MHT CET 2024 10th May Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}𝑀& =\left|\stackrel{\to \leftarrow }{\mathrm{OP}}\right|\\ 𝑀& =\hat{a}\cos 𝑡+\widehat{𝑏}\sin 𝑡\\ & =\sqrt{\surd (\widehat{𝑎}\cos 𝑡{)}^2+(\widehat{𝑏}\sin 𝑡{)}^2+2(\widehat{𝑎}\cos 𝑡)\cdot (\widehat{𝑏}\cdot \sin 𝑡)}\\ & =\sqrt{\surd {\sin }^2𝑡+{\cos }^2𝑡+\widehat{𝑎}\cdot \widehat{𝑏}(2\sin 𝑡\cos 𝑡)}\\ & =\sqrt{\surd 1+\widehat{𝑎}\cdot \widehat{𝑏}(\sin 2𝑡)}\end{array}$

Maximum value of $\sin 2𝑡=1$

$\begin{array}{cc}\therefore & 2t={\sin }^{-1}\left(1\right)\\ \therefore & t=\frac{𝜋}{4}\\ \therefore & 𝑀=\sqrt{\surd 1+\widehat{𝑎}\cdot \widehat{𝑏}\left(1\right)}=(1+\widehat{𝑎}\cdot \widehat{𝑏}{)}^{\frac{1}{2}}\end{array}$

Now, $\hat{u}=\frac{\overline{\mathrm{OP}}}{\left|\overline{\mathrm{OP}}\right|}$

$\begin{array}{cc}& =\frac{\widehat{𝑎}\cos 𝑡+\widehat{𝑏}\sin 𝑡}{|\widehat{𝑎}\sin 𝑡+\widehat{𝑏}\sin 𝑡|}=\frac{\widehat{𝑎}\left(\frac{1}{\sqrt{\surd 2}}\right)+\widehat{𝑏}\left(\frac{1}{\sqrt{\surd 2}}\right)}{\mid \widehat{𝑎}\left(\frac{1}{\sqrt{\surd 2}}\right)+\widehat{𝑏}\left(\frac{1}{\sqrt{\surd 2}}\right)\mid }\\ & \text{ Unit vector of OP is }\\ & \widehat{𝑢}=\frac{\widehat{𝑎}+\widehat{𝑏}}{\left|\right(\widehat{𝑎}+\widehat{𝑏}\left)\right|}\end{array}$

Let $({𝑥}_1,{𝑦}_1,{𝑧}_1)=(1,2,-1)$, $a_1,b_1,c_1=2,1,-1$ and $a_2,b_2,c_2=1,-1,3$

$\therefore$ the equation of required plane is

$\begin{array}{cc}& \mid \begin{array}{ccc}𝑥-{𝑥}_1& 𝑦-{𝑦}_1& 𝑧-{𝑧}_1\\ {𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\end{array}\mid =0\\ & \Rightarrow \mid \begin{array}{ccc}𝑥-1& 𝑦-2& 𝑧+1\\ 2& 1& -1\\ 1& -1& 3\end{array}\mid =0\\ & \Rightarrow 2𝑥-2-7𝑦+14-3𝑧-3=0\\ & \Rightarrow 2𝑥-7𝑦-3𝑧+9=0\\ & \Rightarrow \bar{r}\cdot (2\hat{i}-7\hat{j}-3\hat{k})=-9\end{array}$

Given that:

$m\angle A={45}^{\circ },m\angle B={75}^{\circ },m\angle C={60}^{\circ }$

we know that

$\begin{array}{cc}& \frac{\sin 𝐴}{𝑎}=\frac{\sin 𝐵}{𝑏}=\frac{\sin 𝐶}{𝑐}\\ & \frac{\sin {45}^{\circ }}{𝑎}=\frac{\sin {75}^{\circ }}{𝑏}=\frac{\sin {60}^{\circ }}{𝑐}\\ \therefore & \frac{\sin {45}^{\circ }}{𝑎}=\frac{\sin ({45}^{\circ }+{30}^{\circ })}{𝑏}\end{array}$

$\therefore \frac{1}{\sqrt{\surd 2}𝑎}=\frac{\frac{1}{\sqrt{\surd 2}}\times \frac{1}{2}+\frac{1}{\sqrt{\surd 2}}\times \frac{\sqrt{\surd 3}}{2}}{𝑏}$

$\begin{array}{cc}\therefore & \frac{1}{\sqrt{\surd 2}𝑎}=\frac{1+\sqrt{\surd 3}}{2\sqrt{\surd 2}𝑏}\\ \therefore & 𝑎=\frac{2𝑏}{1+\sqrt{\surd 3}}\\ & \frac{\sin {60}^{\circ }}{𝑐}=\frac{\sin ({45}^{\circ }+{30}^{\circ })}{𝑏}\\ \therefore & \frac{\sqrt{\surd 3}}{2𝑐}=\frac{1+\sqrt{\surd 3}}{2\sqrt{\surd 2}𝑏}\\ \therefore & \sqrt{\surd 2}𝑐=\frac{2\sqrt{\surd 3}𝑏}{(1+\sqrt{\surd 3})}\\ \therefore & 𝑎+\sqrt{\surd 2}𝑐=\frac{2𝑏}{1+\sqrt{\surd 3}}+\frac{2\sqrt{\surd 3}𝑏}{1+\sqrt{\surd 3}}=2𝑏\end{array}$

$\log (𝑥+𝑦)=2𝑥𝑦\text{.... (i)}$

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

$\begin{array}{cc}& \left(\frac{1}{𝑥+𝑦}\right)(1+\frac{d𝑦}{d𝑥})=2(𝑥\frac{d𝑦}{d𝑥}+𝑦)\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{1-2𝑥𝑦-2{𝑦}^2}{2{𝑥}^2+2𝑥𝑦-1}\end{array}$

Putting $𝑥=0$ in (i), we get

$\begin{array}{cc}& 𝑦=1\\ \therefore & {𝑦}^{\prime }\left(0\right)=\frac{1-0-2}{0+0-1}=1\end{array}$

$\begin{array}{cc}& \left|\bar{a}\right|=1,\left|\bar{b}\right|=2,\left|\bar{c}\right|=3\\ & [\bar{a}+\bar{b}+\bar{c}\bar{b}-\bar{a}\bar{c}]\\ & =(\bar{a}+\bar{b}+\bar{c})\cdot \left[\right(\bar{b}-\bar{a})\times \bar{c}]\\ & =(\bar{a}+\bar{b}+\bar{c})\cdot (\bar{b}\times \bar{c}-\bar{a}\times \bar{c})\\ & =\left[\bar{a}\bar{b}\bar{c}\right]-\left[\bar{b}\bar{a}\bar{c}\right]\\ & =2\left[\bar{a}\bar{b}\bar{c}\right]\\ & =2\bar{a}\cdot (\bar{b}\times \bar{c})\\ & =2\left|\bar{a}\right||\bar{b}\times \bar{c}|\cos 0^{\circ }\cdots \left[\begin{array}{c}\because \bar{a}⟂\bar{b}\text{ and }\bar{c}\\ \therefore \bar{a}\Vert \bar{b}\times \bar{c}\end{array}\right]\\ & =2\left|\bar{a}\right||\bar{b}\times \bar{c}|\\ & =2\left|\bar{a}\right|\left|\bar{b}\right|\left|\bar{c}\right|\sin {90}^{\circ }\cdots [\because \bar{b}⟂\bar{c}]\\ & =2\times 1\times 2\times 3\\ & =12\end{array}$

$\begin{array}{cc}& V=\frac{4}{3}𝜋r^3\\ & \Rightarrow 288𝜋=\frac{4}{3}𝜋r^3\\ & \Rightarrow r=6\mathrm{cm}\end{array}$

$\begin{array}{cc}& 𝑉=\frac{4}{3}𝜋{𝑟}^3\\ & \frac{𝑑𝑉}{𝑑𝑡}=4𝜋{𝑟}^2\frac{𝑑𝑟}{𝑑𝑡}\\ & \Rightarrow 4𝜋=4𝜋{𝑟}^2\frac{𝑑𝑟}{𝑑𝑡}\\ & \Rightarrow \frac{𝑑𝑟}{𝑑𝑡}=\frac{1}{{𝑟}^2}=\frac{1}{36}\mathrm{cm}/\sec \end{array}$

$\begin{array}{cc}& \text{ Let }I={\int }_{\frac{𝜋}{4}}^0\log \left(\frac{\sin 𝑥+\cos 𝑥}{\cos 𝑥}\right)\mathrm{dx}\\ & \begin{array}{cc}& ={\int }_{\frac{𝜋}{4}}^0\log (1+\tan 𝜃)d𝜃\\ & ={\int }_{\frac{𝜋}{4}}^0\log [1+\tan (\frac{𝜋}{4}-𝜃)]𝑑𝜃\\ & \dots .[\because {\int }_{𝑎}^{0}f(𝑥)d𝑥={\int }_1^{0}f(a-𝑥\left)d𝑥\right]\end{array}\end{array}$

$\begin{array}{cc}& ={\int }_{\frac{𝜋}{4}}^0\log (1+\frac{1-\tan 𝜃}{1+\tan 𝜃})𝑑𝜃\\ & ={\int }_{\frac{𝜋}{4}}^0\log 2𝑑𝜃-{\int }_{\frac{𝜋}{4}}^0\log (1+\tan 𝜃)𝑑𝜃\\ \therefore 2𝐼& ={\int }_{\frac{𝜋}{4}}^0\log 2𝑑𝜃\Rightarrow 𝐼=\frac{\log 2}{2}[𝜃{]}_{𝜋/4}^0=\frac{𝜋}{8}\log 2\end{array}$

$\begin{array}{cc}& \text{ Given that }P\left(A\right)=\frac{3}{10},P\left(B\right)=\frac{2}{5}\\ & \begin{array}{cc}\therefore & 𝑃\left({𝐴}^{\prime }\right)=\frac{7}{10}\\ & 𝑃({𝐴}^{\prime }\cup 𝐵)\\ & =𝑃\left({𝐴}^{\prime }\right)+𝑃\left(𝐵\right)-𝑃({𝐴}^{\prime }\cap 𝐵)\\ & =𝑃\left({𝐴}^{\prime }\right)+𝑃\left(𝐵\right)-𝑃\left({𝐴}^{\prime }\right)\left(𝐵\right)\end{array}\end{array}$

$\dots [\because A$ and B are independent, $A^{\prime }$ and B are also independent]

$\begin{array}{cc}& =\frac{7}{10}+\frac{2}{5}-\frac{7}{10}\times \frac{2}{5}\\ & =\frac{41}{50}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Let }𝑦=(𝑥-1)(𝑥-2)(𝑥-3)\\ & \log 𝑦=\log (𝑥-1)+\log (𝑥-2)+\log (𝑥-3)\end{array}\\ & \text{ Differentiating w.r.t. }𝑥\text{, we get }\\ & \begin{array}{cc}& \frac{1}{𝑦}\frac{d𝑦}{d𝑥}=\frac{1}{𝑥-1}+\frac{1}{𝑥-2}+\frac{1}{𝑥-3}\\ \therefore & f^{\prime }\left(𝑥\right)=\frac{d𝑦}{d𝑥}=(𝑥-2)(𝑥-3)+(𝑥-1)(𝑥-3)\\ & +(𝑥-1)(𝑥-2)\end{array}\end{array}$

$\begin{array}{cc}\therefore & f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{6-(-6)}{4}=3,\text{ for }c\in [0,4]\\ \therefore & (c-2)(c-3)+(c-1)(c-3)+(𝑐-1)(𝑐-2)=3\\ \therefore & 3c^2-12c+11=3\\ \therefore & 3c^2-12c+8=0\\ \therefore & c=\frac{12\pm \sqrt{\surd 144-96}}{6}=\frac{6\pm 2\sqrt{\surd 3}}{3}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& e^{𝑦-𝑥}\frac{d𝑦}{d𝑥}=𝑦\left(\frac{\sin 𝑥+\cos 𝑥}{1+𝑦\log 𝑦}\right)\\ \therefore & \frac{e^{𝑦}}{e^{𝑥}}\frac{d𝑦}{d𝑥}=\frac{𝑦}{(1+𝑦\log 𝑦)}(\sin 𝑥+\cos 𝑥)\\ \therefore & e^{𝑦}\frac{(1+𝑦\log 𝑦)}{𝑦}d𝑦=e^{𝑥}(\sin 𝑥+\cos 𝑥)d𝑥\\ \therefore & e^{𝑦}(\log 𝑦+\frac{1}{𝑦})d𝑦=e^{𝑥}(\sin 𝑥+\cos 𝑥)d𝑥\end{array}\\ & \text{ Integrating both sides, we get }\\ & \begin{array}{cc}& e^{𝑦}\log 𝑦=e^{𝑥}\sin 𝑥+c\\ & \dots [\because \int e^{𝑥}\left[f\right(𝑥)+f^{\prime }(𝑥\left)\right]d𝑥=e^{𝑥}f(𝑥)+c]\end{array}\end{array}$

Let the coin be tossed ' $𝑛$ ' number of times.

Probability of getting head is $𝑝=\frac{1}{2}$

$\begin{array}{cc}\therefore & q=1-\frac{1}{2}=\frac{1}{2}\\ & P(X\ge 1)>\frac{99}{100}\\ \therefore & 1-P(X=0)>\frac{99}{100}\\ \therefore & 1-{\left(\frac{1}{2}\right)}^n>\frac{99}{100}\\ \therefore & {\left(\frac{1}{2}\right)}^n<\frac{1}{100}\\ \therefore & 100<2^n\end{array}$

$\therefore$ Minimum value of $𝑛$ is 7.

Let the required vector be $\overline{𝑟}=𝑥\widehat{𝑖}+𝑦\widehat{𝑗}+𝑧\widehat{𝑘}$.

Then, $\left|\widehat{𝐻}\right|=6$

$\Rightarrow {𝑥}^2+{𝑦}^2+{𝑧}^2=36\text{.... (i)}$

Now, $\overline{𝑟}$ is perpendicular to vectors $\overline{𝑎}=2\widehat{𝑖}+\widehat{𝑗}-3\widehat{𝑘}$ and $\overline{𝑏}=\widehat{𝑖}-2\widehat{𝑗}+\widehat{𝑘}$.

$\therefore \bar{a}\times \bar{b}=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -3\\ 1& -2& 1\end{array}\mid =-5(\hat{i}+\hat{j}+\hat{k})$

$\therefore 𝑥=𝑦=𝑧$

Let $𝑥=𝑦=𝑧=𝜆$

Let $𝑥=𝑦=𝑧=𝜆$

From (i), we get

$\begin{array}{cc}& 3{𝜆}^2=36\\ \therefore & 𝜆=2\sqrt{\surd 3}\\ \therefore & \text{ Required vector is }2\sqrt{\surd 3}(\widehat{𝑖}+\widehat{𝑗}+\widehat{𝑘}).\end{array}$

$\begin{array}{cc}& \bar{r}=(\hat{i}+2\hat{j}-\hat{k})+𝜆(2\hat{i}+\hat{j}-3\hat{k})\text{ and }\\ & \bar{r}=(2\hat{i}-\hat{j}+2\hat{k})+𝜇(\hat{i}-\hat{j}+\hat{k})\\ & \text{ Let }{a}_1=\hat{i}+2\hat{j}-\hat{k},{𝑎}_2=2\hat{i}-\hat{j}+2\hat{k}\\ & b_1=2\hat{i}+\hat{j}-3\hat{k},b_2=\hat{i}-\hat{j}+\hat{k}\\ & {\bar{a}}_2-{\bar{a}}_1=\hat{i}-3\hat{j}+3\hat{k}\\ & {\bar{b}}_1\times {\bar{b}}_2=-2\hat{i}-5\hat{j}-3\hat{k}\\ & \mid {\bar{b}}_1\times {\bar{b}}_2\mid =\sqrt{\surd 38}\end{array}$

$\begin{array}{cc}\text{ Shortest distance }& =\mid \frac{({\bar{b}}_1\times {\bar{b}}_2)\cdot ({\bar{a}}_2-{\bar{a}}_1)}{\mid {\bar{b}}_1\times {\bar{b}}_2\mid }\mid \\ & =\mid \frac{4}{\sqrt{\surd 38}}\mid =\mid \frac{2\sqrt{\surd 2}}{\sqrt{\surd 19}}\mid \text{ units }\end{array}$

$\begin{array}{cc}& \tan ({\cos }^{-1}\frac{1}{\sqrt{\surd 2}}+{\tan }^{-1}\frac{1}{2})\\ & \text{ Let }𝜃={\cos }^{-1}\frac{1}{\sqrt{\surd 2}}\text{ and }𝜙={\tan }^{-1}\frac{1}{2}\\ & \begin{array}{cc}& \therefore \cos 𝜃=\frac{1}{\sqrt{\surd 2}}\text{ and }\tan 𝜙=\frac{1}{2}\\ & \therefore \tan 𝜃=1\\ & \text{ Given expression }=\tan (𝜃+𝜙)\\ & =\frac{\tan 𝜃+\tan 𝜙}{1-\tan 𝜃\tan 𝜙}\\ & =\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3\end{array}\end{array}$

To find $\frac{d𝑦}{d𝑥}$ at $𝑥=1$ and $𝑦=2$, start by differentiating the given equation ${𝑥}^2{𝑦}^2={\sin }^{-1}𝑥+{\cos }^{-1}𝑥$ with respect to $𝑥$.

First, note the identity:

${\sin }^{-1}𝑥+{\cos }^{-1}𝑥=\frac{𝜋}{2}$

Thus, the equation simplifies to:

${𝑥}^2{𝑦}^2=\frac{𝜋}{2}$

Differentiate both sides with respect to $𝑥$:

$\frac{d}{d𝑥}\left({𝑥}^2{𝑦}^2\right)=0$

Apply the product rule to ${𝑥}^2{𝑦}^2$:

$2𝑥{𝑦}^2+{𝑥}^2\cdot 2𝑦\frac{d𝑦}{d𝑥}=0$

Simplify:

$2𝑥{𝑦}^2+2{𝑥}^2𝑦\frac{d𝑦}{d𝑥}=0$

Isolate $\frac{d𝑦}{d𝑥}$:

$2{𝑥}^2𝑦\frac{d𝑦}{d𝑥}=-2𝑥{𝑦}^2$

${𝑥}^2𝑦\frac{d𝑦}{d𝑥}=-𝑥{𝑦}^2$

Divide both sides by ${𝑥}^2𝑦$:

$\frac{d𝑦}{d𝑥}=-\frac{𝑥{𝑦}^2}{{𝑥}^2𝑦}=-\frac{𝑦}{𝑥}$

Evaluate at $𝑥=1$ and $𝑦=2$:

$\frac{d𝑦}{d𝑥}{\mid }_{𝑥=1,𝑦=2}=-\frac{2}{1}=-2$

Thus, the correct answer is $\overline{)-2}$.

Let's determine the vector $𝐛=({𝑏}_1,{𝑏}_2,{𝑏}_3)$ satisfying the two equations:

$𝐚\times 𝐛+𝐜=\mathrm{𝟎},$

$𝐚\cdot 𝐛=3.$

We are given:

$𝐚=\widehat{𝑗}-\widehat{𝑘}=(0,1,-1),$

$𝐜=\widehat{𝑖}-\widehat{𝑗}-\widehat{𝑘}=(1,-1,-1).$

Step 1. Expressing the cross product:

The cross product $𝐚\times 𝐛$ for $𝐚=(0,1,-1)$ and $𝐛=({𝑏}_1,{𝑏}_2,{𝑏}_3)$ is computed by:

$𝐚\times 𝐛=({𝑎}_{𝑦}{𝑏}_{𝑧}-{𝑎}_{𝑧}{𝑏}_{𝑦},{𝑎}_{𝑧}{𝑏}_{𝑥}-{𝑎}_{𝑥}{𝑏}_{𝑧},{𝑎}_{𝑥}{𝑏}_{𝑦}-{𝑎}_{𝑦}{𝑏}_{𝑥}).$

Plug in the values:

$\text{For }\widehat{𝑖}\text{-component: }1\cdot {𝑏}_3-(-1)\cdot {𝑏}_2={𝑏}_3+{𝑏}_2,$

$\text{For }\widehat{𝑗}\text{-component: }(-1)\cdot {𝑏}_1-0\cdot {𝑏}_3=-{𝑏}_1,$

$\text{For }\widehat{𝑘}\text{-component: }0\cdot {𝑏}_2-1\cdot {𝑏}_1=-{𝑏}_1.$

Thus,

$𝐚\times 𝐛=({𝑏}_3+{𝑏}_2,-{𝑏}_1,-{𝑏}_1).$

Step 2. Use the first given equation:

We have:

$𝐚\times 𝐛+𝐜=\mathrm{𝟎}\Rightarrow 𝐚\times 𝐛=-𝐜.$

Since

$-𝐜=(-1,1,1),$

equate the corresponding components:

${𝑏}_3+{𝑏}_2=-1,$

$-{𝑏}_1=1,$

$-{𝑏}_1=1.$

From the second equation, it follows immediately that:

${𝑏}_1=-1.$

Step 3. Use the second given equation:

We are given:

$𝐚\cdot 𝐛=3.$

Since $𝐚=(0,1,-1)$ and $𝐛=({𝑏}_1,{𝑏}_2,{𝑏}_3),$ the dot product is:

$𝐚\cdot 𝐛=0\cdot {𝑏}_1+1\cdot {𝑏}_2+(-1)\cdot {𝑏}_3={𝑏}_2-{𝑏}_3.$

Thus,

${𝑏}_2-{𝑏}_3=3.$

Step 4. Solve the system of equations for ${𝑏}_2$ and ${𝑏}_3$:

We have:

${𝑏}_2+{𝑏}_3=-1,$

${𝑏}_2-{𝑏}_3=3.$

Add the two equations:

$({𝑏}_2+{𝑏}_3)+({𝑏}_2-{𝑏}_3)=-1+3\Rightarrow 2{𝑏}_2=2\Rightarrow {𝑏}_2=1.$

Substitute ${𝑏}_2=1$ into ${𝑏}_2+{𝑏}_3=-1$:

$1+{𝑏}_3=-1\Rightarrow {𝑏}_3=-2.$

Step 5. Write the vector $𝐛$:

We found:

${𝑏}_1=-1,$

${𝑏}_2=1,$

${𝑏}_3=-2.$

Thus,

$𝐛=-\widehat{𝑖}+\widehat{𝑗}-2\widehat{𝑘}.$

This corresponds to Option A.

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

${𝑦}^2=𝑝{𝑥}^3+𝑞\text{.... (i)}$

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

$\begin{array}{cc}& 2𝑦\cdot \frac{d𝑦}{d𝑥}=3p{𝑥}^2\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\frac{3p}{2}\left(\frac{{𝑥}^2}{𝑦}\right)\\ \therefore & {\left(\frac{d𝑦}{d𝑥}\right)}_{(2,3)}=\frac{3p}{2}\times \frac{4}{3}=2p\end{array}$

Since the line touches the curve, their slopes are equal.

$\therefore 2𝑝=4\Rightarrow 𝑝=2$

Since $(2,3)$ lies on ${𝑦}^2=𝑝{𝑥}^3+𝑞$.

$\therefore 9=2\times 8+𝑞\Rightarrow 𝑞=-7$