MHT CET 2026 18th April Evening Shift (Mathematics) Question Paper with Solutions

For the quadratic equation

${𝑥}^2+2𝑥+4=0$

we use the quadratic formula:

$𝑥=\frac{-𝑏\pm \sqrt{\surd {𝑏}^2-4𝑎𝑐}}{2𝑎}$

Here, $𝑎=1$, $𝑏=2$, and $𝑐=4$.

So,

$𝑥=\frac{-2\pm \sqrt{\surd 2^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}$

$𝑥=\frac{-2\pm \sqrt{\surd 4-16}}{2}$

$𝑥=\frac{-2\pm \sqrt{\surd -12}}{2}$

Since $𝑖=\sqrt{\surd -1}$,

$\sqrt{\surd -12}=\sqrt{\surd 12}\sqrt{\surd -1}=2\sqrt{\surd 3}𝑖$

So the roots are

$𝑥=\frac{-2\pm 2𝑖\sqrt{\surd 3}}{2}$

$𝑥=-1\pm 𝑖\sqrt{\surd 3}$

Thus the two roots are

$-1+𝑖\sqrt{\surd 3}\text{and}-1-𝑖\sqrt{\surd 3}$

Their difference is

$(-1+𝑖\sqrt{\surd 3})-(-1-𝑖\sqrt{\surd 3})$

$=2𝑖\sqrt{\surd 3}$

So the correct option is

$\overline{)2𝑖\sqrt{\surd 3}}$

Option C.

For each question, a student can mark either True or False.

So, for $7$ questions, the total number of possible answer sequences is

$2^7=128$

This includes every possible pattern of answers.

Now, exactly one of these $128$ sequences is the completely correct answer sequence.

But the question says no student got all the answers correct. So, that one completely correct sequence cannot be used by any student.

Also, the sequence of answers for every student is unique, so no two students can have the same answer sequence.

Hence, the maximum number of students can be

$128-1=127$

So, the correct option is

$\overline{)127}$

Option C.

Given

${𝑓}_{𝑘}\left(𝑥\right)=\frac{1}{𝑘}({\cos }^{𝑘}𝑥+{\sin }^{𝑘}𝑥)$

We need to find

${𝑓}_6\left(𝑥\right)-{𝑓}_4\left(𝑥\right)$

Now,

${𝑓}_6\left(𝑥\right)=\frac{1}{6}({\cos }^6𝑥+{\sin }^6𝑥)$

and

${𝑓}_4\left(𝑥\right)=\frac{1}{4}({\cos }^4𝑥+{\sin }^4𝑥)$

Use the identities:

${\cos }^6𝑥+{\sin }^6𝑥=({\cos }^2𝑥+{\sin }^2𝑥{)}^3-3{\sin }^2𝑥{\cos }^2𝑥({\sin }^2𝑥+{\cos }^2𝑥)$

Since

${\sin }^2𝑥+{\cos }^2𝑥=1$

we get

${\cos }^6𝑥+{\sin }^6𝑥=1-3{\sin }^2𝑥{\cos }^2𝑥$

Also,

${\cos }^4𝑥+{\sin }^4𝑥=({\cos }^2𝑥+{\sin }^2𝑥{)}^2-2{\sin }^2𝑥{\cos }^2𝑥$

So,

${\cos }^4𝑥+{\sin }^4𝑥=1-2{\sin }^2𝑥{\cos }^2𝑥$

Now substitute:

${𝑓}_6\left(𝑥\right)-{𝑓}_4\left(𝑥\right)=\frac{1}{6}(1-3{\sin }^2𝑥{\cos }^2𝑥)-\frac{1}{4}(1-2{\sin }^2𝑥{\cos }^2𝑥)$

$=\frac{1}{6}-\frac{1}{2}{\sin }^2𝑥{\cos }^2𝑥-\frac{1}{4}+\frac{1}{2}{\sin }^2𝑥{\cos }^2𝑥$

The variable terms cancel out:

${𝑓}_6\left(𝑥\right)-{𝑓}_4\left(𝑥\right)=\frac{1}{6}-\frac{1}{4}$

$=\frac{2-3}{12}=-\frac{1}{12}$

So the correct option is

$\overline{)-\frac{1}{12}}$

Option B.

The slopes of the lines

$4𝑥-7𝑦+10=0$

and

$7𝑥+4𝑦=15$

are $\frac{4}{7}$ and $-\frac{7}{4}$ respectively.

Their product is

$\frac{4}{7}\times (-\frac{7}{4})=-1.$

Hence, these two sides are perpendicular. So the triangle is right-angled, and its orthocenter is the vertex at which these two lines intersect.

Solving

$4𝑥-7𝑦=-10$

$7𝑥+4𝑦=15,$

we get

$𝑥=1,𝑦=2.$

Thus, the orthocenter is

$\overline{)(1,2)}$

So, Option A is correct.

The given second-degree equation can be factorised as

${𝑥}^2-6𝑥𝑦+5{𝑦}^2+10𝑥-14𝑦+9=0$

$(𝑥-𝑦+1)(𝑥-5𝑦+9)=0$

So, the two lines are

$𝑥-𝑦+1=0$

and

$𝑥-5𝑦+9=0$

From the first line,

$𝑥=𝑦-1$

Substituting this in the second line:

$𝑦-1-5𝑦+9=0$

$-4𝑦+8=0\Rightarrow 𝑦=2$

Hence,

$𝑥=2-1=1$

The point of intersection is $(1,2)$.

The lines through $(1,2)$ parallel to the coordinate axes are

$𝑥=1\text{and}𝑦=2$

Their combined equation is

$(𝑥-1)(𝑦-2)=0$

$\overline{)𝑥𝑦-2𝑥-𝑦+2=0}$

Hence, the correct answer is Option D.

The centre of a circle touching both lines must be equidistant from them.

The given lines are

$𝑥+𝑦-1=0$

and

$𝑥-𝑦-1=0.$

Their angle bisectors satisfy

$\frac{𝑥+𝑦-1}{\sqrt{\surd 2}}=\pm \frac{𝑥-𝑦-1}{\sqrt{\surd 2}}.$

This gives the two possible loci of the centre:

$𝑦=0\text{or}𝑥=1.$

For a centre $(ℎ,0)$, the radius is its distance from either line:

$𝑟=\frac{|ℎ-1|}{\sqrt{\surd 2}}.$

Since the circle passes through the origin,

$𝑟=\left|ℎ\right|.$

Thus,

$\left|ℎ\right|=\frac{|ℎ-1|}{\sqrt{\surd 2}}$

$2{ℎ}^2=(ℎ-1{)}^2$

${ℎ}^2+2ℎ-1=0.$

This equation has two distinct real roots:

$ℎ=-1\pm \sqrt{\surd 2}.$

So, two circles are obtained.

For a centre $(1,𝑘)$, passing through the origin would require

$\sqrt{\surd 1+{𝑘}^2}=\frac{\left|𝑘\right|}{\sqrt{\surd 2}},$

which gives no real value of $𝑘$.

Hence, the number of circles is

$\overline{)2}$

Option B.

Let the focus be $𝐹(𝑎,0)$, where $𝑎>0$. The directrix is the $𝑌$-axis, whose equation is $𝑥=0$.

For any point $(𝑥,𝑦)$ on the parabola, its distance from the focus equals its distance from the directrix:

$\sqrt{\surd (𝑥-𝑎{)}^2+{𝑦}^2}=𝑥$

Squaring,

$(𝑥-𝑎{)}^2+{𝑦}^2={𝑥}^2$

${𝑦}^2=2𝑎𝑥-{𝑎}^2$

Since $𝑃(𝛼,4)$ lies on the parabola,

$16=2𝑎𝛼-{𝑎}^2\left(1\right)$

Differentiating the equation of the parabola,

$2𝑦\frac{𝑑𝑦}{𝑑𝑥}=2𝑎$

At $𝑃(𝛼,4)$, the slope of the tangent is

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑎}{4}$

The tangent passes through the origin and $𝑃(𝛼,4)$, so its slope is also

$\frac{4}{𝛼}$

Thus,

$\frac{𝑎}{4}=\frac{4}{𝛼}$

$𝑎𝛼=16$

Using this in equation $\left(1\right)$,

$16=2\left(16\right)-{𝑎}^2$

${𝑎}^2=16$

Since $𝑎>0$, $𝑎=4$, and hence $𝛼=4$.

Therefore, the distance of $𝑃(4,4)$ from the origin is

$𝑂𝑃=\sqrt{\surd 4^2+4^2}=\sqrt{\surd 32}$

Hence, the correct answer is $\overline{)\sqrt{\surd 32}}$, Option D.

The right-hand side limit is

$\underset{𝑥\to 0}{lim}\frac{1-\cos \left(2𝑥\right)}{𝑥\sin 𝑥}$

Using the identity

$1-\cos 2𝑥=2{\sin }^2𝑥$

we get

$\underset{𝑥\to 0}{lim}\frac{2{\sin }^2𝑥}{𝑥\sin 𝑥}=\underset{𝑥\to 0}{lim}2\cdot \frac{\sin 𝑥}{𝑥}$

Since

$\underset{𝑥\to 0}{lim}\frac{\sin 𝑥}{𝑥}=1$

the right-hand side limit is

$2$

Now consider the left-hand side:

$\underset{𝑥\to 𝑘}{lim}\frac{{𝑥}^3-{𝑘}^3}{{𝑥}^2-{𝑘}^2}$

Factorise numerator and denominator:

${𝑥}^3-{𝑘}^3=(𝑥-𝑘)({𝑥}^2+𝑥𝑘+{𝑘}^2)$

and

${𝑥}^2-{𝑘}^2=(𝑥-𝑘)(𝑥+𝑘)$

So,

$\underset{𝑥\to 𝑘}{lim}\frac{(𝑥-𝑘)({𝑥}^2+𝑥𝑘+{𝑘}^2)}{(𝑥-𝑘)(𝑥+𝑘)}=\underset{𝑥\to 𝑘}{lim}\frac{{𝑥}^2+𝑥𝑘+{𝑘}^2}{𝑥+𝑘}$

Now put $𝑥=𝑘$:

$\frac{{𝑘}^2+{𝑘}^2+{𝑘}^2}{𝑘+𝑘}=\frac{3{𝑘}^2}{2𝑘}=\frac{3𝑘}{2}$

Given that both limits are equal,

$\frac{3𝑘}{2}=2$

So,

$3𝑘=4$

$𝑘=\frac{4}{3}$

Hence, the correct option is

$\overline{)\frac{4}{3}}$

Option A.

The switching expression is

$({𝑆}_1\wedge {𝑆}_2)\vee ({𝑆}_{\prime }^1\wedge {𝑆}_{\prime }^2)$

Here, $\wedge$ represents a series connection, while $\vee$ represents a parallel connection.

The circuit conducts when either:

• ${𝑆}_1$ and ${𝑆}_2$ are both ON, or

• ${𝑆}_{\prime }^1$ and ${𝑆}_{\prime }^2$ are both ON, meaning ${𝑆}_1$ and ${𝑆}_2$ are both OFF.

Thus, the room is lit when both switches are in the same state.

$\overline{)\text{Option C: }{𝑆}_1\text{ and }{𝑆}_2\text{ both ON or both OFF}}$

The given statement pattern is

$(\sim 𝑝\wedge 𝑞)\vee (\sim 𝑝\wedge \sim 𝑞)\vee (𝑝\wedge \sim 𝑞)$

Its truth value is given as $𝐹$.

For an “or” statement to be false, each part joined by $\vee$ must be false.

So,

$\sim 𝑝\wedge 𝑞=𝐹$

$\sim 𝑝\wedge \sim 𝑞=𝐹$

$𝑝\wedge \sim 𝑞=𝐹$

Notice the given pattern contains these three cases:

• $\sim 𝑝\wedge 𝑞$

• $\sim 𝑝\wedge \sim 𝑞$

• $𝑝\wedge \sim 𝑞$

The only possible case not included is

$𝑝\wedge 𝑞$

So the given statement becomes false only when

$𝑝=𝑇,𝑞=𝑇$

Now find the truth values of the two statements.

First,

$𝑝\vee \sim 𝑞$

Since $𝑝=𝑇$ and $𝑞=𝑇$, we have $\sim 𝑞=𝐹$.

So,

$𝑝\vee \sim 𝑞=𝑇\vee 𝐹=𝑇$

Next,

$𝑝\to 𝑞$

A conditional statement $𝑝\to 𝑞$ is false only when $𝑝=𝑇$ and $𝑞=𝐹$.

Here,

$𝑝=𝑇,𝑞=𝑇$

So,

$𝑝\to 𝑞=𝑇$

Hence, the truth values are

$𝑇,𝑇$

So the correct option is Option C.

Let the switches ${𝑆}_1,{𝑆}_2,$ and ${𝑆}_3$ be represented by the Boolean variables $𝑝,𝑞,$ and $𝑟$. Their complementary switches ${𝑆}_{\prime }^1$ and ${𝑆}_{\prime }^2$ are represented by ${𝑝}^{\prime }$ and ${𝑞}^{\prime }$.

In a switching circuit, switches connected in series are represented by multiplication, while parallel branches are represented by addition.

The circuit has three parallel branches. Hence, its Boolean expression is:
$𝐿=({𝑆}_1\cdot {𝑆}_2)+({𝑆}_{\prime }^1\cdot {𝑆}_2)+({𝑆}_3\cdot {𝑆}_{\prime }^2)$

Substituting ${𝑆}_1=𝑝$, ${𝑆}_2=𝑞$, and ${𝑆}_3=𝑟$, we get:
$𝐿=(𝑝\cdot 𝑞)+({𝑝}^{\prime }\cdot 𝑞)+(𝑟\cdot {𝑞}^{\prime })$

Taking $𝑞$ common from the first two terms:
$𝐿=𝑞\cdot (𝑝+{𝑝}^{\prime })+𝑟\cdot {𝑞}^{\prime }$

A Boolean variable and its complement satisfy $𝑝+{𝑝}^{\prime }=1$. So:
$𝐿=𝑞\cdot 1+𝑟\cdot {𝑞}^{\prime }$
$𝐿=𝑞+{𝑞}^{\prime }\cdot 𝑟$

Using the Boolean identity $𝐴+{𝐴}^{\prime }𝐵=𝐴+𝐵$:
$𝐿=(𝑞+{𝑞}^{\prime })\cdot (𝑞+𝑟)$
$𝐿=1\cdot (𝑞+𝑟)$
$𝐿=𝑞+𝑟$

Returning to the switch notation:
$𝐿={𝑆}_2+{𝑆}_3$

This represents the two switches ${𝑆}_2$ and ${𝑆}_3$ connected in parallel. Hence, the minimum number of switches is $2$.

In $△𝐴𝐵𝐶$, we know that

$𝑠=\frac{𝑎+𝑏+𝑐}{2}$

So,

$𝑏+𝑐-𝑎=(𝑎+𝑏+𝑐)-2𝑎=2𝑠-2𝑎=2(𝑠-𝑎)$

Now use the standard half-angle formula:

$\tan \left(\frac{𝐴}{2}\right)=\frac{𝑟}{𝑠-𝑎}$

where $𝑟$ is the inradius of the triangle.

Also, the area of the triangle is

$\Delta =𝑟𝑠$

Now substitute these in the given expression:

$2𝑠(𝑏+𝑐-𝑎)\tan \left(\frac{𝐴}{2}\right)$

$=2𝑠\cdot 2(𝑠-𝑎)\cdot \frac{𝑟}{𝑠-𝑎}$

Cancel $(𝑠-𝑎)$:

$=4𝑠𝑟$

Since $\Delta =𝑟𝑠$,

$4𝑠𝑟=4\Delta$

So the value is

$\overline{)4\Delta }$

Correct option: D.

Since ${𝐴}^{-1}$ is the inverse of $𝐴$,

$𝐴{𝐴}^{-1}=𝐼.$

Using the first row of $𝐴$ and the third column of ${𝐴}^{-1}$, the corresponding off-diagonal entry must be $0$:

$0\left(\frac{1}{2}\right)+1\left(𝑦\right)+2\left(\frac{1}{2}\right)=0.$

So,

$𝑦+1=0\Rightarrow 𝑦=-1.$

Now use the third row of $𝐴$ and the first column of ${𝐴}^{-1}$:

$3\left(\frac{1}{2}\right)+𝑥(-4)+1\left(\frac{5}{2}\right)=0.$

$\frac{3}{2}-4𝑥+\frac{5}{2}=0$

$4-4𝑥=0\Rightarrow 𝑥=1.$

Thus, the Cartesian coordinates are $(1,-1)$.

The polar radius is

$𝑟=\sqrt{\surd {𝑥}^2+{𝑦}^2}=\sqrt{\surd 1^2+(-1{)}^2}=\sqrt{\surd 2}.$

The point $(1,-1)$ lies in the fourth quadrant, with reference angle $\frac{𝜋}{4}$. Hence,

$𝜃=2𝜋-\frac{𝜋}{4}=\frac{7𝜋}{4}.$

The polar coordinates are

$\overline{)(\sqrt{\surd 2},\frac{7𝜋}{4})}.$

Hence, Option C is correct.

We have

$𝐴=\left[\begin{array}{cc}-3& 2\\ 1& 4\end{array}\right]$

First find ${𝐴}^2$:

${𝐴}^2=𝐴\cdot 𝐴$

${𝐴}^2=\left[\begin{array}{cc}-3& 2\\ 1& 4\end{array}\right]\left[\begin{array}{cc}-3& 2\\ 1& 4\end{array}\right]$

Multiplying rows by columns,

${𝐴}^2=\left[\begin{array}{cc}(-3)(-3)+\left(2\right)\left(1\right)& (-3)\left(2\right)+\left(2\right)\left(4\right)\\ \left(1\right)(-3)+\left(4\right)\left(1\right)& \left(1\right)\left(2\right)+\left(4\right)\left(4\right)\end{array}\right]$

${𝐴}^2=\left[\begin{array}{cc}9+2& -6+8\\ -3+4& 2+16\end{array}\right]=\left[\begin{array}{cc}11& 2\\ 1& 18\end{array}\right]$

Now,

${𝐴}^2-2𝐴+𝐼$

Here,

$2𝐴=2\left[\begin{array}{cc}-3& 2\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}-6& 4\\ 2& 8\end{array}\right]$

and

$𝐼=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

So,

${𝐴}^2-2𝐴+𝐼=\left[\begin{array}{cc}11& 2\\ 1& 18\end{array}\right]-\left[\begin{array}{cc}-6& 4\\ 2& 8\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}11+6+1& 2-4+0\\ 1-2+0& 18-8+1\end{array}\right]$

$=\left[\begin{array}{cc}18& -2\\ -1& 11\end{array}\right]$

Given,

${𝐴}^2-2𝐴+𝐼=\left[\begin{array}{cc}18& 𝑝\\ 𝑞& 11\end{array}\right]$

So,

$𝑝=-2,𝑞=-1$

Hence, the correct option is Option A.

We simplify the expression inside ${\cot }^{-1}$ first.

$𝑦={\cot }^{-1}\left(\frac{𝑥-\log {𝑥}^{{𝑥}^2}}{\log {𝑒}^{{𝑥}^2}+\log {𝑥}^{𝑥}}\right)$

Using logarithm rules,

$\log {𝑥}^{{𝑥}^2}={𝑥}^2\log 𝑥$

$\log {𝑒}^{{𝑥}^2}={𝑥}^2$

and

$\log {𝑥}^{𝑥}=𝑥\log 𝑥$

So the expression becomes

$𝑦={\cot }^{-1}\left(\frac{𝑥-{𝑥}^2\log 𝑥}{{𝑥}^2+𝑥\log 𝑥}\right)$

Taking $𝑥$ common from numerator and denominator,

$𝑦={\cot }^{-1}\left(\frac{𝑥(1-𝑥\log 𝑥)}{𝑥(𝑥+\log 𝑥)}\right)$

Since $𝑥>0$,

$𝑦={\cot }^{-1}\left(\frac{1-𝑥\log 𝑥}{𝑥+\log 𝑥}\right)$

Now recall the identity

${\tan }^{-1}𝑎+{\tan }^{-1}𝑏={\tan }^{-1}\left(\frac{𝑎+𝑏}{1-𝑎𝑏}\right)$

when $𝑎𝑏<1$.

Here take

$𝑎=𝑥,𝑏=\log 𝑥$

Given,

$𝑥\log 𝑥<1$

So,

${\tan }^{-1}𝑥+{\tan }^{-1}(\log 𝑥)={\tan }^{-1}\left(\frac{𝑥+\log 𝑥}{1-𝑥\log 𝑥}\right)$

Hence,

$\cot [{\tan }^{-1}𝑥+{\tan }^{-1}(\log 𝑥\left)\right]=\frac{1-𝑥\log 𝑥}{𝑥+\log 𝑥}$

So,

$𝑦={\tan }^{-1}𝑥+{\tan }^{-1}(\log 𝑥)$

Now differentiate:

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑑}{𝑑𝑥}({\tan }^{-1}𝑥)+\frac{𝑑}{𝑑𝑥}({\tan }^{-1}(\log 𝑥\left)\right)$

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{1+{𝑥}^2}+\frac{1}{1+(\log 𝑥{)}^2}\cdot \frac{1}{𝑥}$

Therefore,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{1+{𝑥}^2}+\frac{1}{𝑥[1+(\log 𝑥{)}^2]}$

So the correct option is

$\overline{)\text{Option B}}$

Since ${\cos }^{-1}𝑥$ and ${\sin }^{-1}𝑥$ are both present, and $𝑥>0$, we must have

$0<𝑥\le 1$

For such $𝑥$,

${\sin }^{-1}𝑥+{\cos }^{-1}𝑥=\frac{𝜋}{2}$

Let

${\cos }^{-1}𝑥=𝐴,{\tan }^{-1}𝑥=𝑇$

Then

${\sin }^{-1}𝑥=\frac{𝜋}{2}-𝐴$

The given left side is

$\sin (𝐴+𝑇)-\cos (\frac{𝜋}{2}-𝐴+𝑇)$

Now,

$\cos (\frac{𝜋}{2}-𝐴+𝑇)=\cos (\frac{𝜋}{2}-(𝐴-𝑇\left)\right)=\sin (𝐴-𝑇)$

So the left side becomes

$\sin (𝐴+𝑇)-\sin (𝐴-𝑇)$

Using

$\sin 𝐶-\sin 𝐷=2\cos \frac{𝐶+𝐷}{2}\sin \frac{𝐶-𝐷}{2}$

we get

$\sin (𝐴+𝑇)-\sin (𝐴-𝑇)=2\cos 𝐴\sin 𝑇$

Now,

$\cos 𝐴=𝑥$

and since $𝑇={\tan }^{-1}𝑥$,

$\sin 𝑇=\frac{𝑥}{\sqrt{\surd 1+{𝑥}^2}}$

So the left side is

$2𝑥\cdot \frac{𝑥}{\sqrt{\surd 1+{𝑥}^2}}=\frac{2{𝑥}^2}{\sqrt{\surd 1+{𝑥}^2}}$

Also,

$\sin ({\cot }^{-1}2)=\frac{1}{\sqrt{\surd 5}}$

Hence,

$\frac{2{𝑥}^2}{\sqrt{\surd 1+{𝑥}^2}}=\frac{1}{\sqrt{\surd 5}}$

Squaring both sides,

$\frac{4{𝑥}^4}{1+{𝑥}^2}=\frac{1}{5}$

$20{𝑥}^4=1+{𝑥}^2$

Let ${𝑥}^2=𝑦$.

$20{𝑦}^2-𝑦-1=0$

$20{𝑦}^2-5𝑦+4𝑦-1=0$

$5𝑦(4𝑦-1)+1(4𝑦-1)=0$

$(4𝑦-1)(5𝑦+1)=0$

Since $𝑦={𝑥}^2>0$,

$4𝑦-1=0$

$𝑦=\frac{1}{4}$

So,

${𝑥}^2=\frac{1}{4}$

Since $𝑥>0$,

$𝑥=\frac{1}{2}$

So the correct option is

$\overline{)\frac{1}{2}}$

Option B.

Using the formula

${\tan }^{-1}𝑝+{\tan }^{-1}𝑞={\tan }^{-1}\left(\frac{𝑝+𝑞}{1-𝑝𝑞}\right)$

when $𝑝𝑞<1$.

Here,

$𝑝=𝑎𝑥,𝑞=3𝑥$

So,

$𝑝𝑞=3𝑎{𝑥}^2<1$

which is already given. Hence we can apply the formula directly.

Given,

${\tan }^{-1}𝑎𝑥+{\tan }^{-1}3𝑥=\frac{𝜋}{4}$

Taking tangent on both sides,

$\frac{𝑎𝑥+3𝑥}{1-3𝑎{𝑥}^2}=\tan \frac{𝜋}{4}$

Since $\tan \frac{𝜋}{4}=1$,

$\frac{𝑥(𝑎+3)}{1-3𝑎{𝑥}^2}=1$

Now put

$𝑥=\frac{1}{6}$

So,

$\frac{\frac{1}{6}(𝑎+3)}{1-3𝑎{\left(\frac{1}{6}\right)}^2}=1$

$\frac{\frac{𝑎+3}{6}}{1-\frac{𝑎}{12}}=1$

So,

$\frac{𝑎+3}{6}=1-\frac{𝑎}{12}$

Multiplying by $12$,

$2(𝑎+3)=12-𝑎$

$2𝑎+6=12-𝑎$

$3𝑎=6$

$𝑎=2$

So the correct option is

$\overline{)2}$

Option A.

Let

$𝜃={\tan }^{-1}3$

This means

$\tan 𝜃=3$

Using the identity

${\sec }^2𝜃=1+{\tan }^2𝜃$

we get

${\sec }^2({\tan }^{-1}3)=1+3^2=10$

Now let

$𝜙={\sec }^{-1}3$

This means

$\sec 𝜙=3$

Again using

${\tan }^2𝜙={\sec }^2𝜙-1$

we get

${\tan }^2({\sec }^{-1}3)=3^2-1=8$

So,

${\sec }^2({\tan }^{-1}3)-{\tan }^2({\sec }^{-1}3)=10-8=2$

So the correct option is

$\overline{)2}$

Option C.

For the function

$f(x)=\sqrt{\frac{x}{1+x}}$

the expression inside the square root must be non-negative, because the square root of a real number is defined only when the number is greater than or equal to $0$.

So we need

$\frac{x}{1+x} \geq 0$

Also, the denominator cannot be zero:

$1+x \neq 0$

$x \neq -1$

Now check the sign of

$\frac{x}{1+x}$

The critical points are $x=-1$ and $x=0$.

For $x<-1$, both $x$ and $1+x$ are negative, so

$\frac{x}{1+x}>0$

So $(-\infty,-1)$ is included.

At $x=-1$, the denominator becomes zero, so it is not included.

For $-1<x<0$, $x$ is negative and $1+x$ is positive, so

$\frac{x}{1+x}<0$

So this interval is not included.

At $x=0$,

$\frac{0}{1+0}=0$

and $\sqrt{0}$ is defined, so $0$ is included.

For $x>0$, both $x$ and $1+x$ are positive, so

$\frac{x}{1+x}>0$

So $[0,\infty)$ is included.

Hence, the domain is

$(-\infty,-1)\cup[0,\infty)$

So the correct option is

$\boxed{\text{Option A}}$

For $𝑓\left(𝑥\right)$ to be continuous at $𝑥=0$, we must have

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

So,

$𝑓\left(0\right)=\underset{𝑥\to 0}{lim}\frac{3^{𝑥}+3^{-𝑥}-2}{\tan 𝑥\cdot \log (1+𝑥)}$

Write $3^{𝑥}$ as ${𝑒}^{𝑥\log 3}$. Then

$3^{𝑥}+3^{-𝑥}-2={𝑒}^{𝑥\log 3}+{𝑒}^{-𝑥\log 3}-2$

Using the standard expansion near $0$,

${𝑒}^{𝑢}=1+𝑢+\frac{{𝑢}^2}{2}+\cdots$

and

${𝑒}^{-𝑢}=1-𝑢+\frac{{𝑢}^2}{2}+\cdots$

Adding,

${𝑒}^{𝑢}+{𝑒}^{-𝑢}-2={𝑢}^2+\cdots$

Here $𝑢=𝑥\log 3$, so

$3^{𝑥}+3^{-𝑥}-2\approx {𝑥}^2(\log 3{)}^2$

Also, as $𝑥\to 0$,

$\tan 𝑥\approx 𝑥$

and

$\log (1+𝑥)\approx 𝑥$

So the denominator becomes

$\tan 𝑥\cdot \log (1+𝑥)\approx 𝑥\cdot 𝑥={𝑥}^2$

Hence,

$𝑓\left(0\right)=\underset{𝑥\to 0}{lim}\frac{{𝑥}^2(\log 3{)}^2}{{𝑥}^2}$

$𝑓\left(0\right)=(\log 3{)}^2$

So the correct option is

$\overline{)(\log 3{)}^2}$

Option B.

Given,

$\int {𝑓}^{\prime }\left(𝑥\right){𝑒}^{{𝑥}^2}𝑑𝑥=(𝑥-1){𝑒}^{{𝑥}^2}+𝑘$

To find $𝑓\left(𝑥\right)$, differentiate both sides with respect to $𝑥$.

On the left side, differentiation cancels integration:

$\frac{𝑑}{𝑑𝑥}(\int {𝑓}^{\prime }(𝑥\left){𝑒}^{{𝑥}^2}𝑑𝑥\right)={𝑓}^{\prime }\left(𝑥\right){𝑒}^{{𝑥}^2}$

Now differentiate the right side:

$\frac{𝑑}{𝑑𝑥}\left(\right(𝑥-1){𝑒}^{{𝑥}^2}+𝑘)$

Since $𝑘$ is a constant, its derivative is $0$.

Use product rule on $(𝑥-1){𝑒}^{{𝑥}^2}$:

$\frac{𝑑}{𝑑𝑥}\left(\right(𝑥-1\left){𝑒}^{{𝑥}^2}\right)=(𝑥-1)\frac{𝑑}{𝑑𝑥}\left({𝑒}^{{𝑥}^2}\right)+{𝑒}^{{𝑥}^2}\frac{𝑑}{𝑑𝑥}(𝑥-1)$