MHT CET 2023 12th May Morning Shift (Mathematics) Question Paper with Solutions

Let $\mathrm{p}$ : The surface area increases

$\qquad\mathrm{q}:$ The pressure decreases

Given statement is $\mathrm{p} \rightarrow \mathrm{q}$

$\therefore \quad$ It's inverse is $\sim p \rightarrow \sim q$

$\therefore \quad$ Option (A) is correct.

$$\begin{array}{ll} & \cos ^2 \theta-2 \sin \theta+\frac{1}{4}=0 \\ \therefore \quad & \left(1-\sin ^2 \theta\right)-2 \sin \theta+\frac{1}{4}=0 \\ \therefore \quad & \sin ^2 \theta+2 \sin \theta-\frac{5}{4}=0 \\ \therefore \quad & 4 \sin ^2 \theta+8 \sin \theta-5=0 \\ \therefore \quad & 4 \sin ^2 \theta+10 \sin \theta-2 \sin \theta-5=0 \\ \therefore \quad & 2 \sin \theta(2 \sin \theta+5)-1(2 \sin \theta+5)=0 \\ \therefore \quad & (2 \sin \theta-1)(2 \sin \theta+5)=0 \\ \therefore \quad & \sin \theta=\frac{1}{2} \text { or } \sin \theta=\frac{-5}{2} \end{array}$$

But $\sin \theta=\frac{-5}{2}$ is not possible as $\sin \theta \in[-1,1]$ for all values of $\theta$.

$$\begin{array}{ll} \therefore & \sin \theta=\frac{1}{2} \\ \therefore & \sin \theta=\sin \frac{\pi}{6} \\ \therefore & \theta=\frac{n \pi}{1}+(-1)^n \frac{\pi}{6} \\ \therefore \quad & A=1 \text { and } B=6 \\ & \Rightarrow A+B=7 \end{array}$$

\begin{aligned} \int_\limits0^4|2 x-5| \mathrm{d} x & =\int_\limits0^{\frac{5}{2}}(5-2 x) \mathrm{d} x+\int_\limits{\frac{5}{2}}^4(2 x-5) \mathrm{d} x \\ & =\left[5 x-x^2\right]_0^{\frac{5}{2}}+\left[x^2-5 x\right]_{\frac{5}{2}}^4 \\ & =\frac{25}{4}+\frac{9}{4} \\ & =\frac{34}{4} \\ & =\frac{17}{2} \end{aligned}

Let $\mathrm{a}$ and $\mathrm{b}$ be the lengths of two sides of $\mathrm{a}$ triangle.

$\therefore \quad$ According to the given condition,

$$\begin{aligned} & \mathrm{a}+\mathrm{b}=x \text { and } \mathrm{ab}=y \\ & \therefore \quad x^2-c^2=y \Rightarrow(a+b)^2-c^2=a b \\ & \Rightarrow a^2+b^2+2 a b-c^2=a b \\ & \Rightarrow a^2+b^2-c^2=-a b \\ & \Rightarrow \frac{a^2+b^2-c^2}{2 a b}=-\frac{1}{2} \\ & \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\ & \Rightarrow \mathrm{C}=\frac{2 \pi}{3} \\ & \Rightarrow \text { circumradius }=\frac{c}{2 \sin C}=\frac{c}{2 \sin \left(\frac{2 \pi}{3}\right)}=\frac{c}{\sqrt{3}} \\ & \end{aligned}$$

$y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)$ .... (i)

Taking '$\log$' on both sides, we get

$$\begin{aligned} \log y=\log (1+x)+\log \left(1+x^2\right) & +\log \left(1+x^4\right) \\ & +\ldots+\log \left(1+x^{2 n}\right) \end{aligned}$$

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

$\frac{1}{y} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\ldots+\frac{2 \mathrm{n} \times x^{2 \mathrm{n}-1}}{1+x^{2 \mathrm{n}}}$ ..... (ii)

At $x=0$, (i) $\Rightarrow y=1$

$\therefore\quad$ (ii) $\left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0}=1+0+0+\ldots+0=1$

Let $\left(x_1, y_1\right)$ be the required point

$\therefore \quad 2 x_1-y_1=5$ ..... (i)

Also, $\left(x_1, y_1\right)$ is at the distance of 1 unit from line $3 x+4 y=5$

$$\begin{array}{ll} \therefore & 1=\left|\frac{3 x_1+4 y_1-5}{\sqrt{9+16}}\right| \\ \therefore & \pm 5=3 x_1+4 y_1-5 \\ \therefore & 3 x_1+4 y_1-5=5 \quad \text { or } 3 x_1+4 y_1-5=-5 \\ \therefore & 3 x_1+4 y_1=10 \quad \text{.... (ii)} \end{array}$$

or

$3 x_1+4 y_1=0$ .... (iii)

Solving equations (i) and (ii), we get

$x_1=\frac{30}{11}$ and $y_1=\frac{5}{11}$

Solving equation (i) and (iii), we get

$x_1=\frac{20}{11}$ and $y_1=\frac{-15}{11}$

$\therefore\left(\frac{30}{11}, \frac{5}{11}\right)$ and $\left(\frac{20}{11}, \frac{-15}{11}\right)$ are the required points.

Let $\theta$ be the angle between $\overline{\mathrm{AB}}$ and $\overline{\mathrm{AD}}$

$$\begin{aligned} \therefore \quad \cos \theta & =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{AD}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{AD}}|} \\ & =\frac{(2 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\sqrt{4+100+121} \sqrt{1+4+4}} \\ & =\frac{-2+20+22}{\sqrt{225} \sqrt{9}} \\ & =\frac{40}{45} \\ & =\frac{8}{9} \end{aligned}$$

$\therefore \quad \sin \theta=\sqrt{1-\left(\frac{8}{9}\right)^2}=\frac{\sqrt{17}}{9}$

$\alpha$ is the angle of rotation of AD.

$\therefore \quad$ The angle between side $\mathrm{AB}$ and $\mathrm{AD}$

$$\begin{aligned} & =\alpha+\theta \\ & =90^{\circ} \quad ....[\text{Given}] \end{aligned}$$

$$\begin{aligned} & \therefore \quad \cos (\alpha+\theta)=\cos \left(90^{\circ}\right) \\ & \therefore \quad \cos \alpha \cos \theta-\sin \alpha \sin \theta=0 \\ & \therefore \quad 8 \cos \alpha=\sqrt{17} \sin \alpha \\ & \therefore \quad 64 \cos ^2 \alpha=17\left(1-\cos ^2 \alpha\right) \\ & \therefore \quad 81 \cos ^2 \alpha=17 \\ & \therefore \quad \cos \alpha=\frac{\sqrt{17}}{9} \end{aligned}$$

Equation of the plane passing through $(1,1,1)$ is given as

$\mathrm{a}(x-1)+\mathrm{b}(y-1)+\mathrm{c}(\mathrm{z}-1)=0$ .... (i)

As the plane is parallel to the lines having direction ratios $1,0,-1$ and $-1,1,0$, we get $\mathrm{a}-\mathrm{c}=0$ and $-\mathrm{a}+\mathrm{b}=0$

$\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}$ ..... (ii)

$\therefore \quad$ From (i) and (ii), we get

$$\begin{aligned} & x-1+y-1+z-1=0 \\ & \therefore x+y+z=3 \quad \Rightarrow \frac{x}{3}+\frac{y}{3}+\frac{z}{3}=1 \\ \end{aligned}$$

$\therefore \quad$ Co-ordinates of A, B, C are $(3,0,0),(0,3,0)$ and $(0,0,3)$ respectively.

$\therefore \quad$ Volume of tetrahedron $\mathrm{OABC}$

$$\begin{aligned} & =\frac{1}{6}\left|\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right| \\ & =\frac{1}{6} \times 27 \\ & =\frac{9}{2} \text { cu. units } \end{aligned}$$

$$\begin{aligned} \mathrm{P}(\mathrm{X}) & =\left\{\begin{array}{l} \frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, x=1,2,3, \ldots ., \mathrm{n} \\ 0 \quad, \text { othewise } \end{array}\right. \\ \therefore \quad \mathrm{E}(\mathrm{X}) & =\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}\left(x_{\mathrm{i}}\right) \\ & =\frac{2}{\mathrm{n}(\mathrm{n}+1)}+\frac{8}{\mathrm{n}(\mathrm{n}+1)}+\ldots+\frac{2 \mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)} \\ & =\frac{2\left(1^2+2^2+\ldots+\mathrm{n}^2\right)}{\mathrm{n}(\mathrm{n}+1)} \\ & =\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}(\mathrm{n}+1)} \\ & =\frac{2 \mathrm{n}+1}{3} \end{aligned}$$

$$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\cos \left(\frac{7 \pi}{180}\right) x-\cos \left(\frac{2 \pi}{180}\right) x}{x^2} \\ & =\frac{\left(\frac{2 \pi}{180}\right)^2-\left(\frac{7 \pi}{180}\right)^2}{2} \\ & \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{\cos m x-\cos n x}{x^2}=\frac{n^2-m^2}{2}\right] \\ & =\frac{-\pi^2}{1440} \end{aligned}$$

Let $\mathrm{p}: x$ and $y$ are integers such that $x y$ is odd.

$\mathrm{q}:$ both $x$ and $y$ are odd.

$\therefore \quad$ Given statement is $\mathrm{p} \rightarrow \mathrm{q}$

$\therefore \quad$ Its contrapositive is $\sim \mathrm{q} \rightarrow \mathrm{p}$

$\therefore \quad$ Option (D) is correct.

$$\begin{aligned} & A=\left[\begin{array}{cc} 1 & 2 \\ -5 & 1 \end{array}\right] \\ & \therefore|\mathrm{A}|=11 \\ & A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc} 1 & -2 \\ 5 & 1 \end{array}\right]=\frac{1}{11}\left[\begin{array}{cc} 1 & -2 \\ 5 & 1 \end{array}\right] \\ & \mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I} \text {, we get } \\ & {\left[\begin{array}{cc} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{array}\right]=\left[\begin{array}{cc} x & 2 x \\ -5 x & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 0 & y \end{array}\right]} \\ & \begin{array}{ll} \therefore & {\left[\begin{array}{cc} \frac{1}{11} & \frac{-2}{11} \\ \frac{5}{11} & \frac{1}{11} \end{array}\right]=\left[\begin{array}{cc} x+y & 2 x \\ -5 x & x+y \end{array}\right]} \\ \Rightarrow & x=\frac{-1}{11} \text { and } y=\frac{2}{11} \\ \therefore & 2 x+3 y=2\left(\frac{-1}{11}\right)+3\left(\frac{2}{11}\right)=\frac{4}{11} \end{array} \end{aligned}$$

Experiment succeeds twice as often as it fails.

$\therefore \quad$ According to the given condition, if 'p' is success and 'q' is failure, then p = 2q

$$\begin{aligned} \therefore \quad \mathrm{p}+\mathrm{q}=1 & \Rightarrow 2 \mathrm{q}+\mathrm{q}=1 \\ & \Rightarrow \mathrm{q}=\frac{1}{3} \text { and } \mathrm{p}=\frac{2}{3} \end{aligned}$$

Here, $\mathrm{n}=6$

Let $X$ be the random variable

$$\begin{array}{ll} \therefore \quad & X \sim B(n, p) \\ \therefore \quad & \text { Required probability } \\ & =\mathrm{P}(\mathrm{X} \geq 4) \\ & =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \\ & ={ }^6 \mathrm{C}_4 \mathrm{p}^4 \mathrm{q}^2+{ }^6 \mathrm{C}_5 \mathrm{p}^5 \mathrm{q}+{ }^6 \mathrm{C}_6 \mathrm{p}^6 \\ & =15 \times\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^2+6\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)^6 \\ & =\frac{496}{729} \end{array}$$

According to the given condition,

$$\begin{array}{ll} & \mathrm{f}^{\prime}(1)=0 \text { and } \mathrm{f}^{\prime}(2)=0 \\ & \mathrm{f}(x)=\alpha \log x+\beta x^2+x \\ \therefore \quad & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1\\ \therefore \quad & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 \quad \text{.... (i)}\\ & \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2 \quad \text{.... (ii)} \end{array}$$

$\therefore \quad$ From (i) and (ii), we get

$$\begin{aligned} & \beta=\frac{-1}{2} \text { and } \alpha=2 \\ \therefore \quad & \alpha^2+2 \beta=4-1=3 \end{aligned}$$

$$\begin{aligned} & \quad \cos (x+y) \mathrm{d} y=\mathrm{d} x \\ & \therefore \quad \frac{\mathrm{d} x}{\mathrm{~d} y}=\cos (x+y) \quad \text{..... (i)} \end{aligned}$$

Put $x+y=\mathrm{u}$ .... (ii)

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

$\frac{\mathrm{d} x}{\mathrm{~d} y}+1=\frac{\mathrm{du}}{\mathrm{d} y}$

$\therefore \quad \frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{\mathrm{du}}{\mathrm{d} y}-1$ ..... (iii)

Substituting (ii) and (iii) in (i), we get

$$\begin{aligned} & \quad \frac{\mathrm{du}}{\mathrm{d} y}-1=\cos \mathrm{u} \\ & \therefore \quad \frac{\mathrm{du}}{1+\operatorname{cosu}}=\mathrm{d} y \\ & \therefore \quad \frac{\mathrm{du}}{2 \cos ^2\left(\frac{\mathrm{u}}{2}\right)}=\mathrm{d} y \end{aligned}$$

Integrating on both sides, we get

$$\begin{aligned} & \frac{1}{2} \int \sec ^2\left(\frac{\mathrm{u}}{2}\right) \mathrm{du}=\int \mathrm{d} y \\ \therefore \quad y & =\tan \left(\frac{x+y}{2}\right)+\mathrm{c} \end{aligned}$$

$$\begin{aligned} z & =\frac{(1+i)^2}{a+i} \\ & =\frac{2 i}{a+i} \\ & =\frac{2 i(a-i)}{(a+i)(a-i)} \\ & =\frac{2+2 a i}{a^2+1} \quad \text{.... (i)} \end{aligned}$$

$$\begin{aligned} & \quad|z|=\frac{2}{\sqrt{5}} \Rightarrow \frac{4}{\left(a^2+1\right)^2}+\frac{4 a^2}{\left(a^2+1\right)^2}=\frac{4}{5} \\ & \Rightarrow 20+20 a^2=4\left(a^4+2 a^2+1\right) \\ & \Rightarrow 4 a^4-12 a^2-16=0 \\ & \Rightarrow a^4-3 a^2-4=0 \\ & \Rightarrow\left(a^2-4\right)\left(a^2+1\right)=0 \\ & \Rightarrow a^2=4 ~\text{and}~ a^2=-1 \\ & \Rightarrow a=2 \quad ... [\because a > 0]\\ & \therefore \quad \text { (i) } \Rightarrow z=\frac{2}{5}+\frac{4}{5} i \\ & \therefore \quad \bar{z}=\frac{2}{5}-\frac{4}{5} i \end{aligned}$$

$$\begin{aligned} & x=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1} \mathrm{a}\right)\right)\right)\right)\right) \\ &=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{a}^2}}\right)\right)\right)\right)\right) \\ &=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\frac{1}{\sqrt{1-\mathrm{a}^2}}\right)\right)\right)\right) \\ &=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cos ^{-1} \frac{1}{\sqrt{2-\mathrm{a}^2}}\right)\right)\right) \\ &=\operatorname{cosec}\left(\tan ^{-1}\left(\frac{1}{\sqrt{2-\mathrm{a}^2}}\right)\right) \\ &=\operatorname{cosec}\left(\operatorname{cosec}^{-1}\left(\sqrt{3-\mathrm{a}^2}\right)\right) \\ & \therefore \quad x=\sqrt{3-\mathrm{a}^2} \\ & \therefore \quad x^2+\mathrm{a}^2=3 \end{aligned}$$

Given equation of the circle is

$x^2+y^2+2 x+2 y-3=0$

Which can be written as: $(x+1)^2+(y+1)^2=5$

It is a circle with centre $(-1,-1)$ and radius $\sqrt{5}$

Given line is: $2 x+y+13=0$

To find the required distance, we find the equation of a line perpendicular to the given line, and passing through the centre of the given circle.

$\therefore \quad$ Equation of this line is: $(y+1)=\frac{1}{2}(x+1)$ i.e., $x=2 y+1$

Now, we find the points where line $x=2 y+1$ intersects the circle $x^2+y^2+2 x+2 y-3=0$

$$\begin{array}{ll} \therefore & (2 y+1)^2+y^2+2(2 y+1)+2 y-3=0 \\ \therefore & 4 y^2+4 y+1+y^2+4 y+2+2 y-3=0 \\ \therefore & 5 y^2+10 y=0 \\ \therefore & y(y+2)=0 \\ \therefore & y=0 \text { or } y=-2 \\ \therefore & x=1 \text { or } x=-3 \end{array}$$

$\therefore \quad(1,0)$ and $(-3,-2)$ are the points on the circle, and one of them is at the maximum distance from the given line.

$$\begin{array}{ll} \therefore & \mathrm{d}_1=\left|\frac{2(1)+(0)+13}{\sqrt{4+1}}\right| \text { and } \mathrm{d}_2=\left|\frac{2(-3)+(-2)+13}{\sqrt{4+1}}\right| \\ \therefore & \mathrm{d}_1=\frac{15}{\sqrt{5}}=3 \sqrt{5} \quad \text { and } \mathrm{d}_2=\frac{5}{\sqrt{5}}=\sqrt{5} \\ \therefore & \lambda=3 \sqrt{5} \end{array}$$

Let $\mathrm{X}$ denotes the number of jacks

$\therefore$ Possible values of $\mathrm{X}$ are $0,1,2$

$\therefore \quad P(X=0)=\frac{{ }^{48} C_1 \times{ }^{48} C_1}{{ }^{52} C_1 \times{ }^{52} C_1}=\frac{144}{169}$

$$\begin{aligned} & \mathrm{P}(\mathrm{X}=1)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}+\frac{{ }^4 \mathrm{C}_1 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{24}{169} \\ & \mathrm{P}(\mathrm{X}=2)=\frac{{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{1}{169} \end{aligned}$$

$\therefore \quad$ Option (A) is correct.

$\log (x+y)=2 x y$ .... (i)

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

$$\begin{array}{ll} & \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 y+2 x \frac{\mathrm{d} y}{\mathrm{~d} x} \\ \therefore \quad & \frac{1}{x+y}+\frac{1}{x+y} \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y+2 x \frac{\mathrm{d} y}{\mathrm{~d} x} \quad \text{.... (ii)}\\ & \text { At } x=0, \text { (i) } \Rightarrow y=1 \\ \therefore \quad & \text { (ii) }\left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=0}=1 \end{array}$$