MHT CET 2019 2nd May Evening Shift (Mathematics) Question Paper with Solutions

In a binomial distribution, the mean ($𝜇$) and variance (${𝜎}^2$) are given by the formulas:

$𝜇=𝑛𝑝$

${𝜎}^2=𝑛𝑝(1-𝑝)$

where $𝑛$ is the number of trials and $𝑝$ is the probability of success in each trial.

Given:

Mean, $𝜇=18$

Variance, ${𝜎}^2=12$

Substituting the mean into the formula, we have:

$18=𝑛𝑝$

For the variance, substitute and rearrange:

$12=𝑛𝑝(1-𝑝)$

Now, we solve these two equations simultaneously.

From the mean:

$𝑛=\frac{18}{𝑝}$

Substitute $𝑛$ from Equation 1 into Equation 2:

$12=\left(\frac{18}{𝑝}\right)𝑝(1-𝑝)$

Simplify the expression:

$12=18(1-𝑝)$

$12=18-18𝑝$

Rearrange and solve for $𝑝$:

$18𝑝=18-12$

$18𝑝=6$

$𝑝=\frac{6}{18}$

$𝑝=\frac{1}{3}$

Thus, $𝑝=\frac{1}{3}$.

The correct option is Option B: $\frac{1}{3}$.

Given lines are

$\frac{𝑥-1}{2}=\frac{𝑦+1}{3}=\frac{𝑧-1}{4}={𝜆}_1\text{ (let) }\text{.... (i)}$

and $\frac{𝑥-3}{1}=\frac{𝑦-𝜆}{2}=\frac{𝑧}{1}={𝜆}_2$ (let) ... (ii)

Then, any point on line (i) is $(2{𝜆}_1+1,3{𝜆}_1-1,4{𝜆}_1+1)$ and any point on line (ii) is $({𝜆}_2+3,2{𝜆}_2+𝜆,{𝜆}_2)$

Clearly, then lines (i) and (ii) will intersect if $(2{𝜆}_1+1,3{𝜆}_1-1,4{𝜆}_1+1)=({𝜆}_2+3,2{𝜆}_2+𝜆,{𝜆}_2)$

For some particular value of ${𝜆}_1$ and ${𝜆}_2$

$\begin{array}{cc}& \Rightarrow 2{𝜆}_1+1={𝜆}_2+3,3{𝜆}_1-1=2{𝜆}_2+𝜆\\ & \text{ and }4{𝜆}_1+1={𝜆}_2\\ & \Rightarrow 2{𝜆}_1-{𝜆}_2=2,3{𝜆}_1-2{𝜆}_2=𝜆+1\text{ and }4{𝜆}_1-{𝜆}_2=-1\end{array}$

On solving $2{𝜆}_1-{𝜆}_2=2$ and $4{𝜆}_1-{𝜆}_2=-1$ We get ${𝜆}_1=-\frac{3}{2}$ and ${𝜆}_2=-5$

Now, putting the values of ${𝜆}_1$ and ${𝜆}_2$ in

$\begin{array}{cc}& 3{𝜆}_1-2{𝜆}_2=𝜆+1\\ \Rightarrow & 3\left(\frac{-3}{2}\right)-2(-5)=𝜆+1\Rightarrow \frac{-9}{2}+10=𝜆+1\\ \Rightarrow & 𝜆+1=\frac{11}{2}\Rightarrow 𝜆=\frac{9}{2}\end{array}$

We have, differential equations,

$\begin{array}{cc}& \log \left(\frac{𝑑𝑦}{𝑑𝑥}\right)=𝑥\Rightarrow \frac{𝑑𝑦}{𝑑𝑥}={𝑒}^{𝑥}\\ \Rightarrow & 𝑑𝑦={𝑒}^{𝑥}𝑑𝑥\end{array}$

Integrating on both sides, we get

$\begin{array}{c}\int 𝑑𝑦=\int {𝑒}^{𝑥}𝑑𝑥\\ 𝑦={𝑒}^{𝑥}+𝐶\text{.... (i)}\end{array}$

On putting $𝑥=0,𝑦=1$ is Eq. (i), we get

$1={𝑒}^0+𝐶\Rightarrow 𝐶=0$

Now, particular solution of the given differential is $𝑦={𝑒}^{𝑥}$

We have p. d. $𝑓$ of a random variable $𝑥$ is given by

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{1}{4𝑎},0<𝑥<4𝑎\cdot (𝑎>0)\\ & \text{ and also, }𝑝(𝑥<\frac{3𝑎}{2})=𝑘𝑝(𝑥>\frac{5𝑎}{2})\\ & \Rightarrow {\int }_{\frac{3𝑎}{2}}^0𝑓\left(𝑥\right)𝑑𝑥=𝑘{\int }_{4𝑎}^{\frac{5𝑎}{2}}𝑓\left(𝑥\right)𝑑𝑥\\ & \Rightarrow {\int }_{\frac{3𝑎}{2}}^0\left(\frac{1}{4𝑎}\right)𝑑𝑥=𝑘{\int }_{4𝑎}^{\frac{5𝑎}{2}}\left(\frac{1}{4𝑎}\right)𝑑𝑥\\ & \Rightarrow \frac{1}{4𝑎}{\int }_{\frac{3𝑎}{2}}^0𝑑𝑥=\frac{𝑘}{4𝑎}{\int }_{4𝑎}^{\frac{5𝑎}{2}}𝑑𝑥\\ & \Rightarrow (𝑥{)}_{\frac{3𝑎}{2}}^0=𝑘(𝑥{)}_4^{\frac{5𝑎}{4}}\end{array}$

$\begin{array}{cc}\Rightarrow & (\frac{3𝑎}{2}-0)=𝑘(4𝑎-\frac{5𝑎}{2})\\ \Rightarrow & \frac{3𝑎}{2}=𝑘\left(\frac{3𝑎}{2}\right)\\ \Rightarrow & 𝑘=1\end{array}$

We have function,

$\begin{array}{cc}& 𝑓\left(𝑥\right)=\frac{({𝑒}^{𝑘𝑥}-1)\tan 𝑘𝑥}{4{𝑥}^2},𝑥\ne 0\\ & 16,𝑥=0\end{array}$

$\begin{array}{cc}& \text{ is continuous at }𝑥=0\\ & \begin{array}{cc}& \therefore \underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)=𝑓\left(0\right)\\ & \Rightarrow \underset{𝑥\to 0}{lim}\frac{({𝑒}^{𝑘𝑥}-1)\tan 𝑘𝑥}{4{𝑥}^2}=16\left(\frac{0}{0}\text{ form }\right)\\ & \Rightarrow \underset{𝑥\to 0}{lim}\left(\frac{{𝑒}^{𝑘𝑥}-1}{𝑘𝑥}\right)\left(\frac{\tan 𝑘𝑥}{𝑘𝑥}\right)\times \frac{{𝑘}^2}{4}=16\\ & \Rightarrow \frac{{𝑘}^2}{4}=16\Rightarrow {𝑘}^2=64\Rightarrow 𝑘=\pm 8\end{array}\end{array}$

We have differential equation $𝑦𝑑𝑥-𝑥𝑑𝑦=𝑥𝑦𝑑𝑥$

$\begin{array}{cc}\Rightarrow & \frac{𝑦𝑑𝑥-𝑥𝑑𝑦}{𝑥𝑦}=𝑑𝑥\\ \Rightarrow & 𝑑(\log \frac{𝑥}{𝑦})=𝑑𝑥\end{array}$

On integrating both sides, we get

$\begin{array}{c}\log \left(\frac{𝑥}{𝑦}\right)=𝑥\Rightarrow \frac{𝑥}{𝑦}={𝑒}^{𝑥}\\ 𝑥=𝑦{𝑒}^{𝑥}\end{array}$

$\begin{array}{cc}& \text{We have, }𝑧=6𝑥+8𝑦\\ & \text{ subject to constrants }𝑥-𝑦\ge 0,𝑥+3𝑦\le 12,𝑥\ge 0\text{, }\\ & 𝑦\ge 0\text{. }\end{array}$

On taking given constraints as equations, we get the following graph

Intersecting point of the line $𝑥-𝑦=0$ and $𝑥+3𝑦=12$ is $𝐵(3,3)$.

Here, $𝑂𝐴𝐵𝑂$ is the required feasible region

Whose corner points are $𝑂(0,0),𝐴(12,0)$ and $𝐵(0,4)$ Now,

$\therefore$ Maximum value of Z is 72.

$\begin{array}{cc}& \text{We have, }\sum _{𝑟=1}^{𝑛}(2𝑟+1)=440\\ & \Rightarrow 3+5+7\dots +(2𝑛+1)=440\\ & \Rightarrow \frac{𝑛}{2}[2\times 3+(𝑛-1\left)\right(2\left)\right]=440\\ & \Rightarrow 𝑛(3+𝑛-1)=440\\ & \Rightarrow 𝑛(𝑛+2)=440\\ & \Rightarrow 𝑛=20\end{array}$

We have statements $𝑝,𝑞\to 𝑇$ and $𝑟,𝑠\to 𝐹$

Option $\left(𝑎\right)(𝑞\wedge 𝑟)\vee (\sim 𝑝\wedge 𝑠)\equiv (𝑇\wedge 𝐹)\vee (𝐹\wedge 𝐹)\equiv 𝐹\vee 𝐹\equiv 𝐹$

Option (b) $(\sim 𝑝\to 𝑞)\leftrightarrow (𝑟\wedge 𝑠)\equiv (𝑝\vee 𝑞)\to (𝑟\vee 𝑠)(\because 𝑝\to 𝑞<text>\equiv </text><text>\sim </text>𝑝\vee 𝑞)$

$\equiv (\sim 𝑝\vee 𝑞)\vee (𝑟\wedge 𝑠)\wedge \left(\right(𝑝\vee 𝑞)\vee (\sim 𝑟\wedge 𝑠\left)\right))$

$(\because 𝑝\leftrightarrow 𝑞\equiv (\sim 𝑝\vee 𝑞)\wedge (𝑝\vee -𝑞\left)\right)$

$\equiv (\sim (𝑇\vee 𝑇)\vee (𝐹\wedge 𝐹)\wedge ((𝑇\vee 𝐹)\vee (\sim (𝐹\wedge 𝐹\left)\right))$

$\equiv (𝐹\vee 𝐹)\wedge (𝑇\vee 𝑇)\equiv 𝐹\wedge 𝑇\equiv 𝐹$

$\begin{array}{cc}& \text{ Option }\left(𝑐\right)(𝑝\to 𝑞)\vee (𝑟\leftrightarrow 𝑆)\\ & \equiv (\sim 𝑝\vee 𝑞)\vee \left(\right(\sim 𝑟\vee 𝑠)\wedge (𝑟\vee \sim 𝑠\left)\right)(\because 𝑝\to 𝑞\equiv 𝑝\vee 𝑞)\\ & \equiv (𝐹\vee 𝑇)\vee \left(\right(𝑇\vee 𝐹)\wedge (𝐹\vee 𝑇\left)\right)\end{array}$

$\equiv 𝑇\vee (𝑇\wedge 𝑇)\equiv 𝑇\vee 𝑇\equiv 𝑇$

Option (d) do similar as option (a).

We have standard deviation of $𝑋=\sqrt{\surd 3𝑝𝑞}$

$\Rightarrow \mathrm{Var}\left(𝑋\right)=3𝑝𝑞$

and mean, $𝐸\left(𝑋\right)=3𝑝$

Now, $3𝑝𝑞=𝐸\left({𝑥}^2\right)-\left(3{𝑝}^2\right)$

$(\because \mathrm{Var}(𝑋)=𝐸\left({𝑋}^2\right)-\left(𝐸\right(𝑋{)}^2)$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow 𝐸\left({𝑋}^2\right)& =3𝑝𝑞+9{𝑝}^2\\ & =3𝑝(1-𝑝)+9{𝑝}^2(\because 𝑝+𝑞=1)\\ & =3𝑝-3{𝑝}^2+9{𝑝}^2\\ \Rightarrow 3𝑝+6{𝑝}^2& \Rightarrow 3𝑝(1+2𝑝)\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }𝑓\left(𝑥\right)=\left[𝑥\right]\\ & \begin{array}{cc}\therefore {𝑓}^{\prime }\left(1^{+}\right)=& \underset{𝑥\to 1^{+}}{lim}\frac{𝑓\left(𝑥\right)-𝑓\left(1\right)}{𝑥-1}=\underset{ℎ\to 0}{lim}\frac{𝑓(1+ℎ)-𝑓\left(1\right)}{ℎ}\\ & =\underset{ℎ\to 0}{lim}\frac{[1+ℎ]-\left[1\right]}{ℎ}\Rightarrow \underset{ℎ\to 0}{lim}\frac{1-1}{ℎ}=0\end{array}\end{array}$

$\begin{array}{cc}& \text{ we have, pair of lines represented by }\\ & \begin{array}{cc}& (1+{\sin }^2𝜃){𝑥}^2+2ℎ𝑥𝑦+2\sin 𝜃{𝑦}^2=0,𝜃\in [0,2𝜋]\\ & \therefore (1+{\sin }^2𝜃)+2\sin 𝜃=0\\ & \Rightarrow (1+\sin 𝜃{)}^2=0\\ & \Rightarrow 1+\sin 𝜃=0\\ & \Rightarrow \sin 𝜃=-1=-\sin \frac{𝜋}{2}=\sin (2𝜋-\frac{𝜋}{2})\\ & \Rightarrow \sin 𝜃=\sin \frac{3𝜋}{2}\Rightarrow 𝜃=\frac{3𝜋}{2}\end{array}\end{array}$

We have, $𝐴=\left\{𝑥\right|𝑥\in 𝑁,𝑥|$ is a prime number less than 12$\}=\{2,3,5,7,11\}$

and $𝐵=\{𝑥\mid 𝑥\in 𝑁,𝑥$ is a factor of 10$\}=\{1,2,5,10\}$

$\therefore 𝐴\cap 𝐵=\{2,5\}$

In any $△𝐴𝐵𝐶$ we know that

Area of $\Delta =\frac{1}{2}𝑏𝑐\sin 𝐴$

$\Rightarrow \sin 𝐴=\frac{2\Delta }{𝑏𝑐}\text{.... (i)}$

also, $𝑅=\frac{𝑎}{2\sin 𝐴}\text{... (ii)}$

From Eqs. (i) and (ii), we have

$\begin{array}{cc}& 𝑅=\frac{𝑎}{2\left(\frac{2\Delta }{𝑏𝑐}\right)}\\ \Rightarrow & 𝑅=\frac{𝑎𝑏𝑐}{4\Delta }\\ \Rightarrow & \Delta =𝐴\left(△𝐴𝐵𝐶\right)=\frac{𝑎𝑏𝑐}{4𝑅}\end{array}$

We are given points

$𝐴(3,7,4),𝐵(5,-2,-3),𝐶(-4,5,6),\text{and}𝐷(1,2,3).$

The volume of the parallelepiped formed by the edges at point A, namely $\overrightarrow{𝐴𝐵},\overrightarrow{𝐴𝐶},\text{and}\overrightarrow{𝐴𝐷},$ is given by the absolute value of the scalar triple product:

$\text{Volume}=\mid \overrightarrow{𝐴𝐵}\cdot (\overrightarrow{𝐴𝐶}\times \overrightarrow{𝐴𝐷})\mid .$

Below are the steps to compute the volume:

Compute the vectors:

$\overrightarrow{𝐴𝐵}=𝐵-𝐴=(5-3,-2-7,-3-4)=(2,-9,-7).$

$\overrightarrow{𝐴𝐶}=𝐶-𝐴=(-4-3,5-7,6-4)=(-7,-2,2).$

$\overrightarrow{𝐴𝐷}=𝐷-𝐴=(1-3,2-7,3-4)=(-2,-5,-1).$

Find the cross product $\overrightarrow{𝐴𝐶}\times \overrightarrow{𝐴𝐷}:$

Using the determinant formula:

$\overrightarrow{𝐴𝐶}\times \overrightarrow{𝐴𝐷}=\mid \begin{array}{ccc}𝐢& 𝐣& 𝐤\\ -7& -2& 2\\ -2& -5& -1\end{array}\mid ,$

we compute each component:

$\text{i-component}:(-2)(-1)-\left(2\right)(-5)=2+10=12.$

$\text{j-component}:-[(-7)(-1)-\left(2\right)(-2)]=-[7+4]=-11.$

$\text{k-component}:(-7)(-5)-(-2)(-2)=35-4=31.$

So,

$\overrightarrow{𝐴𝐶}\times \overrightarrow{𝐴𝐷}=(12,-11,31).$

Calculate the scalar triple product:

Take the dot product of $\overrightarrow{𝐴𝐵}$ and $\overrightarrow{𝐴𝐶}\times \overrightarrow{𝐴𝐷}$:

$\overrightarrow{𝐴𝐵}\cdot (\overrightarrow{𝐴𝐶}\times \overrightarrow{𝐴𝐷})=\left(2\right)\left(12\right)+(-9)(-11)+(-7)\left(31\right).$

Compute each term:

$2\times 12=24,$

$-9\times -11=99,$

$-7\times 31=-217.$

Adding these:

$24+99-217=-94.$

Take the absolute value to get the volume:

$\text{Volume}=|-94|=94\text{cubic units}.$

Thus, the volume of the parallelepiped is $\overline{)94}$ cubic units, which corresponds to Option B.

$\begin{array}{cc}& \begin{array}{cc}& \text{We have, }𝑟\cos 𝜃=-\sqrt{\surd 2}\text{ and }𝑟\sin 𝜃=\sqrt{\surd 2}\\ & \therefore {𝑟}^2{\cos }^2𝜃+{𝑟}^2{\sin }^2𝜃=2+2=4\\ & \Rightarrow {𝑟}^2({\cos }^2𝜃+{\sin }^2𝜃)=4\\ & \Rightarrow 𝑟=\pm 2\text{ and }𝜃={\tan }^{-1}\left(\frac{\sqrt{\surd 2}}{-\sqrt{\surd 2}}\right)\\ & 𝜃={\tan }^{-1}(-1)\Rightarrow 𝜃=\frac{3𝜋}{4}\end{array}\\ & \therefore \text{ Required polar co-ordinate is }(𝑟,𝜃)\\ & =(2,\frac{3𝜋}{4})\end{array}$

We have,

$\int \frac{\cos 𝑥-\sin 𝑥}{8-\sin 2𝑥}=\frac{1}{𝑝}\log \left[\frac{3+\sin 𝑥+\cos 𝑥}{3-\sin 𝑥-\cos 𝑥}\right]+𝐶$

Now, $\int \frac{\cos 𝑥-\sin 𝑥}{8-\sin 2𝑥}𝑑𝑥$

$\begin{array}{cc}& =\int \frac{\cos 𝑥-\sin 𝑥}{9-(1+2\sin 𝑥\cos 𝑥)}𝑑𝑥\\ & =\int \frac{\cos 𝑥-\sin 𝑥}{9-({\sin }^2𝑥+{\cos }^2𝑥+2\sin 𝑥\cos 𝑥)}𝑑𝑥\\ & =\int \frac{\cos 𝑥-\sin 𝑥}{(3{)}^2-(\cos 𝑥+\sin 𝑥{)}^2}𝑑𝑥\end{array}$

$\begin{array}{cc}& \text{ put }\cos 𝑥+\sin 𝑥=𝑡\\ & \begin{array}{cc}& (-\sin 𝑥+\cos 𝑥)𝑑𝑥=𝑑𝑡\\ & =\int \frac{𝑑𝑡}{(3{)}^2-(𝑡{)}^2}=\frac{1}{2\left(3\right)}\log \mid \frac{3+𝑡}{3-𝑡}\mid +𝐶\\ & =\frac{1}{6}\log \mid \frac{3+\sin 𝑥+\cos 𝑥}{3-\sin 𝑥-\cos 𝑥}\mid +𝐶\\ & \therefore 𝑝=6\end{array}\end{array}$

$\begin{array}{cc}& \text{ We have, }𝐴\text{ is non-singular matrix }\\ & \begin{array}{cc}& \therefore \left|𝐴\right|=0\\ & \text{ and }(𝐴+1)(𝐴-1)=0\\ & \Rightarrow {𝐴}^2-{𝐼}^2=0\\ & \Rightarrow {𝐴}^2={𝐼}^2=1\\ & \Rightarrow 𝐴.𝐴=𝐼\Rightarrow {𝐴}^{-1}=𝐴\\ & \therefore 𝐴+{𝐴}^{-1}=𝐴+𝐴=2𝐴\end{array}\end{array}$

Given equation of plane

$𝑥-2𝑦+2𝑧+4=0\text{... (i)}$

Equation of plane parallel to plane (i) is

$𝑥-2𝑦+2𝑧+𝑘=0\text{.... (ii)}$

Since, distance of plane (ii) from point $(1,2,3)$ is 1 unit

$\begin{array}{cc}\therefore & \frac{\left|1\right(1)-2(2)+2(3)+𝑘|}{\sqrt{\surd (1{)}^2+(-2{)}^2+(2{)}^2}}=1\\ \Rightarrow & \frac{|1-4+6+𝑘|}{\sqrt{\surd 1+4+4}}=1\\ \Rightarrow & \frac{|3+𝑘|}{\sqrt{\surd 9}}=1\\ \Rightarrow & |3+𝑘|=3\Rightarrow 3+𝑘=\pm 3\\ \Rightarrow & 𝑘=0,-6\end{array}$

$\therefore$ Required equation of plane are

$\begin{array}{cc}& 𝑥-2𝑦+2𝑧+0=0\text{ and }𝑥-2𝑦+2𝑧-6=0\\ & \Rightarrow 𝑥-2𝑦+2𝑧=0\text{ and }𝑥-2𝑦+2𝑧-6=0\end{array}$

Slope of the given line,

$𝑥-2𝑦=4\text{ is }\frac{1}{2}$

Equation of a line passing through $𝐴(6,1)$ and perpendicular to given line is

$\begin{array}{ccc}& & 𝑦-1=(-2)(𝑥-6)\\ \Rightarrow & & 𝑦-1=-2𝑥+12\\ \Rightarrow & & 2𝑥+𝑦=13\\ \Rightarrow & & \frac{𝑥}{\left(\frac{13}{2}\right)}+\frac{𝑦}{13}=1\text{... (i)}\end{array}$

$\therefore 𝑦$-intercept of slope (i) is 13.

Given function,

$\begin{array}{cc}𝑓\left(𝑥\right)& =𝑥-\frac{\left|𝑥\right|}{𝑥},𝑥<0\\ & =𝑥+\frac{\left|𝑥\right|}{𝑥},𝑥>0=1,𝑥=0\end{array}$

Now, at $𝑥=0$

$\begin{array}{cc}& \mathrm{LHL}=\underset{𝑥\to 0^{-}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0^{-}}{lim}(𝑥-\frac{\left|𝑥\right|}{𝑥})\\ & \underset{𝑥\to 0}{lim}(𝑥-\frac{(-𝑥)}{𝑥})=\underset{𝑥\to 0}{lim}(𝑥+1)=1\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \mathrm{RHL}=\underset{𝑥\to 0^{+}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0^{+}}{lim}+(𝑥+\frac{\left|𝑥\right|}{𝑥})\\ & =\underset{𝑥\to 0}{lim}(𝑥+1)=1\text{ and }𝑓\left(0\right)=1\\ & \therefore \underset{𝑥\to 0^{-}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to 0^{+}}{lim}𝑓\left(𝑥\right)=𝑓\left(0\right)=1\end{array}\end{array}$

$\Rightarrow 𝑓\left(𝑥\right)$ is continunus at $𝑥=0$

We have, $\tan 𝐴+\tan 𝐵+\tan 𝐶=6$

$\Rightarrow \tan 𝐴\tan 𝐵\tan 𝐶=6\text{.... (i)}$

and $\tan 𝐴\cdot \tan 𝐵=2\text{.... (ii)}$

From Eqs. (i) and (ii) we get, $\tan 𝐶=3$

$\begin{array}{cc}& \text{ In }△𝑃𝑄𝑅\text{ is right angled, at }𝜃\\ & \begin{array}{cc}& \therefore 𝑃{𝑅}^2=𝑃{𝑄}^2+𝑅{𝑄}^2\\ & \Rightarrow (6-6{)}^2+(-10-10{)}^2+(𝜆-10{)}^2\\ & +\left[\right(1-6{)}^2+(0-10{)}^2+(-5-10{)}^2]\\ & =\left[\right(1-6{)}^2+(0+10{)}^2+(-5-𝜆{)}^2]\\ & \Rightarrow 400+{𝜆}^2+100-20𝜆=(25+100+225)+\\ & (25+100+25+{𝜆}^2+10𝜆)\\ & \Rightarrow {𝜆}^2-20𝜆+500=350+150+10𝜆+{𝜆}^2\\ & \Rightarrow -20𝜆=10𝜆\Rightarrow 30𝜆=0\\ & \Rightarrow 𝜆=0\end{array}\end{array}$

We have, maximise $𝑧=4{𝑥}_1+2{𝑥}_2$

Subject to constracts, $3{𝑥}_1+2{𝑥}_2\ge 9,{𝑥}_1-{𝑥}_2\le 3$, ${𝑥}_1\ge 0,{𝑥}_2\ge 0$

On taking given constraints as equation, we get the following graphs

Here, feasible region is unbounded.

We have, function $𝑓\left(𝑥\right)={𝑥}^3-3𝑥$

$\Rightarrow {𝑓}^{\prime }\left(𝑥\right)=3{𝑥}^2-3=3(𝑥+1)(𝑥-1)$

Here, ${𝑓}^{\prime }\left(𝑥\right)>0$ in $(-\infty ;-1)\cup (1,\infty )$ and ${𝑓}^{\prime }\left(𝑥\right)<0$ in $(-1,1)$.

$\therefore 𝑓\left(𝑥\right)$ is increasing in $(-\infty ,-1),\cup (1,\infty )$ and decreasing in $(-1,1)$.

We have, $𝑥=\sin 𝜃,𝑦={\sin }^3𝜃$ $𝑑𝑦$

$\begin{array}{cc}& \therefore \frac{𝑑𝑦}{𝑑𝑥}=\frac{\frac{𝑑𝑦}{𝑑𝜃}}{\frac{𝑑𝑥}{𝑑𝜃}}=\frac{3{\sin }^2𝜃(\cos 𝜃)}{\cos 𝜃}\\ & \Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=3{\sin }^2𝜃\end{array}$