MHT CET 2024 11th May Morning Shift (Mathematics) Question Paper with Solutions

$\text{ Let }I=\int \frac{𝑥}{(𝑥-1{)}^2(𝑥+2)}d𝑥$

Let $\frac{𝑥}{(𝑥-1{)}^2(𝑥+2)}=\frac{A}{(𝑥-1)}+\frac{B}{(𝑥-1{)}^2}+\frac{C}{(𝑥+2)}$

$\therefore 𝑥=A(𝑥-1)(𝑥+2)+B(𝑥+2)+C(𝑥-1{)}^2$

for $𝑥=1$, we get $𝐵=\frac{1}{3}$

for $𝑥=-2$, we get $𝐶=\frac{-2}{9}$

Equating the coefficients of ${𝑥}^2$, we get

$\begin{array}{cc}& 𝐴+𝐶=0\\ \therefore & =\frac{2}{9}\\ \therefore & =\frac{2}{9}\int \frac{1}{(𝑥-1)}𝑑𝑥+\frac{1}{3}\int \frac{1}{(𝑥-1{)}^2}𝑑𝑥-\frac{2}{9}\int \frac{1}{(𝑥+2)}𝑑𝑥\\ & =\frac{2}{9}\log (𝑥-1)-\frac{1}{3}\times \frac{1}{(𝑥-1)}-\frac{2}{9}\log (𝑥+2)+𝑐\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \text{ Let }𝑧=𝑎+𝑖𝑏\\ \therefore & \left|𝑧\right|=\sqrt{\surd {𝑎}^2+{𝑏}^2}\\ \therefore & \left|𝑧\right|+𝑧=2+𝑖\\ \therefore & \sqrt{\surd {𝑎}^2+{𝑏}^2}+𝑎+𝑖𝑏=2+𝑖\end{array}\\ & \text{ Comparing both sides, we get }\\ & \begin{array}{cc}& \sqrt{\surd {𝑎}^2+{𝑏}^2}+𝑎=2\text{ and }𝑏=1\\ \therefore & \sqrt{\surd 1+{𝑎}^2}+𝑎=2\\ \therefore & 1+{𝑎}^2=(2-𝑎{)}^2\\ \therefore & 1+{𝑎}^2=4-4𝑎+{𝑎}^2\\ \therefore & 4𝑎=3\end{array}\end{array}$

$\begin{array}{cc}\therefore & 𝑎=\frac{3}{4}\\ \therefore & \left|𝑧\right|=\sqrt{\surd {𝑎}^2+{𝑏}^2}=\frac{5}{4}\end{array}$

All seven-digit numbers formed with the given digits are greater than a million.

Note that the digit at millions place cannot be 0. It can be any one of the digits $2,3,4$.

$\therefore$ It can be any one of the digits $2,3,4$.

Case I :

Digit at millions place is ' 2 '.

Remaining 6 digits can be arranged in $\frac{6!}{3!}$ $=120$ ways.

Case II :

Digit at millions place is ' 3 '.

Remaining 6 digits can be arranged in $\frac{6!}{2!\times 2!}$ $=180$ ways.

Case III :

Digit at millions place is ' 4 '.

Remaining 6 digits can be arraigned in $\frac{6!}{31\times 2!}$ $=60$ ways

$\begin{array}{cc}\therefore \text{ Required number of numbers }& =120+180+60\\ & =360\end{array}$

The shaded region lies in the first quadrant, which gives:

$𝑥\ge 0,𝑦\ge 0$

The region lies on and below the line $𝑦=𝑥$, giving:

$𝑦\le 𝑥⟹𝑥-𝑦\ge 0$

The region also lies on and below the line $-𝑥+3𝑦=3$, giving:

$-𝑥+3𝑦\le 3$

Therefore, the linear constraints defining the shaded region are $𝑥-𝑦\ge 0$, $-𝑥+3𝑦\le 3$, $𝑥\ge 0$, and $𝑦\ge 0$.

Hence, the correct option is (C).

$\begin{array}{cc}\text{ Let }L& =\underset{𝑥\to 2}{lim}\left(\frac{5^{𝑥}+5^{3-𝑥}-30}{5^{3-𝑥}-5^{\frac{𝑥}{2}}}\right)\\ & =\underset{𝑥\to 2}{lim}⎛⎝\frac{5^{𝑥}+\frac{5^3}{5^{𝑥}}-30}{\frac{5^3}{5^{𝑥}}-{\left(5^{𝑥}\right)}^{\frac{1}{2}}}⎞⎠\end{array}$

$\begin{array}{cc}& \text{ Let }𝑡=5^{𝑥}\\ & \begin{array}{cc}\therefore 𝑥& \to 2\Rightarrow 𝑡\to 25\\ \therefore 𝐿& =\underset{𝑡\to 25}{lim}\left(\frac{𝑡+\frac{125}{𝑡}-30}{\frac{125}{𝑡}-\sqrt{\surd 𝑡}}\right)\\ & =\underset{𝑡\to 25}{lim}\left(\frac{{𝑡}^2-30𝑡+125}{25\sqrt{\surd 25}-𝑡\sqrt{\surd 𝑡}}\right)\\ & =\underset{𝑡\to 25}{lim}\left(\frac{(𝑡-25)(𝑡-5)}{{25}^{\frac{3}{2}}-{𝑡}^{\frac{3}{2}}}\right)\\ & =\underset{𝑡\to 25}{lim}\frac{𝑡-5}{-\left(\frac{{25}^{\frac{3}{2}}-{𝑡}^{\frac{3}{2}}}{25-𝑡}\right)}\\ & =\frac{25-5}{-\frac{3}{2}(25{)}^{\frac{1}{2}}}\\ & =\frac{-8}{3}\end{array}\end{array}$

Let $I=\int (7𝑥-2)\sqrt{\surd 3𝑥+2}d𝑥$

Let $3𝑥+2=t\Rightarrow 𝑥=\frac{t-2}{3}\Rightarrow \mathrm{dx}=\frac{1}{3}\mathrm{dt}$

$\begin{array}{cc}\therefore 𝐼& =\int [7\left(\frac{t-2}{3}\right)-2]\sqrt{\surd t}\mathrm{dt}\\ & =\frac{1}{3}\int (\frac{7t}{3}-\frac{20}{3})\sqrt{\surd t}\mathrm{dt}\\ & =\frac{7}{9}\int t^{\frac{3}{2}}\mathrm{dt}-\frac{20}{9}\int t^{\frac{1}{2}}\mathrm{dt}\\ & =\frac{14}{45}(3𝑥+2{)}^{\frac{5}{2}}-\frac{40}{27}(3𝑥+2{)}^{\frac{3}{2}}+𝐶\\ \therefore & 𝐴=\frac{14}{45}\text{ and }𝐵=\frac{-40}{27}\end{array}$

$\begin{array}{cc}& 𝑦={\sin }^{-1}\left(\frac{\log {𝑥}^2}{1+(\log 𝑥{)}^2}\right)={\sin }^{-1}\left(\frac{2\log 𝑥}{1+(\log 𝑥{)}^2}\right)\\ & \text{ Let }\log 𝑥=\tan 𝜃\\ \therefore & 𝑦={\sin }^{-1}\left(\frac{2\tan 𝜃}{1+\tan 2𝜃}\right)={\sin }^{-1}(\sin 2𝜃)=2𝜃\\ \therefore & 𝑦=2{\tan }^{-1}(\log 𝑥)\\ \therefore & \frac{𝑑𝑦}{𝑑𝑥}=2\times \frac{1}{1+(\log 𝑥{)}^2}\times \frac{1}{𝑥}\\ \therefore & \left(\frac{d𝑦}{𝑑𝑥}\right){\mid }_{𝑎𝑡𝑥=1}=2\end{array}$

$\begin{array}{cc}& \underset{𝑥\to 0^{-}}{lim}f\left(𝑥\right)=f\left(0\right)=\underset{𝑥\to 0^{+}}{lim}f\left(𝑥\right)\\ \therefore & \underset{𝑥\to 0}{lim}\frac{a}{2}(𝑥-|𝑥\left|\right)=0\\ \therefore & \underset{𝑥\to 0}{lim}\frac{a}{2}[𝑥-(-𝑥\left)\right]=0\dots [\because 𝑥<0\Rightarrow |𝑥|=-𝑥]\\ \therefore & \underset{𝑥\to 0}{lim}𝑥=0\end{array}$

Which is true for any real value of a.

$\underset{𝑥\to 0^{+}}{lim}𝑓\left(𝑥\right)=𝑓\left(0\right)$

$\therefore \underset{𝑥\to 0}{lim}𝑏{𝑥}^2\sin \left(\frac{1}{𝑥}\right)=0$

Note that $𝑥\ne 0$

$\therefore -1\le \sin \left(\frac{1}{𝑥}\right)\le 1$

$\therefore$ for any real value of $𝑏$, above limit will be 0.

$\begin{array}{cc}P(X\le 1)& =P(X=0)+P(X=1)\\ & =\frac{{}^5{C}_0}{2^5}+\frac{{}^5{C}_1}{2^5}=\frac{6}{2^5}\\ P(X\le 2)& =P(X=0)+P(X=1)+P(X=2)\\ & =\frac{{}^5{C}_0}{2^5}+\frac{{}^5{C}_1}{2^5}+\frac{{}^5{C}_2}{2^5}=\frac{16}{2^5}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}P(X\ge 3)& =P(X=3)+P(X=4)+P(X=5)\\ & =\frac{{}^5{C}_3}{2^5}+\frac{{}^5{C}_4}{2^5}+\frac{{}^5{C}_5}{2^5}=\frac{16}{2^5}\\ P(X\le 3)& =P(X=0)+P(X=1)+P(X=2)\\ & =\frac{{}^5{C}_0}{2^5}+\frac{{}^5{C}_1}{2^5}+\frac{{}^5{C}_2}{2^5}+\frac{{}^5{C}_3}{2^5}=\frac{26}{2^5}\\ P(X\ge 4)& =P(X=4)+P(X=5)\\ & =\frac{{}^5{C}_4}{2^5}+\frac{{}^5{C}_5}{2^5}=\frac{6}{2^5}\end{array}\\ \therefore & P(X\le 2)>P(X\ge 3)\text{ is not true. }\end{array}$

$𝑦=(1+𝑥)(1+{𝑥}^2)(1+{𝑥}^4)(1+{𝑥}^8)\text{..... (i)}$

Taking 'log' on both sides, we get

$\begin{array}{cc}& \log 𝑦=\log (1+𝑥)+\log (1+{𝑥}^2)+\log (1+{𝑥}^4)\\ & +\log (1+{𝑥}^8)\end{array}$

Differentiating w.r.t. $𝑥$, we get

$\frac{1}{𝑦}\frac{d𝑦}{d𝑥}=\frac{1}{1+𝑥}+\frac{2𝑥}{1+{𝑥}^2}+\frac{4{𝑥}^3}{1+{𝑥}^4}+\frac{8{𝑥}^7}{1+{𝑥}^8}\text{.... (ii)}$

At $𝑥=1$, (i) $\Rightarrow 𝑦=16$

$\therefore \left(\mathrm{ii}\right)\Rightarrow {\left(\frac{d𝑦}{d𝑥}\right)}_{𝑥=1}=16(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2})=120$

$\begin{array}{cc}{\int }_{35}^{0.2}\left[𝑥\right]d𝑥& ={\int }_1^{0.2}\left(0\right)d𝑥+{\int }_2^1\left(1\right)d𝑥+{\int }_3^{2}2d𝑥+{\int }_{3.5}^{3}3d𝑥\\ & =0+[𝑥{]}_2^1+2[𝑥{]}_3^2+3[𝑥{]}_{3.5}^3\\ & =0+(2-1)+2(3-2)+3(3.5-3)\\ & =0+1+2+1.5\\ & =4.5\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑞\vee \sim 𝑝\\ & <text>\equiv </text><text>\sim </text>𝑞<text>\to </text><text>\sim </text>𝑝\dots [\because 𝑝\to 𝑞<text>\equiv </text><text>\sim </text>𝑝\vee 𝑞]\end{array}\\ & \therefore \text{ Option (D) is correct. }\end{array}$

$\begin{array}{cc}& \text{ Initial number of bacteria }=N\\ \therefore & \text{ After }8\text{ hours number of bacteria }=2N\\ \therefore & \text{ After }16\text{ hours number of bacteria }=4N\\ \therefore & \text{ After }24\text{ hours number of bacteria }=8N\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑓\left(𝑥\right)={𝑥}^3-6{𝑥}^2+9𝑥+3\\ \therefore & {𝑓}^{\prime }\left(𝑥\right)=3{𝑥}^2-12𝑥+9\end{array}\\ & \text{ Since }𝑓\left(𝑥\right)\text{ is monotonically decreasing, }\\ & \begin{array}{cc}& f^{\prime }\left(𝑥\right)<0\\ & \Rightarrow 3{𝑥}^2-12𝑥+9<0\\ & \Rightarrow {𝑥}^2-4𝑥+3<0\\ & \Rightarrow (𝑥-3)(𝑥-1)<0\\ & \Rightarrow 𝑥\in (1,3)\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given: }\\ & \begin{array}{cc}& [𝑥{]}^2-5[𝑥]+6=0\\ & \text{ Let }\left[𝑥\right]=a\\ & \Rightarrow a^2-5a+6=0\\ & \Rightarrow (a-2)(a-3)=0\\ & \Rightarrow a=2\text{ or }a=3\\ & \Rightarrow \left[𝑥\right]=2\text{ or }\left[𝑥\right]=3\\ & \Rightarrow 𝑥\in [2,3)\text{ or }𝑥\in [3,4)\\ & \Rightarrow 𝑥\in [2,4)\end{array}\end{array}$

The formula to calculate the perpendicular distance of a point $({𝑥}_0,{𝑦}_0,{𝑧}_0)$ from a plane $𝐴𝑥+𝐵𝑦+𝐶𝑧+𝐷=0$ is given by:

$\text{Distance}=\frac{|𝐴{𝑥}_0+𝐵{𝑦}_0+𝐶{𝑧}_0+𝐷|}{\sqrt{\surd {𝐴}^2+{𝐵}^2+{𝐶}^2}}$

For the plane $2𝑥+𝑦-2𝑧-18=0$, the coefficients are $𝐴=2$, $𝐵=1$, $𝐶=-2$, and $𝐷=-18$. The point for which we are calculating the distance is the origin $(0,0,0)$.

Substituting these values into the formula:

$\text{Distance}=\frac{\left|2\right(0)+1(0)-2(0)-18|}{\sqrt{\surd 2^2+1^2+(-2{)}^2}}$

Simplify the numerator:

$=\frac{|-18|}{\sqrt{\surd 4+1+4}}$

Further simplification gives:

$=\frac{18}{\sqrt{\surd 9}}$

Since $\sqrt{\surd 9}=3$, the expression simplifies to:

$=\frac{18}{3}=6$

Therefore, the perpendicular distance of the origin from the plane is 6 units.

Option C: 6 units

Inverse of $𝑝\to 𝑞$ is $\sim 𝑝<text>\to </text><text>\sim </text>𝑞$ i.e., If a man is not judge, then he is not honest.

$𝑥=\sqrt{\surd t},𝑦=t-\frac{1}{\sqrt{\surd t}}$

$\therefore 𝑦={𝑥}^2-\frac{1}{𝑥}$, at $t=4,𝑥=2$ and $𝑦=\frac{7}{2}$

$\therefore \frac{d𝑦}{d𝑥}=2𝑥+\frac{1}{{𝑥}^2}$

$\therefore$ Slope of the normal at $t=4$ is $\frac{-1}{{\left(\frac{d𝑦}{d𝑥}\right)}_{t=4}}=\frac{-4}{17}$

$\therefore$ Required equation is $(𝑦-\frac{7}{2})=\frac{-4}{17}(𝑥-2)$

i.e., $8𝑥+34𝑦=135$

Let r.v X denotes the no. of heads obtained

$\therefore$ Probability distribution of X is given as

$\begin{array}{cc}E\left(𝑥\right)& =0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}=\frac{3}{2}\\ E\left({𝑥}^2\right)& =0\times \frac{1}{8}+1\times \frac{3}{8}+4\times \frac{3}{8}+9\times \frac{1}{8}=3\\ \therefore V\left(𝑥\right)& =E\left({𝑥}^2\right)-\left[E\right(𝑥){]}^2\\ & =3-\frac{9}{4}\\ & =\frac{3}{4}\\ & =0.75\end{array}$

$\begin{array}{cc}& \text{ Note that }\\ & \begin{array}{cc}& A\equiv (\frac{1}{2},0,0),B\equiv (0,\frac{1}{3},0),C\equiv (0,0,\frac{1}{4})\\ & \therefore \text{ Cendroid }=(\frac{\frac{1}{2}+0+0}{3},\frac{0+\frac{1}{3}+0}{3},\frac{0+0+\frac{1}{4}}{3})\\ & =(\frac{1}{6},\frac{1}{9},\frac{1}{12})\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& {𝑥}^2+{𝑦}^2=t+\frac{1}{t}\text{ and }{𝑥}^4+{𝑦}^4=t^2+\frac{1}{t^2}\\ \therefore & {({𝑥}^2+{𝑦}^2)}^2={(t+\frac{1}{t})}^2\\ \therefore & {𝑥}^4+{𝑦}^4+2{𝑥}^2{𝑦}^2=t^2+\frac{1}{t^2}+2\\ \therefore & t^2+\frac{1}{t^2}+2{𝑥}^2{𝑦}^2=t^2+\frac{1}{t^2}+2\\ \therefore & {𝑥}^2{𝑦}^2=1\\ \therefore & {𝑦}^2=\frac{1}{{𝑥}^2}\end{array}\\ & \text{ Differentiating w.r.t. }𝑥\text{, we get }\\ & \begin{array}{cc}& 2𝑦\frac{d𝑦}{d𝑥}=\frac{-2}{{𝑥}^3}\\ \therefore & \frac{d𝑦}{d𝑥}=\frac{-1}{{𝑥}^3𝑦}\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \frac{d𝑦}{d𝑥}+\sin \left(\frac{𝑥+𝑦}{2}\right)=\sin \left(\frac{𝑥-𝑦}{2}\right)\\ & \Rightarrow \frac{d𝑦}{d𝑥}=\sin \left(\frac{𝑥-𝑦}{2}\right)-\sin \left(\frac{𝑥+𝑦}{2}\right)\\ & \Rightarrow \frac{d𝑦}{d𝑥}=-2\sin \left(\frac{𝑦}{2}\right)\cdot \cos \left(\frac{𝑥}{2}\right)\end{array}\\ & \text{ Integrating on both sides, we get }\\ & \begin{array}{cc}& \int \mathrm{cosec}\left(\frac{𝑦}{2}\right)𝑑𝑦=-\int 2\cos \left(\frac{𝑥}{2}\right)𝑑𝑥+{𝑐}_1\\ & \Rightarrow \frac{\log \tan \left(\frac{𝑦}{4}\right)}{\frac{1}{2}}=-\frac{2\sin \left(\frac{𝑥}{2}\right)}{\frac{1}{2}}+{𝑐}_1\\ & \Rightarrow \log \tan \left(\frac{𝑦}{4}\right)=𝐶-2\sin \left(\frac{𝑥}{2}\right),\text{ where }𝐶=\frac{1}{2}{𝑐}_1\end{array}\end{array}$

Given that $𝐴$ and $𝐵$ are independent events, the following relationships hold:

$P\left({𝐴}^{\prime }\right)=0.75$

$P(𝐴\cup 𝐵)=0.65$

$P\left(𝐵\right)=𝑝$

We need to find the value of $𝑝$.

First, recall the formula for the probability of the complement event:

$P\left(𝐴\right)=1-P\left({𝐴}^{\prime }\right)=1-0.75=0.25$

Next, apply the formula for the probability of the union of two independent events:

$P(𝐴\cup 𝐵)=P\left(𝐴\right)+P\left(𝐵\right)-P\left(𝐴\right)\cdot P\left(𝐵\right)$

Substitute the given values into the equation:

$0.65=0.25+𝑝-(0.25\cdot 𝑝)$

Simplify:

$0.65=0.25+𝑝-0.25𝑝$

$0.65=0.25+𝑝(1-0.25)$

$0.65=0.25+0.75𝑝$

Rearrange and solve for $𝑝$:

$0.75𝑝=0.65-0.25$

$0.75𝑝=0.40$

$𝑝=\frac{0.40}{0.75}$

$𝑝=\frac{40}{75}=\frac{8}{15}$

Thus, the value of $𝑝$ is $\frac{8}{15}$.

Therefore, the correct option is Option D: $\frac{8}{15}$.

$\begin{array}{cc}& \text{ Since the given lines intersect, }\\ & \begin{array}{cc}& \mid \begin{array}{ccc}{𝑥}_2-{𝑥}_1& {𝑦}_2-{𝑦}_1& {𝑧}_2-{𝑧}_1\\ {𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\end{array}\mid =0\\ & \therefore \mid \begin{array}{ccc}-10+1& -1+𝑘& 1-4\\ -10& -1& 1\\ -1& -3& 4\end{array}\mid =0\\ & \Rightarrow \mid \begin{array}{ccc}-9& -1+𝑘& -3\\ -10& -1& 1\\ -1& -3& 4\end{array}\mid =0\\ & \Rightarrow 39k=117\\ & \Rightarrow k=3\end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }f\left(𝑥\right)={𝑥}^{\frac{3}{2}}\\ & \therefore f^{\prime }\left(𝑥\right)=\frac{3}{2}{𝑥}^{\frac{1}{2}}\\ & \text{ Here, }a=4,h=-0.022\\ & f\left(3.978\right)=(4{)}^{\frac{3}{2}}=-0.022\times \frac{3}{2}\times (4{)}^{\frac{1}{2}}\\ & =8-0.033\times 2\\ & =8-0.066=7.934\end{array}$

$\begin{array}{cc}& \text{ Let }𝑥=\cos 2𝜃\\ & \begin{array}{cc}\therefore & 𝜃=\frac{1}{2}{\cos }^{-1}𝑥\\ \therefore & {\tan }^{-1}\left(\frac{\sqrt{\surd 1+𝑥}-\sqrt{\surd 1-𝑥}}{\sqrt{\surd 1+𝑥}+\sqrt{\surd 1-𝑥}}\right)\\ =& {\tan }^{-1}\left(\frac{\sqrt{\surd 1+\cos 2𝜃}-\sqrt{\surd 1-\cos 2𝜃}}{\sqrt{\surd 1+\cos 2𝜃}+\sqrt{\surd 1-\cos 2𝜃}}\right)\\ =& {\tan }^{-1}\left(\frac{\sqrt{\surd 2}\cos 𝜃-\sqrt{\surd 2}\sin 𝜃}{\sqrt{\surd 2}\cos 𝜃+\sqrt{\surd 2}\sin 𝜃}\right)\\ =& {\tan }^{-1}\left(\frac{1-\tan 𝜃}{1+\tan 𝜃}\right)\\ =& {\tan }^{-1}\left(1\right)-\tan {}^{-1}(\tan 𝜃)\\ & =\frac{𝜋}{4}-𝜃\\ & =\frac{𝜋}{4}-\frac{1}{2}{\cos }^{-1}𝑥\end{array}\end{array}$

$\begin{array}{cc}\therefore & {\tan }^{-1}\left(\frac{\sqrt{\surd 1+𝑥}-\sqrt{\surd 1-𝑥}}{\sqrt{\surd 1+𝑥}+\sqrt{\surd 1-𝑥}}\right)+\frac{1}{2}{\cos }^{-1}𝑥\\ & =\frac{𝜋}{4}-\frac{1}{2}{\cos }^{-1}𝑥+\frac{1}{2}{\cos }^{-1}𝑥=\frac{𝜋}{4}\end{array}$