MHT CET 2025 26th April Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}\text{ Since }𝑝\vee \sim 𝑝\equiv 𝑇,& \\ (\sim 𝑞\wedge 𝑝)\vee (𝑝\vee \sim 𝑝)& \equiv (\sim 𝑞\wedge 𝑝)\vee 𝑇\\ & \equiv 𝑇\dots \left[𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦𝑙𝑎𝑤\right]\end{array}$

$\therefore (\sim 𝑞\wedge 𝑝)\vee (𝑝\vee \sim 𝑝)$ is a tautology.

$𝑓\left(𝑥\right)=\frac{(𝑥-1)(𝑥+1)(𝑥-2)(𝑥+2)}{|𝑥-1||𝑥-2|}$

Since $\underset{𝑥\to 1}{lim}\frac{𝑥-1}{|𝑥-1|}$ does not exist.

Also, $\underset{𝑥\to 2}{lim}\frac{𝑥-2}{|𝑥-2|}$ does not exist

$\therefore f\left(𝑥\right)$ is discontinuous at $𝑥=1,2$.

For any $𝑥\ne 1,2,f\left(𝑥\right)$ is the quotient of two polynomials and a polynomial is everywhere continuous. Therefore, $f\left(𝑥\right)$ is continuous for all $𝑥\ne 1,2$.

$\therefore f\left(𝑥\right)$ is continuous on $R-\{1,2\}$.

Area of the first square $\left({𝐴}_1\right)={𝑥}^2$

Area of the second square $\left(A_2\right)={𝑦}^2$

$A_1={𝑥}^2$ and $A_2={𝑦}^2$

$\therefore \frac{{\mathrm{dA}}_1}{\mathrm{dt}}=2𝑥\frac{d𝑥}{\mathrm{dt}}\text{ and }\frac{{\mathrm{dA}}_2}{\mathrm{dt}}=2𝑦\frac{d𝑦}{\mathrm{dt}}$

$\therefore \frac{{\mathrm{dA}}_2}{{\mathrm{dA}}_1}=\frac{\frac{{\mathrm{dA}}_2}{\mathrm{dt}}}{\frac{\mathrm{dA}{A}_1}{\mathrm{dt}}}=\frac{2𝑦\frac{d𝑦}{\mathrm{dt}}}{2𝑥\frac{d𝑥}{\mathrm{dt}}}$

$=\frac{𝑦}{𝑥}\left(\frac{d𝑦}{d𝑥}\right)$

Given $𝑦=𝑥-{𝑥}^2$

$\therefore \frac{d𝑦}{d𝑥}=1-2𝑥$

$\therefore \frac{{\mathrm{dA}}_2}{{\mathrm{dA}}_1}=\frac{𝑦}{𝑥}(1-2𝑥)$

$=\frac{𝑥-{𝑥}^2}{𝑥}(1-2𝑥)$

$=(1-𝑥)(1-2𝑥)=2{𝑥}^2-3𝑥+1$

Let the equation of the line with equal intercepts be $𝑥+𝑦=a$.

Slope of this line is ${𝑚}_1=-1$

Slope of line $4𝑥-2𝑦+13=0$ is ${𝑚}_2=\frac{-4}{-2}=2$

Let $𝜃$ be the acute angle.

Then,

$\begin{array}{cc}& \tan 𝜃=\mid \frac{-1-2}{1+(-1)\left(2\right)}\mid \\ & \Rightarrow \tan 𝜃=3\\ & \Rightarrow 𝜃={\tan }^{-1}\left(3\right)\end{array}$

$\begin{array}{cc}& \text{ Since }\mid 𝑥-\frac{1}{2}\mid =-(𝑥-\frac{1}{2}),\text{ if }𝑥<\frac{1}{2}\\ & =𝑥-\frac{1}{2},\text{ if }𝑥\ge \frac{1}{2}\\ & \begin{array}{cc}\therefore & {\int }_1^0𝑥\mid 𝑥-\frac{1}{2}\mid d𝑥\\ & =-{\int }_{\frac{1}{2}}^0𝑥(𝑥-\frac{1}{2})d𝑥+{\int }_1^{\frac{1}{2}}𝑥(𝑥-\frac{1}{2})d𝑥\\ & ={\int }_{\frac{1}{2}}^0(\frac{1}{2}𝑥-{𝑥}^2)d𝑥+{\int }_1^{\frac{1}{2}}({𝑥}^2-\frac{1}{2}𝑥)d𝑥\\ & ={[\frac{{𝑥}^2}{4}-\frac{{𝑥}^3}{3}]}_{1/2}^0+{[\frac{{𝑥}^3}{3}-\frac{{𝑥}^2}{4}]}_1^{1/2}\\ & =(\frac{1}{16}-\frac{1}{24})+(\frac{1}{3}-\frac{1}{4}+\frac{1}{16}-\frac{1}{24})=\frac{1}{8}\end{array}\end{array}$

$\begin{array}{cc}f\left(𝑥\right)& =\left[𝑥\right(𝑥-2){]}^2\\ & ={𝑥}^2(𝑥-2{)}^2\\ \therefore f^{\prime }\left(𝑥\right)& =2𝑥(𝑥-2{)}^2+2(𝑥-2){𝑥}^2\\ & =2𝑥(𝑥-2)(𝑥-2+𝑥)\\ & =4𝑥(𝑥-1)(𝑥-2)\end{array}$

For $f\left(𝑥\right)$ to be increasing, $f^{\prime }\left(𝑥\right)>0$

$\begin{array}{cc}& \Rightarrow 4𝑥(𝑥-1)(𝑥-2)>0\\ & \Rightarrow 𝑥(𝑥-1)(𝑥-2)>0\\ & \Rightarrow 𝑥\in (0,1)\cup (2,\infty )\end{array}$

The truth values of $𝑝,𝑞,𝑟$ and $𝑠$ are $𝐹,𝑇,𝐹$ and $𝑇$ respectively.

Consider option (D)

$\begin{array}{cc}& (𝑠\vee \sim 𝑝)\leftrightarrow (\sim 𝑝\wedge \sim 𝑟)\\ & \equiv (𝑇\vee \sim 𝐹)\leftrightarrow (\sim 𝐹\wedge \sim 𝐹)\\ & \equiv (𝑇\vee 𝑇)\leftrightarrow (𝑇\wedge 𝑇)\\ & \equiv 𝑇\leftrightarrow 𝑇\\ & \equiv 𝑇\end{array}$

$\therefore$ Option (D) is the correct answer.

$\begin{array}{cc}𝐿𝑒𝑡p& =a𝑥+b𝑦\\ & =a𝑥+b\left(\frac{c^2}{𝑥}\right)\dots [\because 𝑥𝑦=c^2]\end{array}$

$\begin{array}{cc}\therefore & p=a𝑥+\frac{{\mathrm{bc}}^2}{𝑥}...\left(𝑖\right)\\ \therefore & \frac{\mathrm{dp}}{d𝑥}=a-\frac{{\mathrm{bc}}^2}{{𝑥}^2}\\ \therefore & \frac{d^{2}p}{d{𝑥}^2}=\frac{2{\mathrm{bc}}^2}{{𝑥}^3}\end{array}$

Now, $\frac{\mathrm{dp}}{\mathrm{dx}}=0$

$\begin{array}{cc}& \Rightarrow a-\frac{{\mathrm{bc}}^2}{{𝑥}^2}=0\\ & \Rightarrow {𝑥}^2=\frac{{\mathrm{bc}}^2}{a}\\ & \Rightarrow 𝑥=\pm c\sqrt{\surd \frac{b}{a}}\\ & \text{ At }𝑥=c\sqrt{\surd \frac{b}{a}},\frac{d^{2}p}{d{𝑥}^2}>0\\ & \text{ At }𝑥=-c\sqrt{\surd \frac{b}{a}},\frac{d^{2}p}{d{𝑥}^2}<0\end{array}$

$\therefore p$ is minimum at $𝑥=c\sqrt{\surd \frac{b}{a}}$.

$\begin{array}{cc}\therefore \text{ Minimum value }& =𝑎𝑐\sqrt{\surd \frac{𝑏}{𝑎}}+\frac{𝑏{𝑐}^2}{𝑐\sqrt{\surd \frac{𝑏}{𝑎}}}\dots \left[\text{ From (i) }\right]\\ & =𝑐\sqrt{\surd 𝑎𝑏}+𝑐\sqrt{\surd 𝑎𝑏}\\ & =2𝑐\sqrt{\surd 𝑎𝑏}\end{array}$

We need to find

$\underset{𝑥\to 0}{lim}\frac{{𝑒}^{{𝑥}^2}-\cos 3𝑥}{\sin 𝑥\log (1+2𝑥)}.$

Since both numerator and denominator go to $0$ as $𝑥\to 0$, we expand each term using standard limits/Taylor expansions near $𝑥=0$.

Recall:

${𝑒}^{{𝑥}^2}=1+{𝑥}^2+\frac{{𝑥}^4}{2}+\cdots$

and

$\cos 3𝑥=1-\frac{(3𝑥{)}^2}{2}+\cdots =1-\frac{9{𝑥}^2}{2}+\cdots$

So the numerator becomes

${𝑒}^{{𝑥}^2}-\cos 3𝑥=(1+{𝑥}^2+\cdots )-(1-\frac{9{𝑥}^2}{2}+\cdots )$

$={𝑥}^2+\frac{9{𝑥}^2}{2}+\cdots =\frac{11{𝑥}^2}{2}+\cdots$

Now for the denominator:

$\sin 𝑥=𝑥+\cdots$

and

$\log (1+2𝑥)=2𝑥+\cdots$

Hence,

$\sin 𝑥\log (1+2𝑥)=(𝑥+\cdots )(2𝑥+\cdots )=2{𝑥}^2+\cdots$

Therefore,

$\underset{𝑥\to 0}{lim}\frac{{𝑒}^{{𝑥}^2}-\cos 3𝑥}{\sin 𝑥\log (1+2𝑥)}=\frac{\frac{11}{2}}{2}=\frac{11}{4}.$

So the correct answer is

$\overline{)\frac{11}{4}}$

Hence, Option C.

$\begin{array}{cc}& \text{ Since }{\int }_{\frac{𝜋}{2}}^0\frac{𝑎\sin 𝑥+𝑏\cos 𝑥}{\sin 𝑥+\cos 𝑥}𝑑𝑥=\frac{𝜋}{4}(𝑎+𝑏)\\ & {\int }_{\frac{𝜋}{2}}^0\frac{300\sin 𝑥+100\cos 𝑥}{\sin 𝑥+\cos 𝑥}𝑑𝑥\\ & =\frac{𝜋}{4}(300+100)\\ & =100𝜋\end{array}$

$\begin{array}{cc}& {𝐴}^2-4𝐴+3𝐼=0\\ & \Rightarrow {𝐴}^2+3𝐴-7𝐴+3𝐼=0\\ & \Rightarrow {𝐴}^2+3𝐴-7𝐴-21𝐼+24𝐼=0\\ & \Rightarrow 𝐴(𝐴+3𝐼)-7𝐼(𝐴+3𝐼)=-24𝐼\\ & \Rightarrow (𝐴+3𝐼)(𝐴-7𝐼)=-24𝐼\\ & \Rightarrow (𝐴+3𝐼)(\frac{7𝐼}{24}-\frac{𝐴}{24})=𝐼\end{array}$

$\Rightarrow (𝐴+3𝐼{)}^{-1}=\frac{7𝐼}{24}-\frac{𝐴}{24}$

$\begin{array}{cc}& (1+\frac{𝑎}{𝑐}+\frac{𝑏}{𝑐})(1+\frac{𝑐}{𝑏}-\frac{𝑎}{𝑏})\\ & =\left(\frac{𝑐+𝑎+𝑏}{𝑐}\right)\left(\frac{𝑏+𝑐-𝑎}{𝑏}\right)\\ & =\frac{1}{𝑏𝑐}\left[\right(𝑏+𝑐{)}^2-{𝑎}^2]\\ & =\frac{1}{𝑏𝑐}({𝑏}^2+{𝑐}^2+2𝑏𝑐-{𝑎}^2)\\ & =\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{𝑏𝑐}+2\\ & =2\left(\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}\right)+2\\ & =2\cos 𝐴+2\text{ ...[By cosine rule] }\\ & =2\cos {30}^{\circ }+2\\ & =2\left(\frac{\sqrt{\surd 3}}{2}\right)+2\\ & =\sqrt{\surd 3}+2\end{array}$

$\begin{array}{cc}& \text{ In }△\mathrm{PQR},\\ & \angle P+\angle Q+\angle R={180}^{\circ }\\ \therefore & \angle P+\angle Q+\frac{𝜋}{2}={180}^{\circ }\\ \therefore & \angle P+\angle Q=\frac{𝜋}{2}\\ \therefore & \frac{\angle P}{2}+\frac{\angle Q}{2}=\frac{𝜋}{4}\end{array}$

$\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are the roots of the equation $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$        ...[Given].

$\begin{array}{cc}\therefore & \text{ Sum of roots }=\frac{-𝑏}{𝑎}\\ & \Rightarrow \tan \left(\frac{𝑃}{2}\right)+\tan \left(\frac{𝑄}{2}\right)=\frac{-𝑏}{𝑎}\end{array}$

Also, $\tan \left(\frac{P}{2}\right)\cdot \tan \left(\frac{Q}{2}\right)=\frac{c}{a}$

Using, $\tan (\frac{𝑃}{2}+\frac{𝑄}{2})=\frac{\tan \frac{𝑃}{2}+\tan \frac{𝑄}{2}}{1-\tan \frac{𝑃}{2}\tan \frac{𝑄}{2}}$, we get

$\begin{array}{cc}& \tan \left(\frac{𝜋}{4}\right)=\frac{\frac{-𝑏}{𝑎}}{1-\frac{𝑐}{𝑎}}\\ & \Rightarrow 1=\frac{-𝑏}{𝑎-𝑐}\\ & \Rightarrow 𝑎-𝑐=-𝑏\\ & \Rightarrow 𝑎+𝑏=𝑐\end{array}$

$\begin{array}{cc}& {\cot }^{-1}\left(\sqrt{\surd \cos 𝛼}\right)-{\tan }^{-1}\left(\sqrt{\surd \cos 𝛼}\right)=𝑥\\ \Rightarrow & {\tan }^{-1}\left(\frac{1}{\sqrt{\surd \cos 𝛼}}\right)-{\tan }^{-1}\left(\sqrt{\surd \cos 𝛼}\right)=𝑥\\ \Rightarrow & {\tan }^{-1}⎛⎝⎜\frac{\frac{1}{\sqrt{\surd \cos 𝛼}}-\sqrt{\surd \cos 𝛼}}{1+\left(\frac{1}{\sqrt{\surd \cos 𝛼}}\right)\left(\sqrt{\surd \cos 𝛼}\right)}⎞⎠⎟=𝑥\\ \Rightarrow & \tan 𝑥=\frac{1-\cos 𝛼}{2\sqrt{\surd \cos 𝛼}}\\ \therefore & \sin 𝑥=\frac{1-\cos 𝛼}{1+\cos 𝛼}=\frac{2{\sin }^2\frac{𝛼}{2}}{2{\cos }^2\frac{𝛼}{2}}={\tan }^2\frac{𝛼}{2}\end{array}$

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

$\begin{array}{cc}2{\tan }^{-1}\left(\frac{1}{2}\right)& ={\tan }^{-1}\left(\frac{4}{3}\right)\\ & \dots [\because 2{\tan }^{-1}𝑥={\tan }^{-1}\left(\frac{2𝑥}{1-{𝑥}^2}\right)]\end{array}$

$\begin{array}{cc}& \tan ({\sin }^{-1}\left(\frac{3}{5}\right)-2{\cos }^{-1}\left(\frac{2}{\sqrt{\surd 5}}\right))\\ =& \tan ({\tan }^{-1}\left(\frac{3}{4}\right)-{\tan }^{-1}\left(\frac{4}{3}\right))\\ =& \tan ({\tan }^{-1}\left(\frac{\frac{3}{4}-\frac{4}{3}}{1+\left(\frac{3}{4}\right)\left(\frac{4}{3}\right)}\right))\\ =& \tan ({\tan }^{-1}(-\frac{7}{24}))=-\frac{7}{24}\end{array}$

$\begin{array}{cc}& \text{ According to the given condition, }\\ & \begin{array}{cc}& {𝐴}_1={𝐴}_2\\ & \Rightarrow {\int }_{𝛼}^0\frac{{𝑥}^2}{4}𝑑𝑥={\int }_4^{𝛼}\frac{{𝑥}^2}{4}𝑑𝑥\\ & \Rightarrow \frac{1}{4}{\left[\frac{{𝑥}^3}{3}\right]}_{𝛼}^0=\frac{1}{4}{\left[\frac{{𝑥}^3}{3}\right]}_4^{𝛼}\\ & \Rightarrow \frac{1}{12}({𝛼}^3-0)=\frac{1}{12}(4^3-{𝛼}^3)\\ & \Rightarrow {𝛼}^3=64-{𝛼}^3\\ & \Rightarrow 2{𝛼}^3=64\\ & \Rightarrow {𝛼}^3=32\\ & \Rightarrow 𝛼=(32{)}^{\frac{1}{3}}\end{array}\end{array}$

$\begin{array}{cc}& {𝑝}^3={𝑞}^4={𝑟}^6={𝑡}^7={𝑠}^2\\ & \Rightarrow {𝑝}^3={𝑡}^7,{𝑞}^4={𝑡}^7,{𝑟}^6={𝑡}^7,{𝑠}^2={𝑡}^7\\ & \Rightarrow 𝑝={𝑡}^{\frac{7}{3}},𝑞={𝑡}^{\frac{7}{4}},𝑟={𝑡}^{\frac{7}{6}},𝑠={𝑡}^{\frac{7}{2}}\\ & \Rightarrow 𝑝𝑞𝑟𝑠=\left({𝑡}^{\frac{7}{3}}\right)\left({𝑡}^{\frac{7}{4}}\right)\left({𝑡}^{\frac{7}{6}}\right)\left({𝑡}^{\frac{7}{2}}\right)={𝑡}^{\frac{525}{60}}={𝑡}^{\frac{35}{4}}\\ \therefore & {\log }_{𝑡}\left(𝑝𝑞𝑟𝑠\right)={\log }_{𝑡}\left({𝑡}^{\frac{35}{4}}\right)=\frac{35}{4}{\log }_{𝑡}𝑡=\frac{35}{4}\end{array}$

The slope of the tangent at point $(𝑥,𝑦)$ is $\frac{d𝑦}{d𝑥}$.

According to the given condition,

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}+5=𝑥+𝑦\\ & \Rightarrow \frac{d𝑦}{d𝑥}-𝑦=(𝑥-5)\\ \therefore & \text{ I.F. }=e^{\int -1.d𝑥}=e^{-𝑥}\end{array}$

$\therefore$ Solution of the given equation is

$\begin{array}{cc}& 𝑦e^{-𝑥}=\int (𝑥-5)e^{-𝑥}d𝑥+c\\ & =\int 𝑥e^{-𝑥}d𝑥-5\int e^{-𝑥}d𝑥+c\\ & =-𝑥e^{-𝑥}+\int e^{-𝑥}d𝑥+5e^{-𝑥}+c\\ & =-𝑥e^{-𝑥}-e^{-𝑥}+5e^{-𝑥}+c\\ & \therefore 𝑦e^{-𝑥}=-𝑥e^{-𝑥}+4e^{-𝑥}+c\\ & \therefore 𝑦=-𝑥+4+𝑐e^{𝑥}...\left(𝑖\right)\end{array}$

But the required curve passes through $(0,2)$.

$\begin{array}{cc}\therefore & 2=0+4+𝑐e^0\Rightarrow c=-2\\ \therefore & 𝑦=4-𝑥-2e^{𝑥}\dots [\text{ From (i)] }\end{array}$

$\begin{array}{cc}& \left|z\right|+z=3+i\\ \therefore & z=3-\left|z\right|+i\\ & \text{ Let }3-\left|z\right|=𝑥\Rightarrow \left|z\right|=3-𝑥...\left(𝑖\right)\\ \therefore & z=𝑥+i\\ & \Rightarrow \left|z\right|=\sqrt{\surd {𝑥}^2+1}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow |𝑧{|}^2={𝑥}^2+1...(𝑖𝑖)\\ & \Rightarrow (3-𝑥{)}^2={𝑥}^2+1\dots [\text{ From (i)] }\\ & \Rightarrow 9-6𝑥+{𝑥}^2={𝑥}^2+1\\ & \Rightarrow 6𝑥=8\\ & \Rightarrow 𝑥=\frac{4}{3}\end{array}\\ & \text{ Equation (i) becomes, }\\ & \left|𝑧\right|=3-\frac{4}{3}=\frac{5}{3}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& (1+𝑥)\frac{d𝑦}{d𝑥}-𝑥𝑦=1-𝑥\\ \Rightarrow & \frac{d𝑦}{d𝑥}-\left(\frac{𝑥}{1+𝑥}\right)𝑦=\frac{1-𝑥}{1+𝑥}\\ \therefore & \text{ I. F. }=e^{-\int \frac{𝑥}{1+𝑥}\mathrm{dx}}\\ =& e^{\int (-1+\frac{1}{1+𝑥})\mathrm{dx}}\\ =& e^{-𝑥+\log (1+𝑥)}\\ =& e^{-𝑥}{e}^{\log (1+𝑥)}\\ =& e^{-𝑥}(1+𝑥)\end{array}\\ & \therefore \text{ Solution of the given equation is }\\ & \begin{array}{cc}& 𝑦(1+𝑥)e^{-𝑥}=\int \left(\frac{1-𝑥}{1+𝑥}\right)(1+𝑥)e^{-𝑥}d𝑥+c\\ & \Rightarrow 𝑦(1+𝑥)e^{-𝑥}=\int (1-𝑥)e^{-𝑥}d𝑥+c\\ & \Rightarrow 𝑦(1+𝑥)e^{-𝑥}=\int e^{-𝑥}d𝑥-\int 𝑥e^{-𝑥}d𝑥+c\\ & \Rightarrow 𝑦(1+𝑥)e^{-𝑥}=-e^{-𝑥}-(-𝑥e^{-𝑥}-e^{-𝑥})+c\\ & \Rightarrow 𝑦(1+𝑥)e^{-𝑥}=𝑥e^{-𝑥}+c\\ & \Rightarrow 𝑦(1+𝑥)=𝑥+𝑐e^{𝑥}\end{array}\end{array}$

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

$A,B,C$ are in $A.P$

$\Rightarrow 𝐴+𝐶=2𝐵$

Also, $A+B+C={180}^{\circ }$