MHT CET 2025 21st April Evening Shift (Mathematics) Question Paper with Solutions

Event E: Product of the two digits is 24 .

$\begin{array}{cc}\therefore & 𝐸=\{38,46,64,83\}\\ \therefore & 𝑝=\frac{4}{90}=\frac{1}{25}\\ \Rightarrow & 𝑞=1-\frac{1}{25}=\frac{24}{25}\end{array}$

Here, $n=4$

$\begin{array}{cc}\therefore P(X\ge 3)& =P(X=3)+P(X=4)\\ & ={}^4{C}_3{\left(\frac{1}{25}\right)}^3{\left(\frac{24}{25}\right)}^1+{}^4{C}_4{\left(\frac{1}{25}\right)}^4{\left(\frac{24}{25}\right)}^0\\ & =4{\left(\frac{1}{25}\right)}^3\left(\frac{24}{25}\right)+{\left(\frac{1}{25}\right)}^4\\ & ={\left(\frac{1}{25}\right)}^4(4\times 24+1)\\ & ={\left(\frac{1}{25}\right)}^4(96+1)\\ & =\frac{97}{(25{)}^4}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Given, }3e^{𝑥}\tan 𝑦d𝑥+(1-e^{𝑥}){\sec }^2𝑦d𝑦=0\\ & \Rightarrow 3e^{𝑥}\tan 𝑦d𝑥=-(1-e^{𝑥}){\sec }^2𝑦d𝑦\\ & \Rightarrow \frac{3e^{𝑥}d𝑥}{(1-e^{𝑥})}=-\frac{{\sec }^2𝑦}{\tan 𝑦}d𝑦\end{array}\\ & \text{ Integrating on both sides, we got }\\ & \begin{array}{cc}& \int \frac{3e^{𝑥}}{(1-e^{𝑥})}d𝑥=-\int \frac{{\sec }^2𝑦}{\tan 𝑦}d𝑦\\ & -3\int \frac{-e^{𝑥}}{1-e^{𝑥}}d𝑥=-\int \frac{{\sec }^2𝑦}{\tan 𝑦}d𝑦\end{array}\end{array}$

$\begin{array}{cc}\therefore & -3\log (1-e^{𝑥})=-\log (\tan 𝑦)+\log c\\ & \Rightarrow 3\log (1-e^{𝑥})=\log (\tan 𝑦)-\log c\\ & \Rightarrow \log {(1-e^{𝑥})}^3=\log \left(\frac{\tan 𝑦}{c}\right)\\ & \Rightarrow {(1-e^{𝑥})}^3=\frac{\tan 𝑦}{c}...\left(𝑖\right)\end{array}$

Put, $𝑦=\frac{𝜋}{4}$ and $𝑥=1\dots \left[𝐺𝑖𝑣𝑒𝑛\right]$

We get

$\begin{array}{cc}& (1-𝑒{)}^3=\frac{\tan \frac{𝜋}{4}}{𝑐}\\ & \Rightarrow 𝑐=\frac{1}{(1-𝑒{)}^3}\end{array}$

$\begin{array}{cc}& \text{ Substituting value of ’c’ in equation (i), we get }\\ & \begin{array}{cc}& {(1-{𝑒}^{𝑥})}^3=\frac{\tan 𝑦}{\frac{1}{(1-𝑒{)}^3}}\\ & \Rightarrow \tan 𝑦={\left(\frac{1-{𝑒}^{𝑥}}{1-𝑒}\right)}^3\end{array}\end{array}$

Let $\bar{u}=\bar{a}+\bar{b},\bar{v}=2\bar{b}-\bar{c}$

Vector $\bar{d}$ is coplanar with $\bar{u}$ and $\bar{v}$

$\begin{array}{cc}\therefore \bar{d}& =𝑥\bar{u}+𝑦\bar{v}\\ \bar{d}& =𝑥(\bar{a}+\bar{b})+𝑦(2\bar{b}-\bar{c})\\ & =𝑥\bar{a}+𝑥\bar{b}+2𝑦\bar{b}-𝑦\bar{c}\\ \bar{d}& =(𝑥+2𝑦)\bar{b}+𝑥\bar{a}-𝑦\bar{c}...\left(𝑖\right)\end{array}$

Taking dot product with $\bar{a}$, on both sides, we get

$\bar{d}\cdot \bar{a}=(𝑥+2𝑦)\bar{b}\cdot \bar{a}+𝑥\bar{a}\cdot \bar{a}-𝑦\bar{c}\cdot \bar{a}$

$\begin{array}{cc}& \begin{array}{cc}1=0+𝑥|a{|}^2-0& \dots [\because \bar{a}\text{ is }⟂\text{ to }\bar{b}\text{ and }\bar{c}]\\ 1=𝑥\left(1\right)& \dots \left[\text{ Given }\right|a|=1]\\ \Rightarrow 𝑥=1& \end{array}\\ & \text{ Substituting }𝑥=1\text{ in equation (i), we get }\\ & \begin{array}{cc}& \bar{d}=(1+2𝑦)\bar{b}+\bar{a}-𝑦\bar{c}\\ & \bar{d}=\bar{a}+(1+2𝑦)\bar{b}-𝑦\bar{c}\\ & \left|\bar{d}\right|=\bar{d}\cdot \bar{d}\end{array}\end{array}$

$\begin{array}{cc}& =(\bar{a}+(1+2𝑦)\bar{b}-𝑦\bar{c})[\bar{a}+(1+2𝑦)\bar{b}-𝑦\bar{c}]\\ & =1+(1+2𝑦{)}^2\bar{b}\cdot \bar{b}-𝑦(1+2𝑦)\bar{b}\cdot \bar{c}-𝑦(1+2𝑦)\bar{b}\cdot \bar{c}+{𝑦}^2\\ & =1+(1+4𝑦+4{𝑦}^2)4-2𝑦(1+2𝑦)+{𝑦}^2\\ & =1+4+16𝑦+16{𝑦}^2-2𝑦-4{𝑦}^2-{𝑦}^2\\ & =13{𝑦}^2+14𝑦+5\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& {\cot }^{-1}\left(\frac{1}{𝑥}\right)-{\cot }^{-1}\left(\frac{1}{𝑦}\right)=0...\left[𝐺𝑖𝑣𝑒𝑛\right]\\ & \Rightarrow {\cot }^{-1}\left(\frac{1}{𝑥}\right)={\cot }^{-1}\left(\frac{1}{𝑦}\right)\\ & \Rightarrow \frac{1}{𝑥}=\cot ({\cot }^{-1}\left(\frac{1}{𝑦}\right))\\ & \Rightarrow \frac{1}{𝑥}=\frac{1}{𝑦}\\ & \Rightarrow 𝑥=𝑦...\left(𝑖\right)\end{array}\\ & \end{array}$

${\sin }^{-1}𝑥+{\sin }^{-1}𝑦=\frac{𝜋}{3}\dots \left[𝐺𝑖𝑣𝑒𝑛\right]$

$

Put $𝑥=𝑦$, we get

$\begin{array}{cc}& {\sin }^{-1}𝑥+{\sin }^{-1}𝑥=\frac{𝜋}{3}\\ & 2{\sin }^{-1}𝑥=\frac{𝜋}{3}\\ & 𝑥=\sin \left(\frac{𝜋}{6}\right)\\ & 𝑥=\frac{1}{2}\\ & \Rightarrow 𝑦=\frac{1}{2}\text{...[From (i)] }\end{array}$

Substituting values of $𝑥$ and $𝑦$ in given equation $2{𝑥}^2+{𝑦}^2-𝑥𝑦$, we get

$\begin{array}{cc}{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)& =\frac{2}{4}+\frac{1}{4}-\frac{1}{4}\\ & =\frac{1}{2}\end{array}$

$\begin{array}{cc}& \sin [{\tan }^{-1}\left(\frac{1-{𝑥}^2}{2𝑥}\right)+{\cos }^{-1}\left(\frac{1-{𝑥}^2}{1+{𝑥}^2}\right)]\\ & \text{ Substituting }𝑥=\tan 𝜃\text{, we get }\\ & \begin{array}{cc}& =\sin [{\tan }^{-1}\left(\frac{1-{\tan }^2𝜃}{2\tan 𝜃}\right)+{\cos }^{-1}\left(\frac{1-{\tan }^2𝜃}{1+{\tan }^2𝜃}\right)]\\ & =\sin [{\tan }^{-1}(\cot 2𝜃)+{\cos }^{-1}(\cos 2𝜃\left)\right]\\ & =\sin [{\tan }^{-1}\{\tan (\frac{𝜋}{2}-2𝜃)\}+{\cos }^{-1}(\cos 2𝜃\left)\right]\\ & =\sin \frac{𝜋}{2}=1\end{array}\end{array}$

$\begin{array}{cc}\text{ Required area }& ={\int }_3^1|𝑥-2|d𝑥\\ & ={\int }_2^1(2-𝑥)d𝑥+{\int }_3^2(𝑥-2)d𝑥\\ & ={[2𝑥-\frac{{𝑥}^2}{2}]}_2^1+{[\frac{{𝑥}^2}{2}-2𝑥]}_3^2\\ & =\frac{1}{2}+\frac{1}{2}\\ & =1\text{ sq. unit }\end{array}$

$\begin{array}{cc}𝐿𝑒𝑡𝐼& ={\int }_3^{-1}({\tan }^{-1}\left(\frac{𝑥}{{𝑥}^2+1}\right)+{\tan }^{-1}\left(\frac{{𝑥}^2+1}{𝑥}\right))𝑑𝑥\\ & ={\int }_3^{-1}{\tan }^{-1}\left(\frac{𝑥}{{𝑥}^2+1}\right)+{\cot }^{-1}\left(\frac{𝑥}{{𝑥}^2+1}\right)𝑑𝑥\end{array}$

$\begin{array}{cc}\Rightarrow I& ={\int }_3^{-1}\frac{𝜋}{2}d𝑥\dots [{\tan }^{-1}𝑥+{\cot }^{-1}𝑥=\frac{𝜋}{2}]\\ & =\frac{𝜋}{2}{\int }_3^{-1}1d𝑥\\ & =\frac{𝜋}{2}[𝑥{]}_3^{-1}\\ & =\frac{𝜋}{2}[3-(-1\left)\right]\\ & =\frac{𝜋}{2}\left[4\right]\\ & =2𝜋\end{array}$

$\begin{array}{cc}& \left|𝐴\right|=𝑘\ne 0\dots [𝐺𝑖𝑣𝑒𝑛\\ \therefore & {𝐴}^{-1}=\frac{1}{\left|𝐴\right|}(\mathrm{adj}𝐴)\\ & {𝐴}^{-1}=\frac{(\mathrm{adj}𝐴)}{𝑘}\end{array}$

]

k. $A^{-1}=(\mathrm{adj}A)$

Taking inverse on both sides, we get

$\begin{array}{cc}& {\left(𝑘{𝐴}^{-1}\right)}^{-1}=(\mathrm{adj}𝐴{)}^{-1}\\ & \Rightarrow {𝑘}^{-1}{\left({𝐴}^{-1}\right)}^{-1}=(\mathrm{adj}𝐴{)}^{-1}\end{array}$

$\Rightarrow k^{-1}A=(\mathrm{adj}A{)}^{-1}\dots (\because {\left(A^{-1}\right)}^{-1}=A)$

$\Rightarrow \frac{𝐴}{𝑘}=(\mathrm{adj}𝐴{)}^{-1}$

$f\left(𝑥\right)$ is continuous at $𝑥=0,\dots \left[𝐺𝑖𝑣𝑒𝑛\right]$

$\begin{array}{cc}\therefore & \underset{𝑥\to 0^{-}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0+}{lim}𝑓\left(𝑥\right)=𝑓\left(0\right)\\ & \mathrm{At}𝑥=0\\ & 𝑓\left(𝑥\right)={𝑒}^{𝑥}\sin 𝑥+𝑖𝑥+𝜆\log 4\\ & 𝑓\left(0\right)=𝜆\log 4\end{array}$

Consider,

$\begin{array}{cc}\text{ LHL }& =\underset{𝑥\to 0^{+}}{lim}𝑓\left(𝑥\right)\\ & =\underset{𝑥\to 0}{lim}\frac{8^{𝑥}-4^{𝑥}-2^{𝑥}+1^{𝑥}}{{𝑥}^2}\\ & =\underset{𝑥\to 0}{lim}\frac{4^{𝑥}(2^{𝑥}-1)-1(2^{𝑥}-1)}{{𝑥}^2}\\ & =\underset{𝑥\to 0}{lim}\frac{(4^{𝑥}-1)(2^{𝑥}-1)}{{𝑥}^2}\\ & =\underset{𝑥\to 0}{lim}\frac{4^{𝑥}-1}{𝑥}\times \underset{𝑥\to 0}{lim}\frac{2^{𝑥}-1}{𝑥}\\ & =\log 4\times \log 2\\ & =2\log 2\times \log 2\\ & =2(\log 2{)}^2\\ \therefore 𝑓\left(0\right)& =2(\log 2{)}^2\\ 𝜆\log 4& =2(\log 2{)}^2\end{array}$

$\begin{array}{cc}2𝜆\log 2& =2(\log 2{)}^2\\ 𝜆=\log 2& \\ \therefore 500e^{𝜆}& =500e^{\log 2}\\ & =500\times 2\\ & =1000\end{array}$

Given,

$\begin{array}{cc}& \sqrt{\surd {\log }_3{𝑥}^{16}}+9{\log }_{27}\text{exception 25:}=5\\ & \sqrt{\surd 16{\log }_3𝑥}+9{\log }_{27}{\left(\frac{3}{𝑥}\right)}^{1/3}=5\\ & 4\sqrt{\surd {\log }_3𝑥}+\frac{9}{3}[{\log }_{27}3-{\log }_{27}𝑥]=5\\ & 4\sqrt{\surd {\log }_3𝑥}+\frac{3\log 3}{\log 27}-3\frac{\log 𝑥}{\log 27}=5\\ & 4\sqrt{\surd {\log }_3𝑥}+\frac{3\log 3}{3\log 3}-\frac{3\log 𝑥}{3\log 3}=5\\ & 4\sqrt{\surd {\log }_3𝑥}+1-{\log }_3𝑥=5\\ & \text{ Put }𝑦={\log }_3𝑥...\left(𝑖\right)\\ & \Rightarrow 4\sqrt{\surd 𝑦}+1-𝑦=5\\ & \Rightarrow 4\sqrt{\surd 𝑦}=5-1+𝑦\\ & 4\sqrt{\surd 𝑦}=4+𝑦\end{array}$

Squaring on both sides, we get

$\begin{array}{cc}& 16𝑦=(4+𝑦{)}^2\\ & 16𝑦=16+8𝑦+{𝑦}^2\\ & \Rightarrow {𝑦}^2-8𝑦+16=0\\ & \Rightarrow (𝑦-4{)}^2=0\Rightarrow 𝑦=4\end{array}$

$\begin{array}{cc}& \text{ from (i), }\\ & 4={\log }_3𝑥\end{array}$

$\Rightarrow 3^4=𝑥\dots [\because {\log }_{b}a=n\Rightarrow a=b^n]$

$\Rightarrow 𝑥=81$

The equation of the line passing through ( $5,1,a$ ) and ( $3,b,1$ ) is

$\begin{array}{cc}& \frac{𝑥-5}{3-5}=\frac{𝑦-1}{𝑏-1}=\frac{𝑧-𝑎}{1-𝑎}\\ & \Rightarrow \frac{𝑥-5}{-2}=\frac{𝑦-1}{𝑏-1}=\frac{𝑧-𝑎}{1-𝑎}\end{array}$

It crosses the $𝑦𝑧$-plane at $(0,\frac{17}{2},\frac{-13}{2})$.

This means that this point lies on the line

$\therefore \frac{-5}{-2}=\frac{\frac{17}{2}-1}{𝑏-1}=\frac{-\frac{13}{2}-𝑎}{1-𝑎}$

$\begin{array}{cc}& \Rightarrow \frac{5}{2}=\frac{15}{2(𝑏-1)}\text{ and }\frac{5}{2}=\frac{-13-2𝑎}{2(1-𝑎)}\\ & \Rightarrow 𝑏=4\text{ and }𝑎=6\\ & \Rightarrow 2𝑎+3𝑏=2\left(6\right)+3\left(4\right)\\ & =24\end{array}$

Let $a,b,c$ be the direction ratios of the line.

$\begin{array}{cc}\therefore & 𝑎-𝑏+𝑐=0\text{ and }...\left(𝑖\right)\\ & 𝑎-3𝑏=0...\left(𝑖𝑖\right)\\ \therefore & \frac{𝑎}{3}=\frac{𝑏}{1}=\frac{𝑐}{-2}\end{array}$

$\therefore$ the direction ratios of the line are $3,1,-2$.

Let $\bar{r}=3\hat{i}-2\hat{j}+k+𝜆(\hat{i}+\hat{j}+\hat{k})$

$\Rightarrow$ point on the line is, $(3,-2,1)$

Given equation of plane is

$\bar{r}(2\hat{i}+\hat{j}+\hat{k})=4$

i.e. $2𝑥+𝑦+𝑧-4=0$

Distance between point on the line and plane is given by,

$\begin{array}{cc}𝐷& =\frac{|𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑|}{\sqrt{\surd {𝑎}^2+{𝑏}^2+{𝑐}^2}}\\ & =\frac{\left|2\right(3)+1(-2)+1(1\left)\right|}{\sqrt{\surd 2^2+1^2+1^2}}=\frac{1}{\sqrt{\surd 6}}\text{ units }\end{array}$

Let $𝑦={\sin }^2\{{\cot }^{-1}\left(\sqrt{\surd \frac{1-𝑥}{1+𝑥}}\right)\}$

Put $𝑥=\cos 𝜃$

$\therefore 𝑦={\sin }^2\{{\cot }^{-1}\left(\sqrt{\surd \frac{1-\cos 𝜃}{1+\cos 𝜃}}\right)\}$

$\begin{array}{cc}& ={\sin }^2⎧⎩⎨⎪⎪{\cot }^{-1}⎛⎝⎜\sqrt{⎷\frac{2{\sin }^2\frac{𝜃}{2}}{2{\cos }^2\frac{𝜃}{2}}}⎞⎠⎟⎫⎭⎬⎪⎪\\ & ={\sin }^2\{{\cot }^{-1}(\tan \frac{𝜃}{2})\}\\ & ={\sin }^2\{{\cot }^{-1}(\cot (\frac{𝜋}{2}-\frac{𝜃}{2}))\}\\ & ={\sin }^2(\frac{𝜋}{2}-\frac{𝜃}{2})\\ & \therefore 𝑦={\cos }^2\frac{𝜃}{2}=\frac{1+\cos 𝜃}{2}=\frac{1+𝑥}{2}\\ & \therefore \frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝐿𝑒𝑡,I& =\int \frac{\sin 𝑥}{\sqrt{\surd 5{\sin }^2𝑥+6{\cos }^2𝑥}}d𝑥\\ & =\int \frac{\sin 𝑥}{\sqrt{\surd 5(1-{\cos }^2𝑥)+6{\cos }^2𝑥}}d𝑥\\ & =\int \frac{\sin 𝑥}{\sqrt{\surd 5-5{\cos }^2𝑥+6{\cos }^2𝑥}}d𝑥\\ & =\int \frac{\sin 𝑥}{\sqrt{\surd 5+{\cos }^2𝑥}}d𝑥\end{array}\\ & \text{ Put }\cos 𝑥=t\\ & \begin{array}{cc}& \Rightarrow -\sin 𝑥𝑑𝑥=𝑑𝑡\\ \therefore 𝐼& =-\int \frac{𝑑𝑡}{\sqrt{\surd 5+{𝑡}^2}}\\ & =-\log (𝑡+\sqrt{\surd 5+{𝑡}^2})+𝑐\\ 𝐼& =-\log (\cos 𝑥+\sqrt{\surd 5+{\cos }^2𝑥})+𝑐\end{array}\end{array}$

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{𝑘\sin 𝑥+2\cos 𝑥}{\sin 𝑥+\cos 𝑥}\\ & \text{ Differentiating w.r.t. }𝑥\text{, we get }\end{array}$

$f^{\prime }\left(𝑥\right)=\frac{(\sin 𝑥+\cos 𝑥)\times (𝑘\cos 𝑥-2\sin 𝑥)-(k\sin 𝑥+2\cos 𝑥)\times (\cos 𝑥-\sin 𝑥)}{(\sin 𝑥+\cos 𝑥{)}^2}$

$=\frac{(k\sin 𝑥\cos 𝑥-2{\sin }^2𝑥+k{\cos }^2𝑥-2\sin 𝑥\cos 𝑥)-(k\sin 𝑥\cos 𝑥-k{\sin }^2𝑥+2{\cos }^2𝑥-2\sin 𝑥\cos 𝑥)}{(\sin 𝑥+\cos 𝑥{)}^2}$

$\begin{array}{cc}& =\frac{(k-2){\sin }^2𝑥+(k-2){\cos }^2𝑥}{(\sin 𝑥+\cos 𝑥{)}^2}\\ & =\frac{(k-2)({\sin }^2𝑥+{\cos }^2𝑥)}{(\sin 𝑥+\cos 𝑥{)}^2}\\ & f^{\prime }\left(𝑥\right)=\frac{k-2}{(\sin 𝑥+\cos 𝑥{)}^2}\end{array}$

function $f\left(𝑥\right)$ is strictly increasing       …[Given]

$\begin{array}{cc}\therefore & f^{\prime }\left(𝑥\right)>0\\ \therefore & \frac{k-2}{(\sin 𝑥-\cos 𝑥{)}^2}>0\\ & \Rightarrow k-2>0\\ & \Rightarrow k>2\end{array}$

$\begin{array}{cc}& |z+3|-|z-3|=6\\ & \Rightarrow |𝑥+i𝑦+3|-|𝑥+i𝑦-3|=6\\ & \Rightarrow |𝑥+3+i𝑦|-|𝑥-3+i𝑦|=6\\ & \Rightarrow \sqrt{\surd (𝑥+3{)}^2+{𝑦}^2}-\sqrt{\surd (𝑥-3{)}^2+{𝑦}^2}=6\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow \sqrt{\surd (𝑥+3{)}^2+{𝑦}^2}=6+\sqrt{\surd (𝑥-3{)}^2+{𝑦}^2}\\ & \Rightarrow (𝑥+3{)}^2+{𝑦}^2={(\sqrt{\surd (𝑥-3{)}^2+{𝑦}^2}+6)}^2\\ & \Rightarrow 𝑦=0\end{array}\\ & \text{ It is a equation of }X\text{-axis }\end{array}$

$\text{ Given equation of circle is }$

$\begin{array}{cc}& {𝑥}^2+{𝑦}^2=16\\ & 𝐶\equiv (0,0)\\ & 𝑟=4\text{ units }\\ & \text{ In }△𝐴𝑃𝑂,\\ & \angle 𝑃={30}^{\circ }\end{array}$

$\begin{array}{cc}\therefore & \sin {30}^{\circ }=\frac{\mathrm{OA}}{\mathrm{OP}}\\ & \frac{1}{2}=\frac{4}{\mathrm{OP}}\\ & \Rightarrow \mathrm{OP}=8\text{ units }\\ & \Rightarrow \sqrt{\surd (𝑥-0)+(𝑦-0{)}^2}=8\\ & \Rightarrow {𝑥}^2+{𝑦}^2=8^2\\ & \Rightarrow {𝑥}^2+{𝑦}^2=64\end{array}$

$\begin{array}{cc}& 𝑃(𝑋=0)=\frac{4{𝐶}_0}{2^4}=\frac{1}{16}\\ & 𝑃(𝑋=1)=\frac{4{𝐶}_1}{2^4}=\frac{4}{16}\\ & 𝑃(𝑋=2)=\frac{4{𝐶}_2}{2^4}=\frac{6}{16}\\ & 𝑃(𝑋=3)=\frac{4{𝐶}_3}{2^4}=\frac{4}{16}\\ & 𝑃(𝑋=4)=\frac{4{𝐶}_4}{2^4}=\frac{1}{16}\\ & 𝜇=𝐸\left(𝑋\right)=\sum {𝑥}_{𝑖}𝑝\left({𝑥}_{𝑖}\right)\\ & =0\left(\frac{1}{16}\right)+1\left(\frac{4}{16}\right)+2\left(\frac{6}{16}\right)+3\left(\frac{4}{16}\right)+4\left(\frac{1}{16}\right)\\ & =\frac{32}{16}=2\\ & {𝜎}^2=\mathrm{Var}\left(𝑥\right)=𝐸(𝑋{)}^2-[𝐸\left(𝑥\right){]}^2\\ & =0^2\times \frac{1}{16}+1\times \frac{4}{16}+2^2\times \frac{6}{16}+3^2\times \frac{4}{16}+4^2\times \frac{1}{16}-(2{)}^2\\ & =5-4=1\end{array}$