MHT CET 2026 17th April Evening Shift (Mathematics) Question Paper with Solutions

Using the relation between roots and coefficients for

$5{𝑥}^2-4𝑥+1=0,$

we get

$\tan 𝐴+\tan 𝐵=\frac{4}{5}$

and

$\tan 𝐴\tan 𝐵=\frac{1}{5}.$

Now,

$\tan (𝐴+𝐵)=\frac{\tan 𝐴+\tan 𝐵}{1-\tan 𝐴\tan 𝐵}.$

So,

$\tan (𝐴+𝐵)=\frac{\frac{4}{5}}{1-\frac{1}{5}}=\frac{\frac{4}{5}}{\frac{4}{5}}=1.$

Among the given options, the angle whose tangent is $1$ is

$𝐴+𝐵=\frac{𝜋}{4}.$

Hence, the correct choice is Option B, $\frac{𝜋}{4}$.

Using De Moivre’s theorem,

$(\cos 𝜃+𝑖\sin 𝜃{)}^{𝑛}=\cos 𝑛𝜃+𝑖\sin 𝑛𝜃$

So,

$(\cos 2𝜃+𝑖\sin 2𝜃{)}^7=\cos 14𝜃+𝑖\sin 14𝜃$

and

$(\cos 4𝜃+𝑖\sin 4𝜃{)}^3=\cos 12𝜃+𝑖\sin 12𝜃.$

When two complex numbers in trigonometric form are divided, their arguments are subtracted:

$\frac{\cos 14𝜃+𝑖\sin 14𝜃}{\cos 12𝜃+𝑖\sin 12𝜃}=\cos (14𝜃-12𝜃)+𝑖\sin (14𝜃-12𝜃).$

Thus,

$\overline{)\cos 2𝜃+𝑖\sin 2𝜃}$

Hence, the correct answer is Option A.

Each of the $5$ questions has $2$ possible answers: True or False.

So, the total number of different answer sequences is

$2^5=32$

Only one of these sequences has all the answers correct. Since no student got all the answers correct, that sequence cannot belong to any student.

Also, every student has a unique sequence. Hence, the maximum possible number of students is

$32-1=31$

So, the correct answer is Option B: $\overline{)31}$.

A point equidistant from both coordinate axes has equal distances from the $𝑥$-axis and the $𝑦$-axis.

In the first quadrant, this means

$𝑥=𝑦>0.$

Since the point lies on

$𝑦=2𝑥+𝑐,$

putting $𝑦=𝑥$ gives

$𝑥=2𝑥+𝑐$

$𝑐=-𝑥.$

As $𝑥>0$, we must have

$𝑐<0.$

Among the given options, the only negative value is

$\overline{)𝑐=-2}.$

So, Option D is correct.

For the general second-degree equation

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2+2𝑔𝑥+2𝑓𝑦+𝑐=0,$

the condition for representing a pair of straight lines is

$\mid \begin{array}{ccc}𝑎& ℎ& 𝑔\\ ℎ& 𝑏& 𝑓\\ 𝑔& 𝑓& 𝑐\end{array}\mid =0.$

Here,

$ℎ=2,𝑏=-2,𝑔=2,𝑓=4,𝑐=1.$

So,

$\mid \begin{array}{ccc}𝑎& 2& 2\\ 2& -2& 4\\ 2& 4& 1\end{array}\mid =0.$

Expanding,

$-18𝑎+36=0,$

which gives

$𝑎=2.$

The point of intersection of the two lines satisfies

$𝑎𝑥+ℎ𝑦+𝑔=0$

and

$ℎ𝑥+𝑏𝑦+𝑓=0.$

Putting the values,

$2𝑥+2𝑦+2=0$

and

$2𝑥-2𝑦+4=0.$

Thus,

$𝑥+𝑦+1=0,𝑥-𝑦+2=0.$

Solving these equations,

$𝑥=-\frac{3}{2},𝑦=\frac{1}{2}.$

Hence, the point of intersection is

$\overline{)(-\frac{3}{2},\frac{1}{2})}.$

So, the correct choice is Option B.

The first circle is

${𝑥}^2+{𝑦}^2=16,$

so its centre is $(0,0)$ and radius is $4$.

For the second circle,

${𝑥}^2+{𝑦}^2+2𝑎𝑥+4𝑦+4=0.$

Completing the squares,

$(𝑥+𝑎{)}^2+(𝑦+2{)}^2={𝑎}^2.$

Thus, its centre is $(-𝑎,-2)$ and radius is $𝑎$ (taking $𝑎>0$).

The distance between the centres is

$\sqrt{\surd (-𝑎{)}^2+(-2{)}^2}=\sqrt{\surd {𝑎}^2+4}.$

For internal touching,

$\text{distance between centres}=\text{difference of radii}.$

Hence,

$\sqrt{\surd {𝑎}^2+4}=4-𝑎.$

Squaring both sides,

${𝑎}^2+4=16-8𝑎+{𝑎}^2,$

$8𝑎=12,$

$𝑎=\frac{3}{2}.$

So, the correct option is

$\overline{)\text{Option D: }\frac{3}{2}}.$

Given ellipse:

$9{𝑥}^2+16{𝑦}^2=144$

Dividing by $144$,

$\frac{{𝑥}^2}{16}+\frac{{𝑦}^2}{9}=1$

So, ${𝑎}^2=16$ and ${𝑏}^2=9$.

The line is

$3𝑥+4𝑦+𝑘=0$

or

$𝑦=-\frac{3}{4}𝑥-\frac{𝑘}{4}$

For the line $𝑦=𝑚𝑥+𝑐$ to touch the ellipse $\frac{{𝑥}^2}{{𝑎}^2}+\frac{{𝑦}^2}{{𝑏}^2}=1$, the condition is

${𝑐}^2={𝑎}^2{𝑚}^2+{𝑏}^2$

Here, $𝑚=-\frac{3}{4}$ and $𝑐=-\frac{𝑘}{4}$. Thus,

${(-\frac{𝑘}{4})}^2=16{(-\frac{3}{4})}^2+9$

$\frac{{𝑘}^2}{16}=9+9=18$

${𝑘}^2=288$

$𝑘=\pm 12\sqrt{\surd 2}$

Hence, the correct option is Option D, representing $\overline{)\pm 12\sqrt{\surd 2}}$.

For ${𝑆}_1$, choose $𝑛=1$:

${𝑛}^2+𝑛+2=1^2+1+2=4,$

which is divisible by $4$. So ${𝑆}_1$ is true.

For ${𝑆}_2$,

$𝑥-17<20\Rightarrow 𝑥<37.$

There are many natural numbers satisfying this, such as $𝑥=1$. So ${𝑆}_2$ is true.

In ${𝑆}_3$, the variable should be $𝑥$ instead of $𝑛$. The equation

${𝑥}^2+3𝑥-10=0$

gives

$(𝑥+5)(𝑥-2)=0.$

It is true only for $𝑥=2$ among natural numbers, not for every natural number. Hence ${𝑆}_3$ is false.

For ${𝑆}_4$, every natural number $𝑛\ge 1$, so

${𝑛}^2\ge 1.$

Thus ${𝑆}_4$ is true.

Hence, only ${𝑆}_3$ is false.

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

The given statement is

$(𝑝\wedge 𝑞)\to ((𝑝\vee 𝑟)\to \neg 𝑞).$

We use the rule

$\neg (𝐴\to 𝐵)\equiv 𝐴\wedge \neg 𝐵.$

So its negation is

$(𝑝\wedge 𝑞)\wedge \neg ((𝑝\vee 𝑟)\to \neg 𝑞).$

Again applying the same rule,

$\neg ((𝑝\vee 𝑟)\to \neg 𝑞)\equiv (𝑝\vee 𝑟)\wedge 𝑞.$

Hence,

$(𝑝\wedge 𝑞)\wedge (𝑝\vee 𝑟)\wedge 𝑞.$

Since $𝑝$ is already true, $𝑝\vee 𝑟$ is automatically true. Thus the expression simplifies to

$𝑝\wedge 𝑞.$

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

Let

$𝐴=(𝑝\wedge 𝑞),𝐵=(\sim 𝑝\vee 𝑟).$

The given statement becomes

$(𝐴\to 𝐵)\vee (𝐵\to 𝐴).$

Using $𝐴\to 𝐵<text>\equiv </text><text>\sim </text>𝐴\vee 𝐵$,

$(\sim 𝐴\vee 𝐵)\vee (\sim 𝐵\vee 𝐴).$

Rearranging the terms,

$(\sim 𝐴\vee 𝐴)\vee (𝐵\vee \sim 𝐵).$

Since $𝐴\vee \sim 𝐴$ and $𝐵\vee \sim 𝐵$ are always true, the whole statement is always true.

Hence, the statement pattern is a tautology.

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

Since $\cot 𝐴,\cot 𝐵,\cot 𝐶$ are in arithmetic progression,

$2\cot 𝐵=\cot 𝐴+\cot 𝐶.$

Using the cosine rule and the area $\Delta$ of the triangle,

$\cot 𝐴=\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{4\Delta },$

and similarly,

$\cot 𝐵=\frac{{𝑐}^2+{𝑎}^2-{𝑏}^2}{4\Delta },\cot 𝐶=\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{4\Delta }.$

Substituting these values,

$2({𝑐}^2+{𝑎}^2-{𝑏}^2)=({𝑏}^2+{𝑐}^2-{𝑎}^2)+({𝑎}^2+{𝑏}^2-{𝑐}^2).$

Simplifying,

$2{𝑎}^2+2{𝑐}^2-2{𝑏}^2=2{𝑏}^2,$

so

${𝑎}^2+{𝑐}^2=2{𝑏}^2.$

Thus, ${𝑎}^2,{𝑏}^2,{𝑐}^2$ are in arithmetic progression.

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

Put $𝑡=\tan \frac{𝜃}{2}$. Then

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

Its determinant is

$\left|𝐴\right|=1+{𝑡}^2.$

Hence,

${𝐴}^{-1}=\frac{1}{1+{𝑡}^2}\left[\begin{array}{cc}1& 𝑡\\ -𝑡& 1\end{array}\right]=\frac{1}{1+{𝑡}^2}𝐵.$

So,

${𝐴}^{-1}𝐵=\frac{1}{1+{𝑡}^2}\left[\begin{array}{cc}1& 𝑡\\ -𝑡& 1\end{array}\right]\left[\begin{array}{cc}1& 𝑡\\ -𝑡& 1\end{array}\right]$

$=\frac{1}{1+{𝑡}^2}\left[\begin{array}{cc}1-{𝑡}^2& 2𝑡\\ -2𝑡& 1-{𝑡}^2\end{array}\right].$

Using the identities

$\cos 𝜃=\frac{1-{\tan }^2\frac{𝜃}{2}}{1+{\tan }^2\frac{𝜃}{2}},\sin 𝜃=\frac{2\tan \frac{𝜃}{2}}{1+{\tan }^2\frac{𝜃}{2}},$

we get

${𝐴}^{-1}𝐵=\left[\begin{array}{cc}\cos 𝜃& \sin 𝜃\\ -\sin 𝜃& \cos 𝜃\end{array}\right].$

Thus, the correct answer is $\overline{)\text{Option A}}$.

First, find $2𝐴+3𝐵$:

$2𝐴=⎡⎣⎢\begin{array}{ccc}6& 0& 0\\ 0& -10& 0\\ 0& 0& 14\end{array}⎤⎦⎥,3𝐵=⎡⎣⎢\begin{array}{ccc}-3& 0& 0\\ 0& 6& 0\\ 0& 0& 12\end{array}⎤⎦⎥$

Thus,

$2𝐴+3𝐵=⎡⎣⎢\begin{array}{ccc}3& 0& 0\\ 0& -4& 0\\ 0& 0& 26\end{array}⎤⎦⎥$

The inverse of a diagonal matrix is obtained by taking the reciprocal of each non-zero diagonal entry. Hence,

$(2𝐴+3𝐵{)}^{-1}=⎡⎣⎢\begin{array}{ccc}\frac{1}{3}& 0& 0\\ 0& -\frac{1}{4}& 0\\ 0& 0& \frac{1}{26}\end{array}⎤⎦⎥$

So, the correct answer is Option A.

The domains of the first two inverse trigonometric functions determine $𝑥$.

For ${\sin }^{-1}(𝑥-2)$,

$-1\le 𝑥-2\le 1\Rightarrow 1\le 𝑥\le 3.$

For ${\cos }^{-1}𝑥$,

$-1\le 𝑥\le 1.$

The common value is

$𝑥=1.$

Substituting $𝑥=1$ in the given equation,

${\sin }^{-1}(-1)+{\cos }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)+{\cot }^{-1}\left(5\right)={\sec }^{-1}\left(\sqrt{\surd 𝑘}\right)-\frac{𝜋}{2}.$

Using ${\sin }^{-1}(-1)=-\frac{𝜋}{2}$ and ${\cos }^{-1}\left(1\right)=0$,

${\tan }^{-1}\left(3\right)+{\cot }^{-1}\left(5\right)={\sec }^{-1}\left(\sqrt{\surd 𝑘}\right).$

Since ${\cot }^{-1}\left(5\right)={\tan }^{-1}\left(\frac{1}{5}\right)$,

$\tan [{\tan }^{-1}(3)+{\tan }^{-1}\left(\frac{1}{5}\right)]=\frac{3+\frac{1}{5}}{1-3\cdot \frac{1}{5}}=8.$

Hence, if $𝜃={\sec }^{-1}\left(\sqrt{\surd 𝑘}\right)$, then

$\tan 𝜃=8.$

So,

${\sec }^2𝜃=1+{\tan }^2𝜃=65.$

Thus,

$𝑘=65.$

Now,

$\cos (2𝑐𝑜𝑠𝑒{𝑐}^{-1}\sqrt{\surd 𝑘-1})=\cos (2𝑐𝑜𝑠𝑒{𝑐}^{-1}8).$

Let $𝛼=𝑐𝑜𝑠𝑒{𝑐}^{-1}8$. Then $\sin 𝛼=\frac{1}{8}$. Therefore,

$\cos 2𝛼=1-2{\sin }^2𝛼=1-2{\left(\frac{1}{8}\right)}^2=1-\frac{2}{64}=\frac{31}{32}.$

$\overline{)\frac{31}{32}}$

Option B.

Let

$𝜃={\sin }^{-1}\left(\frac{1}{5}\right)$

Then

$\sin 𝜃=\frac{1}{5}$

Using the identity

$\sin 3𝜃=3\sin 𝜃-4{\sin }^3𝜃,$

we get

$\sin 3𝜃=3\left(\frac{1}{5}\right)-4{\left(\frac{1}{5}\right)}^3$

$=\frac{3}{5}-\frac{4}{125}=\frac{75-4}{125}=\frac{71}{125}.$

Hence, the correct answer is Option B, $\overline{)\frac{71}{125}}$.

The principal value ranges are:

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

Since

$\cos \frac{4𝜋}{3}=-\frac{1}{2},$

we get

${\cos }^{-1}(\cos \frac{4𝜋}{3})={\cos }^{-1}(-\frac{1}{2})=\frac{2𝜋}{3}.$

Also,

$\sin \frac{4𝜋}{3}=-\frac{\sqrt{\surd 3}}{2},$

so

${\sin }^{-1}(\sin \frac{4𝜋}{3})={\sin }^{-1}(-\frac{\sqrt{\surd 3}}{2})=-\frac{𝜋}{3}.$

Thus,

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

Hence, the correct answer is Option C: $\frac{𝜋}{3}$.

Given

$𝑓\left(𝑥\right)=\frac{1-𝑥}{1+𝑥}$

First,

$𝑓(\cos 𝑥)=\frac{1-\cos 𝑥}{1+\cos 𝑥}$

Now apply $𝑓$ again:

$𝑓\left(𝑓\right(\cos 𝑥\left)\right)=\frac{1-\frac{1-\cos 𝑥}{1+\cos 𝑥}}{1+\frac{1-\cos 𝑥}{1+\cos 𝑥}}$

Simplifying,

$𝑓\left(𝑓\right(\cos 𝑥\left)\right)=\frac{\frac{2\cos 𝑥}{1+\cos 𝑥}}{\frac{2}{1+\cos 𝑥}}=\cos 𝑥$

Hence, the correct answer is

$\overline{)\text{Option A: }\cos 𝑥}$

For continuity at $𝑥=\frac{𝜋}{2}$, we must have

$𝑓\left(\frac{𝜋}{2}\right)=\underset{𝑥\to \frac{𝜋}{2}}{lim}𝑓\left(𝑥\right).$

Using $\sin 𝐴+\sin 𝐵=2\sin \frac{𝐴+𝐵}{2}\cos \frac{𝐴-𝐵}{2}$,

$\sin 3𝑥+\sin 𝑥=2\sin 2𝑥\cos 𝑥.$

Also,

$\cos \frac{5𝑥}{2}-\cos \frac{3𝑥}{2}=-2\sin 2𝑥\sin \frac{𝑥}{2}.$

So the denominator becomes

$2\sin 2𝑥\sin \frac{3𝑥}{2}-2\sin 2𝑥\sin \frac{𝑥}{2}.$

Taking $2\sin 2𝑥$ common,

$=2\sin 2𝑥(\sin \frac{3𝑥}{2}-\sin \frac{𝑥}{2}).$

Now,

$\sin \frac{3𝑥}{2}-\sin \frac{𝑥}{2}=2\cos 𝑥\sin \frac{𝑥}{2}.$

Hence,

$𝑓\left(𝑥\right)=\frac{8\sqrt{\surd 2}\sin 2𝑥\cos 𝑥}{4\sin 2𝑥\cos 𝑥\sin \frac{𝑥}{2}}=\frac{2\sqrt{\surd 2}}{\sin \frac{𝑥}{2}}.$

Thus,

$𝑓\left(\frac{𝜋}{2}\right)=\underset{𝑥\to \frac{𝜋}{2}}{lim}\frac{2\sqrt{\surd 2}}{\sin \frac{𝑥}{2}}=\frac{2\sqrt{\surd 2}}{\sin \frac{𝜋}{4}}=\frac{2\sqrt{\surd 2}}{\frac{1}{\sqrt{\surd 2}}}=4=(2{)}^2.$

Hence, the correct answer is Option A: $(2{)}^2$.

At $𝑥=\frac{𝜋}{4}$,

$\cos 2𝑥=\cos \frac{𝜋}{2}=0.$

In the product rule, every term contains this zero factor except the term obtained by differentiating $\cos 2𝑥$.

Since

$\frac{𝑑}{𝑑𝑥}(\cos 2𝑥)=-2\sin 2𝑥,$

we get

${𝑓}^{\prime }\left(\frac{𝜋}{4}\right)=\cos \frac{𝜋}{4}(-2\sin \frac{𝜋}{2})\cos 𝜋\cos 2𝜋\cos 4𝜋.$

Using the standard values,

${𝑓}^{\prime }\left(\frac{𝜋}{4}\right)=\frac{1}{\sqrt{\surd 2}}(-2)\left(1\right)(-1)\left(1\right)\left(1\right)=\sqrt{\surd 2}.$

Also,

$\mathrm{cosec}\frac{𝜋}{4}=\frac{1}{\sin \frac{𝜋}{4}}=\sqrt{\surd 2}.$

Hence, the correct answer is

$\overline{)\text{Option A: }\mathrm{cosec}\left(\frac{𝜋}{4}\right)}.$

Given

$𝑦=\frac{1}{3𝑥+5}=(3𝑥+5{)}^{-1}$

Using the chain rule repeatedly,

$\frac{{𝑑}^{𝑛}}{𝑑{𝑥}^{𝑛}}(3𝑥+5{)}^{-1}=(-1{)}^{𝑛}𝑛!3^{𝑛}(3𝑥+5{)}^{-(𝑛+1)}$

For $𝑛=9$,

$\frac{{𝑑}^9𝑦}{𝑑{𝑥}^9}=\frac{(-1{)}^9\times 9!\times 3^9}{(3𝑥+5{)}^{10}}$

Hence, the correct answer is Option C.

Let

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

We need

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

Rationalising the expression inside $𝑦$,

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

Using the half-angle identity,

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

we get

$𝑦=\frac{1}{2}{\tan }^{-1}𝑥.$

Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{2(1+{𝑥}^2)}.$

Also,

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

so

$\frac{𝑑𝑧}{𝑑𝑥}=\frac{1}{\sqrt{\surd 1-{𝑥}^2}}.$

Thus,

$\frac{𝑑𝑦}{𝑑𝑧}=\frac{\sqrt{\surd 1-{𝑥}^2}}{2(1+{𝑥}^2)}.$

At $𝑥=\frac{1}{2}$,

$\frac{𝑑𝑦}{𝑑𝑧}=\frac{\sqrt{\surd 1-\frac{1}{4}}}{2(1+\frac{1}{4})}=\frac{\frac{\sqrt{\surd 3}}{2}}{\frac{5}{2}}=\frac{\sqrt{\surd 3}}{5}.$

Hence, the correct answer is

$\overline{)\frac{\sqrt{\surd 3}}{5}}$

Option C.

The repeating radical is equal to $𝑦$. So,

$𝑦=\sqrt{\surd \cos {𝑥}^2+𝑦}$

Squaring both sides,

${𝑦}^2=\cos {𝑥}^2+𝑦$

Differentiate with respect to $𝑥$:

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

Hence,

$(2𝑦-1)\frac{𝑑𝑦}{𝑑𝑥}=-2𝑥\sin {𝑥}^2$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{-2𝑥\sin {𝑥}^2}{2𝑦-1}$

Comparing this with

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑓\left(𝑥\right)}{2𝑦-1},$

we get

$𝑓\left(𝑥\right)=-2𝑥\sin {𝑥}^2.$

Thus,

$\int 𝑓\left(𝑥\right)𝑑𝑥=\int -2𝑥\sin {𝑥}^2𝑑𝑥=\cos {𝑥}^2+𝐶.$

Hence, the correct answer is $\overline{)\text{Option C: }\cos {𝑥}^2+𝐶}$.

Put $x=\tan\theta$. Since $0<a<\dfrac12$, for $x$ near $a$ we have

$0<\theta=\tan^{-1}x<\tan^{-1}\left(\frac12\right)<\frac{\pi}{6}.$

Using

$\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta},$

we get

$\frac{3x-x^3}{1-3x^2}=\tan 3\theta.$

As $0<3\theta<\dfrac{\pi}{2}$,

$f(x)=\cot^{-1}(\tan 3\theta) =\frac{\pi}{2}-3\theta =\frac{\pi}{2}-3\tan^{-1}x.$

Hence,

$f'(x)=-\frac{3}{1+x^2}.$

Also,

$\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta} =\frac{1-x^2}{1+x^2}.$

Since $0<2\theta<\pi$,

$g(x)=\cos^{-1}(\cos 2\theta) =2\theta =2\tan^{-1}x.$

Thus,

$g'(x)=\frac{2}{1+x^2}.$

The given limit is the ratio of derivatives at $x=a$:

$\lim\limits_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)} =\frac{f'(a)}{g'(a)} =\frac{-\dfrac{3}{1+a^2}}{\dfrac{2}{1+a^2}} =-\frac32.$

Hence, the correct answer is Option C, $\boxed{-\dfrac32}$.

For a point $(𝑥,𝑦)$ on the curve,

$4𝑦={𝑥}^2.$

Its squared distance from $(0,5)$ is

${𝐷}^2={𝑥}^2+(𝑦-5{)}^2.$

Using ${𝑥}^2=4𝑦$,