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

The given complex numbers are the non-real cube roots of unity:

$𝜔=\frac{-1+𝑖\sqrt{\surd 3}}{2},{𝜔}^2=\frac{-1-𝑖\sqrt{\surd 3}}{2}$

Since ${𝜔}^3=1$,

${𝜔}^{18}=({𝜔}^3{)}^6=1$

Similarly,

$({𝜔}^2{)}^{18}={𝜔}^{36}=({𝜔}^3{)}^{12}=1$

Hence,

${\left(\frac{-1+𝑖\sqrt{\surd 3}}{2}\right)}^{18}+{\left(\frac{-1-𝑖\sqrt{\surd 3}}{2}\right)}^{18}=1+1=2$

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

For $11$ players sitting around a circular table, the total number of arrangements is

$(11-1)!=10!$

Now count the arrangements in which the wicketkeeper and captain sit together.

Treat them as one unit. Then there are $10$ units around the circular table, which can be arranged in

$(10-1)!=9!$

ways. The captain and wicketkeeper can interchange their positions in $2$ ways. So, the number of arrangements in which they sit together is

$2\times 9!$

Hence, the required number of arrangements is

$10!-2\times 9!$

$=10\times 9!-2\times 9!=8\times 9!$

Thus, the correct answer is

$\overline{)8\times 9!}$

Option C.

In a triangle,

$𝐴+𝐵+𝐶=𝜋.$

Given $𝐶=\frac{𝜋}{3}$,

$𝐴+𝐵=𝜋-\frac{𝜋}{3}=\frac{2𝜋}{3}.$

Using the identities

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

and

$\cos 𝐴\cos 𝐵=\frac{\cos (𝐴+𝐵)+\cos (𝐴-𝐵)}{2},$

the given expression becomes

$\begin{array}{cc}{\cos }^2𝐴+{\cos }^2𝐵+\cos 𝐴\cos 𝐵& =1+\cos (𝐴+𝐵)\cos (𝐴-𝐵)\\ & +\frac{\cos (𝐴+𝐵)+\cos (𝐴-𝐵)}{2}.\end{array}$

Since

$\cos (𝐴+𝐵)=\cos \frac{2𝜋}{3}=-\frac{1}{2},$

we get

$\begin{array}{cc}& 1-\frac{1}{2}\cos (𝐴-𝐵)+\frac{-\frac{1}{2}+\cos (𝐴-𝐵)}{2}\\ & =1-\frac{1}{4}=\frac{3}{4}.\end{array}$

Hence, the correct answer is

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

Option A.

Since ${𝐿}_1$ passes through $(5,0)$, substitute $𝑥=5$ and $𝑦=0$ in

$\frac{𝑥}{𝑝}+\frac{𝑦}{2}=1.$

This gives

$\frac{5}{𝑝}=1\Rightarrow 𝑝=5.$

Thus, ${𝐿}_1$ is

$\frac{𝑥}{5}+\frac{𝑦}{2}=1$

or

$2𝑥+5𝑦-10=0.$

Its slope is $-\frac{2}{5}$.

For ${𝐿}_2$,

$\frac{𝑥}{10}+\frac{𝑦}{𝑞}=1,$

the slope is $-\frac{𝑞}{10}$. Since ${𝐿}_2$ is parallel to ${𝐿}_1$,

$-\frac{𝑞}{10}=-\frac{2}{5}\Rightarrow 𝑞=4.$

So, ${𝐿}_2$ becomes

$\frac{𝑥}{10}+\frac{𝑦}{4}=1$

or

$2𝑥+5𝑦-20=0.$

The distance between the parallel lines

$2𝑥+5𝑦-10=0\text{and}2𝑥+5𝑦-20=0$

is

$𝑑=\frac{|-20-(-10\left)\right|}{\sqrt{\surd 2^2+5^2}}=\frac{10}{\sqrt{\surd 29}}.$

Hence, the correct answer is

$\overline{)\frac{10}{\sqrt{\surd 29}}}$

Option A.

The given equation is

${𝑥}^2-4𝑥𝑦+4{𝑦}^2+𝑘𝑥-2𝑘𝑦=0$

The first three terms form a perfect square:

$(𝑥-2𝑦{)}^2+𝑘(𝑥-2𝑦)=0$

Factorising,

$(𝑥-2𝑦)(𝑥-2𝑦+𝑘)=0$

Thus, the two parallel lines are

$𝑥-2𝑦=0$

and

$𝑥-2𝑦+𝑘=0$

The distance between two parallel lines $𝑎𝑥+𝑏𝑦+{𝑐}_1=0$ and $𝑎𝑥+𝑏𝑦+{𝑐}_2=0$ is

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

Hence,

$3=\frac{\left|𝑘\right|}{\sqrt{\surd 1^2+(-2{)}^2}}=\frac{\left|𝑘\right|}{\sqrt{\surd 5}}$

So,

$\left|𝑘\right|=3\sqrt{\surd 5}$

From the given options,

$\overline{)𝑘=3\sqrt{\surd 5}}$

Hence, the correct answer is Option D.

For the circle

${𝑥}^2+{𝑦}^2=4,$

the centre is $(0,0)$. The normal at $(1,\sqrt{\surd 3})$ passes through the centre. Hence, its equation is

$𝑦=\sqrt{\surd 3}𝑥.$

It meets the $𝑥$-axis at $(0,0)$.

The tangent to ${𝑥}^2+{𝑦}^2=4$ at $({𝑥}_1,{𝑦}_1)$ is

$𝑥{𝑥}_1+𝑦{𝑦}_1=4.$

At $(1,\sqrt{\surd 3})$, it becomes

$𝑥+\sqrt{\surd 3}𝑦=4.$

It meets the $𝑥$-axis at $(4,0)$.

Thus, the bounded region is a triangle with base $4$ units and height $\sqrt{\surd 3}$ units.

$\text{Area}=\frac{1}{2}\times 4\times \sqrt{\surd 3}=2\sqrt{\surd 3}.$

Hence, the correct option is $\overline{)\text{D }2\sqrt{\surd 3}}$.

For the circle

${𝑥}^2+{𝑦}^2=10,$

the tangent at $(3,1)$ is

$3𝑥+𝑦=10.$

The second circle is

${𝑥}^2+{𝑦}^2-2\sqrt{\surd 10}𝑥-20𝑦+𝑘=0.$

Its centre is $(\sqrt{\surd 10},10)$, and its radius satisfies

${𝑟}^2=(\sqrt{\surd 10}{)}^2+{10}^2-𝑘=110-𝑘.$

Since the line $3𝑥+𝑦-10=0$ touches this circle, the perpendicular distance from its centre to the line equals its radius:

$𝑟=\frac{|3\sqrt{\surd 10}+10-10|}{\sqrt{\surd 3^2+1^2}}=3.$

Thus,

$110-𝑘=3^2=9,$

so

$\overline{)𝑘=101}.$

Hence, Option D is correct.

A focal chord of an ellipse passes through one of its foci.

The given focal chord is

$3𝑥+2𝑦-6=0.$

The foci lie on the $𝑥$-axis, so put $𝑦=0$:

$3𝑥-6=0⟹𝑥=2.$

Thus, the focus is $(2,0)$ and

$𝑎𝑒=2.$

Given $𝑒=\frac{2}{3}$,

$𝑎\left(\frac{2}{3}\right)=2⟹𝑎=3.$

Hence,

${𝑎}^2=9.$

For an ellipse,

${𝑏}^2={𝑎}^2(1-{𝑒}^2).$

So,

${𝑏}^2=9(1-\frac{4}{9})=5.$

Therefore,

${𝑎}^2+{𝑏}^2=9+5=\overline{)14}.$

Hence, the correct answer is Option D.

Divide ${𝑥}^2+𝑥+1$ by $𝑥+1$:

$\frac{{𝑥}^2+𝑥+1}{𝑥+1}=𝑥+\frac{1}{𝑥+1}$

So,

$\underset{𝑥\to \infty }{lim}\left[\right(1-𝑎)𝑥+\frac{1}{𝑥+1}-𝑏]=3$

For the limit to be finite, the coefficient of $𝑥$ must be zero:

$1-𝑎=0⟹𝑎=1$

Also, as $𝑥\to \infty$,

$\frac{1}{𝑥+1}\to 0$

Hence,

$-𝑏=3⟹𝑏=-3$

Thus,

$𝑎-𝑏=1-(-3)=4$

So, the correct answer is Option D: $4$.

Using $𝑎\to 𝑏<text>\equiv </text><text>\sim </text>𝑎\vee 𝑏$:

For (I),

$𝑝\to (𝑞\to 𝑟)<text>\equiv </text><text>\sim </text>𝑝\vee (\sim 𝑞\vee 𝑟)<text>\equiv </text><text>\sim </text>(𝑝\wedge 𝑞)\vee 𝑟\equiv (𝑝\wedge 𝑞)\to 𝑟$

So, (I) is correct.

For (II),

$(𝑝\to 𝑞)\to 𝑟<text>\equiv </text><text>\sim </text>(\sim 𝑝\vee 𝑞)\vee 𝑟\equiv (𝑝\wedge \sim 𝑞)\vee 𝑟$

But

$𝑝\to (𝑞\vee 𝑟)<text>\equiv </text><text>\sim </text>𝑝\vee 𝑞\vee 𝑟$

These are not equivalent. So, (II) is incorrect.

For (III),

$(𝑝\to 𝑟)\wedge (\sim 𝑞\to 𝑟)\equiv (\sim 𝑝\vee 𝑟)\wedge (𝑞\vee 𝑟)\equiv (\sim 𝑝\wedge 𝑞)\vee 𝑟$

This is different from $(𝑝\wedge \sim 𝑞)\vee 𝑟$. Hence, (III) is incorrect.

For (IV),

$𝑝\to (𝑞\to 𝑟)<text>\equiv </text><text>\sim </text>𝑝\vee \sim 𝑞\vee 𝑟$

Also,

$𝑞\to (𝑝\to 𝑟)<text>\equiv </text><text>\sim </text>𝑞\vee \sim 𝑝\vee 𝑟$

Both expressions are the same because the order in disjunction does not matter. Thus, (IV) is correct.

Hence, only (I) and (IV) are correct.

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

For a statement $𝐴\to 𝐵$, its contrapositive is

$\sim 𝐵\to \sim 𝐴$

Here,

$𝐴=𝑝\vee (𝑝\to 𝑞)$

and

$𝐵=𝑝\wedge \sim 𝑞$

Using De Morgan’s law,

$\sim 𝐵<text>=</text><text>\sim </text>(𝑝\wedge \sim 𝑞)<text>=</text><text>\sim </text>𝑝\vee 𝑞$

Also,

$\sim 𝐴<text>=</text><text>\sim </text>[𝑝\vee (𝑝\to 𝑞\left)\right]<text>=</text><text>\sim </text>𝑝\wedge \sim (𝑝\to 𝑞)$

Since

$\sim (𝑝\to 𝑞)=𝑝\wedge \sim 𝑞,$

we get

$\sim 𝐴<text>=</text><text>\sim </text>𝑝\wedge (𝑝\wedge \sim 𝑞)$

Hence, the contrapositive is

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

So, the correct answer is Option D.

The converse of $𝑝\vee 𝑞$ is $𝑞\vee 𝑝$. Since disjunction is commutative,

$𝑞\vee 𝑝\equiv 𝑝\vee 𝑞$

Using De Morgan’s law, its negation is

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

which is the same as

$\sim 𝑝\wedge \sim 𝑞$

Hence, the correct answer is Option D: $\sim 𝑝\wedge \sim 𝑞$.

Given

$(𝑎+𝑏+𝑐)(𝑎+𝑏-𝑐)=𝑎𝑏$

Using $(𝑥+𝑦)(𝑥-𝑦)={𝑥}^2-{𝑦}^2$,

$(𝑎+𝑏{)}^2-{𝑐}^2=𝑎𝑏$

So,

${𝑎}^2+2𝑎𝑏+{𝑏}^2-{𝑐}^2=𝑎𝑏$

which gives

${𝑐}^2={𝑎}^2+{𝑏}^2+𝑎𝑏$

By the cosine rule,

${𝑐}^2={𝑎}^2+{𝑏}^2-2𝑎𝑏\cos 𝐶$

Comparing the two expressions,

$-2𝑎𝑏\cos 𝐶=𝑎𝑏$

$\cos 𝐶=-\frac{1}{2}$

Hence,

$𝐶=\frac{2𝜋}{3}$

So the correct answer is Option B.

Here, $𝑎=13$, $𝑏=14$, $𝑐=15$.

The semi-perimeter is

$𝑠=\frac{𝑎+𝑏+𝑐}{2}=\frac{13+14+15}{2}=21.$

Using the half-angle formula,

$\sin \frac{𝐴}{2}=\sqrt{\surd \frac{(𝑠-𝑏)(𝑠-𝑐)}{𝑏𝑐}}=\sqrt{\surd \frac{(21-14)(21-15)}{14\cdot 15}}$

$=\sqrt{\surd \frac{7\cdot 6}{210}}=\frac{1}{\sqrt{\surd 5}}=\frac{\sqrt{\surd 5}}{5}.$

By the cosine rule,

$\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}=\frac{{14}^2+{15}^2-{13}^2}{2\cdot 14\cdot 15}=\frac{3}{5}.$

Hence,

$\sin 𝐴=\sqrt{\surd 1-{\cos }^2𝐴}=\sqrt{\surd 1-\frac{9}{25}}=\frac{4}{5}.$

Thus,

$\sin \frac{𝐴}{2}+\sin 𝐴=\frac{\sqrt{\surd 5}}{5}+\frac{4}{5}=\overline{)\frac{\sqrt{\surd 5}+4}{5}}.$

So, the correct choice is Option B.

For the matrix

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

its determinant is

$\left|𝐴\right|=({\cos }^2𝛼+{\sin }^2𝛼)\left(1\right)=1.$

We know that

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

Hence,

$𝐵=\mathrm{adj}𝐴={𝐴}^{-1}.$

Taking inverse on both sides,

${𝐵}^{-1}=({𝐴}^{-1}{)}^{-1}=𝐴.$

Thus, the correct answer is

$\overline{)𝐴}$

So, Option D is correct.

For

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

first find ${𝐴}^{-1}$. Since

$\left|𝐴\right|=3(-1)-(-4)\left(1\right)=1,$

we get

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

Now,

$({𝐴}^{-1}{)}^2=\left[\begin{array}{cc}-1& 4\\ -1& 3\end{array}\right]\left[\begin{array}{cc}-1& 4\\ -1& 3\end{array}\right]=\left[\begin{array}{cc}-3& 8\\ -2& 5\end{array}\right].$

Thus,

$({𝐴}^{-1}{)}^2+𝐵=\left[\begin{array}{cc}-3& 8\\ -2& 5\end{array}\right]+\left[\begin{array}{cc}6& -13\\ 5& -10\end{array}\right]=\left[\begin{array}{cc}3& -5\\ 3& -5\end{array}\right].$

The given equation becomes

$\left[\begin{array}{cc}3& -5\\ 3& -5\end{array}\right]\left[\begin{array}{c}𝑥\\ 𝑦\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right].$

So,

$3𝑥-5𝑦=0\Rightarrow 3𝑥=5𝑦.$

Among the given options, $(𝑥,𝑦)=(5,3)$ satisfies this relation.

$\overline{)(𝑥,𝑦)=(5,3)}$

Hence, Option D is correct.

Since $0\le 𝑥\le 1$, let

$𝜃={\sin }^{-1}𝑥.$

Then $0\le 𝜃\le \frac{𝜋}{2}$ and

$\sqrt{\surd 1-{𝑥}^2}=\sqrt{\surd 1-{\sin }^2𝜃}=\cos 𝜃.$

Hence,

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

So,

${𝐼}_1+{𝐼}_2=\int (\frac{𝜋}{2}-{\sin }^{-1}𝑥)𝑑𝑥+\int {\sin }^{-1}𝑥𝑑𝑥=\int \frac{𝜋}{2}𝑑𝑥.$

Thus,

${𝐼}_1+{𝐼}_2=\frac{𝜋}{2}𝑥+𝐶.$

Ignoring the arbitrary constant, as intended in the options,

$\overline{){𝐼}_1+{𝐼}_2=\frac{𝜋}{2}𝑥}$

Hence, Option C is correct.

Since $0<t<1$,

$\frac{1+t^2}{2t}>1.$

Given

$x=\csc^{-1}\left(\frac{1+t^2}{2t}\right),$

we have

$\sin x=\frac{2t}{1+t^2}.$

Using

$\sin(2\tan^{-1}t)=\frac{2t}{1+t^2},$

and the given range of $t$, we get

$x=2\tan^{-1}t.$

Thus,

$\frac{dx}{dt}=\frac{2}{1+t^2}.$

Also,

$y=\cot^{-1}\left(\frac{\sqrt{1-t^2}}{t}\right).$

So,

$\cot y=\frac{\sqrt{1-t^2}}{t}.$

As $0<t<1$, this gives $\sin y=t$, or

$y=\sin^{-1}t.$

Hence,

$\frac{dy}{dt}=\frac{1}{\sqrt{1-t^2}}.$

Therefore,

$\frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \frac{1+t^2}{2\sqrt{1-t^2}}.$

Thus,

$f(t)=\frac{1+t^2}{2\sqrt{1-t^2}}.$

Putting $t=\tan\alpha$,

$f(\tan\alpha) = \frac{1+\tan^2\alpha}{2\sqrt{1-\tan^2\alpha}}.$

Now,

$1+\tan^2\alpha=\sec^2\alpha$

and, since $0<\alpha<\dfrac{\pi}{4}$,

$\sqrt{1-\tan^2\alpha} = \sqrt{\frac{\cos 2\alpha}{\cos^2\alpha}} = \frac{\sqrt{\cos 2\alpha}}{\cos\alpha}.$

Therefore,

$f(\tan\alpha) = \frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}.$

Hence, the correct option is

$\boxed{\text{Option B}}$

Put $𝑡={\log }_{𝑒}𝑥=\ln 𝑥$. Since $𝑥\in ({𝑒}^{-1/3},{𝑒}^{1/12})$,

$-\frac{1}{3}<𝑡<\frac{1}{12}.$

Using logarithm properties,

$𝑦={\tan }^{-1}\left(\frac{1-3𝑡}{1+3𝑡}\right)+{\tan }^{-1}\left(\frac{4+3𝑡}{1-12𝑡}\right).$

Take

$𝑢=\frac{1-3𝑡}{1+3𝑡},𝑣=\frac{4+3𝑡}{1-12𝑡}.$

Then

$\tan 𝑦=\frac{𝑢+𝑣}{1-𝑢𝑣}.$

Now,

$𝑢+𝑣=\frac{5(1+9{𝑡}^2)}{(1+3𝑡)(1-12𝑡)}$

and

$1-𝑢𝑣=\frac{-3(1+9{𝑡}^2)}{(1+3𝑡)(1-12𝑡)}.$

Hence,

$\tan 𝑦=-\frac{5}{3},$

which is constant. Throughout the given interval, both $𝑢$ and $𝑣$ are positive, so there is no change of branch of ${\tan }^{-1}$. Thus $𝑦$ remains constant.

Therefore,

$\overline{)\frac{𝑑𝑦}{𝑑𝑥}=0}$

So, the correct choice is $\overline{)\text{Option B}}$.

For $0\le 𝑥\le 1$,

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

Put

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

where $0\le 𝜃\le \frac{𝜋}{2}$.

Now,

${𝜃}^3+{(\frac{𝜋}{2}-𝜃)}^3={(𝜃+\frac{𝜋}{2}-𝜃)}^3-3𝜃(\frac{𝜋}{2}-𝜃)\left(\frac{𝜋}{2}\right).$

Since two non-negative numbers with a fixed sum have maximum product when they are equal,

$𝜃(\frac{𝜋}{2}-𝜃)\le {\left(\frac{𝜋}{4}\right)}^2.$

Hence,

$({\sin }^{-1}𝑥{)}^3+({\cos }^{-1}𝑥{)}^3\ge {\left(\frac{𝜋}{2}\right)}^3-3{\left(\frac{𝜋}{4}\right)}^2\left(\frac{𝜋}{2}\right)=\frac{{𝜋}^3}{32}.$

Given that the expression equals $𝑎{𝜋}^3$,

$𝑎{𝜋}^3\ge \frac{{𝜋}^3}{32}\Rightarrow 𝑎\ge \frac{1}{32}.$

Thus, the correct choice is Option A.

We use the identity

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

when the angles are in the proper range.

Here,

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

So,

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

Now the given sum becomes

$\sum _{𝑛=1}^{2026}{\tan }^{-1}\left(\frac{1}{{𝑛}^2+𝑛+1}\right)$

$=\sum _{𝑛=1}^{2026}[{\tan }^{-1}(𝑛+1)-{\tan }^{-1}(𝑛\left)\right]$

This is a telescopic sum:

$={\tan }^{-1}\left(2027\right)-{\tan }^{-1}\left(1\right)$

Now apply the same identity again:

${\tan }^{-1}\left(2027\right)-{\tan }^{-1}\left(1\right)={\tan }^{-1}\left(\frac{2027-1}{1+2027}\right)$

$={\tan }^{-1}\left(\frac{2026}{2028}\right)$

$={\tan }^{-1}\left(\frac{1013}{1014}\right)$

The question says this is equal to

${\tan }^{-1}(1-\frac{1}{𝑥})$

So,

$1-\frac{1}{𝑥}=\frac{1013}{1014}$

$\frac{1}{𝑥}=1-\frac{1013}{1014}$

$\frac{1}{𝑥}=\frac{1}{1014}$

Hence,

$𝑥=1014$

So the correct option is Option C: $1014$.

Given

$𝑓\left(𝑥\right)=𝑎𝑥+𝑏,𝑔\left(𝑥\right)=𝑐𝑥+𝑑$

First find $𝑓\left(𝑔\right(𝑥\left)\right)$.

Since $𝑔\left(𝑥\right)=𝑐𝑥+𝑑$,

$𝑓\left(𝑔\right(𝑥\left)\right)=𝑓(𝑐𝑥+𝑑)=𝑎(𝑐𝑥+𝑑)+𝑏$

$𝑓\left(𝑔\right(𝑥\left)\right)=𝑎𝑐𝑥+𝑎𝑑+𝑏$

Now find $𝑔\left(𝑓\right(𝑥\left)\right)$.

Since $𝑓\left(𝑥\right)=𝑎𝑥+𝑏$,

$𝑔\left(𝑓\right(𝑥\left)\right)=𝑔(𝑎𝑥+𝑏)=𝑐(𝑎𝑥+𝑏)+𝑑$

$𝑔\left(𝑓\right(𝑥\left)\right)=𝑎𝑐𝑥+𝑐𝑏+𝑑$

For $𝑓\left(𝑔\right(𝑥\left)\right)=𝑔\left(𝑓\right(𝑥\left)\right)$ for all $𝑥$, the coefficients and constant terms must be equal.

The coefficient of $𝑥$ is already the same:

$𝑎𝑐𝑥=𝑎𝑐𝑥$

So we only need the constant terms to be equal:

$𝑎𝑑+𝑏=𝑐𝑏+𝑑$

Now notice:

$𝑓\left(𝑑\right)=𝑎𝑑+𝑏$

and

$𝑔\left(𝑏\right)=𝑐𝑏+𝑑$

So the condition becomes

$𝑓\left(𝑑\right)=𝑔\left(𝑏\right)$

Hence the correct option is

$\overline{)\text{Option C }𝑓\left(𝑑\right)=𝑔\left(𝑏\right)}$

The function is

$𝑓\left(𝑥\right)=\frac{1}{{𝑥}^2-5\left|𝑥\right|+6}$

A function of the form $\frac{1}{𝑔\left(𝑥\right)}$ is discontinuous where the denominator becomes zero.

So we need to find where

${𝑥}^2-5\left|𝑥\right|+6=0$

Since

${𝑥}^2=\left(\right|𝑥|{)}^2$

put

$\left|𝑥\right|=𝑡$

Then the equation becomes

${𝑡}^2-5𝑡+6=0$

Factorising,

${𝑡}^2-5𝑡+6=(𝑡-2)(𝑡-3)$

So,

$(𝑡-2)(𝑡-3)=0$

Hence,

$𝑡=2\text{or}𝑡=3$

That means

$\left|𝑥\right|=2\text{or}\left|𝑥\right|=3$

So,

$𝑥=\pm 2,\pm 3$

Thus, the function is discontinuous at $4$ points.

So the correct option is

$\overline{)\text{Option D }4}$

Given

$𝑥{𝑒}^{𝑥𝑦}=𝑦+{\sin }^2𝑥$

First find the value of $𝑦$ when $𝑥=0$.

Put $𝑥=0$ in the equation:

$0\cdot {𝑒}^{0\cdot 𝑦}=𝑦+{\sin }^{2}0$

$0=𝑦+0$

So,

$𝑦=0$

Now differentiate both sides with respect to $𝑥$.

For the left side, use the product rule:

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

Since

$\frac{𝑑}{𝑑𝑥}\left(𝑥𝑦\right)=𝑦+𝑥\frac{𝑑𝑦}{𝑑𝑥}$

we get

$\frac{𝑑}{𝑑𝑥}\left(𝑥{𝑒}^{𝑥𝑦}\right)={𝑒}^{𝑥𝑦}+𝑥{𝑒}^{𝑥𝑦}(𝑦+𝑥\frac{𝑑𝑦}{𝑑𝑥})$