MHT CET 2025 20th April Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \begin{array}{cc}& 𝑥(𝑥-1)\frac{d𝑦}{d𝑥}={𝑥}^3(2𝑥-1)+(𝑥-2)𝑦\\ \therefore & 𝑥(𝑥-1)\frac{d𝑦}{d𝑥}-(𝑥-2)𝑦={𝑥}^3(2𝑥-1)\\ \therefore & \frac{d𝑦}{d𝑥}-\frac{𝑥-2}{𝑥(𝑥-1)}𝑦=\frac{{𝑥}^2(2𝑥-1)}{𝑥-1}\end{array}\\ & \text{ This is of the type }\\ & \begin{array}{cc}& \frac{d𝑦}{d𝑥}+P𝑦=Q\\ & \text{ I.F. }=e^{\int \mathrm{Pd}𝑥}\\ & \text{ P }=\frac{2-𝑥}{𝑥(𝑥-1)}=\frac{A}{𝑥}+\frac{B}{𝑥-1}\\ \therefore & 2-𝑥=A(𝑥-1)+B𝑥\\ \therefore & 2-𝑥=(A+B)𝑥-A\\ \therefore & 𝐴=-2\text{ and }\\ & 𝐴+𝐵=-1\\ & \Rightarrow 𝐵=1\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\text{ I.F. }=e^{\int \mathrm{Pd}𝑥}& =e^{-\int \frac{2}{𝑥}d𝑥+\int \frac{1}{𝑥-1}d𝑥}\\ & =e^{-2\log 𝑥+\log (𝑥-1)}\\ & =e^{-\log {𝑥}^2+\log (𝑥-1)}\\ & =e^{\log \left(\frac{𝑥-1}{{𝑥}^2}\right)}\\ & =\frac{𝑥-1}{{𝑥}^2}\end{array}\\ & \therefore \text{ Solution of the given equation is }\\ & \begin{array}{cc}𝑦\left(\frac{𝑥-1}{{𝑥}^2}\right)& =\int \frac{𝑥-1}{{𝑥}^2}\cdot \frac{{𝑥}^2(2𝑥-1)}{𝑥-1}d𝑥+c\\ & ={𝑥}^2-𝑥+c\\ \therefore 𝑦(𝑥-1)& ={𝑥}^4-{𝑥}^3+{\mathrm{cx}}^2\\ \therefore 𝑦(𝑥-1)& ={𝑥}^3(𝑥-1)+{\mathrm{cx}}^2\end{array}\end{array}$

Given equation of planes are

$\begin{array}{cc}& 𝑥-𝑦+2𝑧=5\\ & 3𝑥+𝑦+𝑧=6\end{array}$

Vectors normal to given planes are

$\begin{array}{cc}& \overline{{𝑛}_1}=\widehat{𝑖}-\widehat{𝑗}+2\widehat{𝑘}\text{ and }\\ \overline{{𝑛}_2}& =3\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘}\\ \overline{𝑛}& =\overline{{𝑛}_1}\times \overline{{𝑛}_2}\\ & =\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 1& -1& 2\\ 3& 1& 1\end{array}\mid \\ & =-3\widehat{𝑖}+5\widehat{𝑗}+4\widehat{𝑘}\\ \therefore \widehat{𝑛}& =\frac{-3}{5\sqrt{\surd 2}}\widehat{𝑖}+\frac{5}{5\sqrt{\surd 2}}\widehat{𝑗}+\frac{4}{5\sqrt{\surd 2}}\widehat{𝑘}\end{array}$

Let $𝑥+𝑦=20\Rightarrow 𝑦=20-𝑥...\left(𝑖\right)$

and ${𝑥}^3{𝑦}^2=z$

$\begin{array}{cc}& \Rightarrow 𝑧={𝑥}^3(20-𝑥{)}^2\Rightarrow 𝑧=400{𝑥}^3+{𝑥}^5-40{𝑥}^4\\ & \therefore \frac{𝑑𝑧}{𝑑𝑥}=1200{𝑥}^2+5{𝑥}^4-160{𝑥}^3\end{array}$

For maximum or minimum,

$\begin{array}{cc}& \frac{\mathrm{dz}}{\mathrm{dx}}=0\\ & \Rightarrow 1200{𝑥}^2+5{𝑥}^4-160{𝑥}^3=0\\ & \Rightarrow 𝑥=12,20\\ & \frac{d^{2}z}{d{𝑥}^2}=2400𝑥+20{𝑥}^3-480{𝑥}^2\\ \therefore & {\left(\frac{d^{2}z}{d{𝑥}^2}\right)}_{𝑥=12}=-5760<0\end{array}$

$\therefore z$ is maximum at $𝑥=12$.

From (i), $𝑦=20-12=8$

$\therefore$ The parts of 20 are 12 and 8.

Here,

Slope of $1^{\text{st }}$ diagonal $\left({𝑚}_1\right)=\frac{3+1}{2-2}=\infty$

Slope of $2^{\text{nd }}$ diagonal $\left({𝑚}_2\right)=\frac{0-2}{4-0}=-\frac{1}{2}$

$\begin{array}{cc}\tan 𝜃& =\mid \frac{{𝑚}_2-{𝑚}_1}{1+{𝑚}_1{𝑚}_2}\mid \\ & =\mid \frac{\frac{{𝑚}_2}{{𝑚}_1}-1}{\frac{1}{{𝑚}_1}+{𝑚}_2}\mid =\mid \frac{1}{\frac{1}{2}}\mid \end{array}$

$\begin{array}{cc}\therefore & \tan 𝜃=2\\ \therefore & 𝜃={\tan }^{-1}2\end{array}$

Given equation of line is

$𝑥-𝑦+1=0...\left(𝑖\right)$

$\therefore \text{ Slope }=1$

$𝑥={𝑦}^2$

Differentiating w.r.t. $𝑥$, we get

$1=2𝑦\frac{d𝑦}{d𝑥}$

$\begin{array}{cc}& \therefore \frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{2𝑦}\\ & \text{ is same as the slope of (i) }\\ & \begin{array}{cc}\therefore & \frac{1}{2𝑦}=1\\ \therefore & 𝑦=\frac{1}{2}\\ \therefore & P(𝑥,𝑦)=(\frac{1}{4},\frac{1}{2})\\ \therefore & \text{ Shortest distance }=\frac{\mid \frac{1}{4}-\frac{1}{2}+1\mid }{\sqrt{\surd 1+1}}=\frac{3\sqrt{\surd 2}}{8}\end{array}\end{array}$

$\begin{array}{cc}& {𝑥}^2-{𝑦}^2-2𝑥+4𝑦-3=0\\ \therefore & {𝑥}^2-2𝑥+1-{𝑦}^2+4𝑦-4=0\\ \therefore & (𝑥-1{)}^2-(𝑦-2{)}^2=0\\ \therefore & (𝑥-1+𝑦-2)(𝑥-1-𝑦+2)=0\\ \therefore & (𝑥+𝑦-3)(𝑥-𝑦+1)=0\end{array}$

$\therefore$ Equations of diameters are

$\begin{array}{cc}& 𝑥+𝑦=3...\left(𝑖\right)\\ & 𝑥-𝑦=-1...\left(𝑖𝑖\right)\end{array}$

On solving, we get

$\text{ Centre }\equiv (𝑥,𝑦)=(1,2)$

As point $(1,1)$ lies on the circle,

$𝑟=\sqrt{\surd (1-1{)}^2+(2-1{)}^2}=1$

$\therefore$ Equation of circle is

$(𝑥-1{)}^2+(𝑦-2{)}^2=1$

$\begin{array}{cc}& \begin{array}{cc}𝐿𝑒𝑡I& =\int {\sin }^5𝑥d𝑥\\ & =\int {\sin }^4𝑥\sin 𝑥d𝑥\\ & =\int {(1-{\cos }^2𝑥)}^2\sin 𝑥d𝑥\\ & =\int (1+{\cos }^4𝑥-2{\cos }^2𝑥)\sin 𝑥d𝑥\end{array}\\ & \text{ Put }\cos 𝑥=t\\ & \begin{array}{cc}\therefore & -\sin 𝑥𝑑𝑥=𝑑𝑡\\ \therefore 𝐼& =-\int (1+{𝑡}^4-2{𝑡}^2)𝑑𝑡\\ & =-(𝑡+\frac{{𝑡}^5}{5}-\frac{2{𝑡}^3}{3})+𝑐\\ & =-(\cos 𝑥-\frac{2{\cos }^3𝑥}{3}+\frac{{\cos }^5𝑥}{5})+𝑐\end{array}\end{array}$

The d.r.s of normal to the given planes are $1,-2,3$ and $1,𝛼,2$

$\begin{array}{cc}& \therefore \cos 𝜃=\frac{1\left(1\right)-2\left(𝛼\right)+3\left(2\right)}{\sqrt{\surd 1+4+9}\cdot \sqrt{\surd 1+{𝛼}^2+4}}\\ & \therefore \frac{1}{14}=\frac{7-2𝛼}{\sqrt{\surd 14(5+{𝛼}^2)}}\end{array}$

$\begin{array}{cc}\therefore & \frac{1}{{14}^2}=\frac{(7-2𝛼{)}^2}{14(5+{𝛼}^2)}\dots \left[𝑆𝑞𝑢𝑎𝑟𝑖𝑛𝑔𝑏𝑜𝑡ℎ𝑠𝑖𝑑𝑒𝑠\right]\\ \therefore & 14(7-2𝛼{)}^2={𝛼}^2+5\end{array}$

On solving, we get

$55{𝛼}^2-392𝛼+681=0$

$\therefore$ Roots are $\frac{-𝑏\pm \sqrt{\surd {𝑏}^2-4𝑎𝑐}}{2𝑎}$

$\begin{array}{cc}\therefore \text{ Difference between roots }& =\frac{\sqrt{\surd {𝑏}^2-4𝑎𝑐}}{𝑎}\\ & =\frac{\sqrt{\surd (-392{)}^2-4(55\left)\right(681)}}{55}\\ & =\frac{\sqrt{\surd 153664-149820}}{55}\\ & =\frac{\sqrt{\surd 3844}}{55}\\ & =\frac{62}{55}\end{array}$

$\begin{array}{cc}& {\overline{𝑟}}_1=(𝑘\widehat{𝑖}+4\widehat{𝑗}+3\widehat{𝑘})+𝛼(2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘})\\ & {\overline{𝑟}}_2=2\widehat{𝑖}+4\widehat{𝑗}+7\widehat{𝑘}+𝛽(2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘})\\ \therefore & \overline{𝑏}=2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘}\\ & {\overline{𝑎}}_1=𝑘\widehat{𝑖}+4\widehat{𝑗}+3\widehat{𝑘}\\ & {\overline{𝑎}}_2=2\widehat{𝑖}+4\widehat{𝑗}+7\widehat{𝑘}\\ & \text{ Shortest distance }=\frac{\mid \overline{𝑏}\times ({\overline{𝑎}}_2-{\overline{𝑎}}_1)\mid }{\left|𝑏\right|}\end{array}$

$\begin{array}{cc}\therefore & \frac{13}{\sqrt{\surd 29}}=\frac{\mid (2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘})\times \left[\right(2-𝑘)\widehat{𝑖}+4\widehat{𝑘}]}{\sqrt{\surd 29}}\\ \therefore & |-8\widehat{𝑗}-3(2-𝑘)\widehat{𝑘}+12\widehat{𝑖}+4(2-𝑘\left)\widehat{𝑗}\right|=13\\ \therefore & |12\widehat{𝑖}-4𝑘\widehat{𝑗}+(3𝑘-6\left)\widehat{𝑘}\right|=13\\ \therefore & 144+16{𝑘}^2+(3𝑘-6{)}^2=169\\ \therefore & 16{𝑘}^2+(3𝑘-6{)}^2=25\\ \therefore & 𝑘=1\end{array}$

$\begin{array}{cc}& \text{ Given equations of lines are }\\ & \begin{array}{cc}& 𝑥=-2+2𝑡,𝑦=3-4𝑡,𝑧=-4+𝑡\\ \therefore & \frac{𝑥+2}{2}=\frac{𝑦-3}{-4}=\frac{𝑧+4}{1}=𝑡\text{ and }\\ & 𝑥=-2-𝑡,𝑦=3+2𝑡,𝑧=-4+3𝑡\\ \therefore & \frac{𝑥+2}{-1}=\frac{𝑦-3}{2}=\frac{𝑧+4}{3}=𝑡\\ & {𝑎}_1,{𝑏}_1,{𝑐}_1=2,-4,1\\ & {𝑎}_2,{𝑏}_2,{𝑐}_2=-1,2,3\\ & \cos 𝜃=\mid \frac{{𝑎}_1{𝑎}_2+{𝑏}_1{𝑏}_2+{𝑐}_1{𝑐}_2}{\sqrt{\surd {𝑎}_1{}^2+{𝑏}_1{}^2+{𝑐}_1{}^2}\sqrt{\surd {𝑎}_2{}^2+{𝑏}_2{}^2+{𝑐}_2{}^2}}\mid \\ & =\mid \frac{-2-8+3}{\sqrt{\surd 4+16+1}\sqrt{\surd 1+4+9}}\mid \end{array}\end{array}$

$\begin{array}{c}=\mid \frac{7}{\sqrt{\surd 21}\sqrt{\surd 14}}\mid \\ =\frac{1}{\sqrt{\surd 6}}\\ \therefore 𝜃={\cos }^{-1}\left(\frac{1}{\sqrt{\surd 6}}\right)\end{array}$

Let the given curves be:

Curve 1: ${𝑦}^2=6𝑥$

Curve 2: $9{𝑥}^2+𝑏{𝑦}^2=16$

Suppose they intersect at point $({𝑥}_1,{𝑦}_1)$.

Step 1: Find slopes of the tangents at the point of intersection

For Curve 1:

Differentiate ${𝑦}^2=6𝑥$ w.r.t $𝑥$:

$2𝑦\frac{𝑑𝑦}{𝑑𝑥}=6⟹\frac{𝑑𝑦}{𝑑𝑥}=\frac{6}{2𝑦}=\frac{3}{𝑦}$

For Curve 2:

Differentiate $9{𝑥}^2+𝑏{𝑦}^2=16$:

$18𝑥+2𝑏𝑦\frac{𝑑𝑦}{𝑑𝑥}=0$

$2𝑏𝑦\frac{𝑑𝑦}{𝑑𝑥}=-18𝑥⟹\frac{𝑑𝑦}{𝑑𝑥}=\frac{-18𝑥}{2𝑏𝑦}=\frac{-9𝑥}{𝑏𝑦}$

Step 2: Condition for orthogonality

The tangents at the intersection are orthogonal, so

${𝑚}_1\cdot {𝑚}_2=-1$

where

${𝑚}_1=\frac{3}{{𝑦}_1},{𝑚}_2=\frac{-9{𝑥}_1}{𝑏{𝑦}_1}$

$\frac{3}{{𝑦}_1}\cdot \frac{-9{𝑥}_1}{𝑏{𝑦}_1}=-1$

$\frac{-27{𝑥}_1}{𝑏{𝑦}_2^1}=-1⟹\frac{27{𝑥}_1}{𝑏{𝑦}_2^1}=1⟹27{𝑥}_1=𝑏{𝑦}_2^1$

Step 3: Use the intersection point

Since the point lies on both curves:

From Curve 1: ${𝑦}_2^1=6{𝑥}_1$

From above: ${𝑦}_2^1=\frac{27}{𝑏}{𝑥}_1$

Set the values of ${𝑦}_2^1$ equal:

$6{𝑥}_1=\frac{27}{𝑏}{𝑥}_1⟹6=\frac{27}{𝑏}⟹𝑏=\frac{27}{6}=\frac{9}{2}$

Final Answer:

$\overline{)\frac{9}{2}}$

So the correct option is Option D: $\frac{9}{2}$.

$\begin{array}{cc}& \sqrt{\surd 3}\cot {20}^{\circ }-4\cos {20}^{\circ }\\ =& \sqrt{\surd 3}\frac{\cos {20}^{\circ }}{\sin {20}^{\circ }}-4\cos {20}^{\circ }\\ =& \frac{\sqrt{\surd 3}\cos {20}^{\circ }-4\sin {20}^{\circ }\cos {20}^{\circ }}{\sin {20}^{\circ }}\\ =& \frac{\frac{2\sqrt{\surd 3}}{2}\cos {20}^{\circ }-2\sin {40}^{\circ }}{\sin {20}^{\circ }}\\ =& \frac{2\sin {60}^{\circ }\cos {20}^{\circ }-2\sin {40}^{\circ }}{\sin {20}^{\circ }}\\ =& \frac{\sin {80}^{\circ }+\sin {40}^{\circ }-2\sin {40}^{\circ }}{\sin {20}^{\circ }}\\ =& \frac{\sin {80}^{\circ }-\sin {40}^{\circ }}{\sin {20}^{\circ }}\\ =& \frac{2\cos {60}^{\circ }\sin {20}^{\circ }}{\sin {20}^{\circ }}\\ =& 2\times \frac{1}{2}\\ =& 1\end{array}$

$\begin{array}{cc}& (\bar{a}\cdot \bar{b}{)}^2+(\bar{a}\times \bar{b}{)}^2=\left|\bar{a}{|}^2\right|\bar{b}{|}^2\\ \therefore & (\bar{a}\times \bar{b}{)}^2=(1{)}^2(4{)}^2-2^2=12\\ & \bar{c}=(2\bar{a}\times \bar{b})-3\bar{b}\end{array}$

$\begin{array}{cccc}\therefore & & \bar{c}\cdot \bar{c}& =\left[\right(2\bar{a}\times \bar{b})-3\bar{b}]\cdot \left[\right(2\bar{a}\times \bar{b})-3\bar{b}]\\ \therefore & & |\bar{c}{|}^2& =|2\bar{a}\times \bar{b}{|}^2+|3\bar{b}{|}^2-2\times 3\bar{b}\cdot (2\bar{a}\times \bar{b})\\ \therefore & & |\bar{c}{|}^2& =4|\bar{a}\times \bar{b}{|}^2+9|\bar{b}{|}^2-0\dots [\because \bar{b}\cdot (\bar{a}\times \bar{b})=0]\\ & & & =4\times 12+9\times 16\\ & & & =48+144\\ & & & =192\\ & & \bar{b}\cdot \bar{c}& =\left|\bar{b}\right|\left|\bar{c}\right|\cos 𝜃\end{array}$

$\begin{array}{cc}\therefore & \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos 𝜃=\bar{b}\cdot \left[\right(2\bar{a}\times \bar{b})-3\bar{b}]\\ \therefore & 4\sqrt{\surd 192}\cos 𝜃=2\bar{b}\cdot (\bar{a}\times \bar{b})-3\bar{b}\cdot \bar{b}\\ \therefore & 4\times 8\sqrt{\surd 3}\cos 𝜃=-3\times 16\\ \therefore & \cos 𝜃=\frac{-3}{2\sqrt{\surd 3}}=\frac{-\sqrt{\surd 3}}{2}\\ \therefore & 𝜃=\frac{5𝜋}{6}\end{array}$

$\begin{array}{cc}& \overline{𝑎}\cdot \overline{𝑏}=\left|\overline{𝑎}\right|\left|\overline{𝑏}\right|\cos 𝜃\\ \therefore & \frac{1}{2}=\cos 𝜃\Rightarrow \sin 𝜃=\frac{\sqrt{\surd 3}}{2}\\ & \overline{𝑐}\cdot \overline{𝑑}=\left|\overline{𝑐}\right|\left|\overline{𝑑}\right|\cos 𝜙\end{array}$

$\begin{array}{cc}\therefore & \frac{1}{2}=\cos 𝜙\Rightarrow \sin 𝜙=\frac{\sqrt{\surd 3}}{2}\\ & \mid \begin{array}{ccc}\overline{𝑎}& \overline{𝑏}& \overline{𝑑}\end{array}]\overline{𝑐}-\left[\begin{array}{ccc}\overline{𝑎}& \overline{𝑏}& \overline{𝑐}\end{array}\right]\overline{𝑑}\mid \\ & =\left|\right((\overline{𝑎}\times \overline{𝑏})\cdot \overline{𝑑})\overline{𝑐}-((\overline{𝑎}\times \overline{𝑏})\cdot \overline{𝑐}\left)\overline{𝑑}\right|\\ & =\left|\right(\overline{𝑎}\times \overline{𝑏})\times (\overline{𝑐}\times \overline{𝑑}\left)\right|\\ & =|\overline{𝑎}\times \overline{𝑏}||\overline{𝑐}\times \overline{𝑑}|\sin \frac{𝜋}{6}\\ & =\left|\overline{𝑎}\right|\left|\overline{𝑏}\right|\sin 𝜃\left|\overline{𝑐}\right|\left|\overline{𝑑}\right|\sin 𝜙\times \frac{1}{2}\\ & =\frac{\sqrt{\surd 3}}{2}\times \frac{\sqrt{\surd 3}}{2}\times \frac{1}{2}\\ & =\frac{3}{8}\end{array}$

Given:

$\overrightarrow{𝑎}=4\widehat{𝑖}+3\widehat{𝑗}+\widehat{𝑘}$

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

We are asked to find:

$\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times \overrightarrow{𝑏})\left)\right)$

Let's proceed step by step.

Step 1: Compute $\overrightarrow{𝑎}\times \overrightarrow{𝑏}$

Use the formula for cross product:

$\overrightarrow{𝑎}\times \overrightarrow{𝑏}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 4& 3& 1\\ 1& -2& 2\end{array}\mid$

Expanding determinant:

$=\widehat{𝑖}\mid \begin{array}{cc}3& 1\\ -2& 2\end{array}\mid -\widehat{𝑗}\mid \begin{array}{cc}4& 1\\ 1& 2\end{array}\mid +\widehat{𝑘}\mid \begin{array}{cc}4& 3\\ 1& -2\end{array}\mid$

$=\widehat{𝑖}[3\times 2-1\times (-2\left)\right]-\widehat{𝑗}[4\times 2-1\times 1]+\widehat{𝑘}[4\times (-2)-3\times 1]$

$=\widehat{𝑖}[6+2]-\widehat{𝑗}[8-1]+\widehat{𝑘}[-8-3]$

$=8\widehat{𝑖}-7\widehat{𝑗}-11\widehat{𝑘}$

Step 2: Compute $\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times \overrightarrow{𝑏})$

Let $\overrightarrow{𝑐}=\overrightarrow{𝑎}\times \overrightarrow{𝑏}=8\widehat{𝑖}-7\widehat{𝑗}-11\widehat{𝑘}$

Now,

$\overrightarrow{𝑎}\times \overrightarrow{𝑐}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 4& 3& 1\\ 8& -7& -11\end{array}\mid$

Expanding:

$=\widehat{𝑖}\mid \begin{array}{cc}3& 1\\ -7& -11\end{array}\mid -\widehat{𝑗}\mid \begin{array}{cc}4& 1\\ 8& -11\end{array}\mid +\widehat{𝑘}\mid \begin{array}{cc}4& 3\\ 8& -7\end{array}\mid$

$=\widehat{𝑖}[3\times (-11)-1\times (-7\left)\right]-\widehat{𝑗}[4\times (-11)-1\times 8]+\widehat{𝑘}[4\times (-7)-3\times 8]$

$=\widehat{𝑖}[-33+7]-\widehat{𝑗}[-44-8]+\widehat{𝑘}[-28-24]$

$=-26\widehat{𝑖}+52\widehat{𝑗}-52\widehat{𝑘}$

Step 3: Compute $\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times \overrightarrow{𝑏}\left)\right)$

Let $\overrightarrow{𝑑}=-26\widehat{𝑖}+52\widehat{𝑗}-52\widehat{𝑘}$

$\overrightarrow{𝑎}\times \overrightarrow{𝑑}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 4& 3& 1\\ -26& 52& -52\end{array}\mid$

Expanding:

$=\widehat{𝑖}\mid \begin{array}{cc}3& 1\\ 52& -52\end{array}\mid -\widehat{𝑗}\mid \begin{array}{cc}4& 1\\ -26& -52\end{array}\mid +\widehat{𝑘}\mid \begin{array}{cc}4& 3\\ -26& 52\end{array}\mid$

$=\widehat{𝑖}[3\times (-52)-1\times 52]-\widehat{𝑗}[4\times (-52)-1\times (-26\left)\right]+\widehat{𝑘}[4\times 52-3\times (-26\left)\right]$

$=\widehat{𝑖}[-156-52]-\widehat{𝑗}[-208+26]+\widehat{𝑘}[208+78]$

$=-208\widehat{𝑖}+182\widehat{𝑗}+286\widehat{𝑘}$

Step 4: Compute $\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times (\overrightarrow{𝑎}\times \overrightarrow{𝑏})\left)\right)$

Let $\overrightarrow{𝑒}=-208\widehat{𝑖}+182\widehat{𝑗}+286\widehat{𝑘}$

$\overrightarrow{𝑎}\times \overrightarrow{𝑒}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 4& 3& 1\\ -208& 182& 286\end{array}\mid$

Expanding:

$=\widehat{𝑖}\mid \begin{array}{cc}3& 1\\ 182& 286\end{array}\mid -\widehat{𝑗}\mid \begin{array}{cc}4& 1\\ -208& 286\end{array}\mid +\widehat{𝑘}\mid \begin{array}{cc}4& 3\\ -208& 182\end{array}\mid$

$=\widehat{𝑖}[3\times 286-1\times 182]-\widehat{𝑗}[4\times 286-1\times (-208\left)\right]+\widehat{𝑘}[4\times 182-3\times (-208\left)\right]$

$=\widehat{𝑖}[858-182]-\widehat{𝑗}[1144+208]+\widehat{𝑘}[728+624]$

$=\widehat{𝑖}\left[676\right]-\widehat{𝑗}\left[1352\right]+\widehat{𝑘}\left[1352\right]$

$=676\widehat{𝑖}-1352\widehat{𝑗}+1352\widehat{𝑘}$

Now, observe that

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

$676\overrightarrow{𝑏}=676\widehat{𝑖}-1352\widehat{𝑗}+1352\widehat{𝑘}$

So, the answer is

$\overline{)676\overrightarrow{𝑏}}$

This matches Option B.

$\begin{array}{cc}& \text{ Here, }𝑛=4\\ & 𝑃(𝑋=3)=3𝑃(𝑋=4)\\ \therefore & {}^4{𝐶}_3{𝑝}^3𝑞=3{}^4{𝐶}_4{𝑝}^4{𝑞}^0\\ \therefore & 4{𝑝}^3𝑞=3{𝑝}^4\\ \therefore & 4(1-𝑝)=3𝑝\dots [𝑝+𝑞=1]\\ \therefore & 4-4𝑝=3𝑝\\ \therefore & 4=7𝑝\\ \therefore & 𝑝=\frac{4}{7}\end{array}$