MHT CET 2026 11th April Evening Shift (Mathematics) Question Paper with Solutions

Since $𝛼$ and $𝛽$ are roots of

${𝑥}^2+𝑥+1=0$

we know that

${𝑥}^3-1=(𝑥-1)({𝑥}^2+𝑥+1)$

So the roots of ${𝑥}^2+𝑥+1=0$ satisfy

${𝛼}^3=1,{𝛽}^3=1$

Also, from the quadratic equation,

$𝛼+𝛽=-\frac{1}{1}=-1$

Now,

$2026=3\times 675+1$

So,

${𝛼}^{2026}={𝛼}^{3\times 675+1}=({𝛼}^3{)}^{675}𝛼=1^{675}𝛼=𝛼$

Similarly,

${𝛽}^{2026}=𝛽$

Hence,

${𝛼}^{2026}+{𝛽}^{2026}=𝛼+𝛽=-1$

So the correct option is

$\overline{)-1}$

Option A.

For $8$ beads of different colours, first arrange them in a circle.

In a circular arrangement, rotations are considered the same. So the number of circular arrangements of $8$ different beads is

$(8-1)!=7!$

$7!=5040$

But in a necklace, we can turn it over. So a clockwise arrangement and its anticlockwise arrangement are also considered the same.

Hence, we divide by $2$:

$\frac{7!}{2}=\frac{5040}{2}=2520$

So, the number of ways is

$\overline{)2520}$

Correct option: Option B.

In $\Delta 𝐴𝐵𝐶$, sides $𝑎,𝑏,𝑐$ are opposite to angles $𝐴,𝐵,𝐶$ respectively.

Given,

$\angle 𝐴={90}^{\circ }$

So $𝐵𝐶=𝑎$ is the hypotenuse, and $𝑏,𝑐$ are the other two sides.

Since

$𝐵+𝐶={90}^{\circ }$

we need to find:

$\sin (𝐵-𝐶)$

Using the identity,

$\sin (𝐵-𝐶)=\sin 𝐵\cos 𝐶-\cos 𝐵\sin 𝐶$

In the right triangle,

$\sin 𝐵=\frac{𝑏}{𝑎},\cos 𝐵=\frac{𝑐}{𝑎}$

and

$\sin 𝐶=\frac{𝑐}{𝑎},\cos 𝐶=\frac{𝑏}{𝑎}$

So,

$\sin (𝐵-𝐶)=\frac{𝑏}{𝑎}\cdot \frac{𝑏}{𝑎}-\frac{𝑐}{𝑎}\cdot \frac{𝑐}{𝑎}$

$\sin (𝐵-𝐶)=\frac{{𝑏}^2-{𝑐}^2}{{𝑎}^2}$

By Pythagoras theorem,

${𝑎}^2={𝑏}^2+{𝑐}^2$

Hence,

$\sin (𝐵-𝐶)=\frac{{𝑏}^2-{𝑐}^2}{{𝑏}^2+{𝑐}^2}$

So the correct option is:

$\overline{)\text{Option C }\frac{{𝑏}^2-{𝑐}^2}{{𝑏}^2+{𝑐}^2}}$

Take

$𝐸={\cos }^2{10}^{\circ }-\cos {10}^{\circ }\cos {50}^{\circ }+{\cos }^2{50}^{\circ }$

Use the identities:

${\cos }^2𝐴+{\cos }^2𝐵=1+\cos (𝐴+𝐵)\cos (𝐴-𝐵)$

and

$\cos 𝐴\cos 𝐵=\frac{1}{2}\{\cos (𝐴+𝐵)+\cos (𝐴-𝐵\left)\right\}$

Here $𝐴={10}^{\circ }$ and $𝐵={50}^{\circ }$.

So,

${\cos }^2{10}^{\circ }+{\cos }^2{50}^{\circ }=1+\cos {60}^{\circ }\cos {40}^{\circ }$

Since $\cos {60}^{\circ }=\frac{1}{2}$,

${\cos }^2{10}^{\circ }+{\cos }^2{50}^{\circ }=1+\frac{1}{2}\cos {40}^{\circ }$

Also,

$\cos {10}^{\circ }\cos {50}^{\circ }=\frac{1}{2}\{\cos {60}^{\circ }+\cos {40}^{\circ }\}$

$=\frac{1}{2}\{\frac{1}{2}+\cos {40}^{\circ }\}$

$=\frac{1}{4}+\frac{1}{2}\cos {40}^{\circ }$

Now substitute:

$𝐸=1+\frac{1}{2}\cos {40}^{\circ }-(\frac{1}{4}+\frac{1}{2}\cos {40}^{\circ })$

$𝐸=1-\frac{1}{4}$

$𝐸=\frac{3}{4}$

So the correct option is

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

Option C.

We need to solve

$\cos 4𝜃=\cos 3𝜃$

Using the NCERT result:

$\cos 𝐴=\cos 𝐵⟹𝐴=2𝑛𝜋\pm 𝐵,𝑛\in 𝑍$

Here,

$𝐴=4𝜃,𝐵=3𝜃$

So,

$4𝜃=2𝑛𝜋\pm 3𝜃$

Now take both cases.

For the positive sign:

$4𝜃=2𝑛𝜋+3𝜃$

$𝜃=2𝑛𝜋$

For the negative sign:

$4𝜃=2𝑛𝜋-3𝜃$

$7𝜃=2𝑛𝜋$

$𝜃=\frac{2𝑛𝜋}{7}$

The values $𝜃=2𝑛𝜋$ are already included in

$𝜃=\frac{2𝑛𝜋}{7}$

when $𝑛$ is a multiple of $7$.

So the general solution is

$𝜃=\frac{2𝑛𝜋}{7},𝑛\in 𝑍$

Correct option: A

The line is

$2𝑥+𝑦=4$

It meets the $𝑋$-axis where $𝑦=0$.

$2𝑥+0=4\Rightarrow 𝑥=2$

So,

$𝐴=(2,0)$

It meets the $𝑌$-axis where $𝑥=0$.

$0+𝑦=4\Rightarrow 𝑦=4$

So,

$𝐵=(0,4)$

Now $𝐴𝐶$ and $𝐵𝐶$ are perpendicular, so

$\angle 𝐴𝐶𝐵={90}^{\circ }$

This means $𝐶$ lies on the circle having $𝐴𝐵$ as diameter. By the circle theorem, angle in a semicircle is a right angle.

So the distance of $𝐶$ from the midpoint of $𝐴𝐵$ is the radius of this circle.

First find the length of $𝐴𝐵$:

$𝐴𝐵=\sqrt{\surd (2-0{)}^2+(0-4{)}^2}$

$𝐴𝐵=\sqrt{\surd 4+16}$

$𝐴𝐵=\sqrt{\surd 20}=2\sqrt{\surd 5}$

Radius $=\frac{𝐴𝐵}{2}$

$\text{Distance of }𝐶\text{ from midpoint of }𝐴𝐵=\frac{2\sqrt{\surd 5}}{2}=\sqrt{\surd 5}$

So the correct option is

$\overline{)\sqrt{\surd 5}}$

Option B.

The given equation is

$(𝑥-2𝑦{)}^2+𝑘(𝑥-2𝑦)=0$

Take $(𝑥-2𝑦)$ common:

$(𝑥-2𝑦)[(𝑥-2𝑦)+𝑘]=0$

So the two lines are

$𝑥-2𝑦=0$

and

$𝑥-2𝑦+𝑘=0$

These are parallel lines because both have the same coefficients of $𝑥$ and $𝑦$.

The distance between two parallel lines

$𝑎𝑥+𝑏𝑦+{𝑐}_1=0$

and

$𝑎𝑥+𝑏𝑦+{𝑐}_2=0$

is

$\frac{|{𝑐}_1-{𝑐}_2|}{\sqrt{\surd {𝑎}^2+{𝑏}^2}}$

Here,

$𝑥-2𝑦=0$

and

$𝑥-2𝑦+𝑘=0$

So $𝑎=1$, $𝑏=-2$, ${𝑐}_1=0$, and ${𝑐}_2=𝑘$.

Given distance is $3$ units:

$\frac{|𝑘-0|}{\sqrt{\surd 1^2+(-2{)}^2}}=3$

$\frac{𝑘}{\sqrt{\surd 5}}=3$

since $𝑘>0$.

So,

$𝑘=3\sqrt{\surd 5}$

Hence the equations of the lines are

$𝑥-2𝑦=0$

and

$𝑥-2𝑦+3\sqrt{\surd 5}=0$

So the correct option is

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

For the pair of straight lines,

$\frac{{𝑥}^2}{𝑎}+\frac{2𝑥𝑦}{ℎ}+\frac{{𝑦}^2}{𝑏}=0$

both lines pass through the origin. So if the slope of a line is $𝑚$, then we put

$𝑦=𝑚𝑥$

Substitute this in the equation:

$\frac{{𝑥}^2}{𝑎}+\frac{2𝑥\left(𝑚𝑥\right)}{ℎ}+\frac{(𝑚𝑥{)}^2}{𝑏}=0$

${𝑥}^2(\frac{1}{𝑎}+\frac{2𝑚}{ℎ}+\frac{{𝑚}^2}{𝑏})=0$

So the slopes are given by

$\frac{{𝑚}^2}{𝑏}+\frac{2𝑚}{ℎ}+\frac{1}{𝑎}=0$

Let the two slopes be $𝑚$ and $2𝑚$.

For the quadratic in slopes,

$\frac{1}{𝑏}{𝑚}^2+\frac{2}{ℎ}𝑚+\frac{1}{𝑎}=0$

Product of roots:

$𝑚\cdot 2𝑚=\frac{\frac{1}{𝑎}}{\frac{1}{𝑏}}$

$2{𝑚}^2=\frac{𝑏}{𝑎}$

So,

${𝑚}^2=\frac{𝑏}{2𝑎}$

Sum of roots:

$𝑚+2𝑚=-\frac{\frac{2}{ℎ}}{\frac{1}{𝑏}}$

$3𝑚=-\frac{2𝑏}{ℎ}$

Squaring both sides,

$9{𝑚}^2=\frac{4{𝑏}^2}{{ℎ}^2}$

Now put ${𝑚}^2=\frac{𝑏}{2𝑎}$:

$9\cdot \frac{𝑏}{2𝑎}=\frac{4{𝑏}^2}{{ℎ}^2}$

${ℎ}^2=\frac{8𝑎𝑏}{9}$

Hence,

${ℎ}^2:𝑎𝑏=8:9$

So the correct option is

$\overline{)8:9}$

Option C.

For a second degree equation to represent a circle, the coefficients of ${𝑥}^2$ and ${𝑦}^2$ must be equal, and the coefficient of $𝑥𝑦$ must be zero.

Given equation:

$3{𝑥}^2+(3-𝑝)𝑥𝑦+𝑞{𝑦}^2-2𝑝𝑥=8𝑝𝑞$

For it to be a circle,

$3-𝑝=0$

So,

$𝑝=3$

Also, the coefficient of ${𝑦}^2$ must be equal to the coefficient of ${𝑥}^2$:

$𝑞=3$

Now substitute $𝑝=3$ and $𝑞=3$ in the equation:

$3{𝑥}^2+3{𝑦}^2-6𝑥=8\left(3\right)\left(3\right)$

$3{𝑥}^2+3{𝑦}^2-6𝑥=72$

Divide by $3$:

${𝑥}^2+{𝑦}^2-2𝑥=24$

Complete the square:

${𝑥}^2-2𝑥+{𝑦}^2=24$

$(𝑥-1{)}^2-1+{𝑦}^2=24$

$(𝑥-1{)}^2+{𝑦}^2=25$

This is a circle with radius $𝑟=5$.

Area of the circle:

$𝜋{𝑟}^2=𝜋(5{)}^2=25𝜋$

So, the correct option is

$\overline{)25𝜋}$

Option C.

The ellipse is

$4{𝑥}^2+9{𝑦}^2=1$

Write it in standard form:

$\frac{{𝑥}^2}{\frac{1}{4}}+\frac{{𝑦}^2}{\frac{1}{9}}=1$

Here,

${𝑎}^2=\frac{1}{4},{𝑏}^2=\frac{1}{9}$

Since $\frac{1}{4}>\frac{1}{9}$, the major axis is along the $𝑥$-axis.

So,

$𝑎=\frac{1}{2},𝑏=\frac{1}{3}$

For an ellipse, the foci are at $(\pm 𝑐,0)$, where

${𝑐}^2={𝑎}^2-{𝑏}^2$

So,

${𝑐}^2=\frac{1}{4}-\frac{1}{9}=\frac{9-4}{36}=\frac{5}{36}$

Hence,

$𝑐=\frac{\sqrt{\surd 5}}{6}$

So the foci are

$𝑆(\frac{\sqrt{\surd 5}}{6},0),{𝑆}^{\prime }(-\frac{\sqrt{\surd 5}}{6},0)$

The ellipse intersects the positive $𝑌$-axis at $𝐴$. On the $𝑌$-axis, $𝑥=0$.

$9{𝑦}^2=1$

$𝑦=\frac{1}{3}$

So,

$𝐴=(0,\frac{1}{3})$

Now, in $△𝑆𝐴{𝑆}^{\prime }$:

Base:

$𝑆{𝑆}^{\prime }=2𝑐=2\cdot \frac{\sqrt{\surd 5}}{6}=\frac{\sqrt{\surd 5}}{3}$

Height from $𝐴$ to $𝑆{𝑆}^{\prime }$ is:

$\frac{1}{3}$

Area of triangle:

$\frac{1}{2}\times \frac{\sqrt{\surd 5}}{3}\times \frac{1}{3}=\frac{\sqrt{\surd 5}}{18}$

So the area is

$\overline{)\frac{\sqrt{\surd 5}}{18}}$

Correct option: B.

Given limit is

$\underset{𝑥\to 1}{lim}[\frac{𝑥-2}{{𝑥}^2-𝑥}-\frac{1}{{𝑥}^3-3{𝑥}^2+2𝑥}]$

First factor the denominators:

${𝑥}^2-𝑥=𝑥(𝑥-1)$

and

${𝑥}^3-3{𝑥}^2+2𝑥=𝑥({𝑥}^2-3𝑥+2)=𝑥(𝑥-1)(𝑥-2)$

So the expression becomes

$\frac{𝑥-2}{𝑥(𝑥-1)}-\frac{1}{𝑥(𝑥-1)(𝑥-2)}$

Take the common denominator $𝑥(𝑥-1)(𝑥-2)$:

$\frac{(𝑥-2{)}^2-1}{𝑥(𝑥-1)(𝑥-2)}$

Now simplify the numerator:

$(𝑥-2{)}^2-1={𝑥}^2-4𝑥+4-1={𝑥}^2-4𝑥+3$

${𝑥}^2-4𝑥+3=(𝑥-1)(𝑥-3)$

So,

$\frac{(𝑥-1)(𝑥-3)}{𝑥(𝑥-1)(𝑥-2)}$

Cancel $(𝑥-1)$:

$\frac{𝑥-3}{𝑥(𝑥-2)}$

Now put $𝑥=1$:

$\frac{1-3}{1(1-2)}=\frac{-2}{-1}=2$

So the value of the limit is

$\overline{)2}$

Correct option: Option C.

For $𝐴=\{1,2,3,4,5,6\}$, we check each statement.

Option A:

$\exists 𝑥\in 𝐴\text{ such that }𝑥+3=9$

This means there is at least one value of $𝑥$ in $𝐴$ for which $𝑥+3=9$.

If $𝑥=6$, then

$6+3=9$

So Option A is true.

Option B:

$\exists 𝑥\in 𝐴\text{ such that }𝑥+2<9$

If $𝑥=1$, then

$1+2=3<9$

So Option B is true.

Option C:

$\forall 𝑥\in 𝐴,𝑥+6\ge 10$

This means for every value of $𝑥$ in $𝐴$, $𝑥+6$ should be greater than or equal to $10$.

But if $𝑥=1$, then

$1+6=7$

and

$7\ngeqq 10$

So Option C is false.

Option D:

$\exists 𝑥\in 𝐴\text{ such that }𝑥+6<10$

If $𝑥=1$, then

$1+6=7<10$

So Option D is true.

Hence, the statement which is not true is:

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

For $𝑝\Rightarrow 𝑞$ to be false, $𝑝$ must be true and $𝑞$ must be false.

So in statement $𝑟$:

$𝑝=𝑇,𝑞=𝐹$

Then

$𝑝\vee 𝑞=𝑇\vee 𝐹=𝑇$

But statement $𝑟$ says $𝑝\vee 𝑞$ is false. That is not true.

So,

$𝑟=𝐹$

Now for $𝑝\leftrightarrow 𝑞$ to be false, $𝑝$ and $𝑞$ must have different truth values.

So either

$𝑝=𝑇,𝑞=𝐹$

or

$𝑝=𝐹,𝑞=𝑇$

In both cases,

$𝑝\vee 𝑞=𝑇$

So the statement “$𝑝\vee 𝑞$ is false” is again not true.

Hence,

$𝑠=𝐹$

Now we find the truth values:

$𝑟\Rightarrow 𝑠=𝐹\Rightarrow 𝐹=𝑇$

and

$𝑠\Rightarrow 𝑟=𝐹\Rightarrow 𝐹=𝑇$

So the truth values are

$𝑇,𝑇$

Correct option: A.

Since the angles $𝐴,𝐵,𝐶$ are in Arithmetic Progression, the middle angle is the average of the other two:

$𝐴+𝐶=2𝐵$

Also, in a triangle,

$𝐴+𝐵+𝐶=𝜋$

Using $𝐴+𝐶=2𝐵$, we get

$2𝐵+𝐵=𝜋$

$3𝐵=𝜋$

$𝐵=\frac{𝜋}{3}$

Now given,

$3{𝑐}^2=2{𝑏}^2$

So,

$\frac{{𝑐}^2}{{𝑏}^2}=\frac{2}{3}$

Using the sine rule,

$\frac{𝑐}{𝑏}=\frac{\sin 𝐶}{\sin 𝐵}$

Squaring both sides,

$\frac{{𝑐}^2}{{𝑏}^2}=\frac{{\sin }^2𝐶}{{\sin }^2𝐵}$

Hence,

$\frac{{\sin }^2𝐶}{{\sin }^2𝐵}=\frac{2}{3}$

Since $𝐵=\frac{𝜋}{3}$,

$\sin 𝐵=\sin \frac{𝜋}{3}=\frac{\sqrt{\surd 3}}{2}$

So,

${\sin }^2𝐵=\frac{3}{4}$

Thus,

${\sin }^2𝐶=\frac{2}{3}\times \frac{3}{4}$

${\sin }^2𝐶=\frac{1}{2}$

So,

$\sin 𝐶=\frac{1}{\sqrt{\surd 2}}$

This gives

$𝐶=\frac{𝜋}{4}$

Now,

$𝐴+𝐵+𝐶=𝜋$

$𝐴+\frac{𝜋}{3}+\frac{𝜋}{4}=𝜋$

$𝐴=𝜋-\frac{𝜋}{3}-\frac{𝜋}{4}$

$𝐴=\frac{12𝜋-4𝜋-3𝜋}{12}$

$𝐴=\frac{5𝜋}{12}$

So the measure of angle $𝐴$ is

$\overline{)\frac{5𝜋}{12}}$

Correct option: A.

Given

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

First find ${𝐴}^2$.

${𝐴}^2=𝐴\cdot 𝐴$

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

Multiplying row by column,

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

So,

${𝐴}^2=𝐼$

This means

$𝐴\cdot 𝐴=𝐼$

By the definition of inverse matrix, if

$𝐴\cdot 𝐵=𝐼$

then $𝐵={𝐴}^{-1}$.

Here, $𝐵=𝐴$ itself. Hence,

${𝐴}^{-1}=𝐴$

So the true option is

$\overline{)\text{Option C: }{𝐴}^{-1}=𝐴}$

The given matrix equation means

$⎡⎣⎢\begin{array}{ccc}2& 1& -1\\ -1& 2& 1\\ 1& -1& 2\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}𝑎\\ 𝑏\\ 𝑐\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}4\\ 0\\ -2\end{array}⎤⎦⎥$

So we get the system of linear equations:

$2𝑎+𝑏-𝑐=4$

$-𝑎+2𝑏+𝑐=0$

$𝑎-𝑏+2𝑐=-2$

Solving these equations gives

$𝑎=1,𝑏=1,𝑐=-1$

So the point is

$𝑃(1,1,-1)$

Now, the distance of a point $({𝑥}_1,{𝑦}_1,{𝑧}_1)$ from the plane

$𝐴𝑥+𝐵𝑦+𝐶𝑧+𝐷=0$

is

$\frac{|𝐴{𝑥}_1+𝐵{𝑦}_1+𝐶{𝑧}_1+𝐷|}{\sqrt{\surd {𝐴}^2+{𝐵}^2+{𝐶}^2}}$

The given plane is

$2𝑥+𝑦+2𝑧=10$

or

$2𝑥+𝑦+2𝑧-10=0$

Here,

$𝐴=2,𝐵=1,𝐶=2,𝐷=-10$

Using $𝑃(1,1,-1)$,

$\text{Distance}=\frac{\left|2\right(1)+1(1)+2(-1)-10|}{\sqrt{\surd 2^2+1^2+2^2}}$

$=\frac{|2+1-2-10|}{\sqrt{\surd 4+1+4}}$

$=\frac{|-9|}{3}$

$=3$

So, the distance is

$\overline{)3\text{ units}}$

Correct option: Option B.

Take

$𝜃={\cos }^{-1}𝑥$

So,

$\cos 𝜃=𝑥$

Using a right triangle, if $\cos 𝜃=𝑥$, then

$\sin 𝜃=\sqrt{\surd 1-{𝑥}^2}$

So,

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

Now take

$𝜙={\tan }^{-1}\left(\frac{𝑎}{\sqrt{\surd {𝑏}^2-{𝑎}^2}}\right)$

So,

$\tan 𝜙=\frac{𝑎}{\sqrt{\surd {𝑏}^2-{𝑎}^2}}$

In a right triangle,

• perpendicular $=𝑎$

• base $=\sqrt{\surd {𝑏}^2-{𝑎}^2}$

Then hypotenuse is

$\sqrt{\surd {𝑎}^2+({𝑏}^2-{𝑎}^2)}=\sqrt{\surd {𝑏}^2}=𝑏$

So,

$\sec 𝜙=\frac{\text{hypotenuse}}{\text{base}}=\frac{𝑏}{\sqrt{\surd {𝑏}^2-{𝑎}^2}}$

Given,

$\cot ({\cos }^{-1}𝑥)=\sec ({\tan }^{-1}\frac{𝑎}{\sqrt{\surd {𝑏}^2-{𝑎}^2}})$

So,

$\frac{𝑥}{\sqrt{\surd 1-{𝑥}^2}}=\frac{𝑏}{\sqrt{\surd {𝑏}^2-{𝑎}^2}}$

Squaring both sides,

$\frac{{𝑥}^2}{1-{𝑥}^2}=\frac{{𝑏}^2}{{𝑏}^2-{𝑎}^2}$

Cross-multiply:

${𝑥}^2({𝑏}^2-{𝑎}^2)={𝑏}^2(1-{𝑥}^2)$

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

${𝑥}^2(2{𝑏}^2-{𝑎}^2)={𝑏}^2$

${𝑥}^2=\frac{{𝑏}^2}{2{𝑏}^2-{𝑎}^2}$

So,

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

Hence, the correct option is

$\overline{)\text{Option D }\frac{𝑏}{\sqrt{\surd 2{𝑏}^2-{𝑎}^2}}}$

Given

$𝑔\left(𝑥\right)={𝑥}^2+𝑥-2$

and

$\frac{1}{2}(𝑔\circ 𝑓)\left(𝑥\right)=2{𝑥}^2-5𝑥+2$

So,

$(𝑔\circ 𝑓)\left(𝑥\right)=2(2{𝑥}^2-5𝑥+2)$

$𝑔\left(𝑓\right(𝑥\left)\right)=4{𝑥}^2-10𝑥+4$

Now, since

$𝑔\left(𝑥\right)={𝑥}^2+𝑥-2$

we have

$𝑔\left(𝑓\right(𝑥\left)\right)=\left(𝑓\right(𝑥){)}^2+𝑓(𝑥)-2$

So we need a function $𝑓\left(𝑥\right)$ such that

$\left(𝑓\right(𝑥){)}^2+𝑓(𝑥)-2=4{𝑥}^2-10𝑥+4$

Now check option B:

$𝑓\left(𝑥\right)=2𝑥-3$

Then

$𝑔\left(𝑓\right(𝑥\left)\right)=𝑔(2𝑥-3)$

$=(2𝑥-3{)}^2+(2𝑥-3)-2$

$=4{𝑥}^2-12𝑥+9+2𝑥-3-2$

$=4{𝑥}^2-10𝑥+4$

This matches the required value.

Hence,

$𝑓\left(𝑥\right)=2𝑥-3$

Correct option: B.

For $𝑥\ne 0$,

$𝑓\left(𝑥\right)=\frac{𝑥-\left|𝑥\right|}{𝑥}$

We check the value of $\left|𝑥\right|$ in two cases.

For $𝑥>0$, we have $\left|𝑥\right|=𝑥$.

So,

$𝑓\left(𝑥\right)=\frac{𝑥-𝑥}{𝑥}=0$

For $𝑥<0$, we have $\left|𝑥\right|=-𝑥$.

So,

$𝑓\left(𝑥\right)=\frac{𝑥-(-𝑥)}{𝑥}=\frac{2𝑥}{𝑥}=2$

Also, it is given that

$𝑓\left(0\right)=2$

So the function can be written as

$𝑓\left(𝑥\right)=⎧⎩⎨\begin{array}{cc}2,& 𝑥<0\\ 2,& 𝑥=0\\ 0,& 𝑥>0\end{array}$

For all $𝑥<0$, the function is constant, so it is continuous.

For all $𝑥>0$, the function is also constant, so it is continuous.

Now check at $𝑥=0$:

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

and

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

Since the left-hand limit and right-hand limit are not equal, the limit at $𝑥=0$ does not exist.

So $𝑓$ is not continuous at $𝑥=0$.

Hence, $𝑓$ is continuous for all $𝑥$ except at $𝑥=0$.

The correct option is:

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

Take

$𝑦={\tan }^{-1}\left(\frac{\sqrt{\surd 1+{𝑥}^2}-1}{𝑥}\right)$

and

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

We need

$\frac{𝑑𝑦}{𝑑𝑧}=\frac{\frac{𝑑𝑦}{𝑑𝑥}}{\frac{𝑑𝑧}{𝑑𝑥}}$

First simplify $𝑦$.

Rationalise the expression inside ${\tan }^{-1}$:

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

Now we use the standard identity

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

So,

$𝑦=\frac{1}{2}{\tan }^{-1}𝑥$

Differentiating,

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

Now consider

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

Using the identity

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

for the usual principal range, we get

$𝑧=2{\tan }^{-1}𝑥$

Differentiating,

$\frac{𝑑𝑧}{𝑑𝑥}=2\cdot \frac{1}{1+{𝑥}^2}=\frac{2}{1+{𝑥}^2}$

Hence,

$\frac{𝑑𝑦}{𝑑𝑧}=\frac{\frac{1}{2(1+{𝑥}^2)}}{\frac{2}{1+{𝑥}^2}}$

$\frac{𝑑𝑦}{𝑑𝑧}=\frac{1}{4}$

So the correct option is

$\overline{)\frac{1}{4}}$

Option B.

The given function is

${\tan }^{-1}(𝑎+𝑏𝑥)$

Its derivative with respect to $𝑥$ is

$\frac{𝑑}{𝑑𝑥}({\tan }^{-1}(𝑎+𝑏𝑥\left)\right)=\frac{𝑏}{1+(𝑎+𝑏𝑥{)}^2}$

At $𝑥=0$, the derivative is given as $1$.

So,

$\frac{𝑏}{1+{𝑎}^2}=1$

This gives

$𝑏=1+{𝑎}^2$

Now we need to find

${𝑎}^6-{𝑏}^3$

Since

$𝑏=1+{𝑎}^2$

we get

${𝑏}^3=(1+{𝑎}^2{)}^3$

Using expansion,

$(1+{𝑎}^2{)}^3=1+3{𝑎}^2+3{𝑎}^4+{𝑎}^6$

So,

${𝑎}^6-{𝑏}^3={𝑎}^6-(1+3{𝑎}^2+3{𝑎}^4+{𝑎}^6)$

$=-1-3{𝑎}^2-3{𝑎}^4$

Now,

$3{𝑎}^2𝑏=3{𝑎}^2(1+{𝑎}^2)=3{𝑎}^2+3{𝑎}^4$

Hence,

${𝑎}^6-{𝑏}^3=-1-3{𝑎}^2𝑏$

So the correct option is

$\overline{)-1-3{𝑎}^2𝑏}$

Option C.

Given

${𝑥}^3+{𝑦}^3=6$

Differentiate both sides with respect to $𝑥$:

$3{𝑥}^2+3{𝑦}^2\frac{𝑑𝑦}{𝑑𝑥}=0$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=-\frac{{𝑥}^2}{{𝑦}^2}$

Now differentiate again with respect to $𝑥$:

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{𝑑}{𝑑𝑥}(-\frac{{𝑥}^2}{{𝑦}^2})$

Writing it as $-{𝑥}^2{𝑦}^{-2}$,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=-2𝑥{𝑦}^{-2}+2{𝑥}^2{𝑦}^{-3}\frac{𝑑𝑦}{𝑑𝑥}$

Substitute $\frac{𝑑𝑦}{𝑑𝑥}=-\frac{{𝑥}^2}{{𝑦}^2}$:

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=-\frac{2𝑥}{{𝑦}^2}-\frac{2{𝑥}^4}{{𝑦}^5}$