MHT CET 2026 20th April Evening Shift (Mathematics) Question Paper with Solutions

Option A: $\sin \left(𝑥\right)$ is an even function.

For an even function, we must have

$𝑓(-𝑥)=𝑓\left(𝑥\right)$

But for $\sin 𝑥$,

$\sin (-𝑥)=-\sin 𝑥$

So $\sin 𝑥$ is an odd function, not an even function.

Option A is false.

Option B: Every square matrix is non-singular.

A square matrix is non-singular only if its determinant is not zero.

But some square matrices have determinant zero, for example,

$\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$

Its determinant is

$1\times 4-2\times 2=4-4=0$

So it is singular.

Option B is false.

Option C: The product of a complex number and its conjugate is purely imaginary.

If $𝑧=𝑎+𝑖𝑏$, then its conjugate is $\overline{𝑧}=𝑎-𝑖𝑏$.

Their product is

$𝑧\overline{𝑧}=(𝑎+𝑖𝑏)(𝑎-𝑖𝑏)={𝑎}^2+{𝑏}^2$

This is a real number, not purely imaginary.

Option C is false.

Option D: A square of any real number is non-negative.

For any real number $𝑥$,

${𝑥}^2\ge 0$

For example, $(-3{)}^2=9$ and $0^2=0$.

So this statement is always true.

Hence, the statement having truth value $𝑇$ is

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

We need to find

${[{𝑖}^{18}+{\left(\frac{1}{𝑖}\right)}^{25}]}^3$

We know powers of $𝑖$ repeat in a cycle of $4$:

${𝑖}^1=𝑖,{𝑖}^2=-1,{𝑖}^3=-𝑖,{𝑖}^4=1$

First,

${𝑖}^{18}={𝑖}^{16+2}={𝑖}^{16}\cdot {𝑖}^2$

Since ${𝑖}^4=1$, ${𝑖}^{16}=({𝑖}^4{)}^4=1$.

So,

${𝑖}^{18}=1\cdot (-1)=-1$

Now,

$\frac{1}{𝑖}=\frac{𝑖}{{𝑖}^2}=\frac{𝑖}{-1}=-𝑖$

So,

${\left(\frac{1}{𝑖}\right)}^{25}=(-𝑖{)}^{25}$

Since $25$ is odd,

$(-𝑖{)}^{25}=-{𝑖}^{25}$

Also,

${𝑖}^{25}={𝑖}^{24+1}={𝑖}^{24}\cdot 𝑖=({𝑖}^4{)}^6\cdot 𝑖=𝑖$

So,

$(-𝑖{)}^{25}=-𝑖$

Now the expression inside the bracket becomes

${𝑖}^{18}+{\left(\frac{1}{𝑖}\right)}^{25}=-1+(-𝑖)=-1-𝑖$

Now cube it:

$(-1-𝑖{)}^3$

First square it:

$(-1-𝑖{)}^2=1+2𝑖+{𝑖}^2$

Since ${𝑖}^2=-1$,

$(-1-𝑖{)}^2=1+2𝑖-1=2𝑖$

So,

$(-1-𝑖{)}^3=(-1-𝑖{)}^2(-1-𝑖)$

$=2𝑖(-1-𝑖)$

$=-2𝑖-2{𝑖}^2$

Again, ${𝑖}^2=-1$:

$=-2𝑖-2(-1)$

$=2-2𝑖$

So the correct option is

$\overline{)2-2𝑖}$

Option D.

The word is INSTITUTION.

First count the total letters:

$\text{INSTITUTION has }11\text{ letters}$

Now count the repeated letters:

• $𝐼$ appears $3$ times

• $𝑇$ appears $3$ times

• $𝑁$ appears $2$ times

• $𝑆,𝑈,𝑂$ appear once each

If all letters were different, the number of arrangements would be:

$11!$

But since some letters are repeated, we divide by the factorials of their repetitions:

$\frac{11!}{3!\times 3!\times 2!}$

Now calculate:

$\frac{11!}{3!\times 3!\times 2!}=\frac{39916800}{6\times 6\times 2}$

$=\frac{39916800}{72}=554400$

So, the number of permutations is:

$\overline{)554400}$

Correct option: D

Given,

$𝑥+𝑦=\frac{𝜋}{4}$

So,

$\tan (𝑥+𝑦)=\tan \frac{𝜋}{4}=1$

Using the formula,

$\tan (𝑥+𝑦)=\frac{\tan 𝑥+\tan 𝑦}{1-\tan 𝑥\tan 𝑦}$

So,

$\frac{\tan 𝑥+\tan 𝑦}{1-\tan 𝑥\tan 𝑦}=1$

Now cross-multiply:

$\tan 𝑥+\tan 𝑦=1-\tan 𝑥\tan 𝑦$

Bring the product term to the left side:

$\tan 𝑥+\tan 𝑦+\tan 𝑥\tan 𝑦=1$

Now we need to find:

$(1+\tan 𝑥)(1+\tan 𝑦)$

Expanding it,

$(1+\tan 𝑥)(1+\tan 𝑦)=1+\tan 𝑥+\tan 𝑦+\tan 𝑥\tan 𝑦$

But we know,

$\tan 𝑥+\tan 𝑦+\tan 𝑥\tan 𝑦=1$

So,

$(1+\tan 𝑥)(1+\tan 𝑦)=1+1=2$

So the correct answer is:

$\overline{)2}$

Option B.

Since the line makes equal negative intercepts on the coordinate axes, suppose each intercept is $-𝑎$, where $𝑎>0$.

So the line cuts the $𝑥$-axis at $(-𝑎,0)$ and the $𝑦$-axis at $(0,-𝑎)$.

Using intercept form,

$\frac{𝑥}{-𝑎}+\frac{𝑦}{-𝑎}=1$

This gives

$-\frac{𝑥+𝑦}{𝑎}=1$

So,

$𝑥+𝑦+𝑎=0$

Now the perpendicular distance from the origin $(0,0)$ to the line

$𝑥+𝑦+𝑎=0$

is

$\frac{\left|𝑎\right|}{\sqrt{\surd 1^2+1^2}}=\frac{𝑎}{\sqrt{\surd 2}}$

Given distance is $4$ units, so

$\frac{𝑎}{\sqrt{\surd 2}}=4$

$𝑎=4\sqrt{\surd 2}$

Hence the equation of the line is

$𝑥+𝑦+4\sqrt{\surd 2}=0$

So the correct option is

$\overline{)\text{Option B }𝑥+𝑦+4\sqrt{\surd 2}=0}$

The two lines pass through the origin, so if they make angles $𝛼$ and $𝛽$ with the X-axis, their slopes are

${𝑚}_1=\tan 𝛼,{𝑚}_2=\tan 𝛽$

For a line through the origin, we can write

$𝑦=𝑚𝑥$

Substitute $𝑦=𝑚𝑥$ in the given equation:

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0$

So,

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

${𝑥}^2(𝑎+2ℎ𝑚+𝑏{𝑚}^2)=0$

Since ${𝑥}^2\ne 0$, we get

$𝑏{𝑚}^2+2ℎ𝑚+𝑎=0$

The roots of this quadratic are ${𝑚}_1$ and ${𝑚}_2$.

So,

${𝑚}_1+{𝑚}_2=-\frac{2ℎ}{𝑏}$

and

${𝑚}_1{𝑚}_2=\frac{𝑎}{𝑏}$

Now,

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

$\tan (𝛼+𝛽)=\frac{{𝑚}_1+{𝑚}_2}{1-{𝑚}_1{𝑚}_2}$

Substitute the values:

$\tan (𝛼+𝛽)=\frac{-\frac{2ℎ}{𝑏}}{1-\frac{𝑎}{𝑏}}$

$=\frac{-\frac{2ℎ}{𝑏}}{\frac{𝑏-𝑎}{𝑏}}$

$=\frac{-2ℎ}{𝑏-𝑎}$

$=\frac{2ℎ}{𝑎-𝑏}$

So the correct option is

$\overline{)\frac{2ℎ}{𝑎-𝑏}}$

Option B.

Let the centre of the circle be $(ℎ,𝑘)$.

It is given that the centre lies on

$𝑥+2𝑦=0$

So,

$ℎ+2𝑘=0$

Now the circle touches the two lines

$3𝑥-4𝑦+8=0$

and

$3𝑥-4𝑦-28=0$

These two lines are parallel. Since the circle touches both of them, its centre must lie exactly midway between these two parallel lines. So the perpendicular distances of the centre from both lines must be equal.

Using the distance formula from a point to a line,

$\frac{|3ℎ-4𝑘+8|}{\sqrt{\surd 3^2+(-4{)}^2}}=\frac{|3ℎ-4𝑘-28|}{\sqrt{\surd 3^2+(-4{)}^2}}$

Since $\sqrt{\surd 3^2+(-4{)}^2}=5$, we get

$|3ℎ-4𝑘+8|=|3ℎ-4𝑘-28|$

This gives the midpoint line:

$3ℎ-4𝑘+8=-(3ℎ-4𝑘-28)$

$3ℎ-4𝑘+8=-3ℎ+4𝑘+28$

$6ℎ-8𝑘=20$

$3ℎ-4𝑘=10$

So the centre satisfies

$ℎ+2𝑘=0$

and

$3ℎ-4𝑘=10$

From $ℎ+2𝑘=0$,

$ℎ=-2𝑘$

Substitute in $3ℎ-4𝑘=10$:

$3(-2𝑘)-4𝑘=10$

$-6𝑘-4𝑘=10$

$-10𝑘=10$

$𝑘=-1$

So,

$ℎ=2$

Hence the centre is

$(2,-1)$

Now the radius is the distance from $(2,-1)$ to either line.

Using $3𝑥-4𝑦+8=0$,

$𝑟=\frac{\left|3\right(2)-4(-1)+8|}{5}$

$𝑟=\frac{|6+4+8|}{5}$

$𝑟=\frac{18}{5}$

So,

${𝑟}^2=\frac{324}{25}$

The equation of the circle is

$(𝑥-2{)}^2+(𝑦+1{)}^2=\frac{324}{25}$

Multiplying by $25$,

$25(𝑥-2{)}^2+25(𝑦+1{)}^2=324$

So the correct option is

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

The ellipse is

$16{𝑥}^2+25{𝑦}^2=400$

Divide by $400$:

$\frac{{𝑥}^2}{25}+\frac{{𝑦}^2}{16}=1$

So,

${𝑎}^2=25,{𝑏}^2=16$

$𝑎=5,𝑏=4$

For an ellipse,

${𝑐}^2={𝑎}^2-{𝑏}^2$

So,

${𝑐}^2=25-16=9$

$𝑐=3$

Hence the foci are

$𝑆(3,0)\text{and}{𝑆}^{\prime }(-3,0)$

So the length of $𝑆{𝑆}^{\prime }$ is

$𝑆{𝑆}^{\prime }=6$

Now the base of triangle $𝑃𝑆{𝑆}^{\prime }$ is $𝑆{𝑆}^{\prime }$, which lies on the $𝑥$-axis.

If $𝑃=(𝑥,𝑦)$, then the height of the triangle from $𝑃$ to $𝑆{𝑆}^{\prime }$ is $\left|𝑦\right|$.

Given area of $△𝑃𝑆{𝑆}^{\prime }$ is $9$ square units:

$\frac{1}{2}\times 6\times \left|𝑦\right|=9$

$3\left|𝑦\right|=9$

$\left|𝑦\right|=3$

So,

${𝑦}^2=9$

Now substitute in the ellipse:

$\frac{{𝑥}^2}{25}+\frac{{𝑦}^2}{16}=1$

$\frac{{𝑥}^2}{25}+\frac{9}{16}=1$

$\frac{{𝑥}^2}{25}=\frac{7}{16}$

${𝑥}^2=\frac{175}{16}$

$𝑥=\pm \frac{5\sqrt{\surd 7}}{4}$

So the abscissa of point $𝑃$ is

$\overline{)\frac{5\sqrt{\surd 7}}{4}}$

Correct option: C.

The expression inside the bracket is

$\frac{1}{1-{𝑛}^2}+\frac{2}{1-{𝑛}^2}+\cdots +\frac{𝑛}{1-{𝑛}^2}$

Here the denominator is the same in every term, so we can take it common:

$\frac{1+2+\cdots +𝑛}{1-{𝑛}^2}$

Now,

$1+2+\cdots +𝑛=\frac{𝑛(𝑛+1)}{2}$

So the bracket becomes

$\frac{\frac{𝑛(𝑛+1)}{2}}{1-{𝑛}^2}$

Since

$1-{𝑛}^2=(1-𝑛)(1+𝑛),$

we get

$\frac{𝑛(𝑛+1)}{2(1-𝑛)(1+𝑛)}$

Cancel $(𝑛+1)$ with $(1+𝑛)$:

$\frac{𝑛}{2(1-𝑛)}$

Now,

$\frac{𝑛}{2(1-𝑛)}=-\frac{𝑛}{2(𝑛-1)}$

As $𝑛\to \infty$,

$-\frac{𝑛}{2(𝑛-1)}\to -\frac{1}{2}$

So the required limit is

${(-\frac{1}{2})}^3=-\frac{1}{8}$

Hence, the correct option is

$\overline{)\text{Option D }\frac{-1}{8}}$

Given

$\sim 𝑝\vee 𝑞$

is false.

For an “or” statement to be false, both parts must be false.

So,

$\sim 𝑝=𝐹$

and

$𝑞=𝐹$

Since $\sim 𝑝$ is false, $𝑝$ must be true.

So we have:

$𝑝=𝑇,𝑞=𝐹$

Now check the options.

Option A:

$𝑝\leftrightarrow 𝑞$

This is true only when both $𝑝$ and $𝑞$ have the same truth value. Here,

$𝑝=𝑇,𝑞=𝐹$

So this is false.

Option B:

$𝑝\to 𝑞$

This is false only when $𝑝$ is true and $𝑞$ is false. Here exactly that happens, so it is false.

Option C:

$𝑞\to 𝑝$

Here,

$𝑞=𝐹,𝑝=𝑇$

An implication with a false first part is true. So,

$𝑞\to 𝑝=𝑇$

Hence, the correct option is:

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

For any implication $𝑟\to 𝑠$, its contrapositive is

$\sim 𝑠<text>\to </text><text>\sim </text>𝑟$

Here the given statement is

$\sim 𝑞\to 𝑝$

So take

$𝑟<text>=</text><text>\sim </text>𝑞,𝑠=𝑝$

Its contrapositive will be

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

Since $\sim (\sim 𝑞)=𝑞$, we get

$\sim 𝑝\to 𝑞$

Now use the logical equivalence

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

So,

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

$=𝑝\vee 𝑞$

So the contrapositive of $\sim 𝑞\to 𝑝$ is equivalent to

$𝑝\vee 𝑞$

Correct option: Option C.

Since the angles of $△𝐴𝐵𝐶$ are in A.P., we take

$𝐴,𝐵,𝐶$

in arithmetic progression.

So the middle angle is the average of the other two:

$𝐴+𝐶=2𝐵$

Also, in any triangle,

$𝐴+𝐵+𝐶={180}^{\circ }$

Using $𝐴+𝐶=2𝐵$,

$2𝐵+𝐵={180}^{\circ }$

$3𝐵={180}^{\circ }$

$𝐵={60}^{\circ }$

Now, by the sine rule,

$\frac{𝑏}{𝑐}=\frac{\sin 𝐵}{\sin 𝐶}$

Given,

$𝑏:𝑐=\sqrt{\surd 3}:\sqrt{\surd 2}$

So,

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

Since $𝐵={60}^{\circ }$,

$\frac{\sin {60}^{\circ }}{\sin 𝐶}=\frac{\sqrt{\surd 3}}{\sqrt{\surd 2}}$

$\frac{\frac{\sqrt{\surd 3}}{2}}{\sin 𝐶}=\frac{\sqrt{\surd 3}}{\sqrt{\surd 2}}$

Cancelling $\sqrt{\surd 3}$,

$\sin 𝐶=\frac{\sqrt{\surd 2}}{2}$

So,

$𝐶={45}^{\circ }$

Now,

$𝐴+𝐵+𝐶={180}^{\circ }$

$𝐴+{60}^{\circ }+{45}^{\circ }={180}^{\circ }$

$𝐴={75}^{\circ }$

So, the correct option is

$\overline{){75}^{\circ }}$

Option D.

The largest angle of a triangle is always opposite the largest side.

Here the sides are $3$, $5$ and $7$ units.

So the largest angle is opposite the side $7$.

Using the cosine rule,

$\cos 𝐶=\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}$

Take $𝑎=3$, $𝑏=5$, and $𝑐=7$.

$\cos 𝐶=\frac{3^2+5^2-7^2}{2\cdot 3\cdot 5}$

$\cos 𝐶=\frac{9+25-49}{30}$

$\cos 𝐶=\frac{-15}{30}$

$\cos 𝐶=-\frac{1}{2}$

We know that

$\cos \frac{2𝜋}{3}=-\frac{1}{2}$

So,

$𝐶=\frac{2𝜋}{3}$

Hence, the largest angle of the triangle is

$\overline{)\frac{2𝜋}{3}}$

Correct option: Option C.

For a $2\times 2$ matrix

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

we use

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

Here,

$𝐴=\left[\begin{array}{cc}1& -5\\ -2& 4\end{array}\right]$

So,

$𝑎=1,𝑏=-5,𝑐=-2,𝑑=4$

First find the determinant:

$\left|𝐴\right|=𝑎𝑑-𝑏𝑐$

$\left|𝐴\right|=\left(1\right)\left(4\right)-(-5)(-2)$

$\left|𝐴\right|=4-10=-6$

Now,

${𝐴}^{-1}=\frac{1}{-6}\left[\begin{array}{cc}4& 5\\ 2& 1\end{array}\right]$

So,

${𝐴}^{-1}=-\frac{1}{6}\left[\begin{array}{cc}4& 5\\ 2& 1\end{array}\right]$

Hence, the correct option is

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

For the matrix

$𝐴=⎡⎣⎢\begin{array}{ccc}3& 1& 2\\ 1& 2& 0\\ 1& 1& 4\end{array}⎤⎦⎥$

we need the cofactors ${𝐶}_{11}$, ${𝐶}_{21}$ and ${𝐶}_{32}$.

Cofactor is given by

${𝐶}_{𝑖𝑗}=(-1{)}^{𝑖+𝑗}{𝑀}_{𝑖𝑗}$

where ${𝑀}_{𝑖𝑗}$ is the minor of ${𝑎}_{𝑖𝑗}$.

First,

${𝐶}_{11}=(-1{)}^{1+1}\mid \begin{array}{cc}2& 0\\ 1& 4\end{array}\mid$

${𝐶}_{11}=1(2\times 4-0\times 1)=8$

Next,

${𝐶}_{21}=(-1{)}^{2+1}\mid \begin{array}{cc}1& 2\\ 1& 4\end{array}\mid$

${𝐶}_{21}=-1(1\times 4-2\times 1)$

${𝐶}_{21}=-2$

Now,

${𝐶}_{32}=(-1{)}^{3+2}\mid \begin{array}{cc}3& 2\\ 1& 0\end{array}\mid$

${𝐶}_{32}=-1(3\times 0-2\times 1)$

${𝐶}_{32}=2$

Given,

$𝑝{𝐶}_{11}+4{𝐶}_{21}-5{𝐶}_{32}=-2$

Substitute the values:

$𝑝\left(8\right)+4(-2)-5\left(2\right)=-2$

$8𝑝-8-10=-2$

$8𝑝-18=-2$

$8𝑝=16$

$𝑝=2$

So, the value of $𝑝$ is

$\overline{)2}$

Correct option: Option B.

Put

$𝑥=\tan 𝜃$

Then,

$\frac{2𝑥}{1+{𝑥}^2}=\frac{2\tan 𝜃}{1+{\tan }^2𝜃}=\sin 2𝜃$

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

and

$\frac{2𝑥}{1-{𝑥}^2}=\frac{2\tan 𝜃}{1-{\tan }^2𝜃}=\tan 2𝜃$

Now check the options.

For option C,

$𝑥=\frac{1}{\sqrt{\surd 3}}$

So,

$𝑥=\tan \frac{𝜋}{6}$

Hence,

$2𝜃=\frac{𝜋}{3}$

Now,

${\sin }^{-1}(\sin \frac{𝜋}{3})=\frac{𝜋}{3}$

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

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

Substitute in the given expression:

$3\left(\frac{𝜋}{3}\right)-4\left(\frac{𝜋}{3}\right)+2\left(\frac{𝜋}{3}\right)$

$=𝜋-\frac{4𝜋}{3}+\frac{2𝜋}{3}$

$=\frac{𝜋}{3}$

This satisfies the given equation.

So,

$\overline{)𝑥=\frac{1}{\sqrt{\surd 3}}}$

Correct option: C.

Let

${\sin }^{-1}𝑥=𝐴\text{and}{\cos }^{-1}𝑥=𝐵$

For the same value of $𝑥$, we use the standard identity

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

So,

$𝐴+𝐵=\frac{𝜋}{2}$

Given,

$2𝐴-3𝐵=4$

From $𝐴+𝐵=\frac{𝜋}{2}$,

$𝐴=\frac{𝜋}{2}-𝐵$

Substitute this in the given equation:

$2(\frac{𝜋}{2}-𝐵)-3𝐵=4$

$𝜋-2𝐵-3𝐵=4$

$𝜋-5𝐵=4$

$𝐵=\frac{𝜋-4}{5}$

Now we need

$2{\sin }^{-1}𝑥+3{\cos }^{-1}𝑥=2𝐴+3𝐵$

Using $𝐴=\frac{𝜋}{2}-𝐵$,

$2𝐴+3𝐵=2(\frac{𝜋}{2}-𝐵)+3𝐵$

$=𝜋-2𝐵+3𝐵$

$=𝜋+𝐵$

Substitute $𝐵=\frac{𝜋-4}{5}$:

$2𝐴+3𝐵=𝜋+\frac{𝜋-4}{5}$

$=\frac{5𝜋+𝜋-4}{5}$

$=\frac{6𝜋-4}{5}$

So the correct option is

$\overline{)\frac{6𝜋-4}{5}}$

Option A.

Given

$𝑓\left(𝑥\right)=\frac{2𝑥-1}{𝑥+5},𝑥\ne -5$

To find ${𝑓}^{-1}\left(𝑥\right)$, first write

$𝑦=\frac{2𝑥-1}{𝑥+5}$

Now interchange $𝑥$ and $𝑦$:

$𝑥=\frac{2𝑦-1}{𝑦+5}$

Now solve for $𝑦$.

$𝑥(𝑦+5)=2𝑦-1$

$𝑥𝑦+5𝑥=2𝑦-1$

Bring the $𝑦$ terms together:

$𝑥𝑦-2𝑦=-1-5𝑥$

Take $𝑦$ common:

$𝑦(𝑥-2)=-(1+5𝑥)$

$𝑦=\frac{-(1+5𝑥)}{𝑥-2}$

This can be written as

$𝑦=\frac{5𝑥+1}{2-𝑥}$

So,

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

Also, the denominator should not be zero:

$2-𝑥\ne 0$

$𝑥\ne 2$

Hence, the correct option is

$\overline{)\text{Option B }\frac{5𝑥+1}{2-𝑥},𝑥\ne 2}$

For the function to be continuous at $𝑥=3$, the value of the function from the left side and the right side must be equal.

For $𝑥\le 3$,

$𝑓\left(𝑥\right)=𝑎𝑥+1$

So,

$𝑓\left(3\right)=3𝑎+1$

For $𝑥>3$,

$𝑓\left(𝑥\right)=𝑏𝑥+3$

So the right-hand limit at $𝑥=3$ is

$3𝑏+3$

Since $𝑓\left(𝑥\right)$ is continuous at $𝑥=3$,

$3𝑎+1=3𝑏+3$

Now simplify:

$3𝑎-3𝑏=3-1$

$3(𝑎-𝑏)=2$

$𝑎-𝑏=\frac{2}{3}$

So the correct option is

$\overline{)\frac{2}{3}}$

Option A.

Given,

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

So,

${𝑓}^{\prime \prime }\left(𝑥\right)-{𝑔}^{\prime \prime }\left(𝑥\right)=0$

This means

$\left(𝑓\right(𝑥)-𝑔(𝑥){)}^{\prime \prime }=0$

Let

$ℎ\left(𝑥\right)=𝑓\left(𝑥\right)-𝑔\left(𝑥\right)$

Then,

${ℎ}^{\prime \prime }\left(𝑥\right)=0$

So $ℎ\left(𝑥\right)$ must be a linear function.

Now from the given condition,

${𝑓}^{\prime }\left(1\right)=2{𝑔}^{\prime }\left(1\right)=4$

So,

${𝑓}^{\prime }\left(1\right)=4$

and

$2{𝑔}^{\prime }\left(1\right)=4\Rightarrow {𝑔}^{\prime }\left(1\right)=2$

Hence,

${ℎ}^{\prime }\left(1\right)={𝑓}^{\prime }\left(1\right)-{𝑔}^{\prime }\left(1\right)=4-2=2$

Since $ℎ\left(𝑥\right)$ is linear, its derivative is constant. So,

${ℎ}^{\prime }\left(𝑥\right)=2$

Thus,

$ℎ\left(𝑥\right)=2𝑥+𝐶$

Now use the second condition:

$𝑓\left(2\right)=3𝑔\left(2\right)=9$

So,

$𝑓\left(2\right)=9$

and

$3𝑔\left(2\right)=9\Rightarrow 𝑔\left(2\right)=3$

Therefore,

$ℎ\left(2\right)=𝑓\left(2\right)-𝑔\left(2\right)=9-3=6$

Using $ℎ\left(𝑥\right)=2𝑥+𝐶$,

$ℎ\left(2\right)=2\left(2\right)+𝐶=6$

$4+𝐶=6$

$𝐶=2$

So,

$ℎ\left(𝑥\right)=2𝑥+2$

At $𝑥=4$,

$ℎ\left(4\right)=2\left(4\right)+2=10$

Hence,

$𝑓\left(4\right)-𝑔\left(4\right)=10$

The correct option is Option B: $10$.

Let

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

Then

$\cot 𝜃=\sqrt{\surd \frac{1-𝑥}{1+𝑥}}$

Squaring both sides,

${\cot }^2𝜃=\frac{1-𝑥}{1+𝑥}$

We need

$𝑦={\cos }^2𝜃$

Using the identity

${\cos }^2𝜃=\frac{{\cot }^2𝜃}{1+{\cot }^2𝜃}$

we get

$𝑦=\frac{\frac{1-𝑥}{1+𝑥}}{1+\frac{1-𝑥}{1+𝑥}}$

Simplify the denominator:

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

So,

$𝑦=\frac{\frac{1-𝑥}{1+𝑥}}{\frac{2}{1+𝑥}}=\frac{1-𝑥}{2}$

Now differentiate with respect to $𝑥$:

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

So the correct option is

$\overline{)-\frac{1}{2}}$

Option A.

Given

$\sqrt{\surd \frac{𝑥}{𝑦}}+\sqrt{\surd \frac{𝑦}{𝑥}}=6$

First combine the two terms. Since

$\sqrt{\surd \frac{𝑥}{𝑦}}=\frac{\sqrt{\surd 𝑥}}{\sqrt{\surd 𝑦}},\sqrt{\surd \frac{𝑦}{𝑥}}=\frac{\sqrt{\surd 𝑦}}{\sqrt{\surd 𝑥}}$

we get

$\frac{\sqrt{\surd 𝑥}}{\sqrt{\surd 𝑦}}+\frac{\sqrt{\surd 𝑦}}{\sqrt{\surd 𝑥}}=6$

Taking LCM,

$\frac{𝑥+𝑦}{\sqrt{\surd 𝑥𝑦}}=6$

So,

$𝑥+𝑦=6\sqrt{\surd 𝑥𝑦}$

Squaring both sides,

$(𝑥+𝑦{)}^2=36𝑥𝑦$

${𝑥}^2+2𝑥𝑦+{𝑦}^2=36𝑥𝑦$

${𝑥}^2-34𝑥𝑦+{𝑦}^2=0$

Now differentiate with respect to $𝑥$:

$\frac{𝑑}{𝑑𝑥}\left({𝑥}^2\right)-34\frac{𝑑}{𝑑𝑥}\left(𝑥𝑦\right)+\frac{𝑑}{𝑑𝑥}\left({𝑦}^2\right)=0$

Using product rule for $𝑥𝑦$,

$2𝑥-34(𝑦+𝑥\frac{𝑑𝑦}{𝑑𝑥})+2𝑦\frac{𝑑𝑦}{𝑑𝑥}=0$

Expand:

$2𝑥-34𝑦-34𝑥\frac{𝑑𝑦}{𝑑𝑥}+2𝑦\frac{𝑑𝑦}{𝑑𝑥}=0$

Group the terms containing $\frac{𝑑𝑦}{𝑑𝑥}$:

$(2𝑦-34𝑥)\frac{𝑑𝑦}{𝑑𝑥}=34𝑦-2𝑥$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{34𝑦-2𝑥}{2𝑦-34𝑥}$