MHT CET 2025 23rd April Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}\arg \left(\frac{13-5𝑖}{4-9𝑖}\right)& =\arg (13-5𝑖)-\arg (4-9𝑖)\\ & =-{\tan }^{-1}\left(\frac{5}{13}\right)+{\tan }^{-1}\left(\frac{9}{4}\right)\\ & ={\tan }^{-1}1=\frac{𝜋}{4}\end{array}$

$\begin{array}{cc}& \text{ Given }\sin 𝜃=\frac{1}{2}(𝑥+\frac{1}{𝑥})\\ & \Rightarrow 𝑥+\frac{1}{𝑥}=2\sin 𝜃...\left(𝑖\right)\\ & \Rightarrow {(𝑥+\frac{1}{𝑥})}^3=8{\sin }^3𝜃\\ & \Rightarrow {𝑥}^3+3(𝑥+\frac{1}{𝑥})+\frac{1}{{𝑥}^3}=8{\sin }^3𝜃\\ & \Rightarrow {𝑥}^3+3(2\sin 𝜃)+\frac{1}{{𝑥}^3}=8{\sin }^3𝜃\dots \text{ [From (i)] }\\ & \Rightarrow {𝑥}^3+\frac{1}{{𝑥}^3}+6\sin 𝜃-8{\sin }^3𝜃=0\\ & \Rightarrow {𝑥}^3+\frac{1}{{𝑥}^3}+2(3\sin 𝜃-4{\sin }^3𝜃)=0\\ & \Rightarrow ({𝑥}^3+\frac{1}{{𝑥}^3})+2\sin 3𝜃=0\\ & \Rightarrow \frac{1}{2}({𝑥}^3+\frac{1}{{𝑥}^3})+\sin 3𝜃=0\end{array}$

Let $\overline{𝑎}=0\widehat{𝑖}+0\widehat{𝑗},\overline{𝑏}=3\widehat{𝑖},\overline{𝑐}=-4\widehat{𝑗}$

$\begin{array}{cc}& \overline{\mathrm{AB}}=3\hat{i},\overline{\mathrm{AC}}=-4\hat{j},\overline{\mathrm{BC}}=-4\hat{j}-3\hat{i}\\ & \left|\overline{\mathrm{AB}}\right|=3,\left|\overline{\mathrm{AC}}\right|=4,\left|\overline{\mathrm{BC}}\right|=5\end{array}$

If $𝐻\left(\overline{ℎ}\right)$ is the incentre of $△𝐴𝐵𝐶$, then

$\begin{array}{cc}\bar{h}& =\frac{\left|\overline{\mathrm{BC}}\right|\bar{a}+\left|\overline{\mathrm{AC}}\right|\bar{b}+|\overline{\mathrm{AB}}{|}^{-}\bar{c}}{\left|\overline{\mathrm{BC}}\right|+\left|\overline{\mathrm{AC}}\right|+\left|\overline{\mathrm{AB}}\right|}\\ & =\frac{5(0\hat{i}+0\hat{k})+4\left(3\hat{i}\right)+3(-4\hat{j})}{3+4+5}\\ & =\hat{i}-\hat{j}\\ \Rightarrow 𝐻& \equiv (1,-1)\end{array}$

Given equation of circle is

${𝑥}^2+{𝑦}^2=36\Rightarrow 𝑎=6$

Slope of $5𝑥+𝑦=2$ is -5

$\therefore$ Slope of required line $=\frac{1}{5}$

Equations of tangents to the circle ${𝑥}^2+{𝑦}^2=a^2$ is

$\begin{array}{cc}& 𝑦=m𝑥\pm \sqrt{\surd a^2(1+m^2)}\\ & \Rightarrow 𝑦=\frac{1}{5}𝑥\pm \sqrt{\surd 36(1+\frac{1}{25})}\\ & \Rightarrow 5𝑦=𝑥\pm 6\sqrt{\surd 26}\\ & \Rightarrow 𝑥-5𝑦\pm 6\sqrt{\surd 26}=0\end{array}$

$𝑔\left(𝑥\right)=\{\begin{array}{cc}-𝑥,& -2\le 𝑥\le 3\\ 𝑥,& 0\le 𝑥\le 1\end{array}$

A function cannot be defined with two different outputs at the same input

$\therefore 𝑔\left(𝑥\right)=⎧⎩⎨⎪⎪\begin{array}{cc}-𝑥& \text{ for }-2<𝑥\le 0\\ 𝑥& \text{ for }0\le 𝑥<1\\ -𝑥& \text{ for }1\le 𝑥\le 3\end{array}$

Case I: $-2<𝑥\le 0$

$\begin{array}{cc}& 𝑔\left(𝑥\right)=-𝑥\\ & 𝑓\left(𝑔\right(𝑥\left)\right)=𝑓(-𝑥),𝑥\in [-2,0]\\ \therefore & 𝑓(-𝑥)=1+\frac{5(-𝑥)}{3}=1-\frac{5𝑥}{3}\end{array}$

$\therefore$ Range of $f\left(g\right(n\left)\right)$ is $[1,\frac{13}{3}]$

Case II: $0<𝑥<1$

$\begin{array}{cc}& g\left(𝑥\right)=𝑥\\ & f\left(g\right(𝑥\left)\right)=f\left(𝑥\right),𝑥\in (0,1)\\ & f\left(g\right(𝑥\left)\right)=1+\frac{5𝑥}{3}\end{array}$

Range of $𝑓\left(𝑔\right(𝑛\left)\right)=[1,\frac{8}{3}]$

Case III: $1\le 𝑥<3$

$\begin{array}{cc}& g\left(𝑥\right)=-𝑥\\ & f\left(g\right(𝑥\left)\right)=f(-𝑥)\\ & f\left(g\right(𝑥\left)\right)=1-\frac{5n}{3}\end{array}$

$\therefore$ Range of $𝑓\left(𝑔\right(𝑥\left)\right)=[-4,\frac{-2}{3}]$

$\therefore$ Combined range of $𝑓\left(𝑔\right(𝑛\left)\right)=[-4,\frac{13}{3}]$

Given equation of conic is

$\begin{array}{cc}& 25{𝑥}^2+16{𝑦}^2-150𝑥=175\\ & \Rightarrow 25({𝑥}^2-6𝑥)+16{𝑦}^2=175\\ & \Rightarrow 25({𝑥}^2-6𝑥+9)+16{𝑦}^2=175+225\\ & \Rightarrow 25(𝑥-3{)}^2+16{𝑦}^2=400\\ & \Rightarrow \frac{(𝑥-3{)}^2}{16}+\frac{{𝑦}^2}{25}=1\\ & a^2=16\Rightarrow a=4\\ & b^2=25\Rightarrow b=5\\ & e=\frac{\sqrt{\surd b^2-a^2}}{b}=\frac{3}{5}\end{array}$

Centre of ellipse is $(3,0)$

$\begin{array}{cc}\therefore \text{ Foci }& =(ℎ,𝑘\pm 𝑏𝑒)\\ & =(3,0\pm 3)\end{array}$

$\begin{array}{cc}𝐴& =\underset{𝑥\to 0^{+}}{lim}{(1+{\tan }^2\sqrt{\surd 𝑥})}^{\frac{1}{2𝑥}}\\ & =e^{\underset{𝑥\to 0}{lim}\frac{1}{2𝑥}{\tan }^2\sqrt{\surd 𝑥}}\end{array}$

$\cdots ⎡⎣⎢\begin{array}{c}\text{ If }\underset{𝑥\to 𝑎}{lim}𝑓\left(𝑥\right)=1\text{ and }\underset{𝑥\to 𝑎}{lim}𝑔\left(𝑥\right)=\infty ,\\ \text{ then }\underset{𝑥\to 𝑎}{lim}\left[𝑓\right(𝑥){]}^{𝑔\left(𝑥\right)}={𝑒}^{\underset{𝑥\to 0}{lim}𝑔\left(𝑥\right)\left[\right(𝑓\left(𝑥\right)-1]}\end{array}⎤⎦⎥$

$\begin{array}{cc}& =e^{\underset{𝑥\to 0}{lim}\frac{1}{2}{\left(\frac{\tan \sqrt{\surd 𝑥}}{\sqrt{\surd 𝑥}}\right)}^2}\\ \therefore A& =e^{\frac{1}{2}}\\ \Rightarrow & {\log }_{e}A=\frac{1}{2}\end{array}$

$\begin{array}{cc}& f\left(𝑥\right)=2𝑥-\mid 𝑥-{𝑥}^2\mid \\ & \text{ For }𝑥-{𝑥}^2>0\\ & f\left(𝑥\right)=2𝑥-𝑥(1-{𝑥}^2)\\ & =2𝑥-𝑥+{𝑥}^2=𝑥+{𝑥}^2\end{array}$

For $𝑥-{𝑥}^2<0$

$\begin{array}{cc}f\left(𝑥\right)& =2𝑥+𝑥(1-{𝑥}^2)\\ & =2𝑥+𝑥-{𝑥}^2\\ & =3𝑥-{𝑥}^2\end{array}$

Continuity at $𝑥=0$

$\begin{array}{cc}& \text{ L.H.L. }=\underset{𝑥\to 0^{-}}{lim}f\left(𝑥\right)=\underset{𝑥\to 0}{lim}𝑥+{𝑥}^2=0\\ & \text{ R.H.L. }=\underset{𝑥\to 0^{+}}{lim}f\left(𝑥\right)=\underset{𝑥\to 0}{lim}3𝑥-{𝑥}^2=0\end{array}$

$\therefore f\left(𝑥\right)$ is continuous at $𝑥=0$

Continuity at $𝑥=1$,

$\begin{array}{cc}& \text{ L.H.L }=\underset{𝑥\to 1^{-}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 1}{lim}{𝑥}^2+𝑥=2\\ & \text{ R.H.L }=\underset{𝑥\to 1^{+}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 1}{lim}3𝑥-{𝑥}^2=3-1=2\end{array}$

$\therefore f\left(𝑥\right)$ is continuous at $𝑥=1$

$\begin{array}{cc}\left(𝐴\right)𝐿𝑒𝑡𝑝:4+3& =8\\ 𝑞:5+3& =9\end{array}$

Truth values of $𝑝$ and $𝑞$ are $𝐹$ and $𝐹$.

$\therefore p\to q\equiv T$

(B) Let $p:6+4=10$

q : Moon is flat.

Truth values of $𝑝$ and $𝑞$ are $𝑇$ and $𝐹$.

$\therefore 𝑝\to 𝑞\equiv 𝐹$

(C) Let $p:$ Both $\left(A\right)$ and $\left(B\right)$ are true.

$𝑞:5+6=17$

Truth values of $𝑝,𝑞$ are false and false.

$\therefore p\to q\equiv T$

(A) and (C) are true, while (B) is false.

$\begin{array}{cc}& 𝐴=⎡⎣⎢\begin{array}{ccc}1& -1& 1\\ 0& 2& -3\\ 2& 1& 0\end{array}⎤⎦⎥\\ & 𝐵=\mathrm{adj}𝐴=⎡⎣⎢\begin{array}{ccc}3& 1& 1\\ -6& -2& 3\\ -4& -3& 2\end{array}⎤⎦⎥\\ & \therefore \mathrm{adj}𝐵=⎡⎣⎢\begin{array}{ccc}5& -5& 5\\ 0& 10& -15\\ 10& 5& 0\end{array}⎤⎦⎥=5𝐴\end{array}$

$\therefore \mathrm{adj}B=C$$\dots .[\because C=5A$ (given) $]$

$\begin{array}{cc}& \Rightarrow |\mathrm{adj}𝐵|=\left|𝐶\right|\\ & \Rightarrow \frac{|\mathrm{adj}𝐵|}{\left|𝐶\right|}=1\end{array}$

$\begin{array}{cc}& 2{\sin }^2𝑥+5\sin 𝑥-3=0\\ & \Rightarrow 2{\sin }^2𝑥+6\sin 𝑥-\sin 𝑥-3=0\\ & \Rightarrow 2\sin 𝑥(\sin 𝑥+3)-1(\sin 𝑥+3)=0\\ & \Rightarrow (2\sin 𝑥-1)(\sin 𝑥+3)=0\\ & \Rightarrow \sin 𝑥=\frac{1}{2}\text{ or }-3\end{array}$

$\begin{array}{cc}& \Rightarrow \sin 𝑥=\frac{1}{2}\dots [\because \sin 𝑥\ne -3]\\ \therefore & 𝑥=\frac{𝜋}{6},\frac{5𝜋}{6},\frac{13𝜋}{6},\frac{17𝜋}{6}\dots [𝑥\in [0,3𝜋\left]\right]\\ \therefore & \text{ Number of solutions }=4\end{array}$

Given equation of pair of lines is

$\begin{array}{cc}& {𝑥}^2-3𝑥𝑦+𝜆{𝑦}^2+3𝑥-5𝑦+2=0\\ & a=1,b=𝜆,c=2,f=\frac{-5}{2},g=\frac{3}{2},h=\frac{-3}{2}\end{array}$

Since the given equation represents pair of straight lines

$\begin{array}{cc}\therefore & 𝑎𝑏𝑐+2𝑓𝑔ℎ-𝑎{𝑓}^2-𝑏{𝑔}^2-𝑐{ℎ}^2=0\\ & \Rightarrow 2𝜆+\frac{45}{4}-\frac{25}{4}-\frac{9𝜆}{4}-\frac{18}{4}=0\\ & \Rightarrow 𝜆=2\end{array}$

$\begin{array}{cc}& \tan 𝜃=\mid \frac{2\sqrt{\surd {ℎ}^2-𝑎𝑏}}{𝑎+𝑏}\mid \\ & =\mid \frac{2\sqrt{\surd \frac{9}{4}-2}}{3}\mid \\ & \Rightarrow \tan 𝜃=\frac{1}{3}\\ & \text{ Now, }1+{\cot }^2𝜃={\mathrm{cosec}}^2𝜃\\ & \Rightarrow {\mathrm{cosec}}^2𝜃=1+\frac{1}{{\tan }^2𝜃}\\ & =1+\frac{1}{{\left(\frac{1}{3}\right)}^2}=10\\ & \begin{array}{cc}\frac{\mathrm{cosec}{}^2𝜃}{\sqrt{\surd 10}}& =\frac{10}{\sqrt{\surd 10}}=\sqrt{\surd 10}\end{array}\end{array}$

Given equations are

$\begin{array}{cc}& 6𝑚𝑛-2𝑛𝑙+5𝑙𝑚=0\\ & 3𝑙+𝑚+5𝑛=0\\ & \Rightarrow 𝑚=-3𝑙-5𝑛\end{array}$

Substituting above value in (i), we get

$6𝑛(-3𝑙-5𝑛)-2𝑛𝑙+5𝑙(-3𝑙-5𝑛)=0$

$\begin{array}{cc}& \Rightarrow -18n𝑙-30n^2-2n𝑙-15{𝑙}^2-25n𝑙=0\\ & \Rightarrow -30n^2-45n𝑙-15{𝑙}^2=0\\ & \Rightarrow 2n^2+3n𝑙+{𝑙}^2=0\\ & \Rightarrow (2n+𝑙)(n+𝑙)=0\\ & \Rightarrow 𝑙=-2n\text{ or }𝑙=-n\dots \left[𝐹𝑟𝑜𝑚\right(𝑖𝑖\left)\right]\end{array}$

When $𝑙=-2n$,

$\begin{array}{cc}𝑚& =-3𝑙-5𝑛\\ & =6𝑛-5𝑛\\ \Rightarrow 𝑚& =𝑛\end{array}$

Direction cosines are $(-2,1,1)$

When $𝑙=-n\dots \left[𝐹𝑟𝑜𝑚\right(𝑖𝑖\left)\right]$

$\begin{array}{cc}& 𝑚=-3𝑙-5𝑛=3𝑛-5𝑛\\ & \Rightarrow 𝑚=-2𝑛\end{array}$

$\begin{array}{cc}& \text{ Direction cosines are }(1,2,-1)\\ & \begin{array}{cc}\cos 𝜃& =\mid \frac{{𝑎}_1{𝑎}_2+{𝑏}_1{𝑏}_2+{𝑐}_1{𝑐}_2}{\sqrt{\surd {𝑎}_2^1{𝑏}_2^1{𝑐}_2^1}\cdot \sqrt{\surd {𝑎}_2^2+{𝑏}_2^2+{𝑐}_2^2}}\mid \\ & =\frac{(-2)\left(1\right)+\left(1\right)\left(2\right)+\left(1\right)(-1)}{\sqrt{\surd (-2{)}^2+1^2+1^2}\cdot \sqrt{\surd 1^2+2^2+(-1{)}^2}}\\ & =\mid \frac{-1}{6}\mid \\ \Rightarrow \cos 𝜃& =\frac{1}{6}\\ \sin 𝜃& =\sqrt{\surd 1-{\cos }^2𝜃}=\sqrt{\surd 1-\frac{1}{36}}=\frac{\sqrt{\surd 35}}{6}\end{array}\end{array}$

$\begin{array}{cc}& X\sim B(6,\frac{1}{2})\\ & \text{ Here }n=6,p=\frac{1}{2},q=\frac{1}{2}\\ & \text{ Consider }|𝑥-2|\le 1\\ \therefore & -1\le 𝑥-2\le 1\\ \therefore & 1\le 𝑥\le 3\end{array}$

$\begin{array}{cc}\therefore P\left(\right|𝑥-2|\le 1)& =P(1\le 𝑥\le 3)\\ & =P(X=1)+P(X=2)+P(X=3)\\ & ={}^6{C}_1{p}^6{q}^5+{}^6{C}_2{p}^2{q}^4+{}^6{C}_3{p}^3{q}^3\\ & =6\times {\left(\frac{1}{2}\right)}^6+15{\left(\frac{1}{6}\right)}^6+20{\left(\frac{1}{6}\right)}^6\\ & ={\left(\frac{1}{2}\right)}^6[6+15+20]\\ & =\frac{41}{64}\end{array}$

$\begin{array}{cc}& \overline{𝑝}\times \overline{𝑏}=\overline{𝑐}\times \overline{𝑏}\\ \Rightarrow & \overline{𝑎}\times (\overline{𝑝}\times \overline{𝑏})=\overline{𝑎}\times (\overline{𝑐}\times \overline{𝑏})\\ \Rightarrow & (\overline{𝑎}\cdot \overline{𝑏})\overline{𝑝}-(\overline{𝑎}\cdot \overline{𝑝})\overline{𝑏}=(\overline{𝑎}\cdot \overline{𝑏})\overline{𝑐}-(\overline{𝑎}\cdot \overline{𝑐})\overline{𝑏}\\ \Rightarrow & (\overline{𝑎}\cdot \overline{𝑏})\overline{𝑝}-0=(\overline{𝑎}\cdot \overline{𝑏})\overline{𝑐}-(\overline{𝑎}\cdot \overline{𝑐})\overline{𝑏}\\ \Rightarrow & \overline{𝑝}=\overline{𝑐}-\left(\frac{\overline{𝑎}\cdot \overline{𝑐}}{\overline{𝑎}\cdot \overline{𝑏}}\right)\overline{𝑏}\\ & \overline{𝑎}\cdot \overline{𝑐}=(\widehat{𝑖}+\widehat{𝑗})\cdot (3\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘})=2\\ & \overline{𝑎}\cdot \overline{𝑏}=(\widehat{𝑖}+\widehat{𝑗})\cdot (2\widehat{𝑖}-\widehat{𝑘})=2\\ \therefore & \overline{𝑝}=3\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘}-\left(\frac{2}{2}\right)(2\widehat{𝑖}-\widehat{𝑘})\\ & =3\widehat{𝑖}-\widehat{𝑗}+\widehat{𝑘}-2\widehat{𝑖}+\widehat{𝑘}\\ & =\widehat{𝑖}-\widehat{𝑗}+2\widehat{𝑘}\end{array}$

Since, $f\left(𝑥\right)$ is a p.d.f. of $𝑥$.

$\begin{array}{cc}\therefore & \int 𝑓\left(𝑥\right)𝑑𝑥=1\\ & \Rightarrow {\int }_1^0𝑓\left(𝑥\right)𝑑𝑥=1\\ & \Rightarrow {\int }_1^0𝑘𝑥(1-𝑥)𝑑𝑥=1\\ & \Rightarrow 𝑘{[\frac{{𝑥}^2}{2}-\frac{{𝑥}^3}{3}]}_1^0=1\Rightarrow 𝑘=6\\ & 𝑃(𝑥>𝑎)=1-𝑃(𝑥\le 𝑎)\\ & \Rightarrow \frac{20}{27}=1-{\int }_{𝑎}^0𝑓\left(𝑥\right)𝑑𝑥\\ & \Rightarrow \frac{20}{27}=1-{\int }_{𝑎}^{0}6𝑥(1-𝑥)𝑑𝑥\\ & \Rightarrow \frac{20}{27}=1-6{[\frac{{𝑥}^2}{2}-\frac{{𝑥}^3}{3}]}_{𝑎}^0\\ & \Rightarrow 6[\frac{{𝑎}^2}{2}-\frac{{𝑎}^3}{3}]=1-\frac{20}{27}\\ & \Rightarrow 3{𝑎}^2-2{𝑎}^3=\frac{7}{27}\end{array}$

On solving, we get

$\Rightarrow 𝑎=\frac{1}{3}$

$\begin{array}{cc}& E\left(X\right)=\Sigma {𝑥}_{i}P\left({𝑥}_i\right)\\ & =0\left(0.4\right)+1\left(0.3\right)+2\left(0.1\right)+3\left(0.1\right)+4\left(0.1\right)\\ & =0.3+0.2+0.3+0.4=1.2\\ & \text{ Variance }=E\left(X^2\right)-\left[E\right(X){]}^2\\ & =0^2\left(0.4\right)+1^2\left(0.3\right)+2^2\left(0.1\right)+3^2\left(0.1\right)+4^2\left(0.1\right)-(1.2{)}^2\\ & =0.3+0.4+0.9+1.6-1.44\\ & =1.76\end{array}$

$\begin{array}{cc}& {({\cos }^{-1}𝑥)}^2-{({\sin }^{-1}𝑥)}^2>0\\ & \Rightarrow ({\cos }^{-1}𝑥+{\sin }^{-1}𝑥)({\cos }^{-1}𝑥-{\sin }^{-1}𝑥)>0\\ & \Rightarrow \frac{𝜋}{2}({\cos }^{-1}𝑥-{\sin }^{-1}𝑥)>0\\ & \dots [{\cos }^{-1}𝑥+{\sin }^{-1}𝑥=\frac{𝜋}{2}]\end{array}$

$\begin{array}{cc}& \Rightarrow {\cos }^{-1}𝑥-{\sin }^{-1}𝑥>0\\ & \Rightarrow \frac{𝜋}{2}-{\sin }^{-1}𝑥-{\sin }^{-1}𝑥>0\\ & \Rightarrow \frac{𝜋}{2}-2{\sin }^{-1}𝑥>0\\ & \Rightarrow \frac{𝜋}{2}>2{\sin }^{-1}𝑥\Rightarrow {\sin }^{-1}𝑥<\frac{𝜋}{4}\\ & \Rightarrow -1\le 𝑥<\frac{1}{\sqrt{\surd 2}}\end{array}$

$\begin{array}{cc}& \bar{a}=2\hat{i}+3\hat{j}+4\hat{k}\\ & \bar{b}=\hat{i}-2\hat{j}-2\hat{k}\\ & \bar{c}=-\hat{i}+4\hat{j}+3\hat{k}\\ & \bar{b}\times \bar{c}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& -2\\ -1& 4& 3\end{array}\mid =2\hat{i}-\hat{j}+2\hat{k}\end{array}$

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

Given, $\overrightarrow{a}\cdot \overrightarrow{d}=18$

$\begin{array}{cc}& \Rightarrow (2\widehat{𝑖}+3\widehat{𝑗}+4\widehat{𝑘})\cdot \left[𝜆\right(2\widehat{𝑖}-\widehat{𝑗}+2\widehat{𝑘}\left)\right]=18\\ & \Rightarrow 9𝜆=18\\ & \Rightarrow 𝜆=2\\ \therefore & \overline{𝑑}=4\widehat{𝑖}-2\widehat{𝑗}+4\widehat{𝑘}\\ & \text{ Now, }\overline{𝑎}\times \overline{𝑑}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 2& 3& 4\\ 4& -2& 4\end{array}\mid =2\hat{0}𝑖+8𝑗-16\widehat{𝑘}\\ \therefore & |\overrightarrow{𝑎}\times \overrightarrow{𝑑}{|}^2={20}^2+8^2+(-16{)}^2=720.\end{array}$

Let the volume and surface area of the spherical balloon be V and A respectively

Given that, $\frac{\mathrm{dV}}{\mathrm{dt}}\propto A$

$\begin{array}{cc}\therefore & \frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{kA}\\ & \Rightarrow \frac{d}{\mathrm{dt}}\left(\frac{4}{3}𝜋r^3\right)=k\times 4𝜋r^2\\ & \Rightarrow 4𝜋r^2\times \frac{\mathrm{dr}}{\mathrm{dt}}=k\times 4𝜋r^2\\ & \Rightarrow k=\frac{\mathrm{dr}}{\mathrm{dt}}\\ & \Rightarrow \mathrm{dr}=\mathrm{kdt}\end{array}$

Integrating on both sides, we get

$r=\mathrm{kt}+c...\left(𝑖\right)$

When $t=0,r=3\mathrm{cm}$

$\Rightarrow 𝑐=3$

When $t=2$ minutes, $r=9\mathrm{cm}$

$\therefore$ From (i)

$\begin{array}{cc}& 9=2𝑘+𝑐\\ & \Rightarrow 9=2𝑘+3\\ & \Rightarrow 𝑘=3\\ \therefore & 𝑟=3𝑡+3\end{array}$