MHT CET 2024 4th May Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& {\sin }^2𝜃-\cos 𝜃=\frac{1}{4}\\ & (1-{\cos }^2𝜃)-\cos 𝜃=\frac{1}{4}\\ & 4-4{\cos }^2𝜃-4\cos 𝜃-1=0\\ & \Rightarrow 4{\cos }^2𝜃+4\cos 𝜃-3=0\\ & \Rightarrow \cos 𝜃=\frac{-1}{2}\text{ or }\cos 𝜃=\frac{1}{2}\\ \therefore & 𝜃=2𝜋-\frac{𝜋}{3},\frac{𝜋}{3}\\ \therefore & 𝜃=\{\frac{5𝜋}{3},\frac{𝜋}{3}\}\end{array}$

Since the points $(1,-1,𝜆)$ and $(-3,0,1)$ are equidistant from the given plane

$\begin{array}{cc}& \mid \frac{3+4-12𝜆+13}{\sqrt{\surd 9+16+144}}\mid =\mid \frac{-9-12+13}{\sqrt{\surd 9+16+144}}\mid \\ & \Rightarrow |3+4-12𝜆+13|=|-9-12+13|\\ & \Rightarrow 20-12𝜆=\pm 8\end{array}$

$\begin{array}{cc}& \Rightarrow 𝜆=1,\frac{7}{3}\\ & \therefore \text{ Sum of all possible values of }𝜆=1+\frac{7}{3}=\frac{10}{3}\end{array}$

To find the angle between two vectors, use the formula for the dot product, given by:

$\overline{𝑢}\cdot \overline{𝑣}=\Vert \overline{𝑢}\Vert \Vert \overline{𝑣}\Vert \cos 𝜃$

where $𝜃$ is the angle between the vectors $\overline{𝑢}$ and $\overline{𝑣}$.

First, calculate the vectors $2\overline{𝑎}+\overline{𝑏}$ and $\overline{𝑎}+2\overline{𝑏}$.

Given:

$\overline{𝑎}=\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘}$

$\overline{𝑏}=2\widehat{𝑖}+3\widehat{𝑗}-\widehat{𝑘}$

Calculate $2\overline{𝑎}+\overline{𝑏}$:

$2\overline{𝑎}=2(\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘})=2\widehat{𝑖}-4\widehat{𝑗}+6\widehat{𝑘}$

$2\overline{𝑎}+\overline{𝑏}=(2\widehat{𝑖}-4\widehat{𝑗}+6\widehat{𝑘})+(2\widehat{𝑖}+3\widehat{𝑗}-\widehat{𝑘})=4\widehat{𝑖}-\widehat{𝑗}+5\widehat{𝑘}$

Calculate $\overline{𝑎}+2\overline{𝑏}$:

$2\overline{𝑏}=2(2\widehat{𝑖}+3\widehat{𝑗}-\widehat{𝑘})=4\widehat{𝑖}+6\widehat{𝑗}-2\widehat{𝑘}$

$\overline{𝑎}+2\overline{𝑏}=(\widehat{𝑖}-2\widehat{𝑗}+3\widehat{𝑘})+(4\widehat{𝑖}+6\widehat{𝑗}-2\widehat{𝑘})=5\widehat{𝑖}+4\widehat{𝑗}+\widehat{𝑘}$

Now, find the dot product $(2\overline{𝑎}+\overline{𝑏})\cdot (\overline{𝑎}+2\overline{𝑏})$:

$(4\widehat{𝑖}-\widehat{𝑗}+5\widehat{𝑘})\cdot (5\widehat{𝑖}+4\widehat{𝑗}+\widehat{𝑘})=4\cdot 5+(-1)\cdot 4+5\cdot 1=20-4+5=21$

Calculate the magnitudes of the vectors:

$\Vert 2\overline{𝑎}+\overline{𝑏}\Vert =\sqrt{\surd 4^2+(-1{)}^2+5^2}=\sqrt{\surd 16+1+25}=\sqrt{\surd 42}$

$\Vert \overline{𝑎}+2\overline{𝑏}\Vert =\sqrt{\surd 5^2+4^2+1^2}=\sqrt{\surd 25+16+1}=\sqrt{\surd 42}$

Using the dot product and magnitudes to find $\cos 𝜃$:

$\cos 𝜃=\frac{21}{\sqrt{\surd 42}\times \sqrt{\surd 42}}=\frac{21}{42}=\frac{1}{2}$

The angle $𝜃$ for which $\cos 𝜃=\frac{1}{2}$ is $\frac{𝜋}{3}$.

Therefore, the angle between the vectors is $\frac{𝜋}{3}$.

Option B: $\frac{𝜋}{3}$

The corner points of feasible region are $𝐴(4,0)$, $B(0,4),C(0,6)$

$\begin{array}{cc}& \begin{array}{cc}& \text{ At }𝐴(4,0),𝑍=4\left(4\right)+6\left(0\right)=16\\ & \text{ At }𝐵(0,4),𝑍=4\left(0\right)+6\left(4\right)=24\\ & \text{ At }𝐶(0,6),𝑍=4\left(0\right)+6\left(6\right)=36\end{array}\\ & \therefore Z\text{ has maximum value at }C(0,6)\text{ which is }36\text{. }\end{array}$

$\begin{array}{cc}& \text{ Let }𝑦={\cos }^{-1}\left(\frac{𝑥-\frac{1}{𝑥}}{𝑥+\frac{1}{𝑥}}\right)\\ & \begin{array}{cc}& ={\cos }^{-1}\left(\frac{{𝑥}^2-1}{{𝑥}^2+1}\right)={\cos }^{-1}\left[\right(-1\left)\left(\frac{1-{𝑥}^2}{1+{𝑥}^2}\right)\right]\\ & 𝑦=𝜋-{\cos }^{-1}\left(\frac{1-{𝑥}^2}{1+{𝑥}^2}\right)\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Put }𝑥=\tan 𝜃\\ \therefore & 𝜃={\tan }^{-1}𝑥\\ \therefore & 𝑦=𝜋-{\cos }^{-1}\left(\frac{1-{\tan }^2𝜃}{1+{\tan }^2𝜃}\right)\\ \Rightarrow & 𝑦=𝜋-{\cos }^{-1}(\cos 2𝜃)\\ \Rightarrow & 𝑦=𝜋-2𝜃\\ \Rightarrow & 𝑦=𝜋-2{\tan }^{-1}𝑥\end{array}\\ & \text{ Differentiating w.r.to }𝑥\text{, we get }\\ & \frac{d𝑦}{d𝑥}=0-\frac{2}{1+{𝑥}^2}=\frac{-2}{1+{𝑥}^2}\end{array}$

$\begin{array}{cc}& 2\bar{a}\cdot \bar{p}=\frac{2(\bar{b}\times \bar{c})\cdot \bar{a}}{\left[\bar{a}\bar{b}\bar{c}\right]}=\frac{2\left[\bar{b}\bar{c}\bar{a}\right]}{\left[\bar{a}\bar{b}\bar{c}\right]}=2\\ & \bar{b}\cdot \bar{q}=\frac{(\bar{c}\times \bar{a})\cdot \bar{b}}{\left[\bar{a}\bar{b}\bar{c}\right]}=\frac{\left[\bar{c}\bar{a}\bar{b}\right]}{\left[\bar{a}\bar{b}\bar{c}\right]}=1\\ & \bar{c}\cdot \bar{r}=\frac{(\bar{a}\times \bar{b})\cdot \bar{c}}{\left[\bar{a}\bar{b}\bar{c}\right]}=\frac{\left[\bar{a}\bar{b}\bar{c}\right]}{\left[\bar{a}\bar{b}\bar{c}\right]}=1\\ \therefore & 2\bar{a}\cdot \bar{p}+\bar{b}\cdot \bar{q}+\bar{c}\cdot \bar{r}=2+1+1=4\end{array}$

$\begin{array}{cc}& \text{ Let }f\left(𝑥\right)={\tan }^{-1}𝑥\\ & \therefore f^{\prime }\left(𝑥\right)=\frac{1}{1+{𝑥}^2}\\ & \begin{array}{cc}& \text{ Here, }a=1\text{ and }h=-0.001\\ & \therefore f(a+h)\approx f\left(a\right)+\mathrm{hf}{}^{\prime }\left(a\right)\\ & \therefore {\tan }^{-1}\left(0.999\right)\approx \frac{𝜋}{4}+\frac{1}{1+1}(-0.001)\\ & \approx \frac{𝜋}{4}-\frac{0.001}{2}\\ & \approx \frac{𝜋}{4}-0.0005\\ & \approx \frac{3.1415}{4}-0.0005\\ & \approx 0.7849\end{array}\end{array}$

Equation of the plane passing through $(2,1,0)$, $(4,1,1)$ and $(5,0,1)$ is

$\begin{array}{cc}& \mid \begin{array}{ccc}𝑥-2& 𝑦-1& 𝑧-0\\ 4-2& 1-1& 1-0\\ 5-2& 0-1& 1-0\end{array}\mid =0\\ & \Rightarrow 𝑥+𝑦-2𝑧=3\end{array}$

$R^{\prime }(𝑥,𝑦,z)$ is image of $R(2,1,6)$ w.r.t. to plane

$\begin{array}{cc}& 𝑥+𝑦-2z=3\\ & \Rightarrow \frac{𝑥-2}{1}=\frac{𝑦-1}{1}=\frac{z-6}{-2}=\frac{-2[2+1-2(6)-3]}{1+1+4}\\ & \Rightarrow \frac{𝑥-2}{1}=\frac{𝑦-1}{1}=\frac{z-6}{-2}=4\\ & \Rightarrow 𝑥=6,𝑦=5,z=-2\\ \therefore & R^{\prime }(𝑥,𝑦,z)\equiv (6,5,-2)\end{array}$

$\begin{array}{cc}& \mathrm{OD}\text{ is the median }\\ & \begin{array}{cc}\therefore & D\equiv (\frac{1+3}{2},\frac{2+4}{2})\\ & \Rightarrow D\equiv (2,3)\end{array}\end{array}$

Equation of OD is $𝑦=𝑚𝑥$

$\begin{array}{cc}& \Rightarrow 𝑦=\frac{3}{2}𝑥\\ & \Rightarrow 3𝑥-2𝑦=0\end{array}$

Slope of line $𝐴𝐵=\frac{2}{2}=1$

Given, $\mathrm{OE}⟂\mathrm{AB}$

$\therefore$ Slope of $\mathrm{OE}=-1$

Equation of OE is $𝑦=\mathrm{mx}$

$\begin{array}{cc}& \Rightarrow 𝑦=-𝑥\\ & \Rightarrow 𝑥+𝑦=0\end{array}$

$\therefore$ Joint equation of median and altitude is

$\begin{array}{cc}& (3𝑥-2𝑦)(𝑥+𝑦)=0\\ & \Rightarrow 3{𝑥}^2+𝑥𝑦-2{𝑦}^2=0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Let }({𝑥}_1,{𝑦}_1,{𝑧}_1)=(-1,2,-3)\\ & a_1,b_1,c_1=3,2,-4\text{ and }\\ & a_2,b_2,c_2=2,-3,2\end{array}\\ & \therefore \text{ The equation of required plane is }\\ & \begin{array}{cc}& \mid \begin{array}{ccc}𝑥-{𝑥}_1& 𝑦-{𝑦}_1& 𝑧-{𝑧}_1\\ {𝑎}_1& {𝑏}_1& {𝑐}_1\\ a_2& b_2& {𝑐}_2\end{array}\mid =0\\ & \Rightarrow \mid \begin{array}{ccc}𝑥+1& 𝑦-2& 𝑧+3\\ 3& 2& -4\\ 2& -3& 2\end{array}\mid =0\\ & \Rightarrow (𝑥+1)(-8)-(𝑦-2)\left(14\right)+(𝑧+3)(-13)=0\\ & \Rightarrow -8𝑥-8-14𝑦+28-13𝑧-39=0\\ & \Rightarrow -8𝑥-14𝑦-13𝑧-19=0\\ & \Rightarrow 8𝑥+14𝑦+13𝑧+19=0\end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }\bar{a}=2\widehat{𝑗}+\widehat{𝑘},\widehat{𝑏}=-2\widehat{𝑖},\widehat{𝑐}=-2\widehat{𝑖}+2\widehat{𝑘}\\ \therefore & \overline{\mathrm{AB}}=-2\widehat{𝑖}-2\widehat{𝑗}-\widehat{𝑘},\\ & \overline{𝐵𝐶}=2\widehat{𝑘}\\ & \overline{\mathrm{AC}}=-2\widehat{𝑖}-2\widehat{𝑗}+\widehat{𝑘}\\ \Rightarrow & \left|\stackrel{\to \leftarrow }{𝐴𝐵}\right|=3,\left|\stackrel{\to \leftarrow }{𝐵𝐶}\right|=2,\left|\stackrel{\to \leftarrow }{𝐴𝐶}\right|=3\end{array}$

$\begin{array}{cc}& \text{ Incentre of }△𝐴𝐵𝐶\text{ is given by }\\ & \begin{array}{cc}& \frac{\left|\stackrel{\to \leftarrow }{\mathrm{AB}}\right|\bar{c}+\left|\stackrel{\to \leftarrow }{\mathrm{BC}}\right|\hat{a}+\left|\stackrel{\to \leftarrow }{\mathrm{AC}}\right|\overrightarrow{b}}{\left|\overline{\mathrm{AB}}\right|+\left|\stackrel{\to \leftarrow }{\mathrm{BC}}\right|+\left|\stackrel{\to \leftarrow }{\mathrm{AC}}\right|}\\ =& \frac{3(-2\hat{i}+2\hat{k})+2(2\widehat{𝑗}+\hat{k})+3(-2\hat{i})}{3+2+3}\\ =& \frac{-12\widehat{𝑖}+4\widehat{𝑗}+8\hat{k}}{8}\\ =& -\frac{3}{2}\hat{i}+\frac{1}{2}\widehat{𝑗}+\hat{k}\end{array}\end{array}$

Given line $3𝑥+𝑦=0\Rightarrow$ slope $=-3$

Let the slope of required line be $𝑚$

$\begin{array}{cc}\therefore & \tan {45}^{\circ }=\mid \frac{m+3}{1-3m}\mid \\ & \Rightarrow 1=\mid \frac{m+3}{1-3m}\mid \\ & \Rightarrow 2m^2-3m-2=0\text{.... (i)}\end{array}$

Since the line passes through origin, its equation is $𝑦=𝑚𝑥$

$\Rightarrow m=\frac{𝑦}{𝑥}$

Substituting the value of $𝑚$ in equation (i), we get

$\begin{array}{cc}& 2{\left(\frac{𝑦}{𝑥}\right)}^2-3\left(\frac{𝑦}{𝑥}\right)-2=0\\ & \Rightarrow 2{𝑦}^2-3𝑥𝑦-2{𝑥}^2=0\\ & \Rightarrow 2{𝑥}^2+3𝑥𝑦-2{𝑦}^2=0\end{array}$

$\begin{array}{cc}& \frac{1}{𝑥}\frac{d𝑦}{d𝑥}={\tan }^{-1}𝑥\\ & d𝑦=𝑥\cdot {\tan }^{-1}𝑥d𝑥\\ & \text{ Integrating on both sides we get, }\\ & 𝑦=\int 𝑥\cdot {\tan }^{-1}𝑥d𝑥\\ & 𝑦={\tan }^{-1}𝑥\int 𝑥d𝑥-\int (\frac{d}{d𝑥}{\tan }^{-1}𝑥\cdot \int 𝑥d𝑥)d𝑥\\ & 𝑦={\tan }^{-1}𝑥\cdot \frac{{𝑥}^2}{2}-\int (\frac{1}{1+{𝑥}^2}\times \frac{{𝑥}^2}{2})d𝑥\\ & 𝑦={\tan }^{-1}𝑥\cdot \frac{{𝑥}^2}{2}-\frac{1}{2}\int \left(\frac{{𝑥}^2}{1+{𝑥}^2}\right)d𝑥\\ & =\frac{{𝑥}^2\cdot {\tan }^{-1}𝑥}{2}-\frac{1}{2}(\int \frac{{𝑥}^2+1}{{𝑥}^2+1}-\int \frac{1}{1+{𝑥}^2}d𝑥)\\ & \Rightarrow 𝑦=\frac{{𝑥}^2{\tan }^{-1}𝑥}{2}-\frac{1}{2}(𝑥-{\tan }^{-1}𝑥)+c\end{array}$

$\begin{array}{cc}\text{ Required area }& =\underset{0}{\overset{3}{\int }}({𝑥}^2+2)-(𝑥+1)𝑑𝑥\\ & =\underset{0}{\overset{3}{\int }}({𝑥}^2-𝑥+1)𝑑𝑥\\ & ={[\frac{{𝑥}^3}{3}-\frac{{𝑥}^2}{2}+𝑥]}_3^0=\frac{15}{2}\text{ sq. units }\end{array}$

Let $A({𝑥}_1,{𝑦}_1,z_1)=A(3,4,1)$

$B({𝑥}_2,{𝑦}_2,z_2)=B(5,1,6)$

The equation of line passing through the points $({𝑥}_1,{𝑦}_1,{𝑧}_1)$ and $({𝑥}_2,{𝑦}_2,{𝑧}_2)$ is given by

$\begin{array}{cc}& \frac{𝑥-{𝑥}_1}{{𝑥}_2-{𝑥}_1}=\frac{𝑦-{𝑦}_1}{{𝑦}_2-{𝑦}_1}=\frac{𝑧-{𝑧}_1}{{𝑧}_2-{𝑧}_1}\\ \therefore & \frac{𝑥-3}{5-3}=\frac{𝑦-4}{1-4}=\frac{𝑧-1}{6-1}\\ \therefore & \frac{𝑥-3}{2}=\frac{𝑦-4}{-3}=\frac{𝑧-1}{5}\end{array}$

Since the line crosses the XY plane, $z=0$

$\therefore \frac{𝑥-3}{2}=\frac{𝑦-4}{-3}=\frac{-1}{5}$

$\therefore \frac{𝑥-3}{2}=\frac{-1}{5}$ and $\frac{𝑦-4}{-3}=\frac{-1}{5}$

$\Rightarrow 𝑥=\frac{13}{5}$ and $𝑦=\frac{23}{5}$

Required point is $(\frac{13}{5},\frac{23}{5},0)$

$\begin{array}{cc}& \frac{{\tan }^{-1}\left(\sqrt{\surd 3}\right)-{\sec }^{-1}(-2)}{{\mathrm{cosec}}^{-1}(-\sqrt{\surd 2})+{\cos }^{-1}\left(\frac{-1}{2}\right)}\\ & =\frac{{\tan }^{-1}\left(\sqrt{\surd 3}\right)-{\cos }^{-1}\left(\frac{-1}{2}\right)}{{\sin }^{-1}\left(\frac{-1}{\sqrt{\surd 2}}\right)+{\cos }^{-1}\left(\frac{-1}{2}\right)}\cdots \left[\begin{array}{c}{\sec }^{-1}𝑥={\cos }^{-1}\frac{1}{𝑥}\\ {\mathrm{cosec}}^{-1}𝑥={\sin }^{-1}\frac{1}{𝑥}\end{array}\right]\end{array}$

$\begin{array}{cc}& =\frac{{\tan }^{-1}\left(\sqrt{\surd 3}\right)-𝜋+{\cos }^{-1}\left(\frac{1}{2}\right)}{-{\sin }^{-1}\left(\frac{1}{\sqrt{\surd 2}}\right)+𝜋-{\cos }^{-1}\left(\frac{1}{2}\right)}\\ & \cdots \left[\begin{array}{c}{\cos }^{-1}(-𝑥)=𝜋-{\cos }^{-1}𝑥\\ {\sin }^{-1}(-𝑥)=-{\sin }^{-1}𝑥\end{array}\right]\end{array}$

$\begin{array}{cc}& =\frac{\frac{𝜋}{3}-𝜋+\frac{𝜋}{3}}{-\frac{𝜋}{4}+𝜋-\frac{𝜋}{3}}\\ & =\frac{-\frac{𝜋}{3}}{\frac{5𝜋}{12}}\\ & =\frac{-4}{5}\end{array}$

$\begin{array}{cc}\text{ Let }I& =\int \frac{d𝑥}{{𝑥}^2{({𝑥}^4+1)}^{\frac{3}{4}}}\\ & =\int \frac{d𝑥}{{𝑥}^2{\left[{𝑥}^4(1+\frac{1}{{𝑥}^4})\right]}^{\frac{3}{4}}}\\ & =\int \frac{d𝑥}{{𝑥}^5{(1+\frac{1}{{𝑥}^4})}^{\frac{3}{4}}}\\ & =\int \frac{{(1+\frac{1}{{𝑥}^4})}^{\frac{-3}{4}}}{{𝑥}^5}\end{array}$

$\begin{array}{cc}& \text{ Let }1+\frac{1}{{𝑥}^4}=𝑡\\ & \begin{array}{cc}\Rightarrow & \frac{-4}{{𝑥}^5}𝑑𝑥=𝑑𝑡\\ & \Rightarrow \frac{𝑑𝑥}{{𝑥}^5}=-\frac{𝑑𝑡}{4}\\ \therefore 𝐼& =\frac{1}{4}\frac{{𝑡}^{\frac{-3}{4}}}{\frac{-3}{4}}𝑑𝑡\\ & =\frac{-1}{4}\frac{{𝑡}^{\frac{-3}{4}}}{\frac{-3}{4}}+1\\ & =-{𝑡}^{\frac{1}{4}}+𝑐\\ \Rightarrow 𝐼& =-{(1+\frac{1}{{𝑥}^4})}^{\frac{1}{4}}+𝑐\\ & =-{\left(\frac{{𝑥}^4+1}{{𝑥}^4}\right)}^{\frac{1}{4}}+𝑐\\ \Rightarrow 𝐼& =\frac{-{({𝑥}^4+1)}^{\frac{1}{4}}}{𝑥}+𝑐\end{array}\end{array}$

$\begin{array}{cc}(𝑎+𝑏)\cos 𝐶+(𝑏+𝑐)& \cos 𝐴+(𝑐+𝑎)\cos 𝐵=72\\ \therefore 𝑎\cos 𝐶+𝑏\cos 𝐶+𝑏\cos 𝐴& +𝑐\cos 𝐴\\ & +𝑐\cos 𝐵+𝑎\cos 𝐵=72\\ \Rightarrow 𝑎\cos 𝐶+𝑐\cos 𝐴+𝑏& \cos 𝐴+𝑎\cos 𝐵\\ & +𝑏\cos 𝐶+𝑐\cos 𝐵=72\end{array}$

$\cdots ⎡⎣⎢\begin{array}{c}\text{ By projection }\\ 𝑎=𝑏\cos 𝐶+𝑐\cos 𝐵\\ 𝑏=𝑐\cos 𝐴+𝑎\cos 𝐶\\ 𝑐=𝑎\cos 𝐵+𝑏\cos 𝐴\end{array}⎤⎦⎥$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow 𝑏+𝑐+𝑎=72\\ & \Rightarrow 𝑎+𝑏+𝑐=72\\ & \Rightarrow 18+24+𝑐=72[𝑎=18,𝑏=24\dots .\text{ Given }]\\ & \Rightarrow 𝑐=30\end{array}\\ & \begin{array}{cc}\text{ Area of }△\mathrm{ABC}& =\sqrt{\surd s(s-a)(s-b)(s-c)}\\ & =\sqrt{\surd 36(36-18)(36-24)(36-30)}\\ & =\sqrt{\surd 36\times 18\times 12\times 6}\\ & =216\text{ sq. units }\end{array}\end{array}$

$\begin{array}{cc}& {\cot }^{-1}7+{\cot }^{-1}8+{\cot }^{-1}18={\cot }^{-1}𝑥\\ & \Rightarrow {\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)={\tan }^{-1}\left(\frac{1}{𝑥}\right)\\ & \dots [{\tan }^{-1}\left(\frac{1}{𝑥}\right)={\cot }^{-1}𝑥]\end{array}$

$\begin{array}{cc}& \Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\left(\frac{1}{7}\right)\left(\frac{1}{8}\right)}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)={\tan }^{-1}\left(\frac{1}{𝑥}\right)\\ & \Rightarrow {\tan }^{-1}\left(\frac{3}{11}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)={\tan }^{-1}\left(\frac{1}{𝑥}\right)\\ & \Rightarrow {\tan }^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\left(\frac{3}{11}\right)\left(\frac{1}{18}\right)}\right)={\tan }^{-1}\left(\frac{1}{𝑥}\right)\\ & \Rightarrow 𝑥=3\end{array}$

Given probability distribution is

$\begin{array}{cc}& E\left(X\right)=\sum {𝑥}_{i}P\left({𝑥}_i\right)\\ & =2\left(0.3\right)+3\left(0.4\right)+4\left(0.3\right)\\ & \begin{array}{cc}& E\left(X\right)=3\\ & \begin{array}{cc}\text{ Variance }& =E\left({𝑥}^2\right)-\left[E\right(𝑥){]}^2\\ & =2^2\left(0.3\right)+3^2\left(0.4\right)+4^2\left(0.3\right)-(3{)}^2\\ & =0.6\end{array}\end{array}.\end{array}$