MHT CET 2025 19th April Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cccccc}p& q& \sim p& \sim q& q\wedge \sim p& p\to (q\wedge \sim p)\\ T& T& F& F& F& F\\ T& F& F& T& F& F\\ F& T& T& F& T& T\\ F& F& T& T& F& T\end{array}$

$\begin{array}{ccc}𝑝\vee \sim 𝑞& (𝑝\vee \sim 𝑞)\wedge 𝑝& [𝑝\to (𝑞\wedge \sim 𝑝\left)\right]\vee \left[\right(𝑝\vee \sim 𝑞)\wedge 𝑝]\\ 𝑇& 𝑇& 𝑇\\ 𝑇& 𝑇& 𝑇\\ 𝐹& 𝐹& 𝑇\\ 𝑇& 𝐹& 𝑇\end{array}$

$\therefore \text{ The entries of the last column are TTTT. }$

Let’s first write the equations of the two lines like this:

${𝑙}_1:4𝑥+3𝑦=10$

${𝑙}_2:8𝑥+6𝑦=-5$

Both lines have the same slope, which is $-\frac{4}{3}$. This means the lines are parallel.

The shortest distance from the origin (O) to a line $𝑎𝑥+𝑏𝑦+𝑐=0$ can be found with the formula:

$𝑑=\mid \frac{𝑐}{\sqrt{\surd {𝑎}^2+{𝑏}^2}}\mid$

Let’s find the distance from the origin to the first line (${𝑙}_1$):

$\text{Distance}=\mid \frac{-10}{\sqrt{\surd 4^2+3^2}}\mid =\mid \frac{-10}{5}\mid =2$

Now, let’s find the distance from the origin to the second line (${𝑙}_2$):

$\text{Distance}=\mid \frac{5}{\sqrt{\surd 8^2+6^2}}\mid =\mid \frac{5}{10}\mid =\frac{1}{2}$

Since the straight line goes through the origin and meets the two parallel lines at points $𝐴$ and $𝐵$, the origin ($𝑂$) is between those two points. $𝑂$ divides the line segment $𝐴𝐵$ in the same ratio as these distances.

So, the ratio is $𝐴𝑂:𝑂𝐵=2:\frac{1}{2}$.

This is the same as $4:1$ when you multiply both sides by 2.

We are asked to find the size of the acceleration (resultant acceleration) for a point that changes its position with time.

Step 1: Write the formulas for position.
The position in the $𝑥$ direction is $𝑥=a+\mathrm{bt}-{\mathrm{ct}}^2$.
The position in the $𝑦$ direction is $𝑦=\mathrm{at}+{\mathrm{bt}}^2$.

Step 2: Find the second derivatives (acceleration components).
To get acceleration, find the second derivative (differentiation twice) of $𝑥$ and $𝑦$ with respect to $𝑡$.

For $𝑥$:
$\frac{d𝑥}{\mathrm{dt}}=b-2c𝑡$ $\frac{d^2𝑥}{{\mathrm{dt}}^2}=-2c$

For $𝑦$:
$\frac{d𝑦}{\mathrm{dt}}=a+2b𝑡$ $\frac{d^2𝑦}{{\mathrm{dt}}^2}=2b$

Step 3: Write the acceleration vector.
The acceleration vector $\overrightarrow{𝑎}=(-2𝑐,2𝑏)$.

Step 4: Calculate the magnitude of acceleration.
To get the actual (resultant) acceleration, use the magnitude formula for a vector: $\left|\overrightarrow{𝑎}\right|=\sqrt{\surd (-2𝑐{)}^2+(2𝑏{)}^2}$ $=\sqrt{\surd 4{𝑐}^2+4{𝑏}^2}$ $=2\sqrt{\surd {𝑏}^2+{𝑐}^2}$

So, the magnitude of the acceleration is $2\sqrt{\surd {𝑏}^2+{𝑐}^2}$ unit/seconds${}^2$.

$\overline{\mathrm{OA}}=\bar{a},\overline{\mathrm{OB}}=\bar{b}$

…[Given]

Direction of the angle bisector of $\angle 𝐴𝑂𝐵$ is given by,

$\overrightarrow{𝑣}=\frac{\overline{𝑎}}{\left|\overline{𝑎}\right|}+\frac{\overline{𝑏}}{\left|\overline{𝑏}\right|}...\left(𝑖\right)$

But, angle bisector is given in the form

$\overrightarrow{v}=𝑥\cdot \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}+𝑦\cdot \frac{\stackrel{\to \leftarrow }{b}}{\left|\stackrel{\to \leftarrow }{b}\right|}...\left(𝑖𝑖\right)$

From (i) and (ii), we get

$\begin{array}{cc}& 𝑥=1\text{ and }𝑦=1\\ & \Rightarrow 𝑥-𝑦=0\end{array}$

Let the vertices of triangle ABC be $A=(0,0),B=(1,0),C=(0,1)𝑃$ divides $𝐵𝐶$ in the ratio $3:4$ and $𝑄$ divides CA in the ratio $5:3$.

By using section formula, we get

$\begin{array}{cc}& 𝑃=(\frac{3\times 0+4\times 1}{3+4},\frac{3\times 1+4\times 0}{3+4})=(\frac{4}{7},\frac{3}{7})\\ & 𝑄=(\frac{5\times 0+3\times 0}{5+3},\frac{5\times 0+3\times 1}{5+3})=(0,\frac{3}{8})\end{array}$

Point on AP is given by,

$\begin{array}{cc}& (𝑥,𝑦)=(1-𝑡)\cdot (0,0)+𝑡\cdot (\frac{4}{7},\frac{3}{7})\\ & \therefore (𝑥,𝑦)=(\frac{4𝑡}{7},\frac{3𝑡}{7})\end{array}$

Point on BQ is given by,

$\begin{array}{cc}& (𝑥,𝑦)=(1-s)\cdot (1,0)+s\cdot (0,\frac{3}{8})\\ & \therefore (𝑥,𝑦)=(1-s,\frac{3s}{8})\\ \Rightarrow & (\frac{4t}{7},\frac{3t}{7})=(1-s,\frac{3s}{8})\\ \Rightarrow & \frac{4t}{7}=1-s\dots \text{ (i) and }\frac{3t}{7}=\frac{3s}{8}...\left(𝑖𝑖\right)\end{array}$

From (i) and (ii), we get

$𝑡=\frac{7}{12}$

Now,

$\begin{array}{cc}𝐴𝐺:𝐺𝑃=𝑡:(1-𝑡)& =\frac{7}{12}:(1-\frac{7}{12})\\ & =\frac{7}{12}:\frac{5}{12}\\ & =7:5\end{array}$

$𝑦=(1-𝑥)(2-𝑥)(3-𝑥)\dots (𝑛-𝑥)$

Let's write $𝑦$ as $𝑦=(1-𝑥)\cdot 𝑓\left(𝑥\right)$,

where $𝑓\left(𝑥\right)=(2-𝑥)(3-𝑥)\dots (𝑛-𝑥)$

We are going to use the product rule to find the derivative of $𝑦$ with respect to $𝑥$:

$\frac{𝑑𝑦}{𝑑𝑥}=(1-𝑥)\cdot {𝑓}^{\prime }\left(𝑥\right)+𝑓\left(𝑥\right)\cdot (-1)$

Now, we find the value at $𝑥=1$:

At $𝑥=1$, $𝑓\left(𝑥\right)=(2-1)(3-1)\dots (𝑛-1)=(𝑛-1)!$

Also, $(1-𝑥)$ becomes $0$ when $𝑥=1$, so $(1-𝑥)\cdot {𝑓}^{\prime }\left(𝑥\right)=0$

So, ${\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}_{𝑥=1}=0+(𝑛-1)!\cdot (-1)=-(𝑛-1)!$

$\begin{array}{cc}\frac{𝑃(𝑋=𝑘)}{𝑃(𝑋=𝑘-1)}& =\frac{{}^{𝑛}{𝐶}_{𝑘}\left(𝑝{)}^{𝑘}\right(𝑞{)}^{𝑛-𝑘}}{{}^{𝑛}{𝐶}_{𝑘-1}\left(𝑝{)}^{𝑘-1}\right(𝑞{)}^{𝑛-𝑘+1}}\\ & =\frac{{}^{𝑛}{𝐶}_{𝑘}}{{}^{𝑛}{𝐶}_{𝑘-1}}\cdot \frac{𝑝}{𝑞}\\ \therefore \frac{𝑃(𝑋=𝑘)}{𝑃(𝑋=𝑘-1)}& =\frac{𝑛-𝑘+1}{𝑘}\cdot \frac{𝑝}{𝑞}\end{array}$

The equation of the family of lines passing through $(1,-1)$ is

$\begin{array}{cc}& 𝑦+1=𝑚(𝑥-1)\\ & \Rightarrow 𝑦=𝑚(𝑥-1)-1...\left(𝑖\right)\\ & \Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=𝑚\end{array}$

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

$𝑦=\frac{𝑑𝑦}{𝑑𝑥}(𝑥-1)-1$

$\begin{array}{cc}& \text{ Let }𝑦=({\cos }^{-1}(\sin \sqrt{\surd \frac{1+𝑥}{2}})+{𝑥}^{𝑥})\\ & \text{ Put }u=\sqrt{\surd \frac{1+𝑥}{2}}\\ & 𝑦={\cos }^{-1}(\sin u)+{𝑥}^{𝑥}\\ & \therefore \frac{d𝑦}{d𝑥}=\frac{-1}{\sqrt{\surd 1-{\sin }^{2}u}}\cdot \frac{d(\sin u)}{d𝑥}+{𝑥}^{𝑥}[1+\log 𝑥]\\ & =\frac{-1}{\sqrt{\surd 1-{\sin }^2\sqrt{\surd \frac{1+𝑥}{2}}}}\times \frac{d}{d𝑥}\sin \sqrt{\surd \frac{1+𝑥}{2}}+{𝑥}^{𝑥}[1+\log 𝑥](\because u=\sqrt{\surd \frac{1+𝑥}{2}})\end{array}$

$\begin{array}{cc}& =\frac{-1}{\sqrt{\surd 1-{\sin }^2\sqrt{\surd \frac{1+𝑥}{2}}}}\times \cos \sqrt{\surd \frac{1+𝑥}{2}}\times \frac{d}{d𝑥}\sqrt{\surd \frac{1+𝑥}{2}}\\ & =\frac{-1}{2\sqrt{\surd \frac{1+𝑥}{2}}}\times \left(\frac{1}{2}\right)+{𝑥}^{𝑥}[1+\log 𝑥]\\ & =\frac{-\sqrt{\surd 2}}{4\sqrt{\surd 1+𝑥}}+{𝑥}^{𝑥}[1+\log 𝑥]\\ & \therefore {\left[\frac{d𝑦}{d𝑥}\right]}_{𝑥=1}=\frac{-\sqrt{\surd 2}}{4\sqrt{\surd 2}}+1.[1+\log 1]\\ & =\frac{-1}{4}+1\\ & =\frac{3}{4}\end{array}$

According to the given condition, Projection of $\overline{𝑣}$ on $\overline{𝑢}=$ projection of $\overline{𝑤}$ on $\overline{𝑢}$

$\begin{array}{cc}& \Rightarrow \frac{\bar{v}\cdot \bar{u}}{\left|\bar{u}\right|}=\frac{\bar{w}\cdot \bar{u}}{\left|\bar{u}\right|}\\ & \Rightarrow \bar{v}\cdot \bar{u}=\bar{w}\cdot \bar{u}...\left(𝑖\right)\end{array}$

Vectors $\bar{v}$ and $\bar{w}$ are perpendicular to each other.

$\therefore \bar{v}\cdot \bar{w}=0...\left(𝑖𝑖\right)$

$\begin{array}{cc}& \text{ Consider, }\\ & \begin{array}{cc}|\bar{u}-\bar{v}+\bar{w}{|}^2=|\bar{u}{|}^2+|\bar{v}{|}^2& +|\bar{w}{|}^2-2\bar{u}\cdot \bar{v}+2\bar{u}\cdot \bar{w}-2\bar{v}\cdot \bar{w}\end{array}\end{array}$

$\begin{array}{cc}& =(1{)}^2+(2{)}^2+(3{)}^2\dots [\text{ From (i) and (ii) }]\\ & =1+4+9\\ \therefore & |\bar{u}-\bar{v}+\bar{w}{|}^2=14\\ \therefore & |\bar{u}-\bar{v}+\bar{w}|=\sqrt{\surd 14}\end{array}$

$\begin{array}{cc}\text{ Required area }& ={\int }_4^0\sqrt{\surd 4𝑥}d𝑥-{\int }_4^0\left|𝑥\right|𝑑𝑥\\ & =2{\int }_4^0\sqrt{\surd 𝑥}d𝑥-{\int }_4^0𝑥𝑑𝑥\\ & =2{\left[\frac{2}{3}{𝑥}^{\frac{3}{2}}\right]}_4^0-{\left[\frac{{𝑥}^2}{2}\right]}_4^0\\ & =\frac{4}{3}(4^{\frac{3}{2}}-0)-\frac{1}{2}(4^2-0)\\ & =\frac{4}{3}\left(8\right)-8\\ & =8(\frac{4}{3}-1)=\frac{8}{3}\text{ sq. units }\end{array}$

$\begin{array}{cc}& \tan (𝐴+𝐵)=\frac{\tan 𝐴+\tan 𝐵}{1-\tan 𝐴\cdot \tan 𝐵}\\ & =\frac{\frac{1}{\sqrt{\surd 𝑥({𝑥}^2+𝑥+1)}}+\frac{\sqrt{\surd 𝑥}}{\sqrt{\surd {𝑥}^2+𝑥+1}}}{1-\frac{1}{\sqrt{\surd 𝑥({𝑥}^2+𝑥+1)}}\times \frac{\sqrt{\surd 𝑥}}{\sqrt{\surd {𝑥}^2+𝑥+1}}}\\ & =\frac{\sqrt{\surd {𝑥}^2+𝑥+1}+\sqrt{\surd 𝑥}\times \sqrt{\surd 𝑥({𝑥}^2+𝑥+1)}}{\sqrt{\surd 𝑥({𝑥}^2+𝑥+1)}\times \sqrt{\surd {𝑥}^2+𝑥+1}-\sqrt{\surd 𝑥}}\\ & =\frac{\sqrt{\surd {𝑥}^2+𝑥+1}}{\sqrt{\surd 𝑥}}\times \left(\frac{1+𝑥}{({𝑥}^2+𝑥+1)-1}\right)\\ & =\frac{\sqrt{\surd {𝑥}^2+𝑥+1}}{\sqrt{\surd 𝑥}}\times \left(\frac{1+𝑥}{𝑥(𝑥+1)}\right)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \therefore \tan (A+B)=\frac{\sqrt{\surd {𝑥}^2+𝑥+1}}{𝑥\sqrt{\surd 𝑥}}...\left(𝑖\right)\\ & \tan C=\sqrt{\surd {𝑥}^{-1}+{𝑥}^{-2}+{𝑥}^{-3}}\dots \text{ [Given] }\\ & =\sqrt{\surd \frac{1}{𝑥}+\frac{1}{{𝑥}^2}+\frac{1}{{𝑥}^3}}\\ & \therefore \tan C=\frac{\sqrt{\surd {𝑥}^2+𝑥+1}}{𝑥\sqrt{\surd 𝑥}}...\left(𝑖𝑖\right)\end{array}\\ & \text{ From (i) and (ii), we get }\\ & \begin{array}{cc}& \tan (𝐴+𝐵)=\tan 𝐶\\ & \Rightarrow 𝐴+𝐵=𝐶\end{array}\end{array}$

$\begin{array}{cc}& \text{ Total probability }=1\\ & \begin{array}{cc}& \Rightarrow P(X=30)+P(X=10)+P(X=-10)=1\\ & \Rightarrow \frac{1}{5}+A+B=1\\ & \Rightarrow A+B=\frac{4}{5}...\left(𝑖\right)\\ & E\left(X\right)=\sum {𝑥}_{i}P\left({𝑥}_i\right)\end{array}\end{array}$

$\begin{array}{cc}\therefore & 4=30\times \frac{1}{5}+10\times 𝐴+(-10)\times 𝐵\\ & \Rightarrow 4=6+10𝐴-10𝐵\\ & \Rightarrow -2=10(𝐴-𝐵)\\ & \Rightarrow 𝐴-𝐵=\frac{-1}{5}...\left(𝑖𝑖\right)\end{array}$

Solving (i) and (ii), we get

$A=\frac{3}{10},B=\frac{1}{2}$

$\therefore$ A. $𝐵=\frac{3}{10}\cdot \frac{1}{2}=\frac{3}{20}$

Let $\overrightarrow{𝐴}=(2,1,-3)$ and $\overrightarrow{𝐵}=(-1,0,2)$

$\stackrel{\to \leftarrow }{\mathrm{AB}}=\overrightarrow{B}-\overrightarrow{A}=(-3,-1,5)$

Let $\overrightarrow{𝑎}=\stackrel{\to \leftarrow }{𝐴𝐵}$ and $\overrightarrow{𝑏}=(3,2,6)$

Projection of $\overrightarrow{𝑎}$ on $\overrightarrow{𝑏}=\frac{\overrightarrow{𝑎}\cdot \overrightarrow{𝑏}}{\left|\overrightarrow{𝑏}\right|}$

$\begin{array}{cc}& =\frac{(-3\hat{i}-\hat{j}+5\hat{k})\cdot (3\hat{i}+2\hat{j}+6\hat{k})}{\sqrt{\surd 3^2+2^2+6^2}}\\ & =\frac{(-3)\left(3\right)+(-1)\left(2\right)+\left(5\right)\left(6\right)}{\sqrt{\surd 9+4+36}}\\ & =\frac{-9-2+30}{\sqrt{\surd 49}}\\ & =\frac{19}{7}\text{ units }\end{array}$

${𝑥}^{\frac{2}{5}}+{𝑦}^{\frac{2}{5}}=a^{\frac{2}{5}}$

Differentiating w.r.t $𝑥$, we get

$\frac{2}{5}{𝑥}^{\frac{2}{5}-1}+\frac{2}{5}{𝑦}^{\frac{2}{5}-1}\cdot \frac{d𝑦}{d𝑥}=0$

$\begin{array}{cc}& \Rightarrow \frac{2}{5}{𝑦}^{\frac{-3}{5}}\frac{d𝑦}{d𝑥}=\frac{-2}{5}{𝑥}^{\frac{-3}{5}}\\ & \Rightarrow \frac{d𝑦}{d𝑥}=-\frac{{𝑥}^{\frac{-3}{5}}}{{𝑦}^{\frac{-3}{5}}}\\ & =-\frac{{𝑦}^{\frac{3}{5}}}{{𝑥}^{\frac{3}{5}}}\\ & =-\text{exception 25:}\end{array}$

$\begin{array}{cc}& (𝑥-2𝑦+1{)}^2+k(𝑥-2𝑦+1)=0\\ & \Rightarrow (𝑥-2𝑦+1)(𝑥-2𝑦+1+k)=0\\ & \Rightarrow 𝑥-2𝑦+1=0\text{ or }𝑥-2𝑦+(1+k)=0\end{array}$

So the two lines are,

$𝑥-2𝑦+1=0\text{ and }𝑥-2𝑦+(1+k)=0$

For two parallel lines $A𝑥+B𝑦+C_1=0$ and $A𝑥+B𝑦+C_2=0$,

perpendicular distance between lines

$=\frac{|{𝐶}_1-{𝐶}_2|}{\sqrt{\surd {𝐴}^2+{𝐵}^2}}$

$\therefore$ Perpendicular distance between lines

$\begin{array}{cc}& =\frac{|1-(1+k\left)\right|}{\sqrt{\surd 1^2+(-2{)}^2}}\\ & \Rightarrow \sqrt{\surd 5}=\frac{|-k|}{\sqrt{\surd 5}}\\ & \Rightarrow k=5\end{array}$

$\begin{array}{cc}{\sin }^{-1}\left(\frac{-1}{\sqrt{\surd 2}}\right)=-{\sin }^{-1}& \left(\frac{1}{\sqrt{\surd 2}}\right)[\because {\sin }^{-1}(-𝑥)=-{\sin }^{-1}𝑥]\\ =& \frac{𝜋}{4}\end{array}$

$\begin{array}{cc}{\cos }^{-1}\left(\frac{-1}{2}\right)=& 𝜋-{\cos }^{-1}\left(\frac{1}{2}\right)[\because {\cos }^{-1}(-𝑥)=𝜋-{\cos }^{-1}𝑥]\\ =& 𝜋-\frac{𝜋}{3}\\ =& \frac{2𝜋}{3}\end{array}$

$\begin{array}{cc}{\cot }^{-1}\left(\frac{-1}{\sqrt{\surd 3}}\right)& =𝜋-{\cot }^{-1}\left(\frac{1}{\sqrt{\surd 3}}\right)\\ & =𝜋-\frac{𝜋}{3}\\ & =\frac{2𝜋}{3}\\ {\tan }^{-1}(-\sqrt{\surd 3})& =-{\tan }^{-1}\left(\sqrt{\surd 3}\right)\\ & =\frac{-𝜋}{3}\end{array}$

$\begin{array}{cc}\therefore & \begin{array}{cc}& \sin {}^{-1}\left(\frac{-1}{\sqrt{\surd 2}}\right)+{\cos }^{-1}\left(\frac{-1}{2}\right)-{\cot }^{-1}\left(\frac{-1}{\sqrt{\surd 3}}\right)+{\tan }^{-1}(-\sqrt{\surd 3})\end{array}\\ & =\frac{-𝜋}{4}+\frac{2𝜋}{3}-\frac{2𝜋}{3}-\frac{𝜋}{3}\\ & =\frac{-7𝜋}{12}\end{array}$

$\begin{array}{cc}& \text{ In }△\mathrm{ABC},\angle A={90}^{\circ }\\ & \Rightarrow \angle B+\angle C={90}^{\circ }\end{array}$

and the roots of the equation $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$ are,

$\begin{array}{cc}& \tan \left(\frac{𝐵}{2}\right)\text{ and }\tan \left(\frac{𝐶}{2}\right)\\ & \Rightarrow \text{ sum of roots }=\frac{-𝑏}{𝑎}\\ & \Rightarrow \tan \left(\frac{𝐵}{2}\right)+\tan \left(\frac{𝐶}{2}\right)=\frac{-𝑏}{𝑎}\end{array}$

and product of roots $=\frac{𝑐}{𝑎}$

$\Rightarrow \tan \left(\frac{B}{2}\right)\cdot \tan \left(\frac{C}{2}\right)=\frac{c}{a}$

Since $𝐵+𝐶={90}^{\circ }$

$\Rightarrow \frac{B}{2}+\frac{C}{2}={45}^0$

Consider,

$\begin{array}{cc}& \tan (\frac{𝐵}{2}+\frac{𝐶}{2})=\frac{\tan \left(\frac{𝐵}{2}\right)+\tan \left(\frac{𝐶}{2}\right)}{1-\tan \left(\frac{𝐵}{2}\right)\cdot \tan \left(\frac{𝐶}{2}\right)}\\ & \Rightarrow \tan \left({45}^{\circ }\right)=\frac{\frac{-𝑏}{𝑎}}{1-\frac{𝑐}{𝑎}}\\ & \Rightarrow 1=\frac{\frac{-𝑏}{𝑎}}{\frac{𝑎-𝑐}{𝑎}}\\ & \Rightarrow 𝑎-𝑐=-𝑏\\ & \Rightarrow 𝑎+𝑏=𝑐\end{array}$

$\begin{array}{cc}& \text{ Let }𝑎=𝑏=𝑐=1\\ & \Rightarrow {𝑟}^2=1^2+1^2+1^2=3\\ & \Rightarrow 𝑟=\sqrt{\surd 3}\\ & {\tan }^{-1}\left(\frac{𝑎𝑏}{𝑐𝑟}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{\surd 3}}\right)=\frac{𝜋}{6}\\ & \text{ Similarly, }{\tan }^{-1}\left(\frac{𝑏𝑐}{𝑎𝑟}\right)={\tan }^{-1}\left(\frac{𝑐𝑎}{𝑏𝑟}\right)=\frac{𝜋}{6}\end{array}$

$\begin{array}{cc}& \Rightarrow {\tan }^{-1}\left(\frac{\mathrm{ab}}{\mathrm{cr}}\right)+{\tan }^{-1}\left(\frac{\mathrm{bc}}{\mathrm{ar}}\right)+{\tan }^{-1}\left(\frac{\mathrm{ca}}{\mathrm{br}}\right)\\ & =\frac{𝜋}{6}+\frac{𝜋}{6}+\frac{𝜋}{6}\\ & =\frac{𝜋}{2}\end{array}$

$\begin{array}{cc}& \text{ Let }I={\int }_1^{\frac{1}{3}}\frac{{(𝑥-{𝑥}^3)}^{\frac{1}{3}}}{{𝑥}^4}d𝑥\\ & ={\int }_1^{\frac{1}{3}}\frac{𝑥{(\frac{1}{{𝑥}^2}-1)}^{\frac{1}{3}}}{{𝑥}^4}d𝑥\\ & ={\int }_1^{\frac{1}{3}}\frac{{(\frac{1}{{𝑥}^2}-1)}^{\frac{1}{3}}}{{𝑥}^3}d𝑥\\ & \begin{array}{c}\text{ Put }\frac{1}{{𝑥}^2}-1=t\\ \Rightarrow \frac{-2}{{𝑥}^3}d𝑥=\mathrm{dt}\\ \therefore \frac{-1}{2}{\int }_0^8{t}^{\frac{1}{3}}\mathrm{dt}\\ =\frac{-1}{2}{\left[\frac{3t^{\frac{4}{3}}}{4}\right]}_0^8\\ =\frac{-1}{2}(0-\frac{3}{4}{8}^{\frac{4}{3}})\\ =\frac{3}{8}\left(2^4\right)\\ =6\mathrm{sq}\cdot \text{ units }\end{array}\end{array}$