MHT CET 2026 13th April Evening Shift (Mathematics) Question Paper with Solutions

For roots $1$ and $𝛼$, the monic quadratic polynomial is

$𝑝\left(𝑥\right)=(𝑥-1)(𝑥-𝛼)$

So,

$𝑝\left(𝑥\right)={𝑥}^2-(1+𝛼)𝑥+𝛼$

Similarly, for roots $1$ and $𝛽$,

$𝑞\left(𝑥\right)=(𝑥-1)(𝑥-𝛽)$

so

$𝑞\left(𝑥\right)={𝑥}^2-(1+𝛽)𝑥+𝛽$

Now,

$𝑟\left(𝑥\right)=𝑝\left(𝑥\right)+𝑞\left(𝑥\right)$

Hence,

$𝑟\left(𝑥\right)=[{𝑥}^2-(1+𝛼)𝑥+𝛼]+[{𝑥}^2-(1+𝛽)𝑥+𝛽]$

$𝑟\left(𝑥\right)=2{𝑥}^2-(2+𝛼+𝛽)𝑥+(𝛼+𝛽)$

Given that $𝛼$ and $𝛽$ are the roots of $𝑟\left(𝑥\right)$.

For $𝑟\left(𝑥\right)=2{𝑥}^2-(2+𝛼+𝛽)𝑥+(𝛼+𝛽)$,

sum of roots is

$𝛼+𝛽=\frac{2+𝛼+𝛽}{2}$

So,

$2𝛼+2𝛽=2+𝛼+𝛽$

$𝛼+𝛽=2$

Also, product of roots is

$𝛼𝛽=\frac{𝛼+𝛽}{2}$

Using $𝛼+𝛽=2$,

$𝛼𝛽=1$

So $𝛼$ and $𝛽$ have sum $2$ and product $1$. Hence,

$𝛼=𝛽=1$

Thus,

$𝑝\left(𝑥\right)=𝑞\left(𝑥\right)=(𝑥-1{)}^2$

Now,

$\sqrt{\surd 𝑝\left(𝑥\right)}-\sqrt{\surd 𝑞\left(𝑥\right)}=\sqrt{\surd (𝑥-1{)}^2}-\sqrt{\surd (𝑥-1{)}^2}=0$

So,

$\underset{𝑥\to \infty }{lim}[\sqrt{\surd 𝑝\left(𝑥\right)}-\sqrt{\surd 𝑞\left(𝑥\right)}]=0$

Correct option: A.

Since the same infinite radical repeats inside itself, take the whole expression as $𝑥$.

So,

$𝑥=\sqrt{\surd -1-𝑥}$

Squaring both sides,

${𝑥}^2=-1-𝑥$

Bring all terms to one side:

${𝑥}^2+𝑥+1=0$

Now, for cube roots of unity, we know

$1+𝜔+{𝜔}^2=0$

So the non-real cube roots of unity $𝜔$ and ${𝜔}^2$ satisfy

${𝑥}^2+𝑥+1=0$

Thus,

$𝑥=𝜔\text{or}𝑥={𝜔}^2$

Among the given options, the matching value is

$\overline{){𝜔}^2}$

So the correct option is Option D.

Given,

$\sin 3𝛼=4\sin 𝛼\cdot \sin (𝑥+𝛼)\cdot \sin (𝑥-𝛼)$

Use the identity

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

So the equation becomes

$\sin 3𝛼=4\sin 𝛼({\sin }^2𝑥-{\sin }^2𝛼)$

Also,

$\sin 3𝛼=3\sin 𝛼-4{\sin }^3𝛼$

Hence,

$3\sin 𝛼-4{\sin }^3𝛼=4\sin 𝛼{\sin }^2𝑥-4{\sin }^3𝛼$

Since $𝛼\ne 𝑛𝜋$, we have $\sin 𝛼\ne 0$.

Cancel $-4{\sin }^3𝛼$ from both sides:

$3\sin 𝛼=4\sin 𝛼{\sin }^2𝑥$

Divide by $\sin 𝛼$:

$3=4{\sin }^2𝑥$

So,

${\sin }^2𝑥=\frac{3}{4}$

Thus,

$\sin 𝑥=\pm \frac{\sqrt{\surd 3}}{2}$

Hence,

$𝑥=𝑛𝜋\pm \frac{𝜋}{3},𝑛\in 𝑍$

So the correct option is

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

Put

$𝑥=𝜃-𝛼\text{and}𝑦=𝜃-𝛽$

Then

$𝛼-𝛽=(𝜃-𝛽)-(𝜃-𝛼)=𝑦-𝑥$

Given,

$\cos 𝑥=𝑎$

and

$\sin 𝑦=𝑏$

We need to find

${\cos }^2(𝛼-𝛽)+2𝑎𝑏\sin (𝛼-𝛽)+{\cos }^2(𝜃-𝛼)$

So the expression becomes

${\cos }^2(𝑦-𝑥)+2𝑎𝑏\sin (𝑦-𝑥)+{\cos }^2𝑥$

Since $𝑎=\cos 𝑥$ and $𝑏=\sin 𝑦$, this is

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

Now use the identities:

$\cos (𝑦-𝑥)=\cos 𝑦\cos 𝑥+\sin 𝑦\sin 𝑥$

and

$\sin (𝑦-𝑥)=\sin 𝑦\cos 𝑥-\cos 𝑦\sin 𝑥$

So,

$\cos (𝑦-𝑥)=𝑎\cos 𝑦+𝑏\sin 𝑥$

and

$\sin (𝑦-𝑥)=𝑎𝑏-\cos 𝑦\sin 𝑥$

Now substitute:

${\cos }^2(𝑦-𝑥)=(𝑎\cos 𝑦+𝑏\sin 𝑥{)}^2$

$={𝑎}^2{\cos }^2𝑦+2𝑎𝑏\cos 𝑦\sin 𝑥+{𝑏}^2{\sin }^2𝑥$

Also,

$2𝑎𝑏\sin (𝑦-𝑥)=2𝑎𝑏(𝑎𝑏-\cos 𝑦\sin 𝑥)$

$=2{𝑎}^2{𝑏}^2-2𝑎𝑏\cos 𝑦\sin 𝑥$

Adding all terms:

${𝑎}^2{\cos }^2𝑦+2𝑎𝑏\cos 𝑦\sin 𝑥+{𝑏}^2{\sin }^2𝑥+2{𝑎}^2{𝑏}^2-2𝑎𝑏\cos 𝑦\sin 𝑥+{𝑎}^2$

The middle terms cancel:

$2𝑎𝑏\cos 𝑦\sin 𝑥-2𝑎𝑏\cos 𝑦\sin 𝑥=0$

So we get

${𝑎}^2{\cos }^2𝑦+{𝑏}^2{\sin }^2𝑥+2{𝑎}^2{𝑏}^2+{𝑎}^2$

Now,

${\cos }^2𝑦=1-{\sin }^2𝑦=1-{𝑏}^2$

and

${\sin }^2𝑥=1-{\cos }^2𝑥=1-{𝑎}^2$

Substitute these:

${𝑎}^2(1-{𝑏}^2)+{𝑏}^2(1-{𝑎}^2)+2{𝑎}^2{𝑏}^2+{𝑎}^2$

$={𝑎}^2-{𝑎}^2{𝑏}^2+{𝑏}^2-{𝑎}^2{𝑏}^2+2{𝑎}^2{𝑏}^2+{𝑎}^2$

The ${𝑎}^2{𝑏}^2$ terms cancel:

$=2{𝑎}^2+{𝑏}^2$

So the correct option is

$\overline{)2{𝑎}^2+{𝑏}^2}$

Option D.

We have

$\cos 𝜃-\sin 𝜃=1$

Use the identity

$\cos 𝜃-\sin 𝜃=\sqrt{\surd 2}\cos (𝜃+\frac{𝜋}{4})$

So,

$\sqrt{\surd 2}\cos (𝜃+\frac{𝜋}{4})=1$

Divide by $\sqrt{\surd 2}$:

$\cos (𝜃+\frac{𝜋}{4})=\frac{1}{\sqrt{\surd 2}}$

Now,

$\cos 𝑥=\frac{1}{\sqrt{\surd 2}}$

when

$𝑥=2𝑛𝜋\pm \frac{𝜋}{4},𝑛\in 𝑍$

Here,

$𝑥=𝜃+\frac{𝜋}{4}$

So,

$𝜃+\frac{𝜋}{4}=2𝑛𝜋+\frac{𝜋}{4}$

or

$𝜃+\frac{𝜋}{4}=2𝑛𝜋-\frac{𝜋}{4}$

From the first equation,

$𝜃=2𝑛𝜋$

From the second equation,

$𝜃=2𝑛𝜋-\frac{𝜋}{2}$

Hence, the general solution is

$𝜃=2𝑛𝜋-\frac{𝜋}{2}$

or

$𝜃=2𝑛𝜋,𝑛\in 𝑍$

So the correct option is

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

Let the line cut the X-axis at $𝐴(𝑎,0)$ and the Y-axis at $𝐵(0,𝑏)$.

The point $𝑃(-5,3)$ divides the intercepted portion $𝐴𝐵$ in the ratio $4:3$ from the X-axis to the Y-axis.

So,

$𝐴𝑃:𝑃𝐵=4:3$

Using section formula, if $𝑃$ divides $𝐴(𝑎,0)$ and $𝐵(0,𝑏)$ in the ratio $4:3$, then

$𝑃=(\frac{4\cdot 0+3\cdot 𝑎}{4+3},\frac{4\cdot 𝑏+3\cdot 0}{4+3})$

So,

$(-5,3)=(\frac{3𝑎}{7},\frac{4𝑏}{7})$

Now compare coordinates:

$\frac{3𝑎}{7}=-5$

$3𝑎=-35$

$𝑎=-\frac{35}{3}$

And,

$\frac{4𝑏}{7}=3$

$4𝑏=21$

$𝑏=\frac{21}{4}$

So the intercepts are

$𝐴(-\frac{35}{3},0),𝐵(0,\frac{21}{4})$

Using intercept form of a line,

$\frac{𝑥}{𝑎}+\frac{𝑦}{𝑏}=1$

Substitute the values of $𝑎$ and $𝑏$:

$\frac{𝑥}{-\frac{35}{3}}+\frac{𝑦}{\frac{21}{4}}=1$

$-\frac{3𝑥}{35}+\frac{4𝑦}{21}=1$

Taking LCM $105$:

$-9𝑥+20𝑦=105$

$9𝑥-20𝑦+105=0$

So the correct option is:

$\overline{)9𝑥-20𝑦+105=0}$

Option C.

For a second degree equation

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2+2𝑔𝑥+2𝑓𝑦+𝑐=0$

if it represents a pair of straight lines, then the point of intersection of the two lines is found from

$𝑎𝑥+ℎ𝑦+𝑔=0$

and

$ℎ𝑥+𝑏𝑦+𝑓=0$

Here the given equation is

$𝑘{𝑥}^2-5𝑥𝑦+6{𝑦}^2+𝑥-3𝑦=0$

Comparing with the general form,

$𝑎=𝑘,2ℎ=-5,𝑏=6,2𝑔=1,2𝑓=-3$

So,

$ℎ=-\frac{5}{2},𝑔=\frac{1}{2},𝑓=-\frac{3}{2}$

The point of intersection satisfies

$𝑘𝑥-\frac{5}{2}𝑦+\frac{1}{2}=0$

and

$-\frac{5}{2}𝑥+6𝑦-\frac{3}{2}=0$

From the second equation,

$-\frac{5}{2}𝑥+6𝑦-\frac{3}{2}=0$

Multiplying by $2$,

$-5𝑥+12𝑦-3=0$

Now check the given options in this equation.

For $(-3,-1)$:

$-5(-3)+12(-1)-3=15-12-3=0$

So $(-3,-1)$ satisfies the required condition.

Hence, the coordinates of the point of intersection are

$\overline{)(-3,-1)}$

Correct option: Option A.

For the pair of straight lines

$\sqrt{\surd 3}{𝑥}^2-4𝑥𝑦+\sqrt{\surd 3}{𝑦}^2=0$

compare it with the general homogeneous equation

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0$

Here,

$𝑎=\sqrt{\surd 3},2ℎ=-4\Rightarrow ℎ=-2,𝑏=\sqrt{\surd 3}$

The angle $𝜃$ between the two lines represented by

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0$

is given by

$\tan 𝜃=\frac{2\sqrt{\surd {ℎ}^2-𝑎𝑏}}{𝑎+𝑏}$

Now,

${ℎ}^2=(-2{)}^2=4$

and

$𝑎𝑏=\sqrt{\surd 3}\cdot \sqrt{\surd 3}=3$

So,

${ℎ}^2-𝑎𝑏=4-3=1$

Hence,

$\tan 𝜃=\frac{2\sqrt{\surd 1}}{\sqrt{\surd 3}+\sqrt{\surd 3}}$

$\tan 𝜃=\frac{2}{2\sqrt{\surd 3}}=\frac{1}{\sqrt{\surd 3}}$

We know that

$\tan {30}^{\circ }=\frac{1}{\sqrt{\surd 3}}$

So,

$𝜃={30}^{\circ }$

Thus, the two straight lines are inclined at ${30}^{\circ }$ to each other.

The correct option is Option C.

The given circle is

${𝑥}^2+{𝑦}^2-6𝑥+7=0$

Complete the square for $𝑥$:

${𝑥}^2-6𝑥+{𝑦}^2+7=0$

$(𝑥-3{)}^2-9+{𝑦}^2+7=0$

$(𝑥-3{)}^2+{𝑦}^2=2$

So, the centre of this circle is

$(3,0)$

A concentric circle has the same centre. So its equation will be of the form

${𝑥}^2+{𝑦}^2-6𝑥+𝑐=0$

Now this circle touches the line

$𝑥+𝑦+3=0$

So, its radius is equal to the perpendicular distance from the centre $(3,0)$ to this line.

Distance of point $({𝑥}_1,{𝑦}_1)$ from line $𝑎𝑥+𝑏𝑦+𝑐=0$ is

$\frac{|𝑎{𝑥}_1+𝑏{𝑦}_1+𝑐|}{\sqrt{\surd {𝑎}^2+{𝑏}^2}}$

Here,

$\frac{|3+0+3|}{\sqrt{\surd 1^2+1^2}}=\frac{6}{\sqrt{\surd 2}}=3\sqrt{\surd 2}$

So radius is $3\sqrt{\surd 2}$.

For the circle

${𝑥}^2+{𝑦}^2-6𝑥+𝑐=0$

we have centre $(3,0)$ and radius squared:

${𝑟}^2=9-𝑐$

Since

${𝑟}^2=(3\sqrt{\surd 2}{)}^2=18$

we get

$9-𝑐=18$

$𝑐=-9$

Hence the required circle is

$\overline{){𝑥}^2+{𝑦}^2-6𝑥-9=0}$

So the correct option is Option B.

For the parabola

${𝑦}^2=16𝑥$

we compare it with the standard form

${𝑦}^2=4𝑎𝑥$

So,

$4𝑎=16$

$𝑎=4$

For ${𝑦}^2=4𝑎𝑥$, the focus is

$(𝑎,0)=(4,0)$

and the directrix is

$𝑥=-𝑎$

So the directrix is

$𝑥=-4$

The distance of the focus $(4,0)$ from the directrix $𝑥=-4$ is

$4-(-4)=8$

So the correct answer is

$\overline{)8}$

Option B.

The given limit is

$\underset{𝑛\to \infty }{lim}\frac{(𝑛+2)!+(𝑛+1)!}{(𝑛+2)!-(𝑛+1)!}$

We know that

$(𝑛+2)!=(𝑛+2)(𝑛+1)!$

So substitute this in the expression:

$\frac{(𝑛+2)!+(𝑛+1)!}{(𝑛+2)!-(𝑛+1)!}=\frac{(𝑛+2)(𝑛+1)!+(𝑛+1)!}{(𝑛+2)(𝑛+1)!-(𝑛+1)!}$

Take $(𝑛+1)!$ common from numerator and denominator:

$\frac{(𝑛+1)!\left[\right(𝑛+2)+1]}{(𝑛+1)!\left[\right(𝑛+2)-1]}$

Cancel $(𝑛+1)!$:

$\frac{𝑛+3}{𝑛+1}$

Now take the limit:

$\underset{𝑛\to \infty }{lim}\frac{𝑛+3}{𝑛+1}=1$

So the value of the limit is

$\overline{)1}$

Correct option: C.

The given compound statement is

$\left[\right(𝑝\vee 𝑞)\wedge (𝑞\Rightarrow 𝑟)\wedge (\sim 𝑟\left)\right]\Rightarrow (𝑝\wedge 𝑞)$

Its truth value is false.

An implication $𝐴\Rightarrow 𝐵$ is false only when

$𝐴\text{ is true and }𝐵\text{ is false}$

So here,

$\left[\right(𝑝\vee 𝑞)\wedge (𝑞\Rightarrow 𝑟)\wedge (\sim 𝑟\left)\right]$

must be true, and

$(𝑝\wedge 𝑞)$

must be false.

Now, since

$\sim 𝑟\text{ is true}$

we get

$𝑟=𝐹$

Also,

$𝑞\Rightarrow 𝑟$

is true. Since $𝑟=𝐹$, the implication $𝑞\Rightarrow 𝑟$ can be true only if

$𝑞=𝐹$

Now,

$𝑝\vee 𝑞$

is true. But $𝑞=𝐹$, so for $𝑝\vee 𝑞$ to be true, we must have

$𝑝=𝑇$

So the truth values are:

$𝑝=𝑇,𝑞=𝐹,𝑟=𝐹$

Now find:

$𝑝\Rightarrow 𝑞=𝑇\Rightarrow 𝐹=𝐹$

and

$𝑞\Rightarrow 𝑝=𝐹\Rightarrow 𝑇=𝑇$

Hence, the truth values of $𝑝\Rightarrow 𝑞$ and $𝑞\Rightarrow 𝑝$ are respectively

$(𝐹,𝑇)$

So the correct option is

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

The statement $𝑝\leftrightarrow 𝑞$ is true when $𝑝$ and $𝑞$ have the same truth value.

So, $\sim (𝑝\leftrightarrow 𝑞)$ will be true when $𝑝$ and $𝑞$ have opposite truth values.

That means:

• $𝑝$ is true and $𝑞$ is false, or

• $𝑝$ is false and $𝑞$ is true.

Now look at Option D:

$𝑝<text>\leftrightarrow </text><text>\sim </text>𝑞$

This means $𝑝$ and $\sim 𝑞$ have the same truth value.

If $𝑝$ is true, then $\sim 𝑞$ must be true, so $𝑞$ is false.

If $𝑝$ is false, then $\sim 𝑞$ must be false, so $𝑞$ is true.

So $𝑝$ and $𝑞$ are opposite in truth value.

Hence,

$\sim (𝑝\leftrightarrow 𝑞)\equiv 𝑝<text>\leftrightarrow </text><text>\sim </text>𝑞$

So the correct option is:

$\overline{)\text{Option D }𝑝<text>\leftrightarrow </text><text>\sim </text>𝑞}$

For a $2\times 2$ matrix

$\left[\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right],$

we know that

$\mathrm{adj}𝐴=\left[\begin{array}{cc}𝑑& -𝑏\\ -𝑐& 𝑎\end{array}\right].$

Here,

$𝐴=\left[\begin{array}{cc}\sec 𝜃& -\tan 𝜃\\ -\tan 𝜃& \sec 𝜃\end{array}\right].$

So,

$\mathrm{adj}𝐴=\left[\begin{array}{cc}\sec 𝜃& \tan 𝜃\\ \tan 𝜃& \sec 𝜃\end{array}\right].$

Now add $𝐴$ and $\mathrm{adj}𝐴$:

$𝐴+\mathrm{adj}𝐴=\left[\begin{array}{cc}\sec 𝜃& -\tan 𝜃\\ -\tan 𝜃& \sec 𝜃\end{array}\right]+\left[\begin{array}{cc}\sec 𝜃& \tan 𝜃\\ \tan 𝜃& \sec 𝜃\end{array}\right]$

$=\left[\begin{array}{cc}2\sec 𝜃& 0\\ 0& 2\sec 𝜃\end{array}\right].$

Given that

$𝐴+\mathrm{adj}𝐴=4𝐼,$

so

$\left[\begin{array}{cc}2\sec 𝜃& 0\\ 0& 2\sec 𝜃\end{array}\right]=\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right].$

Thus,

$2\sec 𝜃=4$

$\sec 𝜃=2$

$\cos 𝜃=\frac{1}{2}.$

Since

$𝜃\in (0,\frac{𝜋}{2}),$

we get

$𝜃=\frac{𝜋}{3}.$

So the correct option is

$\overline{)\text{Option C }\frac{𝜋}{3}}$

For a matrix $𝐴$, we know that

${𝐴}^{-1}=\frac{1}{\left|𝐴\right|}\mathrm{adj}𝐴$

The element in the $1$st row and $2$nd column of ${𝐴}^{-1}$ is

$({𝐴}^{-1}{)}_{12}=\frac{(\mathrm{adj}𝐴{)}_{12}}{\left|𝐴\right|}$

Since adjoint is the transpose of the cofactor matrix,

$(\mathrm{adj}𝐴{)}_{12}={𝐶}_{21}$

So we need the cofactor of the element in the $2$nd row and $1$st column.

Given

$𝐴=⎡⎣⎢\begin{array}{ccc}1& 3& -2\\ -3& 0& -5\\ 2& 5& 0\end{array}⎤⎦⎥$

First find $\left|𝐴\right|$:

$\left|𝐴\right|=1\mid \begin{array}{cc}0& -5\\ 5& 0\end{array}\mid -3\mid \begin{array}{cc}-3& -5\\ 2& 0\end{array}\mid +(-2)\mid \begin{array}{cc}-3& 0\\ 2& 5\end{array}\mid$

$\left|𝐴\right|=1\left(25\right)-3\left(10\right)+(-2)(-15)$

$\left|𝐴\right|=25-30+30=25$

Now find ${𝐶}_{21}$.

Remove the $2$nd row and $1$st column:

$\mid \begin{array}{cc}3& -2\\ 5& 0\end{array}\mid =3\left(0\right)-(-2)\left(5\right)=10$

So,

${𝐶}_{21}=(-1{)}^{2+1}(10)=-10$

Hence,

$({𝐴}^{-1}{)}_{12}=\frac{{𝐶}_{21}}{\left|𝐴\right|}=\frac{-10}{25}=-\frac{2}{5}$

So the correct option is

$\overline{)-\frac{2}{5}}$

Option A.

The denominators follow the pattern

${𝑥}^2+𝑥+1,{𝑥}^2+3𝑥+3,{𝑥}^2+5𝑥+7,\dots$

The $𝑟$th denominator can be written as

${𝑥}^2+(2𝑟-1)𝑥+𝑟(𝑟-1)+1$

or

$(𝑥+𝑟-1)(𝑥+𝑟)+1$

So the $𝑟$th term is

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

Now use the identity

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

Taking

$𝑎=𝑥+𝑟,𝑏=𝑥+𝑟-1$

we get

$𝑎-𝑏=1$

and

$1+𝑎𝑏=1+(𝑥+𝑟)(𝑥+𝑟-1)$

So,

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

Hence the given sum becomes

$𝑦=[{\tan }^{-1}(𝑥+1)-{\tan }^{-1}𝑥]+[{\tan }^{-1}(𝑥+2)-{\tan }^{-1}(𝑥+1\left)\right]+\cdots$

All middle terms cancel, so upto $𝑛$ terms,

$𝑦={\tan }^{-1}(𝑥+𝑛)-{\tan }^{-1}𝑥$

Differentiate with respect to $𝑥$:

${𝑦}^{\prime }=\frac{1}{1+(𝑥+𝑛{)}^2}-\frac{1}{1+{𝑥}^2}$

At $𝑥=0$,

${𝑦}^{\prime }\left(0\right)=\frac{1}{1+{𝑛}^2}-1$

${𝑦}^{\prime }\left(0\right)=\frac{1-({𝑛}^2+1)}{{𝑛}^2+1}$

${𝑦}^{\prime }\left(0\right)=-\frac{{𝑛}^2}{{𝑛}^2+1}$

So the correct option is

$\overline{)-\frac{{𝑛}^2}{{𝑛}^2+1}}$

Option B.

The function is

$𝑓\left(𝑥\right)=({\sin }^{-1}𝑥{)}^2+({\cos }^{-1}𝑥{)}^2$

The domain is

$-1\le 𝑥\le 1$

For every $𝑥\in [-1,1]$, we know the standard identity:

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

Let

${\sin }^{-1}𝑥=𝑡$

Then

${\cos }^{-1}𝑥=\frac{𝜋}{2}-𝑡$

Also, since ${\sin }^{-1}𝑥\in [-\frac{𝜋}{2},\frac{𝜋}{2}]$, we have

$𝑡\in [-\frac{𝜋}{2},\frac{𝜋}{2}]$

Now,

$𝑓\left(𝑥\right)={𝑡}^2+{(\frac{𝜋}{2}-𝑡)}^2$

Expanding,

$𝑓\left(𝑥\right)={𝑡}^2+\frac{{𝜋}^2}{4}-𝜋𝑡+{𝑡}^2$

$𝑓\left(𝑥\right)=2{𝑡}^2-𝜋𝑡+\frac{{𝜋}^2}{4}$

Complete the square:

$𝑓\left(𝑥\right)=2{(𝑡-\frac{𝜋}{4})}^2+\frac{{𝜋}^2}{8}$

Since the square term is always non-negative, the least value is

$\frac{{𝜋}^2}{8}$

This occurs at

$𝑡=\frac{𝜋}{4}$

Now for the greatest value, check the endpoints of

$𝑡\in [-\frac{𝜋}{2},\frac{𝜋}{2}]$

At $𝑡=-\frac{𝜋}{2}$,

$𝑓={(-\frac{𝜋}{2})}^2+{\left(𝜋\right)}^2$

$𝑓=\frac{{𝜋}^2}{4}+{𝜋}^2=\frac{5{𝜋}^2}{4}$

At $𝑡=\frac{𝜋}{2}$,

$𝑓={\left(\frac{𝜋}{2}\right)}^2+0^2=\frac{{𝜋}^2}{4}$

So the greatest value is

$\frac{5{𝜋}^2}{4}$

Hence, the required sum is

$\frac{5{𝜋}^2}{4}+\frac{{𝜋}^2}{8}$

$=\frac{10{𝜋}^2}{8}+\frac{{𝜋}^2}{8}$

$=\frac{11{𝜋}^2}{8}$

So the correct option is

$\overline{)\text{Option B }\frac{11{𝜋}^2}{8}}$

Using the principal values of inverse trigonometric functions,

${\tan }^{-1}\left(1\right)=\frac{𝜋}{4}$

because $\tan \frac{𝜋}{4}=1$.

Also,

${\cos }^{-1}\left(\frac{1}{2}\right)=\frac{𝜋}{3}$

and

${\sin }^{-1}\left(\frac{1}{2}\right)=\frac{𝜋}{6}$

So the numerator is

$\frac{𝜋}{4}+\frac{𝜋}{3}+\frac{𝜋}{6}$

Now,

$\frac{𝜋}{3}+\frac{𝜋}{6}=\frac{𝜋}{2}$

Hence,

$\text{Numerator}=\frac{𝜋}{4}+\frac{𝜋}{2}=\frac{3𝜋}{4}$

For the denominator,

${\tan }^{-1}\left(\sqrt{\surd 3}\right)=\frac{𝜋}{3}$

Now,

${\sec }^{-1}(-2)={\cos }^{-1}(-\frac{1}{2})$

Since

$\cos \frac{2𝜋}{3}=-\frac{1}{2}$

we get

${\sec }^{-1}(-2)=\frac{2𝜋}{3}$

So the denominator is

$\frac{𝜋}{3}-\frac{2𝜋}{3}=-\frac{𝜋}{3}$

Now the given value is

$\frac{\frac{3𝜋}{4}}{-\frac{𝜋}{3}}$

$=\frac{3𝜋}{4}\times \frac{-3}{𝜋}$

$=-\frac{9}{4}$

So the correct option is

$\overline{)-\frac{9}{4}}$

Option D.

Since $𝑔\left(𝑥\right)$ is the inverse function of $𝑓\left(𝑥\right)$, we have

$𝑔\left(1\right)={𝑓}^{-1}\left(1\right)$

This means we need to find the value of $𝑥$ for which

$𝑓\left(𝑥\right)=1$

Given,

$𝑓\left(𝑥\right)=\frac{5𝑥+3}{4𝑥-1}$

So,

$\frac{5𝑥+3}{4𝑥-1}=1$

Now multiply both sides by $4𝑥-1$:

$5𝑥+3=4𝑥-1$

Bring like terms together:

$5𝑥-4𝑥=-1-3$

$𝑥=-4$

So,

$𝑔\left(1\right)=-4$

The correct option is Option A: $-4$.

Since $𝑓\left(𝑥\right)$ is continuous at $𝑥=0$, we must have

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

Given $𝑓\left(0\right)=2$, so

$\underset{𝑥\to 0}{lim}\frac{8^{𝑥}-2^{𝑥}}{{𝑘}^{𝑥}-1}=2$

Now write $8^{𝑥}$ as $(2^3{)}^{𝑥}=2^{3𝑥}$.

Using the standard limit idea,

${𝑎}^{𝑥}\approx 1+𝑥\ln 𝑎\text{when }𝑥\to 0$

So,

$8^{𝑥}-2^{𝑥}\approx (1+𝑥\ln 8)-(1+𝑥\ln 2)$

$=𝑥(\ln 8-\ln 2)$

$=𝑥\ln 4$

Also,

${𝑘}^{𝑥}-1\approx 𝑥\ln 𝑘$

So the limit becomes

$\underset{𝑥\to 0}{lim}\frac{8^{𝑥}-2^{𝑥}}{{𝑘}^{𝑥}-1}=\frac{\ln 4}{\ln 𝑘}$

This must be equal to $2$.

$\frac{\ln 4}{\ln 𝑘}=2$

Since $\ln 4=\ln \left(2^2\right)=2\ln 2$,

$\frac{2\ln 2}{\ln 𝑘}=2$

$\ln 𝑘=\ln 2$

So,

$𝑘=2$

Hence, the correct option is

$\overline{)2}$

Option D.