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

Since $𝑧$ is the largest side of the right angled triangle, $𝑧$ is the hypotenuse.

So by Pythagoras theorem,

${𝑥}^2+{𝑦}^2={𝑧}^2$

Now the given expression is

$\frac{1}{{\log }_{𝑥+𝑧}𝑦}+\frac{1}{{\log }_{𝑧-𝑥}𝑦}$

Using the property

$\frac{1}{{\log }_{𝑎}𝑏}={\log }_{𝑏}𝑎$

we get

$\frac{1}{{\log }_{𝑥+𝑧}𝑦}={\log }_{𝑦}(𝑥+𝑧)$

and

$\frac{1}{{\log }_{𝑧-𝑥}𝑦}={\log }_{𝑦}(𝑧-𝑥)$

So the expression becomes

${\log }_{𝑦}(𝑥+𝑧)+{\log }_{𝑦}(𝑧-𝑥)$

Using ${\log }_{𝑎}𝑚+{\log }_{𝑎}𝑛={\log }_{𝑎}\left(𝑚𝑛\right)$,

${\log }_{𝑦}((𝑥+𝑧)(𝑧-𝑥))$

Now,

$(𝑥+𝑧)(𝑧-𝑥)={𝑧}^2-{𝑥}^2$

From Pythagoras theorem,

${𝑧}^2-{𝑥}^2={𝑦}^2$

So,

${\log }_{𝑦}({𝑧}^2-{𝑥}^2)={\log }_{𝑦}{𝑦}^2$

${\log }_{𝑦}{𝑦}^2=2$

Hence, the value is

$\overline{)2}$

Correct option: Option B.

Since $𝜔$ is a complex cube root of unity, we use the standard properties

${𝜔}^3=1$

and for the non-real cube root of unity,

$1+𝜔+{𝜔}^2=0$

So,

$𝜔+{𝜔}^2=-1$

Now reduce the powers using ${𝜔}^3=1$.

Since $10=3\times 3+1$,

${𝜔}^{10}=𝜔$

Since $23=3\times 7+2$,

${𝜔}^{23}={𝜔}^2$

So,

${𝜔}^{10}+{𝜔}^{23}=𝜔+{𝜔}^2=-1$

Now substitute this in the given expression:

$\sin \left[𝜋\right({𝜔}^{10}+{𝜔}^{23})-\frac{𝜋}{4}]=\sin \left[𝜋\right(-1)-\frac{𝜋}{4}]$

$=\sin (-𝜋-\frac{𝜋}{4})=\sin (-\frac{5𝜋}{4})$

We know

$\sin \left(\frac{5𝜋}{4}\right)=-\frac{1}{\sqrt{\surd 2}}$

So,

$\sin (-\frac{5𝜋}{4})=\frac{1}{\sqrt{\surd 2}}$

Hence the value is

$\overline{)\frac{1}{\sqrt{\surd 2}}}$

Correct option: C.

Total number of ways to seat $6$ boys and $3$ girls in a row is

$9!$

Now we need arrangements where both end seats are occupied by girls and no two girls are side by side.

There are $3$ girls. The two end seats can be filled by any two of the $3$ girls in

${}^3{𝑃}_2=3\times 2=6$

ways.

After placing girls at both ends, $1$ girl is left.

Since no two girls should be side by side, this remaining girl cannot sit next to either end girl.

So she cannot sit in position $2$ or position $8$.

The possible positions for the remaining girl are

$3,4,5,6,7$

So she can be seated in $5$ ways.

The $6$ boys can be seated in the remaining $6$ seats in

$6!$

ways.

So favourable arrangements are

$6\times 5\times 6!$

Hence the required probability is

$\frac{6\times 5\times 6!}{9!}$

Since

$9!=9\times 8\times 7\times 6!$

we get

$\frac{6\times 5\times 6!}{9\times 8\times 7\times 6!}=\frac{30}{504}=\frac{5}{84}$

So the correct option is

$\overline{)\frac{5}{84}}$

Option D.

There are 23 balls in all.

Out of these, 7 balls are identical. So the remaining balls are

$23-7=16$

These 16 balls are different.

We have to select 12 balls. Since the 7 identical balls are alike, only the number of identical balls selected matters.

Suppose we select $𝑟$ identical balls. Then $𝑟$ can be

$0,1,2,3,4,5,6,7$

If we select $𝑟$ identical balls, then we must select $12-𝑟$ balls from the 16 different balls.

So the total number of selections is

${}^{16}{𝐶}_{12}+{}^{16}{𝐶}_{11}+{}^{16}{𝐶}_{10}+{}^{16}{𝐶}_9+{}^{16}{𝐶}_8+{}^{16}{𝐶}_7+{}^{16}{𝐶}_6+{}^{16}{𝐶}_5$

That is,

$\sum _{𝑘=5}^{12}{}^{16}{𝐶}_{𝑘}$

Using the identity

${}^{𝑛}{𝐶}_{𝑟}={}^{𝑛}{𝐶}_{𝑛-𝑟}$

this can be simplified to

${}^{18}{𝐶}_6+{}^{18}{𝐶}_8$

So the correct option is

$\overline{){}^{18}{𝐶}_6+{}^{18}{𝐶}_8}$

Hence, the answer is Option A.

We need to find the approximate value of

$(0.007{)}^{\frac{1}{3}}$

This means we have to find the cube root of $0.007$.

Now,

$0.007=\frac{7}{1000}$

So,

$(0.007{)}^{\frac{1}{3}}={\left(\frac{7}{1000}\right)}^{\frac{1}{3}}$

Since $1000={10}^3$,

$\text{exception 25:}=10$

So the value is approximately

$\frac{\text{exception 25:}}{10}$

Now,

$\text{exception 25:}\approx 1.91$

So,

$(0.007{)}^{\frac{1}{3}}\approx \frac{1.91}{10}=0.191$

Now check the options:

$\frac{21}{120}=0.175$

$\frac{23}{120}\approx 0.1917$

$\frac{29}{120}\approx 0.2417$

$\frac{31}{120}\approx 0.2583$

The closest value to $0.191$ is

$\frac{23}{120}$

So the correct option is

$\overline{)\text{Option B }\frac{23}{120}}$

Given $33𝜃=𝜋$, we have

$𝜃=\frac{𝜋}{33}$

We need to find

$\cos 𝜃\cos 2𝜃\cos 4𝜃\cos 8𝜃\cos 16𝜃$

Use the formula

$\sin 2𝑥=2\sin 𝑥\cos 𝑥$

Applying it repeatedly,

$\sin 32𝜃=32\sin 𝜃\cos 𝜃\cos 2𝜃\cos 4𝜃\cos 8𝜃\cos 16𝜃$

So,

$\cos 𝜃\cos 2𝜃\cos 4𝜃\cos 8𝜃\cos 16𝜃=\frac{\sin 32𝜃}{32\sin 𝜃}$

Now, since

$33𝜃=𝜋$

we get

$32𝜃=𝜋-𝜃$

So,

$\sin 32𝜃=\sin (𝜋-𝜃)=\sin 𝜃$

Hence,

$\cos 𝜃\cos 2𝜃\cos 4𝜃\cos 8𝜃\cos 16𝜃=\frac{\sin 𝜃}{32\sin 𝜃}=\frac{1}{32}$

So the correct option is

$\overline{)\frac{1}{32}}$

Option B.

Let

$𝑥=2𝜃$

Then the given equation becomes

$\sec 2𝑥-\sec 𝑥=2$

Using $\sec 𝐴=\frac{1}{\cos 𝐴}$,

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

$\frac{\cos 𝑥-\cos 2𝑥}{\cos 𝑥\cos 2𝑥}=2$

So,

$\cos 𝑥-\cos 2𝑥=2\cos 𝑥\cos 2𝑥$

Now put $\cos 𝑥=𝑐$.

We know

$\cos 2𝑥=2{𝑐}^2-1$

So,

$𝑐-(2{𝑐}^2-1)=2𝑐(2{𝑐}^2-1)$

$𝑐-2{𝑐}^2+1=4{𝑐}^3-2𝑐$

$4{𝑐}^3+2{𝑐}^2-3𝑐-1=0$

Factorising,

$4{𝑐}^3+2{𝑐}^2-3𝑐-1=(𝑐+1)(4{𝑐}^2-2𝑐-1)$

So,

$𝑐=-1$

or

$4{𝑐}^2-2𝑐-1=0$

From the quadratic,

$𝑐=\frac{1\pm \sqrt{\surd 5}}{4}$

Thus,

$\cos 𝑥=-1,\cos 𝑥=\frac{1+\sqrt{\surd 5}}{4},\cos 𝑥=\frac{1-\sqrt{\surd 5}}{4}$

Now,

$\frac{1+\sqrt{\surd 5}}{4}=\cos \frac{𝜋}{5}$

and

$\frac{1-\sqrt{\surd 5}}{4}=\cos \frac{3𝜋}{5}$

So the solutions for $𝑥$ are odd multiples of $\frac{𝜋}{5}$:

$𝑥=\frac{(2𝑛+1)𝜋}{5}$

Since $𝑥=2𝜃$,

$2𝜃=\frac{(2𝑛+1)𝜋}{5}$

$𝜃=\frac{(2𝑛+1)𝜋}{10}$

This can be written as

$𝜃=\frac{𝑛𝜋}{5}+\frac{𝜋}{10},𝑛\in 𝑍$

This includes values like

$𝜃=𝑛𝜋+\frac{𝜋}{2}$

as well.

So the correct option is

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

The required line passes through the point of intersection of

$𝑥+2𝑦+6=0$

and

$2𝑥-𝑦=2$

First, find their point of intersection.

From

$2𝑥-𝑦=2$

we get

$𝑦=2𝑥-2$

Substitute this in $𝑥+2𝑦+6=0$:

$𝑥+2(2𝑥-2)+6=0$

$𝑥+4𝑥-4+6=0$

$5𝑥+2=0$

$𝑥=-\frac{2}{5}$

Now,

$𝑦=2(-\frac{2}{5})-2$

$𝑦=-\frac{4}{5}-\frac{10}{5}=-\frac{14}{5}$

So, the point of intersection is

$(-\frac{2}{5},-\frac{14}{5})$

The line makes an intercept $5$ on the $𝑦$-axis, so it passes through the point

$(0,5)$

Now find the equation of the line passing through

$(-\frac{2}{5},-\frac{14}{5})$

and

$(0,5)$

Slope:

$𝑚=\frac{5-(-\frac{14}{5})}{0-(-\frac{2}{5})}$

$𝑚=\frac{\frac{25}{5}+\frac{14}{5}}{\frac{2}{5}}$

$𝑚=\frac{\frac{39}{5}}{\frac{2}{5}}=\frac{39}{2}$

Using point-slope form with point $(0,5)$:

$𝑦-5=\frac{39}{2}(𝑥-0)$

$2𝑦-10=39𝑥$

$39𝑥-2𝑦+10=0$

So the correct option is

$\overline{)39𝑥-2𝑦+10=0}$

Option B.

The points A and B lie on the line

$𝑥+𝑦=3$

So we can write

$𝑦=3-𝑥$

Now substitute this in the pair of straight lines

${𝑥}^2-3𝑥𝑦+{𝑦}^2=0$

We get

${𝑥}^2-3𝑥(3-𝑥)+(3-𝑥{)}^2=0$

Simplifying,

${𝑥}^2-9𝑥+3{𝑥}^2+9-6𝑥+{𝑥}^2=0$

$5{𝑥}^2-15𝑥+9=0$

Let the $𝑥$-coordinates of A and B be ${𝑥}_1$ and ${𝑥}_2$.

For the quadratic equation

$5{𝑥}^2-15𝑥+9=0$

sum of roots is

${𝑥}_1+{𝑥}_2=\frac{15}{5}=3$

So the $𝑥$-coordinate of the mid-point is

$\frac{{𝑥}_1+{𝑥}_2}{2}=\frac{3}{2}$

Since both points lie on $𝑥+𝑦=3$, their $𝑦$-coordinates satisfy

${𝑦}_1=3-{𝑥}_1,{𝑦}_2=3-{𝑥}_2$

So,

${𝑦}_1+{𝑦}_2=(3-{𝑥}_1)+(3-{𝑥}_2)=6-({𝑥}_1+{𝑥}_2)$

${𝑦}_1+{𝑦}_2=6-3=3$

Hence the $𝑦$-coordinate of the mid-point is

$\frac{{𝑦}_1+{𝑦}_2}{2}=\frac{3}{2}$

So the mid-point of AB is

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

Correct option: D.

The circle is

$(𝑥-7{)}^2+(𝑦+1{)}^2=25$

So its centre is

$𝐶(7,-1)$

and its radius is

$𝑟=5$

The tangents are drawn from the origin, so take the external point as

$𝑂(0,0)$

Now find the distance from the origin to the centre:

$𝑂𝐶=\sqrt{\surd (7-0{)}^2+(-1-0{)}^2}$

$𝑂𝐶=\sqrt{\surd 49+1}=\sqrt{\surd 50}=5\sqrt{\surd 2}$

If tangents are drawn from an external point to a circle, the line joining the external point to the centre bisects the angle between the two tangents.

Let the angle between $𝑂𝐶$ and one tangent be $𝜃$.

In the right triangle formed by the centre, point of contact, and origin,

$\sin 𝜃=\frac{\text{radius}}{\text{distance from origin to centre}}$

$\sin 𝜃=\frac{5}{5\sqrt{\surd 2}}=\frac{1}{\sqrt{\surd 2}}$

So,

$𝜃={45}^{\circ }$

This is half of the angle between the two tangents.

Hence, the angle between the tangents is

$2𝜃=2\times {45}^{\circ }={90}^{\circ }$

So the correct option is

$\overline{){90}^{\circ }}$

Option B.

For the parabola

${𝑦}^2=4𝑥$

we compare it with the standard form

${𝑦}^2=4𝑎𝑥$

So, $𝑎=1$.

The tangent to the parabola ${𝑦}^2=4𝑎𝑥$ in slope form is

$𝑦=𝑚𝑥+\frac{𝑎}{𝑚}$

Here $𝑎=1$, so the tangent is

$𝑦=𝑚𝑥+\frac{1}{𝑚}$

This same line must also touch the circle

$(𝑥-3{)}^2+{𝑦}^2=9$

The centre of the circle is $(3,0)$ and radius is $3$.

Write the line as

$𝑚𝑥-𝑦+\frac{1}{𝑚}=0$

For this line to be tangent to the circle, the perpendicular distance from the centre $(3,0)$ to the line must be equal to the radius $3$.

So,

$\frac{\mid 3𝑚+\frac{1}{𝑚}\mid }{\sqrt{\surd {𝑚}^2+1}}=3$

Squaring both sides,

${(3𝑚+\frac{1}{𝑚})}^2=9({𝑚}^2+1)$

Expanding,

$9{𝑚}^2+6+\frac{1}{{𝑚}^2}=9{𝑚}^2+9$

So,

$\frac{1}{{𝑚}^2}=3$

Hence,

${𝑚}^2=\frac{1}{3}$

So,

$𝑚=\pm \frac{1}{\sqrt{\surd 3}}$

The tangent above the $𝑥$-axis corresponds to

$𝑚=\frac{1}{\sqrt{\surd 3}}$

Now substitute this in

$𝑦=𝑚𝑥+\frac{1}{𝑚}$

We get

$𝑦=\frac{𝑥}{\sqrt{\surd 3}}+\sqrt{\surd 3}$

Multiplying by $\sqrt{\surd 3}$,

$\sqrt{\surd 3}𝑦=𝑥+3$

So the correct option is

$\overline{)\sqrt{\surd 3}𝑦=𝑥+3}$

Option C.

For very large $𝑥$, the highest power terms decide the value of the limit.

We have

$\underset{𝑥\to \infty }{lim}\frac{(2𝑥-1{)}^{19}(3𝑥+2{)}^{11}}{(6𝑥-5{)}^{30}}$

Take $𝑥$ common from each bracket:

$2𝑥-1=𝑥(2-\frac{1}{𝑥})$

$3𝑥+2=𝑥(3+\frac{2}{𝑥})$

$6𝑥-5=𝑥(6-\frac{5}{𝑥})$

So the expression becomes

$\frac{{𝑥}^{19}{(2-\frac{1}{𝑥})}^{19}\cdot {𝑥}^{11}{(3+\frac{2}{𝑥})}^{11}}{{𝑥}^{30}{(6-\frac{5}{𝑥})}^{30}}$

Since $19+11=30$, the powers of $𝑥$ cancel:

$\underset{𝑥\to \infty }{lim}\frac{{(2-\frac{1}{𝑥})}^{19}{(3+\frac{2}{𝑥})}^{11}}{{(6-\frac{5}{𝑥})}^{30}}$

As $𝑥\to \infty$,

$\frac{1}{𝑥},\frac{2}{𝑥},\frac{5}{𝑥}\to 0$

So the limit is

$\frac{2^{19}\cdot 3^{11}}{6^{30}}$

Now,

$6^{30}=(2\cdot 3{)}^{30}=2^{30}\cdot 3^{30}$

Hence,

$\frac{2^{19}\cdot 3^{11}}{2^{30}\cdot 3^{30}}=2^{19-30}\cdot 3^{11-30}=2^{-11}\cdot 3^{-19}$

So,

$𝑎=-11,𝑏=-19$

Thus,

$𝑎+𝑏=-11-19=-30$

Answer: Option A, $-30$

The given statement is

$(𝑝\wedge \sim 𝑞)⟶(𝑞\wedge \sim 𝑝)$

To find its dual, first write implication in terms of $\vee$ and $\sim$.

We know

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

So,

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

Now take the dual by interchanging $\wedge$ and $\vee$.

So the dual becomes

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

Using De Morgan’s law,

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

Hence the dual is

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

Since $\sim 𝑝\wedge 𝑞$ is already contained in $𝑞\vee \sim 𝑝$, this simplifies to

$\sim 𝑝\wedge 𝑞$

Now check the options.

Option B is

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

Simplify it:

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

and

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

So,

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

Thus Option B becomes

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

which simplifies to

$\sim 𝑝\wedge 𝑞$

So the dual is equivalent to Option B.

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

The given statement is of the form:

$\text{If }𝑝,\text{ then }𝑞$

where

$𝑝:\text{An integer is greater than }4\text{ and less than }5$

and

$𝑞:\text{It is a multiple of }3$

The negation of an implication

$𝑝\Rightarrow 𝑞$

is

$𝑝\text{ and not }𝑞$

So the negation will be:

“An integer is greater than $4$ and less than $5$, but it is not a multiple of $3$.”

This matches Option C.

So the correct answer is:

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

Since $𝐷$ divides $𝐵𝐶$ internally in the ratio $1:3$,

$\frac{𝐵𝐷}{𝐷𝐶}=\frac{1}{3}$

Now compare the areas of triangles $𝐴𝐵𝐷$ and $𝐴𝐶𝐷$.

Both triangles have their bases on the same line $𝐵𝐶$, so their areas are in the ratio of their bases:

$\frac{\left[𝐴𝐵𝐷\right]}{\left[𝐴𝐶𝐷\right]}=\frac{𝐵𝐷}{𝐷𝐶}=\frac{1}{3}$

Using the formula for area with two sides and included angle,

$\left[𝐴𝐵𝐷\right]=\frac{1}{2}𝐴𝐵\cdot 𝐴𝐷\cdot \sin \angle 𝐵𝐴𝐷$

and

$\left[𝐴𝐶𝐷\right]=\frac{1}{2}𝐴𝐶\cdot 𝐴𝐷\cdot \sin \angle 𝐶𝐴𝐷$

So,

$\frac{\left[𝐴𝐵𝐷\right]}{\left[𝐴𝐶𝐷\right]}=\frac{𝐴𝐵\cdot \sin \angle 𝐵𝐴𝐷}{𝐴𝐶\cdot \sin \angle 𝐶𝐴𝐷}$

Hence,

$\frac{1}{3}=\frac{𝐴𝐵}{𝐴𝐶}\cdot \frac{\sin \angle 𝐵𝐴𝐷}{\sin \angle 𝐶𝐴𝐷}$

So,

$\frac{\sin \angle 𝐵𝐴𝐷}{\sin \angle 𝐶𝐴𝐷}=\frac{1}{3}\cdot \frac{𝐴𝐶}{𝐴𝐵}$

Now by the sine rule in $△𝐴𝐵𝐶$,

$\frac{𝐴𝐶}{𝐴𝐵}=\frac{\sin 𝐵}{\sin 𝐶}$

Given,

$𝐵=\frac{𝜋}{3},𝐶=\frac{𝜋}{4}$

So,

$\frac{𝐴𝐶}{𝐴𝐵}=\frac{\sin \frac{𝜋}{3}}{\sin \frac{𝜋}{4}}=\frac{\frac{\sqrt{\surd 3}}{2}}{\frac{\sqrt{\surd 2}}{2}}=\sqrt{\surd \frac{3}{2}}$

Thus,

$\frac{\sin \angle 𝐵𝐴𝐷}{\sin \angle 𝐶𝐴𝐷}=\frac{1}{3}\sqrt{\surd \frac{3}{2}}=\frac{1}{\sqrt{\surd 6}}$

So the correct option is

$\overline{)\frac{1}{\sqrt{\surd 6}}}$

Option C.

Let the sides opposite $𝐴,𝐵,𝐶$ be $𝑎,𝑏,𝑐$ respectively.

Since ${𝑃}_1,{𝑃}_2,{𝑃}_3$ are the altitudes from $𝐴,𝐵,𝐶$,

$△=\frac{1}{2}𝑎{𝑃}_1$

So,

${𝑃}_1=\frac{2△}{𝑎}$

Similarly,

${𝑃}_2=\frac{2△}{𝑏},{𝑃}_3=\frac{2△}{𝑐}$

Now,

$\frac{\cos 𝐴}{{𝑃}_1}+\frac{\cos 𝐵}{{𝑃}_2}+\frac{\cos 𝐶}{{𝑃}_3}=\frac{𝑎\cos 𝐴}{2△}+\frac{𝑏\cos 𝐵}{2△}+\frac{𝑐\cos 𝐶}{2△}$

$=\frac{𝑎\cos 𝐴+𝑏\cos 𝐵+𝑐\cos 𝐶}{2△}$

Using the sine rule,

$𝑎=2𝑅\sin 𝐴,𝑏=2𝑅\sin 𝐵,𝑐=2𝑅\sin 𝐶$

So,

$𝑎\cos 𝐴+𝑏\cos 𝐵+𝑐\cos 𝐶=2𝑅(\sin 𝐴\cos 𝐴+\sin 𝐵\cos 𝐵+\sin 𝐶\cos 𝐶)$

Since $2\sin 𝐴\cos 𝐴=\sin 2𝐴$,

$=𝑅(\sin 2𝐴+\sin 2𝐵+\sin 2𝐶)$

For a triangle,

$\sin 2𝐴+\sin 2𝐵+\sin 2𝐶=4\sin 𝐴\sin 𝐵\sin 𝐶$

Hence,

$𝑎\cos 𝐴+𝑏\cos 𝐵+𝑐\cos 𝐶=4𝑅\sin 𝐴\sin 𝐵\sin 𝐶$

Also,

$△=2{𝑅}^2\sin 𝐴\sin 𝐵\sin 𝐶$

So,

$\frac{𝑎\cos 𝐴+𝑏\cos 𝐵+𝑐\cos 𝐶}{2△}=\frac{4𝑅\sin 𝐴\sin 𝐵\sin 𝐶}{2(2{𝑅}^2\sin 𝐴\sin 𝐵\sin 𝐶)}$

$=\frac{1}{𝑅}$

So the correct option is

$\overline{)\frac{1}{𝑅}}$

Option B.

Since $𝐴𝐵=𝐶$, we need the inverse of $𝐵$.

From

$𝐴𝐵=𝐶$

Taking inverse on both sides:

$(𝐴𝐵{)}^{-1}={𝐶}^{-1}$

Using the rule $(𝐴𝐵{)}^{-1}={𝐵}^{-1}{𝐴}^{-1}$,

${𝐵}^{-1}{𝐴}^{-1}={𝐶}^{-1}$

Multiplying both sides on the right by $𝐴$,

${𝐵}^{-1}={𝐶}^{-1}𝐴$

Now,

$𝐶=\left[\begin{array}{cc}7& 3\\ 0& 6\end{array}\right]$

Its determinant is

$\left|𝐶\right|=7\times 6-0\times 3=42$

So,

${𝐶}^{-1}=\frac{1}{42}\left[\begin{array}{cc}6& -3\\ 0& 7\end{array}\right]$

Now multiply ${𝐶}^{-1}$ with $𝐴$:

${𝐵}^{-1}=\frac{1}{42}\left[\begin{array}{cc}6& -3\\ 0& 7\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$

${𝐵}^{-1}=\frac{1}{42}\left[\begin{array}{cc}18+3& 6-6\\ 0-7& 0+14\end{array}\right]$

${𝐵}^{-1}=\frac{1}{42}\left[\begin{array}{cc}21& 0\\ -7& 14\end{array}\right]$

Taking $7$ common from the matrix,

${𝐵}^{-1}=\frac{7}{42}\left[\begin{array}{cc}3& 0\\ -1& 2\end{array}\right]$

${𝐵}^{-1}=\frac{1}{6}\left[\begin{array}{cc}3& 0\\ -1& 2\end{array}\right]$

So the correct answer is

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

For a matrix $𝐴$, the inverse is given by

${𝐴}^{-1}=\frac{1}{\left|𝐴\right|}\mathrm{adj}𝐴$

We need the element in the third row and second column of ${𝐴}^{-1}$, that is $({𝐴}^{-1}{)}_{32}$.

In the inverse matrix,

$({𝐴}^{-1}{)}_{32}=\frac{{𝐶}_{23}}{\left|𝐴\right|}$

because the adjoint is the transpose of the cofactor matrix.

Given

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

First find the determinant:

$\left|𝐴\right|=1(1\cdot 4-3\cdot 3)-3(3\cdot 4-3\cdot 3)+3(3\cdot 3-1\cdot 3)$

$=1(4-9)-3(12-9)+3(9-3)$

$=-5-9+18=4$

Now find ${𝐶}_{23}$.

Remove row $2$ and column $3$:

$\mid \begin{array}{cc}1& 3\\ 3& 3\end{array}\mid =1\cdot 3-3\cdot 3=3-9=-6$

So,

${𝐶}_{23}=(-1{)}^{2+3}(-6)$

${𝐶}_{23}=6$

Hence,

$({𝐴}^{-1}{)}_{32}=\frac{{𝐶}_{23}}{\left|𝐴\right|}=\frac{6}{4}=\frac{3}{2}$

So the correct option is

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

Option A.

Given

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

Use the identities:

$1+\sin 2𝑥=(\sin 𝑥+\cos 𝑥{)}^2$

and

$1-\sin 2𝑥=(\cos 𝑥-\sin 𝑥{)}^2$

So,

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

At $𝑥=\frac{𝜋}{6}$, we have $\cos 𝑥>\sin 𝑥$, so the square root becomes

$\frac{\sin 𝑥+\cos 𝑥}{\cos 𝑥-\sin 𝑥}$

Now,

$\frac{\sin 𝑥+\cos 𝑥}{\cos 𝑥-\sin 𝑥}=\tan (\frac{𝜋}{4}+𝑥)$

Hence,

$𝑦={\tan }^{-1}(\tan (\frac{𝜋}{4}+𝑥))$

At $𝑥=\frac{𝜋}{6}$,

$\frac{𝜋}{4}+𝑥=\frac{𝜋}{4}+\frac{𝜋}{6}=\frac{5𝜋}{12}$

This lies in the principal range of ${\tan }^{-1}$, so

$𝑦=\frac{𝜋}{4}+𝑥$

Differentiating,

$\frac{𝑑𝑦}{𝑑𝑥}=1$

So the value of $\frac{𝑑𝑦}{𝑑𝑥}$ at $𝑥=\frac{𝜋}{6}$ is

$\overline{)1}$

Correct option: B.

We need to find

${\tan }^{-1}\left(\frac{\cos \left(\frac{19𝜋}{4}\right)-1}{\sin \left(\frac{𝜋}{4}\right)}\right)$

First simplify the angle inside cosine.

$\frac{19𝜋}{4}=4𝜋+\frac{3𝜋}{4}$

Since cosine has period $2𝜋$,

$\cos (4𝜋+\frac{3𝜋}{4})=\cos \frac{3𝜋}{4}$

Now,

$\cos \frac{3𝜋}{4}=-\frac{1}{\sqrt{\surd 2}}$

and

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

So the expression inside ${\tan }^{-1}$ becomes

$\frac{-\frac{1}{\sqrt{\surd 2}}-1}{\frac{1}{\sqrt{\surd 2}}}$

Simplify it:

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

$=-(1+\sqrt{\surd 2})$

So we need

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

We know that

$\tan \frac{3𝜋}{8}=1+\sqrt{\surd 2}$

Hence,

$\tan (-\frac{3𝜋}{8})=-(1+\sqrt{\surd 2})$

Also, $-\frac{3𝜋}{8}$ lies in the principal range of ${\tan }^{-1}𝑥$, which is

$(-\frac{𝜋}{2},\frac{𝜋}{2})$

So,

${\tan }^{-1}(-(1+\sqrt{\surd 2}\left)\right)=-\frac{3𝜋}{8}$

Hence the correct option is

$\overline{)-\frac{3𝜋}{8}}$

Option C.

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

$𝑓\left(0\right)=\underset{𝑥\to 0}{lim}𝑓\left(𝑥\right)$

Given,

$𝑓\left(𝑥\right)=\frac{(256-8𝑥{)}^{\frac{1}{4}}-4}{16-4(64+3𝑥{)}^{\frac{1}{3}}}$

At $𝑥=0$, the numerator becomes $4-4=0$ and the denominator becomes $16-4\left(4\right)=0$, so it is a $\frac{0}{0}$ form.

Let

$𝑎=(256-8𝑥{)}^{\frac{1}{4}}$

Then

${𝑎}^4=256-8𝑥$

So,

${𝑎}^4-4^4=-8𝑥$

Using ${𝑝}^4-{𝑞}^4=(𝑝-𝑞)({𝑝}^3+{𝑝}^2𝑞+𝑝{𝑞}^2+{𝑞}^3)$,

$(𝑎-4)({𝑎}^3+4{𝑎}^2+16𝑎+64)=-8𝑥$