MHT CET 2023 10th May Morning Shift (Mathematics) Question Paper with Solutions

First six prime numbers greater than 5 are 7,11, 13, 17, 19, 23.

$$\begin{aligned} & \text { Variance }\left(\sigma^2\right)=\frac{1}{\mathrm{n}}\left(\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^2\right)-(\bar{x})^2 \\ & \therefore \quad \sigma^2=\frac{1}{6}\left(7^2+11^2+13^2+17^2+19^2+23^2\right) \\ & -\left(\frac{7+11+13+17+19+23}{6}\right)^2 \\ & =\frac{1518}{6}-\left(\frac{90}{6}\right)^2 \\ & =253-225 \\ & =28 \\ \end{aligned}$$

Let $\overline{\mathrm{d}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}$

$$\begin{aligned} \overline{\mathrm{d}} & =(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+2 \lambda \hat{\mathrm{i}}+\lambda \hat{\mathrm{j}}-\lambda \hat{\mathrm{k}} \\ & =(2 \lambda+2) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(2-\lambda) \hat{\mathrm{k}} \end{aligned}$$

Now, $\overline{\mathrm{d}}$ is perpendicular to $\overline{\mathrm{c}}$.

$$\begin{aligned} \therefore & \quad \bar{c} \cdot \bar{d}=0 \\ & \quad \Rightarrow(\hat{i}+3 \hat{j}) \cdot[(2 \lambda+2) \hat{i}+(3+\lambda) \hat{j}+(2-\lambda) \hat{k}]=0 \\ & \quad \Rightarrow 1(2 \lambda+2)+3(3+\lambda)=0 \\ & \quad \Rightarrow 2 \lambda+2+9+3 \lambda=0 \\ & \quad \Rightarrow 5 \lambda+11=0 \\ & \Rightarrow \lambda=\frac{-11}{5} \end{aligned}$$

When player tosses 2 fair coins, then

$\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$

Let $\mathrm{X}$ be a random variable that denotes the amount received by player.

Then, $\mathrm{X}$ can take values 5, 2 and 1

Now, $P(X=5)=\frac{1}{4}, P(X=2)=\frac{1}{2}$ and $\mathrm{P}(\mathrm{X}=1)=\frac{1}{4}$

$\therefore \quad$ The probability distribution of $\mathrm{X}$ is as follows:

$$\begin{aligned} \text { Variance of } X & =\sum X^2 P(X)-\left[\sum X P(X)\right]^2 \\ & =\left[\frac{25}{4}+2+\frac{1}{4}\right]-\left[\frac{5}{4}+1+\frac{1}{4}\right]^2 \\ & =\frac{34}{4}-\left(\frac{10}{4}\right)^2 \\ & =\frac{34}{4}-\frac{25}{4} \\ & =\frac{9}{4} \end{aligned}$$

Line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$

The direction ratios of the line are $3,-5,2$.

The direction ratios of the normal to the plane are $1,3,-\alpha$.

The given line is perpendicular to the normal of plane.

$$\begin{aligned} \therefore \quad & 3(1)+(-5)(3)+2(-\alpha)=0 \\ & \Rightarrow 3-15-2 \alpha=0 \\ & \Rightarrow-12-2 \alpha=0 \\ & \Rightarrow \alpha=-6 \end{aligned}$$

Also, point $(2,1,-2)$ lies on the plane

$$\begin{aligned} & x+3 y-\alpha z+\beta=0 \\ & \Rightarrow 2+3-(-6)(-2)+\beta=0 \\ & \Rightarrow 2+3-12+\beta=0 \\ & \Rightarrow \beta=7 \\ \therefore \quad & \alpha^2+\alpha \beta+\beta^2=36-42+49=43 \end{aligned}$$

Given that angle between $\overline{\mathrm{a}} \times \overline{\mathrm{b}}$ and ' in math mode at position 22: …ine{\mathrm{c}}̲ is \frac{\pi}…" style="color:#cc0000">\overline{\mathrm{c}} is $\frac{\pi}{4}

$\therefore \quad|(\bar{a} \times \bar{b}) \times \bar{c}|=|(\bar{a} \times \bar{b})||\bar{c}| \sin \frac{\pi}{4}$ ..... (i)

Now, $$\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|$$

$$\begin{aligned} & =\hat{\mathrm{i}}(0+2)-\hat{\mathrm{j}}(0+2)+\hat{\mathrm{k}}(2-1) \\ & =2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \therefore \quad |\overrightarrow{\mathrm{a}} \times \overline{\mathrm{b}}|= & \sqrt{2^2+(-2)^2+1}=3 \end{aligned}$$

Given, $\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

$a=\sqrt{2^2+1^2+(-2)^2}=3$

Given, $|\bar{c}-\bar{a}|=2 \sqrt{2}$

Squaring on both sides, we get

$$\begin{aligned} & |\bar{c}|^2+|\bar{a}|^2-2 \bar{a} \cdot \bar{c}=8 \\ & \Rightarrow|\bar{c}|^2+3^2-2|\bar{c}|=8 \quad ...[\because \bar{a} . \bar{c}=|\bar{c}|] \end{aligned}$$

$$\begin{aligned} & \Rightarrow|\bar{c}|^2-2|\bar{c}|+1=0 \\ & \Rightarrow(|\bar{c}|-1)^2=0 \\ & \Rightarrow|\bar{c}|=1 \end{aligned}$$

From (i),

$$\begin{aligned} |(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}| & =|(\overline{\mathrm{a}} \times \overline{\mathrm{b}})||\overline{\mathrm{c}}| \cdot \sin \frac{\pi}{4} \\ & =3 \times 1 \times \frac{1}{\sqrt{2}} \\ & =\frac{3}{\sqrt{2}} \end{aligned}$$

We have, $\mathrm{f}(x)=2 x-3, \mathrm{~g}(x)=x^3+5$

$$\begin{aligned} \mathrm{fog}(x) & =\mathrm{f}(\mathrm{g}(x)) \\ & =2 \mathrm{~g}(x)-3 \\ & =2\left(x^3+5\right)-3=2 x^3+7 \end{aligned}$$

Let $(\mathrm{fog})(x)=y=2 x^3+7$

$$\begin{aligned} & y=2 x^3+7 \\ & \Rightarrow y-7=2 x^3 \\ & \Rightarrow x^3=\frac{y-7}{2} \\ & \Rightarrow x=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ \therefore \quad & (\text { fog })^{-1}(y)=\left(\frac{y-7}{2}\right)^{\frac{1}{3}} \\ \therefore \quad & (\text { fog })^{-1}(-9)=\left(\frac{-9-7}{2}\right)^{\frac{1}{3}}=(-8)^{\frac{1}{3}}=-2 \end{aligned}$$

Let the length, breadth and depth of open tank be $x, x$ and $y$ respectively.

Volume $(\mathrm{V})=x^2 y$

$\therefore \quad 500=x^2 y$ ..... (i)

Total surface area of open tank is given by

$\mathrm{S}=x^2+4 x y$ ..... (ii)

From (i), $y=\frac{500}{x^2}$

From (ii), $\mathrm{S}=x^2+4 x \times \frac{500}{x^2}$

$=x^2+\frac{2000}{x}$

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

$\frac{\mathrm{dS}}{\mathrm{d} x}=2 x-\frac{2000}{x^2}$

For minimum area, $\frac{\mathrm{dS}}{\mathrm{d} x}=0$

$$\begin{aligned} \therefore \quad & 2 x-\frac{2000}{x^2}=0 \\ & \Rightarrow 2000=2 x^3 \\ & \Rightarrow x^3=1000 \end{aligned}$$

$$\begin{aligned} & \Rightarrow x=10 \mathrm{~m} \\ & \frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{~d} x^2}=2+\frac{4000}{x^3} \\ & \Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{~d} x^2}\right)_{x=10}>0 \end{aligned}$$

$\mathrm{S}$ is minimum when $x=10 \mathrm{~m}$ and $y=5 \mathrm{~m}\quad$ ...[From (i)]

The probability that the first critic favors the book is $\mathrm{P}(\mathrm{A})=\frac{2}{2+5}=\frac{2}{7}$

$\therefore \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\frac{2}{7}=\frac{5}{7}$

The probability that the second critic favors the book is $P(B)=\frac{3}{3+4}=\frac{3}{7}$

$\therefore \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{3}{7}=\frac{4}{7}$

The probability that the third critic favors the book is $\mathrm{P}(\mathrm{C})=\frac{4}{4+3}=\frac{4}{7}$

$\therefore \quad \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{4}{7}=\frac{3}{7}$

$\therefore \quad$ Majority will be in favor of the book if at least two critics favor the book.

Hence, the probability is

$$\begin{aligned} & \mathrm{P}\left(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}^{\prime}\right) \quad+\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime} \cap \mathrm{C}\right) \\ & \quad+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B} \cap \mathrm{C}\right)+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}) \\ & \begin{aligned} = & \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}(\mathrm{C}) \\ & \quad+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{C}) \\ = & \frac{2}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7} \\ = & \frac{18}{343}+\frac{32}{343}+\frac{60}{343}+\frac{24}{343}=\frac{134}{343} \end{aligned} \end{aligned}$$

A committee of 5 is to be formed from 8 boys and 5 girls so that it consists of at least 2 girls and at most 2 boys.

Total number of ways

$$\begin{aligned} & =(\text { Five girls })+(4 \text { girls })(1 \text { boy })+(3 \text { girls })(2 \text { boys }) \\ & ={ }^5 C_5+{ }^5 C_4 \times{ }^8 C_1+{ }^5 C_3 \times{ }^8 C_2 \\ & =1+(5 \times 8)+(10 \times 28) \\ & =1+40+280 \\ & =321 \end{aligned}$$

$$\begin{aligned} & \text { Given differential equation is } \\ & \frac{\mathrm{dP}}{\mathrm{dt}}=0.5 \mathrm{P}-450 \\ & =\frac{P}{2}-\frac{900}{2} \\ & \therefore \quad \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{P}-900}{2} \\ & \therefore \quad \frac{2 \mathrm{dP}}{\mathrm{P}-900}=\mathrm{dt} \\ & \text { Integrating on both sides, we get } \\ & 2 \log |\mathrm{P}-900|=\mathrm{t}+\mathrm{c} \\ & \mathrm{P}(0)=850 \text { i.e., } \mathrm{P}=850 \text { when } \mathrm{t}=0 \\ & \therefore \quad c=2 \log 50 \\ & \therefore \quad 2 \log |\mathrm{P}-900|=\mathrm{t}+2 \log 50 \\ & \text { When } \mathrm{P}=0 \text {, } \\ & 2 \log 900=\mathrm{t}+2 \log 50 \\ & \Rightarrow \mathrm{t}=2(\log 900-\log 50) \\ & =2 \log \frac{900}{50}=2 \log 18 \\ \end{aligned}$$

$$\begin{aligned} \text { Required area } & =2 \int_0^7 \sqrt{49-x^2} \mathrm{~d} x \\ & =2\left[\frac{x}{2} \sqrt{49-x^2}+\frac{49}{2} \sin ^{-1}\left(\frac{x}{7}\right)\right]_0^7 \\ & =2\left[\frac{7}{2} \sqrt{49-49}+\frac{49}{2} \sin ^{-1}\left(\frac{7}{7}\right)\right]-0 \\ & =2 \times \frac{49}{2} \times \frac{\pi}{2} \\ & =\frac{49 \pi}{2} \text { sq. units } \end{aligned}$$

In $\triangle \mathrm{ABC}, \mathrm{AC}$ represents ladder

$\mathrm{AB} \rightarrow$ vertical wall

Let $\mathrm{AB}=x, \mathrm{BC}=y$

$\therefore \quad \angle \mathrm{ABC}=90^{\circ}$

By Pythagoras theorem,

$$\begin{aligned} & \mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2 \\ & \Rightarrow x^2+y^2=17^2 \\ & \Rightarrow x^2=289-y^2 \\ & \Rightarrow x^2=289-64 \\ & \Rightarrow x^2=225 \\ & \Rightarrow x=15 \mathrm{~m} \end{aligned}$$

Consider equation (i),

$x^2=289-y^2$

Differentiating w.r.t. t, we get

$$\begin{aligned} & 2 x \frac{\mathrm{d} x}{\mathrm{dt}}=-2 y \frac{\mathrm{d} y}{\mathrm{dt}} \\ & \Rightarrow 15 \frac{\mathrm{d} x}{\mathrm{dt}}=-8(1) \\ & \Rightarrow \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{-8}{15} \mathrm{~m} / \mathrm{s} \end{aligned}$$

Negative sign shows that the ladder is moving down. i.e., vertical length is decreasing

$\therefore \quad$ Upper end is coming down at the rate of $\frac{8}{15} \mathrm{~m} / \mathrm{s}$.

$$\begin{aligned} \text { Let I } \quad & =\int \frac{1}{7-6 x-x^2} d x \\ & =\int \frac{1}{7-6 x-x^2-9+9} d x \\ & =\int \frac{1}{16-\left(x^2+6 x+9\right)} d x \\ & =\int \frac{1}{(4)^2-(x+3)^2} d x \\ & =\frac{1}{8} \log \left|\frac{4+x+3}{4-(x+3)}\right|+c \\ & =\frac{1}{8} \log \left|\frac{7+x}{1-x}\right|+c \end{aligned}$$

Feasible region lies on origin side of $x+y=4, x=2, y=1$ and non-origin side of $x+y=1$ and in first quadrant.

$\therefore$ Vertices of feasible region are $A(1,0), B(2,0), C(2,1), D(0,1)$

$[\sim(\sim p \vee q) \vee(p \wedge r)] \wedge(\sim q \wedge r)$

$\equiv[(p \wedge \sim q) \vee(p \wedge r)] \wedge(\sim q \wedge r)$ ...[De Morgan's law]

$\equiv p \wedge(\sim q \vee r) \wedge(\sim q \wedge r)$ ...[Distributive law]

$\equiv p \wedge[(\sim q \vee r) \wedge \sim q] \wedge r$ ...[Associative law]

$\equiv p \wedge(\sim q) \wedge r$ ....[Absorption law]

$\equiv(p \wedge r) \wedge \sim q$ ...[Commutative law]

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

Put $\tan \frac{x}{2}=\mathrm{t}$

$$\begin{aligned} & \Rightarrow x=2 \tan ^{-1} \mathrm{t} \\ & \Rightarrow \mathrm{d} x=\frac{2}{1+\mathrm{t}^2} \mathrm{dt} \end{aligned}$$

and $\cos x=\frac{1-t^2}{1+t^2}, \sin x=\frac{2 t}{1+t^2}$

$$\begin{aligned} \therefore \quad I & =\int \frac{\frac{2}{1+t^2}}{\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} d t \\ & =\int \frac{2}{2 t+1-t^2} d t \\ & =-2 \int \frac{1}{t^2-2 t-1} d t \\ & =-2 \int \frac{1}{t^2-2 t+1-1-1} \\ & =-2 \int \frac{1}{(t-1)^2-(\sqrt{2})^2} \end{aligned}$$

$$\begin{aligned} & =-2 \times \frac{1}{2 \sqrt{2}} \log \left|\frac{\mathrm{t}-1-\sqrt{2}}{\mathrm{t}-1+\sqrt{2}}\right|+\mathrm{c} \\ & =\frac{-1}{\sqrt{2}} \log \left|\frac{\tan \frac{x}{2}-1-\sqrt{2}}{\tan \frac{x}{2}-1+\sqrt{2}}\right|+\mathrm{c} \\ & =\frac{-1}{\sqrt{2}} \log \left|\frac{\tan \frac{x}{2}-(\sqrt{2}+1)}{\tan \frac{x}{2}+\sqrt{2}-1}\right|+\mathrm{c} \end{aligned}$$

Using $|\operatorname{Adj} A|=|A|^{n-1}$

But $\mathrm{B}=\operatorname{Adj}(\mathrm{A})\quad$ .... [Given]

$\therefore \quad|\mathrm{B}|=|\mathrm{A}|^2$

$$\begin{aligned} & \Rightarrow\left|\begin{array}{lll} 1 & \alpha & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3 \end{array}\right|=|\mathrm{A}|^2 \\ & \Rightarrow \alpha-2=5^2 \\ & \Rightarrow \alpha-2=25 \\ & \Rightarrow \alpha=27 \end{aligned}$$

$$\begin{array}{ll} & x y=1 \\ \therefore \quad & y=\frac{1}{x} \\ \therefore \quad & y^{\prime}=\frac{-1}{x^2} \\ \therefore \quad & \text { Slope of the normal }=x^2 \\ & \text { Slope of the line } \mathrm{a} x+\mathrm{b} y+\mathrm{c}=0 \text { is } \frac{-\mathrm{a}}{\mathrm{b}} . \end{array}$$

Since the line $\mathrm{a} x+\mathrm{b} y+\mathrm{c}=0$ is a normal to the curve $x y=1$,

$x^2=-\frac{a}{b}$

For this condition to hold true, either $\mathrm{a} < 0, \mathrm{~b} > 0$ or $\mathrm{b} < 0, \mathrm{a} > 0$

Let $\overline{\mathrm{r}}$ be the vector coplanar to $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$. Then,

$$\begin{aligned} \overline{\mathrm{r}} & =\overline{\mathrm{a}}+\mathrm{m} \overline{\mathrm{b}} \\ & =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{m}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =\hat{\mathrm{i}}(1+\mathrm{m})+\hat{\mathrm{j}}(2-\mathrm{m})+\hat{\mathrm{k}}(1+\mathrm{m}) \quad \text{.... (i)} \end{aligned}$$

Since the projection of $\overline{\mathrm{r}}$ along $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}, \frac{\overline{\mathrm{r}} \cdot \overline{\mathrm{c}}}{|\overline{\mathrm{c}}|}= \pm \frac{1}{\sqrt{3}}$

$$\begin{aligned} & \Rightarrow \frac{(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})}{\sqrt{3}}= \pm \frac{1}{\sqrt{3}} \\ & \Rightarrow(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})= \pm 1 \\ \therefore \quad & \mathrm{m}=3 \text { or } \mathrm{m}=1 \end{aligned}$$

Substituting $\mathrm{m}=3$ in equation (i), we get

$$\begin{aligned} & \bar{r}=\hat{i}(1+3)+\hat{j}(2-3)+\hat{k}(1+3) \\ & \Rightarrow \bar{r}=4 \hat{i}-\hat{j}+4 \hat{k} \end{aligned}$$

Consider, $\tan ^{-1} x+\tan ^{-1} 6 x=\frac{\pi}{4}$

$$\begin{aligned} & \Rightarrow \tan ^{-1}\left(\frac{7 x}{1-6 x^2}\right)=\frac{\pi}{4} \\ & \quad \ldots\left[\tan ^{-1}(x)+\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \\ & \Rightarrow \frac{7 x}{1-6 x^2}=1 \\ & \Rightarrow 7 x=1-6 x^2 \\ & \Rightarrow 6 x^2+7 x-1=0 \\ & \Rightarrow x=\frac{-7 \pm \sqrt{73}}{12} \\ & \text { Since } x \geq 0 \\ & x=\frac{-7+\sqrt{73}}{12} \end{aligned}$$

$\mathrm{A}$ is a singleton set.