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

The height is given by

$ℎ=3+14𝑡-5{𝑡}^2$

This is a quadratic polynomial in $𝑡$.

Since the coefficient of ${𝑡}^2$ is negative, that is $-5$, the graph opens downwards. So it has a maximum value.

For a quadratic expression

$𝑎{𝑡}^2+𝑏𝑡+𝑐$

the maximum or minimum occurs at

$𝑡=\frac{-𝑏}{2𝑎}$

Here,

$𝑎=-5,𝑏=14$

So,

$𝑡=\frac{-14}{2(-5)}$

$𝑡=\frac{14}{10}=1.4$

Now put $𝑡=1.4$ in the height formula:

$ℎ=3+14\left(1.4\right)-5(1.4{)}^2$

$ℎ=3+19.6-5\left(1.96\right)$

$ℎ=3+19.6-9.8$

$ℎ=12.8$

So, the maximum height is

$\overline{)12.8}$

Correct option: Option B.

Given,

$(1+𝑖)(1+2𝑖)(1+3𝑖)\cdots (1+𝑛𝑖)=𝑥+𝑖𝑦$

We need the value of

$\left(2\right)\left(5\right)\left(10\right)\cdots (1+{𝑛}^2)$

Notice that

$2=1+1^2,5=1+2^2,10=1+3^2$

So the given product is

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

Now take modulus on both sides of the complex product.

For each factor,

$|1+𝑘𝑖{|}^2=1^2+{𝑘}^2=1+{𝑘}^2$

So,

$\left|\right(1+𝑖\left)\right(1+2𝑖)\cdots (1+𝑛𝑖){|}^2=(1+1^2\left)\right(1+2^2)\cdots (1+{𝑛}^2)$

But the product is given as $𝑥+𝑖𝑦$.

Hence,

$|𝑥+𝑖𝑦{|}^2={𝑥}^2+{𝑦}^2$

So,

$\left(2\right)\left(5\right)\left(10\right)\cdots (1+{𝑛}^2)={𝑥}^2+{𝑦}^2$

The correct option is

$\overline{){𝑥}^2+{𝑦}^2}$

Option B.

The required numbers are strictly between $10$ and $1000$.

So they can be either:

• $2$-digit numbers

• $3$-digit numbers

The available digits are

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

and no digit can be repeated.

For $2$-digit numbers, the tens place cannot be $0$.

So the tens place has $6$ choices:

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

After choosing the tens digit, the units place has $6$ remaining choices.

So total $2$-digit numbers are

$6\times 6=36$

But $10$ is included in this count, and the number must be strictly greater than $10$.

So valid $2$-digit numbers are

$36-1=35$

For $3$-digit numbers, the hundreds place cannot be $0$.

Hundreds place: $6$ choices

Tens place: $6$ choices

Units place: $5$ choices

So total $3$-digit numbers are

$6\times 6\times 5=180$

Total required arrangements:

$35+180=215$

So the correct option is

$\overline{)215}$

Option B.

Use the identities

$\sin 3𝐴=3\sin 𝐴-4{\sin }^3𝐴$

and

$\cos 3𝐴=4{\cos }^3𝐴-3\cos 𝐴$

So,

$\sin 3𝐴=\sin 𝐴(3-4{\sin }^2𝐴)$

and

$\cos 3𝐴=\cos 𝐴(4{\cos }^2𝐴-3)$

Now substitute in the given expression:

$\frac{{\sin }^{2}3𝐴}{{\sin }^2𝐴}-\frac{{\cos }^{2}3𝐴}{{\cos }^2𝐴}$

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

$=(3-4{\sin }^2𝐴{)}^2-(4{\cos }^2𝐴-3{)}^2$

Now write these in terms of $\cos 2𝐴$.

Since

${\sin }^2𝐴=\frac{1-\cos 2𝐴}{2}$

we get

$3-4{\sin }^2𝐴=3-4\left(\frac{1-\cos 2𝐴}{2}\right)=1+2\cos 2𝐴$

Also,

${\cos }^2𝐴=\frac{1+\cos 2𝐴}{2}$

so

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

Thus the expression becomes

$(1+2\cos 2𝐴{)}^2-(2\cos 2𝐴-1{)}^2$

Using ${𝑎}^2-{𝑏}^2=(𝑎-𝑏)(𝑎+𝑏)$,

$=[(1+2\cos 2𝐴)-(2\cos 2𝐴-1)][(1+2\cos 2𝐴)+(2\cos 2𝐴-1)]$

$=2\times 4\cos 2𝐴$

$=8\cos 2𝐴$

So the correct option is

$\overline{)\text{Option B }8\cos 2𝐴}$

Given,

$\sin 𝑥-\sin 2𝑥+\sin 3𝑥=0$

Group $\sin 𝑥$ and $\sin 3𝑥$ together:

$\sin 𝑥+\sin 3𝑥-\sin 2𝑥=0$

Using the identity

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

we get

$\sin 𝑥+\sin 3𝑥=2\sin 2𝑥\cos 𝑥$

So the equation becomes

$2\sin 2𝑥\cos 𝑥-\sin 2𝑥=0$

Take $\sin 2𝑥$ common:

$\sin 2𝑥(2\cos 𝑥-1)=0$

So,

$\sin 2𝑥=0$

or

$2\cos 𝑥-1=0$

Now $0\le 𝑥\le \frac{𝜋}{2}$.

For $\sin 2𝑥=0$:

$2𝑥=0,𝜋$

So,

$𝑥=0,\frac{𝜋}{2}$

For $2\cos 𝑥-1=0$:

$\cos 𝑥=\frac{1}{2}$

So,

$𝑥=\frac{𝜋}{3}$

Thus the values of $𝑥$ are

$0,\frac{𝜋}{3},\frac{𝜋}{2}$

So the number of values is

$3$

Hence, the correct option is Option C.

Suppose the line cuts the $𝑥$-axis at $𝑎$ and the $𝑦$-axis at $𝑏$.

Since the intercepts are non-zero, we can write the line in intercept form:

$\frac{𝑥}{𝑎}+\frac{𝑦}{𝑏}=1$

It is given that the sum of the non-zero intercepts is zero.

So,

$𝑎+𝑏=0$

Hence,

$𝑏=-𝑎$

Now substitute $𝑏=-𝑎$ in the intercept form:

$\frac{𝑥}{𝑎}+\frac{𝑦}{-𝑎}=1$

$\frac{𝑥-𝑦}{𝑎}=1$

$𝑥-𝑦=𝑎$

This line passes through $𝐴(3,4)$, so put $𝑥=3$ and $𝑦=4$:

$3-4=𝑎$

$𝑎=-1$

So there is only one such line:

$𝑥-𝑦=-1$

Hence, the number of lines is

$\overline{)1}$

Correct option: Option A.

For the pair of lines

$𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0,$

put $𝑦=𝑚𝑥$, where $𝑚$ is the slope. Then

$𝑎+2ℎ𝑚+𝑏{𝑚}^2=0.$

Thus,

$𝑚=\frac{-ℎ\pm \sqrt{\surd {ℎ}^2-𝑎𝑏}}{𝑏}.$

Given

$4𝑎𝑏=3{ℎ}^2\Rightarrow 𝑎𝑏=\frac{3{ℎ}^2}{4}.$

So,

$\sqrt{\surd {ℎ}^2-𝑎𝑏}=\sqrt{\surd {ℎ}^2-\frac{3{ℎ}^2}{4}}=\frac{\left|ℎ\right|}{2}.$

The two slopes are proportional to

$-ℎ+\frac{ℎ}{2}=-\frac{ℎ}{2}$

and

$-ℎ-\frac{ℎ}{2}=-\frac{3ℎ}{2}.$

Hence, their ratio is

${𝑚}_1:{𝑚}_2=1:3.$

So, the correct answer is Option D: $\overline{)1:3}$.

The curve

${𝑥}^2+{𝑦}^2=10$

is a circle with centre $(0,0)$ and radius $\sqrt{\surd 10}$.

A line parallel to

$2𝑥+𝑦-1=0$

has the form

$2𝑥+𝑦=𝑐.$

For this line to be tangent to the circle, the perpendicular distance from the centre to the line must equal the radius:

$\frac{\left|𝑐\right|}{\sqrt{\surd 2^2+1^2}}=\sqrt{\surd 10}.$

Thus,

$\frac{\left|𝑐\right|}{\sqrt{\surd 5}}=\sqrt{\surd 10}$

$\left|𝑐\right|=\sqrt{\surd 50}=5\sqrt{\surd 2}.$

Hence, the two tangents are

$\overline{)2𝑥+𝑦=\pm 5\sqrt{\surd 2}}.$

So, the correct answer is Option B.

Substitute the line $𝑦=2𝑥+𝜆$ into the hyperbola:

$36{𝑥}^2-25{𝑦}^2=3600$

So,

$36{𝑥}^2-25(2𝑥+𝜆{)}^2=3600$

On simplifying,

$64{𝑥}^2+100𝜆𝑥+25{𝜆}^2+3600=0$

For the line to be a tangent, this quadratic equation must have equal roots. Hence, its discriminant is zero:

$(100𝜆{)}^2-4(64\left)\right(25{𝜆}^2+3600)=0$

$10000{𝜆}^2-6400{𝜆}^2-921600=0$

$3600{𝜆}^2=921600$

${𝜆}^2=256$

Thus,

$\overline{)𝜆=\pm 16}$

Hence, Option C is correct.

Using the standard limits

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

$\underset{𝜃\to 0}{lim}\frac{\tan 𝜃}{𝜃}=1,\underset{𝑢\to 0}{lim}\frac{\log (1+𝑢)}{𝑢}=1,$

we have, as $𝑥\to 0$,

$5^{𝑥}-1\sim 𝑥\log 5,$

$\csc (𝑥\log 5)\sim \frac{1}{𝑥\log 5},\tan (𝑥\log 5)\sim 𝑥\log 5,$

and

$\log (1+{𝑥}^2\log 25)\sim {𝑥}^2\log 25=2{𝑥}^2\log 5.$

Thus,

$\begin{array}{cc}\underset{𝑥\to 0}{lim}\frac{(5^{𝑥}-1{)}^4\csc (𝑥\log 5)}{\tan (𝑥\log 5)\log (1+{𝑥}^2\log 25)}& =\frac{(𝑥\log 5{)}^4\cdot \frac{1}{𝑥\log 5}}{(𝑥\log 5)(2{𝑥}^2\log 5)}\\ & =\frac{\log 5}{2}\\ & =\log \sqrt{\surd 5}.\end{array}$

Hence, the correct option is

$\overline{)\text{Option B: }\log \sqrt{\surd 5}}.$

Let $𝑝$ represent the switch ${𝑆}_1$ and $𝑞$ represent the switch ${𝑆}_2$.

In a switching circuit, switches in series mean “and” $\wedge$, and switches in parallel mean “or” $\vee$.

The given circuit has three parallel branches. So, the logical expression for the circuit is:

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

Now group the first two terms, because both of them have $𝑝$ common:

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

Here, $𝑞\vee \sim 𝑞$ means either $𝑞$ is true or $𝑞$ is not true. This is always true, so it is denoted by $𝑇$.

$𝐸=(𝑝\wedge 𝑇)\vee (\sim 𝑝\wedge 𝑞)$

Since $𝑝\wedge 𝑇=𝑝$, we get:

$𝐸=𝑝\vee (\sim 𝑝\wedge 𝑞)$

Now use the distributive law:

$𝐸=(𝑝\vee \sim 𝑝)\wedge (𝑝\vee 𝑞)$

Again, $𝑝\vee \sim 𝑝$ is always true, so it becomes $𝑇$.

$𝐸=𝑇\wedge (𝑝\vee 𝑞)$

Since $𝑇\wedge (𝑝\vee 𝑞)=𝑝\vee 𝑞$, we get:

$𝐸=𝑝\vee 𝑞$

So, the simplest form is $𝑝\vee 𝑞$. This means switches ${𝑆}_1$ and ${𝑆}_2$ are connected in parallel.

Using

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

the given condition becomes

$\frac{𝑏(1+\cos 𝐶)+𝑐(1+\cos 𝐵)}{2}=\frac{3𝑎}{2}.$

By the cosine rule,

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

and

$𝑐\cos 𝐵=\frac{{𝑎}^2+{𝑐}^2-{𝑏}^2}{2𝑎}.$

Adding these,

$𝑏\cos 𝐶+𝑐\cos 𝐵=𝑎.$

Hence,

$\frac{𝑏+𝑐+𝑎}{2}=\frac{3𝑎}{2}.$

So,

$𝑏+𝑐=2𝑎.$

Thus, $𝑎$ is the arithmetic mean of $𝑏$ and $𝑐$. Therefore, $𝑏,𝑎,𝑐$ are in A.P.

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

Using the sine rule,

$\frac{𝑎}{𝑏}=\frac{\sin 𝐴}{\sin 𝐵}$

Given,

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

Also,

$𝐴+𝐵={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

For $𝐴={105}^{\circ }$ and $𝐵={15}^{\circ }$,

$\frac{\sin {105}^{\circ }}{\sin {15}^{\circ }}=\frac{\frac{\sqrt{\surd 6}+\sqrt{\surd 2}}{4}}{\frac{\sqrt{\surd 6}-\sqrt{\surd 2}}{4}}=2+\sqrt{\surd 3}$

This matches $\frac{𝑎}{𝑏}$.

Hence,

$\overline{)𝐴={105}^{\circ },𝐵={15}^{\circ }}$

So, Option A is correct.

For a matrix

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

its inverse is

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

Here,

$𝐴=\left[\begin{array}{cc}3& -2\\ 1& 4\end{array}\right].$

Its determinant is

$\left|𝐴\right|=\left(3\right)\left(4\right)-(-2)\left(1\right)=12+2=14.$

Thus,

${𝐴}^{-1}=\frac{1}{14}\left[\begin{array}{cc}4& 2\\ -1& 3\end{array}\right]=⎡⎣⎢\begin{array}{cc}\frac{2}{7}& \frac{1}{7}\\ -\frac{1}{14}& \frac{3}{14}\end{array}⎤⎦⎥.$

Hence, the correct answer is Option A.

For any square matrix $𝐴$,

$𝐴(\mathrm{adj}𝐴)=\left|𝐴\right|𝐼$

Given

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

its determinant is

$\left|𝐴\right|=\left(5𝑎\right)\left(2\right)-(-𝑏)\left(3\right)=10𝑎+3𝑏.$

Also,

${𝐴}^{𝑇}=\left[\begin{array}{cc}5𝑎& 3\\ -𝑏& 2\end{array}\right].$

Thus,

$𝐴{𝐴}^{𝑇}=\left[\begin{array}{cc}5𝑎& -𝑏\\ 3& 2\end{array}\right]\left[\begin{array}{cc}5𝑎& 3\\ -𝑏& 2\end{array}\right]=\left[\begin{array}{cc}25{𝑎}^2+{𝑏}^2& 15𝑎-2𝑏\\ 15𝑎-2𝑏& 13\end{array}\right].$

Since $𝐴(\mathrm{adj}𝐴)=𝐴{𝐴}^{𝑇}$, we have

$\left[\begin{array}{cc}10𝑎+3𝑏& 0\\ 0& 10𝑎+3𝑏\end{array}\right]=\left[\begin{array}{cc}25{𝑎}^2+{𝑏}^2& 15𝑎-2𝑏\\ 15𝑎-2𝑏& 13\end{array}\right].$

Comparing corresponding entries,

$15𝑎-2𝑏=0\Rightarrow 𝑏=\frac{15𝑎}{2},$

and

$10𝑎+3𝑏=13.$

Substituting $𝑏=\frac{15𝑎}{2}$,

$10𝑎+\frac{45𝑎}{2}=13$

$\frac{65𝑎}{2}=13\Rightarrow 𝑎=\frac{2}{5}.$

Hence,

$𝑏=\frac{15}{2}\times \frac{2}{5}=3.$

Therefore,

$5𝑎+𝑏=5\left(\frac{2}{5}\right)+3=5.$

The correct option is $\overline{)\text{C }5}$.

Put

$𝑥=\sin 𝜃,\sqrt{\surd 1-{𝑥}^2}=\cos 𝜃,$

where $-\frac{𝜋}{2}\le 𝜃\le \frac{𝜋}{2}$.

Then

$\frac{𝑥-\sqrt{\surd 1-{𝑥}^2}}{𝑥+\sqrt{\surd 1-{𝑥}^2}}=\frac{\sin 𝜃-\cos 𝜃}{\sin 𝜃+\cos 𝜃}=\tan (𝜃-\frac{𝜋}{4}).$

Thus, on each interval where the given function is defined,

$𝑦=𝜃-\frac{𝜋}{4}$

up to an additive constant $𝜋$. An additive constant does not affect the derivative. Hence,

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{𝑑𝜃}{𝑑𝑥}.$

Since $𝑥=\sin 𝜃$,

$\frac{𝑑𝑥}{𝑑𝜃}=\cos 𝜃=\sqrt{\surd 1-{𝑥}^2}.$

Therefore,

$\overline{)\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{\sqrt{\surd 1-{𝑥}^2}}}.$

So, the correct answer is Option B.

The principal value range of ${\cos }^{-1}𝑥$ is $[0,𝜋]$.

Since

$\cos \left(\frac{2𝜋}{3}\right)=-\frac{1}{2},$

we have

${\cos }^{-1}(-\frac{1}{2})=\frac{2𝜋}{3}.$

Thus,

$\cos ({\cos }^{-1}(-\frac{1}{2})+\frac{𝜋}{3})=\cos (\frac{2𝜋}{3}+\frac{𝜋}{3})=\cos 𝜋=-1.$

Hence, the correct answer is Option B: $-1$.

Let

$𝐴={\sin }^{-1}\left(\frac{12}{13}\right),𝐵={\cos }^{-1}\left(\frac{4}{5}\right),𝐶={\tan }^{-1}\left(\frac{63}{16}\right).$

Since $𝐴$ is acute, using a right triangle,

$\sin 𝐴=\frac{12}{13}\Rightarrow \cos 𝐴=\frac{5}{13}\Rightarrow \tan 𝐴=\frac{12}{5}.$

Similarly,

$\cos 𝐵=\frac{4}{5}\Rightarrow \sin 𝐵=\frac{3}{5}\Rightarrow \tan 𝐵=\frac{3}{4}.$

Now,

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

$=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot \frac{3}{4}}=\frac{\frac{63}{20}}{-\frac{4}{5}}=-\frac{63}{16}.$

Both $𝐴$ and $𝐵$ are acute, and $\tan (𝐴+𝐵)<0$. Hence $𝐴+𝐵$ lies in the second quadrant.

Also,

$\tan 𝐶=\frac{63}{16},$

where $𝐶$ is acute. Thus,

$𝐴+𝐵=𝜋-𝐶.$

So,

$𝐴+𝐵+𝐶=𝜋.$

Hence, the correct answer is

$\overline{)𝜋}$

Option C.

Given

$𝑓\left(𝑥\right)=\frac{4𝑥+3}{6𝑥-4},$

we find $𝑔\left(𝑥\right)=(𝑓\circ 𝑓)\left(𝑥\right)=𝑓\left(𝑓\right(𝑥\left)\right)$:

$𝑓\left(𝑓\right(𝑥\left)\right)=\frac{4\left(\frac{4𝑥+3}{6𝑥-4}\right)+3}{6\left(\frac{4𝑥+3}{6𝑥-4}\right)-4}.$

Simplifying the numerator and denominator,

$𝑓\left(𝑓\right(𝑥\left)\right)=\frac{\frac{34𝑥}{6𝑥-4}}{\frac{34}{6𝑥-4}}=𝑥.$

Thus,

$𝑔\left(𝑥\right)=𝑥,$

so $𝑔$ is the identity function. Repeated application of $𝑔$ does not change the input:

$(𝑔\circ 𝑔\circ 𝑔\circ 𝑔\circ 𝑔)\left(3\right)=3.$

Hence, the correct answer is

$\overline{)3}$

Option A.

For ${𝑓}^{\prime }\left(𝑥\right)$ to be continuous everywhere, we only need to check the joining point $𝑥=-1$.

First, $𝑓$ must be continuous at $𝑥=-1$:

$𝑎(-1{)}^2+𝑏=𝑏(-1{)}^2+𝑎(-1)+4$

$𝑎+𝑏=𝑏-𝑎+4$

$2𝑎=4⟹𝑎=2$

Now differentiate both parts:

${𝑓}^{\prime }\left(𝑥\right)=\{\begin{array}{cc}2𝑎𝑥,& 𝑥<-1\\ 2𝑏𝑥+𝑎,& 𝑥>-1\end{array}$

For continuity of ${𝑓}^{\prime }$ at $𝑥=-1$:

$-2𝑎=-2𝑏+𝑎$

$2𝑏=3𝑎$

Using $𝑎=2$:

$2𝑏=6⟹𝑏=3$

Hence,

$\overline{)𝑎=2,𝑏=3}$

So, the correct answer is Option A.

Let

$𝑦={\log }_8({\log }_5𝑥).$

Using

$\frac{𝑑}{𝑑𝑥}({\log }_{𝑎}𝑢)=\frac{{𝑢}^{\prime }}{𝑢\log 𝑎},$

we get

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{{\log }_5𝑥\log 8}\cdot \frac{1}{𝑥\log 5}.$

Since

${\log }_5𝑥=\frac{\log 𝑥}{\log 5},$

the factor $\log 5$ cancels:

$\frac{𝑑𝑦}{𝑑𝑥}=\frac{1}{𝑥\log 8\log 𝑥}.$

Hence, the correct answer is Option D.

Given

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

Differentiating with respect to $𝑥$,

$\frac{1}{𝑚}{𝑦}^{\frac{1}{𝑚}-1}{𝑦}_1-\frac{1}{𝑚}{𝑦}^{-\frac{1}{𝑚}-1}{𝑦}_1=2$

Taking $\frac{{𝑦}_1}{𝑚𝑦}$ common,

$\frac{{𝑦}_1}{𝑚𝑦}({𝑦}^{\frac{1}{𝑚}}-{𝑦}^{-\frac{1}{𝑚}})=2$

Hence,

${𝑦}_2^1{({𝑦}^{\frac{1}{𝑚}}-{𝑦}^{-\frac{1}{𝑚}})}^2=4{𝑚}^2{𝑦}^2$

Also,

$4{𝑥}^2={({𝑦}^{\frac{1}{𝑚}}+{𝑦}^{-\frac{1}{𝑚}})}^2$

Since ${𝑦}^{\frac{1}{𝑚}}\cdot {𝑦}^{-\frac{1}{𝑚}}=1$,

$4({𝑥}^2-1)={({𝑦}^{\frac{1}{𝑚}}-{𝑦}^{-\frac{1}{𝑚}})}^2$

Using (2) in (1),

$4({𝑥}^2-1){𝑦}_2^1=4{𝑚}^2{𝑦}^2$

Thus,

$({𝑥}^2-1){𝑦}_2^1={𝑚}^2{𝑦}^2$

So, the correct answer is Option B: $\overline{){𝑚}^2{𝑦}^2}$.

Since $𝐶$ moves along $𝑥=𝑦$, its coordinates are

$𝐶(𝑥,𝑥).$

The area of triangle $𝐴𝐵𝐶$ is

$𝑠=\frac{1}{2}\mid \mid \begin{array}{ccc}𝑎& 0& 1\\ 0& 𝑏& 1\\ 𝑥& 𝑥& 1\end{array}\mid \mid .$

Taking the usual orientation assumed in the question,

$𝑠=\frac{1}{2}[𝑎𝑏-(𝑎+𝑏\left)𝑥\right].$

Differentiating with respect to $𝑥$,

$\frac{𝑑𝑠}{𝑑𝑥}=-\frac{𝑎+𝑏}{2}.$

Hence, the correct answer is

$\overline{)-\left(\frac{𝑎+𝑏}{2}\right)}$

Option B.

Given

$f(x)=\tan^{-1}(\sin x+\cos x)$

Differentiate:

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

The denominator is always positive. So, $f(x)$ is increasing when

$\cos x-\sin x>0.$

Using

$\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right),$

we need

$\cos\left(x+\frac{\pi}{4}\right)>0.$

This is true for

$-\frac{\pi}{2}<x<\frac{\pi}{4}.$

Hence, the correct answer is

$\boxed{\text{Option D: }\left(-\frac{\pi}{2},\frac{\pi}{4}\right)}.$

Let

$𝑓\left(𝑥\right)=\frac{\log 𝑥}{𝑥},𝑥\in (2,\infty ).$

For every finite $𝑥>2$, both $\log 𝑥>0$ and $𝑥>0$. Hence,

$\frac{\log 𝑥}{𝑥}>0.$

Also,

$\underset{𝑥\to \infty }{lim}\frac{\log 𝑥}{𝑥}=0.$

Thus, the function approaches $0$ but never attains it for any $𝑥\in (2,\infty )$. So $0$ is the infimum, not the minimum value.

Hence, the correct answer is

$\overline{)\text{Option D: Does not exist}}$