MHT CET 2026 19th April Evening Shift (Mathematics) Question Paper with Solutions

The powers of $𝑖$ repeat in a cycle of $4$:

$1,𝑖,-1,-𝑖$

Their sum is

$1+𝑖-1-𝑖=0.$

From $𝑛=0$ to $𝑛=2023$, there are $2024$ terms, forming $506$ complete cycles. So their sum is $0$.

The remaining terms are

${𝑖}^{2024}+{𝑖}^{2025}+{𝑖}^{2026}=1+𝑖-1=𝑖.$

Thus,

$𝑧=𝑖={𝑒}^{𝑖𝜋/2}.$

Hence, one square root is

$\sqrt{\surd 𝑧}={𝑒}^{𝑖𝜋/4}.$

So the correct answer is

$\overline{)\text{Option A: }{𝑒}^{𝑖𝜋/4}}.$

For an anticlockwise rotation through an angle $𝜃$ about the origin, the new ordinate is

${𝑦}^{\prime }=𝑥\sin 𝜃+𝑦\cos 𝜃.$

Here, $𝑥=5$, $𝑦=12$, and $𝜃={30}^{\circ }$.

$\begin{array}{cc}{𝑦}^{\prime }& =5\sin {30}^{\circ }+12\cos {30}^{\circ }\\ & =5\left(\frac{1}{2}\right)+12\left(\frac{\sqrt{\surd 3}}{2}\right)\\ & =\frac{5}{2}+6\sqrt{\surd 3}.\end{array}$

Thus, the ordinate of point $𝐵$ is

$\overline{)6\sqrt{\surd 3}+\frac{5}{2}}$

Hence, Option A is correct.

Since ${}^{𝑛}{𝐶}_4,{}^{𝑛}{𝐶}_5,{}^{𝑛}{𝐶}_6$ are in A.P., the middle term is the average of the other two:

$2{}^{𝑛}{𝐶}_5={}^{𝑛}{𝐶}_4+{}^{𝑛}{𝐶}_6$

Dividing by ${}^{𝑛}{𝐶}_5$,

$2=\frac{{}^{𝑛}{𝐶}_4}{{}^{𝑛}{𝐶}_5}+\frac{{}^{𝑛}{𝐶}_6}{{}^{𝑛}{𝐶}_5}$

Using the relations

$\frac{{}^{𝑛}{𝐶}_4}{{}^{𝑛}{𝐶}_5}=\frac{5}{𝑛-4},\frac{{}^{𝑛}{𝐶}_6}{{}^{𝑛}{𝐶}_5}=\frac{𝑛-5}{6},$

we get

$2=\frac{5}{𝑛-4}+\frac{𝑛-5}{6}.$

Multiplying by $6(𝑛-4)$,

$12(𝑛-4)=30+(𝑛-5)(𝑛-4).$

On simplifying,

$12𝑛-48={𝑛}^2-9𝑛+50,$

${𝑛}^2-21𝑛+98=0,$

$(𝑛-7)(𝑛-14)=0.$

Thus,

$𝑛=7\text{or}𝑛=14.$

Hence, the correct answer is Option B: $7$ or $14$.

Given

$4𝑎𝑥+3𝑏𝑦+𝑐=0$

and

$𝑎+𝑏+𝑐=0,$

we have

$𝑐=-𝑎-𝑏.$

Substituting this in the equation of the line,

$4𝑎𝑥+3𝑏𝑦-𝑎-𝑏=0.$

Rearranging,

$𝑎(4𝑥-1)+𝑏(3𝑦-1)=0.$

For every line of the family to pass through one fixed point, this equation must hold for all values of $𝑎$ and $𝑏$. Hence,

$4𝑥-1=0,3𝑦-1=0.$

Thus,

$𝑥=\frac{1}{4},𝑦=\frac{1}{3}.$

So the lines are concurrent at

$\overline{)(\frac{1}{4},\frac{1}{3})}.$

Hence, Option C is correct.

For a line through the origin, take $𝑦=𝑚𝑥$, where $𝑚$ is its slope.

Substituting in

${𝑥}^2-(1+\sqrt{\surd 3})𝑥𝑦+\sqrt{\surd 3}{𝑦}^2=0,$

we get

${𝑥}^2-(1+\sqrt{\surd 3})𝑥\left(𝑚𝑥\right)+\sqrt{\surd 3}(𝑚𝑥{)}^2=0.$

Dividing by ${𝑥}^2$,

$\sqrt{\surd 3}{𝑚}^2-(1+\sqrt{\surd 3})𝑚+1=0.$

If the slopes are ${𝑚}_1=\tan 𝛼$ and ${𝑚}_2=\tan 𝛽$, then

${𝑚}_1+{𝑚}_2=\frac{1+\sqrt{\surd 3}}{\sqrt{\surd 3}},{𝑚}_1{𝑚}_2=\frac{1}{\sqrt{\surd 3}}.$

Using

$\tan (𝛼+𝛽)=\frac{\tan 𝛼+\tan 𝛽}{1-\tan 𝛼\tan 𝛽},$

we get

$\tan (𝛼+𝛽)=\frac{\frac{1+\sqrt{\surd 3}}{\sqrt{\surd 3}}}{1-\frac{1}{\sqrt{\surd 3}}}=\frac{\sqrt{\surd 3}+1}{\sqrt{\surd 3}-1}.$

Hence, the correct answer is

$\overline{)\frac{\sqrt{\surd 3}+1}{\sqrt{\surd 3}-1}}$

Option C.

Since $𝑃=(6,8)$ and $𝑂=(0,0)$,

$𝑂𝑃=\sqrt{\surd 6^2+8^2}=10$

A radius is perpendicular to the tangent at the point of contact. Hence,

$𝑂𝐴⟂𝑃𝐴,𝑂𝐵⟂𝑃𝐵$

Let the radius be $𝑟$. From right triangle $𝑂𝐴𝑃$,

$𝑃𝐴=\sqrt{\surd 𝑂{𝑃}^2-𝑂{𝐴}^2}=\sqrt{\surd 100-{𝑟}^2}$

The quadrilateral $𝑃𝐴𝑂𝐵$ consists of two congruent right triangles, $𝑃𝐴𝑂$ and $𝑃𝐵𝑂$. Its area is

$2(\frac{1}{2}\times 𝑟\times \sqrt{\surd 100-{𝑟}^2})=𝑟\sqrt{\surd 100-{𝑟}^2}$

To maximise this area, maximise its square:

${𝑟}^2(100-{𝑟}^2)$

Put $𝑥={𝑟}^2$. Then

$𝑥(100-𝑥)$

is maximum when the two factors are equal:

$𝑥=100-𝑥$

$2𝑥=100⟹𝑥=50$

Thus,

${𝑟}^2=50⟹𝑟=5\sqrt{\surd 2}$

So the correct answer is Option B, $\overline{)5\sqrt{\surd 2}}$.

Since the circle passes through $(0,1)$ and $(1,1)$, its centre must lie on the perpendicular bisector of the line segment joining these points.

The segment is horizontal, so its perpendicular bisector is

$𝑥=\frac{1}{2}.$

For the parabola $𝑦={𝑥}^2$,

$\frac{𝑑𝑦}{𝑑𝑥}=2𝑥.$

At $(1,1)$, the slope of the tangent is $2$. Hence, the slope of the normal is

$-\frac{1}{2}.$

Since the circle touches the parabola at $(1,1)$, its centre lies on this normal:

$𝑦-1=-\frac{1}{2}(𝑥-1).$

Putting $𝑥=\frac{1}{2}$,

$𝑦-1=-\frac{1}{2}(\frac{1}{2}-1)=\frac{1}{4},$

so

$𝑦=\frac{5}{4}.$

Thus, the centre of the circle is

$\overline{)(\frac{1}{2},\frac{5}{4})}.$

Hence, Option C is correct.

Given ellipse:

$7{𝑥}^2+16{𝑦}^2-14𝑥+64𝑦-377=0$

Group the $𝑥$ and $𝑦$ terms and complete the squares:

$7({𝑥}^2-2𝑥)+16({𝑦}^2+4𝑦)=377$

Since

${𝑥}^2-2𝑥=(𝑥-1{)}^2-1$

and

${𝑦}^2+4𝑦=(𝑦+2{)}^2-4,$

we get

$7(𝑥-1{)}^2+16(𝑦+2{)}^2=448.$

Dividing by $448$:

$\frac{(𝑥-1{)}^2}{64}+\frac{(𝑦+2{)}^2}{28}=1.$

Thus,

${𝑎}^2=64,{𝑏}^2=28.$

The eccentricity of an ellipse is

$𝑒=\sqrt{\surd 1-\frac{{𝑏}^2}{{𝑎}^2}}=\sqrt{\surd 1-\frac{28}{64}}=\sqrt{\surd \frac{36}{64}}=\frac{3}{4}.$

Hence, the correct answer is Option A: $\frac{3}{4}$.

Since

${𝑥}^2-2𝑥+1=(𝑥-1{)}^2,$

the numerator must also contain $(𝑥-1{)}^2$ as a factor for the limit to be finite.

As the numerator is a monic cubic polynomial, write

${𝑥}^3+𝑎{𝑥}^2+𝑏𝑥+𝑐=(𝑥-1{)}^2(𝑥+𝑘).$

Then

$\underset{𝑥\to 1}{lim}\frac{(𝑥-1{)}^2(𝑥+𝑘)}{(𝑥-1{)}^2}=\underset{𝑥\to 1}{lim}(𝑥+𝑘)=1+𝑘.$

Given that the limit is $2026$,

$1+𝑘=2026⟹𝑘=2025.$

Now,

$\begin{array}{cc}(𝑥-1{)}^2(𝑥+2025)& =({𝑥}^2-2𝑥+1)(𝑥+2025)\\ & ={𝑥}^3+2023{𝑥}^2-4049𝑥+2025.\end{array}$

Thus,

$𝑎=2023,𝑐=2025.$

Hence,

$𝑎-𝑐=2023-2025=-2.$

So the correct answer is Option D, $-2$.

An implication is false only when its antecedent is true and its consequent is false.

Thus, in

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

we must have

$(𝑝\vee 𝑞)\wedge (𝑞\to 𝑟)\wedge (\sim 𝑟)=𝑇.$

Since $\sim 𝑟=𝑇$, we get $𝑟=𝐹$.

Also, $𝑞\to 𝑟=𝑇$. As $𝑟=𝐹$, this implication can be true only when $𝑞=𝐹$.

Now $𝑝\vee 𝑞=𝑇$. Since $𝑞=𝐹$, we get $𝑝=𝑇$.

So,

$𝑝=𝑇,𝑞=𝐹,𝑟=𝐹.$

Checking the options:

• $𝑝\to 𝑞=𝑇\to 𝐹=𝐹$, so Option A is true.

• $𝑝\to 𝑟=𝑇\to 𝐹=𝐹$, so Option B is true.

• $(\sim 𝑞)\to 𝑝=𝑇\to 𝑇=𝑇$, so the claim that it is false is not true.

• $(\sim 𝑝)\wedge 𝑟=𝐹\wedge 𝐹=𝐹$, so Option D is true.

Hence, the statement which is NOT true is

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

Given:

$𝑝=𝑇,𝑞=𝐹,𝑟=𝑇$

Hence,

$\sim 𝑝=𝐹,\sim 𝑞=𝑇,\sim 𝑟=𝐹$

For Option A:

$[𝑞\wedge (𝑝\to 𝑞\left)\right]\to 𝑝$

Since $𝑝\to 𝑞=𝑇\to 𝐹=𝐹$,

$[𝐹\wedge 𝐹]\to 𝑇=𝐹\to 𝑇=𝑇$

For Option B:

$(𝑝\wedge 𝑞)\to (𝑞\vee \sim 𝑝)$

$=(𝑇\wedge 𝐹)\to (𝐹\vee 𝐹)=𝐹\to 𝐹=𝑇$

For Option C:

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

$=\left[\right(𝐹\vee 𝐹)\wedge 𝐹]\leftrightarrow 𝑇$

$=𝐹\leftrightarrow 𝑇=𝐹$

A biconditional is false when the two propositions have different truth values.

For Option D:

$(𝑝\wedge 𝑞)\vee (\sim 𝑞\vee 𝑟)$

$=(𝑇\wedge 𝐹)\vee (𝑇\vee 𝑇)=𝐹\vee 𝑇=𝑇$

Thus, the statement that is not true is:

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

An implication is false only when its antecedent is true and consequent is false.

Since $\sim 𝑝\to 𝑞$ is false,

$\sim 𝑝=𝑇,𝑞=𝐹$

If $\sim 𝑝=𝑇$, then $𝑝=𝐹$.

Also, $𝑞\leftrightarrow 𝑟$ is false when $𝑞$ and $𝑟$ have different truth values. Since $𝑞=𝐹$, we must have $𝑟=𝑇$.

Thus,

$(𝑝,𝑞,𝑟)=(𝐹,𝐹,𝑇)$

So, Option C is correct.

Using Heron’s formula,

$16\Delta^2=(a+b+c)(b+c-a)(c+a-b)(a+b-c).$

Given,

$16\Delta^2=3b^2c^2.$

Also, the area of the triangle in terms of two sides and the included angle is

$\Delta=\frac12bc\sin A.$

Thus,

$16\left(\frac12bc\sin A\right)^2=3b^2c^2.$

$4b^2c^2\sin^2 A=3b^2c^2$

$\sin^2 A=\frac34.$

Since $0^\circ<A<180^\circ$,

$A=60^\circ \quad \text{or} \quad 120^\circ.$

Hence, the correct answer is Option A: $60^\circ$ or $120^\circ$.

Given

$𝑎=2,𝑏=\frac{7}{2},𝑐=\frac{5}{2}.$

First simplify the given expression using $\sin 2𝐴=2\sin 𝐴\cos 𝐴$:

$\frac{2\sin 𝐴-\sin 2𝐴}{2\sin 𝐴+\sin 2𝐴}=\frac{2\sin 𝐴(1-\cos 𝐴)}{2\sin 𝐴(1+\cos 𝐴)}=\frac{1-\cos 𝐴}{1+\cos 𝐴}.$

By the cosine rule,

$\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}.$

So,

$\cos 𝐴=\frac{{\left(\frac{7}{2}\right)}^2+{\left(\frac{5}{2}\right)}^2-2^2}{2\left(\frac{7}{2}\right)\left(\frac{5}{2}\right)}=\frac{29}{35}.$

Thus,

$\frac{1-\cos 𝐴}{1+\cos 𝐴}=\frac{1-\frac{29}{35}}{1+\frac{29}{35}}=\frac{6}{64}=\frac{3}{32}.$

Now, using Heron’s formula, the semiperimeter is

$𝑠=\frac{2+\frac{7}{2}+\frac{5}{2}}{2}=4.$

Hence,

${\Delta }^2=𝑠(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)=4\left(2\right)\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)=6.$

Therefore,

${\left(\frac{3}{4\Delta }\right)}^2=\frac{9}{16{\Delta }^2}=\frac{9}{16\times 6}=\frac{3}{32}.$

So the correct answer is

$\overline{){\left(\frac{3}{4\Delta }\right)}^2}$

Option B.

For a square matrix $𝐵$,

$\mathrm{adj}\left(𝐵\right)=\left|𝐵\right|{𝐵}^{-1}.$

Taking $𝐵={𝐴}^{-1}$,

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

Since $𝐴$ is an upper triangular matrix, its determinant is the product of its diagonal entries:

$\left|𝐴\right|=(-1)\left(2\right)(-1)=2.$

Thus,

$\mathrm{adj}\left({𝐴}^{-1}\right)=\frac{1}{2}𝐴.$

The sum of all elements of $𝐴$ is

$-1+2025+2026+0+2+2027+0+0-1=6078.$

Hence, the required sum is

$\frac{6078}{2}=\overline{)3039}.$

So, the correct answer is Option C.

Since $𝐵={𝐴}^{-1}$, we have

$𝐴𝐵=𝐼.$

Consider the third column of $𝐵$:

$\frac{1}{37}⎡⎣⎢\begin{array}{c}11\\ 14\\ 𝑘\end{array}⎤⎦⎥.$

The product of the first row of $𝐴$ and the third column of $𝐵$ must be $0$, because it is an off-diagonal entry of the identity matrix.

$\frac{1}{37}\left(2\right(11)+(-1\left)\right(14)+4𝑘)=0$

$22-14+4𝑘=0$

$8+4𝑘=0$

$𝑘=-2$

Hence, the correct answer is

$\overline{)-2}$

Option D.

Using the principal value ranges,

${\sin }^{-1}𝑥\in [-\frac{𝜋}{2},\frac{𝜋}{2}],{\cos }^{-1}𝑥\in [0,𝜋],{\tan }^{-1}𝑥\in (-\frac{𝜋}{2},\frac{𝜋}{2}).$

Hence,

${\sin }^{-1}(\sin \frac{5𝜋}{6})={\sin }^{-1}(\sin \frac{𝜋}{6})=\frac{𝜋}{6},$

${\cos }^{-1}(\cos \frac{7𝜋}{6})={\cos }^{-1}(\cos \frac{5𝜋}{6})=\frac{5𝜋}{6},$

and

${\tan }^{-1}(\tan \frac{2𝜋}{3})={\tan }^{-1}(\tan (-\frac{𝜋}{3}))=-\frac{𝜋}{3}.$

Adding,

$\frac{𝜋}{6}+\frac{5𝜋}{6}-\frac{𝜋}{3}=𝜋-\frac{𝜋}{3}=\frac{2𝜋}{3}.$

So the correct answer is Option A, $\overline{)\frac{2𝜋}{3}}$.

Given

$𝑓\left(𝑥\right)=1-\frac{1}{𝑥}$

Successive compositions give

${𝑔}_2\left(𝑥\right)=𝑓\left(𝑓\right(𝑥\left)\right)=\frac{1}{1-𝑥}$

and

${𝑔}_3\left(𝑥\right)=𝑓\left({𝑔}_2\right(𝑥\left)\right)=𝑥.$

Thus, the functions repeat after every $3$ compositions. Since $2025$ is divisible by $3$,

${𝑔}_{2025}\left(𝑥\right)=𝑥$

and

${𝑔}_{2026}\left(𝑥\right)=𝑓\left(𝑥\right)=1-\frac{1}{𝑥}.$

Now,

$\int 𝑥{𝑔}_{2026}\left(𝑥\right)𝑑𝑥=\int 𝑥(1-\frac{1}{𝑥})𝑑𝑥=\int (𝑥-1)𝑑𝑥.$

Also,

$\int {𝑔}_{2025}\left(𝑥\right)𝑑𝑥=\int 𝑥𝑑𝑥.$

Hence,

$\int (𝑥-1)𝑑𝑥=\int 𝑥𝑑𝑥-𝑥+𝑐.$

Therefore,

$\overline{)ℎ\left(𝑥\right)=-𝑥}$

So, the correct choice is Option B.

Let

$𝑔\left(𝑥\right)=1-\sqrt{\surd 𝑥}=𝑦$

Then

$\sqrt{\surd 𝑥}=1-𝑦$

and

$𝑥=(1-𝑦{)}^2.$

Now,

$𝑓\left(𝑦\right)=5+4(1-𝑦)+(1-𝑦{)}^2.$

Simplifying,

$𝑓\left(𝑦\right)=5+4-4𝑦+1-2𝑦+{𝑦}^2$

$𝑓\left(𝑦\right)={𝑦}^2-6𝑦+10.$

Hence,

$𝑓\left(6\right)=6^2-6\left(6\right)+10=10.$

So the correct answer is Option B: $10$.

For continuity at $𝑥=2$, we must have

$𝑓\left(2\right)=\underset{𝑥\to 2}{lim}{\left(\frac{5𝑥-8}{8-3𝑥}\right)}^{\frac{3}{2𝑥-4}}.$

This limit is of the form $1^{\infty }$. Put $ℎ=𝑥-2$, so $ℎ\to 0$. Then

$\frac{5𝑥-8}{8-3𝑥}=\frac{2+5ℎ}{2-3ℎ}=\frac{1+\frac{5ℎ}{2}}{1-\frac{3ℎ}{2}}.$

Let the limit be $𝐿$. Taking logarithms,

$\ln 𝐿=\underset{ℎ\to 0}{lim}\frac{3}{2ℎ}\ln \left(\frac{1+\frac{5ℎ}{2}}{1-\frac{3ℎ}{2}}\right).$

Using $\underset{𝑡\to 0}{lim}\frac{\ln (1+𝑡)}{𝑡}=1$,

$\begin{array}{cc}\ln 𝐿& =\frac{3}{2}(\frac{5}{2}+\frac{3}{2})\\ & =\frac{3}{2}\times 4\\ & =6.\end{array}$

Hence,

$𝐿={𝑒}^6.$

Thus, for continuity,

$\overline{)𝑓\left(2\right)={𝑒}^6}$

So, the correct choice is Option B.

Given

${𝑥}^{𝑚}+{𝑦}^{𝑚}=𝑘.$

Differentiating with respect to $𝑥$,

$𝑚{𝑥}^{𝑚-1}+𝑚{𝑦}^{𝑚-1}{𝑦}^{\prime }=0,$

so

${𝑦}^{\prime }=-\frac{{𝑥}^{𝑚-1}}{{𝑦}^{𝑚-1}}.$

Differentiating again and simplifying,

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

Taking the common denominator,

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

Since ${𝑥}^{𝑚}+{𝑦}^{𝑚}=𝑘$,

${𝑦}^{\prime \prime }=\frac{(1-𝑚)𝑘{𝑥}^{𝑚-2}}{{𝑦}^{2𝑚-1}}.$

Comparing this with

${𝑦}^{\prime \prime }=\frac{𝑎{𝑥}^{𝑏}}{{𝑦}^{𝑐}},$

we get

$𝑎=(1-𝑚)𝑘,𝑏=𝑚-2,𝑐=2𝑚-1.$

Now $𝑎+𝑏+𝑐=0$, so

$(1-𝑚)𝑘+(𝑚-2)+(2𝑚-1)=0,$

$(1-𝑚)𝑘+3𝑚-3=0,$

$(1-𝑚)(𝑘-3)=0.$

As $𝑚\ne 1$,

$\overline{)𝑘=3}.$

Hence, Option C is correct.

Given

$𝑦={\sin }^{-1}\left(\frac{25-{𝑥}^2}{25+{𝑥}^2}\right)$

Using

$\frac{𝑑}{𝑑𝑥}({\sin }^{-1}𝑢)=\frac{{𝑢}^{\prime }}{\sqrt{\surd 1-{𝑢}^2}},$

take

$𝑢=\frac{25-{𝑥}^2}{25+{𝑥}^2}.$

By the quotient rule,

${𝑢}^{\prime }=\frac{(-2𝑥)(25+{𝑥}^2)-(25-{𝑥}^2)\left(2𝑥\right)}{(25+{𝑥}^2{)}^2}=-\frac{100𝑥}{(25+{𝑥}^2{)}^2}.$

Also,

$\sqrt{\surd 1-{𝑢}^2}=\sqrt{\surd 1-{\left(\frac{25-{𝑥}^2}{25+{𝑥}^2}\right)}^2}=\frac{10\left|𝑥\right|}{25+{𝑥}^2}.$

At $𝑥=1$, we have $\left|𝑥\right|=1$. Hence,

${𝑦}^{\prime }\left(1\right)=\frac{-\frac{100}{{26}^2}}{\frac{10}{26}}=-\frac{5}{13}.$

Thus, the correct answer is

$\overline{)-\frac{5}{13}}$

Option A.

Since $𝑓$ and $𝑔$ are inverse functions,

$𝑓\left(𝑔\right(𝑥\left)\right)=𝑥.$

Differentiating both sides with respect to $𝑥$,

${𝑓}^{\prime }\left(𝑔\right(𝑥\left)\right){𝑔}^{\prime }\left(𝑥\right)=1.$

Given

$𝑓\left(𝑥\right)=𝑥+{𝑒}^{𝑥},$

we have

${𝑓}^{\prime }\left(𝑥\right)=1+{𝑒}^{𝑥}.$

Thus,

${𝑓}^{\prime }\left(𝑔\right(𝑥\left)\right)=1+{𝑒}^{𝑔\left(𝑥\right)}.$

Hence,

${𝑔}^{\prime }\left(𝑥\right)=\frac{1}{1+{𝑒}^{𝑔\left(𝑥\right)}}.$

So, the correct answer is Option A.

Given

$3{𝑦}^2-2𝑥𝑦-𝑥=0$

At $𝑦=2$,

$3(2{)}^2-2𝑥(2)-𝑥=0$

$12-5𝑥=0$

So,

$𝑥=\frac{12}{5}$

Differentiating the given equation implicitly with respect to $𝑥$,

$6𝑦\frac{𝑑𝑦}{𝑑𝑥}-2(𝑥\frac{𝑑𝑦}{𝑑𝑥}+𝑦)-1=0$

$(6𝑦-2𝑥)\frac{𝑑𝑦}{𝑑𝑥}=2𝑦+1$

Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{2𝑦+1}{6𝑦-2𝑥}$

Putting $𝑦=2$ and $𝑥=\frac{12}{5}$,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{2\left(2\right)+1}{6\left(2\right)-2\left(\frac{12}{5}\right)}=\frac{5}{12-\frac{24}{5}}=\frac{5}{\frac{36}{5}}=\frac{25}{36}$

So, the correct answer is Option C, $\overline{)\frac{25}{36}}$.

Since $𝑓\left(𝑥\right)$ is a monic quadratic and Rolle’s theorem holds on $[𝛼,𝛼+3]$,

$𝑓\left(𝛼\right)=𝑓(𝛼+3)=0.$

Hence,

$𝑓\left(𝑥\right)=(𝑥-𝛼)(𝑥-𝛼-3).$

For a quadratic polynomial, the point where ${𝑓}^{\prime }\left(𝑐\right)=0$ lies midway between its roots. Thus,

$𝑐=\frac{𝛼+(𝛼+3)}{2}=𝛼+\frac{3}{2}.$

Now $𝑔\left(𝑥\right)=𝑓\left(𝑥\right)+2$ and $𝑔\left(3\right)=0$, so

$𝑓\left(3\right)=-2.$

Using the expression for $𝑓\left(𝑥\right)$,

$(3-𝛼)(-𝛼)=-2,$

which gives

${𝛼}^2-3𝛼+2=0,$

$(𝛼-1)(𝛼-2)=0.$

Thus, $𝛼=1$ or $𝛼=2$.

For $𝑔\left(𝑥\right)$, Rolle’s theorem holds on $[𝛽,3]$. Since the same value of $𝑐$ applies,

$\frac{𝛽+3}{2}=𝛼+\frac{3}{2},$

so

$𝛽=2𝛼.$