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

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

Option A:

$\exists 𝑥\in 𝐴\text{ such that }𝑥+4\le 9$

This means at least one element of $𝐴$ should satisfy the inequality.

If $𝑥=1$,

$1+4=5\le 9$

So Option A is true.

Option B:

$\exists 𝑥\in 𝐴\text{ such that }12-𝑥=7$

Solving,

$12-𝑥=7$

$𝑥=5$

Since $5\in 𝐴$, Option B is true.

Option C:

$\forall 𝑥\in 𝐴,𝑥+3\ge 4$

This means every element of $𝐴$ should satisfy the inequality.

The smallest element of $𝐴$ is $1$.

$1+3=4\ge 4$

So all other elements will also satisfy it. Hence Option C is true.

Option D:

$\exists 𝑥\in 𝐴\text{ such that }𝑥+2<3$

Solving,

$𝑥+2<3$

$𝑥<1$

But there is no element in $𝐴=\{1,2,3,4,5\}$ which is less than $1$.

So Option D is not true.

Hence, the correct answer is:

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

Given,

$-1+\sqrt{\surd -3}=𝑟{𝑒}^{𝑖𝜃}$

Since

$\sqrt{\surd -3}=𝑖\sqrt{\surd 3}$

we get

$-1+\sqrt{\surd -3}=-1+𝑖\sqrt{\surd 3}$

This complex number has real part $-1$ and imaginary part $\sqrt{\surd 3}$.

So the point is

$(-1,\sqrt{\surd 3})$

It lies in the second quadrant.

Now,

$\tan 𝜃=\frac{\text{imaginary part}}{\text{real part}}=\frac{\sqrt{\surd 3}}{-1}=-\sqrt{\surd 3}$

The reference angle is

$\frac{𝜋}{3}$

Since the point lies in the second quadrant,

$𝜃=𝜋-\frac{𝜋}{3}=\frac{2𝜋}{3}$

So the value of $𝜃$ is

$\overline{)\frac{2𝜋}{3}}$

Correct option: Option D.

There are $5$ boys and $3$ girls, so total people are

$5+3=8$

So there are $8$ places in the row:

$1,2,3,4,5,6,7,8$

The odd places are:

$1,3,5,7$

So there are $4$ odd places.

The condition says that the girls occupy only odd places. Since there are $3$ girls, we first choose $3$ odd places out of these $4$ odd places for the girls:

${}^4{𝐶}_3=4$

Now arrange the $3$ girls in those selected places:

$3!=6$

The remaining $5$ places will be occupied by the $5$ boys. They can be arranged in:

$5!=120$

So, total number of ways is

${}^4{𝐶}_3\times 3!\times 5!$

$=4\times 6\times 120$

$=2880$

So the correct option is:

$\overline{)2880}$

Option B.

We know that the tangent function has period $𝜋$.

That means

$\tan (𝜃-𝑛𝜋)=\tan 𝜃$

for any integer $𝑛$.

Here,

$\tan (𝑥-9𝜋)$

Since $9𝜋$ is an integral multiple of $𝜋$, the value of tangent will remain the same.

So,

$\tan (𝑥-9𝜋)=\tan 𝑥$

Hence, the correct option is

$\overline{)\text{Option C }\tan 𝑥}$

Given,

$𝑐𝑜𝑠𝑒𝑐𝜃=-2$

Since $𝑐𝑜𝑠𝑒𝑐𝜃=\frac{1}{\sin 𝜃}$, we get

$\sin 𝜃=-\frac{1}{2}$

Now, $\sin 𝜃$ is negative in the third and fourth quadrants.

The reference angle for $\sin 𝜃=\frac{1}{2}$ is

$\frac{𝜋}{6}$

So the principal solutions in the interval $0\le 𝜃<2𝜋$ are:

In the third quadrant:

$𝜃=𝜋+\frac{𝜋}{6}=\frac{7𝜋}{6}$

In the fourth quadrant:

$𝜃=2𝜋-\frac{𝜋}{6}=\frac{11𝜋}{6}$

So the principal solutions are

$\frac{7𝜋}{6},\frac{11𝜋}{6}$

Hence, the correct option is

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

The given line passes through the point of intersection of

$𝑥-𝑦+1=0$

and

$2𝑥+3𝑦-8=0$

First find their point of intersection.

From

$𝑥-𝑦+1=0$

we get

$𝑦=𝑥+1$

Substitute this in the second equation:

$2𝑥+3(𝑥+1)-8=0$

$2𝑥+3𝑥+3-8=0$

$5𝑥-5=0$

$𝑥=1$

Now,

$𝑦=𝑥+1=1+1=2$

So, the point of intersection is $(1,2)$.

The line has $𝑥$-intercept $3$, so it passes through the point $(3,0)$.

Now find the equation of the line passing through $(1,2)$ and $(3,0)$.

Slope:

$𝑚=\frac{0-2}{3-1}=\frac{-2}{2}=-1$

Using point-slope form:

$𝑦-2=-1(𝑥-1)$

$𝑦-2=-𝑥+1$

$𝑥+𝑦-3=0$

So, the correct option is

$\overline{)𝑥+𝑦-3=0}$

Option B.

For a second degree equation representing a pair of straight lines, the point of intersection of the two lines is obtained from

$\frac{𝜕𝑆}{𝜕𝑥}=0\text{and}\frac{𝜕𝑆}{𝜕𝑦}=0$

Here,

$𝑆={𝑥}^2-𝑘{𝑦}^2-4𝑥+6𝑦-5$

Differentiate with respect to $𝑥$:

$\frac{𝜕𝑆}{𝜕𝑥}=2𝑥-4$

So,

$2𝑥-4=0$

$𝑥=2$

Differentiate with respect to $𝑦$:

$\frac{𝜕𝑆}{𝜕𝑦}=-2𝑘𝑦+6$

So,

$-2𝑘𝑦+6=0$

$𝑘𝑦=3$

$𝑦=\frac{3}{𝑘}$

Now the equation represents a pair of straight lines, so the determinant condition must be satisfied:

$\mid \begin{array}{ccc}1& 0& -2\\ 0& -𝑘& 3\\ -2& 3& -5\end{array}\mid =0$

Expanding,

$1(5𝑘-9)+(-2)(-2𝑘)=0$

$5𝑘-9+4𝑘=0$

$9𝑘-9=0$

$𝑘=1$

So,

$𝑦=\frac{3}{1}=3$

Hence the point of intersection is

$(2,3)$

Correct option: B $(2,3)$

The required circle passes through $(2,3)$ and $(4,5)$.

The centre of a circle passing through two points must lie on the perpendicular bisector of the line joining those two points.

For the points $𝐴(2,3)$ and $𝐵(4,5)$, the midpoint is

$(\frac{2+4}{2},\frac{3+5}{2})=(3,4)$

The slope of $𝐴𝐵$ is

$\frac{5-3}{4-2}=1$

So the slope of the perpendicular bisector is $-1$.

Its equation through $(3,4)$ is

$𝑦-4=-1(𝑥-3)$

$𝑦=-𝑥+7$

The centre also lies on the line

$𝑦-4𝑥+3=0$

So,

$𝑦=4𝑥-3$

Now solve both centre lines:

$-𝑥+7=4𝑥-3$

$5𝑥=10$

$𝑥=2$

Then

$𝑦=4\left(2\right)-3=5$

So the centre is $(2,5)$.

The radius is the distance from $(2,5)$ to $(2,3)$:

$𝑟=2$

Hence the circle is

$(𝑥-2{)}^2+(𝑦-5{)}^2=2^2$

$(𝑥-2{)}^2+(𝑦-5{)}^2=4$

Expanding,

${𝑥}^2-4𝑥+4+{𝑦}^2-10𝑦+25=4$

${𝑥}^2+{𝑦}^2-4𝑥-10𝑦+25=0$

So the correct option is

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

For the greatest integer function $\left[𝑥\right]$, the value changes at every integer.

Here we need

$\underset{𝑥\to 3}{lim}\frac{\left[𝑥\right]-3}{𝑥-3}$

Check the left-hand limit and right-hand limit separately.

For $𝑥\to 3^{-}$, the values of $𝑥$ are just less than $3$, like $2.9,2.99,2.999$.

So,

$\left[𝑥\right]=2$

Hence,

$\frac{\left[𝑥\right]-3}{𝑥-3}=\frac{2-3}{𝑥-3}=\frac{-1}{𝑥-3}$

Since $𝑥-3$ is a very small negative number,

$\frac{-1}{𝑥-3}\to +\infty$

So,

$\underset{𝑥\to 3^{-}}{lim}\frac{\left[𝑥\right]-3}{𝑥-3}=+\infty$

For $𝑥\to 3^{+}$, the values of $𝑥$ are just greater than $3$, like $3.1,3.01,3.001$.

So,

$\left[𝑥\right]=3$

Hence,

$\frac{\left[𝑥\right]-3}{𝑥-3}=\frac{3-3}{𝑥-3}=0$

So,

$\underset{𝑥\to 3^{+}}{lim}\frac{\left[𝑥\right]-3}{𝑥-3}=0$

The left-hand limit and right-hand limit are not equal.

Hence, the limit does not exist.

Correct option: D.

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 $𝐴$ is true and $𝐵$ is false.

So,

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

must be true, and

$(𝑝\wedge 𝑞)$

must be false.

Now, since

$\sim 𝑟$

is true, we get

$𝑟=𝐹$

Also,

$𝑞\Rightarrow 𝑟$

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

$𝑞=𝐹$

Now,

$𝑝\vee 𝑞$

is true. Since $𝑞=𝐹$, we must have

$𝑝=𝑇$

So the truth values are:

$𝑝=𝑇,𝑞=𝐹,𝑟=𝐹$

Now find $𝑝\Rightarrow 𝑞$:

$𝑝\Rightarrow 𝑞=𝑇\Rightarrow 𝐹=𝐹$

And find $𝑞\Rightarrow 𝑝$:

$𝑞\Rightarrow 𝑝=𝐹\Rightarrow 𝑇=𝑇$

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

$(𝐹,𝑇)$

Hence, the correct option is

$\overline{)\text{Option C }(𝐹,𝑇)}$

The given statement pattern is

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

We use the standard logical equivalence:

$𝑎\Rightarrow 𝑏<text>\equiv </text><text>\sim </text>𝑎\vee 𝑏$

Here,

$𝑎=(𝑝\vee 𝑞),𝑏<text>=</text><text>\sim </text>𝑟$

So,

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

Now apply De Morgan’s law:

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

So the expression becomes

$(\sim 𝑝\wedge \sim 𝑞)\vee \sim 𝑟$

This matches Option C.

So, the correct answer is

$\overline{)(\sim 𝑝\wedge \sim 𝑞)\vee \sim 𝑟}$

Since $\cos 𝐴=\frac{1}{2}$, we know

$𝐴=\frac{𝜋}{3}$

Also,

$𝐵=\frac{𝜋}{6}$

Now using the angle sum property of a triangle,

$𝐴+𝐵+𝐶=𝜋$

So,

$\frac{𝜋}{3}+\frac{𝜋}{6}+𝐶=𝜋$

$\frac{𝜋}{2}+𝐶=𝜋$

$𝐶=\frac{𝜋}{2}$

Now use the sine rule:

$\frac{𝑎}{\sin 𝐴}=\frac{𝑏}{\sin 𝐵}=\frac{𝑐}{\sin 𝐶}$

Given $𝑎=3$, so

$\frac{3}{\sin \frac{𝜋}{3}}=\frac{𝑏}{\sin \frac{𝜋}{6}}=\frac{𝑐}{\sin \frac{𝜋}{2}}$

We know

$\sin \frac{𝜋}{3}=\frac{\sqrt{\surd 3}}{2},\sin \frac{𝜋}{6}=\frac{1}{2},\sin \frac{𝜋}{2}=1$

So,

$\frac{3}{\frac{\sqrt{\surd 3}}{2}}=\frac{6}{\sqrt{\surd 3}}=2\sqrt{\surd 3}$

Now,

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

and

$𝑐=2\sqrt{\surd 3}\times 1=2\sqrt{\surd 3}$

So,

$𝑏=\sqrt{\surd 3},𝑐=2\sqrt{\surd 3}$

Hence, the correct option is Option C.

Given

$𝐴=\frac{1}{2}\left[\begin{array}{cc}-1& -\sqrt{\surd 3}\\ \sqrt{\surd 3}& -1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}& -\frac{\sqrt{\surd 3}}{2}\\ \frac{\sqrt{\surd 3}}{2}& -\frac{1}{2}\end{array}\right]$

This is of the form

$\left[\begin{array}{cc}\cos 𝜃& -\sin 𝜃\\ \sin 𝜃& \cos 𝜃\end{array}\right]$

where

$\cos 𝜃=-\frac{1}{2},\sin 𝜃=\frac{\sqrt{\surd 3}}{2}$

So,

$𝜃={120}^{\circ }$

This matrix represents rotation through ${120}^{\circ }$. Applying it three times gives rotation through

$3\times {120}^{\circ }={360}^{\circ }$

So,

${𝐴}^3=𝐼$

Now,

${𝐴}^3=𝐼$

Multiplying both sides by ${𝐴}^{-1}$, we get

${𝐴}^2={𝐴}^{-1}$

So,

${𝐴}^{-1}-{𝐴}^2=0$

That means

${𝐴}^{-1}-{𝐴}^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

This is a null matrix. It is also a diagonal matrix and a scalar matrix, since

$\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0𝐼$

But it is not a unit matrix.

Hence the correct option is:

$\overline{)\text{Option B: a unit matrix}}$

Given

$𝐴=\left[\begin{array}{cc}2𝑖& {𝑖}^3\\ {𝑖}^2& 1\end{array}\right]$

We know

${𝑖}^2=-1,{𝑖}^3={𝑖}^2\cdot 𝑖=-𝑖$

So,

$𝐴=\left[\begin{array}{cc}2𝑖& -𝑖\\ -1& 1\end{array}\right]$

For a $2\times 2$ matrix

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

its inverse is

$\frac{1}{𝑎𝑑-𝑏𝑐}\left[\begin{array}{cc}𝑑& -𝑏\\ -𝑐& 𝑎\end{array}\right]$

Here,

$𝑎=2𝑖,𝑏=-𝑖,𝑐=-1,𝑑=1$

So the determinant is

$𝑎𝑑-𝑏𝑐=\left(2𝑖\right)\left(1\right)-(-𝑖)(-1)$

$=2𝑖-𝑖=𝑖$

Now,

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

Since

$\frac{1}{𝑖}=-𝑖$

we get

${𝐴}^{-1}=-𝑖\left[\begin{array}{cc}1& 𝑖\\ 1& 2𝑖\end{array}\right]$

${𝐴}^{-1}=\left[\begin{array}{cc}-𝑖& 1\\ -𝑖& 2\end{array}\right]$

So the correct answer is

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

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

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

Let

${\sin }^{-1}𝑥=𝜃$

Then

${\cos }^{-1}𝑥=\frac{𝜋}{2}-𝜃$

So the given expression becomes

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

Now simplify:

${𝜃}^2+{(\frac{𝜋}{2}-𝜃)}^2={𝜃}^2+\frac{{𝜋}^2}{4}-𝜋𝜃+{𝜃}^2$

$=2{𝜃}^2-𝜋𝜃+\frac{{𝜋}^2}{4}$

This quadratic expression is minimum when

$𝜃=\frac{𝜋}{4}$

Putting $𝜃=\frac{𝜋}{4}$:

${𝜃}^2+{(\frac{𝜋}{2}-𝜃)}^2={\left(\frac{𝜋}{4}\right)}^2+{\left(\frac{𝜋}{4}\right)}^2$

$=\frac{{𝜋}^2}{16}+\frac{{𝜋}^2}{16}=\frac{{𝜋}^2}{8}$

So the minimum value is

$\overline{)\frac{{𝜋}^2}{8}}$

Correct option: A

Let

$𝑆={\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(5\right)+{\tan }^{-1}\left(\frac{1}{4}\right)$

We use the formula

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

First combine ${\tan }^{-1}\left(1\right)$ and ${\tan }^{-1}\left(3\right)$:

$\tan ({\tan }^{-1}(1)+{\tan }^{-1}(3\left)\right)=\frac{1+3}{1-1\cdot 3}=\frac{4}{-2}=-2$

Now add ${\tan }^{-1}\left(5\right)$:

$\tan ({\tan }^{-1}(1)+{\tan }^{-1}(3)+{\tan }^{-1}(5\left)\right)=\frac{-2+5}{1-(-2)\left(5\right)}$

$=\frac{3}{11}$

Now add ${\tan }^{-1}\left(\frac{1}{4}\right)$:

$\tan 𝑆=\frac{\frac{3}{11}+\frac{1}{4}}{1-\frac{3}{11}\cdot \frac{1}{4}}$

$=\frac{\frac{12+11}{44}}{1-\frac{3}{44}}=\frac{\frac{23}{44}}{\frac{41}{44}}=\frac{23}{41}$

Given,

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

Taking tangent on both sides,

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

Since $\tan (𝜋+𝜃)=\tan 𝜃$,

$\frac{23}{41}=\frac{𝛼}{2}$

So,

$𝛼=\frac{46}{41}$

Hence, the correct option is

$\overline{)\frac{46}{41}}$

Option A.

For the function

$𝑓\left(𝑥\right)=\sqrt{\surd 𝑥-1}+\sqrt{\surd 3-𝑥}$

to be defined, both square roots must have non-negative quantities inside them.

So,

$𝑥-1\ge 0$

which gives

$𝑥\ge 1$

Also,

$3-𝑥\ge 0$

which gives

$𝑥\le 3$

Both conditions must be true together, so

$1\le 𝑥\le 3$

Hence, the domain is

$[1,3]$

So the correct answer is Option C: $[1,3]$.

For $𝑓\left(𝑥\right)$ to be continuous at $𝑥=0$, we need

$𝑘=\underset{𝑥\to 0}{lim}{[\tan (\frac{𝜋}{4}+𝑥)]}^{\frac{1}{𝑥}}$

This is of the form $1^{\infty }$, because as $𝑥\to 0$,

$\tan (\frac{𝜋}{4}+𝑥)\to \tan \frac{𝜋}{4}=1$

Let

$𝐿=\underset{𝑥\to 0}{lim}{[\tan (\frac{𝜋}{4}+𝑥)]}^{\frac{1}{𝑥}}$

Taking logarithm,

$\log 𝐿=\underset{𝑥\to 0}{lim}\frac{1}{𝑥}\log [\tan (\frac{𝜋}{4}+𝑥)]$

Now this is like finding

$\underset{𝑥\to 0}{lim}\frac{𝑔\left(𝑥\right)-𝑔\left(0\right)}{𝑥}$

where

$𝑔\left(𝑥\right)=\log [\tan (\frac{𝜋}{4}+𝑥)]$

Since

$𝑔\left(0\right)=\log (\tan \frac{𝜋}{4})=\log 1=0$

we get

$\log 𝐿={𝑔}^{\prime }\left(0\right)$

Now,

${𝑔}^{\prime }\left(𝑥\right)=\frac{1}{\tan (\frac{𝜋}{4}+𝑥)}\cdot {\sec }^2(\frac{𝜋}{4}+𝑥)$

At $𝑥=0$,

${𝑔}^{\prime }\left(0\right)=\frac{{\sec }^2\frac{𝜋}{4}}{\tan \frac{𝜋}{4}}$

We know

${\sec }^2\frac{𝜋}{4}=2,\tan \frac{𝜋}{4}=1$

So,

${𝑔}^{\prime }\left(0\right)=2$

Thus,

$\log 𝐿=2$

Hence,

$𝐿={𝑒}^2$

So for continuity at $𝑥=0$,

$𝑘={𝑒}^2$

Correct option: Option B.

Given

$𝑦=𝑓\left(\frac{3+2𝑥}{3-2𝑥}\right)$

and

$𝑓\left(𝑥\right)=\tan (\log 𝑥)$

So,

$𝑦=\tan (\log \left(\frac{3+2𝑥}{3-2𝑥}\right))$

Now differentiate with respect to $𝑥$.

Using the chain rule,

$\frac{𝑑𝑦}{𝑑𝑥}={\sec }^2(\log \left(\frac{3+2𝑥}{3-2𝑥}\right))\cdot \frac{𝑑}{𝑑𝑥}[\log \left(\frac{3+2𝑥}{3-2𝑥}\right)]$

Now simplify the derivative of the logarithmic part:

$\log \left(\frac{3+2𝑥}{3-2𝑥}\right)=\log (3+2𝑥)-\log (3-2𝑥)$

So,

$\frac{𝑑}{𝑑𝑥}[\log (3+2𝑥)-\log (3-2𝑥\left)\right]=\frac{2}{3+2𝑥}+\frac{2}{3-2𝑥}$

Taking LCM,

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

$=\frac{6-4𝑥+6+4𝑥}{9-4{𝑥}^2}$

$=\frac{12}{9-4{𝑥}^2}$

Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=\left(\frac{12}{9-4{𝑥}^2}\right){\sec }^2(\log \left(\frac{3+2𝑥}{3-2𝑥}\right))$

Comparing with

$\frac{𝑑𝑦}{𝑑𝑥}=\left(\frac{𝐴}{𝐵+𝐶{𝑥}^2}\right){\sec }^2(\log \left(\frac{3+2𝑥}{3-2𝑥}\right))$

we get

$𝐴=12,𝐵=9,𝐶=-4$

So the correct option is

$\overline{)\text{Option D }(12,9,-4)}$

Given

$𝑦=\sqrt{\surd {𝑥}^2+8𝑥+3}$

We can write it as

$𝑦=({𝑥}^2+8𝑥+3{)}^{\frac{1}{2}}$

Differentiate once with respect to $𝑥$:

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{2}({𝑥}^2+8𝑥+3{)}^{-\frac{1}{2}}(2𝑥+8)$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑥+4}{\sqrt{\surd {𝑥}^2+8𝑥+3}}$

Now differentiate again:

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{𝑑}{𝑑𝑥}\left[\right(𝑥+4\left)\right({𝑥}^2+8𝑥+3{)}^{-\frac{1}{2}}]$

Using product rule,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=({𝑥}^2+8𝑥+3{)}^{-\frac{1}{2}}+(𝑥+4)[-\frac{1}{2}({𝑥}^2+8𝑥+3{)}^{-\frac{3}{2}}(2𝑥+8)]$

Since $2𝑥+8=2(𝑥+4)$,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{1}{\sqrt{\surd {𝑥}^2+8𝑥+3}}-\frac{(𝑥+4{)}^2}{({𝑥}^2+8𝑥+3{)}^{\frac{3}{2}}}$

Taking common denominator,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{{𝑥}^2+8𝑥+3-(𝑥+4{)}^2}{({𝑥}^2+8𝑥+3{)}^{\frac{3}{2}}}$

Now,

$(𝑥+4{)}^2={𝑥}^2+8𝑥+16$

So,

${𝑥}^2+8𝑥+3-({𝑥}^2+8𝑥+16)=-13$

Hence,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{-13}{({𝑥}^2+8𝑥+3{)}^{\frac{3}{2}}}$

At $𝑥=3$,

${𝑥}^2+8𝑥+3=9+24+3=36$

So,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=\frac{-13}{{36}^{\frac{3}{2}}}$

Now,

${36}^{\frac{3}{2}}=(\sqrt{\surd 36}{)}^3=6^3=216$

Therefore,

$\frac{{𝑑}^2𝑦}{𝑑{𝑥}^2}=-\frac{13}{216}$

So the correct option is

$\overline{)\text{Option A }\frac{-13}{216}}$

Given

$𝑦=\sqrt{\surd \tan 𝑥+\sqrt{\surd \tan 𝑥+\sqrt{\surd \tan 𝑥+\cdots }}}$

The expression inside the square root repeats again and again. So the repeated part is again equal to $𝑦$.

So,

$𝑦=\sqrt{\surd \tan 𝑥+𝑦}$

Squaring both sides,

${𝑦}^2=\tan 𝑥+𝑦$

Now differentiate both sides with respect to $𝑥$:

$\frac{𝑑}{𝑑𝑥}\left({𝑦}^2\right)=\frac{𝑑}{𝑑𝑥}(\tan 𝑥+𝑦)$

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

Bring the $\frac{𝑑𝑦}{𝑑𝑥}$ terms together:

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

$(2𝑦-1)\frac{𝑑𝑦}{𝑑𝑥}={\sec }^2𝑥$

So,

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

At

$𝑥=\frac{𝜋}{4}$

we have

$\tan \frac{𝜋}{4}=1$

So from

${𝑦}^2=\tan 𝑥+𝑦$

we get

${𝑦}^2=1+𝑦$

${𝑦}^2-𝑦-1=0$

Since $𝑦$ is a square root expression, $𝑦$ is positive.

$𝑦=\frac{1+\sqrt{\surd 5}}{2}$

Now,

$2𝑦-1=2\left(\frac{1+\sqrt{\surd 5}}{2}\right)-1=\sqrt{\surd 5}$

Also,

${\sec }^2\frac{𝜋}{4}=2$

Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{2}{\sqrt{\surd 5}}$

So,

${\left(\frac{𝑑𝑦}{𝑑𝑥}\right)}^2={\left(\frac{2}{\sqrt{\surd 5}}\right)}^2=\frac{4}{5}$

The correct option is

$\overline{)\frac{4}{5}}$

Option C.

Given,

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

Write it as

$𝑦={(𝑥+\sqrt{\surd {𝑥}^2+1})}^{1/2}$

Differentiate both sides with respect to $𝑥$ using chain rule:

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{2}{(𝑥+\sqrt{\surd {𝑥}^2+1})}^{-1/2}(1+\frac{𝑑}{𝑑𝑥}\sqrt{\surd {𝑥}^2+1})$

Now,

$\frac{𝑑}{𝑑𝑥}\sqrt{\surd {𝑥}^2+1}=\frac{1}{2\sqrt{\surd {𝑥}^2+1}}\cdot 2𝑥=\frac{𝑥}{\sqrt{\surd {𝑥}^2+1}}$

So,

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

Since

$𝑦=\sqrt{\surd 𝑥+\sqrt{\surd {𝑥}^2+1}},$

we get

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

Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{\sqrt{\surd {𝑥}^2+1}+𝑥}{2𝑦\sqrt{\surd {𝑥}^2+1}}$