MHT CET 2025 5th May Evening Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& \begin{array}{cc}& \sim \left[\right(𝑝\wedge \sim 𝑞)\to (𝑝\vee \sim 𝑞\left)\right]\\ & <text>\equiv </text><text>\sim </text>[\sim (𝑝\wedge \sim 𝑞)\vee (𝑝\vee \sim 𝑞\left)\right]...\left[𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙𝑙𝑎𝑤\right]\end{array}\end{array}$

$\begin{array}{cc}& \equiv (𝑝\wedge \sim 𝑞)\wedge \sim (𝑝\vee \sim 𝑞)...\left[𝐷𝑒𝑀𝑜𝑟𝑔𝑎{𝑛}^{\prime }𝑠𝑙𝑎𝑤\right]\end{array}$

$\begin{array}{cc}\equiv (𝑝\wedge \sim 𝑞)\wedge (\sim 𝑝\wedge 𝑞)& \dots \left[\text{ De Morgan’s law }\right]\\ \equiv 𝑝\wedge \sim 𝑞\wedge \sim 𝑝\wedge 𝑞& \dots \left[\text{ Associative law }\right]\\ \equiv (𝑝\wedge \sim 𝑝)\wedge (\sim 𝑞\wedge 𝑞)& \dots \left[\text{ Commutative law }\right]\\ \equiv 𝐹\wedge 𝐹& \dots \left[\text{ Complement law }\right]\\ \equiv 𝐹& \end{array}$

For a triangle with side lengths $a,b$ and c opposite to the vertices $({𝑥}_1,{𝑦}_1),({𝑥}_2,{𝑦}_2)$ and $({𝑥}_3,{𝑦}_3)$ respectively, the incentre is given by

$\begin{array}{cc}& (ℎ,𝑘)=(\frac{𝑎{𝑥}_1+𝑏{𝑥}_2+𝑐{𝑥}_3}{𝑎+𝑏+𝑐},\frac{𝑎{𝑦}_1+𝑏{𝑦}_2+𝑐{𝑦}_3}{𝑎+𝑏+𝑐})\\ & 𝑎=𝐴𝐵=\sqrt{\surd 8^2+6^2}=10\text{ units }\\ & 𝑏=𝑂𝐵=6\text{ units }\\ & 𝑐=𝑂𝐴=8\text{ units }\end{array}$

$\therefore$ Incentre

$\begin{array}{cc}& =(\frac{10\left(0\right)+6\left(8\right)+8\left(0\right)}{10+6+8},\frac{10\left(0\right)+6\left(0\right)+8\left(6\right)}{10+6+8})\\ & =(\frac{48}{24},\frac{48}{24})\\ & =(2,2)\end{array}$

Since $({𝑥}_1,{𝑦}_1)$ lies on the curve $𝑥𝑦=a^2$,

${𝑥}_1{𝑦}_1=a^2...\left(𝑖\right)$

Now, $𝑥𝑦=a^2$

Differentiating w.r.t. $𝑥$, we get

$\begin{array}{cc}& 𝑥\frac{d𝑦}{d𝑥}+𝑦=0\\ & \Rightarrow \frac{d𝑦}{d𝑥}=-\frac{𝑦}{𝑥}\\ \therefore & {\left(\frac{d𝑦}{d𝑥}\right)}_{({𝑥}_1,{𝑦}_1)}=-\frac{{𝑦}_1}{{𝑥}_1}\end{array}$

$\therefore$ Equation of the tangent at $({𝑥}_1,{𝑦}_1)$ is

$\begin{array}{cc}& 𝑦-{𝑦}_1=-\frac{{𝑦}_1}{{𝑥}_1}(𝑥-{𝑥}_1)\\ & \Rightarrow 𝑥{𝑦}_1+𝑦{𝑥}_1=2{𝑥}_1{𝑦}_1\\ & \Rightarrow 𝑥{𝑦}_1+𝑦{𝑥}_1=2a^2\dots .[\text{ From (i)] }\end{array}$

This tangent meets the coordinate axes at

$(\frac{2a^2}{{𝑦}_1},0)\text{ and }(0,\frac{2a^2}{{𝑥}_1}).$

$\begin{array}{cc}\therefore \text{ Required area }& =\frac{1}{2}\left(\frac{2a^2}{{𝑦}_1}\right)\left(\frac{2a^2}{{𝑥}_1}\right)\\ & =\frac{2a^4}{{𝑥}_1{𝑦}_1}\\ & =2a^2\text{ ....[From (i)] }\end{array}$

The equation of the line joining $(1,1,1)$ and ( $2,2,2$ ) is

$\begin{array}{cc}& \frac{𝑥-1}{2-1}=\frac{𝑦-1}{2-1}=\frac{𝑧-1}{2-1}\\ & \Rightarrow 𝑥-1=𝑦-1=𝑧-1=𝜆\left(\text{ say }\right)\end{array}$

Then, any point on the line is

$P\equiv (𝜆+1,𝜆+1,𝜆+1)$

The point P lies on the plane $𝑥+𝑦+z=9$.

$\begin{array}{cc}\therefore & 𝜆+1+𝜆+1+𝜆+1=9\\ & \Rightarrow 3𝜆=6\\ & \Rightarrow 𝜆=2\\ \therefore & 𝑃\equiv (3,3,3)\end{array}$

Step 1: Find the slope of the tangent

The slope of the tangent to the curve is found by differentiating $𝑦={𝑥}^3-3{𝑥}^2+2𝑥+93$ with respect to $𝑥$. This means we find $\frac{𝑑𝑦}{𝑑𝑥}$:

So, $\frac{𝑑𝑦}{𝑑𝑥}=3{𝑥}^2-6𝑥+2$ Let’s call $\frac{𝑑𝑦}{𝑑𝑥}$ as $𝑇$.

Step 2: Find where the slope is the smallest

To find the minimum slope, we need to find where $𝑇$ is smallest. We do this by setting the derivative of $𝑇$ with respect to $𝑥$ equal to zero.

First, find $\frac{𝑑𝑇}{𝑑𝑥}$: $\frac{𝑑𝑇}{𝑑𝑥}=6𝑥-6$

Step 3: Solve for $𝑥$ where the slope is minimum

Set $\frac{𝑑𝑇}{𝑑𝑥}=0$:

$6𝑥-6=0⟹𝑥=1$

Step 4: Make sure it is a minimum

Now, check the second derivative to make sure this is a minimum point: $\frac{{𝑑}^2𝑇}{𝑑{𝑥}^2}=6$ Since $6>0$, the function has a minimum at $𝑥=1$.

Step 5: Find the minimum value of the slope

Plug $𝑥=1$ back into the slope formula: $𝑇=3(1{)}^2-6(1)+2=3-6+2=-1$

So, the minimum value of the slope of the tangent is $-1$.

$\begin{array}{cc}& f\left(𝑥\right)\text{ is continuous at }𝑥=0\text{. }\\ & \begin{array}{cc}\therefore 𝑓\left(0\right)& =\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)\\ & =\underset{𝑥\to 0}{lim}\frac{9^{𝑥}-{2.3}^{𝑥}+1}{\log (1+3𝑥)\cdot \tan 2𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{{(3^{𝑥}-1)}^2}{\log (1+3𝑥)\tan 2𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{{\left(\frac{3^{𝑥}-1}{𝑥}\right)}^2}{\frac{\log (1+3𝑥)}{3𝑥}\times 3\times \frac{\tan 2𝑥}{2𝑥}\times 2}\\ & =\frac{(\log 3{)}^2}{1\times 3\times 1\times 2}\end{array}\end{array}$

$\begin{array}{cc}\therefore & 𝑎(\log 𝑏{)}^{𝑐}=\frac{(\log 3{)}^2}{6}\\ & \Rightarrow 𝑎=\frac{1}{6},𝑏=3,𝑐=2\\ & \Rightarrow 𝑎+𝑏+𝑐=\frac{31}{6}\end{array}$

$\begin{array}{cc}& 𝑥={\tan }^{-1}\left(\frac{\sqrt{\surd 1+t^2}-1}{t}\right)\\ & \text{ Put }t=\tan 𝜃\Rightarrow 𝜃={\tan }^{-1}t\\ & \therefore 𝑥={\tan }^{-1}\left(\frac{\sqrt{\surd 1+{\tan }^2𝜃}-1}{\tan 𝜃}\right)\\ & ={\tan }^{-1}\left(\frac{\sec 𝜃-1}{\tan 𝜃}\right)\\ & ={\tan }^{-1}\left(\frac{1-\cos 𝜃}{\sin 𝜃}\right)\\ & ={\tan }^{-1}(\tan \frac{𝜃}{2})\\ & =\frac{𝜃}{2}=\frac{1}{2}{\tan }^{-1}t\end{array}$

$\begin{array}{cc}& \therefore \frac{d𝑥}{\mathrm{dt}}=\frac{1}{2(1+t^2)}\\ & 𝑦={\cos }^{-1}\left(\frac{1-t^2}{1+t^2}\right)\\ & \text{ Put }t=\tan 𝜃\Rightarrow 𝜃={\tan }^{-1}t\\ & \therefore 𝑦={\cos }^{-1}\left(\frac{1-{\tan }^2𝜃}{1+{\tan }^2𝜃}\right)\\ & ={\cos }^{-1}(\cos 2𝜃)\\ & =2𝜃\\ & =2{\tan }^{-1}t\end{array}$

$\begin{array}{cc}\therefore & \frac{d𝑦}{\mathrm{dt}}=\frac{2}{1+t^2}\\ \therefore & \frac{d𝑦}{d𝑥}=\frac{\frac{d𝑦}{\mathrm{dt}}}{\frac{d𝑥}{\mathrm{dt}}}=\frac{\frac{2}{1+t^2}}{\frac{1}{2(1+t^2)}}=4\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\underset{𝑥\to 2^{-}}{lim}𝑓\left(𝑥\right)& =\underset{𝑥\to 2^{-}}{lim}(𝑏-𝑎𝑥)\\ & =𝑏-2𝑎\\ \underset{𝑥\to 2^{+}}{lim}𝑓\left(𝑥\right)& =\underset{𝑥\to 2^{+}}{lim}(𝑎+2𝑏𝑥)=𝑎+4𝑏\end{array}\\ & \text{ Since }\underset{𝑥\to 2}{lim}f\left(𝑥\right)\text{ exists, }\\ & \begin{array}{cc}& \underset{𝑥\to 2^{-}}{lim}f\left(𝑥\right)=\underset{𝑥\to 2^{+}}{lim}f\left(𝑥\right)\\ & \Rightarrow b-2a=a+4b\\ & \Rightarrow 3a=-3b\\ & \Rightarrow \frac{a}{b}=-1\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}𝑓(-𝑥)& =\sec [\log (-𝑥+\sqrt{\surd 1+(-𝑥{)}^2})]\\ & =\sec [\log (-𝑥+\sqrt{\surd 1+{𝑥}^2})]\\ & =\sec [\log (\sqrt{\surd 1+{𝑥}^2}-𝑥)]\\ & =\sec [\log \left(\frac{1+{𝑥}^2-{𝑥}^2}{\sqrt{\surd 1+{𝑥}^2}+𝑥}\right)]\\ & =\sec [\log \left(\frac{1}{\sqrt{\surd 1+{𝑥}^2}+𝑥}\right)]\\ & =\sec [-\log (\sqrt{\surd 1+{𝑥}^2}+𝑥)]\\ & =\sec [\log (\sqrt{\surd 1+{𝑥}^2}+𝑥)]\\ & =f\left(𝑥\right)\end{array}\\ & \therefore f\left(𝑥\right)\text{ is an even function. }\end{array}$

$\begin{array}{cc}& {}^{47}{C}_4+\sum _{j=1}^5{}^{(52-j)}{C}_3\\ & ={}^{47}{C}_4+{}^{51}{C}_3+{}^{50}{C}_3+{}^{49}{C}_3+{}^{48}{C}_3+{}^{47}{C}_3\\ & =({}^{47}{C}_4+{}^{47}{C}_3)+{}^{48}{C}_3+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3\\ & =({}^{48}{C}_4+{}^{48}{C}_3)+{}^{49}{C}_3+{}^{50}{C}_3+{}^{51}{C}_3\\ & \dots [{}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r]\end{array}$

$\begin{array}{cc}& =({}^{49}{C}_4+{}^{49}{C}_3)+{}^{50}{C}_3+{}^{51}{C}_3\\ & =({}^{50}{C}_4+{}^{50}{C}_3)+{}^{51}{C}_3\\ & ={}^{51}{C}_4+{}^{51}{C}_3\\ & ={}^{52}{C}_4\end{array}$

If $𝑦=\sqrt{\surd f\left(𝑥\right)+\sqrt{\surd f\left(𝑥\right)+\sqrt{\surd f\left(𝑥\right)+\dots \infty }}}$, then

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{{𝑓}^{\prime }\left(𝑥\right)}{2𝑦-1}$

Here, $f\left(𝑥\right)=\sin 𝑥\Rightarrow f^{\prime }\left(𝑥\right)=\cos 𝑥$

$\therefore \frac{d𝑦}{d𝑥}=\frac{\cos 𝑥}{2𝑦-1}$

Given the system of inequalities:

$𝑥-𝑦\ge 0$

$𝑥-5𝑦\le -5$

$𝑥\ge 0,𝑦\ge 0$

Rewriting the first two inequalities in terms of $𝑦$:

$𝑦\le 𝑥$

$𝑦\ge \frac{𝑥+5}{5}$

The inequality $𝑦\le 𝑥$ represents the region on and below the line $𝑥-𝑦=0$, while $𝑦\ge \frac{𝑥+5}{5}$ represents the region on and above the line $𝑥-5𝑦=-5$.

The intersection of the two boundary lines is found by substituting $𝑦=𝑥$ into $𝑥-5𝑦=-5$:

$𝑥-5𝑥=-5⟹𝑥=\frac{5}{4}=1.25$

$𝑦=\frac{5}{4}=1.25$

Both inequalities hold simultaneously; that is,

$\frac{𝑥+5}{5}\le 𝑦\le 𝑥$

which holds only for $𝑥\ge \frac{5}{4}$.

Therefore, the feasible region is the unbounded region lying to the right of the intersection point $(1.25,1.25)$, bounded above by the line $𝑥-𝑦=0$ and below by the line $𝑥-5𝑦=-5$.

Hence, the correct option is (C).

$\begin{array}{cc}& 𝜔\text{ is the complex cube root of unity other than }1\text{. }\\ & \begin{array}{cc}\therefore & 𝜔=\frac{-1}{2}+\frac{\sqrt{\surd 3}}{2}i\\ & {𝜔}^2=\frac{-1}{2}-\frac{\sqrt{\surd 3}}{2}i\\ & \text{ Let }A=(0,1),B=(\frac{-1}{2},\frac{\sqrt{\surd 3}}{2}),C=(\frac{-1}{2},-\frac{\sqrt{\surd 3}}{2})\end{array}\end{array}$

$\begin{array}{cc}\therefore A\left(\Delta \mathrm{ABC}\right)& =\frac{1}{2}\mid \begin{array}{ccc}0& 1& 1\\ \frac{-1}{2}& \frac{\sqrt{\surd 3}}{2}& 1\\ \frac{-1}{2}& \frac{-\sqrt{\surd 3}}{2}& 1\end{array}\mid \\ & =\frac{1}{2}[0-1(0)+1\left(\frac{\sqrt{\surd 3}}{2}\right)]\\ & =\frac{\sqrt{\surd 3}}{4}\text{ sq. units }\end{array}$

Out of 20 numbers from 1 to 20 , three numbers can be chosen in ${}^{20}{C}_3$ ways.

Three consecutive numbers can be chosen in one of the following ways:

$\left\{\right(1,2,3),(2,3,4),\dots (18,19,20\left)\right\}=18$ ways

$\therefore$ Probability that the numbers are consecutive

$=\frac{18}{{}^{20}{C}_3}=\frac{18}{1140}=\frac{3}{190}$

We want to estimate the value of $\frac{1}{(2.002{)}^2}$.

Let’s use a simple approximation. Let $𝑓\left(𝑥\right)=\frac{1}{{𝑥}^2}$.

The derivative of $𝑓\left(𝑥\right)$ is ${𝑓}^{\prime }\left(𝑥\right)=-2{𝑥}^{-3}=\frac{-2}{{𝑥}^3}$.

Let $𝑎=2$ and $ℎ=0.002$, so $𝑎+ℎ=2.002$.

Using linear approximation: $𝑓(𝑎+ℎ)\approx 𝑓\left(𝑎\right)+ℎ{𝑓}^{\prime }\left(𝑎\right)$.

Substitute the values into the formula: $𝑓\left(2.002\right)\approx 𝑓\left(2\right)+0.002\times {𝑓}^{\prime }\left(2\right)$ We know $𝑓\left(2\right)=\frac{1}{2^2}=\frac{1}{4}$ and ${𝑓}^{\prime }\left(2\right)=\frac{-2}{2^3}=\frac{-2}{8}=-\frac{1}{4}$.

So, $𝑓\left(2.002\right)\approx \frac{1}{4}+0.002\times (-\frac{1}{4})=\frac{1}{4}-\frac{0.002}{4}$

This means $\frac{1}{(2.002{)}^2}\approx \frac{0.998}{4}\approx 0.2495$

Let the d.r.s of the normal to the plane be proportional to $a,b,c$.

It passes through $(1,0,0)$.

$\therefore$ The equation of the plane is

$𝑎(𝑥-1)+𝑏(𝑦-0)+𝑐(𝑧-0)=0...\left(𝑖\right)$

Also, the plane passes through $(0,1,0)$.

$\begin{array}{cc}\therefore & 𝑎(-1)+𝑏\left(1\right)+𝑐\left(0\right)=0\\ & \Rightarrow 𝑎=𝑏...\left(𝑖𝑖\right)\end{array}$

Now, the angle between the required plane and the plane $𝑥+𝑦=3$ is $\frac{𝜋}{4}$.

$\begin{array}{cc}\therefore & \cos \frac{𝜋}{4}=\frac{𝑎\left(1\right)+𝑏\left(1\right)+𝑐\left(0\right)}{\sqrt{\surd {𝑎}^2+{𝑏}^2+{𝑐}^2}\sqrt{\surd 1+1}}\\ & \Rightarrow \frac{1}{\sqrt{\surd 2}}=\frac{𝑎+𝑏}{\sqrt{\surd {𝑎}^2+{𝑏}^2+{𝑐}^2}\sqrt{\surd 2}}\end{array}$

Squaring on both sides, we get

$\begin{array}{cc}& {𝑎}^2+{𝑏}^2+{𝑐}^2={𝑎}^2+{𝑏}^2+2𝑎𝑏\\ & \Rightarrow {𝑐}^2=2𝑎𝑏...\left(𝑖𝑖𝑖\right)\end{array}$

From (ii) and (iii), we get

$𝑎:𝑏:𝑐=𝑎:𝑎:\pm \sqrt{\surd 2}𝑎=1:1:\pm \sqrt{\surd 2}$

$\therefore$ Equation of the required plane is

$𝑥+𝑦\pm \sqrt{\surd 2}𝑧-1=0$

Equation of the plane passing through $(2,1,0)$, ( $3,-2,4$ ) and ( $1,-3,3$ ) is

$\begin{array}{cc}& \mid \begin{array}{ccc}𝑥-2& 𝑦-1& 𝑧-0\\ 3-2& -2-1& 4-0\\ 1-2& -3-1& 3-0\end{array}\mid =0\\ & \Rightarrow \mid \begin{array}{ccc}𝑥-2& 𝑦-1& 𝑧\\ 1& -3& 4\\ -1& -4& 3\end{array}\mid =0\\ & \Rightarrow (𝑥-2)\left(7\right)-(𝑦-1)\left(7\right)+𝑧(-7)=0\\ & \Rightarrow 𝑥-2-𝑦+1-𝑧=0\\ & \Rightarrow 𝑥-𝑦-𝑧-1=0\end{array}$

$\therefore$ The distance of this plane from $(5,3-1)$ is $d=\mid \frac{5-3+1-1}{\sqrt{\surd 1+1+1}}\mid =\frac{2}{\sqrt{\surd 3}}$ units

$\begin{array}{cc}& \int \frac{𝑑𝑥}{\cos 𝑥(1+\cos 𝑥)}\\ =& \int (\frac{1}{\cos 𝑥}-\frac{1}{1+\cos 𝑥})𝑑𝑥\\ =& \int (\sec 𝑥-\frac{1}{2{\cos }^2\frac{𝑥}{2}})𝑑𝑥\\ =& \int (\sec 𝑥-\frac{1}{2}{\sec }^2\frac{𝑥}{2})𝑑𝑥\\ =& \log |\sec 𝑥+\tan 𝑥|-\tan \left(\frac{𝑥}{2}\right)+𝑐\end{array}$

Let X denote the number of kings.

Since the card is drawn twice, 0,1 and 2 are possible values of X.

The probability of getting a king in a single draw of a card is

$\begin{array}{cc}& p=\frac{4}{52}=\frac{1}{13},q=\frac{12}{13}\\ & P(X=0)=P\left(\text{ no king }\right)=\mathrm{qq}\\ & =q^2=\frac{144}{169}\\ & \begin{array}{cc}P(X=1)& =𝑃\left(\text{ one king }\right)=\mathrm{pq}+\mathrm{qp}\\ & =2𝑝𝑞=\frac{24}{169}\end{array}\end{array}$

$\begin{array}{cc}P(X=2)& =P\left(\text{ two kings }\right)\\ & =\mathrm{pp}=p^2=\frac{1}{169}\\ E\left(X^2\right)=& \sum {𝑥}_2^{i}P\left({𝑥}_i\right)\\ =& 0+1^2\left(\frac{24}{169}\right)+2^2\left(\frac{1}{169}\right)=\frac{28}{169}\end{array}$

Co-ordinates of any point on the ellipse $\frac{{𝑥}^2}{a^2}+\frac{{𝑦}^2}{b^2}=1$ whose eccentric angle is $𝜃$ are ( $a\cos 𝜃,b\sin 𝜃$ ). $\frac{{𝑥}^2}{48}+\frac{{𝑦}^2}{16}=1$

$\begin{array}{cc}\therefore & {𝑎}^2=48\Rightarrow 𝑎=4\sqrt{\surd 3}\text{ and }\\ & {𝑏}^2=16\Rightarrow 𝑏=4\\ & 𝑥=𝑎\cos 𝜃,𝑦=𝑏\sin 𝜃\\ \Rightarrow & -6=4\sqrt{\surd 3}\cos 𝜃,2=4\sin 𝜃\\ \Rightarrow & \cos 𝜃=\frac{-3}{2\sqrt{\surd 3}},\sin 𝜃=\frac{1}{2}\\ \Rightarrow & \cos 𝜃=\frac{-\sqrt{\surd 3}}{2},\sin 𝜃=\frac{1}{2}\\ \Rightarrow & 𝜃={150}^{\circ }\dots [\cos {150}^{\circ }=\frac{-\sqrt{\surd 3}}{2},\sin {150}^{\circ }=\frac{1}{2}]\end{array}$

The centre of the given circle is $C_1(2,-3)$ and radius

$\begin{array}{cc}& =\sqrt{\surd (-2{)}^2+3^2-(-12)}\\ & =5\\ & C_1{C}_2=\sqrt{\surd (-3-2{)}^2+(2+3{)}^2}\\ & =\sqrt{\surd 25+25}\\ & =\sqrt{\surd 50}\\ & \therefore \text{ Radius of }S\text{ is }{C}_{2}A=\sqrt{\surd (\sqrt{\surd 50}{)}^2+5^2}\\ & \\ & =\sqrt{\surd 75}\\ & =5\sqrt{\surd 3}\end{array}$

$\begin{array}{cc}& \text{ Required area }\\ & \begin{array}{cc}& =2{\int }_1^0𝑦𝑑𝑥\\ & =2{\int }_1^0\sqrt{\surd 27𝑥}𝑑𝑥\\ & =6\sqrt{\surd 3}{\int }_1^0{𝑥}^{\frac{1}{2}}𝑑𝑥\end{array}\end{array}$