MHT CET 2026 13th April Morning Shift (Mathematics) Question Paper with Solutions

Given

$𝑧=\frac{1}{1+𝑖}$

First rationalise the denominator:

$𝑧=\frac{1}{1+𝑖}\cdot \frac{1-𝑖}{1-𝑖}$

$𝑧=\frac{1-𝑖}{(1+𝑖)(1-𝑖)}$

Using $(𝑎+𝑏)(𝑎-𝑏)={𝑎}^2-{𝑏}^2$,

$(1+𝑖)(1-𝑖)=1-{𝑖}^2=1-(-1)=2$

So,

$𝑧=\frac{1-𝑖}{2}$

$𝑧=\frac{1}{2}-\frac{𝑖}{2}$

Now compare with $𝑥+𝑖𝑦$.

Here,

$𝑥=\frac{1}{2},𝑦=-\frac{1}{2}$

The modulus is

$𝑟=\sqrt{\surd {𝑥}^2+{𝑦}^2}$

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

$𝑟=\sqrt{\surd \frac{1}{4}+\frac{1}{4}}=\sqrt{\surd \frac{1}{2}}=\frac{1}{\sqrt{\surd 2}}$

Since $𝑥>0$ and $𝑦<0$, the point lies in the fourth quadrant.

Also,

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

So the argument is

$𝜃=\frac{7𝜋}{4}$

Hence the polar form is

$𝑧=\frac{1}{\sqrt{\surd 2}}(\cos \frac{7𝜋}{4}+𝑖\sin \frac{7𝜋}{4})$

So the correct answer is

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

A cubic equation has the form

$𝑎{𝑥}^3+𝑏{𝑥}^2+𝑐𝑥+𝑑=0$

Here, the coefficients have to be chosen from

$\{0,2,3,4,5,6\}$

Repetition of coefficients is allowed.

For the equation to be cubic, the coefficient of ${𝑥}^3$ must not be $0$.

So, for $𝑎$, we cannot choose $0$.

Number of choices for $𝑎$:

$5$

For $𝑏$, $𝑐$, and $𝑑$, we can choose any of the $6$ given coefficients, including $0$.

So the total number of cubic equations is

$5\times 6\times 6\times 6$

$=5\times 216$

$=1080$

So, the correct option is

$\overline{)1080}$

Option D.

${105}^{\circ }$ can be written as ${60}^{\circ }+{45}^{\circ }$.

Using the formula

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

we get

$\tan {105}^{\circ }=\tan ({60}^{\circ }+{45}^{\circ })$

$=\frac{\tan {60}^{\circ }+\tan {45}^{\circ }}{1-\tan {60}^{\circ }\tan {45}^{\circ }}$

Now,

$\tan {60}^{\circ }=\sqrt{\surd 3},\tan {45}^{\circ }=1$

So,

$\tan {105}^{\circ }=\frac{\sqrt{\surd 3}+1}{1-\sqrt{\surd 3}}$

This matches Option C.

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

Given equation is

$\cos 3𝑥\cdot \tan 5𝑥=\sin 7𝑥$

Since

$\tan 5𝑥=\frac{\sin 5𝑥}{\cos 5𝑥},$

we must remember that $\cos 5𝑥\ne 0$.

So,

$\cos 3𝑥\cdot \frac{\sin 5𝑥}{\cos 5𝑥}=\sin 7𝑥$

Multiplying both sides by $\cos 5𝑥$,

$\cos 3𝑥\sin 5𝑥=\sin 7𝑥\cos 5𝑥$

Now use the product-to-sum identities:

$\sin 𝐴\cos 𝐵=\frac{1}{2}[\sin (𝐴+𝐵)+\sin (𝐴-𝐵\left)\right]$

So,

$\cos 3𝑥\sin 5𝑥=\frac{1}{2}[\sin 8𝑥+\sin 2𝑥]$

and

$\sin 7𝑥\cos 5𝑥=\frac{1}{2}[\sin 12𝑥+\sin 2𝑥]$

Thus,

$\frac{1}{2}[\sin 8𝑥+\sin 2𝑥]=\frac{1}{2}[\sin 12𝑥+\sin 2𝑥]$

Cancel common terms:

$\sin 8𝑥=\sin 12𝑥$

So,

$\sin 8𝑥-\sin 12𝑥=0$

Using

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

we get

$2\cos 10𝑥\sin (-2𝑥)=0$

So,

$\sin 2𝑥=0$

or

$\cos 10𝑥=0$

Now $𝑥\in [0,\frac{𝜋}{2})$.

From

$\sin 2𝑥=0$

we get

$𝑥=0$

This gives $1$ solution.

From

$\cos 10𝑥=0$

we get

$10𝑥=\frac{(2𝑘+1)𝜋}{2}$

So,

$𝑥=\frac{(2𝑘+1)𝜋}{20}$

Now,

$0\le 𝑥<\frac{𝜋}{2}$

So possible values are

$𝑥=\frac{𝜋}{20},\frac{3𝜋}{20},\frac{5𝜋}{20},\frac{7𝜋}{20},\frac{9𝜋}{20}$

These give $5$ solutions.

None of these make $\cos 5𝑥=0$, so all are valid.

Total number of solutions:

$1+5=6$

So the correct answer is

$\overline{)6}$

Option D.

The line is

$𝑝𝑥+𝑞𝑦=𝑟$

To find the triangle formed with the coordinate axes, we first find its intercepts.

For the $𝑥$-intercept, put $𝑦=0$:

$𝑝𝑥=𝑟$

$𝑥=\frac{𝑟}{𝑝}$

For the $𝑦$-intercept, put $𝑥=0$:

$𝑞𝑦=𝑟$

$𝑦=\frac{𝑟}{𝑞}$

So, the triangle formed with the coordinate axes has base $\frac{𝑟}{𝑝}$ and height $\frac{𝑟}{𝑞}$.

Its area is

$\frac{1}{2}\times \frac{𝑟}{𝑝}\times \frac{𝑟}{𝑞}$

Given area is $\frac{1}{54}$, so

$\frac{1}{2}\cdot \frac{{𝑟}^2}{𝑝𝑞}=\frac{1}{54}$

$\frac{{𝑟}^2}{2𝑝𝑞}=\frac{1}{54}$

Cross-multiplying,

$54{𝑟}^2=2𝑝𝑞$

$𝑝𝑞=27{𝑟}^2$

Now, three positive numbers $𝑝,3\sqrt{\surd 3}𝑟,𝑞$ will be in G.P. if

$(3\sqrt{\surd 3}𝑟{)}^2=𝑝𝑞$

Now,

$(3\sqrt{\surd 3}𝑟{)}^2=9\cdot 3{𝑟}^2=27{𝑟}^2$

And we already got

$𝑝𝑞=27{𝑟}^2$

So,

$(3\sqrt{\surd 3}𝑟{)}^2=𝑝𝑞$

Hence, $𝑝,3\sqrt{\surd 3}𝑟,𝑞$ are in G.P.

So the correct option is:

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

The given line is

$3𝑥+𝑦-6=0$

Writing it in slope form,

$𝑦=-3𝑥+6$

So its slope is

${𝑚}_1=-3$

Let the slope of a required line through the origin be $𝑚$.

The angle between two lines with slopes $𝑚$ and ${𝑚}_1$ is given by

$\tan 𝜃=\mid \frac{𝑚-{𝑚}_1}{1+𝑚{𝑚}_1}\mid$

Here,

$𝜃=\frac{𝜋}{4}$

So,

$\tan \frac{𝜋}{4}=1$

Using ${𝑚}_1=-3$,

$\mid \frac{𝑚+3}{1-3𝑚}\mid =1$

So,

$\frac{𝑚+3}{1-3𝑚}=\pm 1$

First,

$𝑚+3=1-3𝑚$

$4𝑚=-2$

$𝑚=-\frac{1}{2}$

Second,

$𝑚+3=-(1-3𝑚)$

$𝑚+3=-1+3𝑚$

$2𝑚=4$

$𝑚=2$

So the two required lines through the origin are

$𝑦=2𝑥$

and

$𝑦=-\frac{𝑥}{2}$

Their equations are

$𝑦-2𝑥=0$

and

$2𝑦+𝑥=0$

The joint equation is

$(𝑦-2𝑥)(2𝑦+𝑥)=0$

Expanding,

$2{𝑦}^2+𝑥𝑦-4𝑥𝑦-2{𝑥}^2=0$

$2{𝑦}^2-3𝑥𝑦-2{𝑥}^2=0$

Multiplying by $-1$,

$2{𝑥}^2+3𝑥𝑦-2{𝑦}^2=0$

So the correct option is

$\overline{)\text{Option A }2{𝑥}^2+3𝑥𝑦-2{𝑦}^2=0}$

The point of contact will lie on the given line, and the radius drawn to the point of contact is perpendicular to the tangent line.

So, we need the foot of the perpendicular from the centre $(-1,1)$ to the line

$𝑥+2𝑦+4=0$

For a line $𝑎𝑥+𝑏𝑦+𝑐=0$, the foot of perpendicular from $({𝑥}_1,{𝑦}_1)$ is

$({𝑥}_1-\frac{𝑎(𝑎{𝑥}_1+𝑏{𝑦}_1+𝑐)}{{𝑎}^2+{𝑏}^2},{𝑦}_1-\frac{𝑏(𝑎{𝑥}_1+𝑏{𝑦}_1+𝑐)}{{𝑎}^2+{𝑏}^2})$

Here,

$𝑎=1,𝑏=2,𝑐=4$

and the centre is

$({𝑥}_1,{𝑦}_1)=(-1,1)$

Now,

$𝑎{𝑥}_1+𝑏{𝑦}_1+𝑐=1(-1)+2\left(1\right)+4$

$=-1+2+4=5$

Also,

${𝑎}^2+{𝑏}^2=1^2+2^2=5$

So the point of contact is

$(-1-\frac{1\left(5\right)}{5},1-\frac{2\left(5\right)}{5})$

$=(-1-1,1-2)$

$=(-2,-1)$

So the correct option is

$\overline{)\text{Option A }(-2,-1)}$

The ordinate means the $𝑦$-coordinate, and the abscissa means the $𝑥$-coordinate.

Given parabola:

${𝑦}^2=\frac{36}{5}𝑥$

The ordinate increases at thrice the rate of the abscissa means

$\frac{𝑑𝑦}{𝑑𝑥}=3$

Now differentiate the equation with respect to $𝑥$:

$\frac{𝑑}{𝑑𝑥}\left({𝑦}^2\right)=\frac{𝑑}{𝑑𝑥}\left(\frac{36}{5}𝑥\right)$

$2𝑦\frac{𝑑𝑦}{𝑑𝑥}=\frac{36}{5}$

So,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{18}{5𝑦}$

Since $\frac{𝑑𝑦}{𝑑𝑥}=3$,

$3=\frac{18}{5𝑦}$

$15𝑦=18$

$𝑦=\frac{6}{5}$

Now substitute $𝑦=\frac{6}{5}$ in the parabola:

${\left(\frac{6}{5}\right)}^2=\frac{36}{5}𝑥$

$\frac{36}{25}=\frac{36}{5}𝑥$

$𝑥=\frac{1}{5}$

So the required point is

$(\frac{1}{5},\frac{6}{5})$

Hence, the correct option is Option B.

For the parabola

${𝑥}^2=48𝑦$

compare it with the standard form

${𝑥}^2=4𝑎𝑦$

So,

$4𝑎=48$

$𝑎=12$

The vertex of the parabola is

$(0,0)$

For the parabola ${𝑥}^2=4𝑎𝑦$, the latus rectum is the line

$𝑦=𝑎$

So here,

$𝑦=12$

Now put $𝑦=12$ in the parabola:

${𝑥}^2=48\left(12\right)$

${𝑥}^2=576$

$𝑥=\pm 24$

So the ends of the latus rectum are

$(-24,12)\text{and}(24,12)$

The triangle is formed by these two points and the vertex $(0,0)$.

The base of the triangle is the distance between $(-24,12)$ and $(24,12)$:

$\text{Base}=48$

The height is the perpendicular distance from $(0,0)$ to the line $𝑦=12$:

$\text{Height}=12$

So the area is

$\frac{1}{2}\times 48\times 12=288$

So the correct option is

$\overline{)288\text{ sq. units}}$

Option B.

For $𝑥$ close to $0$, both $2+𝑥$ and $2-𝑥$ are positive, so

$\log |2+𝑥|=\log (2+𝑥),\log |2-𝑥|=\log (2-𝑥)$

The limit becomes

$\underset{𝑥\to 0}{lim}\frac{\log (2+𝑥)-\log (2-𝑥)}{\tan 𝑥}$

Use the property of logarithms:

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

As $𝑥\to 0$, the numerator and denominator both tend to $0$, so we can use L’Hospital’s rule.

Differentiate numerator:

$\frac{𝑑}{𝑑𝑥}[\log (2+𝑥)-\log (2-𝑥\left)\right]=\frac{1}{2+𝑥}+\frac{1}{2-𝑥}$

Differentiate denominator:

$\frac{𝑑}{𝑑𝑥}(\tan 𝑥)={\sec }^2𝑥$

So the limit is

$\underset{𝑥\to 0}{lim}\frac{\frac{1}{2+𝑥}+\frac{1}{2-𝑥}}{{\sec }^2𝑥}$

Put $𝑥=0$:

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

So,

$\underset{𝑥\to 0}{lim}\left[\frac{\log |2+𝑥|-\log |2-𝑥|}{\tan 𝑥}\right]=1$

Correct option: D $1$

For A,

$(𝑞\Rightarrow 𝑝)\vee (𝑝\Rightarrow 𝑞)$

We know that

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

and

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

So,

$(𝑞\Rightarrow 𝑝)\vee (𝑝\Rightarrow 𝑞)$

becomes

$(\sim 𝑞\vee 𝑝)\vee (\sim 𝑝\vee 𝑞)$

This contains both $𝑝\vee \sim 𝑝$ and $𝑞\vee \sim 𝑞$, so it is always true.

Hence, A is a tautology.

For B,

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

By De Morgan’s law,

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

So both sides of the biconditional are the same.

Hence, B is also a tautology.

For C,

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

This can be written as

$(𝑝\vee 𝑞)\wedge \sim 𝑝\wedge \sim 𝑞$

Now $\sim 𝑝\wedge \sim 𝑞$ means both $𝑝$ and $𝑞$ are false.

But if both $𝑝$ and $𝑞$ are false, then $𝑝\vee 𝑞$ is false.

So the whole statement is always false.

Hence, C is a contradiction.

For D,

$(𝑝\wedge 𝑞)\wedge (\sim 𝑝\vee \sim 𝑞)$

Here $𝑝\wedge 𝑞$ means both $𝑝$ and $𝑞$ are true.

But if both are true, then $\sim 𝑝\vee \sim 𝑞$ is false.

So the whole statement is always false.

Hence, D is a contradiction.

So the correct option is:

$\overline{)\text{Option B: A and B are tautologies, C and D are contradictions.}}$

The statement is

$(𝑝\wedge 𝑞)\Rightarrow (𝑟\vee \sim 𝑠)$

An implication of the form $𝐴\Rightarrow 𝐵$ is false only in one case:

$𝐴\text{ is true and }𝐵\text{ is false}$

So here,

$(𝑝\wedge 𝑞)$

must be true, and

$(𝑟\vee \sim 𝑠)$

must be false.

For $𝑝\wedge 𝑞$ to be true, both $𝑝$ and $𝑞$ must be true.

So,

$𝑝=𝑇,𝑞=𝑇$

Now, for $𝑟\vee \sim 𝑠$ to be false, both parts must be false.

So,

$𝑟=𝐹$

and

$\sim 𝑠=𝐹$

This means

$𝑠=𝑇$

Hence the truth values are

$𝑝=𝑇,𝑞=𝑇,𝑟=𝐹,𝑠=𝑇$

So the correct option is

$\overline{)\text{Option B: T, T, F, T}}$

Let

$\frac{𝑠-𝑎}{11}=\frac{𝑠-𝑏}{12}=\frac{𝑠-𝑐}{13}=𝑘$

So,

$𝑠-𝑎=11𝑘,𝑠-𝑏=12𝑘,𝑠-𝑐=13𝑘$

Now we know that

$(𝑠-𝑎)+(𝑠-𝑏)+(𝑠-𝑐)=𝑠$

because

$3𝑠-(𝑎+𝑏+𝑐)=3𝑠-2𝑠=𝑠$

So,

$𝑠=11𝑘+12𝑘+13𝑘=36𝑘$

For a triangle,

${\tan }^2\frac{𝐴}{2}=\frac{(𝑠-𝑏)(𝑠-𝑐)}{𝑠(𝑠-𝑎)}$

Substitute the values:

${\tan }^2\frac{𝐴}{2}=\frac{\left(12𝑘\right)\left(13𝑘\right)}{\left(36𝑘\right)\left(11𝑘\right)}$

$=\frac{156}{396}=\frac{13}{33}$

Given,

$𝜆{\tan }^2\frac{𝐴}{2}=455$

So,

$𝜆\cdot \frac{13}{33}=455$

$𝜆=455\times \frac{33}{13}$

Since $455=35\times 13$,

$𝜆=35\times 33=1155$

So the correct option is

$\overline{)1155}$

Option A.

In usual notation, $𝑎$ is opposite to angle $𝐴$.

Given,

$2{𝑎}^2={𝑏}^2+{𝑐}^2$

So,

${𝑎}^2=\frac{{𝑏}^2+{𝑐}^2}{2}$

Using cosine rule,

${𝑎}^2={𝑏}^2+{𝑐}^2-2𝑏𝑐\cos 𝐴$

Substitute the given value of ${𝑎}^2$:

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

So,

$2𝑏𝑐\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2}{2}$

$\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2}{4𝑏𝑐}$

Now use the identity

$\cos 3𝐴=4{\cos }^3𝐴-3\cos 𝐴$

Dividing by $\cos 𝐴$,

$\frac{\cos 3𝐴}{\cos 𝐴}=4{\cos }^2𝐴-3$

So,

$\frac{\cos 3𝐴}{\cos 𝐴}+2=4{\cos }^2𝐴-1$

Substitute $\cos 𝐴=\frac{{𝑏}^2+{𝑐}^2}{4𝑏𝑐}$:

$4{\cos }^2𝐴-1=4{\left(\frac{{𝑏}^2+{𝑐}^2}{4𝑏𝑐}\right)}^2-1$

$=\frac{({𝑏}^2+{𝑐}^2{)}^2}{4{𝑏}^2{𝑐}^2}-1$

$=\frac{({𝑏}^2+{𝑐}^2{)}^2-4{𝑏}^2{𝑐}^2}{4{𝑏}^2{𝑐}^2}$

$=\frac{({𝑏}^2-{𝑐}^2{)}^2}{4{𝑏}^2{𝑐}^2}$

$={\left(\frac{{𝑏}^2-{𝑐}^2}{2𝑏𝑐}\right)}^2$

So the correct option is

$\overline{){\left(\frac{{𝑏}^2-{𝑐}^2}{2𝑏𝑐}\right)}^2}$

Option B.

In $△𝐴𝐵𝐶$, using the half-angle formulae,

${\cos }^2\frac{𝐴}{2}=\frac{1+\cos 𝐴}{2}$

and

${\sin }^2\frac{𝐴}{2}=\frac{1-\cos 𝐴}{2}$

Given expression is

$(𝑏-𝑐{)}^2{\cos }^2\frac{𝐴}{2}+(𝑏+𝑐{)}^2{\sin }^2\frac{𝐴}{2}$

Substitute the formulae:

$=(𝑏-𝑐{)}^2\cdot \frac{1+\cos 𝐴}{2}+(𝑏+𝑐{)}^2\cdot \frac{1-\cos 𝐴}{2}$

$=\frac{1}{2}\left[\right(𝑏-𝑐{)}^2(1+\cos 𝐴)+(𝑏+𝑐{)}^2(1-\cos 𝐴\left)\right]$

Open the brackets:

$=\frac{1}{2}\left[\right(𝑏-𝑐{)}^2+(𝑏+𝑐{)}^2+\cos 𝐴\{(𝑏-𝑐{)}^2-(𝑏+𝑐{)}^2\left\}\right]$

Now,

$(𝑏-𝑐{)}^2+(𝑏+𝑐{)}^2=2{𝑏}^2+2{𝑐}^2$

and

$(𝑏-𝑐{)}^2-(𝑏+𝑐{)}^2=-4𝑏𝑐$

So the expression becomes

$=\frac{1}{2}[2{𝑏}^2+2{𝑐}^2-4𝑏𝑐\cos 𝐴]$

$={𝑏}^2+{𝑐}^2-2𝑏𝑐\cos 𝐴$

By cosine rule,

${𝑎}^2={𝑏}^2+{𝑐}^2-2𝑏𝑐\cos 𝐴$

So,

$(𝑏-𝑐{)}^2{\cos }^2\frac{𝐴}{2}+(𝑏+𝑐{)}^2{\sin }^2\frac{𝐴}{2}={𝑎}^2$

Hence, the correct option is

$\overline{)\text{Option A }{𝑎}^2}$

After applying the row transformation ${𝑅}_1\Rightarrow {𝑅}_1+3{𝑅}_2$, only the first row changes.

Given,

$𝐴=\left[\begin{array}{cc}-5& -3\\ 2& 1\end{array}\right]$

Now,

${𝑅}_1={𝑅}_1+3{𝑅}_2$

So,

${𝑅}_1=[-5-3]+3\left[21\right]$

${𝑅}_1=[-5-3]+\left[63\right]$

${𝑅}_1=\left[10\right]$

The second row remains the same.

So the new matrix is

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

This matrix has zero above the main diagonal. So it is a lower triangular matrix.

Hence, the correct option is

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

Since ${𝐴}^{-1}$ is given, we use the fact that

$𝐴{𝐴}^{-1}=𝐼$

The first column of ${𝐴}^{-1}$ is

$⎡⎣⎢\begin{array}{c}𝛾\\ 𝛼\\ 𝛽\end{array}⎤⎦⎥$

So when we multiply $𝐴$ by this column, we should get the first column of the identity matrix:

$𝐴⎡⎣⎢\begin{array}{c}𝛾\\ 𝛼\\ 𝛽\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}1\\ 0\\ 0\end{array}⎤⎦⎥$

Now,

$⎡⎣⎢\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}𝛾\\ 𝛼\\ 𝛽\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{c}2𝛾-𝛽\\ 5𝛾+𝛼\\ 𝛼+3𝛽\end{array}⎤⎦⎥$

So we get

$2𝛾-𝛽=1$

$5𝛾+𝛼=0$

$𝛼+3𝛽=0$

From

$5𝛾+𝛼=0$

we get

$𝛼=-5𝛾$

From

$𝛼+3𝛽=0$

we get

$𝛼=-3𝛽$

So,

$-5𝛾=-3𝛽$

$𝛽=\frac{5𝛾}{3}$

Now use

$2𝛾-𝛽=1$

Substitute $𝛽=\frac{5𝛾}{3}$:

$2𝛾-\frac{5𝛾}{3}=1$

$\frac{6𝛾-5𝛾}{3}=1$

$\frac{𝛾}{3}=1$

$𝛾=3$

Then,

$𝛼=-5𝛾=-15$

and

$𝛽=\frac{5𝛾}{3}=5$

Now,

$|𝛼\cdot 𝛽\cdot 𝛾|=\left|\right(-15\left)\right(5\left)\right(3\left)\right|$

$=|-225|$

$=225$

So the correct option is

$\overline{)225}$

Option C.

Put

$𝑎=\sqrt{\surd 1+{𝑥}^2},𝑏=\sqrt{\surd 1-{𝑥}^2}$

Then

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

So,

$\tan 𝑦=\frac{𝑎-𝑏}{𝑎+𝑏}$

Now use the formula

$\tan 2𝑦=\frac{2\tan 𝑦}{1-{\tan }^2𝑦}$

Substitute $\tan 𝑦=\frac{𝑎-𝑏}{𝑎+𝑏}$:

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

Simplify the denominator:

$1-{\left(\frac{𝑎-𝑏}{𝑎+𝑏}\right)}^2=\frac{(𝑎+𝑏{)}^2-(𝑎-𝑏{)}^2}{(𝑎+𝑏{)}^2}=\frac{4𝑎𝑏}{(𝑎+𝑏{)}^2}$

So,

$\tan 2𝑦=\frac{2(𝑎-𝑏)}{𝑎+𝑏}\cdot \frac{(𝑎+𝑏{)}^2}{4𝑎𝑏}$

$\tan 2𝑦=\frac{(𝑎-𝑏)(𝑎+𝑏)}{2𝑎𝑏}=\frac{{𝑎}^2-{𝑏}^2}{2𝑎𝑏}$

Now,

${𝑎}^2=1+{𝑥}^2,{𝑏}^2=1-{𝑥}^2$

So,

${𝑎}^2-{𝑏}^2=(1+{𝑥}^2)-(1-{𝑥}^2)=2{𝑥}^2$

Also,

$𝑎𝑏=\sqrt{\surd 1+{𝑥}^2}\sqrt{\surd 1-{𝑥}^2}=\sqrt{\surd 1-{𝑥}^4}$

Hence,

$\tan 2𝑦=\frac{2{𝑥}^2}{2\sqrt{\surd 1-{𝑥}^4}}$

$\tan 2𝑦=\frac{{𝑥}^2}{\sqrt{\surd 1-{𝑥}^4}}$

Now recall that if

$𝜃={\sin }^{-1}\left({𝑥}^2\right)$

then

$\tan 𝜃=\frac{{𝑥}^2}{\sqrt{\surd 1-{𝑥}^4}}$

So,

$2𝑦={\sin }^{-1}\left({𝑥}^2\right)$

Thus,

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

Differentiate with respect to $𝑥$:

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{2}\cdot \frac{1}{\sqrt{\surd 1-{𝑥}^4}}\cdot 2𝑥$

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑥}{\sqrt{\surd 1-{𝑥}^4}}$

So the correct option is

$\overline{)\text{Option B }\frac{𝑥}{\sqrt{\surd 1-{𝑥}^4}}}$

The given expression is

${\tan }^{-1}\left(\sqrt{\surd 3}\right)+{\sec }^{-1}(-2)-{\sin }^{-1}(-\frac{1}{2})$

Using the principal value ranges of inverse trigonometric functions:

${\tan }^{-1}\left(\sqrt{\surd 3}\right)=\frac{𝜋}{3}$

because

$\tan \frac{𝜋}{3}=\sqrt{\surd 3}$

Now,

${\sec }^{-1}(-2)$

means we need an angle whose secant is $-2$.

Since

$\sec 𝜃=-2$

we get

$\cos 𝜃=-\frac{1}{2}$

In the principal range of ${\sec }^{-1}𝑥$, the angle is

$\frac{2𝜋}{3}$

So,

${\sec }^{-1}(-2)=\frac{2𝜋}{3}$

Also,

${\sin }^{-1}(-\frac{1}{2})=-\frac{𝜋}{6}$

because

$\sin (-\frac{𝜋}{6})=-\frac{1}{2}$

Now substitute these values:

$\frac{𝜋}{3}+\frac{2𝜋}{3}-(-\frac{𝜋}{6})$

$=𝜋+\frac{𝜋}{6}$

$=\frac{7𝜋}{6}$

So the value is

$\overline{)\frac{7𝜋}{6}}$

Correct option: Option C.

For the function

$𝑓\left(𝑥\right)={𝑒}^{\sqrt{\surd 5𝑥-3-2{𝑥}^2}},$

the exponential function ${𝑒}^{𝑥}$ is defined for all real numbers.

So we only need to make sure that the expression inside the square root is non-negative, because square root is defined for real values only when the radicand is $\ge 0$.

So,

$5𝑥-3-2{𝑥}^2\ge 0$

Rearrange it:

$-2{𝑥}^2+5𝑥-3\ge 0$

Multiply by $-1$. The inequality sign changes:

$2{𝑥}^2-5𝑥+3\le 0$

Now factorise:

$2{𝑥}^2-5𝑥+3=(2𝑥-3)(𝑥-1)$

So,

$(2𝑥-3)(𝑥-1)\le 0$

The zeroes are

$𝑥=1$

and

$𝑥=\frac{3}{2}$

Since the quadratic expression is less than or equal to zero between its roots, we get

$𝑥\in [1,\frac{3}{2}]$

So, the domain is

$\overline{)[1,\frac{3}{2}]}$

Correct option: Option C.

For continuity at $𝑥=0$, we must have

$𝑘=\underset{𝑥\to 0}{lim}\frac{(8-2𝑥{)}^{\frac{1}{3}}-2}{3-(243+5𝑥{)}^{\frac{1}{5}}}$

At $𝑥=0$, the numerator and denominator both become $0$, so we simplify using identities.

Let

$𝑎=(8-2𝑥{)}^{\frac{1}{3}}$

Then ${𝑎}^3=8-2𝑥$ and $2^3=8$.

So,

${𝑎}^3-2^3=(8-2𝑥)-8=-2𝑥$

Using

${𝑎}^3-{𝑏}^3=(𝑎-𝑏)({𝑎}^2+𝑎𝑏+{𝑏}^2)$

we get

$(8-2𝑥{)}^{\frac{1}{3}}-2=\frac{-2𝑥}{{\left(\right(8-2𝑥{)}^{\frac{1}{3}})}^2+2(8-2𝑥{)}^{\frac{1}{3}}+4}$