MHT CET 2019 3rd May Morning Shift (Mathematics) Question Paper with Solutions

Given, equation of plane

$𝐫=(2\widehat{𝐢}+\widehat{𝐤})+𝜆\widehat{𝐢}+𝜇(\widehat{𝐢}+2\widehat{𝐣}-3\widehat{𝐤})\text{.... (i)}$

Here, plane (i) passing through a (let) $=2\widehat{𝐢}+\widehat{𝐤}$

and parallel to vector $𝐛($ let $)=\widehat{𝐢}$ and

$𝐜=\widehat{𝐢}+2\widehat{𝐣}-3\widehat{𝐤}$

We know that equation of plane passing through a point a and parallel to non-parallel vectors band $𝐜$ is

$𝑟\cdot (𝑏\times 𝑐)=𝑎\cdot (𝑏\times 𝑐)=\left[𝑎𝑏𝑐\right]$

Now,

$\begin{array}{cc}& \left[\begin{array}{ccc}𝑎& 𝑏& 𝑐\end{array}\right]=\mid \begin{array}{ccc}2& 0& 1\\ 1& 0& 0\\ 1& 2& -3\end{array}\mid \\ & =2\left(0\right)-0+1(2-0)=2\end{array}$

and $𝐛\times 𝐜=\mid \begin{array}{ccc}\widehat{𝐢}& \widehat{𝐣}& \widehat{𝐤}\\ 1& 0& 0\\ 1& 2& -3\end{array}\mid =3\widehat{𝐢}+2\widehat{𝐤}$

$\therefore 𝐫\cdot (3\widehat{𝐢}+2\widehat{𝐤})=2$

Therefore, $𝛼=2$

$\begin{array}{cc}\text{ Required area }& =2{\int }_{\frac{𝜋}{2}}^0𝑦𝑑𝑥\\ & =2{\int }_{\frac{𝜋}{2}}^0\cos 𝑥𝑑𝑥\\ & =2(\sin 𝑥{)}_{𝜋/2}^0\\ & =2(1-0)\\ & =2\text{ sq. units }\end{array}$

Given, $\frac{2{𝑏}^2}{𝑎}=\frac{18}{5}\Rightarrow \frac{{𝑏}^2}{𝑎}=\frac{9}{5}$

$\Rightarrow {𝑏}^2=\frac{9}{5}𝑎\text{.... (i)}$

and, $𝑒=\sqrt{\surd 1-\frac{{𝑏}^2}{{𝑎}^2}}=\frac{4}{5}\Rightarrow 1-\frac{{𝑏}^2}{{𝑎}^2}=\frac{16}{25}$

$\Rightarrow \frac{{𝑏}^2}{{𝑎}^2}=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow \frac{\left(\frac{9}{5}\right)𝑎}{{𝑎}^2}=\frac{9}{25}\Rightarrow \frac{9}{5𝑎}=\frac{9}{25}$

$\Rightarrow 𝑎=5$

$\therefore {𝑏}^2=\frac{9}{5}\left(5\right)=9$ [using Eq. (i)]

$\therefore$ Required equation of ellipse,

$\frac{{𝑥}^2}{25}+\frac{{𝑦}^2}{9}=1$

$\begin{array}{cc}& \text{ Given, }𝑝,𝑞\to 𝑇\text{ and }𝑟,𝑠\to 𝐹\\ & \therefore 𝑎<text>:</text><text>\sim </text>(𝑝\wedge \sim 𝑟)\vee (\sim 𝑞\vee 𝑠)\\ & <text>\equiv </text><text>\sim </text>(𝑇\wedge 𝑇)\vee (𝐹\vee 𝐹)\\ & <text>\equiv </text><text>\sim </text>\left(𝑇\right)\vee \left(𝐹\right)\\ & \equiv 𝐹\vee 𝐹\equiv 𝐹\\ & \text{ and }𝑏:(𝑝\vee 𝑠)\leftrightarrow (𝑞\wedge 𝑟)\\ & \equiv (\sim (𝑝\vee 𝑠)\vee (𝑞\wedge 𝑟\left)\right)\wedge \left(\right(𝑝\vee 𝑠)\vee \sim (𝑞\wedge 𝑟\left)\right)\\ & \because 𝑝\leftrightarrow 𝑞\equiv (\sim 𝑝\vee 𝑞)\wedge (𝑝\vee \sim 𝑞)\\ & \equiv (\sim (𝑇\vee 𝐹)\vee (𝑇\wedge 𝐹\left)\right)\wedge \left(\right(𝑇\vee 𝐹)\vee \sim (𝑇\wedge 𝐹\left)\right)\\ & \equiv (𝐹\vee 𝐹)\wedge (𝑇\vee 𝑇)\\ & \equiv 𝐹\wedge 𝑇\equiv 𝐹\end{array}$

The inverse of the given statement: "If two triangles are not congruent, then their areas are not equal."

$\therefore$ The contrapositive of the inverse of the given statement: If areas of two triangles are equal, then they are congruent.

$\begin{array}{cc}\text{Let }𝐼& =\int \log 𝑥[\log (𝑒𝑥){]}^{-2}𝑑𝑥\\ & =\int \frac{\log 𝑥}{(\log 𝑒𝑥{)}^2}𝑑𝑥\\ & =\int \frac{\log 𝑥}{(\log 𝑒+\log 𝑥{)}^2}𝑑𝑥\\ & =\int \frac{\log 𝑥}{(1+\log 𝑥{)}^2}𝑑𝑥\end{array}$

$\begin{array}{cccc}\text{ Put }& & \log 𝑥& =𝑡\Rightarrow 𝑥={𝑒}^{𝑡}\\ \Rightarrow & & 𝑑𝑥& ={𝑒}^{𝑡}𝑑𝑡\end{array}$

$\begin{array}{cc}& =\int \frac{𝑡}{(1+𝑡{)}^2}{𝑒}^{𝑡}𝑑𝑡\\ & =\int {𝑒}^{𝑡}(\frac{1}{1+𝑡}-\frac{1}{(1+𝑡{)}^2})𝑑𝑡\\ & =\frac{{𝑒}^{𝑡}}{1+𝑡}+𝐶\\ & =\frac{𝑥}{(1+\log 𝑥)}+𝐶\end{array}$

$\begin{array}{cc}& \text{ We have, }𝑦=\log \left[\frac{𝑥+\sqrt{\surd {𝑥}^2+25}}{\sqrt{\surd {𝑥}^2+25}-𝑥}\right]\\ & 𝑦=\log ⎡⎣⎢\frac{(𝑥+\sqrt{\surd {{𝑥}^2+25)}^2}}{{𝑥}^2+25-{𝑥}^2}⎤⎦⎥\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow 𝑦=\log ⎡⎣⎢\frac{{(𝑥+\sqrt{\surd {𝑥}^2+25})}^2}{25}⎤⎦⎥\\ & \Rightarrow 𝑦=2\log (𝑥+\sqrt{\surd {𝑥}^2+25})-\log 25\end{array}\\ & \text{ On differentiating both sides w.r.t. }𝑥\text{, we get }\\ & \begin{array}{cc}\frac{𝑑𝑦}{𝑑𝑥}& =\frac{2}{𝑥+\sqrt{\surd {𝑥}^2+25}}(1+\frac{1}{2\sqrt{\surd {𝑥}^2+25}}(2𝑥\left)\right)\\ & =\frac{2}{𝑥+\sqrt{\surd {𝑥}^2+25}}\left(\frac{\sqrt{\surd {𝑥}^2+25}+𝑥}{\sqrt{\surd {𝑥}^2+25}}\right)\\ & =\frac{2}{\sqrt{\surd {𝑥}^2+25}}\end{array}\end{array}$

Scalar triple product of the given vectors is 272.

$\therefore \mid \begin{array}{ccc}-3& 7& -3\\ 3& -7& 𝜆\\ 7& -5& -3\end{array}\mid =272(\because$ scalar triple product of the vectors $𝐚,𝐛$ and $𝐜𝐢𝐬\left[𝐚𝐛𝐜\right]$)

$\begin{array}{cc}& \Rightarrow -3(21+5𝜆)-7(-9-7𝜆)-3(-15+49)=272\\ & \Rightarrow -63-15𝜆+63+49𝜆-102=272\\ & \Rightarrow 34𝜆-102=272\\ & \Rightarrow 34𝜆=374\\ & \Rightarrow 𝜆=11\end{array}$

Let edge of a cube be $𝑥\mathrm{cm}$, then surface area of the cube, $𝐴=6{𝑥}^2$

It is given that, $\frac{𝑑𝑥}{𝑑𝑡}=-0.04\mathrm{cm}/\sec$

$\begin{array}{cc}\text{Now,}\frac{𝑑𝐴}{𝑑𝑡}& =12𝑥\frac{𝑑𝑥}{𝑑𝑡}\\ & =12\times (-0.04)\\ & =-0.48𝑥\end{array}$

when, $𝑥=10$, then $\frac{𝑑𝐴}{𝑑𝑡}=-0.48\times 10=-4.8{\mathrm{cm}}^2/\sec$

It is given that slopes of the lines passing through origin are

$\begin{array}{cc}{𝑚}_1\left(\text{ let }\right)=(1+\sqrt{\surd 2})\text{ and }{𝑚}_2\left(\text{ let }\right)& =\frac{-1}{1+\sqrt{\surd 2}}\\ & =-(\sqrt{\surd 2}-1)\end{array}$

$\therefore$ Required joint equation of lines passing through origin is

$\begin{array}{cc}& \text{ origin is }\\ & [𝑦-(1+\sqrt{\surd 2}\left)𝑥\right][𝑦+(\sqrt{\surd 2}-1\left)𝑥\right]=0\\ & \Rightarrow {𝑦}^2+(\sqrt{\surd 2}-1)𝑥𝑦-(1+\sqrt{\surd 2})𝑥𝑦-(2-1){𝑥}^2=0\\ & \Rightarrow {𝑦}^2-2𝑥𝑦-{𝑥}^2=0\\ & \Rightarrow {𝑥}^2+2𝑥𝑦-{𝑦}^2=0\end{array}$

According to question, $\frac{𝑑}{𝑑𝑡}\left(4𝜋{𝑟}^2\right)=𝐾(\because$ surface area of spherical ballon with radius $𝑟$ is $4𝜋{𝑟}^2$ )

$\Rightarrow 4𝜋\left(2𝑟\right)\frac{𝑑𝑟}{𝑑𝑡}=𝐾\Rightarrow 8𝜋𝑟𝑑𝑟=𝐾𝑑𝑡$

$\begin{array}{cc}& \text{ On Integrating both sides, we get }\\ & 8𝜋\int 𝑟𝑑𝑟=𝐾\int 𝑑𝑡\\ & \begin{array}{cc}\Rightarrow & 8𝜋\frac{{𝑟}^2}{2}=𝐾𝑡+𝐶\\ \Rightarrow & 4𝜋{𝑟}^2=𝐾𝑡+𝐶\end{array}\end{array}$

$\begin{array}{cc}\text{ Let }𝐼& ={\int }_{\frac{𝜋}{2}}^0\sqrt{\surd \cos 𝜃}\cdot {\sin }^3𝜃𝑑𝜃\\ & ={\int }_{\frac{𝜋}{2}}^0\sqrt{\surd \cos 𝜃}\cdot \sin 𝜃(1-{\cos }^2𝜃)𝑑𝜃\end{array}$

Put $\cos 𝜃=𝑡$

$\begin{array}{cc}\Rightarrow & -\sin 𝜃𝑑𝜃=𝑑𝑡\\ \Rightarrow & \sin 𝜃𝑑𝜃=-𝑑𝑡\end{array}$

If $𝜃=0,𝑡=1$ and $𝜃=\frac{𝜋}{2},𝑡=0$

$\begin{array}{cc}\therefore 𝐼& ={\int }_0^1\sqrt{\surd 𝑡}(1-{𝑡}^2)(-𝑑𝑡)\\ & ={\int }_1^0({𝑡}^{1/2}-{𝑡}^{5/2})𝑑𝑡\\ & ={(\frac{{𝑡}^{3/2}}{\frac{3}{2}}-\frac{{𝑡}^{7/2}}{\frac{7}{2}})}_1^0\\ & ={(\frac{2}{3}{𝑡}^{3/2}-\frac{2}{7}{𝑡}^{7/2})}_1^0\\ & =(\frac{2}{3}-\frac{2}{7})-(0-0)\\ & =\frac{14-6}{21}=\frac{8}{21}\end{array}$

Given, $𝜔$ is a complex cube root of unity

${𝜔}^3=1$

Given matrix $𝐴$ can be written as $𝐴=𝐼𝐴$

$\Rightarrow ⎡⎣⎢\begin{array}{ccc}𝜔& 0& 0\\ 0& {𝜔}^2& 0\\ 0& 0& 1\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥𝐴$

On apply ${𝑅}_1\to {𝜔}^2{𝑅}_1,{𝑅}_2\to 𝜔{𝑅}_2$, we get

$\Rightarrow ⎡⎣⎢\begin{array}{ccc}{𝜔}^3& 0& 0\\ 0& {𝜔}^3& 0\\ 0& 0& 1\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}{𝜔}^2& 0& 0\\ 0& 𝜔& 0\\ 0& 0& 1\end{array}⎤⎦⎥𝐴$

$\begin{array}{c}\Rightarrow ⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}{𝜔}^2& 0& 0\\ 0& 𝜔& 0\\ 0& 0& 1\end{array}⎤⎦⎥𝐴\\ {𝐴}^{-1}=⎡⎣⎢\begin{array}{ccc}{𝜔}^2& 0& 0\\ 0& 𝜔& 0\\ 0& 0& 1\end{array}⎤⎦⎥\end{array}$

We have, $𝐴$ and $𝐵$ are square matrices of order 3 such that<?p>

$\begin{array}{cc}& \text{ that }\left|𝐴\right|=2,\left|𝐵\right|=4\\ & \begin{array}{c}\text{Now,}\left|𝐴\right(\mathrm{adj}\left)𝐵\right|=\left|𝐴\right||\mathrm{adj}𝑗𝐵|(\because |𝐴𝐵|=|𝐴\left|\right|𝐵\left|\right)\\ {\left|𝐴\right|𝐵|}^{3-1}\end{array}\\ & =\left|𝐴\right||𝐵{|}^2=(2\left)\right(4{)}^2=32\end{array}$

$\begin{array}{cc}\text{ Let }𝐼& =\int \frac{1}{1-\cot 𝑥}𝑑𝑥\\ & =\int \frac{1}{1-\frac{\cos 𝑥}{\sin 𝑥}}𝑑𝑥\\ & =\int \frac{\sin 𝑥}{\sin 𝑥-\cos 𝑥}𝑑𝑥\\ & =\frac{1}{2}\int \frac{2\sin 𝑥}{\sin 𝑥-\cos 𝑥}𝑑𝑥\\ & =\frac{1}{2}\int \frac{\sin 𝑥+\sin 𝑥+\cos 𝑥-\cos 𝑥}{\sin 𝑥-\cos 𝑥}𝑑𝑥\\ & =\frac{1}{2}\int \frac{(\sin 𝑥-\cos 𝑥)+(\sin 𝑥+\cos 𝑥)}{\sin 𝑥-\cos 𝑥}𝑑𝑥\\ & =\frac{1}{2}[\int 𝑑𝑥+\int \frac{\sin 𝑥+\cos 𝑥}{\sin 𝑥-\cos 𝑥}𝑑𝑥\end{array}$

$\begin{array}{cc}& \text{ Let }\sin 𝑥-\cos 𝑥=𝑡\\ & \Rightarrow (\cos 𝑥+\sin 𝑥)𝑑𝑥=𝑑𝑡\\ & \therefore 𝐼=\frac{1}{2}[\int 𝑑𝑥+\int \frac{1}{𝑡}𝑑𝑡]\\ & =\frac{1}{2}[𝑥+\log |𝑡\left|\right]+𝐶\\ & =\frac{1}{2}[𝑥+\log |\sin 𝑥-\cos 𝑥\left|\right]+𝐶\\ & =\frac{𝑥}{2}+\frac{1}{2}\log |\sin 𝑥-\cos 𝑥|+𝐶\\ & \text{ Here, }𝐴=\frac{1}{2},𝐵=\frac{1}{2}\\ & \therefore 𝐴+𝐵=\frac{1}{2}+\frac{1}{2}=1\end{array}$

We have, polar coordinates of $𝑃$ are $(2,\frac{𝜋}{6})$

If $𝑄$ is the image of $𝑃$ about the $𝑋$-axis

$\begin{array}{cc}\therefore & 𝑄\equiv (2,-\frac{𝜋}{6})\\ \Rightarrow & 𝑄\equiv (2,\frac{11𝜋}{6})\end{array}$

$\begin{array}{cc}& \text{ Given, }𝑐=(𝑥-2)𝑎+𝑏\text{ and }\\ & \begin{array}{c}𝑑=(2𝑥+1)𝑎-𝑏\text{ are collinear }\\ \therefore 𝑐=𝜆𝑑\\ \Rightarrow (𝑥-2)𝐚+𝐛=𝜆\left[\right(2𝑥+1)𝑎-𝑏]\\ \Rightarrow \frac{𝑥-2}{2𝑥+1}=\frac{1}{-1}=𝜆\\ \Rightarrow 2𝑥+1=-𝑥+2\\ \Rightarrow 3𝑥=1\\ \Rightarrow 𝑥=\frac{1}{3}\end{array}\end{array}$

We have, $𝑝=\frac{3}{4},𝑞=1-𝑝=\frac{1}{4}$

It is given that $𝑃(𝑋\ge 1)\ge 0.9375$

$\begin{array}{cc}& =1-𝑃(𝑋=0)\ge 0.9375\\ & =1-{}^{𝑛}{𝐶}_0\left({𝑝}^0\right)(𝑔{)}^{𝑛-0}\ge 0.9375\\ & =1-{\left(\frac{1}{4}\right)}^{𝑛}\ge 0.9375\\ & =1-0.9375\ge {\left(\frac{1}{4}\right)}^{𝑛}\\ & =0.0625\ge {\left(\frac{1}{4}\right)}^{𝑛}\\ & =\frac{625}{10000}\ge {\left(\frac{1}{4}\right)}^{𝑛}\\ & =\frac{1}{16}\ge {\left(\frac{1}{4}\right)}^{𝑛}\\ & =16\le 4^{𝑛}\end{array}$ $\Rightarrow 𝑛=2$

We have, $𝑥=\sqrt{\surd 𝑡}$ and $𝑦=𝑡-\frac{1}{\sqrt{\surd 𝑡}}$

Now, $\frac{𝑑𝑥}{𝑑𝑡}=\frac{1}{2\sqrt{\surd 𝑡}},\frac{𝑑𝑦}{𝑑𝑡}=1+\frac{1}{2}{𝑡}^{-3/2}$

$\begin{array}{cc}& \therefore \frac{𝑑𝑦}{𝑑𝑥}=\frac{\frac{𝑑𝑦}{𝑑𝑡}}{\frac{𝑑𝑥}{𝑑𝑡}}=\frac{1+\frac{1}{2}{𝑡}^{-3/2}}{\frac{1}{2\sqrt{\surd 𝑡}}}\\ & =2\sqrt{\surd 𝑡}(1+\frac{1}{2}{𝑡}^{-3/2})\\ & =2\sqrt{\surd 𝑡}(1+\frac{1}{2{𝑡}^{3/2}})\\ & =\frac{2{𝑡}^{3/2}+1}{𝑡}\\ & \text{ At }𝑡=4,\frac{𝑑𝑦}{𝑑𝑥}=\frac{2(4{)}^{3/2}+1}{4}=\frac{17}{4}\end{array}$

$\therefore$ Slope of normal at $𝑡=4,-\frac{1}{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}=-\frac{4}{17}$

Option (a), $(𝑝\to 𝑞)\vee 𝑞$

$\begin{array}{cc}& \equiv (\sim 𝑝\vee 𝑞)\vee 𝑞\\ & \equiv (\sim 𝑝\vee 𝑞)\end{array}$

It is not a tautology since if $𝑝$ is true and $𝑞$ is false then $(\sim 𝑝\vee 𝑞)$ is false.

$\begin{array}{cc}& \text{ Option (b), }𝑝\to (𝑞\vee 𝑝)\\ & <text>\equiv </text><text>\sim </text>𝑝\vee (𝑞\vee 𝑝)\\ & \equiv (\sim 𝑝\vee 𝑝)\vee 𝑞\\ & \equiv 𝑇\vee 𝑞(<text>\because </text><text>\sim </text>𝑝\vee 𝑝\equiv 𝑇)\end{array}$

It is a tautology since if $𝑞$ is true or false then $𝑇\vee 𝑞$ must be true.

Similarly, check other options.

Given lines are

$𝑥-3=0\text{.... (i)}$

Slope of line (i), ${𝑚}_1=\infty \Rightarrow \tan {𝜃}_1=\infty \Rightarrow {𝜃}_1={90}^{\circ }$ and $𝑥+𝑦=19$

$\Rightarrow 𝑦=-𝑥+19\text{.... (ii)}$

Slope of line (ii), ${𝑚}_2=\tan {𝜃}_2=-1$

$\Rightarrow {𝜃}_2={135}^{\circ }$

$\therefore$ Acute angle between given lines is ${45}^{\circ }$.

$\begin{array}{cc}& \text{ We have, In }△𝐴𝐵𝐶\\ & \begin{array}{cc}& (\tan \frac{𝐴}{2})(\tan \frac{𝐵}{2})=\frac{3}{4}\\ & \Rightarrow \sqrt{\surd \frac{(𝑠-𝑏)(𝑠-𝑐)}{𝑠(𝑠-𝑎)}}\sqrt{\surd \frac{(𝑠-𝑎)(𝑠-𝑐)}{𝑠(𝑠-𝑏)}}=\frac{3}{4}\\ & \Rightarrow \sqrt{\surd \frac{(𝑠-𝑏)(𝑠-𝑐)(𝑠-𝑎)(𝑠-𝑐)}{𝑠(𝑠-𝑎)\cdot 𝑠(𝑠-𝑏)}}=\frac{3}{4}\\ & \Rightarrow \frac{(𝑠-𝑐)}{𝑠}=\frac{3}{4}\Rightarrow \frac{\frac{𝑎+𝑏+𝑐}{2}-𝑐}{\frac{𝑎+𝑏+𝑐}{2}}=\frac{3}{4}\\ & \Rightarrow \frac{𝑎+𝑏-𝑐}{𝑎+𝑏+𝑐}=\frac{3}{4}\\ & \Rightarrow 4𝑎+4𝑏-4𝑐=3𝑎+3𝑏+3𝑐\\ & \Rightarrow 𝑎+𝑏=7𝑐\end{array}\end{array}$

Given, pair of lines, ${𝑥}^2-4𝑝𝑥𝑦+8{𝑦}^2=0$

Which is in the form of $𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0$

According to question,

Sum of the slopes $=3\times$ product of the slopes

$\begin{array}{cccc}\Rightarrow & & {𝑚}_1+{𝑚}_2& =3\times \left({𝑚}_1{𝑚}_2\right)\\ \Rightarrow & & \frac{-(-4𝑝)}{8}& =3\times \frac{1}{8}\\ \Rightarrow & & 4𝑝=3\Rightarrow 𝑝=\frac{3}{4}\end{array}$

$\begin{array}{cc}& \text{ We have, }\frac{1-2(\cos {60}^{\circ }-\cos {80}^{\circ })}{2\sin {10}^{\circ }}\\ & \begin{array}{cc}& =\frac{1-2(\frac{1}{2}-\cos {80}^{\circ })}{2\sin {10}^{\circ }}\\ & =\frac{2\cos {80}^{\circ }}{2\sin {10}^{\circ }}\\ & =\frac{\cos ({90}^{\circ }-{10}^{\circ })}{\sin {10}^{\circ }}\\ & =\frac{\sin {10}^{\circ }}{\sin {10}^{\circ }}=1\end{array}\end{array}$

Equation of the circle touching both axes is ${𝑥}^2+{𝑦}^2-2𝑎𝑥-2𝑎𝑦+{𝑎}^2=0$

Here, number of arbitrary constant is one which is equal to the order of its differential equation.

$\begin{array}{cc}& \begin{array}{c}\text{ We have, }𝑓\left(𝑥\right)\text{ is continuous at }𝑥=3\\ \text{where}𝑓\left(𝑥\right)=\{\begin{array}{cc}𝑎𝑥+1& \text{ for }𝑥\le 3\\ 𝑏𝑥+3& \text{ for }𝑥>3\end{array}\end{array}\\ & \therefore \underset{𝑥\to 3^{-}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 3^{+}}{lim}𝑓\left(𝑥\right)=𝑓\left(3\right)\text{.... (i)}\\ & \text{ Now, }\underset{𝑥\to 3^{-}}{lim}𝑓\left(𝑥\right)=\underset{ℎ\to 0}{lim}𝑓(3-ℎ)\\ & =\underset{ℎ\to 0}{lim}𝑎(3-ℎ)+1\\ & =3𝑎+1\text{.... (ii)}\\ & \underset{𝑥\to 3^{+}}{lim}𝑓\left(𝑥\right)=\underset{ℎ\to 0}{lim}𝑓(3+ℎ)\\ & =\underset{ℎ\to 0}{lim}𝑏(3+ℎ)+3\\ & =3𝑏+3\text{.... (iii)}\\ & \text{ and }𝑓\left(3\right)=3𝑎+1\text{... (iv)}\end{array}$

Since, $𝑓\left(𝑥\right)$ is continuous at $𝑥=3$

$\therefore$ From Eqs. (i), (ii), (iii) and (iv) we get

$\begin{array}{c}3𝑎+1=3𝑏+3\\ \Rightarrow 3𝑎-3𝑏=2\\ \Rightarrow 𝑎-𝑏=\frac{2}{3}\end{array}$

Total number of ways that three card drawn from a pack of 52 cards are ${}^{52}{𝐶}_3$ and number of ways that three red. cards drawn are ${}^{26}{C}_3$.

$\begin{array}{cc}\therefore \text{ Required probability }& =\frac{{}^{26}{𝐶}_3}{{}^{52}{𝐶}_3}=\frac{26\times 25\times 24}{52\times 51\times 50}\\ & =\frac{2}{17}\end{array}$

$\begin{array}{cc}& \text{We have, }𝑎\cdot \left[\right(𝑏+𝑐)\times (𝑎+𝑏+𝑐\left)\right]\\ & \begin{array}{cc}& =𝐚\cdot \left[\right(𝐛\times 𝐚+𝐛\times 𝐛+𝐛\times 𝐜+𝐜\times 𝐚+𝐜\times 𝐛+𝐜\times 𝐜\left)\right]\\ & =𝐚\cdot \left[\right(𝐛\times 𝐚)+(𝐛\times 𝐜)+(𝐜\times 𝐚)+(𝑐\times 𝑏\left)\right]\\ & =𝐚\left[\right(𝐛\times 𝐚)+(𝐛\times 𝐜)+(𝐜\times 𝐚)-(𝐛\times 𝐜\left)\right]\\ & =𝑎\left[\right(𝑏\times 𝑎)+(𝑐\times 𝑎\left)\right]\\ & =\left[𝐚\mathrm{𝐛𝐚}\right]+\left[𝐚𝐜𝑎\right]=0+0=0\end{array}\end{array}$