MHT CET 2020 19th October Evening Shift (Mathematics) Question Paper with Solutions

In $△𝐴𝐵𝐶$,

$\angle 𝐴,\angle 𝐵,\angle 𝐶$ are in A.P

$\therefore 2\angle 𝐵=\angle 𝐴+\angle 𝐶\text{... (i)}$

and $\angle 𝐴+\angle 𝐵+\angle 𝐶={180}^{\circ }\text{.... (ii)}$

From Eqs. (i) and (ii), $𝐵={60}^{\circ }$

$\begin{array}{cc}& \therefore \frac{𝑏}{\sin 𝐵}=\frac{𝑐}{\sin 𝐶}\\ & \Rightarrow \frac{\sin 𝐵}{\sin 𝐶}=\frac{𝑏}{𝑐}\\ & \Rightarrow \frac{\sin {60}^{\circ }}{\sin 𝐶}=\frac{\sqrt{\surd 3}}{\sqrt{\surd 2}}[\because \text{ given }𝑏:𝑐=\sqrt{\surd 3}:\sqrt{\surd 2}]\\ & \Rightarrow \sin 𝐶=\frac{\sqrt{\surd 2}\times \sqrt{\surd 3}}{\sqrt{\surd 3}\times 2}=\frac{1}{\sqrt{\surd 2}}\\ & \therefore \sin 𝐶=\sin {45}^{\circ }\\ & \Rightarrow 𝐶={45}^{\circ }\\ & \angle 𝐴=2\angle 𝐵-\angle 𝐶={120}^{\circ }-{45}^{\circ }={75}^{\circ }\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& 𝐼=\int {\sin }^{-1}𝑥𝑑𝑥\\ & 𝐼={\sin }^{-1}𝑥\int 𝑑𝑥-\int \frac{𝑑({\sin }^{-1}𝑥)}{𝑑𝑥}\int 𝑑𝑥\cdot 𝑑𝑥+𝐶\\ & 𝐼=𝑥{\sin }^{-1}𝑥-\int \frac{1}{\sqrt{\surd 1-{𝑥}^2}}\times 𝑥𝑑𝑥+𝐶\\ & 𝐼=𝑥{\sin }^{-1}𝑥-\int \frac{2𝑥}{2\sqrt{\surd 1-{𝑥}^2}}𝑑𝑥+𝐶\\ & 𝐼=𝑥{\sin }^{-1}𝑥+\sqrt{\surd 1-{𝑥}^2}+𝐶\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& 𝑥\frac{𝑑𝑦}{𝑑𝑥}+𝑦\log 𝑥={𝑥}^2\\ & \frac{𝑑𝑦}{𝑑𝑥}+𝑦\left(\frac{\log 𝑥}{𝑥}\right)=𝑥\\ 𝐼𝐹& ={𝑒}^{\int \frac{\log 𝑥}{𝑥}𝑑𝑥}\\ & ={𝑒}^{\frac{(\log 𝑥{)}^2}{2}}\\ & ={𝑒}^{\frac{1}{2}\log 𝑥\cdot \log 𝑥}\\ & ={𝑒}^{\log \sqrt{\surd 𝑥}\log 𝑥}\\ =& {𝑒}^{\log 𝑥\log \sqrt{\surd 𝑥}}={𝑥}^{\log \sqrt{\surd 𝑥}}\end{array}\end{array}$

Given, vectors

$𝐚=3\widehat{𝐢}+\widehat{𝐣}-\widehat{𝐤},𝐛=2\widehat{𝐢}-\widehat{𝐣}+7\widehat{𝐤}$

$\begin{array}{cc}\text{and}& 𝐜=7\widehat{𝐢}-\widehat{𝐣}+23\widehat{𝐤}\\ & 𝐚\ne 𝜆𝐛\end{array}$

$\therefore 𝐚$ and $𝐛$ are not collinear

$\therefore$ Option (a) is false

$𝐚\cdot 𝐛=6-1-7\ne 0$

$\therefore$ a and $𝑏$ are not perpendicular

$\therefore 𝐚,𝐛,𝐜$ are not mutually perpendicular

$\therefore$ Option (b) is also incorrect.

$\begin{array}{cc}& 𝐚\cdot (𝐛\times 𝐜)=\mid \begin{array}{ccc}3& 1& -1\\ 2& -1& 7\\ 7& -1& 23\end{array}\mid \\ & =3(-23+7)-1(46-49)-1(-2+7)\\ & =-48+3-5=-50\ne 0\end{array}$

$\therefore 𝐚,𝐛$ and $𝐜$ are non-coplanar

$\therefore$ Option (d) IV is true.

In right circular cylinder

$𝑟+ℎ=6$

By $\mathrm{AM}\ge \mathrm{GM}$

$\begin{array}{cc}& \frac{\frac{𝑟}{2}+\frac{𝑟}{2}+ℎ}{3}\ge \text{exception 25:}\\ & \Rightarrow \frac{6}{3}\ge {\left(\frac{{𝑟}^2ℎ}{4}\right)}^{\frac{1}{3}}\\ & \Rightarrow (2{)}^3\ge \frac{{𝑟}^2ℎ}{4}\\ & \therefore {𝑟}^2ℎ\le 32\\ & \therefore \text{ Maximum value of }{𝑟}^2ℎ=32\\ & \text{ Volume of cylinder }=𝜋{𝑟}^2ℎ\\ & \therefore \text{ Maximum volume }=𝜋\times 32=32𝜋m^3\end{array}$

$\begin{array}{cc}& \text{ Given line, }\\ & \begin{array}{cc}& 𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0\\ & \therefore {𝑚}_1+{𝑚}_2=-\frac{2ℎ}{𝑏}\\ & {𝑚}_1{𝑚}_2=\frac{𝑎}{𝑏}\\ & \because {({𝑚}_1+{𝑚}_2)}^2-{({𝑚}_1-{𝑚}_2)}^2=4{𝑚}_1{𝑚}_2\\ & \therefore \frac{4{ℎ}^2}{{𝑏}^2}-{({𝑚}_1-{𝑚}_2)}^2=\frac{4𝑎}{𝑏}\end{array}\end{array}$

$\begin{array}{cc}& \therefore \frac{4{ℎ}^2}{{𝑏}^2}-{({𝑚}_1-{𝑚}_2)}^2=\frac{4𝑎}{𝑏}\\ & \Rightarrow {𝑚}_1-{𝑚}_2=\frac{2\sqrt{\surd {ℎ}^2-𝑎𝑏}}{𝑏}\\ & \text{ Given, }\frac{{𝑚}_1}{{𝑚}_2}=\frac{5}{3}\\ & \Rightarrow \frac{{𝑚}_1+{𝑚}_2}{{𝑚}_1-{𝑚}_2}=\frac{5+3}{5-3}\\ & \Rightarrow \frac{-2ℎ/𝑏}{\frac{2\sqrt{\surd {ℎ}^2-𝑎𝑏}}{𝑏}}=4\\ & \Rightarrow -ℎ=4\sqrt{\surd {ℎ}^2-𝑎𝑏}\\ & \Rightarrow {ℎ}^2=16({ℎ}^2-𝑎𝑏)\\ & 15{ℎ}^2=16𝑎𝑏\\ & \Rightarrow \frac{{ℎ}^2}{𝑎𝑏}=\frac{16}{15}\end{array}$

Equation of line passing through $(7,-4)$ and perpendicular to the line passing through $(2,3)$ and $(1,-2)$ is

$\begin{array}{cc}𝑦+4& =-\left(\frac{1-2}{-2-3}\right)(𝑥-7)\\ 𝑦+4& =-\frac{1}{5}(𝑥-7)\\ 5𝑦+20& =-𝑥+7\\ \Rightarrow 𝑥+5𝑦+13& =0\end{array}$

$\begin{array}{cc}& 𝐼=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{2𝑥-1}{1+𝑥-{𝑥}^2}\right)𝑑𝑥\\ & 𝐼=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{2(1-𝑥)-1}{1+(1-𝑥)-(1-𝑥{)}^2}\right)𝑑𝑥\\ & ⎡⎣⎢\because \underset{0}{\overset{𝑎}{\int }}𝑓\left(𝑥\right)𝑑𝑥=\underset{0}{\overset{𝑎}{\int }}𝑓(𝑎-𝑥)𝑑𝑥⎤⎦⎥\\ & 𝐼=\underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(\frac{1-2𝑥}{1+𝑥-{𝑥}^2}\right)𝑑𝑥\\ & 𝐼=\underset{0}{\overset{1}{\int }}-{\tan }^{-1}\left(\frac{2𝑥-1}{1+𝑥-{𝑥}^2}\right)𝑑𝑥\\ & 2𝐼=0\\ & 𝐼=0\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{1-\sin 𝑥+\cos 𝑥}{1+\sin 𝑥+\cos 𝑥}\\ & 𝑓\left(𝑥\right)=\frac{1+\cos 𝑥-\sin 𝑥}{1+\cos 𝑥+\sin 𝑥}\\ & 𝑓\left(𝑥\right)=\frac{2{\cos }^2\frac{𝑥}{2}-2\sin \frac{𝑥}{2}\cos \frac{𝑥}{2}}{2{\cos }^2\frac{𝑥}{2}+2\sin \frac{𝑥}{2}\cos \frac{𝑥}{2}}\\ & 𝑓\left(𝑥\right)=\frac{2\cos \frac{𝑥}{2}(\cos \frac{𝑥}{2}-\sin \frac{𝑥}{2})}{2\cos \frac{𝑥}{2}(\cos \frac{𝑥}{2}+\sin \frac{𝑥}{2})}\\ & 𝑓\left(𝑥\right)=\frac{\cos \frac{𝑥}{2}-\sin \frac{𝑥}{2}}{\cos \frac{𝑥}{2}+\sin \frac{𝑥}{2}}\\ & 𝑓\left(𝜋\right)=\frac{\cos \frac{𝜋}{2}-\sin \frac{𝜋}{2}}{\cos \frac{𝜋}{2}+\sin \frac{𝜋}{2}}=\frac{0-1}{0+1}=-1\end{array}\end{array}$

Given, a and b are non-collinear vector and $𝐩=(2𝑥+1)𝐚-𝐛$ and $𝐪=(𝑥-2)𝐚+𝐛$ are collinear vector.

$\begin{array}{cc}\therefore & \frac{2𝑥+1}{𝑥-2}=\frac{-1}{1}\\ \Rightarrow & 2𝑥+1=-𝑥+2\\ \Rightarrow & 3𝑥=1\\ \Rightarrow & 𝑥=\frac{1}{3}\end{array}$

We have, $𝐴=\left[\begin{array}{cc}4& 5\\ 2& 1\end{array}\right]$

$\begin{array}{c}\text{and}{𝐴}^2-5𝐴-6𝐼=0\\ 6𝐼={𝐴}^2-5𝐴\end{array}$

$\begin{array}{cc}& \text{ Multiply by }{𝐴}^{-1}\text{, we get }\\ & \begin{array}{cc}6{𝐴}^{-1}& =𝐴-5𝐼\\ 6{𝐴}^{-1}& =\left[\begin{array}{cc}4& 5\\ 2& 1\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]\\ 6{𝐴}^{-1}& =\left[\begin{array}{cc}-1& 5\\ 2& -4\end{array}\right]\\ {𝐴}^{-1}& =\frac{1}{6}\left[\begin{array}{cc}-1& 5\\ 2& -4\end{array}\right]\end{array}\end{array}$

To find the quadratic equation with roots having an arithmetic mean of 34 and a geometric mean of 16, let's denote the roots by $𝑎$ and $𝑏$.

The arithmetic mean of the roots is given by:

$\frac{𝑎+𝑏}{2}=34$

which implies:

$𝑎+𝑏=68$

The geometric mean of the roots is given by:

$\sqrt{\surd 𝑎𝑏}=16$

which implies:

$𝑎𝑏={16}^2=256$

A quadratic equation with roots $𝑎$ and $𝑏$ can be expressed as:

${𝑥}^2-(𝑎+𝑏)𝑥+𝑎𝑏=0$

Substituting the values of $𝑎+𝑏$ and $𝑎𝑏$ into the equation, we have:

${𝑥}^2-68𝑥+256=0$

Therefore, the quadratic equation is:

Option C:

${𝑥}^2-68𝑥+256=0$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& 𝑥=𝑎\sin 𝑡-𝑏\cos 𝑡\text{... (i)}\\ & 𝑦=𝑎\cos 𝑡+𝑏\sin 𝑡\text{... (ii)}\end{array}\end{array}$

Squaring and adding Eqs. (i) and (ii), we get

${𝑥}^2+{𝑦}^2={𝑎}^2+{𝑏}^2$

On differentiating w.r.t. ' $𝑥$ ', we get

$\begin{array}{cc}2𝑥+2𝑦\frac{𝑑𝑦}{𝑑𝑥}& =0\\ \frac{𝑑𝑦}{𝑑𝑥}& =\frac{-𝑥}{𝑦}\end{array}$

$\begin{array}{cc}& \text{ Again differentiating w.r.t. ’ }𝑥\text{ ’, we get }\\ & \begin{array}{cc}& \frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=-\left(\frac{𝑦-𝑥𝑑𝑦/𝑑𝑥}{{𝑦}^2}\right)[\because \frac{𝑑𝑦}{𝑑𝑥}=-\frac{𝑥}{𝑦}]\\ & \Rightarrow {𝑦}^2\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=-(𝑦+\frac{𝑥}{𝑦})\\ & \Rightarrow {𝑦}^3\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}+{𝑥}^2+{𝑦}^2=0\end{array}\end{array}$

Equation of parabola, ${𝑥}^2=12𝑦$

$\begin{array}{cc}& \text{ Length of latus rectum }=12\\ & \therefore 𝐴𝐵=12\\ & \text{ Focus of parabola }=(0,3)\\ & \therefore 𝑂𝐷=3\end{array}$

$\begin{array}{cc}\text{ Area of }△𝑂𝐴𝐵& =\frac{1}{2}\times 𝑂𝐷\times 𝐴𝐵\\ & =\frac{1}{2}\times 3\times 12=18\text{ sq units }\end{array}$

In Binomial distribution,

Mean $\left(𝜇\right)=𝑛𝑝$

and variance $\left({𝜎}^2\right)=𝑛𝑝𝑞$

Given, $𝜇+{𝜎}^2=1.8$

$\begin{array}{cc}& 𝑛𝑝+𝑛𝑝𝑞=1.8\\ & 𝑛𝑝+𝑛𝑝(1-𝑝)=1.8\\ & 5𝑝+5𝑝-5{𝑝}^2=1.8[\because 𝑛=5]\\ & 5{𝑝}^2-10𝑝+1.8=0\\ & 5{𝑝}^2-9𝑝-𝑝+1.8=0\\ & \Rightarrow 5𝑝(𝑝-1.8)-1(𝑝-1\cdot 8)=0\\ & \Rightarrow (5𝑝-1)(𝑝-1.8)=0\\ & \Rightarrow 𝑝=0\cdot 2(𝑝\ne 1\cdot 8)(\because 𝑝\le 1)\end{array}$

Getting sum multiple of 3, when pair of dice are thrown

$\begin{array}{cc}& 𝐴=\left\{\right(1,2),(2,1\left)\right(1,5),(5,1\left)\right(2,4\left)\right(4,2),(3,6)\\ & (6,3)(4,5)(5,4),(6,6)(3,3)\}\\ & 𝑛\left(𝐴\right)=12\end{array}$

Not getting multiple of $3=36-12=24$

$\therefore$ Odd in favour of getting sum multiple of 3

$=\frac{\text{ favourable cases }}{\text{ Non }-\text{ favourable cases }}=\frac{12}{24}=\frac{1}{2}=1:2$

Let $𝑀$ be the mass of radioactive element at any time ' $𝑡$ '

Given,

$\begin{array}{cc}& \frac{𝑑𝑀}{𝑑𝑡}=𝑘𝑀\\ & \frac{𝑑𝑀}{𝑀}=𝑘𝑑𝑡\end{array}$

On integrating, we get

$\log 𝑀=𝑘𝑡+𝐶$

$\begin{array}{cc}& \text{When,}𝑡=0,𝑀=1.5\text{gm}\\ & \therefore 𝐶=\log \left(1.5\right)\\ & \therefore \log 𝑀=𝑘𝑡+\log \left(1.5\right)\end{array}$

When, $𝑡=𝑡,𝑀=0.5\mathrm{gm}$

$\begin{array}{cc}\therefore & \log \left(0.5\right)=𝑘𝑡+\log \left(1.5\right)\\ & \log \left(\frac{0.5}{1.5}\right)=𝑘𝑡\Rightarrow \log \frac{1}{3}=𝑘𝑡\\ \Rightarrow & 𝑡=(\log 3)𝑘𝑡\end{array}$

$\therefore$ Time is proportional to $\log 3$

$\begin{array}{cc}& \text{ We have, }p=\frac{𝐛\times 𝐜}{\left[\mathrm{abc}\right]}\\ & \begin{array}{cc}& 𝑞=\frac{𝑐\times 𝑎}{\left[𝑎𝑏𝑐\right]}\\ & 𝐫=\frac{𝐚\times 𝐛}{\left[𝐚𝐛𝐜\right]}\\ & 𝑎\cdot 𝑝+𝑏\cdot 𝑞+𝑐\cdot 𝑟\\ & =\frac{𝐚\cdot (𝐛\times 𝐜)}{\left[𝐚𝐛𝐜\right]}+\frac{𝐛\cdot (𝐜\times 𝐚)}{\left[𝐚𝐛\right]}+\frac{𝐜\cdot (𝐚\times 𝐛)}{\left[𝐚𝐛𝐜\right]}\\ & =1+1+1=3\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \sim 𝑝\vee (𝑞<text>\to </text><text>\sim </text>𝑟)\\ & \sim 𝑝\vee (\sim 𝑞\vee \sim 𝑟)[\because 𝑝\to 𝑞<text>=</text><text>\sim </text>𝑝\vee 𝑞]\\ & \sim 𝑝\vee \sim (𝑞\wedge 𝑟)\\ & \sim (𝑝\wedge (𝑞\wedge 𝑟\left)\right)\\ \therefore & \sim (\sim (𝑝\wedge (𝑞\wedge 𝑟))\\ & 𝑝\wedge (𝑞\wedge 𝑟)\end{array}\end{array}$

We have,

$𝐴\equiv (2,4,5)\text{ and }𝐵\equiv (1,2,3)$

Equation of $𝐴𝐵$ is,

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

Putting $𝑧=3$, we get

$\begin{array}{cc}𝜆=\frac{3-5}{-2}& =1\\ \therefore \frac{𝑦-4}{-2}& =1\\ \Rightarrow 𝑦-4& =-2\\ 𝑦& =2\end{array}$

For the relation $𝑅=\left\{\right(𝑎,𝑏)\mid 𝑏=𝑎-1,𝑎\in \mathbb{Z},5<𝑎<9\}$, first identify the integers $𝑎$ that satisfy $5<𝑎<9$. The integer values for $𝑎$ are $6,7,$ and $8$.

Calculating $𝑏$ for each value of $𝑎$:

If $𝑎=6$, then $𝑏=6-1=5$.

If $𝑎=7$, then $𝑏=7-1=6$.

If $𝑎=8$, then $𝑏=8-1=7$.

Thus, the range of $𝑅$ is the set of all possible values of $𝑏$, which is $\{5,6,7\}$.

Therefore, the correct option is Option B: $\{5,6,7\}$.

We have,

$\begin{array}{cc}𝐼& =\int \log 𝑥(\log 𝑥+2)𝑑𝑥\\ 𝐼& =\int (\log 𝑥{)}^2𝑑𝑥+\int 2\log 𝑥𝑑𝑥\\ 𝐼& =(\log 𝑥{)}^2\int 𝑑𝑥-\int (\frac{𝑑(\log 𝑥{)}^2}{𝑑𝑥}\int 𝑑𝑥)𝑑𝑥+\int 2\log 𝑥𝑑𝑥+𝐶\end{array}$

$\begin{array}{cc}& 𝐼=𝑥(\log 𝑥{)}^2-\int \frac{2\log 𝑥}{𝑥}\cdot 𝑥𝑑𝑥+\int 2\log 𝑥𝑑𝑥+𝑐\\ & 𝐼=𝑥(\log 𝑥{)}^2-\int 2\log 𝑥𝑑𝑥+\int 2\log 𝑥𝑑𝑥+𝑐\\ & 𝐼=(𝑥\log 𝑥{)}^2+𝑐\end{array}$

$\begin{array}{cc}& \text{ We have, }𝑓\left(𝑥\right)=\{\begin{array}{cc}3(1-2{𝑥}^2),& 0<𝑥<1\\ 0,& \text{ otherwise }\end{array}\\ & \begin{array}{cc}𝐹(𝑋& =𝑛)=𝑃(𝑋\le 𝑥)=\underset{0}{\overset{𝑥}{\int }}𝑓(𝑥)𝑑𝑥\\ & =\underset{0}{\overset{𝑥}{\int }}3(1-2{𝑥}^2)𝑑𝑥\\ & ={\left[3(𝑥-\frac{2{𝑥}^3}{3})\right]}_{𝑥}^0=3(𝑥-\frac{2{𝑥}^3}{3})\end{array}\\ & \therefore 𝑘=3\end{array}$

$\begin{array}{cc}& \text{ We have, }𝑓\left(𝑥\right)=\frac{𝑥-3}{5-𝑥}\\ & \begin{array}{cc}& \text{ Put, }𝑓\left(𝑥\right)=𝑦\\ & \therefore 𝑦=\frac{𝑥-3}{5-𝑥}\\ & \Rightarrow 5𝑦-𝑥𝑦=𝑥-3\\ & 𝑥(𝑦+1)=5𝑦+3\\ & 𝑥=\frac{5𝑦+3}{𝑦+1}\\ & \therefore 𝑦\in 𝐑-\{-1\}\end{array}\\ & \therefore \text{ Range of }𝑓\left(𝑥\right)=𝑅-\{-1\}\end{array}$

We have, $𝑦={\sin }^{-1}\left[\frac{\sqrt{\surd 1+𝑥}+\sqrt{\surd 1-𝑥}}{2}\right]$

Put, $𝑥=\cos 2𝜃$

$\begin{array}{cc}\therefore 𝑦& ={\sin }^{-1}\left[\frac{\sqrt{\surd 1+\cos 2𝜃}+\sqrt{\surd 1-\cos 2𝜃}}{2}\right]\\ 𝑦& ={\sin }^{-1}\left(\frac{\cos 𝜃+\sin 𝜃}{\sqrt{\surd 2}}\right)\\ 𝑦& ={\sin }^{-1}(\sin (\frac{𝜋}{4}+𝜃))\\ 𝑦& =\frac{𝜋}{4}+𝜃\\ 𝑦& =\frac{𝜋}{4}+\frac{1}{2}{\cos }^{-1}𝑥\\ \frac{𝑑𝑦}{𝑑𝑥}& =\frac{-1}{2\sqrt{\surd 1-{𝑥}^2}}\end{array}$

We have,

Area of triangle $𝐴𝐵𝐶=10\sqrt{\surd 3}{\mathrm{cm}}^2$

Perimeter of triangle $𝐴𝐵𝐶=20\mathrm{cm}$

$\angle 𝐵={60}^{\circ }$

$\begin{array}{cc}& \text{ Area of }△𝐴𝐵𝐶=\frac{1}{2}𝑎𝑐\sin 𝐵\\ & 10\sqrt{\surd 3}=\frac{1}{2}𝑎𝑐\times \sin {60}^{\circ }\\ & 10\sqrt{\surd 3}=\frac{1}{2}𝑎𝑐\times \frac{\sqrt{\surd 3}}{2}\\ & 𝑎𝑐=40\\ & \cos \frac{𝐵}{2}=\sqrt{\surd \frac{𝑠(𝑠-𝑏)}{𝑎𝑐}}\\ & \therefore \cos {30}^{\circ }=\sqrt{\surd \frac{10(10-𝑏)}{40}}\\ & [\because 𝑠=\frac{𝑎+𝑏+𝑐}{2}=\frac{20}{2}=10]\\ & \Rightarrow \frac{\sqrt{\surd 3}}{2}=\frac{\sqrt{\surd (10-𝑏)}}{2}\\ & \Rightarrow 3=(10-𝑏)\\ & \Rightarrow 𝑏=10-3=7\mathrm{cm}\end{array}$

Given curve, $2{𝑥}^2+{𝑦}^2=12$

$\begin{array}{cc}\Rightarrow 4𝑥+2𝑦\frac{𝑑𝑦}{𝑑𝑥}& =0\\ \frac{𝑑𝑦}{𝑑𝑥}& =\frac{-2𝑥}{𝑦}\end{array}$

Slope of tangent at $(2,2)$ is

${\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}_{(2,2)}=\frac{-2\left(2\right)}{2}=-2$

$\therefore$ Slope of Normal $=-\frac{𝑑𝑥}{𝑑𝑦}=\frac{1}{2}$

Equation of normal at $(2,2)$ is

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