MHT CET 2024 3rd May Evening Shift (Mathematics) Question Paper with Solutions

Required equation passes through point of intersection of lines $3𝑥-𝑦=5$ and $𝑥+3𝑦=1$

$\therefore$ Point of intersection $=(\frac{16}{10},\frac{-2}{10})$

Equation of line in double intercept form is $\frac{𝑥}{𝑎}+\frac{𝑦}{𝑏}=1$

But $a=b$ ...[Given]

So, equation of line is $𝑥+𝑦=a$

Since line passes through $(\frac{16}{10},\frac{-2}{10})$

$\begin{array}{cc}\therefore & \frac{16}{10}+\frac{-2}{10}=𝑎\\ & \Rightarrow \frac{14}{10}=𝑎\\ & \Rightarrow 𝑎=\frac{7}{5}\end{array}$

$\therefore$ The required equation of line is $𝑥+𝑦=\frac{7}{5}$

$\Rightarrow 5𝑥+5𝑦-7=0$

Let $I=\int \frac{\mathrm{cosec}𝑥d𝑥}{{\cos }^2(1+\log \tan \frac{𝑥}{2})}d𝑥$

Let $1+\log (\tan \frac{𝑥}{2})=t$

Differentiating both sides w.r.t. t, we get

$\begin{array}{cc}& \frac{1}{\tan \frac{𝑥}{2}}{\sec }^2\frac{𝑥}{2}\times \frac{1}{2}d𝑥=\mathrm{dt}\\ \therefore & \frac{1}{2\sin \frac{𝑥}{2}\cos \frac{𝑥}{2}}d𝑥=\mathrm{dt}\\ \therefore & \mathrm{cosec}𝑥d𝑥=\mathrm{dt}\\ \therefore & =\int \frac{1}{{\cos }^2𝑡}\mathrm{dt}\\ & =\int {\sec }^{2}t𝑑𝑡\\ & =\tan \left(t\right)+c\\ & =\tan (1+\log (\tan \frac{𝑥}{2}))+c\end{array}$

To solve the problem, begin by examining the condition given:

$\bar{b}\times \bar{c}=\bar{b}\times \bar{d}$

This indicates that the vector difference $\bar{c}-\bar{d}$ must be parallel to $\bar{b}$. Therefore, we can express:

$\bar{c}-\bar{d}=𝑘\bar{b}$

for some scalar $𝑘$. Rearranging gives:

$\bar{d}=\bar{c}-𝑘\bar{b}$

Additionally, the second condition provided is:

$\bar{a}\cdot \bar{d}=0$

Substitute $\bar{d}=\bar{c}-𝑘\bar{b}$ into this equation:

$\bar{a}\cdot (\bar{c}-𝑘\bar{b})=0$

This can be expanded to:

$\bar{a}\cdot \bar{c}-𝑘(\bar{a}\cdot \bar{b})=0$

Rearranging to solve for $𝑘$ results in:

$𝑘=\frac{\bar{a}\cdot \bar{c}}{\bar{a}\cdot \bar{b}}$

Substitute $𝑘$ back into the expression for $\bar{d}$:

$\bar{d}=\bar{c}-\left(\frac{\bar{a}\cdot \bar{c}}{\bar{a}\cdot \bar{b}}\right)\bar{b}$

Thus, the correct choice is:

Option B:

$\bar{c}-\left(\frac{\bar{a}\cdot \bar{c}}{\bar{a}\cdot \bar{b}}\right)\bar{b}$

$\begin{array}{cc}& \text{ Given, }f\left(𝑥\right)\text{ is continuous at }𝑥=2\\ & \begin{array}{cc}& \therefore f\left(2\right)=\underset{𝑥\to 2}{lim}f\left(𝑥\right)\\ & k=\underset{𝑥\to 2}{lim}{\left(\frac{5𝑥-8}{8-3𝑥}\right)}^{\frac{3}{2𝑥-4}}\\ & \text{ Put }𝑥-2=h\\ & \Rightarrow 𝑥=2+h\\ & \text{ As }𝑥\to 2,h\to 0\end{array}\end{array}$

$\begin{array}{cc}\therefore 𝑘& =\underset{ℎ\to 0}{lim}{\left[\frac{5(2+ℎ)-8}{8-3(2+ℎ)}\right]}^{\frac{3}{2(2+ℎ)-4}}\\ 𝑘& =\underset{ℎ\to 0}{lim}{\left[\frac{10+5ℎ-8}{8-6-3ℎ}\right]}^{\frac{3}{2ℎ}}\\ 𝑘& =\underset{ℎ\to 0}{lim}{\left[\frac{2+5ℎ}{2-3ℎ}\right]}^{\frac{3}{2ℎ}}\\ & =\underset{ℎ\to 0}{lim}\frac{{(1+\frac{5}{2}ℎ)}^{\frac{3}{2ℎ}}}{{(1-\frac{3ℎ}{2})}^{\frac{3}{2ℎ}}}\end{array}$

$=\frac{\underset{ℎ\to 0}{lim}{\left[{(1+\frac{5}{2}ℎ)}^{\frac{2}{5ℎ}}\right]}^{\frac{5}{2}\times \frac{3}{2}}}{\underset{ℎ\to 0}{lim}{\left[{(1-\frac{3ℎ}{2})}^{\frac{-2}{3ℎ}}\right]}^{\frac{-3}{2}\times \frac{3}{2}}}$

$\begin{array}{cc}& =\frac{{𝑒}^{\frac{15}{4}}}{{𝑒}^{\frac{-9}{4}}}\cdots \left[\begin{array}{c}ℎ\to 0,\frac{5ℎ}{2}\to 0,\frac{-3ℎ}{2}\to 0\\ \text{ and }\underset{ℎ\to 0}{lim}(1+𝑥{)}^{\frac{1}{𝑥}}=𝑒\end{array}\right]\\ & ={𝑒}^{\frac{24}{4}}\\ & ={𝑒}^6\end{array}$

Given, $f\left(𝑥\right)=\frac{{\log }_2(𝑥+3)}{{𝑥}^2+3𝑥+2}$

Now, $f\left(𝑥\right)$ is defined if ${𝑥}^2+3𝑥+2\ne 0$

$\Rightarrow (𝑥+1)(𝑥+2)\ne 0$

$\Rightarrow 𝑥\ne -1,-2$.

Also, $𝑥+3>0$

$𝑥>-3$

$\therefore 𝑥\in (-3,\infty )-\{-1,-2\}$

Equation of angle bisector of two lines whose general equation is $𝑎{𝑥}^2+2h𝑥𝑦+𝑏{𝑦}^2=0$ is $\frac{{𝑥}^2-{𝑦}^2}{a-b}=\frac{𝑥𝑦}{h}$

$\therefore$ Comparing given equation ${𝑥}^2-4𝑥𝑦-5{𝑦}^2=0$ with $𝑎{𝑥}^2+2h𝑥𝑦+𝑏{𝑦}^2=0$, we get $a=1,b=-5,h=-2$

$\therefore$ Equation of angle bisectors is

$\begin{array}{cc}& \frac{{𝑥}^2-{𝑦}^2}{1-(-5)}=\frac{𝑥𝑦}{-2}\\ & \Rightarrow {𝑥}^2-{𝑦}^2=-3𝑥𝑦\\ & \Rightarrow {𝑥}^2+3𝑥𝑦-{𝑦}^2=0\end{array}$

Given equation is

$(\cos p-1){𝑥}^2+(\cos p)𝑥+\sin p=0$

Comparing with $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$, we get

$𝑎=\cos 𝑝-1,𝑏=\cos 𝑝,𝑐=\sin 𝑝$

It has real roots.

$\begin{array}{cc}\therefore & {𝑏}^2-4𝑎𝑐\ge 0\\ & \Rightarrow {\cos }^2𝑝-4(\cos 𝑝-1)(\sin 𝑝)\ge 0\\ & \Rightarrow {\cos }^2𝑝-4\sin 𝑝\cos 𝑝+4\sin 𝑝\ge 0\\ & \Rightarrow {\cos }^2𝑝-4\sin 𝑝\cos 𝑝+4{\sin }^2𝑝\\ & +4\sin 𝑝-4{\sin }^2𝑝\ge 0\\ & \Rightarrow (\cos 𝑝-2\sin 𝑝{)}^2+4\sin 𝑝(1-\sin 𝑝)\ge 0\end{array}$

$\therefore (\cos p-2\sin p)$ is always positive

$\therefore 1-\sin 𝑝\ge 0$ for all values of $𝑝,𝑝\in (0,𝜋)$

Since $\overline{𝐚}\cdot \overline{𝐛}=0$

$\therefore \bar{a}$ and $\bar{b}$ are perpendicular unit vectors.

Now, $(2\bar{a}-\bar{b})\cdot \left\{\right(\bar{a}\times \bar{b})\times (\bar{a}+2\bar{b}\left)\right\}$

$\begin{array}{cc}& =[2\overline{𝑎}-\overline{𝑏}\overline{𝑎}\times \overline{𝑏}\overline{𝑎}+2\overline{𝑏}]\\ & =-[\overline{𝑎}\times \overline{𝑏}2\overline{𝑎}-\overline{𝑏}\overline{𝑎}+2\overline{𝑏}]\\ & =-(\overline{𝑎}\times \overline{𝑏})\cdot \left\{\right(2\overline{𝑎}-\overline{𝑏})\times (\overline{𝑎}+2\overline{𝑏}\left)\right\}\\ & =-(\overline{𝑎}\times \overline{𝑏})\cdot 5(\overline{𝑎}\times \overline{𝑏})\\ & =-5|\overline{𝑎}\times \overline{𝑏}|=-5|\overline{𝑎}{|}^2\mid {\overline{𝑏}}^2\dots .[\because \overline{𝑎}⟂\overline{𝑏}]\\ & =-5\dots .[\because |\overline{𝑎}|=|\overline{𝑏}|=1]\end{array}$

Let $𝑋\sim 𝐵(𝑛,𝑝)$

According to the given conditions, mean $=𝑛𝑝=8$ and variance $=𝑛𝑝𝑞=4$

$\Rightarrow p=q=\frac{1}{2}$ and $n=16$

$P(X\le 2)=\frac{K}{2^{16}}$

$\Rightarrow 𝑃(𝑋=0)+𝑃(𝑋=1)+𝑃(𝑋=2)=\frac{𝐾}{2^{16}}$

$\begin{array}{cc}\therefore {}^{16}{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{16}+{}^{16}{C}_1{\left(\frac{1}{2}\right)}^1& {\left(\frac{1}{2}\right)}^{15}\\ & +{}^{16}{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{14}=\frac{K}{2^{16}}\end{array}$

$\therefore \frac{1+16+120}{2^{16}}=\frac{𝐾}{2^{16}}$

$\therefore 𝐾=137$.

$\begin{array}{cc}& \left[\begin{array}{ccc}m\bar{a}+\bar{b}& m\bar{b}+\bar{c}& m\bar{c}+\bar{a}\end{array}\right]=28\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & \Rightarrow (m\bar{a}+\bar{b})\left[\right(m\bar{b}+\bar{c})\times (\mathrm{mc}+\bar{a}\left)\right]=28\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & \Rightarrow (m\bar{a}+\bar{b})\\ & \left[m^2\right(\bar{b}\times \bar{c})+m(\bar{b}\times \bar{a})+m(\bar{c}\times \bar{c})+(\bar{c}\times \bar{a}\left)\right]\\ & =28\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & \Rightarrow (\bar{a}+\bar{b})\left[m^2\right(\bar{b}\times \bar{c})+m(\bar{b}\times \bar{a})+(\bar{c}\times \bar{a}\left)\right]\\ & =28\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & \Rightarrow \mathrm{ma}\cdot m^2(\bar{b}\times \bar{c})+m^2\bar{a}(\bar{b}\times \bar{a})+\mathrm{ma}(\bar{c}\times \bar{a})\\ & +\bar{b}\cdot m^2(\bar{b}\times \bar{c})+\bar{b}\cdot m(\bar{b}\times \bar{a})+\bar{b}\cdot (\bar{c}\times \bar{a})\\ & =28\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & \Rightarrow m^3\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]+\left[\begin{array}{ccc}\bar{b}& \bar{a}& \bar{c}\end{array}\right]=28\left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]\\ & \Rightarrow m^3+1=28\\ & \Rightarrow m^3=27\\ & \Rightarrow m=3\end{array}$

$\begin{array}{cc}\text{ Required area }& =\underset{0}{\overset{3}{\int }}({𝑥}^2+2-𝑥)d𝑥\\ & ={[\frac{{𝑥}^3}{3}+2𝑥-\frac{{𝑥}^2}{2}]}_3^0\\ & =9+6-\frac{9}{2}-0\\ & =\frac{21}{2}\text{ sq. unit }\end{array}$

Given,

$\begin{array}{cc}& \sqrt{\surd 1+{𝑥}^2}{[{\{𝑥\cos ({\cot }^{-1}𝑥)+\sin ({\cot }^{-1}𝑥)\}}^2-1]}^{\frac{1}{2}}\\ & \text{ Let }{\cot }^{-1}𝑥=𝜃\\ & \begin{array}{cc}\therefore 𝑥& =\cot 𝜃\\ & \Rightarrow \sin 𝜃=\frac{1}{\sqrt{\surd 1+{𝑥}^2}}\dots \left(i\right)\\ & \Rightarrow \cos 𝜃=\frac{𝑥}{\sqrt{\surd 1+{𝑥}^2}}\dots \left(\mathrm{ii}\right)\\ \therefore & \sqrt{\surd 1+{𝑥}^2}{[{\{𝑥\cos ({\cot }^{-1}𝑥)+\sin ({\cot }^{-1}𝑥)\}}^2-1]}^{\frac{1}{2}}\\ & =\sqrt{\surd 1+{𝑥}^2}{\left[\right\{𝑥\cos 𝜃+\sin 𝜃{\}}^2-1]}^{\frac{1}{2}}\\ & =\sqrt{\surd 1+{𝑥}^2}{[{\{𝑥\times \frac{𝑥}{\sqrt{\surd 1+{𝑥}^2}}+\frac{1}{\sqrt{\surd 1+{𝑥}^2}}\}}^2-1]}^{\frac{1}{2}}\\ & =\sqrt{\surd 1+{𝑥}^2}{[{\left(\frac{{𝑥}^2+1}{\sqrt{\surd 1+{𝑥}^2}}\right)}^2-1]}^{\frac{1}{2}}\\ & =\sqrt{\surd 1+{𝑥}^2}{[{\left(\sqrt{\surd 1+{𝑥}^2}\right)}^2-1]}^{\frac{1}{2}}\\ & =\sqrt{\surd 1+{𝑥}^2}{(1+{𝑥}^2-1)}^{\frac{1}{2}}\\ & =𝑥\sqrt{\surd 1+{𝑥}^2}\end{array}\end{array}$

To find the variance of the random variable $𝑋$, denoted as $\mathrm{Var}\left(𝑋\right)$, we can use the formula for variance:

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

where:

$𝐸\left(𝑋\right)$ is the expected value or mean of $𝑋$.

$𝐸\left({𝑋}^2\right)$ is the expected value of ${𝑋}^2$.

Step 1: Calculate $𝐸\left(𝑋\right)$

The expected value $𝐸\left(𝑋\right)$ is given by:

$𝐸\left(𝑋\right)=\sum [𝑥\cdot 𝑝(𝑥\left)\right]$

Calculating for each value of $𝑋$:

For $𝑥=-2$: $𝑥\cdot 𝑝\left(𝑥\right)=-2\cdot 0.1=-0.2$

For $𝑥=-1$: $𝑥\cdot 𝑝\left(𝑥\right)=-1\cdot 0.2=-0.2$

For $𝑥=0$: $𝑥\cdot 𝑝\left(𝑥\right)=0\cdot 0.2=0.0$

For $𝑥=1$: $𝑥\cdot 𝑝\left(𝑥\right)=1\cdot 0.3=0.3$

For $𝑥=2$: $𝑥\cdot 𝑝\left(𝑥\right)=2\cdot 0.15=0.3$

For $𝑥=3$: $𝑥\cdot 𝑝\left(𝑥\right)=3\cdot 0.05=0.15$

Summing these, the expected value $𝐸\left(𝑋\right)$ is:

$𝐸\left(𝑋\right)=-0.2-0.2+0.0+0.3+0.3+0.15=0.35$

Step 2: Calculate $𝐸\left({𝑋}^2\right)$

The expected value $𝐸\left({𝑋}^2\right)$ is given by:

$𝐸\left({𝑋}^2\right)=\sum [{𝑥}^2\cdot 𝑝(𝑥\left)\right]$

Calculating for each value of $𝑋$:

For $𝑥=-2$: ${𝑥}^2\cdot 𝑝\left(𝑥\right)=4\cdot 0.1=0.4$

For $𝑥=-1$: ${𝑥}^2\cdot 𝑝\left(𝑥\right)=1\cdot 0.2=0.2$

For $𝑥=0$: ${𝑥}^2\cdot 𝑝\left(𝑥\right)=0\cdot 0.2=0.0$

For $𝑥=1$: ${𝑥}^2\cdot 𝑝\left(𝑥\right)=1\cdot 0.3=0.3$

For $𝑥=2$: ${𝑥}^2\cdot 𝑝\left(𝑥\right)=4\cdot 0.15=0.6$

For $𝑥=3$: ${𝑥}^2\cdot 𝑝\left(𝑥\right)=9\cdot 0.05=0.45$

Summing these, the expected value $𝐸\left({𝑋}^2\right)$ is:

$𝐸\left({𝑋}^2\right)=0.4+0.2+0.0+0.3+0.6+0.45=1.95$

Step 3: Calculate $\mathrm{Var}\left(𝑋\right)$

Now, use these values to find the variance:

$\mathrm{Var}\left(𝑋\right)=𝐸\left({𝑋}^2\right)-\left[𝐸\right(𝑋){]}^2=1.95-(0.35{)}^2$

Using the given approximation:

$(0.35{)}^2=0.1225$

Thus,

$\mathrm{Var}\left(𝑋\right)=1.95-0.1225=1.8275$

The variance $\mathrm{Var}\left(𝑋\right)$ is 1.8275. Therefore, the correct answer is:

Option C: 1.8275

$\begin{array}{cc}& (1-\tan 𝜃)(1+\tan 𝜃){\sec }^2𝜃+2{\tan }^2𝜃=0\\ & \Rightarrow (1-{\tan }^2𝜃)(1+{\tan }^2𝜃)+2{\tan }^2𝜃=0\end{array}$

Put ${\tan }^2𝜃=𝑥$

$\begin{array}{cc}& \text{ Put }(1-𝑥)(1+𝑥)+2𝑥=0\\ & \Rightarrow (1-{𝑥}^2+2𝑥=0\\ & \Rightarrow {𝑥}^2-1=2𝑥\end{array}$

Let us draw graph of $𝑦={𝑥}^2-1$ and $𝑦=2𝑥$

From the graph the two curves are intersecting at 2 points.

$\therefore$ There are 2 values of $𝑥$.

Only one value of $𝑥$ exists, for $𝜃\in (\frac{-𝜋}{2},\frac{𝜋}{2})$

$\therefore 𝑥={\tan }^2𝜃$

$\therefore$ Two values of $𝜃$ satisfies above equation

In this problem, we are interested in determining the probability of drawing exactly 1 or exactly 2 kings when two cards are drawn with replacement from a standard deck of 52 cards. A standard deck contains 4 kings.

Since the cards are drawn with replacement, the probability of drawing a king in each draw is always $\frac{4}{52}=\frac{1}{13}$, and the probability of not drawing a king is $\frac{48}{52}=\frac{12}{13}$.

The random variable $𝑋$ follows a binomial distribution with parameters $𝑛=2$ (the number of trials or draws) and $𝑝=\frac{1}{13}$ (the probability of success, i.e., drawing a king).

Probability of drawing exactly 1 king, $𝑃(𝑋=1)$:

$𝑃(𝑋=1)=(\frac{2}{1}){\left(\frac{1}{13}\right)}^1{\left(\frac{12}{13}\right)}^{2-1}$

$=2\times \frac{1}{13}\times \frac{12}{13}$

$=\frac{24}{169}$

Probability of drawing exactly 2 kings, $𝑃(𝑋=2)$:

$𝑃(𝑋=2)=(\frac{2}{2}){\left(\frac{1}{13}\right)}^2{\left(\frac{12}{13}\right)}^{2-2}$

$=1\times \frac{1}{13}\times \frac{1}{13}$

$=\frac{1}{169}$

Therefore, the probability that either 1 or 2 kings are drawn is:

$𝑃(𝑋=1)+𝑃(𝑋=2)=\frac{24}{169}+\frac{1}{169}$

$=\frac{25}{169}$

Thus, the answer is Option D: $\frac{25}{169}$.

To solve this problem, it's necessary to apply the conditions given for the curve and use them to form equations relating to the constants $𝑎$, $𝑏$, and $𝑐$.

The curve touches the $𝑥$-axis at $(-2,0)$:

Since it touches the $𝑥$-axis at this point, the value of the function must be zero there:

$𝑎(-2{)}^3+𝑏(-2{)}^2+𝑐(-2)+5=0.$

Simplifying this equation gives:

$-8𝑎+4𝑏-2𝑐+5=0.$

Also, being a point of tangency implies the first derivative at this point is zero:

$3𝑎{𝑥}^2+2𝑏𝑥+𝑐=0$ at $𝑥=-2$.

Thus:

$3𝑎(-2{)}^2+2𝑏(-2)+𝑐=0$

Simplifying:

$12𝑎-4𝑏+𝑐=0.$

The gradient at the Y-intercept is 3:

The gradient of the curve is given by its first derivative, so at the point where the curve cuts the $𝑦$-axis (i.e., $𝑥=0$), the derivative is:

$3𝑎{𝑥}^2+2𝑏𝑥+𝑐=3.$

At $𝑥=0$:

$𝑐=3.$

Using these three equations:

From the first equation:

$-8𝑎+4𝑏-2𝑐+5=0.$

Substituting $𝑐=3$:

$-8𝑎+4𝑏-2\left(3\right)+5=0⟹-8𝑎+4𝑏-6+5=0$

$-8𝑎+4𝑏-1=0.\left(1\right)$

From the second equation:

$12𝑎-4𝑏+𝑐=0.$

Again, substituting $𝑐=3$:

$12𝑎-4𝑏+3=0.\left(2\right)$

Solve these equations simultaneously:

From equation (1):

$-8𝑎+4𝑏=1$

Then:

$𝑏=2𝑎+\frac{1}{4}$

Substitute in equation (2):

$12𝑎-4(2𝑎+\frac{1}{4})+3=0$

Expanding and simplifying:

$12𝑎-8𝑎-1+3=0$

$4𝑎+2=0$

$4𝑎=-2$

$𝑎=-\frac{1}{2}$

Now find $𝑏$:

$𝑏=2(-\frac{1}{2})+\frac{1}{4}$

$𝑏=-1+\frac{1}{4}$

$𝑏=-\frac{3}{4}$

Now, sum $𝑎+𝑏+𝑐$:

$𝑎+𝑏+𝑐=-\frac{1}{2}-\frac{3}{4}+3$

Finding a common denominator:

$𝑎+𝑏+𝑐=-\frac{2}{4}-\frac{3}{4}+\frac{12}{4}$

$𝑎+𝑏+𝑐=\frac{7}{4}$

Thus, the value of $𝑎+𝑏+𝑐$ is $\overline{)\frac{7}{4}}$.

To find the unit vector perpendicular to two lines in space, we first need the direction vectors of these lines.

Direction Vectors:

For line $L_1:\frac{𝑥+1}{3}=\frac{𝑦+2}{2}=\frac{𝑧+1}{1}$, the direction vector is derived from the coefficients:

${𝐝}_1=3\widehat{𝑖}+2\widehat{𝑗}+1\widehat{𝑘}$

For line $L_2:\frac{𝑥-2}{2}=\frac{𝑦+2}{1}=\frac{𝑧-3}{3}$, the direction vector is:

${𝐝}_2=2\widehat{𝑖}+1\widehat{𝑗}+3\widehat{𝑘}$

Cross Product:

The vector perpendicular to both direction vectors ${𝐝}_1$ and ${𝐝}_2$ is given by the cross product ${𝐝}_1\times {𝐝}_2$.

${𝐝}_1\times {𝐝}_2=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 3& 2& 1\\ 2& 1& 3\end{array}\mid$

Calculating the determinant, we find:

${𝐝}_1\times {𝐝}_2=\widehat{𝑖}(2\cdot 3-1\cdot 1)-\widehat{𝑗}(3\cdot 3-1\cdot 2)+\widehat{𝑘}(3\cdot 1-2\cdot 2)$

$=\widehat{𝑖}(6-1)-\widehat{𝑗}(9-2)+\widehat{𝑘}(3-4)$

$=5\widehat{𝑖}-7\widehat{𝑗}-\widehat{𝑘}$

Unit Vector:

The magnitude of the vector $𝐯=5\widehat{𝑖}-7\widehat{𝑗}-\widehat{𝑘}$ can be calculated as:

$\left|𝐯\right|=\sqrt{\surd 5^2+(-7{)}^2+(-1{)}^2}=\sqrt{\surd 25+49+1}=\sqrt{\surd 75}=5\sqrt{\surd 3}$

The unit vector $\widehat{𝐮}$ perpendicular to both lines is:

$\widehat{𝐮}=\frac{5\widehat{𝑖}-7\widehat{𝑗}-\widehat{𝑘}}{5\sqrt{\surd 3}}$

Conclusion:

The unit vector perpendicular to both $L_1$ and $L_2$ is:

$\overline{)\frac{5\widehat{𝑖}-7\widehat{𝑗}-\widehat{𝑘}}{5\sqrt{\surd 3}}}$

This corresponds to option C.

To find the value of $𝑎+𝑏$, we need to determine the conditions for the given function $𝑦=𝑎\log 𝑥+𝑏{𝑥}^2+𝑥$ to have extreme values at $𝑥=-1$ and $𝑥=2$.

First, find the derivative of the function $𝑦$.

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑑}{𝑑𝑥}(𝑎\log 𝑥+𝑏{𝑥}^2+𝑥)=\frac{𝑎}{𝑥}+2𝑏𝑥+1$

Set the derivative equal to zero to find the critical points:

$\frac{𝑎}{𝑥}+2𝑏𝑥+1=0$

At the critical points $𝑥=-1$ and $𝑥=2$:

Substitute $𝑥=-1$:

$\frac{𝑎}{-1}+2𝑏(-1)+1=0⟹-𝑎-2𝑏+1=0⟹𝑎+2𝑏=1\text{(Equation 1)}$

Substitute $𝑥=2$:

$\frac{𝑎}{2}+2𝑏\left(2\right)+1=0⟹\frac{𝑎}{2}+4𝑏+1=0$

Multiply this equation by 2 to eliminate the fraction:

$𝑎+8𝑏+2=0⟹𝑎+8𝑏=-2\text{(Equation 2)}$

Now solve the system of equations formed by Equation 1 and Equation 2:

Equation 1: $𝑎+2𝑏=1$

Equation 2: $𝑎+8𝑏=-2$

Subtract Equation 1 from Equation 2 to eliminate $𝑎$:

$(𝑎+8𝑏)-(𝑎+2𝑏)=-2-1$

$6𝑏=-3⟹𝑏=-\frac{1}{2}$

Substitute $𝑏=-\frac{1}{2}$ into Equation 1:

$𝑎+2(-\frac{1}{2})=1$

$𝑎-1=1⟹𝑎=2$

Finally, calculate $𝑎+𝑏$:

$𝑎+𝑏=2-\frac{1}{2}=\frac{4}{2}-\frac{1}{2}=\frac{3}{2}$

Thus, the value of $𝑎+𝑏$ is $\frac{3}{2}$.

Option A: $\frac{3}{2}$