MHT CET 2020 16th October Evening Shift (Mathematics) Question Paper with Solutions

Given, equation of line

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2+2𝑔𝑥+2𝑓𝑦=0$

Let the slope of line is ${𝑚}_1$ and ${𝑚}_2$

$\begin{array}{cc}\therefore {𝑚}_1+{𝑚}_2& =\frac{-2ℎ}{𝑏}\\ {𝑚}_1{𝑚}_2& =\frac{𝑎}{𝑏}\end{array}$

Given, one line is bisector of angle between coordinate axes.

$\begin{array}{cc}& \therefore {𝑚}_1=\tan \frac{𝜋}{4}=1\\ & \Rightarrow 1+{𝑚}_2=-2ℎ/𝑏\text{... (i)}\\ & {𝑚}_2=\frac{𝑎}{𝑏}\text{... (ii)}\end{array}$

$\begin{array}{c}\text{ From Eqs. (i) and (ii), we get }\\ \begin{array}{cc}& 1+\frac{𝑎}{𝑏}=\frac{-2ℎ}{𝑏}\\ & \Rightarrow 𝑎+𝑏=-2ℎ\\ & (𝑎+𝑏{)}^2=4{ℎ}^2\end{array}\end{array}$

$\begin{array}{cc}& {[\sin ({\tan }^{-1}\frac{3}{4})]}^2+{[\sin ({\tan }^{-1}\frac{4}{3})]}^2\\ & \text{ Let }{\tan }^{-1}\frac{3}{4}=𝜃\Rightarrow \tan 𝜃=\frac{3}{4}\Rightarrow \sin 𝜃=\frac{3}{5}\\ & \text{ and }{\tan }^{-1}\frac{4}{3}=𝜙\Rightarrow \tan 𝜙=\frac{4}{3}\Rightarrow \sin 𝜙=\frac{4}{5}\\ & \therefore {[\sin ({\tan }^{-1}\frac{3}{4})]}^2+{[\sin ({\tan }^{-1}\frac{4}{3})]}^2\\ & ={\left(\frac{3}{5}\right)}^2+{\left(\frac{4}{5}\right)}^2=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1\end{array}$

Given plane,

$2𝑥+3𝑦+5𝑧=1$ intersect the co-ordinate axes at $𝐴,𝐵,𝐶$

$\therefore$ Equation of plane in intercept form is,

$\begin{array}{cc}& \frac{𝑥}{1/2}+\frac{𝑦}{1/3}+\frac{𝑧}{1/5}=1\\ & \therefore 𝐴(\frac{1}{2},0,0),𝐵(0,\frac{1}{3},0),𝐶(0,0,\frac{1}{5})\end{array}$

$\therefore$ Centroid of $△𝐴𝐵𝐶$ is $(\frac{1}{6},\frac{1}{9},\frac{1}{15})$

Given ellipse,

$\begin{array}{cc}& {𝑦}^2+4{𝑥}^2-12𝑥+6𝑦+14=0\\ & 4{𝑥}^2-12𝑥+{𝑦}^2+6𝑦+14=0\end{array}$

$\begin{array}{c}4({𝑥}^2-3𝑥+\frac{9}{4})+{𝑦}^2+6𝑦+9=-14+9+9\\ 4{(𝑥-\frac{3}{2})}^2+(𝑦+3{)}^2=4\\ \frac{(𝑥-3/2{)}^2}{1}+\frac{(𝑦+3{)}^2}{4}=1\end{array}$

Here, ${𝑎}^2=1,{𝑏}^2=4$

$\begin{array}{cc}\therefore & 𝑒=\sqrt{\surd 1-\frac{{𝑎}^2}{{𝑏}^2}}\\ & 𝑒=\sqrt{\surd 1-\frac{1}{4}}=\frac{\sqrt{\surd 3}}{2}\end{array}$

We have,

$𝑓\left(𝑥\right)=\left[𝑥\right]$

We know that, $𝑥=\left[𝑥\right]+\left\{𝑥\right\}$

Where $\left[𝑥\right]$ is greatest integer function and $\left\{𝑥\right\}$ is fractional part

$\begin{array}{cc}& =\left\{𝑥\right\}=𝑥-\left[𝑥\right]\\ & 0\le \left\{𝑥\right\}<1\\ \therefore & 0\le 𝑥-\left[𝑥\right]<1\\ \Rightarrow & \left[𝑥\right]\le 𝑥<\left[𝑥\right]+1\\ \Rightarrow & \left[𝑥\right]+1>𝑥\end{array}$

We have,

$\begin{array}{cc}& {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)\\ & ={\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3}\times \frac{1}{5}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\times \frac{1}{8}}\right)\\ & ={\tan }^{-1}\left(\frac{8}{14}\right)+{\tan }^{-1}\left(\frac{15}{55}\right)\\ & ={\tan }^{-1}\left(\frac{4}{7}\right)+{\tan }^{-1}\left(\frac{3}{11}\right)\\ & ={\tan }^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7}\times \frac{3}{11}}\right)\\ & ={\tan }^{-1}\left(\frac{65}{65}\right)={\tan }^{-1}1=\frac{𝜋}{4}\end{array}$

We have,

$\begin{array}{cc}& \sin 𝑦\frac{𝑑𝑦}{𝑑𝑥}=\cos 𝑦(1-𝑥\cos 𝑦)\\ & \tan 𝑦\frac{𝑑𝑦}{𝑑𝑥}=1-𝑥\cos 𝑦\\ & \tan 𝑦\frac{𝑑𝑦}{𝑑𝑥}-1=-𝑥\cos 𝑦\end{array}$

$\begin{array}{cc}& \sec 𝑦\tan 𝑦\frac{𝑑𝑦}{𝑑𝑥}-\sec 𝑦=-𝑥\\ \text{ Put, }& \sec 𝑦=𝑧\\ \therefore & \sec 𝑦\tan 𝑦\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑑𝑧}{𝑑𝑥}\\ \therefore & \frac{𝑑𝑧}{𝑑𝑥}-𝑧=-𝑥\\ \therefore & \text{ IF }={𝑒}^{\int -𝑑𝑥}={𝑒}^{-𝑥}\end{array}$

A point lie on the line which bisects the angle between positive direction of $𝑌$ and $𝑍$ axes on $(0,𝑎,𝑎)$

Direction ratio of line passing through origin and point $(0,𝑎,𝑎)$ is $(0,𝑎,𝑎)$

Direction cosine of line is

$\begin{array}{cc}& \frac{0}{\sqrt{\surd 0+{𝑎}^2+{𝑎}^2}},\frac{𝑎}{\sqrt{\surd 2{𝑎}^2}},\frac{𝑎}{\sqrt{\surd 2{𝑎}^2}}\\ & =0,\frac{1}{\sqrt{\surd 2}},\frac{1}{\sqrt{\surd 2}}\end{array}$

Given curve,

$\begin{array}{cc}𝑌& =\sin \left(\frac{𝜋}{4}𝑥\right)\\ \frac{𝑑𝑦}{𝑑𝑥}& =\frac{𝜋}{4}\cos \left(\frac{𝜋}{4}𝑥\right)\\ {\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}_{𝑥=2}& =\frac{𝜋}{4}\cos \left(\frac{𝜋}{2}\right)=0\end{array}$

$\begin{array}{cc}& \text{ Equation of normal of curve at }(2,5)\text{ is, }\\ & \begin{array}{c}𝑥-2=0\\ 𝑥=2\end{array}\end{array}$

The negation of the statement "He is poor but happy" is "He is not poor or not happy"

$\begin{array}{cc}& \text{ We have, }\\ & 𝐴=⎡⎣⎢\begin{array}{ccc}𝑎& -1& 4\\ -3& 0& 1\\ -1& 1& 2\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& 𝐴\text{ is not invertible }\\ & \begin{array}{cc}& \therefore \left|𝐴\right|=0\\ & \mid \begin{array}{ccc}𝑎& -1& 4\\ -3& 0& 1\\ -1& 1& 2\end{array}\mid =0\end{array}\end{array}$

$\begin{array}{c}𝑎(0-1)+1(-6+1)+4(-3+0)=0\\ \Rightarrow -𝑎-5-12=0\\ \Rightarrow 𝑎=-17\end{array}$

Given,

Maximize $𝑧=3𝑥+5𝑦$

Subject to constraints

$\begin{array}{cc}& 3𝑥+2𝑦\le 18\\ & 𝑥\le 4,𝑦\le 6,𝑥,𝑦\ge 0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\text{ Corner point }& 𝑧=3𝑥+5𝑦\\ 𝐴(4,0)& 12\\ 𝐵(4,3)& 27\\ 𝐶(2,6)& 36\\ 𝐷(0,6)& 30\end{array}\\ & \text{ Maximum value of }𝑧\text{ is }36\text{. }\end{array}$

$𝐼=\int {\left(\frac{\log 𝑥-1}{1+(\log 𝑥{)}^2}\right)}^2𝑑𝑥$

$\text{ Put, }\log 𝑥=𝑡\Rightarrow 𝑥={𝑒}^{𝑡}\Rightarrow 𝑑𝑥={𝑒}^{𝑡}𝑑𝑡$

$\begin{array}{cc}& \begin{array}{cc}\therefore 𝐼& =\int \frac{(𝑡-1{)}^2}{{(1+{𝑡}^2)}^2}{𝑒}^{𝑡}𝑑𝑡\\ 𝐼& =\int {𝑒}^{𝑡}\left(\frac{1+{𝑡}^2-2𝑡}{{(1+{𝑡}^2)}^2}\right)𝑑𝑡\\ 𝐼& =\int {𝑒}^{𝑡}(\frac{1}{1+{𝑡}^2}-\frac{2𝑡}{{(1+{𝑡}^2)}^2})𝑑𝑡\end{array}\\ & \text{ Here, }𝑓\left(𝑡\right)=\frac{1}{1+{𝑡}^2}\text{ or }{𝑓}^{\prime }\left(𝑡\right)=\frac{-2𝑡}{{(1+{𝑡}^2)}^2}\end{array}$

$\begin{array}{cc}\therefore 𝐼=\frac{{𝑒}^{𝑡}}{1+{𝑡}^2}+& 𝑐\\ & [\because \int {𝑒}^{𝑥}\left(𝑓\right(𝑥)+{𝑓}^{\prime }(𝑥)𝑑𝑥={𝑒}^{𝑥}𝑓(𝑥)+𝑐]\end{array}$

$\Rightarrow 𝐼=\frac{𝑥}{1+(\log 𝑥{)}^2}+𝑐$

We have,

$\begin{array}{cc}& 𝐼=\int \frac{𝑑𝑥}{\cos 2𝑥-{\cos }^2𝑥}\\ & 𝐼=\int \frac{𝑑𝑥}{2{\cos }^2𝑥-1-{\cos }^2𝑥}\\ & 𝐼=\int \frac{𝑑𝑥}{-{\sin }^2𝑥}\\ & 𝐼=-\int {\mathrm{cosec}}^2𝑥𝑑𝑥\\ & 𝐼=\cot 𝑥+𝑐\end{array}$

We know that, area included between ${𝑥}^2=𝑎𝑦$ and ${𝑦}^2=𝑎𝑥$ is $\frac{{𝑎}^2}{3}$ sq units

$\therefore$ Area included between ${𝑥}^2=5𝑦$ and ${𝑦}^2=5𝑥$ is $\frac{(5{)}^2}{3}=\frac{25}{3}$ sq units.

Given line,

$\begin{array}{cc}& \begin{array}{cc}& 9{𝑥}^2-12𝑥𝑦+4{𝑦}^2=0\\ \Rightarrow & (3𝑥-2𝑦{)}^2=0\\ \Rightarrow & 3𝑥-2𝑦=0\text{ or }3𝑥-2𝑦=0\end{array}\\ & \therefore \text{ both lines are coincident }\end{array}$

In a playing of 52 cards

4 are king and 48 are not king

$\begin{array}{cc}\therefore \text{ odd in favour of king }& =\frac{\text{ Favour in king }}{\text{ Not favour in king }}\\ & =\frac{4}{48}=\frac{1}{12}\\ \therefore & =1:12\end{array}$

$\text{ Let }𝐼={\int }_{3𝜋/10}^{𝜋/5}\frac{\tan 𝑥}{\tan 𝑥+\cot 𝑥}𝑑𝑥$

$𝐼={\int }_{3𝜋/10}^{𝜋/5}\frac{\tan (\frac{3𝜋}{10}+\frac{𝜋}{5}-𝑥)}{\tan (\frac{3𝜋}{10}+\frac{𝜋}{5}-𝑥)+\cot (\frac{3𝜋}{10}+\frac{𝜋}{5}-𝑥)}𝑑𝑥$

$𝐼={\int }_{3𝜋/10}^{𝜋/5}\frac{\cot 𝑥}{\cot 𝑥,\tan 𝑥}$

$\begin{array}{cc}2𝐼& ={\int }_{3𝜋/10}^{𝜋/5}\left(\frac{\tan 𝑥+\cot 𝑥}{\tan 𝑥+\cot 𝑥}\right)𝑑𝑥={\int }_{3𝜋/10}^{𝜋/5}𝑑𝑥\\ 𝐼& =\frac{1}{2}(\frac{3𝜋}{10}-\frac{𝜋}{5})=\frac{𝜋}{20}\end{array}$

Given,

$\begin{array}{cc}\frac{𝑥-1}{4}& =\frac{𝑦-3}{1}=\frac{𝑧}{8}\\ \text{ and }\frac{𝑥-2}{2}& =\frac{𝑦+1}{2}=\frac{𝑧-4}{1}\\ \text{ Here, }{𝑎}_1& =4,{𝑏}_1=1,{𝑐}_1=8\\ {𝑎}_2& =2,{𝑏}_2=2,{𝑐}_2=1\end{array}$

$\begin{array}{cc}\therefore & \cos 𝜃=\frac{{𝑎}_1{𝑎}_2+{𝑏}_1{𝑏}_2+{𝑐}_1{𝑐}_2}{\left(\sqrt{\surd {𝑎}_2^1+{𝑏}_2^1+{𝑐}_2^1}\right)\left(\sqrt{\surd {𝑎}_2^2+{𝑏}_2^2+{𝑐}_2^2}\right)}\\ \Rightarrow & \cos 𝜃=\frac{8+2+8}{\left(\sqrt{\surd 16+1+64}\right)\left(\sqrt{\surd 4+4+1}\right)}\\ \Rightarrow & \cos 𝜃=\frac{18}{9\times 3}=\frac{2}{3}\\ \Rightarrow & 𝜃={\cos }^{-1}\left(\frac{2}{3}\right)\end{array}$

We have,

$\begin{array}{cc}& 2\cos 𝐶=\sin 𝐵\mathrm{cosec}𝐴\\ & \Rightarrow 2\sin 𝐴\cos 𝐶=\sin 𝐵\\ & \Rightarrow \sin (𝐴+𝐶)+\sin (𝐴-𝐶)=\sin 𝐵\\ & \Rightarrow \sin (𝜋-𝐵)+\sin (𝐴-𝐶)=\sin 𝐵[\because 𝐴+𝐵+𝐶=𝑥]\\ & \Rightarrow \sin 𝐵+\sin (𝐴-𝐶)=\sin 𝐵\\ & \Rightarrow \sin (𝐴-𝐶)=0\end{array}$

$\begin{array}{cc}𝐴& =𝐶\\ \therefore \angle 𝐴& =\angle 𝐶\\ \therefore 𝑎& =𝑐\end{array}$

Given differential equation,

$\begin{array}{cc}& {[1+{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^3]}^{7/3}=7\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\\ & {(1+{\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^3)}^7=7^3{\left(\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}\right)}^3\end{array}$

Clearly order $=2$ degree $=3$

Let $𝑇$ be temperature of metal at any time $𝑡$ and ${𝑇}_0$ is temperature of air

According to given condition,

$\frac{𝑑𝑇}{𝑑𝑡}=𝑘(𝑇-{𝑇}_0)$

$\begin{array}{cccc}& & \frac{𝑑𝑇}{𝑇-{𝑇}_0}& =𝑘𝑑𝑡\\ \Rightarrow & & \int \frac{𝑑𝑇}{𝑇-{𝑇}_0}& =𝑘\int 𝑑𝑡\\ \Rightarrow & & \log (𝑇-{𝑇}_0)& =𝑘𝑡+𝐶\\ \text{ Here, }& & {𝑇}_0& =290\end{array}$

$\therefore \log (𝑇-290)=𝑘𝑡+𝐶$

When, $𝑡=0,𝑇=370$

When, $𝑡=10,𝑇=330$

$\begin{array}{cc}& \therefore 𝐶=\log 80\\ & \log \left(\frac{𝑇-290}{80}\right)=𝑘𝑡\end{array}$

$\begin{array}{cc}& 𝑘=\frac{1}{10}\log \left(\frac{1}{2}\right)\\ & \therefore \log \left(\frac{𝑇-290}{80}\right)=\frac{𝑡}{10}\log \frac{1}{2}\end{array}$

$\begin{array}{cc}& \text{ When, }\\ & \begin{array}{cc}\therefore & \log \frac{1}{16}=\frac{𝑡}{10}\log \frac{1}{2}\\ \Rightarrow & 10\times 4\log \frac{1}{2}=𝑡\log \frac{1}{2}\\ \Rightarrow & 𝑡=40\end{array}\end{array}$

We have,

$\begin{array}{cc}& \begin{array}{cc}& 𝐴=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\\ \text{ and }& 𝐵=\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]\end{array}\end{array}$

$\begin{array}{cc}𝐴𝐵& =\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 3& 1\end{array}\right]\\ 𝐴𝐵& =\left[\begin{array}{cc}11& 3\\ 7& 2\end{array}\right]\\ \left|𝐴𝐵\right|& =22-21=1\\ \therefore (𝐴𝐵{)}^{-1}& =\left[\begin{array}{cc}2& -3\\ -7& 11\end{array}\right]\end{array}$

$\begin{array}{cc}& \therefore \text{ We know that, }\\ & \begin{array}{cc}& {𝐵}^{-1}{𝐴}^{-1}=(𝐴𝐵{)}^{-1}\\ \therefore & {𝐵}^{-1}{𝐴}^{-1}=\left[\begin{array}{cc}2& -3\\ -7& 11\end{array}\right]\end{array}\end{array}$

$\begin{array}{cc}& 𝑓\left(𝑥\right)\text{ is defined when }𝑥\ge 0\\ & \therefore 𝑥\in [0,\infty )\\ & \therefore \text{ Division of }𝑓\left(𝑥\right)={𝑅}^{+}\cup \left\{0\right\}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \sin (𝑥+𝑦)+\cos (𝑥+𝑦)=\sin [{\cos }^{-1}\left(\frac{1}{3}\right)]\end{array}$

$\begin{array}{cc}& {\text{ on differentiating w.r.t. }}^{\prime }{𝑥}^{\prime }\text{, we get }\\ & \begin{array}{c}\cos (𝑥+𝑦)(1+\frac{𝑑𝑦}{𝑑𝑥})-\sin (𝑥+𝑦)(1+\frac{𝑑𝑦}{𝑑𝑥})=0\\ \Rightarrow (1+\frac{𝑑𝑦}{𝑑𝑥})(\cos (𝑥+𝑦)-\sin (𝑥+𝑦\left)\right)=0\\ \Rightarrow 1+\frac{𝑑𝑦}{𝑑𝑥}=0\\ \Rightarrow \frac{𝑑𝑦}{𝑑𝑥}=-1\end{array}\end{array}$

Let the number of micro-organism in initial is $𝑛$

Here, $𝑛$ will be $2𝑛$ in 3 yr

$\therefore$ Number of micro-organism in 18 yr

$=\left(2^6\right)𝑛=64𝑛$

$\therefore 64$ time of micro-organism in 18 yr .

$\begin{array}{cc}& \text{ Here, }𝑛=5\\ & \begin{array}{cc}𝑟& =0,1\\ 𝑝& =\frac{10}{100}=\frac{1}{10}\\ 𝑞& =1-\frac{1}{10}=\frac{9}{10}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Required probability, }\\ & \begin{array}{cc}𝑃(𝑋\le 1)& =𝑃(𝑋=0)+𝑃(𝑋=1)\\ & ={}^5{𝐶}_0{\left(\frac{9}{10}\right)}^5+{}^5{𝐶}_{(}{\left(\frac{1}{10}\right)}^1{\left(\frac{9}{10}\right)}^4\\ & ={\left(\frac{9}{10}\right)}^5+\frac{5\times 9^4}{{10}^5}\\ & =\frac{9^4}{{10}^5}(9+5)=\frac{9^4\times 14}{{10}^5}\\ & =\frac{91854}{100000}=0.91854=0.9185\end{array}\end{array}$

Given line,

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

is parallel to plane $𝐫\cdot (3\widehat{𝐢}-2\widehat{𝐣}+𝑚\widehat{𝐤})=0$

We know that, if line $𝐫=𝐚+𝜆𝐛$ is parallel to plane $𝐫\cdot 𝐧=𝑑$

Then, $𝐛\cdot 𝐧=\mathrm{𝟎}$

$\begin{array}{cc}\therefore (2\widehat{𝐢}+\widehat{𝐣}+2\widehat{𝐤})\cdot (3\widehat{𝐢}-2\widehat{𝐣}+𝑚\widehat{𝐤})& =0\\ 6-2+2𝑚& =0\\ 𝑚& =-2\end{array}$

Given,

$𝐺$ be the centroid of a triangle $𝐴𝐵𝐶$ and $𝑂$ be any other point in that plane

$\begin{array}{cc}\therefore & 𝑂𝐺=\frac{𝑂𝐴+𝑂𝐵+𝑂𝐶}{3}\\ \Rightarrow & 𝑂𝐴+𝑂𝐵+𝑂𝐶=3𝑂𝐺\\ \therefore & 𝑂𝐴+𝑂𝐵+𝑂𝐶+𝑂𝐺\\ \Rightarrow & 3𝑂𝐺+𝑂𝐺=4𝑂𝐺\end{array}$

We have,

$𝐴(-𝑎,-𝑏),𝐵(0,0),𝐶(𝑎,𝑏)$ and $𝐷({𝑎}^2,𝑎𝑏)$

Slope of $𝐴𝐵=\frac{𝑏}{𝑎}$

Slope of $𝐵𝐶=𝑏/𝑎$

Slope of $𝐶𝐷=𝑏/𝑎$

$\therefore$ Slope of $𝐴𝐵=$

Slope of $𝐵𝐶=$ slope of $𝐶𝐷$

Hence, $𝐴,𝐵,𝐶,𝐷$ are collinears.

We have,

$\begin{array}{cc}& 𝑦\frac{𝑑𝑥}{𝑑𝑦}=𝑥\log 𝑥\\ \Rightarrow & \frac{𝑑𝑦}{𝑦}=\frac{𝑑𝑥}{𝑥\log 𝑥}\end{array}$

On integrating both sides, we get

$\begin{array}{cc}& \log 𝑦=\log (\log 𝑥)+\log 𝑐\\ & \Rightarrow 𝑦=𝑐\log 𝑥\end{array}$

When, $𝑥=𝑒\text{ and }𝑦=1\Rightarrow 𝑐=1$

$\begin{array}{c}\therefore 𝑦=\log 𝑥\\ 𝑥={𝑒}^{𝑦}\end{array}$