MHT CET 2025 19th April Morning Shift (Mathematics) Question Paper with Solutions

The equation of plane through $𝐴(7,8,6)$ is

$𝑎(𝑥-7)+𝑏(𝑦-8)+𝑐(𝑧-6)=0\text{...... (i)}$

Since, the required plane is parallel to XY -plane

$\therefore$ Direction ratios are $0,0,1$

$\therefore$ Equation of plane is

$\begin{array}{cc}& 0(𝑥-7)+0(𝑦-0)+1(𝑧-6)=0\\ & \Rightarrow 𝑧-6=0\\ & \Rightarrow 𝑧=6\end{array}$

6 boys can seat at a round table in $(6-1)!=5!$ ways. Now, no two girls sit together, i.e., 5 girls are to be arranged in 6 empty seats between two consecutive boys and number of such arrangements will be ${}^6{P}_5=6$ !

Hence, required number of ways

$\begin{array}{cc}& =5!\times 6!\\ & =120\times 720\\ & =86400\end{array}$

$\begin{array}{cc}& \left[\begin{array}{ccc}2\overline{𝑝}-3\overline{𝑟}& \overline{𝑞}& \overline{𝑠}\end{array}\right]+\left[\begin{array}{ccc}3\overline{𝑝}+2\overline{𝑞}& \overline{𝑟}& \overline{𝑠}\end{array}\right]\\ & =m\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+n\left[\begin{array}{ccc}\overline{𝑞}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+t\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑞}& \overline{𝑠}\end{array}\right]\\ & \left[\begin{array}{ccc}2\overline{𝑝}& \overline{𝑞}& \overline{𝑠}\end{array}\right]-\left[\begin{array}{ccc}\overline{3𝑟}& \overline{𝑞}& \overline{𝑠}\end{array}\right]+\left[\begin{array}{ccc}3\overline{𝑝}& \overline{𝑟}& \overline{𝑠}\end{array}\right]\\ & +\left[\begin{array}{ccc}2\overline{𝑞}& \overline{𝑟}& \overline{𝑠}\end{array}\right]\end{array}$

$\begin{array}{c}=m\left[\begin{array}{ccc}\bar{p}& \bar{r}& \bar{s}\end{array}\right]+n\left[\begin{array}{ccc}\bar{q}& \bar{r}& \bar{s}\end{array}\right]+t\left[\begin{array}{ccc}\bar{p}& \bar{q}& \bar{s}\end{array}\right]\\ 3\left[\begin{array}{ccc}\bar{p}& \bar{r}& \bar{s}\end{array}\right]+3\left[\begin{array}{ccc}\bar{q}& \bar{r}& \bar{s}\end{array}\right]+2\left[\begin{array}{ccc}\bar{q}& \bar{r}& \bar{s}\end{array}\right]\\ +\left[\begin{array}{ccc}2\bar{p}& \bar{q}& \bar{s}\end{array}\right]\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& =𝑚\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+𝑛\left[\begin{array}{ccc}\overline{𝑞}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+𝑡\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑞}& \overline{𝑠}\end{array}\right]\\ & 3\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+5\left[\begin{array}{ccc}\overline{𝑞}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+2\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑞}& \overline{𝑠}\end{array}\right]\\ & =𝑚\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+𝑛\left[\begin{array}{ccc}\overline{𝑞}& \overline{𝑟}& \overline{𝑠}\end{array}\right]+𝑡\left[\begin{array}{ccc}\overline{𝑝}& \overline{𝑞}& \overline{𝑠}\end{array}\right]\end{array}\\ & \text{ Comparing above equations, we get }\\ & m=3,n=5,t=2\end{array}$

Let $P\equiv (-3,2,3)$

Equation of line passing through $(4,6,-2)$ and having d.r.'s $-1,2,3$ are

$\frac{𝑥-4}{-1}=\frac{𝑦-6}{2}=\frac{𝑧+2}{3}$

Let $\frac{𝑥-4}{-1}=\frac{𝑦-6}{2}=\frac{𝑧+2}{3}=𝜆$

$\Rightarrow 𝑥=4-𝜆,𝑦=2𝜆+6,z=3𝜆-2$

Co-ordinates of point

$𝑄=(4-𝜆,2𝜆+6,3𝜆-2)$

$\therefore$ the d.r's of PQ are

$\begin{array}{cc}& (4-𝜆-(-3),6+2𝜆-2,-2+3𝜆-3)\\ & \equiv 7-𝜆,4+2𝜆,-5+3𝜆\end{array}$

Since PQ is perpendicular to given lines

$\begin{array}{cc}\therefore & -1(7-𝜆)+\left(2\right)(4+2𝜆)+3(-5+3𝜆)=0\\ & \Rightarrow -7+𝜆+8+4𝜆-15+9𝜆=0\\ & \Rightarrow 𝜆=1\end{array}$

$\therefore$ Co-ordinates of Q are $(3,8,1)$

$\therefore \mathrm{PQ}=\sqrt{\surd (3-(-3){)}^2+(8-2{)}^2+(1-3{)}^2}$

$=\sqrt{\surd 76}=2\sqrt{\surd 19}$

The equation of a plane passing through $(1,-2,1)$ is

$𝑎(𝑥-1)+𝑏(𝑦+2)+𝑐(𝑧-1)=0\text{..... (i)}$

Plane (i) is perpendicular to planes

$2𝑥-2𝑦+𝑧=0\text{ and }𝑥-𝑦+2𝑧=4$

$\therefore 2a-2b+c=0$, and $\text{..... (ii)}$

$𝑎-𝑏+2𝑐=0\text{..... (iii)}$

Solving (ii) and (iii), we get

$𝑎=-3,𝑏=-3,𝑐=0$

Substituting the values of $a,b,c$ in equation (i), we get

$𝑥+𝑦+1=0$

$\therefore$ The distance of this plane from $(1,2,2)$ is

$𝑑=\mid \frac{1+2+1}{\sqrt{\surd 1+1}}\mid =2\sqrt{\surd 2}$

Equation of diagonal AC

$\begin{array}{cc}& \frac{𝑦-{𝑦}_1}{{𝑦}_1-{𝑦}_2}=\frac{𝑥-{𝑥}_1}{{𝑥}_1-{𝑥}_2}\\ & \Rightarrow \frac{𝑦-11}{11-12}=\frac{𝑥-8}{8-10}\\ & \Rightarrow \frac{𝑦-11}{-1}=\frac{𝑥-8}{-2}\\ & \Rightarrow 𝑥-2𝑦+14=0\text{...... (i)}\end{array}$

Equation of diagonal BD is

$\begin{array}{cc}& \frac{𝑦-{𝑦}_1}{{𝑦}_1-{𝑦}_2}=\frac{𝑥-{𝑥}_1}{{𝑥}_1-{𝑥}_2}\\ & \Rightarrow \frac{𝑦-11}{11-12}=\frac{𝑥-10}{10-8}\\ & \Rightarrow \frac{𝑦-11}{-1}=\frac{𝑥-10}{2}\\ & \Rightarrow 𝑥+2𝑦-32=0\text{..... (ii)}\end{array}$

Solving equations (i) and (ii) we get

$𝑥=9,𝑦=\frac{23}{2}$

Number of tickets $=9$

Number of odd numbered tickets $=5$

Number of even numbered tickets $=4$

Required probability

$=P$ {odd, even, odd} + $P$ {even, odd, even}

$\begin{array}{cc}& =\frac{{}^5{C}_1}{{}_9{C}_1}\times \frac{{}^4{C}_1}{{}^8{C}_1}\times \frac{{}^4{C}_1}{{}^7{C}_1}+\frac{{}^4{C}_1}{{}^9{C}_1}\times \frac{{}^5{C}_1}{{}^8{C}_1}\times \frac{{}^3{C}_1}{{}^7{C}_1}\\ & =\frac{5}{9}\times \frac{4}{8}\times \frac{4}{7}+\frac{4}{9}\times \frac{5}{8}\times \frac{3}{7}=\frac{80}{504}+\frac{60}{504}=\frac{140}{504}=\frac{5}{18}\end{array}$

$\underset{𝑥\to 3}{lim}\frac{(84-𝑥{)}^{\frac{1}{4}}-3}{𝑥-3}$

Let $(84-𝑥{)}^{\frac{1}{4}}=t$

$\begin{array}{cc}& \Rightarrow (84-𝑥)={𝑡}^4\\ & \Rightarrow 𝑥=84-{𝑡}^4\end{array}$

As $𝑥\to 3$ then $t\to 3$

$\begin{array}{cc}\therefore \underset{𝑥\to 3}{lim}\frac{(84-𝑥{)}^{\frac{1}{4}}-3}{𝑥-3}& =\underset{𝑡\to 3}{lim}\frac{𝑡-3}{84-{𝑡}^4-3}\\ & =\underset{𝑡\to 3}{lim}\frac{𝑡-3}{81-{𝑡}^4}\\ & =\underset{𝑡\to 0}{lim}\frac{-1}{\frac{({𝑡}^4-81)}{𝑡-3}}\\ & =\frac{-1}{4\times (3{)}^3}=\frac{-1}{108}\end{array}$

$\therefore \left[\right(𝑝\to 𝑞)\wedge \sim 𝑞]\to 𝑟$ is a tautology when all the entries in the last column are $𝑇$, which is only possible when $r<text>\equiv </text><text>\sim </text>q$

Let $\frac{𝛼+𝛽}{2}=𝜃,\frac{𝛼-𝛽}{2}=𝜙$

$\Rightarrow 𝛼+𝛽=2𝜃\text{..... (i)}$

$𝛼-𝛽=2𝜙\text{..... (ii)}$

Solving (i) and (ii) we get

$𝛼=𝜃+𝜙,𝛽=𝜃-𝜙$

Now,

$\begin{array}{cc}& \sin (𝜃+𝜙)=\sin 𝜃\cos 𝜙+\cos 𝜃\sin 𝜙\\ & \sin (𝜃-𝜙)=\sin 𝜃\cos 𝜙-\cos 𝜃\sin 𝜙\end{array}$

Given equation is

$3\sin 𝛼=5\sin 𝛽$

$\begin{array}{cc}& \Rightarrow 3\sin (𝜃+𝜙)=5\sin (𝜃-𝜙)\\ & \Rightarrow 3\sin 𝜃\cos 𝜙+3\cos 𝜃\sin 𝜙\\ & =5\sin 𝜃\cos 𝜙-5\cos 𝜃\sin 𝜙\\ & \Rightarrow -2\sin 𝜃\cos 𝜙+8\cos 𝜃\sin 𝜙=0\\ & \Rightarrow 4\cos 𝜃\sin 𝜙=\sin 𝜃\cos 𝜙\\ & \Rightarrow \frac{\sin 𝜙}{\cos 𝜙}=\frac{\sin 𝜃}{4\cos 𝜃}\\ & \Rightarrow \tan 𝜙=\frac{1}{4}\tan 𝜃\\ & \Rightarrow \frac{\tan 𝜃}{\tan 𝜙}=4\\ & \begin{array}{cc}& \Rightarrow \frac{\tan \left(\frac{𝛼+𝛽}{2}\right)}{\tan \left(\frac{𝛼-𝛽}{2}\right)}=4\\ & \text{ ...[From (i) and (ii)] }\end{array}\end{array}$

$\begin{array}{cc}& \text{ Let }I=\int \frac{d𝑥}{2e^{2𝑥}+3e^{𝑥}+1}\\ \therefore & \text{ Let }{e}^{𝑥}=t\\ & e^{𝑥}d𝑥=\mathrm{dt}\\ & \Rightarrow d𝑥=\frac{\mathrm{dt}}{e^{𝑥}}=\frac{\mathrm{dt}}{t}\end{array}$

$\begin{array}{cc}\therefore 𝐼& =\int \frac{𝑑𝑡}{𝑡(2{𝑡}^2+3𝑡+1)}\\ & =\int \frac{𝑑𝑡}{𝑡(2𝑡+1)(𝑡+1)}\\ & \text{ Let }\frac{1}{𝑡(2𝑡+1)(𝑡+1)}=\frac{𝐴}{𝑡}+\frac{𝐵}{(2𝑡+1)}+\frac{𝐶}{𝑡+1}\\ & \Rightarrow 1=𝐴(2𝑡+1)(𝑡+1)+𝐵𝑡(𝑡+1)+𝐶𝑡(2𝑡+1)\text{..... (i)}\end{array}$

Put $𝑡=-1$ in (i), we get

$𝐶=1$

Put $𝑡=\frac{-1}{2}$ in (i), we get

$𝐵=-4$

Put $𝑡=0$ in (i), we get

$𝐴=1$

$\begin{array}{cc}\therefore & \int \frac{𝑑𝑡}{𝑡(2𝑡+1)(𝑡+1)}\\ & =\int \frac{1}{𝑡}𝑑𝑡-4\int \frac{1}{2𝑡+1}𝑑𝑡+\int \frac{1}{𝑡+1}𝑑𝑡\\ & =\log 𝑡-\frac{1}{2}\times 4\log |2𝑡+1|+\log |𝑡+1|+𝑐\\ & =\log {𝑒}^{𝑥}+\log |{𝑒}^{𝑥}+1|-2\log |2{𝑒}^{𝑥}+1|+𝑐\\ & =𝑥+\log ({𝑒}^{𝑥}+1)-2\log (2{𝑒}^{𝑥}+1)+𝑐\end{array}$

$\begin{array}{cc}\text{ Let }I& =\int \frac{e^{2030\log 𝑥}-e^{2029\log 𝑥}}{e^{2028\log 𝑥}-e^{2027\log 𝑥}}d𝑥\\ & =\int \frac{e^{\log {𝑥}^{2030}}-e^{\log {𝑥}^{2029}}}{e^{\log {𝑥}^{2028}}-e^{\log {𝑥}^{2027}}}d𝑥\\ & =\int \frac{{𝑥}^{2030}-{𝑥}^{2029}}{{𝑥}^{2028}-{𝑥}^{2027}}d𝑥\\ & =\int \frac{{𝑥}^{2027}({𝑥}^3-{𝑥}^2)}{{𝑥}^{2027}(𝑥-1)}d𝑥\\ & =\int \frac{{𝑥}^2(𝑥-1)}{(𝑥-1)}d𝑥\\ & =\int {𝑥}^{2}d𝑥\\ & =\frac{{𝑥}^3}{3}+c\end{array}$

$\begin{array}{cc}& \text{ Let }𝐼={\int }_4^1\log \left[𝑥\right]𝑑𝑥\\ & \begin{array}{cc}{\int }_4^1\log \left[𝑥\right]𝑑𝑥& ={\int }_2^{1}0𝑑𝑥+{\int }_3^2\log 2𝑑𝑥+{\int }_4^3\log 3𝑑𝑥\\ & =0+\log 2[3-2]+\log 3[4-3]\\ & =\log 2+\log 3\\ & =\log (2\times 3)=\log 6\end{array}\end{array}$

Given equation of parabola

${𝑥}^2=4𝑦$

The equation of tangent to parabola ${𝑥}^2=4𝑦$ is

$𝑥=m𝑦+\frac{a}{m}$

$\begin{array}{cc}& \Rightarrow 𝑥=m𝑦+\frac{1}{m}\dots [\because 𝑎=1]\\ & \Rightarrow m𝑥=m^2𝑦+1\\ & \Rightarrow \frac{d𝑦}{d𝑥}𝑥={\left(\frac{d𝑦}{d𝑥}\right)}^2𝑦+1\end{array}$

The order and degree of above differential equation is 1 and 2.

First, write the given probability distribution:

$\begin{array}{cccccc}𝑋& 0& 1& 2& 3& 4\\ 𝑃(𝑋=𝑥)& 2𝑘& 𝑘& 2𝑘& 4𝑘& 𝑘\end{array}$

Step 1: Using the sum of probabilities

The sum of all probabilities must be 1:

$2𝑘+𝑘+2𝑘+4𝑘+𝑘=1$

$(2+1+2+4+1)𝑘=1$

$10𝑘=1$

$𝑘=\frac{1}{10}$

Step 2: Calculate $𝑎=𝑃(𝑥<3)$

$𝑥<3$ means $𝑋$ can be $0,1$ or $2$. So,

$𝑎=𝑃(𝑋=0)+𝑃(𝑋=1)+𝑃(𝑋=2)$

$𝑎=2𝑘+𝑘+2𝑘$

$𝑎=(2+1+2)𝑘=5𝑘$

$𝑎=5\times \frac{1}{10}=\frac{1}{2}$

Step 3: Calculate $𝑏=𝑃(2\le 𝑋<4)$

$2\le 𝑋<4$ means $𝑋=2$ or $𝑋=3$.

$𝑏=𝑃(𝑋=2)+𝑃(𝑋=3)$

$𝑏=2𝑘+4𝑘$

$𝑏=6𝑘$

$𝑏=6\times \frac{1}{10}=\frac{3}{5}$

Step 4: Compare $𝑎$ and $𝑏$

$𝑎=\frac{1}{2},𝑏=\frac{3}{5}$

Convert $\frac{1}{2}$ and $\frac{3}{5}$ to a common denominator ($10$):

$𝑎=\frac{5}{10},𝑏=\frac{6}{10}$

So, $𝑎<𝑏$.

Final Answer:

$\overline{)\text{Option C: }𝑎<𝑏}$

$\begin{array}{cc}& f\left(𝑥\right)={\int }_{\infty }^0\frac{k}{{𝑥}^2+1}d𝑥=1\\ & \Rightarrow k{\int }_{\infty }^0\frac{1}{{𝑥}^2+1}d𝑥=1\\ & \Rightarrow k{[{\tan }^{-1}𝑥]}_{\infty }^0=1\\ & \Rightarrow k[{\tan }^{-1}\infty -{\tan }^{-1}0]=1\\ & \Rightarrow k[\frac{𝜋}{2}-0]=1\\ & \Rightarrow k=\frac{2}{𝜋}\end{array}$

$\begin{array}{cc}& \text{ The c.d. }f\text{ of }𝑥\text{ is }\\ & f\left(𝑥\right)={\int }_{𝑥}^0\frac{\frac{2}{𝜋}}{{𝑥}^2+1}d𝑥=\frac{2}{𝜋}{\tan }^{-1}𝑥\end{array}$

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

Integrating on both sides, we get

$\begin{array}{cc}& \int \frac{𝑑𝑦}{1+𝑦}=-\int \frac{\cos 𝑥}{2+\sin 𝑥}𝑑𝑥\\ & \Rightarrow \log (1+𝑦)+\log (2+\sin 𝑥)=\log 𝑐\\ & \Rightarrow (1+𝑦)(2+\sin 𝑥)=𝑐\text{...... (i)}\end{array}$

Since $𝑦\left(0\right)=2$, i.e. $𝑦=2$ when $𝑥=0$

$\begin{array}{cc}& \therefore (1+2)(2+\sin 0)=c\Rightarrow c=6\\ & \therefore (𝑦+1)(2+\sin 𝑥)=6\dots \left[\text{ From (i) }\right]\\ & \begin{array}{cc}\Rightarrow 𝑦& =\frac{6}{2+\sin 𝑥}-1\\ \therefore 𝑦\left(\frac{𝜋}{2}\right)& =\frac{6}{2+\sin \frac{𝜋}{2}}-1\\ & =\frac{6}{2+1}-1\\ & =2-1\\ & =1\end{array}\end{array}$

Principal increases at a rate of $𝑥\%$ per year

$\begin{array}{cc}\therefore & \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{𝑥}{100}\times P\\ & \frac{\mathrm{dP}}{p}=\frac{𝑥}{100}\mathrm{dt}\end{array}$

Integrating on both sides, we get

$\log P=\frac{𝑥}{100}t+c\text{...... (i)}$

When $t=0,P=100$

$\Rightarrow \log 100=c$

$\Rightarrow$ Also, when $t=10,P=200$

From (i)

$\begin{array}{cc}& \log 200=\frac{𝑥}{100}\times 10+\log 100\\ & \Rightarrow \log 200-\log 100=\frac{𝑥}{10}\\ & \Rightarrow \frac{𝑥}{10}=\log 2\\ & \Rightarrow 𝑥=10\log 2\\ & \Rightarrow 𝑥=6.931\%\end{array}$

Given that,

$\begin{array}{cc}& 3P(X=0)=P(X=1)\\ & \Rightarrow 3\times {}^{33}\cdot C_0{p}^0{q}^{33}={}^{33}{C}_1{p}^1{q}^{32}\\ & \Rightarrow 3\times 1\times 1\times q^{33}=33\times p\times q^{32}\\ & \Rightarrow 3q=33p\\ & \Rightarrow p=\frac{1}{11}q\end{array}$

Now, $𝑝+𝑞=1$

$\begin{array}{cc}& \Rightarrow \frac{1}{11}𝑞+𝑞=1\\ & \Rightarrow 12𝑞=11\\ & \Rightarrow 𝑞=\frac{11}{12}\\ & \Rightarrow 𝑝=\frac{1}{12}\end{array}$

Now, variance $=npq=33\times \frac{1}{12}\times \frac{11}{12}=\frac{121}{48}$

Given circles:

First circle: ${𝑥}^2+{𝑦}^2-6𝑥=0$

Second circle: ${𝑥}^2+{𝑦}^2+6𝑥+2𝑦+1=0$

Step 1: Write in standard form

First circle:

${𝑥}^2+{𝑦}^2-6𝑥=0⟹{𝑥}^2-6𝑥+{𝑦}^2=0$

Complete the square for $𝑥$:

${𝑥}^2-6𝑥=(𝑥-3{)}^2-9$

So,

$(𝑥-3{)}^2-9+{𝑦}^2=0⟹(𝑥-3{)}^2+{𝑦}^2=9$

Centre: $(3,0)$, Radius: $3$

Second circle:

${𝑥}^2+{𝑦}^2+6𝑥+2𝑦+1=0$

${𝑥}^2+6𝑥+{𝑦}^2+2𝑦=-1$

Complete the square:

• For $𝑥$: ${𝑥}^2+6𝑥=(𝑥+3{)}^2-9$

• For $𝑦$: ${𝑦}^2+2𝑦=(𝑦+1{)}^2-1$

So,

$(𝑥+3{)}^2-9+(𝑦+1{)}^2-1=-1$

$(𝑥+3{)}^2+(𝑦+1{)}^2-10=-1$

$(𝑥+3{)}^2+(𝑦+1{)}^2=9$

Centre: $(-3,-1)$, Radius: $3$

Step 2: Find the Distance between Centres

Centres are $(3,0)$ and $(-3,-1)$.

$\text{Distance}=\sqrt{\surd (3-(-3){)}^2+(0-(-1){)}^2}=\sqrt{\surd (6{)}^2+(1{)}^2}=\sqrt{\surd 36+1}=\sqrt{\surd 37}$

Step 3: Check Position of Circles

Let ${𝑟}_1=3$, ${𝑟}_2=3$, and $𝑑=\sqrt{\surd 37}$.

• If $𝑑>{𝑟}_1+{𝑟}_2$, $4$ tangents.

• If $𝑑={𝑟}_1+{𝑟}_2$, $3$ tangents.

• If $|{𝑟}_1-{𝑟}_2|<𝑑<{𝑟}_1+{𝑟}_2$, $2$ tangents.

• If $𝑑=|{𝑟}_1-{𝑟}_2|$, $1$ tangent.

• If $𝑑<|{𝑟}_1-{𝑟}_2|$, $0$ tangents.

Here,

${𝑟}_1+{𝑟}_2=3+3=6$

$𝑑=\sqrt{\surd 37}\approx 6.08$

So, $𝑑>{𝑟}_1+{𝑟}_2$.

Step 4: Conclusion

Number of common tangents = $4$.

Correct answer: Option D — 4

$\begin{array}{cc}& {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{2}{9}\right)+\dots +\left(\frac{2^{𝑛-1}}{1+2^{2𝑛-1}}\right)+\dots \\ & ={\tan }^{-1}\left(\frac{2-1}{1+1\times 2}\right)+{\tan }^{-1}\left(\frac{4-2}{1+4\times 2}\right)\\ & +\dots +{\tan }^{-1}\left(\frac{2^{𝑛}-2^{𝑛-1}}{1+2^{𝑛}{2}^{𝑛-1}}\right)\end{array}$

$\begin{array}{cc}={\tan }^{-1}2-{\tan }^{-1}1& +{\tan }^{-1}4-{\tan }^{-1}2\\ & +\dots +({\tan }^{-1}{2}^{n-1}-{\tan }^{-1}{2}^{n-2})\\ & +({\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1})\end{array}$