MHT CET 2026 16th April Evening Shift (Mathematics) Question Paper with Solutions

We write each complex number in polar form.

For $1+𝑖\sqrt{\surd 3}$,

$|1+𝑖\sqrt{\surd 3}|=\sqrt{\surd 1^2+(\sqrt{\surd 3}{)}^2}=2$

and its argument is

$\tan 𝜃=\frac{\sqrt{\surd 3}}{1}=\sqrt{\surd 3}$

so

$𝜃=\frac{𝜋}{3}$

Thus,

$1+𝑖\sqrt{\surd 3}=2(\cos \frac{𝜋}{3}+𝑖\sin \frac{𝜋}{3})$

Similarly,

$1-𝑖\sqrt{\surd 3}=2(\cos (-\frac{𝜋}{3})+𝑖\sin (-\frac{𝜋}{3}))$

Now using De Moivre’s theorem,

$(1+𝑖\sqrt{\surd 3}{)}^{2𝑛}=2^{2𝑛}(\cos \frac{2𝑛𝜋}{3}+𝑖\sin \frac{2𝑛𝜋}{3})$

and

$(1-𝑖\sqrt{\surd 3}{)}^{2𝑛}=2^{2𝑛}(\cos (-\frac{2𝑛𝜋}{3})+𝑖\sin (-\frac{2𝑛𝜋}{3}))$

Adding them,

$(1+𝑖\sqrt{\surd 3}{)}^{2𝑛}+(1-𝑖\sqrt{\surd 3}{)}^{2𝑛}$

$=2^{2𝑛}[\cos \frac{2𝑛𝜋}{3}+𝑖\sin \frac{2𝑛𝜋}{3}+\cos (-\frac{2𝑛𝜋}{3})+𝑖\sin (-\frac{2𝑛𝜋}{3})]$

Since $\cos (-𝑥)=\cos 𝑥$ and $\sin (-𝑥)=-\sin 𝑥$, the imaginary parts cancel.

So,

$=2^{2𝑛}[2\cos \frac{2𝑛𝜋}{3}]$

$=2^{2𝑛+1}\cos \frac{2𝑛𝜋}{3}$

Hence the correct option is

$\overline{)\text{Option A }{2}^{2𝑛+1}\cos \left(\frac{2𝑛𝜋}{3}\right)}$

Choose the 2 places out of 4 where the vowels will be placed:

${}^4{𝐶}_2$

Now, there are 5 vowels: $𝑎,𝑒,𝑖,𝑜,𝑢$.

Since no vowel is repeated, we have to arrange 2 different vowels in the selected 2 places.

So the number of ways is:

${}^5{𝑃}_2$

Now the remaining 2 places are for consonants.

There are 21 consonants, and consonants may be repeated.

So each consonant place has 21 choices.

Hence the number of ways to fill the consonant places is:

$21\times 21=(21{)}^2$

So the total number of 4-letter words is:

${}^4{𝐶}_2\times {}^5{𝑃}_2\times (21{)}^2$

Thus, the correct option is:

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

Given expression is

$\cos ({60}^{\circ }-𝐴)\cdot \cos 𝐴\cdot \cos ({60}^{\circ }+𝐴)$

Use the identity

$\cos (𝑥-𝑦)\cos (𝑥+𝑦)={\cos }^2𝑥-{\sin }^2𝑦$

Here, take $𝑥={60}^{\circ }$ and $𝑦=𝐴$.

So,

$\cos ({60}^{\circ }-𝐴)\cos ({60}^{\circ }+𝐴)={\cos }^2{60}^{\circ }-{\sin }^2𝐴$

Since

$\cos {60}^{\circ }=\frac{1}{2}$

we get

${\cos }^2{60}^{\circ }=\frac{1}{4}$

So the expression becomes

$(\frac{1}{4}-{\sin }^2𝐴)\cos 𝐴$

Now use

${\sin }^2𝐴=1-{\cos }^2𝐴$

So,

$(\frac{1}{4}-(1-{\cos }^2𝐴\left)\right)\cos 𝐴$

$=({\cos }^2𝐴-\frac{3}{4})\cos 𝐴$

$={\cos }^3𝐴-\frac{3}{4}\cos 𝐴$

$=\frac{1}{4}(4{\cos }^3𝐴-3\cos 𝐴)$

Using the identity

$\cos 3𝐴=4{\cos }^3𝐴-3\cos 𝐴$

we get

$\cos ({60}^{\circ }-𝐴)\cos 𝐴\cos ({60}^{\circ }+𝐴)=\frac{1}{4}\cos 3𝐴$

So the correct option is

$\overline{)\text{Option A }\frac{1}{4}\cos 3𝐴}$

Given,

$\cot 𝜃\cdot \cot 2𝜃=1$

Use the identity

$\cot 2𝜃=\frac{{\cot }^2𝜃-1}{2\cot 𝜃}$

So,

$\cot 𝜃\cdot \frac{{\cot }^2𝜃-1}{2\cot 𝜃}=1$

Cancel $\cot 𝜃$:

$\frac{{\cot }^2𝜃-1}{2}=1$

So,

${\cot }^2𝜃-1=2$

${\cot }^2𝜃=3$

Hence,

$\cot 𝜃=\pm \sqrt{\surd 3}$

We know,

$\cot \frac{𝜋}{6}=\sqrt{\surd 3}$

So the general solution is

$𝜃=𝑛𝜋\pm \frac{𝜋}{6},𝑛\in \mathbb{Z}$

Thus, the correct option is

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

The given line is

$(2+𝑘)𝑥+(1+𝑘)𝑦=5+7𝑘$

For a fixed point, the same point must lie on this line for every value of $𝑘$.

First, separate the terms with $𝑘$ and without $𝑘$:

$2𝑥+𝑘𝑥+𝑦+𝑘𝑦=5+7𝑘$

$(2𝑥+𝑦)+𝑘(𝑥+𝑦)=5+7𝑘$

Bring everything to one side:

$(2𝑥+𝑦-5)+𝑘(𝑥+𝑦-7)=0$

Since this must be true for every value of $𝑘$, both parts must be zero:

$2𝑥+𝑦-5=0$

and

$𝑥+𝑦-7=0$

So,

$2𝑥+𝑦=5$

$𝑥+𝑦=7$

Subtract the second equation from the first:

$(2𝑥+𝑦)-(𝑥+𝑦)=5-7$

$𝑥=-2$

Now put $𝑥=-2$ in $𝑥+𝑦=7$:

$-2+𝑦=7$

$𝑦=9$

So the fixed point is $(-2,9)$.

Distance from the origin is

$𝑑=\sqrt{\surd {𝑥}^2+{𝑦}^2}$

So,

${𝑑}^2={𝑥}^2+{𝑦}^2$

${𝑑}^2=(-2{)}^2+9^2$

${𝑑}^2=4+81=85$

So the correct option is

$\overline{)85}$

Option D.

Given equation:

$2{𝑥}^2+7𝑥𝑦+3{𝑦}^2=0$

Factorising,

$2{𝑥}^2+7𝑥𝑦+3{𝑦}^2=(2𝑥+𝑦)(𝑥+3𝑦)$

So the two lines are

$2𝑥+𝑦=0$

and

$𝑥+3𝑦=0$

Their slopes are:

${𝑚}_1=-2,{𝑚}_2=-\frac{1}{3}$

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,

$\tan 𝜃=\mid \frac{-2-(-\frac{1}{3})}{1+(-2)(-\frac{1}{3})}\mid$

$=\mid \frac{-2+\frac{1}{3}}{1+\frac{2}{3}}\mid$

$=\mid \frac{-\frac{5}{3}}{\frac{5}{3}}\mid =1$

Since $𝜃$ is acute,

$𝜃={45}^{\circ }$

So,

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

Now substitute in the given expression:

$\frac{2\cos 𝜃-3\sin 𝜃}{4\sin 𝜃+5\cos 𝜃}=\frac{2\cdot \frac{1}{\sqrt{\surd 2}}-3\cdot \frac{1}{\sqrt{\surd 2}}}{4\cdot \frac{1}{\sqrt{\surd 2}}+5\cdot \frac{1}{\sqrt{\surd 2}}}$

$=\frac{\frac{2-3}{\sqrt{\surd 2}}}{\frac{4+5}{\sqrt{\surd 2}}}=\frac{-\frac{1}{\sqrt{\surd 2}}}{\frac{9}{\sqrt{\surd 2}}}=-\frac{1}{9}$

So the correct option is

$\overline{)-\frac{1}{9}}$

Option C.

The given line is

$3𝑥+𝑦-6=0$

Writing it in slope form:

$𝑦=-3𝑥+6$

So its slope is

$𝑚=-3$

We need the pair of lines passing through the origin and making an angle $\frac{𝜋}{4}$ with this line.

Let the slope of one required line be ${𝑚}_1$.

Using the formula for angle between two lines,

$\tan 𝜃=\mid \frac{{𝑚}_1-𝑚}{1+{𝑚}_1𝑚}\mid$

Here, $𝜃=\frac{𝜋}{4}$ and $𝑚=-3$.

Since

$\tan \frac{𝜋}{4}=1$

we get

$\mid \frac{{𝑚}_1+3}{1-3{𝑚}_1}\mid =1$

So,

$\frac{{𝑚}_1+3}{1-3{𝑚}_1}=\pm 1$

For the positive case:

${𝑚}_1+3=1-3{𝑚}_1$

$4{𝑚}_1=-2$

${𝑚}_1=-\frac{1}{2}$

For the negative case:

${𝑚}_1+3=-(1-3{𝑚}_1)$

${𝑚}_1+3=-1+3{𝑚}_1$

$2{𝑚}_1=4$

${𝑚}_1=2$

So the two lines through the origin are

$𝑦=2𝑥$

and

$𝑦=-\frac{𝑥}{2}$

Their combined equation is

$(𝑦-2𝑥)(𝑦+\frac{𝑥}{2})=0$

Multiplying,

${𝑦}^2-\frac{3}{2}𝑥𝑦-{𝑥}^2=0$

Multiplying by $-2$,

$2{𝑥}^2+3𝑥𝑦-2{𝑦}^2=0$

So the correct option is

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

The circle is

${𝑥}^2+{𝑦}^2-2𝑥-2𝑦=2$

Complete the squares:

$(𝑥-1{)}^2+(𝑦-1{)}^2=4$

So the center of the circle is

$𝐶(1,1)$

and the radius is

$𝑟=2$

The line is

$𝑥+𝑦=4$

The points $𝐴$ and $𝐵$ lie on this line, so $𝐴𝐵$ is a chord of the circle.

The perpendicular distance from the center $𝐶(1,1)$ to the line $𝑥+𝑦-4=0$ is

$𝑑=\frac{|1+1-4|}{\sqrt{\surd 1^2+1^2}}$

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

This distance is the height of $△𝐴𝐵𝐶$ from $𝐶$ to base $𝐴𝐵$.

Now, for a chord at distance $𝑑$ from the center,

$𝐴𝐵=2\sqrt{\surd {𝑟}^2-{𝑑}^2}$

$𝐴𝐵=2\sqrt{\surd 2^2-(\sqrt{\surd 2}{)}^2}$

$𝐴𝐵=2\sqrt{\surd 4-2}=2\sqrt{\surd 2}$

So the area of $△𝐴𝐵𝐶$ is

$\frac{1}{2}\times 𝐴𝐵\times 𝑑$

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

$=2$

So the correct option is

$\overline{)2}$

Option B.

Given limit is

$\underset{𝑥\to 0}{lim}\frac{{45}^{𝑥}-9^{𝑥}-5^{𝑥}+1}{({𝑘}^{𝑥}-1)(3^{𝑥}-1)}=2$

Notice that $45=9\times 5$, so

${45}^{𝑥}=(9\times 5{)}^{𝑥}=9^{𝑥}\cdot 5^{𝑥}$

So the numerator becomes

${45}^{𝑥}-9^{𝑥}-5^{𝑥}+1=9^{𝑥}{5}^{𝑥}-9^{𝑥}-5^{𝑥}+1$

This can be factorised as

$(9^{𝑥}-1)(5^{𝑥}-1)$

So the limit becomes

$\underset{𝑥\to 0}{lim}\frac{(9^{𝑥}-1)(5^{𝑥}-1)}{({𝑘}^{𝑥}-1)(3^{𝑥}-1)}$

Using the standard result

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

we get

$\underset{𝑥\to 0}{lim}\frac{9^{𝑥}-1}{3^{𝑥}-1}=\frac{\ln 9}{\ln 3}$

and

$\underset{𝑥\to 0}{lim}\frac{5^{𝑥}-1}{{𝑘}^{𝑥}-1}=\frac{\ln 5}{\ln 𝑘}$

So the given limit is

$\frac{\ln 9}{\ln 3}\cdot \frac{\ln 5}{\ln 𝑘}=2$

Now,

$\ln 9=\ln 3^2=2\ln 3$

Hence,

$\frac{2\ln 3}{\ln 3}\cdot \frac{\ln 5}{\ln 𝑘}=2$

$2\cdot \frac{\ln 5}{\ln 𝑘}=2$

So,

$\frac{\ln 5}{\ln 𝑘}=1$

$\ln 𝑘=\ln 5$

Thus,

$𝑘=5$

The correct option is Option C, $5$.

For an implication $𝐴\to 𝐵$ to be false, the first part $𝐴$ must be true and the second part $𝐵$ must be false.

Here,

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

is false.

So,

$(𝑝\leftrightarrow 𝑞)\wedge (𝑞\to 𝑟)\wedge \sim 𝑟$

is true, and

$𝑝\wedge \sim 𝑞$

is false.

Since $\sim 𝑟$ is true, we get

$𝑟=𝐹$

Also, $𝑞\to 𝑟$ is true. Since $𝑟=𝐹$, the implication $𝑞\to 𝑟$ can be true only when

$𝑞=𝐹$

Now $𝑝\leftrightarrow 𝑞$ is true, so $𝑝$ and $𝑞$ must have the same truth value. Since $𝑞=𝐹$,

$𝑝=𝐹$

So we have

$𝑝=𝐹,𝑞=𝐹,𝑟=𝐹$

Now check the first statement pattern:

$(𝑝\to 𝑞)\leftrightarrow (𝑞\to 𝑟)$

Since $𝑝=𝐹$ and $𝑞=𝐹$,

$𝑝\to 𝑞=𝐹\to 𝐹=𝑇$

Since $𝑞=𝐹$ and $𝑟=𝐹$,

$𝑞\to 𝑟=𝐹\to 𝐹=𝑇$

So,

$(𝑝\to 𝑞)\leftrightarrow (𝑞\to 𝑟)=𝑇\leftrightarrow 𝑇=𝑇$

Now check the second statement pattern:

$\sim (𝑝\vee 𝑟)\to (𝑞\wedge 𝑝)$

Since $𝑝=𝐹$ and $𝑟=𝐹$,

$𝑝\vee 𝑟=𝐹\vee 𝐹=𝐹$

So,

$\sim (𝑝\vee 𝑟)=𝑇$

Also,

$𝑞\wedge 𝑝=𝐹\wedge 𝐹=𝐹$

Thus,

$\sim (𝑝\vee 𝑟)\to (𝑞\wedge 𝑝)=𝑇\to 𝐹=𝐹$

Hence the truth values are

$(𝑇,𝐹)$

So the correct option is Option B.

Given statement is

$(𝑝\vee \sim 𝑞)\to (𝑝\wedge \sim 𝑞)$

For a statement of the form $𝐴\to 𝐵$, its contrapositive is

$\sim 𝐵\to \sim 𝐴$

Here,

$𝐴=𝑝\vee \sim 𝑞$

and

$𝐵=𝑝\wedge \sim 𝑞$

So the contrapositive is

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

Now we need the negation of this contrapositive.

The negation of an implication $𝑟\to 𝑠$ is

$𝑟\wedge \sim 𝑠$

So,

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

becomes

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

Using double negation,

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

Now apply De Morgan’s law:

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

So the expression becomes

$(\sim 𝑝\vee 𝑞)\wedge (𝑝\vee \sim 𝑞)$

Hence, the correct option is

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

Using the usual notation, $𝑎,𝑏,𝑐$ are the sides opposite to angles $𝐴,𝐵,𝐶$ respectively.

Given:

$𝑎=4,𝑏=5,𝑐=6$

By the cosine rule,

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

So,

$\cos 𝐴=\frac{5^2+6^2-4^2}{2\cdot 5\cdot 6}=\frac{25+36-16}{60}=\frac{45}{60}=\frac{3}{4}$

Now,

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

So,

$\cos 𝐶=\frac{4^2+5^2-6^2}{2\cdot 4\cdot 5}=\frac{16+25-36}{40}=\frac{5}{40}=\frac{1}{8}$

Now check $\cos 2𝐴$:

$\cos 2𝐴=2{\cos }^2𝐴-1$

Since $\cos 𝐴=\frac{3}{4}$,

$\cos 2𝐴=2{\left(\frac{3}{4}\right)}^2-1=2\cdot \frac{9}{16}-1=\frac{18}{16}-1=\frac{2}{16}=\frac{1}{8}$

So,

$\cos 𝐶=\cos 2𝐴$

Hence,

$𝐶=2𝐴$

So the correct option is:

$\overline{)\text{Option B }2𝐴}$

We know the property of inverse of product of two matrices:

$(𝐴𝐵{)}^{-1}={𝐵}^{-1}{𝐴}^{-1}$

Here ${𝐵}^{-1}$ is already given, so we only need ${𝐴}^{-1}$.

Given,

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

For a $2\times 2$ matrix,

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

So,

$\left|𝐴\right|=2(-2)-3\left(5\right)=-4-15=-19$

Thus,

${𝐴}^{-1}=\frac{1}{-19}\left[\begin{array}{cc}-2& -3\\ -5& 2\end{array}\right]$

${𝐴}^{-1}=\frac{1}{19}\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$

Now,

$(𝐴𝐵{)}^{-1}={𝐵}^{-1}{𝐴}^{-1}$

$=⎡⎣⎢\begin{array}{cc}\frac{1}{5}& \frac{2}{5}\\ \frac{2}{5}& -\frac{1}{5}\end{array}⎤⎦⎥\cdot \frac{1}{19}\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$

Take $\frac{1}{5}$ and $\frac{1}{19}$ together:

$=\frac{1}{95}\left[\begin{array}{cc}1& 2\\ 2& -1\end{array}\right]\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$

Now multiply:

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

$=\left[\begin{array}{cc}12& -1\\ -1& 8\end{array}\right]$

So,

$(𝐴𝐵{)}^{-1}=\frac{1}{95}\left[\begin{array}{cc}12& -1\\ -1& 8\end{array}\right]$

Hence, the correct option is Option B.

The matrix $𝐴$ is given by

${𝑎}_{𝑖𝑗}=|2𝑖-5𝑗|$

For a $3\times 3$ matrix, $𝑖,𝑗=1,2,3$.

So,

$𝐴=⎡⎣⎢\begin{array}{ccc}\left|2\right(1)-5(1\left)\right|& \left|2\right(1)-5(2\left)\right|& \left|2\right(1)-5(3\left)\right|\\ \left|2\right(2)-5(1\left)\right|& \left|2\right(2)-5(2\left)\right|& \left|2\right(2)-5(3\left)\right|\\ \left|2\right(3)-5(1\left)\right|& \left|2\right(3)-5(2\left)\right|& \left|2\right(3)-5(3\left)\right|\end{array}⎤⎦⎥$

$𝐴=⎡⎣⎢\begin{array}{ccc}3& 8& 13\\ 1& 6& 11\\ 1& 4& 9\end{array}⎤⎦⎥$

We need the element in the $2^{\text{nd}}$ row and $3^{\text{rd}}$ column of ${𝐴}^{-1}$, that is $({𝐴}^{-1}{)}_{23}$.

Using the adjoint formula,

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

The element $({𝐴}^{-1}{)}_{23}$ is

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

because cofactors get transposed in the adjoint.

Now find $\left|𝐴\right|$:

$\left|𝐴\right|=\mid \begin{array}{ccc}3& 8& 13\\ 1& 6& 11\\ 1& 4& 9\end{array}\mid$

Expanding along the first row,

$\left|𝐴\right|=3(54-44)-8(9-11)+13(4-6)$

$\left|𝐴\right|=3\left(10\right)-8(-2)+13(-2)$

$\left|𝐴\right|=30+16-26=20$

Now find ${𝐶}_{32}$.

Delete the $3^{\text{rd}}$ row and $2^{\text{nd}}$ column:

${𝐶}_{32}=(-1{)}^{3+2}\mid \begin{array}{cc}3& 13\\ 1& 11\end{array}\mid$

${𝐶}_{32}=-(33-13)$

${𝐶}_{32}=-20$

So,

$({𝐴}^{-1}{)}_{23}=\frac{-20}{20}=-1$

Hence, the correct option is

$\overline{)-1}$

Option D.

Given

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

First find ${𝐴}^2$:

${𝐴}^2=\left[\begin{array}{cc}2& -1\\ 0& 2\end{array}\right]\left[\begin{array}{cc}2& -1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}4& -4\\ 0& 4\end{array}\right]$

Also,

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

and

$𝑦{𝐼}_2=\left[\begin{array}{cc}𝑦& 0\\ 0& 𝑦\end{array}\right]$

Now use the given condition:

${𝐴}^2+𝑥𝐴+𝑦{𝐼}_2={𝑂}_2$

So,

$\left[\begin{array}{cc}4& -4\\ 0& 4\end{array}\right]+\left[\begin{array}{cc}2𝑥& -𝑥\\ 0& 2𝑥\end{array}\right]+\left[\begin{array}{cc}𝑦& 0\\ 0& 𝑦\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Adding corresponding entries,

$\left[\begin{array}{cc}4+2𝑥+𝑦& -4-𝑥\\ 0& 4+2𝑥+𝑦\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

So,

$-4-𝑥=0$

$𝑥=-4$

Now substitute $𝑥=-4$ in

$4+2𝑥+𝑦=0$

$4+2(-4)+𝑦=0$

$4-8+𝑦=0$

$𝑦=4$

Hence,

$𝑥=-4,𝑦=4$

So the correct option is Option B.

For $𝑥\in (0,\frac{𝜋}{2})$, we have $\frac{𝑥}{2}\in (0,\frac{𝜋}{4})$.

Use the identities:

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

and

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

Since $\frac{𝑥}{2}$ lies in the first quadrant, both $\cos \frac{𝑥}{2}$ and $\sin \frac{𝑥}{2}$ are positive, and

$\cos \frac{𝑥}{2}>\sin \frac{𝑥}{2}$

So,

$\sqrt{\surd 1+\sin 𝑥}=\cos \frac{𝑥}{2}+\sin \frac{𝑥}{2}$

and

$\sqrt{\surd 1-\sin 𝑥}=\cos \frac{𝑥}{2}-\sin \frac{𝑥}{2}$

Now substitute:

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

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

$=-\cot \frac{𝑥}{2}$

So the given expression becomes

${\cot }^{-1}(-\cot \frac{𝑥}{2})$

Now,

$\cot (𝜋-\frac{𝑥}{2})=-\cot \frac{𝑥}{2}$

Also, ${\cot }^{-1}$ takes values in $(0,𝜋)$, and $𝜋-\frac{𝑥}{2}$ lies in that interval.

Hence,

${\cot }^{-1}(-\cot \frac{𝑥}{2})=𝜋-\frac{𝑥}{2}$

So the correct option is

$\overline{)𝜋-\frac{𝑥}{2}}$

Option D.

Let

${\tan }^{-1}𝑥=𝐴$

and

${\cot }^{-1}𝑥=𝐵$

Using the NCERT principal values,

${\tan }^{-1}𝑥+{\cot }^{-1}𝑥=\frac{𝜋}{2}$

So,

$𝐴+𝐵=\frac{𝜋}{2}$

Given,

${𝐴}^2+{𝐵}^2=\frac{5{𝜋}^2}{8}$

Now,

$(𝐴+𝐵{)}^2={𝐴}^2+{𝐵}^2+2𝐴𝐵$

Substitute the values:

${\left(\frac{𝜋}{2}\right)}^2=\frac{5{𝜋}^2}{8}+2𝐴𝐵$

$\frac{{𝜋}^2}{4}=\frac{5{𝜋}^2}{8}+2𝐴𝐵$

$2𝐴𝐵=\frac{{𝜋}^2}{4}-\frac{5{𝜋}^2}{8}$

$2𝐴𝐵=-\frac{3{𝜋}^2}{8}$

$𝐴𝐵=-\frac{3{𝜋}^2}{16}$

So $𝐴$ and $𝐵$ have sum $\frac{𝜋}{2}$ and product $-\frac{3{𝜋}^2}{16}$.

The two values are

$-\frac{𝜋}{4}\text{and}\frac{3𝜋}{4}$

Since

${\tan }^{-1}𝑥\in (-\frac{𝜋}{2},\frac{𝜋}{2})$

we take

${\tan }^{-1}𝑥=-\frac{𝜋}{4}$

Hence,

$𝑥=\tan (-\frac{𝜋}{4})$

$𝑥=-1$

So the correct option is

$\overline{)-1}$

Option A.

Given

$𝑓\left(\frac{𝑥-1}{𝑥+1}\right)=𝑥+1$

We need to find $𝑓\left(𝑥\right)$ first.

Put

$𝑦=\frac{𝑥-1}{𝑥+1}$

Now express $𝑥$ in terms of $𝑦$:

$𝑦(𝑥+1)=𝑥-1$

$𝑦𝑥+𝑦=𝑥-1$

$𝑦𝑥-𝑥=-1-𝑦$

$𝑥(𝑦-1)=-(1+𝑦)$

$𝑥=\frac{1+𝑦}{1-𝑦}$

So,

$𝑥+1=\frac{1+𝑦}{1-𝑦}+1$

$𝑥+1=\frac{1+𝑦+1-𝑦}{1-𝑦}$

$𝑥+1=\frac{2}{1-𝑦}$

Hence,

$𝑓\left(𝑦\right)=\frac{2}{1-𝑦}$

So,

$𝑓\left(𝑥\right)=\frac{2}{1-𝑥}$

Now integrate:

$\int 𝑓\left(𝑥\right)𝑑𝑥=\int \frac{2}{1-𝑥}𝑑𝑥$

Since

$\frac{𝑑}{𝑑𝑥}(1-𝑥)=-1$

we get

$\int \frac{2}{1-𝑥}𝑑𝑥=-2\log |1-𝑥|+𝑐$

So the correct option is

$\overline{)-2\log |1-𝑥|+𝑐}$

Option B.

Given,

${𝑒}^{𝑥}+{𝑒}^{𝑓\left(𝑥\right)}=𝑒$