MHT CET 2026 15th April Morning Shift (Mathematics) Question Paper with Solutions

Since $𝜔$ is a complex cube root of unity, we use the standard NCERT results:

${𝜔}^3=1,1+𝜔+{𝜔}^2=0$

Also,

$\frac{1}{𝜔}={𝜔}^2,\frac{1}{{𝜔}^2}=𝜔$

So the given expression becomes

$2(1+{𝜔}^2)(1+𝜔)+3(2+{𝜔}^2)(2+𝜔)+\cdots +(𝑛+1)(𝑛+{𝜔}^2)(𝑛+𝜔)$

The general term is

$(𝑘+1)(𝑘+{𝜔}^2)(𝑘+𝜔)$

Now simplify the product:

$(𝑘+{𝜔}^2)(𝑘+𝜔)={𝑘}^2+𝑘(𝜔+{𝜔}^2)+{𝜔}^3$

Since $𝜔+{𝜔}^2=-1$ and ${𝜔}^3=1$,

$(𝑘+{𝜔}^2)(𝑘+𝜔)={𝑘}^2-𝑘+1$

So the general term becomes

$(𝑘+1)({𝑘}^2-𝑘+1)$

Using the identity

$(𝑘+1)({𝑘}^2-𝑘+1)={𝑘}^3+1$

So the whole expression is

$\sum _{𝑘=1}^{𝑛}({𝑘}^3+1)$

That is,

$\sum _{𝑘=1}^{𝑛}{𝑘}^3+\sum _{𝑘=1}^{𝑛}1$

We know

$\sum _{𝑘=1}^{𝑛}{𝑘}^3={\left[\frac{𝑛(𝑛+1)}{2}\right]}^2$

and

$\sum _{𝑘=1}^{𝑛}1=𝑛$

Hence the value is

${\left[\frac{𝑛(𝑛+1)}{2}\right]}^2+𝑛$

So the correct option is

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

The digits are $2,3,5,7,9$ and we have to form five-digit numbers without repetition.

Since there are exactly $5$ digits given and the number is also five-digit, each number will use all the digits once.

Total possible numbers are

$5!=120$

Now we need numbers greater than $24000$.

Any number starting with $3,5,7,$ or $9$ will definitely be greater than $24000$.

Number of such numbers:

$4\times 4!=4\times 24=96$

Now consider numbers starting with $2$.

To be greater than $24000$, the second digit must be greater than $4$.

From the remaining digits $3,5,7,9$, the possible second digits are $5,7,9$.

So number of such numbers:

$3\times 3!=3\times 6=18$

Total numbers greater than $24000$ are

$96+18=114$

So the correct option is

$\overline{)114}$

Option C.

In $△𝐴𝐵𝐶$, usual notation means:

• $𝑎$ is opposite $\angle 𝐴$

• $𝑏$ is opposite $\angle 𝐵$

• $𝑐$ is opposite $\angle 𝐶$

Given $\angle 𝐶={90}^{\circ }$, so $𝑐$ is the hypotenuse.

We need to find $\sin (𝐴-𝐵)$.

Using the formula,

$\sin (𝐴-𝐵)=\sin 𝐴\cos 𝐵-\cos 𝐴\sin 𝐵$

In the right triangle,

$\sin 𝐴=\frac{𝑎}{𝑐},\cos 𝐴=\frac{𝑏}{𝑐}$

and

$\sin 𝐵=\frac{𝑏}{𝑐},\cos 𝐵=\frac{𝑎}{𝑐}$

So,

$\sin (𝐴-𝐵)=\frac{𝑎}{𝑐}\cdot \frac{𝑎}{𝑐}-\frac{𝑏}{𝑐}\cdot \frac{𝑏}{𝑐}$

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

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

Since $\angle 𝐶={90}^{\circ }$, by Pythagoras theorem,

${𝑐}^2={𝑎}^2+{𝑏}^2$

Hence,

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

So the correct option is

$\overline{)\text{Option D }\frac{{𝑎}^2-{𝑏}^2}{{𝑎}^2+{𝑏}^2}}$

Given $𝑃=\tan {20}^{\circ }$.

The expression is

$\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }\tan {110}^{\circ }}$

We know the identity

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

So here,

$\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }\tan {110}^{\circ }}=\tan ({160}^{\circ }-{110}^{\circ })$

$=\tan {50}^{\circ }$

Now,

$\tan {50}^{\circ }=\tan ({90}^{\circ }-{40}^{\circ })=\cot {40}^{\circ }$

So,

$\tan {50}^{\circ }=\frac{1}{\tan {40}^{\circ }}$

Using the double angle formula,

$\tan {40}^{\circ }=\tan (2\times {20}^{\circ })$

$\tan {40}^{\circ }=\frac{2\tan {20}^{\circ }}{1-{\tan }^2{20}^{\circ }}$

Since $𝑃=\tan {20}^{\circ }$,

$\tan {40}^{\circ }=\frac{2𝑃}{1-{𝑃}^2}$

Hence,

$\tan {50}^{\circ }=\frac{1}{\tan {40}^{\circ }}$

$=\frac{1}{\frac{2𝑃}{1-{𝑃}^2}}$

$=\frac{1-{𝑃}^2}{2𝑃}$

So the value is

$\overline{)\frac{1-{𝑃}^2}{2𝑃}}$

Correct option: D.

Using the identity

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

we take

$𝐴=𝑝𝜃,𝐵=𝑞𝜃$

So,

$\cos \left(𝑝𝜃\right)+\cos \left(𝑞𝜃\right)=2\cos \frac{(𝑝+𝑞)𝜃}{2}\cos \frac{(𝑝-𝑞)𝜃}{2}$

Given,

$\cos \left(𝑝𝜃\right)+\cos \left(𝑞𝜃\right)=0$

So,

$2\cos \frac{(𝑝+𝑞)𝜃}{2}\cos \frac{(𝑝-𝑞)𝜃}{2}=0$

This means either

$\cos \frac{(𝑝+𝑞)𝜃}{2}=0$

or

$\cos \frac{(𝑝-𝑞)𝜃}{2}=0$

Now we know,

$\cos 𝑥=0\Rightarrow 𝑥=\frac{(2𝑛+1)𝜋}{2}$

So,

$\frac{(𝑝+𝑞)𝜃}{2}=\frac{(2𝑛+1)𝜋}{2}$

which gives

$𝜃=\frac{(2𝑛+1)𝜋}{𝑝+𝑞}$

Similarly,

$\frac{(𝑝-𝑞)𝜃}{2}=\frac{(2𝑛+1)𝜋}{2}$

which gives

$𝜃=\frac{(2𝑛+1)𝜋}{𝑝-𝑞}$

Hence the general value is

$\overline{)𝜃=\frac{(2𝑛+1)𝜋}{𝑝\pm 𝑞}}$

So the correct option is

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

The orthocenter is the point where the three altitudes of a triangle meet.

For triangle $𝐴𝐵𝐶$,

$𝐴(1,-6),𝐵(2,-3),𝐶(3,-2)$

First find the slope of $𝐴𝐵$:

${𝑚}_{𝐴𝐵}=\frac{-3-(-6)}{2-1}=\frac{3}{1}=3$

So the altitude from $𝐶$ will be perpendicular to $𝐴𝐵$.

The slope of a line perpendicular to a line of slope $3$ is:

$-\frac{1}{3}$

Equation of altitude from $𝐶(3,-2)$:

$𝑦+2=-\frac{1}{3}(𝑥-3)$

$𝑦=-\frac{𝑥}{3}-1$

Now find the slope of $𝐵𝐶$:

${𝑚}_{𝐵𝐶}=\frac{-2-(-3)}{3-2}=\frac{1}{1}=1$

So the altitude from $𝐴$ will be perpendicular to $𝐵𝐶$.

The slope of this altitude is:

$-1$

Equation of altitude from $𝐴(1,-6)$:

$𝑦+6=-(𝑥-1)$

$𝑦=-𝑥-5$

Now solve the two altitude equations:

$𝑦=-\frac{𝑥}{3}-1$

and

$𝑦=-𝑥-5$

Equating them,

$-\frac{𝑥}{3}-1=-𝑥-5$

$-\frac{𝑥}{3}+𝑥=-5+1$

$\frac{2𝑥}{3}=-4$

$𝑥=-6$

Substitute $𝑥=-6$ in $𝑦=-𝑥-5$:

$𝑦=-(-6)-5=6-5=1$

So the orthocenter is:

$(-6,1)$

Hence, the correct option is Option A.

The given equation is

${𝑥}^2-3𝑥𝑦+2{𝑦}^2=0$

Factorising it,

${𝑥}^2-3𝑥𝑦+2{𝑦}^2=(𝑥-𝑦)(𝑥-2𝑦)=0$

So the two lines are

$𝑥-𝑦=0\text{and}𝑥-2𝑦=0$

That is,

$𝑦=𝑥$

and

$𝑦=\frac{𝑥}{2}$

Their slopes are

${𝑚}_1=1,{𝑚}_2=\frac{1}{2}$

The angle $𝜃$ between two lines with slopes ${𝑚}_1$ and ${𝑚}_2$ is given by

$\tan 𝜃=\mid \frac{{𝑚}_1-{𝑚}_2}{1+{𝑚}_1{𝑚}_2}\mid$

So,

$\tan 𝜃=\mid \frac{1-\frac{1}{2}}{1+1\cdot \frac{1}{2}}\mid$

$\tan 𝜃=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{3}$

Since $𝜃$ is acute,

$\tan 𝜃=\frac{\sin 𝜃}{\cos 𝜃}=\frac{1}{3}$

So we can take

$\sin 𝜃=\frac{1}{\sqrt{\surd 10}},\cos 𝜃=\frac{3}{\sqrt{\surd 10}}$

Now,

$\frac{3\sin 𝜃+2\cos 𝜃}{3\sin 𝜃-2\cos 𝜃}=\frac{3\cdot \frac{1}{\sqrt{\surd 10}}+2\cdot \frac{3}{\sqrt{\surd 10}}}{3\cdot \frac{1}{\sqrt{\surd 10}}-2\cdot \frac{3}{\sqrt{\surd 10}}}$

$=\frac{\frac{3}{\sqrt{\surd 10}}+\frac{6}{\sqrt{\surd 10}}}{\frac{3}{\sqrt{\surd 10}}-\frac{6}{\sqrt{\surd 10}}}=\frac{\frac{9}{\sqrt{\surd 10}}}{-\frac{3}{\sqrt{\surd 10}}}=-3$

So the correct option is

$\overline{)-3}$

Option C.

The given pair of lines is

$2{𝑥}^2-3𝑥𝑦-2{𝑦}^2=0$

Factorising,

$2{𝑥}^2-3𝑥𝑦-2{𝑦}^2=(2𝑥+𝑦)(𝑥-2𝑦)$

So the two lines are

$2𝑥+𝑦=0$

and

$𝑥-2𝑦=0$

That means

$𝑦=-2𝑥$

and

$𝑦=\frac{𝑥}{2}$

Their slopes are $-2$ and $\frac{1}{2}$.

Since

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

the two lines are perpendicular.

So the triangle has a right angle at the origin.

Now check whether the two sides from the origin are equal.

The third line is

$3𝑥-𝑦=7$

Intersection with $𝑦=-2𝑥$:

$3𝑥-(-2𝑥)=7$

$5𝑥=7$

$𝑥=\frac{7}{5},𝑦=-\frac{14}{5}$

Distance from origin:

$\sqrt{\surd {\left(\frac{7}{5}\right)}^2+{(-\frac{14}{5})}^2}=\sqrt{\surd \frac{49}{25}+\frac{196}{25}}=\sqrt{\surd \frac{245}{25}}=\frac{7\sqrt{\surd 5}}{5}$

Intersection with $𝑦=\frac{𝑥}{2}$:

$3𝑥-\frac{𝑥}{2}=7$

$\frac{5𝑥}{2}=7$

$𝑥=\frac{14}{5},𝑦=\frac{7}{5}$

Distance from origin:

$\sqrt{\surd {\left(\frac{14}{5}\right)}^2+{\left(\frac{7}{5}\right)}^2}=\sqrt{\surd \frac{196}{25}+\frac{49}{25}}=\sqrt{\surd \frac{245}{25}}=\frac{7\sqrt{\surd 5}}{5}$

The two perpendicular sides are equal.

So the triangle is right angled isosceles.

Hence, the correct option is

$\overline{)\text{Option D: Right angled isosceles}}$

For the given equation to represent a circle, the coefficients of ${𝑥}^2$ and ${𝑦}^2$ must be equal.

The equation is

$(𝑎+1){𝑥}^2+3{𝑦}^2-6𝑥+9𝑦+𝑎+4=0$

So,

$𝑎+1=3$

$𝑎=2$

Substitute $𝑎=2$ in the equation:

$3{𝑥}^2+3{𝑦}^2-6𝑥+9𝑦+6=0$

Divide the whole equation by $3$:

${𝑥}^2+{𝑦}^2-2𝑥+3𝑦+2=0$

Compare this with the standard form

${𝑥}^2+{𝑦}^2+2𝑔𝑥+2𝑓𝑦+𝑐=0$

Here,

$2𝑔=-2\Rightarrow 𝑔=-1$

and

$2𝑓=3\Rightarrow 𝑓=\frac{3}{2}$

Also,

$𝑐=2$

The centre is

$(-𝑔,-𝑓)$

So the centre is

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

The radius is

$\sqrt{\surd {𝑔}^2+{𝑓}^2-𝑐}$

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

$=\sqrt{\surd 1+\frac{9}{4}-2}$

$=\sqrt{\surd \frac{5}{4}}$

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

So the correct option is

$\overline{)\text{Option C }(1,-\frac{3}{2}),\frac{\sqrt{\surd 5}}{2}}$

The point of contact lies on both the line and the hyperbola.

Given line:

$4𝑥+3𝑦=7$

So,

$3𝑦=7-4𝑥$

$𝑦=\frac{7-4𝑥}{3}$

The hyperbola is:

${𝑥}^2-{𝑦}^2=7$

Substitute $𝑦=\frac{7-4𝑥}{3}$ in the hyperbola:

${𝑥}^2-{\left(\frac{7-4𝑥}{3}\right)}^2=7$

${𝑥}^2-\frac{(7-4𝑥{)}^2}{9}=7$

Multiply by $9$:

$9{𝑥}^2-(7-4𝑥{)}^2=63$

Now expand:

$9{𝑥}^2-(49-56𝑥+16{𝑥}^2)=63$

$9{𝑥}^2-49+56𝑥-16{𝑥}^2=63$

$-7{𝑥}^2+56𝑥-49=63$

$-7{𝑥}^2+56𝑥-112=0$

Divide by $-7$:

${𝑥}^2-8𝑥+16=0$

$(𝑥-4{)}^2=0$

So,

$𝑥=4$

Now put $𝑥=4$ in the line:

$4\left(4\right)+3𝑦=7$

$16+3𝑦=7$

$3𝑦=-9$

$𝑦=-3$

So the point of contact is:

$(4,-3)$

Sum of the co-ordinates:

$4+(-3)=1$

So the correct option is:

$\overline{)1}$

Option C.

We use the standard limits:

$\underset{𝑥\to 0}{lim}\frac{{𝑎}^{𝑥}-1}{𝑥}=\log 𝑎$

$\underset{𝑥\to 0}{lim}\frac{\tan 𝑥}{𝑥}=1$

$\underset{𝑥\to 0}{lim}\frac{\log (1+𝑥)}{𝑥}=1$

Given limit is

$\underset{𝑥\to 0}{lim}\frac{(4^{𝑥}-1{)}^3}{\tan \left(\frac{𝑥}{4}\right)\log (1+\frac{{𝑥}^2}{3})}$

For small $𝑥$,

$4^{𝑥}-1\sim 𝑥\log 4$

So,

$(4^{𝑥}-1{)}^3\sim {𝑥}^3(\log 4{)}^3$

Also,

$\tan \left(\frac{𝑥}{4}\right)\sim \frac{𝑥}{4}$

and

$\log (1+\frac{{𝑥}^2}{3})\sim \frac{{𝑥}^2}{3}$

So the denominator becomes

$\frac{𝑥}{4}\cdot \frac{{𝑥}^2}{3}=\frac{{𝑥}^3}{12}$

Hence the limit is

$\frac{{𝑥}^3(\log 4{)}^3}{\frac{{𝑥}^3}{12}}$

$=12(\log 4{)}^3$

Now,

$\log 4=\log \left(2^2\right)=2\log 2$

So,

$12(\log 4{)}^3=12(2\log 2{)}^3$

$=12\cdot 8(\log 2{)}^3$

$=96(\log 2{)}^3$

Comparing with

$96(\log 𝑎{)}^{𝑏}$

we get

$𝑎=2,𝑏=3$

So,

$𝑎+𝑏=2+3=5$

Answer: Option A, $5$

The given statement is

$𝑝\to (𝑞\to 𝑟)$

First, rewrite the implication in a simpler equivalent form.

We know that

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

So,

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

Also,

$𝑝\to 𝑠<text>\equiv </text><text>\sim </text>𝑝\vee 𝑠$

Hence,

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

Now, notice that this is also equivalent to

$(𝑝\wedge 𝑞)\to 𝑟$

Now take the inverse.

The inverse of

$(𝑝\wedge 𝑞)\to 𝑟$

is

$\sim (𝑝\wedge 𝑞)\to \sim 𝑟$

Now take its converse.

The converse becomes

$\sim 𝑟\to \sim (𝑝\wedge 𝑞)$

Using De Morgan’s law,

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

So we get

$\sim 𝑟\to (\sim 𝑝\vee \sim 𝑞)$

Convert implication into OR form:

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

Now take the dual.

In the dual, $\vee$ changes to $\wedge$, while negations remain unchanged.

So the dual is

$𝑟\wedge \sim 𝑝\wedge \sim 𝑞$

This can be written as

$\sim (𝑝\vee 𝑞\vee \sim 𝑟)$

Now,

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

So,

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

Thus,

$\sim [𝑝\vee (𝑟\to 𝑞\left)\right]<text>\equiv </text><text>\sim </text>(𝑝\vee 𝑞\vee \sim 𝑟)$

So the required statement is

$\overline{)\sim [𝑝\vee (𝑟\to 𝑞\left)\right]}$

Hence, the correct option is

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

Given truth values are:

$𝑝=𝑇,𝑞=𝐹,𝑟=𝐹$

The statement is:

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

First find the needed negations:

$\sim 𝑞=𝑇,\sim 𝑟=𝑇$

Now,

$𝑝\wedge \sim 𝑞=𝑇\wedge 𝑇=𝑇$

So,

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

Also,

$𝑞\vee \sim 𝑟=𝐹\vee 𝑇=𝑇$

Hence the given statement becomes:

$𝐹\vee 𝑇=𝑇$

So, the truth value of the given statement is True.

Now find its dual.

To form the dual, interchange $\wedge$ and $\vee$.

So the dual of

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

is

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

Now evaluate it.

$𝑝\vee \sim 𝑞=𝑇\vee 𝑇=𝑇$

So,

$\sim (𝑝\vee \sim 𝑞)<text>=</text><text>\sim </text>𝑇=𝐹$

Also,

$𝑞\wedge \sim 𝑟=𝐹\wedge 𝑇=𝐹$

Thus the dual becomes:

$𝐹\wedge 𝐹=𝐹$

So the truth values are:

$\text{True, False}$

Hence, the correct option is Option B.

In $△𝐴𝐵𝐶$, using the standard half-angle formula,

$\tan \frac{𝐴}{2}=\frac{𝑟}{𝑠-𝑎}$

and

$\tan \frac{𝐵}{2}=\frac{𝑟}{𝑠-𝑏}$

where $𝑟$ is the inradius and $𝑠$ is the semi-perimeter.

So,

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

Now, from Heron’s formula,

${\Delta }^2=𝑠(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)$

Also, $\Delta =𝑟𝑠$, so

${𝑟}^2=\frac{{\Delta }^2}{{𝑠}^2}=\frac{𝑠(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)}{{𝑠}^2}=\frac{(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)}{𝑠}$

Thus,

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

Hence,

$1+\tan \frac{𝐴}{2}\tan \frac{𝐵}{2}=1+\frac{𝑠-𝑐}{𝑠}$

$=\frac{𝑠+𝑠-𝑐}{𝑠}=\frac{2𝑠-𝑐}{𝑠}$

Since

$2𝑠=𝑎+𝑏+𝑐$

we get

$2𝑠-𝑐=𝑎+𝑏$

So,

$1+\tan \frac{𝐴}{2}\tan \frac{𝐵}{2}=\frac{𝑎+𝑏}{𝑠}$

Hence,

$𝑘=𝑎+𝑏$

The correct option is C.

For a square matrix of order $𝑛$,

$det(\mathrm{adj}𝐴)=(det𝐴{)}^{𝑛-1}$

Here both matrices are of order $3$, so

$det(\mathrm{adj}𝐴)=(det𝐴{)}^2$

and

$det(\mathrm{adj}𝐵)=(det𝐵{)}^2$

Matrix $𝐴$ is upper triangular, so its determinant is the product of diagonal elements.

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

So,

$det𝐴=(2𝑘-1{)}^3$

Hence,

$det(\mathrm{adj}𝐴)=\left\{\right(2𝑘-1{)}^3{\}}^2=(2𝑘-1{)}^6$

Now look at matrix $𝐵$:

$𝐵=⎡⎣⎢\begin{array}{ccc}0& 2𝑘-1& 1\\ 1-2𝑘& 0& 𝑘\\ -1& -𝑘& 0\end{array}⎤⎦⎥$

This is a skew-symmetric matrix because ${𝐵}^{𝑇}=-𝐵$.

For a skew-symmetric matrix of odd order, the determinant is $0$.

Since $𝐵$ is of order $3$,

$det𝐵=0$

So,

$det(\mathrm{adj}𝐵)=(det𝐵{)}^2=0$

Given,

$det(\mathrm{adj}𝐴)+det(\mathrm{adj}𝐵)={11}^6$

So,

$(2𝑘-1{)}^6+0={11}^6$

Thus,

$(2𝑘-1{)}^6={11}^6$

So,

$2𝑘-1=\pm 11$

This gives

$𝑘=6\text{or}𝑘=-5$

Among the given options, if $𝑘=6$, then

$𝑘-5=6-5=1$

So the correct option is

$\overline{)1}$

Option A.

The given matrix is

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

This is a rotation matrix. For this type of matrix,

${𝐴}^{𝑛}=\left[\begin{array}{cc}\cos 𝑛𝜃& -\sin 𝑛𝜃\\ \sin 𝑛𝜃& \cos 𝑛𝜃\end{array}\right]$

So,

${𝐴}^{-3}=\left[\begin{array}{cc}\cos (-3𝜃)& -\sin (-3𝜃)\\ \sin (-3𝜃)& \cos (-3𝜃)\end{array}\right]$

Given

$𝜃=\frac{𝜋}{6}$

So,

$-3𝜃=-3\cdot \frac{𝜋}{6}=-\frac{𝜋}{2}$

Hence,

${𝐴}^{-3}=\left[\begin{array}{cc}\cos (-\frac{𝜋}{2})& -\sin (-\frac{𝜋}{2})\\ \sin (-\frac{𝜋}{2})& \cos (-\frac{𝜋}{2})\end{array}\right]$

Now,

$\cos (-\frac{𝜋}{2})=0$

and

$\sin (-\frac{𝜋}{2})=-1$

So,

${𝐴}^{-3}=\left[\begin{array}{cc}0& -(-1)\\ -1& 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$

So the correct option is

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

We are given

${\tan }^{-1}\left[\frac{\sqrt{\surd 5-2\sqrt{\surd 6}}}{1+\sqrt{\surd 6}}\right]=\frac{𝜋}{3}-{\tan }^{-1}\left(𝑘\right)$

First simplify the square root.

$5-2\sqrt{\surd 6}=(\sqrt{\surd 3}-\sqrt{\surd 2}{)}^2$

So,

$\sqrt{\surd 5-2\sqrt{\surd 6}}=\sqrt{\surd 3}-\sqrt{\surd 2}$

Hence,

${\tan }^{-1}\left[\frac{\sqrt{\surd 3}-\sqrt{\surd 2}}{1+\sqrt{\surd 6}}\right]=\frac{𝜋}{3}-{\tan }^{-1}\left(𝑘\right)$

This means

${\tan }^{-1}\left(𝑘\right)=\frac{𝜋}{3}-{\tan }^{-1}\left[\frac{\sqrt{\surd 3}-\sqrt{\surd 2}}{1+\sqrt{\surd 6}}\right]$

Taking tangent on both sides,

$𝑘=\tan (\frac{𝜋}{3}-{\tan }^{-1}\left[\frac{\sqrt{\surd 3}-\sqrt{\surd 2}}{1+\sqrt{\surd 6}}\right])$

Using

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

we get

$𝑘=\frac{\sqrt{\surd 3}-\frac{\sqrt{\surd 3}-\sqrt{\surd 2}}{1+\sqrt{\surd 6}}}{1+\sqrt{\surd 3}\cdot \frac{\sqrt{\surd 3}-\sqrt{\surd 2}}{1+\sqrt{\surd 6}}}$

Simplifying,

$𝑘=\sqrt{\surd 2}$

Now,

${\sec }^{-1}\left(𝑘\right)={\sec }^{-1}\left(\sqrt{\surd 2}\right)$

Since

$\sec \frac{𝜋}{4}=\sqrt{\surd 2}$

we get

${\sec }^{-1}\left(𝑘\right)=\frac{𝜋}{4}$

So the correct option is

$\overline{)\frac{𝜋}{4}}$

Option B.

Put $𝑦=0$ in the given equation:

$𝑓(𝑥-0)+𝑓(𝑥+0)=2𝑓\left(𝑥\right)𝑓\left(0\right)$

So,

$2𝑓\left(𝑥\right)=2𝑓\left(𝑥\right)𝑓\left(0\right)$

$𝑓\left(𝑥\right)\{1-𝑓(0\left)\right\}=0$

Now, if $𝑓\left(𝑥\right)$ is not the zero function, then we must have

$𝑓\left(0\right)=1$

Now put $𝑥=0$ in the original equation:

$𝑓(0-𝑦)+𝑓(0+𝑦)=2𝑓\left(0\right)𝑓\left(𝑦\right)$

Using $𝑓\left(0\right)=1$,

$𝑓(-𝑦)+𝑓\left(𝑦\right)=2𝑓\left(𝑦\right)$

So,

$𝑓(-𝑦)=𝑓\left(𝑦\right)$

This is exactly the condition for an even function.

If $𝑓\left(𝑥\right)=0$ for all $𝑥$, then it is also an even function.

So $𝑓\left(𝑥\right)$ is an even function.

Correct option: B an even function.

For $𝑓$ to be continuous at $𝑥=1$, we must have

$𝑓\left(1\right)=\underset{𝑥\to 1}{lim}𝑓\left(𝑥\right)$

Given,

$𝑓\left(𝑥\right)=\frac{1+\cos \left(𝜋𝑥\right)}{𝜋(1-𝑥{)}^2},𝑥\ne 1$

So,

$𝑓\left(1\right)=\underset{𝑥\to 1}{lim}\frac{1+\cos \left(𝜋𝑥\right)}{𝜋(1-𝑥{)}^2}$

Put $𝑥=1+ℎ$. Then as $𝑥\to 1$, $ℎ\to 0$.

$1-𝑥=1-(1+ℎ)=-ℎ$

So,

$(1-𝑥{)}^2={ℎ}^2$

Also,

$\cos \left(𝜋𝑥\right)=\cos \left(𝜋\right(1+ℎ\left)\right)$

Using $\cos (𝜋+𝜃)=-\cos 𝜃$,

$\cos \left(𝜋\right(1+ℎ\left)\right)=-\cos \left(𝜋ℎ\right)$

Hence,

$1+\cos \left(𝜋𝑥\right)=1-\cos \left(𝜋ℎ\right)$

Now the limit becomes

$𝑓\left(1\right)=\underset{ℎ\to 0}{lim}\frac{1-\cos \left(𝜋ℎ\right)}{𝜋{ℎ}^2}$

Using the standard limit,

$1-\cos 𝜃\sim \frac{{𝜃}^2}{2}$

near $𝜃=0$, we get

$1-\cos \left(𝜋ℎ\right)\sim \frac{(𝜋ℎ{)}^2}{2}$

Therefore,

$𝑓\left(1\right)=\underset{ℎ\to 0}{lim}\frac{\frac{{𝜋}^2{ℎ}^2}{2}}{𝜋{ℎ}^2}$

$𝑓\left(1\right)=\frac{𝜋}{2}$

So the correct option is

$\overline{)\frac{𝜋}{2}}$

Option A.