MHT CET 2025 22nd April Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \text{ Let }{\cos }^{-1}\frac{1}{5\sqrt{\surd 2}}=𝜃\Rightarrow \cos 𝜃=\frac{1}{5\sqrt{\surd 2}}\\ & \Rightarrow \sin 𝜃=\frac{7}{5\sqrt{\surd 2}}\Rightarrow \tan 𝜃=7\\ & \text{ Let }{\sin }^{-1}\frac{4}{\sqrt{\surd 17}}=𝜙\Rightarrow \sin 𝜙=\frac{4}{\sqrt{\surd 17}}\\ & \Rightarrow \cos 𝜙=\frac{1}{\sqrt{\surd 17}}\Rightarrow \tan 𝜙=4\\ & \tan ({\cos }^{-1}\frac{1}{5\sqrt{\surd 2}}-{\sin }^{-1}\frac{4}{\sqrt{\surd 17}})\\ & =\tan (𝜃-𝜙)=\frac{\tan 𝜃-\tan 𝜙}{1+\tan 𝜃\tan 𝜙}\\ & =\frac{7-4}{1+28}=\frac{3}{29}\end{array}$

Given equation of line is

$\begin{array}{cc}& \bar{r}=(3𝜆-1)\hat{i}+(-2𝜆+3)\hat{j}+(4+𝜆)\hat{k}\\ & \Rightarrow \bar{r}=-\hat{i}+3\hat{j}+4\hat{k}+𝜆(3\hat{i}-2\hat{j}+\hat{k})...\left(𝑖\right)\end{array}$

$\therefore$ d.r.s. of the line are $3,-2,1$

The required line $\left(L\right)$ is parallel to the line (i).

Since $𝐿$ passes through $(1,2,3)$.

$\therefore$ Equation of L is

$\frac{𝑥-1}{3}=\frac{𝑦-2}{-2}=\frac{𝑧-3}{1}=𝜇\left(\text{ say }\right)$

$\therefore$ Any point on line $𝐿$ will be of the form

$\equiv (3𝜇+1,-2𝜇+2,𝜇+3)$

Consider option (D)

$\begin{array}{cc}& 3𝜇+1=-5\Rightarrow 𝜇=-2\\ & -2𝜇+2=6\Rightarrow 𝜇=-2\\ & 𝜇+3=2\Rightarrow 𝜇=-1\ne -2\end{array}$

$\therefore (-5,6,2)$ does not lie on the line L .

Given equation of plane is

$\begin{array}{cc}& \bar{r}=(\hat{i}-\hat{j})+𝜆(\hat{i}+\hat{j}+\hat{k})+𝜇(\hat{i}-2\hat{j}+3\hat{k})\\ & \text{ Vector normal to the plane }=\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -2& 3\end{array}\mid \\ & =\hat{i}(3+2)-\hat{j}(3-1)+\hat{k}(-2-1)\\ & =5\hat{i}-2\hat{j}-3\hat{k}\end{array}$

Since the plane passes through $(1,-1,0)$ and vector normal to the plane is $5\hat{i}-2\hat{j}-3\hat{k}$, the equation of plane is

$\begin{array}{cc}& 5(𝑥-1)-2(𝑦+1)-3(𝑧-0)=0\\ & \Rightarrow 5𝑥-2𝑦-3𝑧=7\end{array}$

$\therefore$ Distance from origin

$\begin{array}{cc}& =\frac{\left|𝑑\right|}{\sqrt{\surd {𝑎}^2+{𝑏}^2+{𝑐}^2}}\\ & =\frac{7}{\sqrt{\surd 25+4+9}}\\ & =\frac{7}{\sqrt{\surd 38}}\text{ units }\end{array}$

$\begin{array}{c}\text{ Here }{𝑎}_1=1,{𝑏}_1=2,{𝑐}_1=𝜆\\ {𝑎}_2=1,{𝑏}_2=2,{𝑐}_2=3\\ \sin 𝜃=\mid \frac{{𝑎}_1{𝑎}_2+{𝑏}_1{𝑏}_2+{𝑐}_1{𝑐}_2}{\sqrt{\surd {𝑎}_2^1+{𝑏}_2^1+{𝑐}_2^1}\sqrt{\surd {𝑎}_2^2+{𝑏}_2^2+{𝑐}_2^2}}\mid \\ =\frac{1+4+3𝜆}{\sqrt{\surd 1+4+{𝜆}^2}\sqrt{\surd 1+4+9}}\\ \therefore \sin 𝜃=\frac{5+3𝜆}{\sqrt{\surd 14}\cdot \sqrt{\surd 5+{𝜆}^2}}...\left(𝑖\right)\end{array}$

Given that $𝜃={\cos }^{-1}\sqrt{\surd \frac{5}{14}}$

$\begin{array}{cc}& \Rightarrow \cos 𝜃=\sqrt{\surd \frac{5}{14}}\Rightarrow \sin 𝜃=\frac{3}{\sqrt{\surd 14}}\\ & \therefore \frac{5+3𝜆}{\sqrt{\surd 14}\sqrt{\surd 5+{𝜆}^2}}=\frac{3}{\sqrt{\surd 14}}\dots \left[\text{ From (i) }\right]\\ & \Rightarrow 5+3𝜆=3\sqrt{\surd 5+{𝜆}^2}\\ & \Rightarrow 9{𝜆}^2+30𝜆+25=9(5+{𝜆}^2)\end{array}$

                                                               …[Squaring both sides]

$\begin{array}{cc}& \Rightarrow 30𝜆=20\\ & \Rightarrow 𝜆=\frac{2}{3}\end{array}$

$\begin{array}{cc}& \frac{𝑑}{𝑑𝑥}\{\log \left[𝑓({𝑒}^{𝑥}+2𝑥)\right]\}\\ & =\frac{1}{𝑓({𝑒}^{𝑥}+2𝑥)}{𝑓}^{\prime }({𝑒}^{𝑥}+2𝑥)\cdot ({𝑒}^{𝑥}+2)\\ & \frac{𝑑}{𝑑𝑥}{\{\log \left[𝑓({𝑒}^{𝑥}+2𝑥)\right]\}}_{𝑥=0}\\ & =\frac{1}{𝑓\left(1\right)}\cdot {𝑓}^{\prime }\left(1\right)\cdot (1+2)\\ & =\frac{1}{3}\cdot 2\cdot 3=2\end{array}$

$\begin{array}{cc}& \text{ Since }\frac{1}{6}\sin 𝜃,\cos 𝜃,\tan 𝜃\text{ are in G.P. }\\ & \begin{array}{cc}& \therefore {\cos }^2𝜃=\frac{1}{6}\sin 𝜃\cdot \tan 𝜃\\ & \Rightarrow {\cos }^2𝜃=\frac{{\sin }^2𝜃}{6\cos 𝜃}\\ & \Rightarrow 6{\cos }^3𝜃={\sin }^2𝜃\\ & \Rightarrow 6{\cos }^3𝜃=1-{\cos }^2𝜃\\ & \Rightarrow 6{\cos }^3𝜃+{\cos }^2𝜃-1=0\\ & \Rightarrow (\cos 𝜃-\frac{1}{2})(6{\cos }^2𝜃+4\cos 𝜃+2)=0\\ & \Rightarrow \cos 𝜃=\frac{1}{2}\\ & \Rightarrow 𝜃=2𝑛𝜋\pm \frac{𝜋}{3}\end{array}\end{array}$

$\begin{array}{cc}& \text{ By Lagrange’s mean value theorem, }\\ & \begin{array}{cc}& {𝑓}^{\prime }\left(𝑐\right)=\frac{𝑓\left(𝑏\right)-𝑓\left(𝑎\right)}{𝑏-𝑎}\\ & \Rightarrow {𝑓}^{\prime }\left(𝑥\right)=\frac{𝑓\left(5\right)-𝑓\left(1\right)}{5-1}\\ & \Rightarrow \frac{𝑓\left(5\right)-1}{4}\le 5\\ & \Rightarrow 𝑓\left(5\right)-1\le 20\\ & \Rightarrow 𝑓\left(5\right)\le 21\end{array}\end{array}$

$\begin{array}{cc}& \cot \frac{𝐴}{2}=\frac{𝑏+𝑐}{𝑎}\\ & \Rightarrow \frac{\cos \frac{𝐴}{2}}{\sin \frac{𝐴}{2}}=\frac{𝑘(\sin 𝐵+\sin 𝐶)}{𝑘\sin 𝐴}\cdots [𝐵𝑦\sin 𝑒𝑟𝑢𝑙𝑒]\\ & \Rightarrow \frac{\cos \frac{𝐴}{2}}{\sin \frac{𝐴}{2}}=\frac{2\sin \left(\frac{𝐵+𝐶}{2}\right)\cos \left(\frac{𝐵-𝐶}{2}\right)}{2\sin \frac{𝐴}{2}\cos \frac{𝐴}{2}}\end{array}$

$\begin{array}{cc}& \Rightarrow \cos \frac{𝐴}{2}=\frac{\sin (\frac{𝜋}{2}-\frac{𝐴}{2})\cos \left(\frac{𝐵-𝐶}{2}\right)}{\cos \frac{𝐴}{2}}\\ & \Rightarrow \cos \frac{𝐴}{2}=\cos \left(\frac{𝐵-𝐶}{2}\right)\\ & \Rightarrow \frac{𝐴}{2}=\frac{𝐵-𝐶}{2}\Rightarrow 𝐴=𝐵-𝐶...\left(𝑖\right)\end{array}$

Now, $A+B+C=𝜋$

$\Rightarrow B-C+B+C=𝜋\dots \left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]$

$\Rightarrow 2B=𝜋\Rightarrow B=\frac{𝜋}{2}$

$\begin{array}{cc}& \begin{array}{cc}& {\mathrm{AA}}^{-1}=I\\ & \Rightarrow \frac{1}{11}⎡⎣⎢\begin{array}{ccc}-1& 7& -24\\ 2& a& 4\\ 2& -3& 15\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}3& 3& 4\\ 2& -3& 4\\ b& -1& c\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥\\ & \Rightarrow \frac{1}{11}⎡⎣⎢\begin{array}{ccc}11-24b& 0& 24-24c\\ 6+2a+4b& 2-3a& 8+4a+4c\\ 15b& 0& -4+15c\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥\end{array}\\ & \text{ By equality of matrices, we get }\\ & \begin{array}{cc}& 15𝑏=0\Rightarrow 𝑏=0\\ & 2-3𝑎=11\Rightarrow 𝑎=-3\\ & 24-24𝑐=0\Rightarrow 𝑐=1\end{array}\end{array}$

p : If 7 is an odd number then 7 is divisible by 2 . Contrapositive of $𝑝$ is

$V_1$ : If 7 is not divisible by 2 then 7 is not an odd number.

Truth value of $V_1\equiv T\to F\equiv F$

q : If 7 is prime number then 7 is an odd number. Contrapositive of q is

$V_2$ : If 7 is not an odd number then 7 is not prime number.

Truth value of $V_2\equiv F\to F\equiv T$

$\begin{array}{cc}& \underset{𝑥\to 1}{lim}{({\log }_{3}3𝑥)}^{\log 𝑥}\\ & =\underset{𝑥\to 1}{lim}{\left(\frac{\log 3+\log 𝑥}{\log 3}\right)}^{\frac{\log 8}{\log 𝑥}}\\ & =\underset{𝑥\to 1}{lim}{(1+\frac{\log 𝑥}{\log 3})}^{\frac{\log 8}{\log 𝑥}}\\ & =\underset{𝑥\to 1}{lim}{\left[{(1+\frac{\log 𝑥}{\log 3})}^{\frac{\log 3}{\log 𝑥}}\right]}^{\frac{\log 8}{\log 3}}\end{array}$

$\begin{array}{cc}& =e^{\frac{\log 8}{\log 3}}\cdots \left[\begin{array}{c}\mathrm{as}𝑥\to 1,\log 𝑥\to 0,\\ \underset{𝑥\to 0}{lim}(1+𝑥{)}^{\frac{1}{𝑥}}=e\end{array}\right]\\ & =e^{{\log }_{3}8}\end{array}$

$\begin{array}{cc}& f\left(𝑥\right)={\sin }^4𝑥+{\cos }^4𝑥\\ & \Rightarrow f^{\prime }\left(𝑥\right)=4{\sin }^3𝑥\cos 𝑥-4{\cos }^3𝑥\sin 𝑥\\ & =-4\sin 𝑥\cos 𝑥({\cos }^2𝑥-{\sin }^2𝑥)\\ & =-2\sin 2𝑥\cos 2𝑥\\ & =-\sin 4𝑥\end{array}$

$\begin{array}{cc}& \text{ Function is increasing when }{f}^{\prime }\left(𝑥\right)>0\\ & \begin{array}{cc}& \Rightarrow -\sin 4𝑥>0\\ & \Rightarrow \sin 4𝑥<0\\ & \Rightarrow 𝜋<4𝑥<2𝜋\\ & \Rightarrow \frac{𝜋}{4}<𝑥<\frac{𝜋}{2}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given that }f\left(𝑥\right)=b{𝑥}^2+c𝑥+d\\ & \begin{array}{cc}& f(𝑥+1)-f\left(𝑥\right)=8𝑥+3\\ & \Rightarrow b(𝑥+1{)}^2+c(𝑥+1)+d-(b{𝑥}^2+c𝑥+d)=8𝑥+1\\ & \Rightarrow 2b𝑥+b+c=8𝑥+3\\ & \Rightarrow 2b=8\Rightarrow b=4\\ & \text{ Also, }b+c=3\Rightarrow c=-1\end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }𝐼=\int \frac{{𝑥}^3}{{𝑥}^4+5{𝑥}^2+4}𝑑𝑥\\ & \text{ Put }{𝑥}^2=𝑡\Rightarrow 2𝑥𝑑𝑥=𝑑𝑡\\ \therefore 𝐼& =\frac{1}{2}\int \frac{𝑡}{{𝑡}^2+5𝑡+4}𝑑𝑡\\ & =\frac{1}{2}\int \frac{𝑡}{(𝑡+1)(𝑡+4)}𝑑𝑡\\ & \text{ Let }\frac{𝑡}{(𝑡+1)(𝑡+4)}=\frac{𝐴}{𝑡+1}+\frac{𝐵}{𝑡+4}\\ \therefore 𝑡& =𝐴(𝑡+4)+𝐵(𝑡+1)\\ \therefore 𝐴& 𝐼=\frac{-1}{3},𝐵=\frac{4}{3}\\ \therefore & =\frac{1}{2}[-\frac{1}{3}\int \frac{1}{𝑡+1}𝑑𝑡+\frac{4}{3}\int \frac{1}{𝑡+4}𝑑𝑡]\\ & =\frac{1}{3}[-\frac{1}{2}\log (𝑡+1)+\frac{4}{3}\log (𝑡+4\left)\right]+𝑐\\ & =\frac{1}{3}[-\log \sqrt{\surd {𝑥}^2+1}+\log {({𝑥}^2+4)}^2]+𝑐\\ & =\frac{1}{3}\log \left[\frac{{({𝑥}^2+4)}^2}{\sqrt{\surd {𝑥}^2+1}}\right]+𝑐\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝑧& =\frac{3+2𝑖\sin 𝜃}{1-2𝑖\sin 𝜃}\\ & =\frac{(3+2𝑖\sin 𝜃)(1+2𝑖\sin 𝜃)}{1+4{\sin }^2𝜃}\\ & =\frac{3-4{\sin }^2𝜃+8𝑖\sin 𝜃}{1+4{\sin }^2𝜃}\end{array}\\ & \text{ Since }𝑧\text{ is purely imaginary. }\\ & \begin{array}{cc}& \therefore \frac{3-4{\sin }^2𝜃}{1+4{\sin }^2𝜃}=0\\ & \Rightarrow 3-4{\sin }^2𝜃=0\\ & \Rightarrow {\sin }^2𝜃=\frac{3}{4}\\ & \Rightarrow {\sin }^2𝜃={\sin }^2\frac{𝜋}{3}\\ & \Rightarrow 𝜃=n𝜋\pm \frac{𝜋}{3},\text{ where }n\in Z\end{array}\end{array}$

Given equation of circle is

${𝑥}^2+{𝑦}^2=36\Rightarrow 𝑎=6$

Slope of $5𝑥+𝑦=2$ is -5

$\therefore$ Slope of required line $=\frac{1}{5}$

Equations of tangents to the circle ${𝑥}^2+{𝑦}^2=a^2$ is

$\begin{array}{cc}& 𝑦=m𝑥\pm \sqrt{\surd a^2(1+m^2)}\\ & \Rightarrow 𝑦=\frac{1}{5}𝑥\pm \sqrt{\surd 36(1+\frac{1}{25})}\\ & \Rightarrow 5𝑦=𝑥\pm 6\sqrt{\surd 26}\\ & \Rightarrow 𝑥-5𝑦\pm 6\sqrt{\surd 26}=0\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \sin 𝐴=𝑛\sin (𝐴+2𝐵)\\ & \Rightarrow \frac{\sin (𝐴+2𝐵)}{\sin 𝐴}=\frac{1}{𝑛}\\ & \Rightarrow \frac{\sin (𝐴+2𝐵)+\sin 𝐴}{\sin (𝐴+2𝐵)-\sin 𝐴}=\frac{1+𝑛}{1-𝑛}\end{array}\\ & & \text{ ...[By componendo-dividendo] }\\ & \begin{array}{cc}& \Rightarrow \frac{2\sin (𝐴+𝐵)\cos 𝐵}{2\cos (𝐴+𝐵)\sin 𝐵}=\frac{1+𝑛}{1-𝑛}\\ & \Rightarrow \frac{\tan (𝐴+𝐵)}{\tan 𝐵}=\frac{1+𝑛}{1-𝑛}\\ & \Rightarrow \tan (𝐴+𝐵)=\left(\frac{1+𝑛}{1-𝑛}\right)\tan 𝐵\end{array}\end{array}$

Since the given vectors are linearly dependent.

$\begin{array}{cc}\therefore & \mid \begin{array}{ccc}𝑝+1& -3& 𝑝\\ 𝑝& 𝑝+1& -3\\ -3& 𝑝& 𝑝+1\end{array}\mid =0\\ & \Rightarrow (𝑝+1)\left[\right(𝑝+1{)}^2+3𝑝]+3\left[𝑝\right(𝑝+1)-9]+𝑝[{𝑝}^2+3(𝑝+1\left)\right]=0\\ \Rightarrow & (𝑝+1)({𝑝}^2+5𝑝+1)+3({𝑝}^2+𝑝-9)+𝑝({𝑝}^2+3𝑝+3)=0\\ \Rightarrow & 2{𝑝}^3+12{𝑝}^2+12𝑝-26=0\\ \Rightarrow & {𝑝}^3+6{𝑝}^2+6𝑝-13=0\\ \Rightarrow & (𝑝-1)({𝑝}^2+7𝑝+13)=0\end{array}$

Since p has integral values.

$\therefore p=1$

$\begin{array}{cc}& \text{ Let }I={\int }_{\frac{𝜋}{4}}^0(\sqrt{\surd \tan 𝑥}+\sqrt{\surd \cot 𝑥})d𝑥\\ & ={\int }_{\frac{𝜋}{4}}^0(\frac{\sqrt{\surd \sin 𝑥}}{\sqrt{\surd \cos 𝑥}}+\frac{\sqrt{\surd \cos 𝑥}}{\sqrt{\surd \sin 𝑥}})d𝑥\\ & ={\int }_{\frac{𝜋}{4}}^0\frac{\sin 𝑥+\cos 𝑥}{\sqrt{\surd \sin 𝑥\cos 𝑥}}d𝑥\\ & =\sqrt{\surd 2}{\int }_{\frac{𝜋}{4}}^0\frac{\sin 𝑥+\cos 𝑥}{\sqrt{\surd 2\sin 𝑥\cos 𝑥}}d𝑥\\ & =\sqrt{\surd 2}{\int }_{\frac{𝜋}{4}}^0\frac{\sin 𝑥+\cos 𝑥}{\sqrt{\surd 1-(\sin 𝑥-\cos 𝑥{)}^2}}d𝑥\\ & \begin{array}{cc}\text{ Put }& \sin 𝑥-\cos 𝑥=t\\ \Rightarrow & (\sin 𝑥+\cos 𝑥)d𝑥=\mathrm{dt}\\ \therefore I& =\sqrt{\surd 2}{\int }_0^{-1}\frac{\mathrm{dt}}{\sqrt{\surd 1-t^2}}\\ & =\sqrt{\surd 2}{[{\sin }^{-1}t]}_0^{-1}\\ & =\sqrt{\surd 2}[{\sin }^{-1}(0)-{\sin }^{-1}(-1\left)\right]\\ & =\sqrt{\surd 2}\cdot \frac{𝜋}{2}=\frac{𝜋}{\sqrt{\surd 2}}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Since the given curve passes through }(1,2)\text{. }\\ & \begin{array}{c}\therefore 2=𝑎+𝑏...\left(𝑖\right)\\ \text{ Area }={\int }_4^0𝑦𝑑𝑥\\ \Rightarrow 8={\int }_4^0(𝑎\sqrt{\surd 𝑥}+𝑏𝑥)𝑑𝑥\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow 8={[a\frac{{𝑥}^{\frac{3}{2}}}{\frac{3}{2}}+b\frac{{𝑥}^2}{2}]}_4^0\\ & \Rightarrow 8=\frac{2a}{3}\cdot 8+8b\\ & \Rightarrow 2a+3b=3...\left(𝑖𝑖\right)\end{array}$

On solving (i) and (ii), we get $a=3$ and $b=-1$

$\Rightarrow 𝑎-𝑏=4$

Given equation of ellipse is

$\begin{array}{cc}& \frac{{𝑥}^2}{25}+\frac{{𝑦}^2}{9}=1\\ & \Rightarrow 𝑎=5,𝑏=3\\ & 𝑒=\sqrt{\surd 1-\frac{{𝑏}^2}{{𝑎}^2}}=\sqrt{\surd 1-\frac{9}{25}}=\frac{4}{5}\end{array}$

Foci $\equiv (\pm \mathrm{ae},0)\equiv (\pm 4,0)$

For hyperbola, $e^{\prime }=2$

$\begin{array}{cc}& \text{ Foci }\equiv (\pm 4,0)\equiv (\pm 𝑎{𝑒}^{\prime },0)\\ & \Rightarrow 𝑎{𝑒}^{\prime }=4\Rightarrow 𝑎=2\\ & {𝑏}^2={𝑎}^2({𝑒}^{\prime 2}-1)=4(4-1)=12\end{array}$

$\therefore$ Equation of required hyperbola is

$\frac{{𝑥}^2}{4}-\frac{{𝑦}^2}{12}=1$

$𝑦=𝑋\sin (6𝑡+5)+𝑌\cos (6𝑡+5)$

Differentiating w.r.t. t, we get

$\frac{𝑑𝑦}{𝑑𝑡}=6𝑋\cos (6𝑡+5)-6𝑌\sin (6𝑡+5)$

Differentiating w.r.t. t, we get

$\begin{array}{cc}& \frac{d^2𝑦}{{\mathrm{dt}}^2}=-36X\sin (6t+5)-36Y\cos (6t+5)\\ & \Rightarrow \frac{d^2𝑦}{{\mathrm{dt}}^2}=-36𝑦\\ & \Rightarrow \frac{d^2𝑦}{{\mathrm{dt}}^2}+36𝑦=0\end{array}$