MHT CET 2024 16th May Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& (\sim 𝑝)\vee (𝑝\wedge \sim 𝑞)\\ & \equiv (\sim 𝑝\vee 𝑝)\wedge (\sim 𝑝\vee \sim 𝑞)\text{... [Distributive law]}\\ & \equiv 𝑇\wedge (\sim 𝑝\vee \sim 𝑞)\text{... [Complement law]}\\ & <text>\equiv </text><text>\sim </text>𝑝\vee \sim 𝑞\text{... [Identity law]}\\ & \equiv 𝑝\to (\sim 𝑞)\dots [\because 𝑝\to 𝑞<text>\equiv </text><text>\sim </text>𝑝\vee 𝑞]\end{array}$

$\begin{array}{cc}2\mathrm{ac}\sin \frac{A-B+C}{2}& =2\mathrm{ac}\sin \frac{𝜋-2B}{2}\\ & =2\mathrm{ac}\cos B\\ & =2\mathrm{ac}\frac{c^2+a^2-b^2}{2\mathrm{ca}}\text{.... [By cosine rule]}\\ & =c^2+a^2-b^2\end{array}$

$m_1:m_2=1:2$

If the slopes of the lines given by $𝑎{𝑥}^2+2h𝑥𝑦+𝑏{𝑦}^2=0$ are in the ratio m:n, then $(𝑚+𝑛{)}^2𝑎𝑏=4𝑚𝑛{ℎ}^2$

$\therefore (1+2{)}^2\left(\frac{1}{𝑎}\right)\left(\frac{1}{𝑏}\right)=4(1\left)\right(2){\left(\frac{1}{ℎ}\right)}^2\Rightarrow \frac{𝑎𝑏}{{ℎ}^2}=\frac{9}{8}$

$\bar{a},\bar{b},\bar{c}$ are coplanar vectors.

$\Rightarrow \left[\begin{array}{ccc}\bar{a}& \bar{b}& \bar{c}\end{array}\right]=0$

Let $\overline{𝛼}=2\bar{a}-\bar{b},\overline{𝛽}=2\bar{b}-\bar{c}$ and $𝛾=2\bar{c}-\bar{a}$. Then,

$\begin{array}{cc}& \left[\begin{array}{ccc}\overline{𝛼}& \overline{𝛽}& \overline{𝛾}\end{array}\right]=\mid \begin{array}{ccc}2& -1& 0\\ 0& 2& -1\\ -1& 0& 2\end{array}\mid \cdot \left[\begin{array}{ccc}\overline{𝑎}& \overline{𝑏}& \overline{𝑐}\end{array}\right]\\ & \Rightarrow \left[\begin{array}{ccc}\overline{𝛼}& \overline{𝛽}& \overline{𝛾}\end{array}\right]=7\left[\begin{array}{ccc}\bar{a}& \bar{b}& \overline{𝑐}\end{array}\right]=7\left(0\right)=0\end{array}$

$\begin{array}{cc}& \text{ Since }\sum _{𝑥=0}^{7}P(X=𝑥)=1,\\ & \begin{array}{cc}& 0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1\\ & \begin{array}{cc}& \Rightarrow 10k^2+9k-1=0\\ & \Rightarrow (k+1)(10k-1)=0\\ & \Rightarrow k=\frac{1}{10}\dots [\because 𝑘\ge 0]\end{array}\\ & \begin{array}{cc}P(X\ge 6)& =P(X=6)+P(X=7)\\ & =2{\left(\frac{1}{10}\right)}^2+7{\left(\frac{1}{10}\right)}^2+\frac{1}{10}\\ & =\frac{2}{100}+\frac{7}{100}+\frac{10}{100}\\ & =\frac{19}{100}\end{array}\end{array}\end{array}$

$\left|\overrightarrow{𝑎}\right|=2,\left|\overrightarrow{𝑏}\right|=4,\left|\overline{𝑐}\right|=4$

According to the given condition,

Projection of $\overline{𝑏}$ on $\overline{𝑎}=$ projection of $\overline{𝑐}$ on $\overline{𝑎}$

$\begin{array}{cc}& \Rightarrow \frac{\bar{b}\cdot \bar{a}}{\left|\overrightarrow{a}\right|}=\frac{\bar{c}\cdot \bar{a}}{\left|\bar{a}\right|}\\ & \Rightarrow \bar{b}\cdot \bar{a}=\bar{c}\cdot \bar{a}\\ & \Rightarrow (\bar{b}-\bar{c})\cdot \bar{a}=0\text{.... (i)}\end{array}$

$\begin{array}{cc}& \text{ Now, }|\overrightarrow{𝑎}+\overline{𝑏}-\overline{𝑐}|=\sqrt{\surd |\overline{𝑎}+\overline{𝑏}-\overline{𝑐}{|}^2}\\ & =\sqrt{\surd |\overline{𝑎}{|}^2+|\overrightarrow{𝑏}-\overline{𝑐}{|}^2+2\overline{𝑎}\cdot (\overline{𝑏}-\overline{𝑐})}\\ & =\sqrt{\surd (2{)}^2+|\overline{𝑏}-\overline{𝑐}{|}^2+2\left(0\right)}\text{.... [From (i)]}\\ & =\sqrt{\surd 4+|\overline{𝑏}-\overline{𝑐}{|}^2}\\ & =\sqrt{\surd 4+|\overline{𝑏}{|}^2+|\overline{𝑐}{|}^2-2(\overline{𝑏}\cdot \overline{𝑐})}\\ & =\sqrt{\surd 4+4^2+4^2-2\left(0\right)}\dots \left[\bar{b}\text{ is perpendicular to }\bar{c}\right]\\ & =\sqrt{\surd 36}=6\end{array}$

$\begin{array}{cc}& \text{ Let the observations be }\\ & \begin{array}{cc}& {𝑥}_1,{𝑥}_2,{𝑥}_3,\dots {𝑥}_{50}\\ & \text{ Then, }\sum _{i=1}^{50}({𝑥}_i-30)=50\\ & \Rightarrow \sum _{i=1}^{50}{𝑥}_i-(30\times 50)=50\\ & \Rightarrow \sum _{i=1}^{50}{𝑥}_i=1550\end{array}\\ & \text{ Mean }=\frac{\sum _{i=1}^{50}{𝑥}_{𝑖}}{𝑛}=\frac{1550}{50}=31\end{array}$

$\begin{array}{cc}& \sqrt{\surd 3}\cos 𝜃+\sin 𝜃=\sqrt{\surd 2}\\ & \Rightarrow \frac{\sqrt{\surd 3}}{2}\cos 𝜃+\frac{1}{2}\sin 𝜃=\frac{\sqrt{\surd 2}}{2}\\ & \Rightarrow \sin \frac{𝜋}{3}\cos 𝜃+\cos \frac{𝜋}{3}\sin 𝜃=\frac{1}{\sqrt{\surd 2}}\\ & \Rightarrow \sin (𝜃+\frac{𝜋}{3})=\sin \frac{𝜋}{4}\\ & \Rightarrow 𝜃+\frac{𝜋}{3}=n𝜋+(-1{)}^n\frac{𝜋}{4},n\in Z\\ & \Rightarrow 𝜃=n𝜋+(-1{)}^n\frac{𝜋}{4}-\frac{𝜋}{3},n\in Z\end{array}$

$\begin{array}{cc}& {\sin }^{-1}\left(\frac{𝑥}{5}\right)+{\mathrm{cosec}}^{-1}\left(\frac{5}{4}\right)=\frac{𝜋}{2}\\ & \Rightarrow {\sin }^{-1}\left(\frac{𝑥}{5}\right)+{\sin }^{-1}\left(\frac{4}{5}\right)=\frac{𝜋}{2}\\ & \Rightarrow {\sin }^{-1}\left(\frac{𝑥}{5}\right)=\frac{𝜋}{2}-{\sin }^{-1}\left(\frac{4}{5}\right)\\ & \Rightarrow {\sin }^{-1}\left(\frac{𝑥}{5}\right)={\cos }^{-1}\left(\frac{4}{5}\right)\dots [\because {\sin }^{-1}𝑥+{\cos }^{-1}𝑥=\frac{𝜋}{2}]\\ & \Rightarrow {\sin }^{-1}\left(\frac{𝑥}{5}\right)={\sin }^{-1}\left(\frac{3}{5}\right)\dots [{\cos }^{-1}𝑥={\sin }^{-1}\sqrt{\surd 1-{𝑥}^2}]\\ & \Rightarrow 𝑥=3\end{array}$

Given,

$P(X=3)=2P(X=1)\text{ and }P(X=2)=0.3\text{.... (i)}$

Now, mean $=1.3$

$\begin{array}{cc}\therefore 0\times P(X=0)+1\times P(X=1)& +2\times P(X=2)\\ & +3\times P(X=3)=1.3\end{array}$

$\Rightarrow 7P(X=1)=0.7\text{...[From (i)]}$

$\Rightarrow 𝑃(𝑋=1)=0.1$

Also, $P(X=0)+P(X=1)+P(X=2)$ $+P(X=3)=1$

$\begin{array}{cc}& \Rightarrow P(X=0)+3P(X=1)=0.7\dots \left[\text{ From }\right(i\left)\right]\\ & \Rightarrow P(X=0)+0.3=0.7\\ & \Rightarrow P(X=0)=0.4\end{array}$

Given, $(\bar{a}\times \bar{b})\times \bar{c}=\frac{1}{3}\left|\bar{b}\right|\left|\bar{c}\right|\bar{a}$

We know that,

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

On comparing, we get

$\begin{array}{cc}& \frac{1}{3}\left|\bar{b}\right|\left|\bar{c}\right|=-\bar{b}\cdot \bar{c}\\ & \Rightarrow \frac{1}{3}\left|\bar{b}\right|\left|\bar{c}\right|=-\left|\bar{b}\right|\left|\bar{c}\right|\cos 𝜃\\ & \Rightarrow \cos 𝜃=\frac{-1}{3}\\ & \Rightarrow {\cos }^2𝜃=\frac{1}{9}\end{array}$

$\begin{array}{cc}{\sin }^2𝜃& =1-{\cos }^2𝜃\\ & =1-\frac{1}{9}\\ \therefore {\sin }^2𝜃& =\frac{8}{9}\\ \therefore \sin 𝜃& =\sqrt{\surd \frac{8}{9}}=\frac{2\sqrt{\surd 2}}{3}\end{array}$

Equation of plane passing through the intersection of given planes is

$\begin{array}{cc}& (𝑥+𝑦+𝑧-1)+𝜆(2𝑥+3𝑦-𝑧+4)=0\\ & \Rightarrow (1+2𝜆)𝑥+(1+3𝜆)𝑦+(1-𝜆)𝑧+4𝜆-1=0\end{array}$

Since the plane is parallel to Y -axis.

$\begin{array}{cc}\therefore & 1+3𝜆=0\\ & \Rightarrow 𝜆=\frac{-1}{3}\end{array}$

$\therefore$ Equation of the required plane is $𝑥+4𝑧-7=0$

$\begin{array}{cc}& \text{ Let }𝑦=f\left(f\right(f\left(𝑥\right)\left)\right)+\left(f\right(𝑥){)}^2\\ & \therefore \frac{d𝑦}{d𝑥}=f^{\prime }\left(f\right(f\left(𝑥\right)\left)\right)\cdot f^{\prime }\left(f\right(𝑥\left)\right)\cdot f^{\prime }\left(𝑥\right)+2f\left(𝑥\right)f^{\prime }\left(𝑥\right)\\ & \begin{array}{cc}{\frac{d𝑦}{d𝑥}\mid }_{𝑥=1}& =f^{\prime }\left(f\right(f\left(1\right)\left)\right)\cdot f^{\prime }\left(f\right(1\left)\right)\cdot f^{\prime }\left(1\right)+2f\left(1\right)f^{\prime }\left(1\right)\\ & =3\cdot 3\cdot 3+2\cdot 1\cdot 3\\ & =33\end{array}\end{array}$

$\begin{array}{cc}& [\bar{a}\times \bar{b}\bar{b}\times \bar{c}\bar{c}\times \bar{a}]\\ & =(\bar{a}\times \bar{b})\cdot \left\{\right(\bar{b}\times \bar{c})\times (\bar{c}\times \bar{a}\left)\right\}\\ & =(\bar{a}\times \bar{b})\cdot \left\{\right((\bar{b}\times \bar{c})\cdot \bar{a})\bar{c}-((\bar{b}\times \bar{c})\cdot \bar{c}\left)\bar{a}\right\}\\ & =(\bar{a}\times \bar{b})\cdot \left\{\right((\bar{b}\times \bar{c})\cdot \bar{a}\left)\bar{c}\right\}\dots [\because (\bar{b}\times \bar{c})\cdot \bar{c}=0]\\ & =\left\{\right(\bar{a}\times \bar{b})\cdot \bar{c}\}\left(\right(\bar{b}\times \bar{c})\cdot \bar{a})\\ & =\left[\bar{a}\bar{b}\bar{c}\right]\left[\bar{a}\bar{b}\bar{c}\right]\dots \dots [\because [\bar{a}\bar{b}\bar{c}]=[\bar{b}\bar{c}\bar{a}\left]\right]\\ & =[\bar{a}\bar{b}\bar{c}{]}^2\end{array}$

$\therefore 𝜆=1$

$\begin{array}{cc}& \underset{𝑥\to 0}{lim}\left\{\right(\sin 𝑥{)}^{\frac{1}{𝑥}}+{\left(\frac{1}{𝑥}\right)}^{\sin 𝑥}\}\\ & =\underset{𝑥\to 0}{lim}(\sin 𝑥{)}^{\frac{1}{𝑥}}+\underset{𝑥\to 0}{lim}{\left(\frac{1}{𝑥}\right)}^{\sin 𝑥}\\ & =0+\underset{𝑥\to 0}{lim}{\left(\frac{1}{𝑥}\right)}^{\sin 𝑥}\\ & =\underset{𝑥\to 0}{lim}{\left(\frac{1}{𝑥}\right)}^{\sin 𝑥}\end{array}$

$\begin{array}{cc}& \text{ Let }𝑙=\underset{𝑥\to 0}{lim}{\left(\frac{1}{𝑥}\right)}^{\sin 𝑥}\text{. Then, }\\ & \begin{array}{cc}& \log 𝑙=\log \underset{𝑥\to 0}{lim}{\left(\frac{1}{𝑥}\right)}^{\sin 𝑥}\\ & \Rightarrow \log 𝑙=\underset{𝑥\to 0}{lim}(-\sin 𝑥\log 𝑥)\\ & \Rightarrow \log 𝑙=-\underset{𝑥\to 0}{lim}\frac{\log 𝑥}{\mathrm{cosec}𝑥}\\ & \Rightarrow \log 𝑙=-\underset{𝑥\to 0}{lim}\frac{\frac{1}{𝑥}}{-\mathrm{cosec}𝑥\cot 𝑥}=\underset{𝑥\to 0}{lim}\frac{{\sin }^2𝑥}{𝑥\cos 𝑥}\\ & \Rightarrow \log 𝑙=\underset{𝑥\to 0}{lim}\frac{\tan 𝑥}{𝑥}\times \sin 𝑥=1\times 0=0\\ & \Rightarrow 𝑙=e^0=1\end{array}\end{array}$

$\begin{array}{cc}& 𝑦=10-{𝑥}^2\text{.... (i)}\\ & 𝑦=2+{𝑥}^2\text{.... (ii)}\end{array}$

Solving (i) and (ii), we get

$𝑦=6$

From (i), $6=10-{𝑥}^2\Rightarrow 𝑥=\pm 2$

Differentiating (i) w.r.t. $𝑥$, we get

$\frac{d𝑦}{d𝑥}=-2𝑥\Rightarrow {\left(\frac{d𝑦}{d𝑥}\right)}_{(-2,6)}=4$

Differentiating (ii) w.r.t. $𝑥$, we get

$\begin{array}{cc}& \frac{𝑑𝑦}{𝑑𝑥}=2𝑥\Rightarrow {\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}_{(-2,6)}=-4\\ \therefore & \tan 𝜃=\mid \frac{4-(-4)}{1+4(-4)}\mid =\mid \frac{8}{-15}\mid \Rightarrow |\tan 𝜃|=\frac{8}{15}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑦=a\log 𝑥+𝑏{𝑥}^2+𝑥\\ & \therefore \frac{d𝑦}{d𝑥}=\frac{a}{𝑥}+2b𝑥+1\\ & \Rightarrow {\left(\frac{d𝑦}{d𝑥}\right)}_{𝑥=-1}=-a-2b+1=0\\ & \Rightarrow a+2b=1\text{.... (i)}\\ & \text{ and }{\left(\frac{d𝑦}{d𝑥}\right)}_{𝑥=2}=\frac{a}{2}+4b+1=0\\ & \Rightarrow a+8b+2=0\text{.... (ii)}\end{array}\\ & \text{ Solving (i) and (ii), we get }\\ & 𝑎=2,𝑏=-\frac{1}{2}\end{array}$

$\begin{array}{cc}& \sin 𝜃-\cos 𝜃=\sqrt{\surd 2}\cos 𝜃\\ & \Rightarrow \sin 𝜃=(\sqrt{\surd 2}+1)\cos 𝜃\\ & \Rightarrow \frac{\sin 𝜃}{\sqrt{\surd 2}+1}\times \frac{\sqrt{\surd 2}-1}{\sqrt{\surd 2}-1}=\cos 𝜃\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{\sin 𝜃(\sqrt{\surd 2}-1)}{(\sqrt{\surd 2}{)}^2-1^2}=\cos 𝜃\\ & \Rightarrow \frac{\sin 𝜃(\sqrt{\surd 2}-1)}{2-1}=\cos 𝜃\\ & \Rightarrow \sqrt{\surd 2}\sin 𝜃-\sin 𝜃=\cos 𝜃\\ & \Rightarrow \sin 𝜃+\cos 𝜃=\sqrt{\surd 2}\sin 𝜃\\ \therefore & 𝑃=𝑄\end{array}$

Since direction cosines of PQ are equal and positive.

$\therefore$ The d.r.s. of PQ are $\frac{1}{\sqrt{\surd 3}},\frac{1}{\sqrt{\surd 3}},\frac{1}{\sqrt{\surd 3}}$

$\therefore$ The equation of the line PQ is

$\begin{array}{cc}& \frac{𝑥-2}{\frac{1}{\sqrt{\surd 3}}}=\frac{𝑦+1}{\frac{1}{\sqrt{\surd 3}}}=\frac{𝑧-2}{\frac{1}{\sqrt{\surd 3}}}\\ & \Rightarrow 𝑥-2=𝑦+1=z-2=k,\text{ say }\end{array}$

$\therefore$ Co-ordinates of the point Q are

$(k+2,k-1,k+2)$

The point Q lies on the plane $2𝑥+𝑦+𝑧=9$

$\begin{array}{c}\therefore 2(𝑘+2)+𝑘-1+𝑘+2=9\\ \Rightarrow 4𝑘+5=9\Rightarrow 𝑘=1\\ \therefore 𝑄\equiv (3,0,3)\\ \begin{array}{c}\therefore 𝑃𝑄\end{array}=\sqrt{\surd (3-2{)}^2+(0+1{)}^2+(3-2{)}^2}\\ =\sqrt{\surd 1+1+1}\\ =\sqrt{\surd 3}\text{ units }\end{array}$

Since $𝑓\left(𝑥\right)$ is continuous in $[0,\frac{𝜋}{2}]$.

$\therefore$ it is continuous at $𝑥=\frac{𝜋}{4}$.

$\therefore 𝑓\left(\frac{𝜋}{4}\right)=\underset{𝑥\to \frac{𝜋}{4}}{lim}𝑓\left(𝑥\right)=\underset{𝑥\to \frac{𝜋}{4}}{lim}\frac{1-\tan 𝑥}{4𝑥-𝜋}$

Applying L'Hospital rule on R.H.S., we get

$\begin{array}{cc}& 𝑓\left(\frac{𝜋}{4}\right)=\underset{𝑥\to \frac{𝜋}{4}}{lim}\frac{-{\sec }^2𝑥}{4}\\ & \Rightarrow 𝑓\left(\frac{𝜋}{4}\right)=\frac{-2}{4}=\frac{-1}{2}\end{array}$

$\begin{array}{cc}& \text{ Let }𝐼=\underset{-\frac{𝜋}{2}}{\overset{\frac{𝜋}{2}}{\int }}\frac{{\sin }^2𝑥}{1+2^{𝑥}}𝑑𝑥\text{.... (i)}\\ \therefore & 𝐼={\int }_{\frac{𝜋}{2}}^{-\frac{𝜋}{2}}\frac{{\sin }^2𝑥}{1+2^{-𝑥}}𝑑𝑥\text{.... (ii)}\end{array}$

$\dots [\because {\int }_{𝑏}^{𝑎}f(𝑥)d𝑥={\int }_{𝑏}^{𝑎}f(a+b-𝑥\left)d𝑥\right]$

$\begin{array}{cc}& \text{ Adding (i) and (ii), we get }\\ & \begin{array}{cc}& 2I={\int }_{\frac{𝜋}{2}}^{-\frac{𝜋}{2}}{\sin }^2𝑥d𝑥\\ & \Rightarrow 2I=2{\int }_{\frac{𝜋}{2}}^0{\sin }^2𝑥d𝑥\end{array}\end{array}$

$\dots .\because {\int }_{𝑎}^{-𝑎}f\left(𝑥\right)d𝑥=2{\int }_{𝑎}^{0}f\left(𝑥\right)d𝑥{}^{\text{if }}f\left(𝑥\right)$ is an even function $]$

$\begin{array}{cc}\Rightarrow 𝐼& ={\int }_{\frac{𝜋}{2}}^0\left(\frac{1-\cos 2𝑥}{2}\right)𝑑𝑥\\ & =\frac{1}{2}{[𝑥-\frac{\sin 2𝑥}{2}]}_{\frac{𝜋}{2}}^0\\ & =\frac{1}{2}(\frac{𝜋}{2}-0)=\frac{𝜋}{4}\end{array}$

$\begin{array}{cc}& \text{ Putting }𝑥=\frac{{𝑦}^2}{2}\text{ in }𝑦=4𝑥-1\text{, we get }\\ & \begin{array}{cc}& 𝑦=4\left(\frac{{𝑦}^2}{2}\right)-1\Rightarrow 2{𝑦}^2-𝑦-1=0\\ & \Rightarrow (𝑦-1)(2𝑦+1)=0\\ & \Rightarrow 𝑦=1,\frac{-1}{2}\end{array}\end{array}$

$\begin{array}{cc}\therefore & \text{ Required area }\\ & ={\int }_1^{-1/2}\left(\frac{𝑦+1}{4}\right)d𝑦-{\int }_1^{-1/2}\frac{{𝑦}^2}{2}d𝑦\\ & =\frac{1}{4}{[\frac{{𝑦}^2}{2}+𝑦]}_1^{-1/2}-\frac{1}{2}{\left[\frac{{𝑦}^3}{3}\right]}_1^{-1/2}\\ & =\frac{1}{4}[(\frac{1}{2}-\frac{1}{8})+(1+\frac{1}{2})]-\frac{1}{2}(\frac{1}{3}+\frac{1}{24})\\ & =\frac{1}{4}\left(\frac{15}{8}\right)-\frac{1}{2}\left(\frac{9}{24}\right)\\ & =\frac{15}{32}-\frac{3}{16}\\ & =\frac{9}{32}\end{array}$

$\begin{array}{cc}& \text{ Given, }(𝑥\log 𝑥)\frac{d𝑦}{d𝑥}+𝑦=2𝑥\log 𝑥\\ & \text{ When }𝑥=1,𝑦=0\\ & (𝑥\log 𝑥)\frac{d𝑦}{d𝑥}+𝑦=2𝑥\log 𝑥\\ \therefore & \frac{d𝑦}{d𝑥}+\frac{𝑦}{𝑥\log 𝑥}=2\\ & \\ \therefore & \text{ I.F. }=e^{\int \frac{1}{𝑥\log 𝑥}d𝑥}=e^{\log (\log 𝑥)}=\log 𝑥\\ \therefore & \text{ Solution of the given equation is }\\ & 𝑦\cdot \log 𝑥=\int 2\log 𝑥d𝑥+c\\ \therefore & 𝑦\log 𝑥=2(𝑥\log 𝑥-𝑥)+c\\ \therefore & \text{ When }𝑥=1,𝑦=0\\ \therefore & 0=-2+c\Rightarrow c=2\\ \therefore & 𝑦\log 𝑥=2(𝑥\log 𝑥-𝑥)+2\\ \therefore & 𝑦\left(e\right)=2(e-e)+2=2\end{array}$