, \mathrm{M}, \mathrm{N}$.

Now $\overline{\mathrm{PS}}+\overline{\mathrm{QR}}$

$=\overline{\mathrm{s}}-\overline{\mathrm{p}}+\overline{\mathrm{r}}-\overline{\mathrm{q}}=(\overline{\mathrm{s}}+\overline{\mathrm{r}})-(\overline{\mathrm{p}}+\overline{\mathrm{q}})$

Since $M$ and $N$ are mid-points of $P Q$ and RS respectively, we write

$\overline{\mathrm{m}}=\frac{\overline{\mathrm{p}}+\overline{\mathrm{q}}}{2} \text { and } \overline{\mathrm{n}}=\frac{\overline{\mathrm{r}}+\overline{\mathrm{s}}}{2}$

$\therefore$ eq. (1) becomes

$\overline{\mathrm{PS}}+\overline{\mathrm{QR}}=2 \overline{\mathrm{n}}-2 \overline{\mathrm{m}}=2(\overline{\mathrm{n}}-\overline{\mathrm{m}})=2 \overline{\mathrm{MN}}$

From given data, $t=2$

$$\begin{aligned} & 2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right) \\ & 2\left[\sin \theta \cos \frac{\pi}{3}+\cos \theta \sin \frac{\pi}{3}\right]=\left(\cos \theta \cos \frac{\pi}{6}+\sin \theta \sin \frac{\pi}{6}\right) \\ & \therefore 2\left(\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right)=\left(\frac{\sqrt{3} \cos \theta}{2}+\frac{\sin \theta}{2}\right) \\ & \therefore\left(\frac{\sin \theta}{2}+\frac{\sqrt{3} \cos \theta}{2}\right)=0 \Rightarrow \sin \theta+\sqrt{3} \cos \theta=0 \Rightarrow \tan \theta=-\sqrt{3}\end{aligned}$$

Lines given by $ax^2+2hxy+by^2=0$ make inclinations $\alpha$ and $\beta$.

$\therefore$ $\tan \alpha + \tan \beta = {{ - 2h} \over b}$ and $\tan \alpha \tan \beta = {a \over b}$

Now $\tan (\alpha + \beta ) = {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }} = {{\left( { - {{2h} \over b}} \right)} \over {1 - \left( {{a \over b}} \right)}} = {{ - 2h} \over b} \times {b \over {(b - a)}}$

$\therefore$ $\tan (\alpha + \beta ) = {{ - 2h} \over {b - a}} = {{2h} \over {a - b}}$

$$\begin{aligned} & \text { Let } \quad=(\log x)^x \\ & \therefore \log u=x \log [\log (x)] \\ & \therefore \frac{1}{u} \frac{d u}{d x}=\frac{x}{\log (x)} \times \frac{1}{x}+\log (\log x)=\log (\log x)+\frac{1}{\log x} \\ & \therefore \frac{d u}{d x}=(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right] \quad \text{.....(1)}\\ & \text { Let } v=\log x \Rightarrow \frac{d v}{d x}=\frac{1}{x} \quad \text{.....(2)}\\ & \therefore \frac{d u}{d v}=(\log x)^x(x)\left[\frac{1}{\log x}+\log (\log x)\right]\quad \text{.....(From (1), (2))}\end{aligned}$$

$\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$

By Componendo - Dividendo, we wirte

$$\begin{aligned} & \frac{\cos (A+B)+\cos (A-B)}{\cos (A+B)-\cos (A-B)}=\frac{\sin (C+D)+\sin (C-D)}{\sin (C+D)-\sin (C-D)} \\ & \frac{2 \cos A \cos B}{-2 \sin A \sin B}=\frac{2 \sin C \cos D}{2 \cos C \sin D} \\ & \therefore \frac{1}{-\tan A \tan B}=\frac{\tan C}{\tan D} \Rightarrow \tan A \tan B \tan C=-\tan D \end{aligned}$$

We have $\mathrm{n}_1=5, \sigma_1^2=4, \overline{\mathrm{x}}_1=2$ and $\mathrm{n}_2=5, \sigma_2^2=5, \overline{\mathrm{x}}_2=4$

Here combined mean $=\bar{x}_c=\frac{(5)(2)+(5)(4)}{(5)+(5)}=3$

$$\begin{aligned} & \mathrm{d}_1=\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_{\mathrm{c}}=2-3=1 \text { and } \mathrm{d}_2=\overline{\mathrm{x}}_2-\overline{\mathrm{x}}_{\mathrm{c}}=4-3=1 \\ & \text { Combined variance }=\frac{\mathrm{n}_1\left(\sigma_1^2+\mathrm{d}_1^2\right)+\mathrm{n}_2\left(\sigma_2^2+\mathrm{d}_2^2\right)}{\mathrm{n}_1+\mathrm{n}_2}=\frac{5(4+1)+5(5+1)}{5+5}=\frac{11}{2} \end{aligned}$$

Let $I=\int e^{\tan x}\left(\sec ^2 x+\sec ^3 x \sin x\right) d x=\int e^{\tan x}\left(\sec ^2 x\right)(1+\tan x) d x$

Put $\tan \mathrm{x}=\mathrm{t} \Rightarrow \sec ^2 \mathrm{xdx}=\mathrm{dt} \quad \therefore I=\int \mathrm{e}^{\mathrm{t}}(1+\mathrm{t}) \mathrm{dt}=\mathrm{e}^{\mathrm{t}}(\mathrm{t})+\mathrm{c}=\mathrm{e}^{\tan \mathrm{x}}(\tan \mathrm{x})+\mathrm{c}$

Let $(\mathrm{h}, 0)$ be the centre of the circle and '$r$' be the radius.

Let $(\mathrm{h}, \mathrm{h}-\mathrm{h})^2+(\mathrm{y}-0)^2=\mathrm{r}^2 \Rightarrow \quad(\mathrm{x}-\mathrm{h})^2+\mathrm{y}^2=\mathrm{r}^2$ ....... (1)

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

$\therefore 2(x-h)(1)+2 y \frac{d y}{d x}=0 \Rightarrow h=x+y \frac{d y}{d x}$

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

$y^2\left(\frac{d y}{d x}\right)^2+y^2=r^2 \Rightarrow y^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=r^2$

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

$$\begin{aligned} & y^2\left(2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}\right)+\left[1+\left(\frac{d y}{d x}\right)^2\right](2 y) \frac{d y}{d x}=0 \\ & \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0 \end{aligned}$$

Equation of plane passing through $(-2,2,2)$ is $a(x+2)+b(y-2)+c(z-2)=0$

Since this plane also passes through $(2,-2,-2)$, we get

$4 a-4 b-4 c=0 \Rightarrow a-b-c=0$ ....... (1)

Normal of the plane is parallel to

$9 x-13 y-3 z=0 \Rightarrow 9 a-13 b-3 c=0$ ....... (2)

Solving (1) and (2), we write

$$\begin{aligned} & \frac{\mathrm{a}}{\left|\begin{array}{ll} -1 & -1 \\ -13 & -3 \end{array}\right|}=\frac{-\mathrm{b}}{\left|\begin{array}{cc} 1 & -1 \\ 9 & -3 \end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc} 1 & -1 \\ 9 & -13 \end{array}\right|} \\ & \therefore \frac{\mathrm{a}}{-10}=\frac{-\mathrm{b}}{6}=\frac{\mathrm{c}}{-4} \Rightarrow \frac{\mathrm{a}}{5}=\frac{\mathrm{b}}{3}=\frac{\mathrm{c}}{2} \end{aligned}$$

Hence equation of required plane is $5(x+2)+3(y-2)+2(z-2)=0$ i.e. $5 x+3 y+2 z=0$

From the given data, we write

Variance $=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=0.6-(0.6)^2=0.6-0.36=0.24$

$$\begin{aligned} & \bar{c}=\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0\end{array}\right|=\hat{i}(0)-\hat{j}(0)+\hat{k}(9+30)=39 \hat{k} \\ & |\bar{a}|=\sqrt{(3)^2+(-5)^2}=\sqrt{34} \text { and }|\bar{b}|=\sqrt{(6)^2+(3)^2}=\sqrt{45} \text { and }|\bar{c}|=\sqrt{39^2}=39 \\ & \therefore a: b: c=\sqrt{34}: \sqrt{45}: 39\end{aligned}$$

$$\begin{aligned} & e^{-y} \cdot y=x \\ & \therefore \frac{y}{e^y}=x \Rightarrow y=x e^y \quad \text {....(1) and } \quad e^y=\frac{y}{x} \quad \text {....(2)}\\ & \quad \text { Now } y=x e^y \\ & \therefore \frac{d y}{d x}=x e^y \frac{d y}{d x}+e^y \\ & \therefore \frac{d y}{d x}\left(x e^y-1\right)=-e^y \Rightarrow \frac{d y}{d x}=\frac{-e^y}{x e^y-1} \end{aligned}$$

From (1) and (2), we write

$\frac{d y}{d x}=-\left(\frac{y}{x}\right) \times \frac{1}{y-1}=\frac{-y}{x(y-1)}=\frac{y}{x(1-y)}$

$$\begin{aligned} & \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\ = & \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}=\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)} \\ = & \lim _{x \rightarrow 1} \frac{2 x-3}{(\sqrt{x}+1)(2 x+3)}=\frac{-1}{2(5)}=\frac{-1}{10}\end{aligned}$$

$$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\ & \therefore \frac{1}{a^2} 2 x+\frac{1}{4} 2 y \frac{d y}{d x}=0 \quad \Rightarrow \quad \frac{d y}{d x}=\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right) \quad \text{.....(1)}\\ \end{aligned}$$

Also $\mathrm{y}^3=16 \mathrm{x}$

$\therefore 3 y^2 \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2}$ ...... (2)

Since curves intersect orthogonally, from (1) and (2), we write

$$\begin{aligned} & \left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)\left(\frac{16}{3 y^2}\right)=-1 \\ & \therefore \frac{64 x}{3 a^2 y^3}=1 \text { and we have } y^3=16 x \\ & \therefore \frac{64 x}{3 a^2(16 x)}=1 \Rightarrow a^2=\frac{4}{3} \end{aligned}$$

Let $I=\int[1+2 \tan x(\tan x+\sec x)]^{1 / 2} d x=\int\left(1+2 \tan ^2 x+2 \tan x \sec x\right)^{1 / 2} d x$

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

Let $I=\int \frac{x^3}{\sqrt{1+x^2}} d x=\int \frac{x^2 \cdot x}{\sqrt{1+x^2}} d x$

Put $\sqrt{1+x^2}=t \quad \Rightarrow 1+x^2=t^2 \quad \Rightarrow 2 x d x=2 t d t$

$$\begin{aligned} & \text { Put } \sqrt{1+\mathrm{x}^2}=\mathrm{t} \Rightarrow 1+\mathrm{x}^2=\mathrm{t}^2 \Rightarrow 2 \mathrm{xdx}=2 \mathrm{td} \\ & \therefore \mathrm{I}=\int \frac{\left(\mathrm{t}^2-1\right) \mathrm{tdt}}{\mathrm{t}}=\int\left(\mathrm{t}^2-1\right) d \mathrm{t} \quad=\frac{\mathrm{t}^3}{3}-\mathrm{t}+\mathrm{c}=\frac{\left(1+\mathrm{x}^2\right)^{\frac{3}{2}}}{3}-\sqrt{1+\mathrm{x}^2}+\mathrm{c} \end{aligned}$$

Comparing with given data, $a=\frac{1}{3}, b=-1 \Rightarrow a+b=\frac{-2}{3}$

Let $(h, k) b$ the centre of the circle. It lies on the line $x-4 y=1 \Rightarrow h=1+4 k$

$\therefore$ centre $\equiv(4 \mathrm{k}+1, \mathrm{k})$

Circle passes through points $(3,7)$ and $(5,5)$

$$\begin{aligned} & \therefore(4 \mathrm{k}+1-5)^2+(\mathrm{k}-5)^2 \quad=(4 \mathrm{k}+1-3)^2+(\mathrm{k}-7)^2 \\ & \therefore(4 \mathrm{k}-4)^2+(\mathrm{k}-5)^2=(4 \mathrm{k}-2)^2+(\mathrm{k}-7)^2 \\ & \therefore 16 \mathrm{k}^2+16-32 \mathrm{k}+\mathrm{k}^2+25-10 \mathrm{k}=16 \mathrm{k}^2+4-16 \mathrm{k}+\mathrm{k}^2+49-14 \mathrm{k} \\ & \therefore-42 \mathrm{k}+41=-30 \mathrm{k}+53 \Rightarrow 12 \mathrm{k}=-12 \Rightarrow \mathrm{k}=-1 \\ & \therefore \text { Centre } \equiv(-3,-1) \\ & \therefore \text { Radius } \quad=\sqrt{(-3-5)^2+(-1-5)^2}=10 \end{aligned}$$

Hence equation of required circle is

$(x+3)^2+(y+1)^2=(10)^2 \text { i.e. } x^2+y^2+6 x+2 y-90=0$

Let $z=a+i b$ and we have

$$\begin{aligned} & \frac{b}{a}=\tan \left(\frac{5 \pi}{6}\right) \text { and } \sqrt{a^2+b^2}=2 \\ & \therefore \frac{b}{a}=\tan \left(\pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=\frac{-1}{\sqrt{3}} \text { and } a^2+b^2=4 \\ & \therefore b=\frac{-a}{\sqrt{3}} \text { and } a^2+b^2=4 \\ & \therefore a^2+\frac{a^2}{3}=4 \quad \Rightarrow \quad 4 a^2=12 \Rightarrow a^2=3 \Rightarrow a= \pm \sqrt{3} \\ & \text { Also } b=\frac{-a}{\sqrt{3}}=\frac{ \pm \sqrt{3}}{\sqrt{3}}= \pm 1 \end{aligned}$$

Since complex number lies in $2^{\text {nd }}$ quadrant, $\mathrm{a}=-\sqrt{3}$ and $\mathrm{b}=1 \Rightarrow \mathrm{z}=-\sqrt{3}+\mathrm{i}$

Refer figure

Let AD be the angle bisector of angle A which divides BC in the ratio AB : AC.

Here AB = $\sqrt{9+16}=\sqrt{25}$ and

AC = $\sqrt{9+25+16}$

$=\sqrt{50}$

$\therefore$ D divides BC in the ratio $\sqrt{25} : \sqrt{50}$ i.e. 1 : 2

$\therefore$ Position vector of $D=\frac{(4)(2) \hat{k}+5 \hat{j}+4 \hat{k}}{1+2}=\frac{5 j+12 \hat{k}}{3}$