MHT CET 2023 13th May Evening Shift (Mathematics) Question Paper with Solutions

$(3 x+2)-(5 y-3) i$ and $(6 x+3)+(2 y-4) i$ are conjugates of each other.

$(3 x+2)-(5 y-3) i=(6 x+3)-(2 y-4) i$

On comparing real parts and imaginary parts, we get

$$\begin{aligned} 3 x+2 & =6 x+3, \quad 5 y-3 =2 y-4 \\ 3 x & =-1, \qquad 3 y =-1 \\ x & =-1 / 3, \quad y =-1 / 3 \\ \therefore & \quad \frac{x-y}{x+y} =\frac{-\frac{1}{3}+\frac{1}{3}}{-\frac{1}{3}-\frac{1}{3}}=0 & \end{aligned}$$

Given, $A=(1,-3)$ and $B(4,3)$ and $y=x-\frac{4}{x}$

We know that,

Slope of tangent $=$ Slope of $A B$

$m_{A B}=\frac{3-(-3)}{4-1}=\frac{6}{3}=2$

$\Rightarrow \quad m_T =2$

$y =x-\frac{4}{x}$

$\frac{d y}{d x} =1+\frac{4}{x^2} \quad \text{[slope of tangent]}$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}=1-4\left(-\frac{1}{x^2}\right)=2 \\ \Rightarrow & 1+\frac{4}{x^2}=2 \\ \Rightarrow & \frac{4}{x^2}=1 \\ \Rightarrow & x^2=4 \Rightarrow x= \pm 2 \end{array}$$

So, $y=0$, at $x=2$ and $y=0$, at $x=-2$

$P(2,0), Q(-2,0)$

Since, $S$ is mid-point of $Q$ and $R$.

$\therefore \quad S=\left(\frac{7+6}{2}, \frac{3-1}{2}\right)=\left(\frac{13}{2}, 1\right)$

Slope of $P S=m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-1}{2-\frac{13}{2}}=\frac{-2}{9}$

$\therefore$ Equation of line passing through point $(1,-1)$ and parallel to $P S$ is

$$\begin{aligned} y+1 & =-\frac{2}{9}(x-1) \\ 9 y+9 & =-2 x+2 \Rightarrow 2 x+9 y=-7 \end{aligned}$$

$\Rightarrow \quad \frac{x}{\left(\frac{-7}{2}\right)}+\frac{y}{\left(\frac{-7}{9}\right)}=1$

$\text { Hence, } x \text {-intercept }=-\frac{7}{2} \text { and } y \text {-intercept }=-\frac{7}{9}$

Let $I=\int \frac{\log \left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}} d t$

Now assume $y=\log \left(t+\sqrt{1+t^2}\right)$

$$\begin{aligned} \Rightarrow \quad d y & =\frac{1}{t+\sqrt{1+t^2}}\left(1+\frac{t}{\sqrt{1+t^2}}\right) d t \\ & =\frac{1}{\sqrt{1+t^2}} d t \end{aligned}$$

$$\begin{aligned} \therefore \quad I & =\int y d y=\frac{y^2}{2}+c \\ & =\frac{\left(\log \left(t+\sqrt{1+t^2}\right)\right)^2}{2}+c \end{aligned}$$

$$\begin{aligned} & \text { Thus, } g(t)=\log \left(t+\sqrt{1+t^2}\right) \\ & \Rightarrow \quad g(2)=\log (2+\sqrt{5}) \end{aligned}$$

$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{2(a+b+c)}{36}$

or $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{a+b+c}{18}=k$

$\therefore \quad b+c=11 k, c+a=12 k, a+b=13 k$

$a+b+c=18 k$

Substituting $b+c=11 k$ in $a+b+c=18 k$, we get $a+11 k=18 k \text { or } a=7 k$

Substituting $a=7 k$ in $c+a=12 k$, we get

$c+7 k=12 k \text { or } c=5 k$

Substituting $c=5 k$ in $b+c=11 k$, we get

$b+5 k=11 k$ or $b=6 k$

$\therefore \quad a=7 k, b=6 k$ and $c=5 k$

Using cosine rule, we get

$$\begin{aligned} \cos A & =\frac{b^2+c^2-a^2}{2 b c}=\frac{(6 k)^2+(5 k)^2-(7 k)^2}{2 \times 6 k \times 5 k}=\frac{1}{5}, \\ \cos B & =\frac{c^2+a^2-b^2}{2 c a} \\ & =\frac{(5 k)^2+(7 k)^2-(6 k)^2}{2 \times 5 k \times 7 k}=\frac{19}{35} \\ \cos C & =\frac{a^2+b^2-c^2}{2 a b}=\frac{(7 k)^2+(6 k)^2-(5 k)^2}{2 \times 7 k \times 6 k}=\frac{5}{7} \\ \cos A & +\cos B+\cos C=\frac{1}{5}+\frac{19}{35}+\frac{5}{7} \\ & =\frac{7+19+25}{35}=\frac{51}{35} \end{aligned}$$

Let the equation of the plane through $(2,2,2)$ be $a(x-2)+b(y-2)+c(z-2)=0$

Since, it is parallel to the straight lines having Dr's $(1,0,-1)$ and $(-1,1,0)$, therefore $a-c=0$ and $-a+b=0$

$\Rightarrow \quad a=b=c$

Therefore, equation of plane is

$$\begin{aligned} & x-2+y-2+z-2=0 \\ & x+y+z=6 \Rightarrow \frac{x}{6}+\frac{y}{6}+\frac{z}{6}=1 \end{aligned}$$

Its intercepts on coordinate axes are $A(6,0,0), B(0,6,0)$ and $C(0,0,6)$.

Hence, the volume of tetrahedron $O A B C$

$$\begin{aligned} & =\frac{1}{6}[a b c]=\frac{1}{6}\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\ & =\frac{1}{6} \times 6 \times 36=36 \text { cube units } \end{aligned}$$

$$\begin{aligned} & \text { Mean }=n p \\ & \text { Variance }=n p(1-p) \\ & \text { So, } \quad n p+n p(1-p)=1.8 \\ & 5 p+5 p(1-p)=1.8 \\ & \Rightarrow \quad 10 p-5 p^2-1.8=0 \\ & \Rightarrow \quad 5 p^2-10 p+1.8=0 \\ & \Rightarrow \quad 5 p^2-9 p-p+1.8=0 \\ & \Rightarrow \quad 5 p(p-1.8)-1(p-1.8)=0 \\ & \Rightarrow \quad(p-1.8)(5 p-1)=0 \\ & \therefore \quad p=1.8,0.2 \\ & \because \quad p=1.8 \\ \end{aligned}$$

(It is not possible because it is not greater than 1)

Hence, p = 0.2

Two dice are rolled.

The total number of outcomes in sample space $n(S)$ are given as

$\Rightarrow \quad n(S)=6^2=36$

The numbers on the faces of the dice are $1,2,3,5, 7, 11$

The possible scenarios where the sum of the numbers on upper most face is prime $(1,1),(1,2), (2,1),(2,3),(2,5),(2,11),(3,2),(5,2),(11,2)$

Number of ordered pairs $=9$

Required probability $=\frac{9}{36}=\frac{1}{4}$

$$\begin{aligned} \sin ^{-1}\left(\sin \frac{3 \pi}{4}\right) & =\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ = & \sin ^{-1}\left(\sin \frac{\pi}{4}\right)=\frac{\pi}{4} \quad\left[\because \frac{-\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}\right] \end{aligned}$$

$\therefore \text { Principle value of } \sin ^{-1}\left(\sin \frac{3 \pi}{4}\right) \text { is } \frac{\pi}{4} \text {. }$

$y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}$

Taking log on both sides,

$$\begin{aligned} \log y= & \frac{1}{3} \log (1+3 x)+\frac{1}{4} \log (1+4 x) \\ & +\frac{1}{5} \log (1+5 x)-\frac{1}{7} \log (1+7 x)-\frac{1}{8} \log (1+8 x) \end{aligned}$$

Differentiating both sides w.r.t. $x$,

$\frac{1}{y} \times \frac{d y}{d x}=\frac{1}{1+3 x}+\frac{1}{1+4 x}+\frac{1}{1+5 x}-\frac{1}{1+7 x}-\frac{1}{1+8 x}$

At $x=0, y=1$

$\therefore \frac{d y}{d x}=1+1+1-1-1=1$

The given point is $P(1,2,3)$.

and given line is $\mathbf{r}=(6 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$

$\Rightarrow \quad \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=k \text { (say) }$ ..... (i)

So, any point on this line is given as $Q(3 k+6,2 k+7,-2 k+7)$

Now, direction ratios of $P Q$ are $3 k+5,2 k+5,-2 k+4$

Also, $P Q \perp \mathbf{r}$

$$\begin{array}{ll} \Rightarrow & 3(3 k+5)+2(2 k+5)-2(-2 k+4)=0 \\ \Rightarrow & 17 k+17=0 \Rightarrow k=-1 \end{array}$$

Hence, foot of the perpendicular drawn on the given line is $(3,5,9)$.

We have, equation of line.

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and plane $: x-y+z=16$

any point on line is $(3 t+2,4 t-1,12 t+2)$ and this point will satisfy the plane.

$\therefore 3 t+2-4 t+1+12 t+2=16 \Rightarrow 11 t=11 \Rightarrow t=1$

So, point will be $(5,3,14)$

Hence, distance between $(5,3,14)$ and $(1,6,2)$ is

$$\begin{aligned} & =\sqrt{(1-5)^2+(6-3)^2+(2-14)^2} \\ & =\sqrt{16+9+144}=\sqrt{169}=13 \text { units } \end{aligned}$$

Let $I=\int \frac{2+\cos \frac{x}{2}}{x+\sin \frac{x}{2}} d x$

Put $x+\sin \frac{x}{2}=t$

$$\begin{aligned} & \left(1+\cos \frac{x}{2} \times \frac{1}{2}\right) d x=d t \\ & \Rightarrow \quad\left(2+\cos \frac{x}{2}\right) d x=2 d t \\ & \therefore \quad I=\int \frac{2 d t}{t}=2 \log t+c=2 \log \left(x+\sin \frac{x}{2}\right)+c \end{aligned}$$

$$\begin{aligned} & 2^x+2^y=2 \\ & 2^y=2-2^x \end{aligned}$$

$$\begin{array}{l} \because & \quad 2^y>0 \quad \text{[exponential function]}\\ \therefore & \quad 2-2^x>0 \\ & \quad 2^x<2 \\ & \quad x<1 \\ \therefore & x \in(-\infty, 1) \end{array}$$

$x d y=y d x+y^2 d y$

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

On integrating both sides w.r.t. $x$,

$y=-\frac{x}{y}+c_1$

Now, $y(1)=1$, If $x=1, y=1$

$$\begin{aligned} \Rightarrow \quad 1 & =\frac{-1}{1}+c_1 \Rightarrow c_1=2 \\ y & =\frac{-x}{y}+2 \end{aligned}$$

If $x=-3, y=\frac{3}{y}+2$

$\Rightarrow \quad y^2-2 y-3=0$

$$\begin{array}{ll} \Rightarrow & (y-3)(y+1)=0 \\ \Rightarrow & y=3,-1 \end{array}$$

We have, $\triangle A B C$ with $A(0,2,1), B(-2,0,0)$ and $C(-2,0,2)$.

$$\begin{aligned} \text { So, } A B & =\sqrt{9}=3=c, \\ B C & =\sqrt{4}=2=a \\ \text { and } A C & =\sqrt{9}=3=b \end{aligned}$$

We know that,

$$\begin{aligned} & \text { incentre }=\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c},\right. \\ & \left.\frac{a z_1+b z_2+c z_3}{a+b+c}\right) \\ & =\left(\frac{0-6-6}{8}, \frac{4+0+0}{8}, \frac{2+0+6}{8}\right) \\ & =\left(\frac{-3}{2}, \frac{1}{2}, 1\right) \end{aligned}$$

The equations of $P Q$ and $P R$ are given by

$y-1=\frac{-2 \mp \tan 45^{\circ}}{1 \pm(-2) \tan 45^{\circ}}(x-2)$

$$\begin{array}{ll} \Rightarrow & y-1=\left(\frac{-2 \mp 1}{1 \pm 2}\right)(x-2) \\ \Rightarrow & y-1=-\frac{1}{3}(x-2) \\ \text { and } & y-1=3(x-2) \\ \Rightarrow & x+3 y=5 \text { and } 3 x-y=5 \end{array}$$

The combined equation of these two lines is

$$\begin{aligned} (x+3 y-5)(3 x-y-5) & =0 \\ \Rightarrow 3 x^2-3 y^2+8 x y-20 x-10 y+25 & =0 \end{aligned}$$

Given curve $y=a \sqrt{x}+b x$ passes through $(1,2)$.

The area formed by curve line $x=4$ and $X$-axis is 8 sq units.

$\because$ Since, the curves $y=a \sqrt{x}+b x$ passes through $(1,2)$.

$$\begin{aligned} \therefore \quad 2 & =a \sqrt{1}+b \cdot 1 \\ & a+b=2 \quad \text{.... (i)} \end{aligned}$$

Now, the area enclosed by the curve line $x=4$ and $X$-axis

\begin{aligned} & =\int_\limits0^4 y d x=\int_\limits0^4(a \sqrt{x}+b x) d x \\ & =\left[a \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}+b \cdot \frac{x^2}{2}\right]_0^4 \\ & =\frac{2 a}{3} \cdot 4^{\frac{3}{2}}+\frac{b}{2} \cdot 4^2=\frac{16 a}{3}+8 b \end{aligned}

From the problem

$$\begin{aligned} & \frac{16 a}{3}+8 b =8 \\ \therefore \quad & 2 a+3 b =3 \quad \text{.... (ii)} \end{aligned}$$

From Eqs. (i) and (ii), we get

a = 3, b = $-$1

$$\text { Here, } A=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]$$

$$\begin{aligned} & \Rightarrow \quad|A|=\sec ^2 x \\ & \operatorname{adj} A=C^T=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]^T \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad \operatorname{adj} A & =\left[\begin{array}{cc} \cos ^2 x & -\sin x \cos x \\ \sin x \cdot \cos x & \cos ^2 x \end{array}\right] \\ & A^{-1}=\frac{\operatorname{adj} A}{|A|}=\left[\begin{array}{cc} \cos ^2 x & -\sin x \cos x \\ \sin x \cdot \cos x & \cos ^2 x \end{array}\right] \end{aligned}$$

Now,

$$\begin{aligned} & A^T A^{-1}=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ & {\left[\begin{array}{cc} \cos ^2 x & -\sin x \cdot \cos x \\ \sin x \cdot \cos x & \cos ^2 x \end{array}\right] } \end{aligned}$$

$$A^T A^{-1}=\left[\begin{array}{cc} \cos 2 x & -\sin 2 x \\ \sin 2 x & \cos 2 x \end{array}\right]$$

$$\begin{aligned} x & =\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right) \\ & =\log _e\left(\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}}\right) \end{aligned}$$

[On dividing $\cos y / 2$ in both numerator and denominator]

$$\begin{aligned} & =\log _e\left(\frac{1-\sqrt{\frac{1-t}{1+t}}}{1+\sqrt{\frac{1-t}{1+t}}}\right)=\log _e\left(\frac{\sqrt{1+t}-\sqrt{1-t}}{\sqrt{1+t}+\sqrt{1-t}}\right) \\ & =\log _e\left(\frac{1+t+1-t-2 \sqrt{1-t^2}}{1+t-1+t}\right) \end{aligned}$$

$x=\log _e\left(\frac{1-\sqrt{1-t^2}}{t}\right)$

$$\begin{aligned} & \frac{d x}{d t}=\frac{t}{1-\sqrt{1-t^2}}\left[\frac{\frac{t^2}{\sqrt{1-t^2}}-\left(1-\sqrt{1-t^2}\right)}{t^2}\right] \\ & \frac{d x}{d t}=\frac{t}{1-\sqrt{1-t^2}}\left[\frac{t^2-\sqrt{1-t^2}+1-t^2}{\sqrt{1-t^2} t^2}\right] \\ & \frac{d x}{d t}=\frac{1}{t \sqrt{1-t^2}} \quad \text{..... (i)} \end{aligned}$$

$$\begin{aligned} & \because \quad \tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}} \Rightarrow \frac{y}{2}=\tan ^{-1}\left(\sqrt{\frac{1-t}{1+t}}\right) \\ & \Rightarrow \quad \frac{1}{2} \frac{d y}{d t}=\left(\frac{t+1}{2}\right)\left(\frac{\sqrt{1+t}}{2 \sqrt{1-t}}\right)\left(\frac{-1-t-1+t}{(1+t)^2}\right) \\ & \Rightarrow \quad \frac{1}{2} \frac{d y}{d t}=\frac{-1}{2 \sqrt{1-t^2}} \\ & \Rightarrow \quad \frac{d y}{d t}=-\frac{1}{\sqrt{1-t^2}} \quad \text{..... (ii)} \end{aligned}$$