AP EAMCET 2020 Mathematics Question Paper with Answer and Solution

(B) The set $X$ represents a grid of points $(a, b)$ where $a$ ranges from $0$ to $20$ ($21$ points) and $b$ ranges from $0$ to $15$ ($16$ points).
To form a rectangle with sides parallel to the axes,we need to choose $2$ distinct values for $a$ from the set $\{0, 1, 2, \dots, 20\}$ and $2$ distinct values for $b$ from the set $\{0, 1, 2, \dots, 15\}$.
The number of ways to choose $2$ values for $a$ is $\binom{21}{2} = \frac{21 \times 20}{2} = 210$.
The number of ways to choose $2$ values for $b$ is $\binom{16}{2} = \frac{16 \times 15}{2} = 120$.
The total number of such rectangles is $210 \times 120 = 25200$.

(C) Given that the angles $A, B, C$ of $\triangle ABC$ are in $AP$,we have $A+C = 2B$.
Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Using the cosine rule in $\triangle ABC$:
$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$
Substituting the given values $AB=6, BC=7, B=60^{\circ}$ and letting $AC=x$:
$\cos 60^{\circ} = \frac{6^2 + 7^2 - x^2}{2 \times 6 \times 7}$
$\frac{1}{2} = \frac{36 + 49 - x^2}{84}$
$42 = 85 - x^2$
$x^2 = 85 - 42 = 43$
$x = \sqrt{43}$
Therefore,$AC = \sqrt{43}$.

(B) The coefficients of the $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(1+x)^n$ are given by $\binom{n}{1}$,$\binom{n}{2}$,and $\binom{n}{3}$ respectively.
Since these are in $A$.$P$.,we have $2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3}$.
Expanding the binomial coefficients: $2 \times \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.
Dividing by $n$ (since $n > 0$): $n-1 = 1 + \frac{(n-1)(n-2)}{6}$.
Multiplying by $6$: $6n - 6 = 6 + (n^2 - 3n + 2)$.
Rearranging the terms: $n^2 - 9n + 14 = 0$.
Factoring the quadratic equation: $(n-7)(n-2) = 0$.
Thus,$n = 7$ or $n = 2$.
For the $4^{\text{th}}$ term to exist,$n$ must be at least $3$,so $n = 7$.

(B) Let $P = \tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \ldots \cdot \tan 89^{\circ}$.
Using the property $\tan(90^{\circ} - \theta) = \cot \theta$,we can write the product as:
$P = (\tan 1^{\circ} \cdot \tan 89^{\circ}) \cdot (\tan 2^{\circ} \cdot \tan 88^{\circ}) \cdot \ldots \cdot (\tan 44^{\circ} \cdot \tan 46^{\circ}) \cdot \tan 45^{\circ}$.
Since $\tan \theta \cdot \tan(90^{\circ} - \theta) = \tan \theta \cdot \cot \theta = 1$,each pair equals $1$.
Thus,$P = 1 \cdot 1 \cdot \ldots \cdot 1 \cdot \tan 45^{\circ} = 1 \cdot 1 = 1$.
The geometric mean of $n$ terms $a_1, a_2, \ldots, a_n$ is $(a_1 \cdot a_2 \cdot \ldots \cdot a_n)^{1/n}$.
Here,$n = 89$,so the geometric mean is $(P)^{1/89} = (1)^{1/89} = 1$.
Hence,option $B$ is correct.

(B) The given series is $x = \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$
We know the binomial expansion $(1-y)^{-n} = 1 + ny + \frac{n(n+1)}{2!}y^2 + \frac{n(n+1)(n+2)}{3!}y^3 + \ldots$
Let $1+x = 1 + \frac{1}{5} + \frac{1 \cdot 3}{5 \cdot 10} + \frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15} + \ldots$
This matches the form $(1-y)^{-n}$ where $ny = \frac{1}{5}$ and $\frac{n(n+1)}{2}y^2 = \frac{3}{50}$.
Solving for $n$ and $y$,we find $n = \frac{1}{2}$ and $y = \frac{2}{5}$.
Thus,$1+x = (1 - \frac{2}{5})^{-1/2} = (\frac{3}{5})^{-1/2} = \sqrt{\frac{5}{3}}$.
So,$x = \sqrt{\frac{5}{3}} - 1$.
Now,calculate $3x^2 + 6x = 3(x^2 + 2x) = 3((x+1)^2 - 1) = 3(x+1)^2 - 3$.
Substituting $x+1 = \sqrt{\frac{5}{3}}$,we get $3(\frac{5}{3}) - 3 = 5 - 3 = 2$.

(C) We know that the sum of the first $n$ squares is given by $1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$.
Given $P(n): 1^2+2^2+3^2+\ldots+n^2 = \frac{6(n-1)(n-2) \ldots(n-2020) + 2n^3+3n^2+n}{6}$.
Substituting the standard sum formula into the given equation,we get:
$\frac{n(n+1)(2n+1)}{6} = (n-1)(n-2) \ldots(n-2020) + \frac{n(n+1)(2n+1)}{6}$.
This equality holds if and only if $(n-1)(n-2) \ldots(n-2020) = 0$.
This product is zero when $n \in \{1, 2, 3, \ldots, 2020\}$.
Therefore,$P(n)$ is true for all $n \leq 2020$.

(B) The given expression is $\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}$.
We can factor out $2^2 = 4$ from the numerator:
$= \lim _{n \rightarrow \infty} \frac{4(1^2+2^2+3^2+\ldots+n^2)}{n^3}$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:
$= 4 \lim _{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3}$.
$= \frac{4}{6} \lim _{n \rightarrow \infty} \frac{n \cdot n(1+\frac{1}{n}) \cdot n(2+\frac{1}{n})}{n^3}$.
$= \frac{2}{3} \lim _{n \rightarrow \infty} (1+\frac{1}{n})(2+\frac{1}{n})$.
As $n \rightarrow \infty$,$\frac{1}{n} \rightarrow 0$,so the limit is $\frac{2}{3} \times (1)(2) = \frac{4}{3}$.
Thus,the correct option is $B$.

(C) Let the three consecutive natural numbers be $(n-1), n, (n+1)$ where $n \geq 2$.
The sum of the cubes of these numbers is:
$(n-1)^3 + n^3 + (n+1)^3 = (n^3 - 3n^2 + 3n - 1) + n^3 + (n^3 + 3n^2 + 3n + 1) = 3n^3 + 6n$.
Factoring the expression,we get $3n(n^2 + 2)$.
If $n$ is a multiple of $3$,then $3n$ is divisible by $9$.
If $n$ is not a multiple of $3$,then $n^2 \equiv 1 \pmod{3}$,so $n^2 + 2 \equiv 1 + 2 \equiv 3 \equiv 0 \pmod{3}$.
Thus,$n^2 + 2$ is divisible by $3$,making $3n(n^2 + 2)$ divisible by $3 \times 3 = 9$.
Therefore,the sum is always divisible by $9$.

(C) The given series is $1+4+7+\cdots+(3n-2)$.
This is an arithmetic progression where the first term $a = 1$ and the common difference $d = 4-1 = 3$.
The number of terms is $n$.
The sum of the first $n$ terms of an arithmetic progression is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values $a=1$ and $d=3$ into the formula:
$S_n = \frac{n}{2}[2(1) + (n-1)3]$
$S_n = \frac{n}{2}[2 + 3n - 3]$
$S_n = \frac{n(3n-1)}{2}$.

(A) Given,$P(n): 2+2^2+2^3+\ldots+2^n = 2^{n+1}-2$.
Assume $P(m)$ is true,so $2+2^2+\ldots+2^m = 2^{m+1}-2$.
Now,consider $P(m+1)$:
$P(m+1) = (2+2^2+\ldots+2^m) + 2^{m+1}$
$= (2^{m+1}-2) + 2^{m+1}$
$= 2 \cdot 2^{m+1} - 2 = 2^{m+2}-2$.
Since $P(m+1)$ holds true whenever $P(m)$ is true,the implication $P(m) \Rightarrow P(m+1)$ is correct.

(A) We know that $\sin(180^{\circ} + \theta) = -\sin \theta$ and $\sin(540^{\circ} + \theta) = \sin(360^{\circ} + 180^{\circ} + \theta) = \sin(180^{\circ} + \theta) = -\sin \theta$.
Step $1$: Simplify $\sin 210^{\circ} = \sin(180^{\circ} + 30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2}$.
Step $2$: Simplify $\sin 585^{\circ} = \sin(540^{\circ} + 45^{\circ}) = -\sin 45^{\circ} = -\frac{1}{\sqrt{2}}$.
Step $3$: Multiply the values: $(-\frac{1}{2}) \times (-\frac{1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}}$.
Thus,the correct option is $A$.

(D) Given $\sin(\theta) + \operatorname{cosec}(\theta) = 2$.
Since $\operatorname{cosec}(\theta) = \frac{1}{\sin(\theta)}$,we have $\sin(\theta) + \frac{1}{\sin(\theta)} = 2$.
Let $x = \sin(\theta)$,then $x + \frac{1}{x} = 2$,which implies $x^2 - 2x + 1 = 0$,so $(x - 1)^2 = 0$.
Thus,$\sin(\theta) = 1$.
Therefore,$\sin^{2020}(\theta) + \operatorname{cosec}^{2020}(\theta) = (1)^{2020} + (1)^{2020} = 1 + 1 = 2$.

(B) Given,$\sec \theta = m$ and $\tan \theta = n$.
We know that $\sec^2 \theta - \tan^2 \theta = 1$,so $m^2 - n^2 = 1$,which implies $(m - n)(m + n) = 1$.
Therefore,$\frac{1}{m + n} = m - n$.
Now,substitute this into the expression:
$\frac{1}{m} [m + n + (m - n)] = \frac{1}{m} [2m] = 2$.

(C) Given expression: $\frac{1-\cos(2x)+\sin(x)}{\sin(2x)+\cos(x)}$
Using the identity $\cos(2x) = 1 - 2\sin^2(x)$,the numerator becomes: $1 - (1 - 2\sin^2(x)) + \sin(x) = 2\sin^2(x) + \sin(x) = \sin(x)(2\sin(x) + 1)$
Using the identity $\sin(2x) = 2\sin(x)\cos(x)$,the denominator becomes: $2\sin(x)\cos(x) + \cos(x) = \cos(x)(2\sin(x) + 1)$
Substituting these back into the expression: $\frac{\sin(x)(2\sin(x) + 1)}{\cos(x)(2\sin(x) + 1)} = \frac{\sin(x)}{\cos(x)} = \tan(x)$

(A) Given,$4 \cos x + 3 \sin x = 5$.
Divide both sides by $\cos x$ (assuming $\cos x \neq 0$):
$4 + 3 \tan x = 5 \sec x$.
Squaring both sides:
$(4 + 3 \tan x)^2 = (5 \sec x)^2$.
$16 + 9 \tan^2 x + 24 \tan x = 25 \sec^2 x$.
Using the identity $\sec^2 x = 1 + \tan^2 x$:
$16 + 9 \tan^2 x + 24 \tan x = 25(1 + \tan^2 x)$.
$16 + 9 \tan^2 x + 24 \tan x = 25 + 25 \tan^2 x$.
Rearranging the terms:
$16 \tan^2 x - 24 \tan x + 9 = 0$.
$(4 \tan x - 3)^2 = 0$.
Therefore,$4 \tan x = 3$,which gives $\tan x = \frac{3}{4}$.

(C) Using the allied angle formulas:
$\sin(\pi+x) = -\sin x$
$\cos(\frac{\pi}{2}+x) = -\sin x$
$\tan(\frac{3\pi}{2}-x) = \cot x$
$\cot(2\pi-x) = -\cot x$
$\sin(2\pi-x) = -\sin x$
$\cos(2\pi+x) = \cos x$
$\operatorname{cosec}(-x) = -\operatorname{cosec} x$
$\sin(\frac{3\pi}{2}+x) = -\cos x$
Substituting these values into the expression:
$= \frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x)}$
$= \frac{-\sin^2 x \cot^2 x}{\sin x \cos x \operatorname{cosec} x \cos x}$
$= \frac{-\sin^2 x \cot^2 x}{\sin x \cos^2 x \cdot \frac{1}{\sin x}}$
$= \frac{-\sin^2 x \cot^2 x}{\cos^2 x} = -\tan^2 x \cdot \cot^2 x = -1$
Wait,let us re-evaluate the signs carefully:
Numerator: $(-\sin x)(-\sin x)(\cot x)(-\cot x) = -\sin^2 x \cot^2 x$
Denominator: $(-\sin x)(\cos x)(-\operatorname{cosec} x)(-\cos x) = -\sin x \cos^2 x \operatorname{cosec} x = -\cos^2 x$
Result: $\frac{-\sin^2 x \cot^2 x}{-\cos^2 x} = \frac{\sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x}}{\cos^2 x} = \frac{\cos^2 x}{\cos^2 x} = 1$
Thus,the correct option is $C$.

(D) Given that $\theta$ lies in the third quadrant,where $\tan \theta$ is positive.
We know that $\sin^2 \theta + \cos^2 \theta = 1$.
Substituting $\cos \theta = -\frac{3}{5}$,we get $\sin^2 \theta + (-\frac{3}{5})^2 = 1$.
$\sin^2 \theta + \frac{9}{25} = 1 \implies \sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$.
Since $\theta$ is in the third quadrant,$\sin \theta$ must be negative,so $\sin \theta = -\frac{4}{5}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-4/5}{-3/5} = \frac{4}{3}$.
Hence,option $D$ is correct.

(C) We know that the formula for $\tan(2x)$ is $\frac{2 \tan(x)}{1-\tan^2(x)}$.
Given the inequality $\frac{2 \tan(x)}{1-\tan^2(x)} > 0$,we have $\tan(2x) > 0$.
Since $x \in \left(0, \frac{\pi}{2}\right)$,it follows that $2x \in (0, \pi)$.
For $\tan(2x) > 0$ in the interval $(0, \pi)$,$2x$ must lie in the first quadrant,i.e.,$0 < 2x < \frac{\pi}{2}$.
Dividing by $2$,we get $0 < x < \frac{\pi}{4}$.
Thus,the range of $x$ is $\left(0, \frac{\pi}{4}\right)$.

(D) Given that $\sec \theta + \tan \theta = 2/3$ $(i)$
We know that $\sec^2 \theta - \tan^2 \theta = 1$,which implies $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting the value from $(i)$,we get $\sec \theta - \tan \theta = 3/2$ (ii)
Adding $(i)$ and (ii): $2 \sec \theta = 2/3 + 3/2 = (4 + 9)/6 = 13/6$,so $\sec \theta = 13/12$.
Subtracting (ii) from $(i)$: $2 \tan \theta = 2/3 - 3/2 = (4 - 9)/6 = -5/6$,so $\tan \theta = -5/12$.
Since $\sec \theta > 0$ and $\tan \theta < 0$,$\theta$ must lie in the $IV$ quadrant.

(B) We know that $\tanh \left(\frac{x}{2}\right) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}$.
Substituting this into the expression:
$\frac{1 + \tanh \left(\frac{x}{2}\right)}{1 - \tanh \left(\frac{x}{2}\right)} = \frac{1 + \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}{1 - \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}$
$= \frac{\frac{e^{x/2} + e^{-x/2} + e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}{\frac{e^{x/2} + e^{-x/2} - (e^{x/2} - e^{-x/2})}{e^{x/2} + e^{-x/2}}}$
$= \frac{2e^{x/2}}{2e^{-x/2}}$
$= e^{x/2 - (-x/2)} = e^x$.

(B) Given $\operatorname{cosec} \theta + \cot \theta = \frac{1}{3}$.
Since $\operatorname{cosec} \theta - \cot \theta = \frac{1}{\operatorname{cosec} \theta + \cot \theta}$,we have $\operatorname{cosec} \theta - \cot \theta = 3$.
Adding the two equations: $(\operatorname{cosec} \theta + \cot \theta) + (\operatorname{cosec} \theta - \cot \theta) = \frac{1}{3} + 3 = \frac{10}{3}$.
$2 \operatorname{cosec} \theta = \frac{10}{3}$ $\Rightarrow \operatorname{cosec} \theta = \frac{5}{3}$ $\Rightarrow \sin \theta = \frac{3}{5}$.
Subtracting the equations: $(\operatorname{cosec} \theta + \cot \theta) - (\operatorname{cosec} \theta - \cot \theta) = \frac{1}{3} - 3 = -\frac{8}{3}$.
$2 \cot \theta = -\frac{8}{3}$ $\Rightarrow \cot \theta = -\frac{4}{3}$ $\Rightarrow \cos \theta = -\frac{4}{5}$.
Since $\sin \theta > 0$ and $\cos \theta < 0$,$\theta$ lies in the $2^{\text{nd}}$ quadrant.

(C) We know that $\cos(\pi - \theta) = -\cos \theta$,so $\cos^2(\pi - \theta) = \cos^2 \theta$.
Also,$\cos(\frac{\pi}{2} - \theta) = \sin \theta$,so $\cos^2(\frac{\pi}{2} - \theta) = \sin^2 \theta$.
Given expression: $E = \cos^2(\frac{7\pi}{8}) + \cos^2(\frac{5\pi}{8}) + \cos^2(\frac{3\pi}{8}) + \cos^2(\frac{\pi}{8})$.
Note that $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$,so $\cos^2(\frac{7\pi}{8}) = \cos^2(\frac{\pi}{8})$.
Note that $\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$,so $\cos^2(\frac{5\pi}{8}) = \cos^2(\frac{3\pi}{8})$.
Thus,$E = 2\cos^2(\frac{\pi}{8}) + 2\cos^2(\frac{3\pi}{8})$.
Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,we have $\cos^2(\frac{3\pi}{8}) = \sin^2(\frac{\pi}{8})$.
Substituting this into $E$: $E = 2\cos^2(\frac{\pi}{8}) + 2\sin^2(\frac{\pi}{8})$.
$E = 2(\cos^2(\frac{\pi}{8}) + \sin^2(\frac{\pi}{8}))$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get $E = 2(1) = 2$.

(A) We use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.
Substituting this into the expression:
$= \frac{1 + \cos 2x}{2} + \frac{1 + \cos(2x + \frac{2\pi}{3})}{2} + \frac{1 + \cos(2x - \frac{2\pi}{3})}{2}$
$= \frac{1}{2} [3 + \cos 2x + \cos(2x + \frac{2\pi}{3}) + \cos(2x - \frac{2\pi}{3})]$
Using the formula $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$:
$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x \cos(\frac{2\pi}{3})]$
Since $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$:
$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x(-\frac{1}{2})]$
$= \frac{1}{2} [3 + \cos 2x - \cos 2x] = \frac{3}{2}$

(D) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.
$\cos 48^{\circ} \cdot \cos 12^{\circ} = \frac{1}{2} [2 \cos 48^{\circ} \cos 12^{\circ}]$
$= \frac{1}{2} [\cos(48^{\circ}+12^{\circ}) + \cos(48^{\circ}-12^{\circ})]$
$= \frac{1}{2} [\cos 60^{\circ} + \cos 36^{\circ}]$
Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$,
$= \frac{1}{2} [\frac{1}{2} + \frac{\sqrt{5}+1}{4}]$
$= \frac{1}{2} [\frac{2 + \sqrt{5} + 1}{4}]$
$= \frac{3+\sqrt{5}}{8}$

(B) We know that $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin(2\theta)}$.
Substituting $\theta = \frac{\pi}{24}$,we get $\tan \frac{\pi}{24} + \cot \frac{\pi}{24} = \frac{2}{\sin(\frac{\pi}{12})}$.
Since $\frac{\pi}{12} = 15^\circ$,$\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Thus,$\frac{1}{2} \left(\tan \frac{\pi}{24} + \cot \frac{\pi}{24}\right) = \frac{1}{2} \cdot \frac{2}{\sin(15^\circ)} = \frac{1}{\sin(15^\circ)} = \frac{2\sqrt{2}}{\sqrt{3}-1}$.
Rationalizing the denominator: $\frac{2\sqrt{2}(\sqrt{3}+1)}{3-1} = \sqrt{6} + \sqrt{2}$.
We are given $\sqrt{a^2+a} + \sqrt{a} = \sqrt{6} + \sqrt{2}$.
Comparing the terms,$\sqrt{a} = \sqrt{2} \implies a = 2$.
Checking the first term: $\sqrt{2^2+2} = \sqrt{6}$,which matches. Therefore,$a = 2$.

(C) Given the equation: $\sin(2x) = \frac{\sqrt{5}-1}{4}$.
We know that $\sin(18^\circ) = \sin(\frac{\pi}{10}) = \frac{\sqrt{5}-1}{4}$.
So,$\sin(2x) = \sin(\frac{\pi}{10})$.
The general solution for $\sin(\theta) = \sin(\alpha)$ is $\theta = n\pi + (-1)^n\alpha$.
Applying this to $2x = \frac{\pi}{10}$:
$2x = n\pi + (-1)^n(\frac{\pi}{10})$.
Dividing by $2$:
$x = \frac{n\pi}{2} + (-1)^n(\frac{\pi}{20})$.
Comparing this with the given form $x = \frac{n\pi}{2} + (-1)^n(m)$,we find $m = \frac{\pi}{20}$.
Thus,option $C$ is correct.

(A) We have the expression: $\tan 81^{\circ} + \tan 9^{\circ} - \tan 63^{\circ} - \tan 27^{\circ}$.
Using the identity $\tan(90^{\circ}-\theta) = \cot \theta$,we can write:
$= (\cot 9^{\circ} + \tan 9^{\circ}) - (\cot 27^{\circ} + \tan 27^{\circ})$.
Using $\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$,we get:
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.
$= 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} \right)$.
Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} \right) = 4 \frac{\cos 36^{\circ}}{\sin 54^{\circ}}$.
Since $\sin 54^{\circ} = \cos(90^{\circ}-54^{\circ}) = \cos 36^{\circ}$,the expression simplifies to $4(1) = 4$.

(B) We know that $\tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B}$.
Applying this to the numerator: $\tan 52^{\circ} - \tan 38^{\circ} = \frac{\sin(52^{\circ} - 38^{\circ})}{\cos 52^{\circ} \cos 38^{\circ}} = \frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ}}$.
Now,the expression becomes $\frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ} \tan 14^{\circ}} = \frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ} \frac{\sin 14^{\circ}}{\cos 14^{\circ}}} = \frac{\cos 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ}}$.
Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have $2 \cos 52^{\circ} \cos 38^{\circ} = \cos(52^{\circ} + 38^{\circ}) + \cos(52^{\circ} - 38^{\circ}) = \cos 90^{\circ} + \cos 14^{\circ} = 0 + \cos 14^{\circ} = \cos 14^{\circ}$.
Thus,the expression is $\frac{\cos 14^{\circ}}{\frac{1}{2} \cos 14^{\circ}} = 2$.

(B) We are given $\tanh(x) = \frac{1}{3}$.
Using the formula $\tanh(3x) = \frac{3\tanh(x) + \tanh^3(x)}{1 + 3\tanh^2(x)}$:
Substitute $\tanh(x) = \frac{1}{3}$ into the formula:
$\tanh(3x) = \frac{3(\frac{1}{3}) + (\frac{1}{3})^3}{1 + 3(\frac{1}{3})^2}$
$\tanh(3x) = \frac{1 + \frac{1}{27}}{1 + 3(\frac{1}{9})}$
$\tanh(3x) = \frac{\frac{28}{27}}{1 + \frac{1}{3}}$
$\tanh(3x) = \frac{\frac{28}{27}}{\frac{4}{3}}$
$\tanh(3x) = \frac{28}{27} \times \frac{3}{4} = \frac{7}{9}$

(B) We know that $\tanh^2\left(\frac{\theta}{2}\right) = \frac{\cosh(\theta) - 1}{\cosh(\theta) + 1}$.
Given that $\cosh(\theta) = \sec(x)$,we substitute this into the expression:
$\tanh^2\left(\frac{\theta}{2}\right) = \frac{\sec(x) - 1}{\sec(x) + 1}$.
Converting $\sec(x)$ to $\frac{1}{\cos(x)}$,we get:
$\frac{\frac{1}{\cos(x)} - 1}{\frac{1}{\cos(x)} + 1} = \frac{1 - \cos(x)}{1 + \cos(x)}$.
Using the half-angle identities $1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right)$ and $1 + \cos(x) = 2\cos^2\left(\frac{x}{2}\right)$:
$= \frac{2\sin^2\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)} = \tan^2\left(\frac{x}{2}\right)$.

(A) We have the expression $\operatorname{cosec} 750^{\circ} - 2 \cot 765^{\circ}$.
Using the periodicity of trigonometric functions,$\operatorname{cosec}(n \times 360^{\circ} + \theta) = \operatorname{cosec} \theta$ and $\cot(n \times 360^{\circ} + \theta) = \cot \theta$.
$\operatorname{cosec} 750^{\circ} = \operatorname{cosec}(2 \times 360^{\circ} + 30^{\circ}) = \operatorname{cosec} 30^{\circ} = 2$.
$\cot 765^{\circ} = \cot(2 \times 360^{\circ} + 45^{\circ}) = \cot 45^{\circ} = 1$.
Substituting these values into the expression:
$2 - 2(1) = 2 - 2 = 0$.
Thus,the correct option is $A$.

(D) Given that $\tan \left(\frac{x}{2}\right) = \frac{m}{n}$.
We use the half-angle formulas: $\sin (x) = \frac{2 \tan (x/2)}{1 + \tan^2 (x/2)}$ and $\cos (x) = \frac{1 - \tan^2 (x/2)}{1 + \tan^2 (x/2)}$.
Substituting these into the expression $m \sin (x) + n \cos (x)$:
$m \left( \frac{2 \tan (x/2)}{1 + \tan^2 (x/2)} \right) + n \left( \frac{1 - \tan^2 (x/2)}{1 + \tan^2 (x/2)} \right)$
$= m \left( \frac{2(m/n)}{1 + (m^2/n^2)} \right) + n \left( \frac{1 - (m^2/n^2)}{1 + (m^2/n^2)} \right)$
$= m \left( \frac{2m/n}{(n^2 + m^2)/n^2} \right) + n \left( \frac{(n^2 - m^2)/n^2}{(n^2 + m^2)/n^2} \right)$
$= \frac{2m^2 n}{n^2 + m^2} + \frac{n(n^2 - m^2)}{n^2 + m^2}$
$= \frac{2m^2 n + n^3 - nm^2}{m^2 + n^2}$
$= \frac{m^2 n + n^3}{m^2 + n^2} = \frac{n(m^2 + n^2)}{m^2 + n^2} = n$.

(B) Given $\sin A + \sin B = \frac{1}{2}$ and $\cos A + \cos B = 1$.
Squaring and adding both equations:
$(\sin A + \sin B)^2 + (\cos A + \cos B)^2 = (\frac{1}{2})^2 + (1)^2$
$\sin^2 A + \sin^2 B + 2 \sin A \sin B + \cos^2 A + \cos^2 B + 2 \cos A \cos B = \frac{1}{4} + 1$
$(\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\cos A \cos B + \sin A \sin B) = \frac{5}{4}$
$1 + 1 + 2 \cos(A-B) = \frac{5}{4}$
$2 + 2 \cos(A-B) = \frac{5}{4}$
$2 \cos(A-B) = \frac{5}{4} - 2 = -\frac{3}{4}$
$\cos(A-B) = -\frac{3}{8}$
Using the identity $\cos \theta = 1 - 2 \sin^2(\frac{\theta}{2})$:
$1 - 2 \sin^2(\frac{A-B}{2}) = -\frac{3}{8}$
$2 \sin^2(\frac{A-B}{2}) = 1 + \frac{3}{8} = \frac{11}{8}$
$\sin^2(\frac{A-B}{2}) = \frac{11}{16}$
$\sin(\frac{A-B}{2}) = \pm \frac{\sqrt{11}}{4}$

(D) Given,$\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) + \cos(\theta_4) = -4$.
Since the range of the cosine function is $[-1, 1]$,the only way their sum can be $-4$ is if each term is equal to $-1$.
Thus,$\cos(\theta_1) = \cos(\theta_2) = \cos(\theta_3) = \cos(\theta_4) = -1$.
This implies $\theta_1 = \theta_2 = \theta_3 = \theta_4 = (2n+1)\pi$ for some integer $n$.
Taking $\theta_i = \pi$,we have $\frac{\theta_i}{2} = \frac{\pi}{2}$.
Then,$\cot(\frac{\theta_i}{2}) = \cot(\frac{\pi}{2}) = 0$.
Therefore,$\cot(\frac{\theta_1}{2}) + \cot(\frac{\theta_2}{2}) + \cot(\frac{\theta_3}{2}) + \cot(\frac{\theta_4}{2}) = 0 + 0 + 0 + 0 = 0$.

(C) The expression is of the form $A \cos \theta + B \sin \theta$,where $\theta = x + \frac{\pi}{3}$,$A = 1$,and $B = 2 \sqrt{2}$.
The range of $A \cos \theta + B \sin \theta$ is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.
Here,$A^2 + B^2 = (1)^2 + (2 \sqrt{2})^2 = 1 + 8 = 9$.
Thus,$\sqrt{A^2 + B^2} = \sqrt{9} = 3$.
The minimum value is $-3$ and the maximum value is $3$.
Therefore,the correct option is $C$.

(D) Given equation: $3 \sin^4 \theta + \cos^4 \theta = 1$
Using $\cos^4 \theta = (1 - \sin^2 \theta)^2 = 1 - 2\sin^2 \theta + \sin^4 \theta$,we get:
$3 \sin^4 \theta + (1 - 2\sin^2 \theta + \sin^4 \theta) = 1$
$4 \sin^4 \theta - 2\sin^2 \theta = 0$
$2 \sin^2 \theta (2 \sin^2 \theta - 1) = 0$
This implies $\sin^2 \theta = 0$ or $\sin^2 \theta = \frac{1}{2}$.
Case $1$: $\sin^2 \theta = 0$ $\Rightarrow \sin \theta = 0$ $\Rightarrow \theta = n\pi$.
Case $2$: $\sin^2 \theta = \frac{1}{2} = \sin^2(\frac{\pi}{4}) \Rightarrow \theta = n\pi \pm \frac{\pi}{4}$.
Combining both cases,the general solution is $\theta = n\pi, n\pi \pm \frac{\pi}{4}$ where $n \in \mathbb{Z}$.

(C) Given the equation $\cos(x) - \sin(x) = 0$.
Rearranging the terms,we get $\cos(x) = \sin(x)$.
Dividing both sides by $\cos(x)$ (assuming $\cos(x) \neq 0$),we get $\tan(x) = 1$.
We know that $\tan(x) = 1 = \tan(\frac{\pi}{4})$.
The general solution for $\tan(x) = \tan(\alpha)$ is $x = n\pi + \alpha$,where $n \in Z$.
Therefore,the general solution is $x = n\pi + \frac{\pi}{4}$,where $n \in Z$.

(D) Given the equation: $(\sin x)(\cos x) = \frac{1}{4}$
Multiply both sides by $2$: $2 \sin x \cos x = \frac{2}{4}$
Using the identity $\sin 2x = 2 \sin x \cos x$,we get: $\sin 2x = \frac{1}{2}$
Since $x \in \left(0, \frac{\pi}{2}\right)$,then $2x \in (0, \pi)$.
The values of $2x$ for which $\sin 2x = \frac{1}{2}$ are $2x = \frac{\pi}{6}$ and $2x = \frac{5\pi}{6}$.
Therefore,$x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$.
Both values lie in the interval $\left(0, \frac{\pi}{2}\right)$.
Comparing with the given options,$\frac{\pi}{12}$ is the correct value.

(D) Given that $\sin \alpha = \sin \beta$ and $\cos \alpha = \cos \beta$.
Squaring and adding both equations:
$(\sin \alpha)^2 + (\cos \alpha)^2 = (\sin \beta)^2 + (\cos \beta)^2$
$1 = 1$.
Alternatively,consider the complex numbers $z_1 = \cos \alpha + i \sin \alpha$ and $z_2 = \cos \beta + i \sin \beta$.
Since $\cos \alpha = \cos \beta$ and $\sin \alpha = \sin \beta$,we have $z_1 = z_2$.
This implies $e^{i \alpha} = e^{i \beta}$,which means $e^{i(\alpha - \beta)} = 1$.
Therefore,$\alpha - \beta = 2 n \pi$ for some integer $n$.
Hence,option $D$ is correct.

(C) We are given the expression: $\frac{\sqrt{3} \sin \theta + \cos \theta}{\sin \left(\theta + \frac{\pi}{6}\right)}$.
Using the expansion formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we expand the denominator:
$\sin \left(\theta + \frac{\pi}{6}\right) = \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6}$.
Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{6} = \frac{1}{2}$,the denominator becomes:
$\frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = \frac{1}{2} (\sqrt{3} \sin \theta + \cos \theta)$.
Substituting this back into the original expression:
$\frac{\sqrt{3} \sin \theta + \cos \theta}{\frac{1}{2} (\sqrt{3} \sin \theta + \cos \theta)} = \frac{1}{1/2} = 2$.
Thus,the correct option is $C$.

(D) Given equation: $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$
Grouping terms: $(\sin 3x + \sin x) + \sin 2x = (\cos 3x + \cos x) + \cos 2x$
Using sum-to-product formulas: $2\sin 2x \cos x + \sin 2x = 2\cos 2x \cos x + \cos 2x$
Factoring out common terms: $\sin 2x(2\cos x + 1) = \cos 2x(2\cos x + 1)$
Rearranging: $(\sin 2x - \cos 2x)(2\cos x + 1) = 0$
Case $1$: $\sin 2x - \cos 2x = 0 \implies \tan 2x = 1 = \tan \frac{\pi}{4}$
$2x = n\pi + \frac{\pi}{4} \implies x = \frac{n\pi}{2} + \frac{\pi}{8}$
Case $2$: $2\cos x + 1 = 0 \implies \cos x = -\frac{1}{2} = \cos \frac{2\pi}{3}$
$x = 2n\pi \pm \frac{2\pi}{3}$
Combining both,the general solution is $x = 2n\pi \pm \frac{2\pi}{3}$ and $x = \frac{n\pi}{2} + \frac{\pi}{8}$ for $n \in Z$.
Hence,option $(D)$ is correct.

(D) We have the expression: $\tan \left(-\frac{23 \pi}{3}\right)-\cot \left(\theta-\frac{13 \pi}{3}\right)$
Since $\tan(-x) = -\tan(x)$,we have $\tan \left(-\frac{23 \pi}{3}\right) = -\tan \left(\frac{23 \pi}{3}\right) = -\tan \left(8 \pi - \frac{\pi}{3}\right) = -(-\tan \frac{\pi}{3}) = \sqrt{3}$.
For the second term,$\cot \left(\theta-\frac{13 \pi}{3}\right) = \cot \left(\theta - (4 \pi + \frac{\pi}{3})\right) = \cot \left(\theta - \frac{\pi}{3}\right) = -\cot \left(\frac{\pi}{3} - \theta\right)$.
Substituting these back into the expression: $\sqrt{3} - (-\cot \left(\frac{\pi}{3} - \theta\right)) = \sqrt{3} + \cot \left(\frac{\pi}{3} - \theta\right)$.
Thus,the correct option is $D$.

(C) Given the equation: $4 \sin^2(x) - 4 \sin(x) + 1 = 0$
This is a quadratic equation in terms of $\sin(x)$,which can be written as $(2 \sin(x) - 1)^2 = 0$.
Taking the square root on both sides,we get: $2 \sin(x) - 1 = 0$
$\sin(x) = \frac{1}{2}$
We know that $\sin(x) = \sin(\frac{\pi}{6})$.
The general solution for $\sin(x) = \sin(\alpha)$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.
Therefore,$x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$.
Hence,option $C$ is correct.

(A) Given,$\cos A + \cos B + \cos C = \frac{3}{2}$.
We know that for any triangle $ABC$,the maximum value of $\cos A + \cos B + \cos C$ is $\frac{3}{2}$,which occurs if and only if $A = B = C = 60^{\circ}$ or $\frac{\pi}{3}$ radians.
Since $A = B = C = \frac{\pi}{3}$,all three angles are equal.
Therefore,$\triangle ABC$ is an equilateral triangle.
Hence,option $A$ is correct.

(A) We have $\sin \left(\frac{5 \pi}{3}\right) + \sec \left(\frac{13 \pi}{3}\right)$.
First,simplify $\sin \left(\frac{5 \pi}{3}\right) = \sin \left(2 \pi - \frac{\pi}{3}\right) = -\sin \left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.
Next,simplify $\sec \left(\frac{13 \pi}{3}\right) = \sec \left(4 \pi + \frac{\pi}{3}\right) = \sec \left(\frac{\pi}{3}\right) = 2$.
Adding these values,we get $-\frac{\sqrt{3}}{2} + 2 = 2 - \frac{\sqrt{3}}{2}$.
Thus,the correct option is $A$.

(C) Given that, $\tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma$.
This implies $\tan \alpha + \tan \beta = -\tan \gamma (1 - \tan \alpha \tan \beta)$.
Dividing by $(1 - \tan \alpha \tan \beta)$, we get $\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \tan(-\gamma)$.
Thus, $\tan(\alpha + \beta) = \tan(-\gamma)$, which means $\alpha + \beta = n\pi - \gamma$ for some integer $n \in \mathbb{Z}$.
Therefore, $\alpha + \beta + \gamma = n\pi$.
Using Euler's formula, $x = e^{i\alpha}$, $y = e^{i\beta}$, and $z = e^{i\gamma}$.
Then $xyz = e^{i\alpha} \cdot e^{i\beta} \cdot e^{i\gamma} = e^{i(\alpha + \beta + \gamma)} = e^{in\pi}$.
Since $e^{in\pi} = \cos(n\pi) + i\sin(n\pi) = \cos(n\pi)$, and $\cos(n\pi)$ is $1$ if $n$ is even and $-1$ if $n$ is odd.
Thus, $xyz = \pm 1$.

(B) We use the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
Let $a = \sin^2 \theta$ and $b = \cos^2 \theta$.
Then $\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3$.
$= (\sin^2 \theta + \cos^2 \theta)((\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2)$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$= 1 \cdot (\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta)$.
We know that $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting this back:
$\sin^6 \theta + \cos^6 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta - \sin^2 \theta \cos^2 \theta = 1 - 3 \sin^2 \theta \cos^2 \theta$.
Therefore,$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = (1 - 3 \sin^2 \theta \cos^2 \theta) + 3 \sin^2 \theta \cos^2 \theta = 1$.

(B) Given the equation: $\tan(x) + \sec(x) = \sqrt{3}$.
We know the identity $\sec^2(x) - \tan^2(x) = 1$,which implies $(\sec(x) - \tan(x))(\sec(x) + \tan(x)) = 1$.
Substituting the given value: $(\sec(x) - \tan(x))(\sqrt{3}) = 1$,so $\sec(x) - \tan(x) = \frac{1}{\sqrt{3}}$.
Adding the two equations:
$(\tan(x) + \sec(x)) + (\sec(x) - \tan(x)) = \sqrt{3} + \frac{1}{\sqrt{3}}$
$2\sec(x) = \frac{3+1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$ $\Rightarrow \sec(x) = \frac{2}{\sqrt{3}}$ $\Rightarrow \cos(x) = \frac{\sqrt{3}}{2}$.
For $x \in [0, 2\pi]$,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{11\pi}{6}$.
Checking the original equation: $\tan(\frac{\pi}{6}) + \sec(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Thus,$x = \frac{\pi}{6}$ is a valid solution.

(B) Given that,$\cos(x) + \cos^2(x) = 1$.
Since $\sin^2(x) = 1 - \cos^2(x)$,we can write $\sin^2(x) = \cos(x)$.
Squaring both sides,we get $\sin^4(x) = \cos^2(x)$.
Substituting these into the expression $\sin^2(x) + \sin^4(x)$,we get $\cos(x) + \cos^2(x)$.
Since $\cos(x) + \cos^2(x) = 1$,the value of the expression is $1$.
Thus,option $B$ is correct.

(B) Let $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)} = k$.
Then,$x = k \cos \alpha$,$y = k \cos \left(\frac{2 \pi}{3} - \alpha\right)$,and $z = k \cos \left(\frac{2 \pi}{3} + \alpha\right)$.
Now,$x + y + z = k \left[ \cos \alpha + \cos \left(\frac{2 \pi}{3} - \alpha\right) + \cos \left(\frac{2 \pi}{3} + \alpha\right) \right]$.
Using the formula $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$,we get:
$x + y + z = k \left[ \cos \alpha + 2 \cos \left(\frac{2 \pi}{3}\right) \cos \alpha \right]$.
Since $\cos \left(\frac{2 \pi}{3}\right) = -\frac{1}{2}$,we have:
$x + y + z = k \left[ \cos \alpha + 2 \left(-\frac{1}{2}\right) \cos \alpha \right] = k [\cos \alpha - \cos \alpha] = 0$.

(B) Given the equation: $\log(x+y) - 2xy = 0$.
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \cdot (1 + y') - 2(x y' + y) = 0$.
Rearranging the terms to isolate $y'$:
$\frac{1}{x+y} + \frac{y'}{x+y} - 2xy' - 2y = 0$.
$y' \left( \frac{1}{x+y} - 2x \right) = 2y - \frac{1}{x+y}$.
$y' = \frac{2y - \frac{1}{x+y}}{\frac{1}{x+y} - 2x}$.
Now,evaluate at $x = 0$:
At $x = 0$,the original equation becomes $\log(y) - 0 = 0$,which implies $\log(y) = 0$,so $y = e^0 = 1$.
Substituting $x = 0$ and $y = 1$ into the expression for $y'$:
$y'(0) = \frac{2(1) - \frac{1}{0+1}}{\frac{1}{0+1} - 2(0)} = \frac{2 - 1}{1 - 0} = 1$.
Note: The provided options suggest the answer is in terms of $y$. Evaluating the expression $\frac{2y - \frac{1}{y}}{\frac{1}{y}} = 2y^2 - 1$.
Thus,$y'(0) = 2y^2 - 1$.

(D) Given,$y = \frac{e^x \log x}{x^2}$.
Applying the quotient rule for differentiation,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = e^x \log x$ and $v = x^2$.
$\frac{du}{dx} = e^x \log x + e^x \cdot \frac{1}{x} = e^x \left( \log x + \frac{1}{x} \right)$.
$\frac{dv}{dx} = 2x$.
Therefore,$\frac{dy}{dx} = \frac{x^2 \left[ e^x \left( \log x + \frac{1}{x} \right) \right] - (e^x \log x)(2x)}{(x^2)^2}$.
$\frac{dy}{dx} = \frac{x^2 e^x \log x + x e^x - 2x e^x \log x}{x^4}$.
$\frac{dy}{dx} = \frac{x e^x \log x + e^x - 2 e^x \log x}{x^3}$.
$\frac{dy}{dx} = \frac{e^x [x \log x + 1 - 2 \log x]}{x^3}$.
$\frac{dy}{dx} = \frac{e^x [1 + (x - 2) \log x]}{x^3}$.
Thus,option $D$ is correct.

(A) Given $y = \log(\cosh x)$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\cosh x} \cdot \frac{d}{dx}(\cosh x)$
Since $\frac{d}{dx}(\cosh x) = \sinh x$,we get:
$\frac{dy}{dx} = \frac{\sinh x}{\cosh x} = \tanh x$
Now,differentiate again with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(\tanh x)$
Since $\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$,we have:
$\frac{d^2 y}{dx^2} = \operatorname{sech}^2 x$
Thus,the correct option is $A$.

(C) Given $f(x) = \log_{x^2} (\ln x)$.
Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we have:
$f(x) = \frac{\ln(\ln x)}{\ln x^2} = \frac{\ln(\ln x)}{2 \ln x}$.
Now,differentiate with respect to $x$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{1}{2} \left[ \frac{(\frac{1}{x \ln x})(\ln x) - (\ln(\ln x))(\frac{1}{x})}{(\ln x)^2} \right]$.
$f'(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{\ln(\ln x)}{x}}{(\ln x)^2} \right] = \frac{1 - \ln(\ln x)}{2x(\ln x)^2}$.
At $x = e$,$\ln x = 1$ and $\ln(\ln x) = \ln(1) = 0$.
$f'(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

(C) Let $y = \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$ and $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$.
Substitute $x = \sin \theta$,then $\sqrt{1-x^2} = \cos \theta$.
$y = \tan ^{-1}\left(\frac{\sin \theta}{1+\cos \theta}\right) = \tan ^{-1}\left(\frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)}\right) = \tan ^{-1}(\tan(\theta/2)) = \frac{\theta}{2}$.
Now,$u = \sec ^{-1}\left(\frac{1}{2 \sin^2 \theta - 1}\right) = \sec ^{-1}\left(\frac{1}{-(1-2 \sin^2 \theta)}\right) = \sec ^{-1}\left(\frac{1}{-\cos 2 \theta}\right) = \sec ^{-1}(-\sec 2 \theta) = \sec ^{-1}(\sec(\pi - 2 \theta)) = \pi - 2 \theta$.
Since $x = \sin \theta$,$\theta = \sin^{-1} x$.
Thus,$y = \frac{1}{2} \sin^{-1} x$ and $u = \pi - 2 \sin^{-1} x$.
Then $\frac{dy}{dx} = \frac{1}{2 \sqrt{1-x^2}}$ and $\frac{du}{dx} = \frac{-2}{\sqrt{1-x^2}}$.
Therefore,$\frac{dy}{du} = \frac{dy/dx}{du/dx} = \frac{1/(2 \sqrt{1-x^2})}{-2/\sqrt{1-x^2}} = -\frac{1}{4}$.

(C) Let $y = \cos ^{-1}\left(\frac{4 x^3}{27}-x\right)$.
We can rewrite the expression as $y = \cos ^{-1}\left(4\left(\frac{x}{3}\right)^3 - 3\left(\frac{x}{3}\right)\right)$.
Let $\frac{x}{3} = \cos A$,then $A = \cos ^{-1}\left(\frac{x}{3}\right)$.
Substituting this into the expression,we get $y = \cos ^{-1}(4 \cos ^3 A - 3 \cos A)$.
Using the trigonometric identity $\cos(3A) = 4 \cos ^3 A - 3 \cos A$,we have $y = \cos ^{-1}(\cos 3A) = 3A$.
Thus,$y = 3 \cos ^{-1}\left(\frac{x}{3}\right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \times \left(-\frac{1}{\sqrt{1 - (x/3)^2}}\right) \times \frac{d}{dx}\left(\frac{x}{3}\right)$.
$\frac{dy}{dx} = 3 \times \left(-\frac{1}{\sqrt{(9-x^2)/9}}\right) \times \frac{1}{3} = 3 \times \left(-\frac{3}{\sqrt{9-x^2}}\right) \times \frac{1}{3} = -\frac{3}{\sqrt{9-x^2}}$.

(C) Given,$f(x) = \cos ^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right)+x^x$.
Using the identity $\cos ^{-1}(y) = \frac{\pi}{2} - \sin ^{-1}(y)$,we have:
$f(x) = \frac{\pi}{2} - \sin ^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right) + x^x$.
Since $\sin ^{-1}(\sin \theta) = \theta$ for the given range,we get:
$f(x) = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$.
Now,differentiating with respect to $x$:
$f'(x) = 0 - \frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + x^x(1 + \ln x)$.
At $x=1$:
$f'(1) = -\frac{1}{2\sqrt{2}\sqrt{2}} + 1^1(1 + \ln 1) = -\frac{1}{4} + 1(1+0) = -\frac{1}{4} + 1 = \frac{3}{4}$.
Thus,the correct option is $C$.

(D) Given $y = \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right]$.
We know that $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.
Thus,$\sqrt{1 + \sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$ and $\sqrt{1 - \sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$.
Substituting these into the expression for $y$:
$y = \tan^{-1} \left[ \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})} \right]$ (assuming $0 < x < \frac{\pi}{2}$).
$y = \tan^{-1} \left( \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} \right) = \tan^{-1} (\cot \frac{x}{2}) = \tan^{-1} \left( \tan (\frac{\pi}{2} - \frac{x}{2}) \right) = \frac{\pi}{2} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{2} - \frac{x}{2}) = -\frac{1}{2}$.
Depending on the interval of $x$,the derivative can be $\pm \frac{1}{2}$.
Therefore,the correct option is $(D)$.

(C) The derivative of $y = \operatorname{cosec}^{-1}(x)$ is given by $\frac{dy}{dx} = \frac{-1}{|x| \sqrt{x^2-1}}$.
For the derivative to be defined,the expression inside the square root must be strictly positive,i.e.,$x^2 - 1 > 0$,which implies $|x| > 1$.
However,the range of $y = \operatorname{cosec}^{-1}(x)$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$.
Since the derivative $\frac{dy}{dx}$ is defined for all $x$ in the domain of the function excluding the endpoints where the derivative might not exist or is undefined,we consider the open interval for $y$.
The values of $y$ corresponding to the domain of the derivative are $y \in \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right)$.

(B) Given,$y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$.
Let $x=\tan \theta$,then $\theta = \tan ^{-1} x$.
Substituting $x$ in the expression:
$y = \tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)$
$y = \tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)$
$y = \tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) = \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$
Using trigonometric identities $1-\cos \theta = 2\sin ^2(\theta/2)$ and $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$:
$y = \tan ^{-1}\left(\frac{2\sin ^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan ^{-1}(\tan(\theta/2)) = \theta/2$.
Since $\theta = \tan ^{-1} x$,we have $y = \frac{1}{2} \tan ^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$.
Hence,option $B$ is correct.

(A) Given,$\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\sin ^{-1}(a)$.
Taking $\cos$ on both sides,we get $\frac{x^2-y^2}{x^2+y^2} = \cos(\sin^{-1} a)$.
Let $\cos(\sin^{-1} a) = k$,where $k$ is a constant.
So,$\frac{x^2-y^2}{x^2+y^2} = k$.
$x^2 - y^2 = k(x^2 + y^2) \Rightarrow x^2(1-k) = y^2(1+k)$.
$y^2 = x^2 \left(\frac{1-k}{1+k}\right)$.
Let $C = \sqrt{\frac{1-k}{1+k}}$,which is a constant.
Then $y^2 = C^2 x^2 \Rightarrow y = Cx$ (assuming $y, x > 0$ for simplicity).
Differentiating with respect to $x$,we get $\frac{dy}{dx} = C$.
Since $y = Cx$,we have $C = \frac{y}{x}$.
Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

(D) $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$
Let $x = \cos 2\theta$,then $2\theta = \cos^{-1} x$ or $\theta = \frac{1}{2} \cos^{-1} x$.
Substituting $x$ in the expression:
$y = \sin^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}}{2}\right)$
Using trigonometric identities $1+\cos 2\theta = 2\cos^2 \theta$ and $1-\cos 2\theta = 2\sin^2 \theta$:
$y = \sin^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{2}\right) = \sin^{-1}\left(\frac{\cos \theta - \sin \theta}{\sqrt{2}}\right)$
$y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\cos \theta - \frac{1}{\sqrt{2}}\sin \theta\right)$
$y = \sin^{-1}\left(\sin \frac{\pi}{4} \cos \theta - \cos \frac{\pi}{4} \sin \theta\right)$
$y = \sin^{-1}\left(\sin\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta$
Substituting $\theta = \frac{1}{2} \cos^{-1} x$:
$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{2} \left(-\frac{1}{\sqrt{1-x^2}}\right) = \frac{1}{2\sqrt{1-x^2}}$
Thus,option $D$ is correct.

(A) Given equation: $\log \sqrt{x^2+y^2}=\tan ^{-1}\left(\frac{x}{y}\right)$
Differentiating both sides with respect to $x$:
$\frac{1}{\sqrt{x^2+y^2}} \cdot \frac{d}{dx}(\sqrt{x^2+y^2}) = \frac{1}{1+(\frac{x}{y})^2} \cdot \frac{d}{dx}(\frac{x}{y})$
$\frac{1}{\sqrt{x^2+y^2}} \cdot \frac{1}{2\sqrt{x^2+y^2}} \cdot (2x + 2y \frac{dy}{dx}) = \frac{y^2}{x^2+y^2} \cdot \frac{y - x \frac{dy}{dx}}{y^2}$
$\frac{2(x + y \frac{dy}{dx})}{2(x^2+y^2)} = \frac{y - x \frac{dy}{dx}}{x^2+y^2}$
$x + y \frac{dy}{dx} = y - x \frac{dy}{dx}$
$y \frac{dy}{dx} + x \frac{dy}{dx} = y - x$
$\frac{dy}{dx}(x + y) = y - x$
$\frac{dy}{dx} = \frac{y-x}{x+y}$
Thus,the correct option is $A$.

(C) To find the derivative of the product $(1+x^2) \tan^{-1}(x)$,we use the product rule: $\frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}$.
Let $u = (1+x^2)$ and $v = \tan^{-1}(x)$.
Then,$\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \frac{1}{1+x^2}$.
Applying the product rule:
$\frac{d}{dx} \left\{ (1+x^2) \tan^{-1}(x) \right\} = (1+x^2) \cdot \frac{d}{dx}(\tan^{-1}(x)) + \tan^{-1}(x) \cdot \frac{d}{dx}(1+x^2)$
$= (1+x^2) \cdot \frac{1}{1+x^2} + \tan^{-1}(x) \cdot (2x)$
$= 1 + 2x \tan^{-1}(x)$.
Thus,the correct option is $C$.

(B) Given $y = \tan^{-1} \left( \frac{5x - x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} \right)$.
Using the formula $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right)$,the first term is $\tan^{-1}(5x) - \tan^{-1}(x)$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right)$,the second term is $\tan^{-1}(2/3) + \tan^{-1}(x)$.
Thus,$y = \tan^{-1}(5x) - \tan^{-1}(x) + \tan^{-1}(2/3) + \tan^{-1}(x) = \tan^{-1}(5x) + \tan^{-1}(2/3)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\tan^{-1}(5x)) + \frac{d}{dx} (\tan^{-1}(2/3))$.
Since $\tan^{-1}(2/3)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = \frac{1}{1 + (5x)^2} \times \frac{d}{dx}(5x) + 0 = \frac{5}{1 + 25x^2}$.

(C) Given $y = \operatorname{Sin}^{-1}\left(\frac{4+5 \sin x}{5+4 \sin x}\right)$.
Let $f(x) = \frac{4+5 \sin x}{5+4 \sin x}$.
To find $\frac{dy}{dx}$,we use the chain rule: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - [f(x)]^2}} \cdot f'(x)$.
First,calculate $f'(x)$ using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:
$u = 4+5 \sin x \implies u' = 5 \cos x$
$v = 5+4 \sin x \implies v' = 4 \cos x$
$f'(x) = \frac{(5 \cos x)(5+4 \sin x) - (4+5 \sin x)(4 \cos x)}{(5+4 \sin x)^2} = \frac{25 \cos x + 20 \sin x \cos x - 16 \cos x - 20 \sin x \cos x}{(5+4 \sin x)^2} = \frac{9 \cos x}{(5+4 \sin x)^2}$.
Next,calculate $1 - [f(x)]^2 = 1 - \left(\frac{4+5 \sin x}{5+4 \sin x}\right)^2 = \frac{(5+4 \sin x)^2 - (4+5 \sin x)^2}{(5+4 \sin x)^2} = \frac{(25 + 16 \sin^2 x + 40 \sin x) - (16 + 25 \sin^2 x + 40 \sin x)}{(5+4 \sin x)^2} = \frac{9 - 9 \sin^2 x}{(5+4 \sin x)^2} = \frac{9 \cos^2 x}{(5+4 \sin x)^2}$.
Thus,$\sqrt{1 - [f(x)]^2} = \frac{3 |\cos x|}{5+4 \sin x}$.
Substituting back: $\frac{dy}{dx} = \frac{5+4 \sin x}{3 |\cos x|} \cdot \frac{9 \cos x}{(5+4 \sin x)^2} = \frac{3 \cos x}{|\cos x| (5+4 \sin x)}$.
This simplifies to $\frac{3}{5+4 \sin x}$ when $\cos x > 0$ and $\frac{-3}{5+4 \sin x}$ when $\cos x < 0$. Given the options,the general derivative is $\frac{\pm 3}{5+4 \sin x}$.

(A) Let $y = \sin^{-1} \left( \frac{3}{5} \cdot \frac{1}{\sqrt{1+x^2}} + \frac{4}{5} \cdot \frac{x}{\sqrt{1+x^2}} \right)$.
Let $\cos \alpha = \frac{3}{5}$,then $\sin \alpha = \frac{4}{5}$.
Also,let $\sin \theta = \frac{x}{\sqrt{1+x^2}}$,then $\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
Substituting these,we get $y = \sin^{-1} (\cos \alpha \cos \theta + \sin \alpha \sin \theta) = \sin^{-1} (\cos(\alpha - \theta)) = \sin^{-1} (\sin(\frac{\pi}{2} - (\alpha - \theta))) = \frac{\pi}{2} - \alpha + \theta$.
Since $\theta = \tan^{-1} x$,we have $y = \frac{\pi}{2} - \alpha + \tan^{-1} x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 0 - 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.

(B) Given the equation: $3 f(x)-2 f\left(\frac{1}{x}\right)=x$
Differentiating both sides with respect to $x$,we get:
$3 f^{\prime}(x)-2 f^{\prime}\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)=1$
$3 f^{\prime}(x)+\frac{2}{x^2} f^{\prime}\left(\frac{1}{x}\right)=1$
For $x=2$:
$3 f^{\prime}(2)+\frac{2}{4} f^{\prime}\left(\frac{1}{2}\right)=1 \Rightarrow 3 f^{\prime}(2)+\frac{1}{2} f^{\prime}\left(\frac{1}{2}\right)=1$ ...$(i)$
For $x=\frac{1}{2}$:
$3 f^{\prime}\left(\frac{1}{2}\right)+\frac{2}{(1/4)} f^{\prime}(2)=1 \Rightarrow 3 f^{\prime}\left(\frac{1}{2}\right)+8 f^{\prime}(2)=1$ ...(ii)
From $(i)$,$f^{\prime}\left(\frac{1}{2}\right)=2-6 f^{\prime}(2)$.
Substituting this into (ii):
$3(2-6 f^{\prime}(2))+8 f^{\prime}(2)=1$
$6-18 f^{\prime}(2)+8 f^{\prime}(2)=1$
$-10 f^{\prime}(2)=-5$
$f^{\prime}(2)=\frac{1}{2}$

(A) Given the differential equation: $\frac{dt}{dx} = \frac{t}{x + t e^{-2x/t}}$.
Taking the reciprocal of both sides,we get:
$\frac{dx}{dt} = \frac{x + t e^{-2x/t}}{t}$.
Dividing each term in the numerator by $t$,we obtain:
$\frac{dx}{dt} = \frac{x}{t} + \frac{t e^{-2x/t}}{t}$.
Simplifying the expression,we get:
$\frac{dx}{dt} = \frac{x}{t} + e^{-2(x/t)}$.
This is in the form $\frac{dx}{dt} = \phi\left(\frac{x}{t}\right)$,where $\phi(v) = v + e^{-2v}$.
Thus,the correct option is $A$.

(C) Given,$f(x)=x^4-x^3+7x^2+14$.
First,we find the first derivative $f^{\prime}(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(x^4-x^3+7x^2+14) = 4x^3-3x^2+14x$.
Next,we find the second derivative $f^{\prime \prime}(x)$ by differentiating $f^{\prime}(x)$:
$f^{\prime \prime}(x) = \frac{d}{dx}(4x^3-3x^2+14x) = 12x^2-6x+14$.
Now,substitute $x=5$ into the expression for $f^{\prime \prime}(x)$:
$f^{\prime \prime}(5) = 12(5)^2 - 6(5) + 14$.
$f^{\prime \prime}(5) = 12(25) - 30 + 14$.
$f^{\prime \prime}(5) = 300 - 30 + 14 = 284$.
Thus,the value is $284$,which corresponds to option $C$.

(A) Given,$y = \sqrt{\frac{1+\tan x}{1-\tan x}} = \sqrt{\tan(\frac{\pi}{4}+x)}$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan(\frac{\pi}{4}+x)}} \cdot \sec^2(\frac{\pi}{4}+x) \cdot \frac{d}{dx}(\frac{\pi}{4}+x)$.
Since $\frac{d}{dx}(\frac{\pi}{4}+x) = 1$,we have:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{\tan(\frac{\pi}{4}+x)}} \cdot \sec^2(\frac{\pi}{4}+x)$.
Substituting $\sqrt{\tan(\frac{\pi}{4}+x)} = \sqrt{\frac{1+\tan x}{1-\tan x}}$,we get:
$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \sec^2(\frac{\pi}{4}+x)$.
Thus,option $A$ is correct.

(A) Given equation is $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0$
$4x - 3(y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$
$4x - 3y - 3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$
Grouping the $\frac{dy}{dx}$ terms:
$(2y - 3x + 2) \frac{dy}{dx} = 3y - 4x - 1$
Therefore,$\frac{dy}{dx} = \frac{3y - 4x - 1}{2y - 3x + 2}$.
Thus,option $A$ is correct.

(B) Given $f(x) = 3 e^{x^2}$.
First,find the derivative $f^{\prime}(x)$ using the chain rule:
$f^{\prime}(x) = 3 \cdot e^{x^2} \cdot \frac{d}{dx}(x^2) = 3 e^{x^2} \cdot 2x = 6x e^{x^2}$.
Now,calculate the terms:
$2x f(x) = 2x(3 e^{x^2}) = 6x e^{x^2}$.
$f(0) = 3 e^{0^2} = 3(1) = 3$.
$f^{\prime}(0) = 6(0) e^{0^2} = 0$.
Substitute these into the expression $f^{\prime}(x) - 2x f(x) + \frac{1}{3} f(0) - f^{\prime}(0)$:
$= 6x e^{x^2} - 6x e^{x^2} + \frac{1}{3}(3) - 0$
$= 0 + 1 - 0 = 1$.

(D) Given,$f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right|$.
Using the property of differentiation of a determinant,$f'(x)$ is the sum of three determinants where each row is differentiated one at a time:
$f'(x) = \left| \begin{array}{ccc} -\sin x & 1 & 0 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right| + \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \cos x & 2x & 2 \\ \tan x & x & 1 \end{array} \right| + \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \sec^2 x & 1 & 0 \end{array} \right|$.
At $x = 0$:
$f'(0) = \left| \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right| + \left| \begin{array}{ccc} 1 & 0 & 1 \\ 2 & 0 & 2 \\ 0 & 0 & 1 \end{array} \right| + \left| \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \end{array} \right|$.
In the first determinant,the second row is all zeros,so its value is $0$.
In the second determinant,the first and third columns are identical,so its value is $0$.
In the third determinant,the second row is all zeros,so its value is $0$.
Therefore,$f'(0) = 0 + 0 + 0 = 0$.

(A) Given the curve equation: $\left(\frac{x}{31}\right)^n + \left(\frac{y}{1209}\right)^n = 2$
Differentiating both sides with respect to $x$:
$\frac{n}{31} \left(\frac{x}{31}\right)^{n-1} + \frac{n}{1209} \left(\frac{y}{1209}\right)^{n-1} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{\frac{n}{31} \left(\frac{x}{31}\right)^{n-1}}{\frac{n}{1209} \left(\frac{y}{1209}\right)^{n-1}} = -\frac{1209}{31} \left(\frac{x}{31}\right)^{n-1} \left(\frac{1209}{y}\right)^{n-1}$
Since $1209 / 31 = 39$,we have:
$\frac{dy}{dx} = -39 \left(\frac{x}{31}\right)^{n-1} \left(\frac{1209}{y}\right)^{n-1}$
At the point $(31, 1209)$,substitute $x = 31$ and $y = 1209$:
$\frac{dy}{dx} = -39 \left(\frac{31}{31}\right)^{n-1} \left(\frac{1209}{1209}\right)^{n-1} = -39(1)^{n-1}(1)^{n-1} = -39$
Thus,the slope of the tangent is $-39$.

(C) Given the curve equation: $y=5x^2-3x+7$.
First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = \frac{d}{dx}(5x^2-3x+7) = 10x-3$.
The slope of the tangent $m_T$ at the point $(-1, 4)$ is:
$m_T = \left. \frac{dy}{dx} \right|_{x=-1} = 10(-1)-3 = -10-3 = -13$.
Now,using the point-slope form of the line equation $y-y_1 = m(x-x_1)$:
$y-4 = -13(x-(-1))$
$y-4 = -13(x+1)$
$y-4 = -13x-13$
$13x+y-4+13 = 0$
$13x+y+9 = 0$.

(A) The equation of the given curve is $3y^2 = 2ax^2 + 6b$ . . . $(i)$
Since the curve passes through the point $P(3, -1)$,we substitute $x = 3$ and $y = -1$ into equation $(i)$:
$3(-1)^2 = 2a(3)^2 + 6b$
$3 = 18a + 6b$
Dividing by $3$,we get $6a + 2b = 1$ . . . $(ii)$
Now,we differentiate equation $(i)$ with respect to $x$ to find the gradient:
$6y \frac{dy}{dx} = 4ax$
Given that the gradient at $P(3, -1)$ is $-1$,we substitute $\frac{dy}{dx} = -1$,$x = 3$,and $y = -1$:
$6(-1)(-1) = 4a(3)$
$6 = 12a$
$a = 1/2$
Substituting $a = 1/2$ into equation $(ii)$:
$6(1/2) + 2b = 1$
$3 + 2b = 1$
$2b = -2$
$b = -1$
Thus,the values are $a = 1/2$ and $b = -1$.

(D) The given curve is $y = \frac{x-7}{(x-2)(x-3)}$. The curve cuts the $X$-axis where $y = 0$. Setting $y = 0$,we get $x - 7 = 0$,so $x = 7$. Thus,the point of intersection is $P(7, 0)$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{[(x-2)(x-3)] \cdot 1 - (x-7) \cdot \frac{d}{dx}[(x-2)(x-3)]}{[(x-2)(x-3)]^2}$
$\frac{dy}{dx} = \frac{(x^2 - 5x + 6) - (x-7)(2x - 5)}{(x-2)^2(x-3)^2}$
At $x = 7$,the denominator is $(7-2)^2(7-3)^2 = 5^2 \cdot 4^2 = 25 \cdot 16 = 400$. The numerator is $(49 - 35 + 6) - (0) = 20$.
So,$\left. \frac{dy}{dx} \right|_{x=7} = \frac{20}{400} = \frac{1}{20}$.
The slope of the normal is $m_n = -\frac{1}{dy/dx} = -20$.
The equation of the normal at $(7, 0)$ is $y - 0 = -20(x - 7)$,which simplifies to $y = -20x + 140$,or $20x + y - 140 = 0$.

(B) Given the equation of the curve is $y^n = ax$.
Differentiating both sides with respect to $x$,we get:
$n y^{n-1} \frac{dy}{dx} = a$
$\frac{dy}{dx} = \frac{a}{n} y^{1-n}$
The length of the subnormal is defined as $|y \frac{dy}{dx}|$.
Substituting the value of $\frac{dy}{dx}$:
$\text{Length of subnormal} = |y \cdot \frac{a}{n} y^{1-n}| = |\frac{a}{n} y^{2-n}|$.
For the subnormal to be constant,it must be independent of the variable $y$.
This occurs when the exponent of $y$ is $0$,i.e.,$2 - n = 0$.
Therefore,$n = 2$.
Hence,the correct option is $B$.

(C) Given the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ ...$(i)$
Let a point on the curve be $P(a \cos^3 \theta, a \sin^3 \theta)$.
Differentiating the curve $(i)$ with respect to $x$,we get:
$\frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3}$.
Substituting the point $P$,the slope of the tangent $m$ is:
$m = -\left(\frac{a \sin^3 \theta}{a \cos^3 \theta}\right)^{1/3} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
The equation of the tangent at point $P$ is:
$y - a \sin^3 \theta = -\tan \theta (x - a \cos^3 \theta)$.
$y \cos \theta - a \sin^3 \theta \cos \theta = -x \sin \theta + a \cos^3 \theta \sin \theta$.
$x \sin \theta + y \cos \theta = a \sin \theta \cos \theta (\cos^2 \theta + \sin^2 \theta) = a \sin \theta \cos \theta$.
For point $A$ (where $y=0$),$x \sin \theta = a \sin \theta \cos \theta \Rightarrow x = a \cos \theta$. So $A = (a \cos \theta, 0)$.
For point $B$ (where $x=0$),$y \cos \theta = a \sin \theta \cos \theta \Rightarrow y = a \sin \theta$. So $B = (0, a \sin \theta)$.
The distance $AB = \sqrt{(a \cos \theta - 0)^2 + (0 - a \sin \theta)^2} = \sqrt{a^2 \cos^2 \theta + a^2 \sin^2 \theta} = \sqrt{a^2} = a$.
Thus,$AB = a$.

(A) The equation of the curve is $y=ax^3+bx^2+cx+5$ ...$(i)$
Since the curve touches the $X$-axis at $P(-2,0)$,it passes through $P(-2,0)$,so $-8a+4b-2c+5=0$ ...(ii)
Also,since it touches the $X$-axis at $x=-2$,the derivative $\frac{dy}{dx}$ at $x=-2$ is $0$.
$\frac{dy}{dx} = 3ax^2+2bx+c$. At $x=-2$,$12a-4b+c=0$ ...(iii)
The curve cuts the $Y$-axis at $Q(0,5)$ where the gradient is $3$,so $\left.\frac{dy}{dx}\right|_{x=0} = 3$,which gives $c=3$.
Substituting $c=3$ into (ii) and (iii):
$-8a+4b-6+5=0 \Rightarrow -8a+4b=1$ ...(iv)
$12a-4b+3=0 \Rightarrow 12a-4b=-3$ ...$(v)$
Adding (iv) and $(v)$: $4a = -2 \Rightarrow a=-\frac{1}{2}$.
Substituting $a=-\frac{1}{2}$ into (iv): $-8(-\frac{1}{2})+4b=1 \Rightarrow 4+4b=1 \Rightarrow 4b=-3 \Rightarrow b=-\frac{3}{4}$.
Thus,$a=-\frac{1}{2}, b=-\frac{3}{4}, c=3$.

(C) Given the curve $y = \sin x$.
First,find the derivative $\frac{dy}{dx} = \cos x$.
The slope of the tangent at $(0, 0)$ is $\left(\frac{dy}{dx}\right)_{(0,0)} = \cos(0) = 1$.
The slope of the normal at $(0, 0)$ is $m = -\frac{1}{\text{slope of tangent}} = -\frac{1}{1} = -1$.
The equation of the normal passing through $(0, 0)$ with slope $m = -1$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - 0 = -1(x - 0)$.
This simplifies to $y = -x$,or $x + y = 0$.

(A) Given point $P=(1,2)$.
The slope of the tangent is given by $\frac{dy}{dx} = \tan(\theta) = \tan(\tan^{-1}(2x+3y)) = 2x+3y$.
This is a linear differential equation: $\frac{dy}{dx} - 3y = 2x$.
Integrating factor $IF = e^{\int -3 dx} = e^{-3x}$.
Multiplying both sides by $IF$: $y e^{-3x} = \int 2x e^{-3x} dx + c$.
Using integration by parts: $\int 2x e^{-3x} dx = 2x \left(\frac{e^{-3x}}{-3}\right) - \int 2 \left(\frac{e^{-3x}}{-3}\right) dx = -\frac{2}{3}x e^{-3x} - \frac{2}{9} e^{-3x} + c$.
Thus,$y e^{-3x} = -\frac{2}{3}x e^{-3x} - \frac{2}{9} e^{-3x} + c$.
Multiplying by $e^{3x}$: $y = -\frac{2}{3}x - \frac{2}{9} + c e^{3x}$.
Since the curve passes through $(1,2)$: $2 = -\frac{2}{3}(1) - \frac{2}{9} + c e^3 \implies 2 = -\frac{8}{9} + c e^3 \implies c e^3 = \frac{26}{9} \implies c = \frac{26}{9} e^{-3}$.
Substituting $c$: $y = -\frac{2}{3}x - \frac{2}{9} + \frac{26}{9} e^{-3} e^{3x} \implies 9y = -6x - 2 + 26 e^{3x-3} \implies 6x + 9y + 2 = 26 e^{3x-3}$.

(C) Given line is $2x+18y=9$. The slope of this line is $m = -\frac{2}{18} = -\frac{1}{9}$.
Given curve is $y=x^3-3x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3x^2-3$.
The slope of the normal is $-\frac{1}{\frac{dy}{dx}} = -\frac{1}{3x^2-3}$.
Since the normal is parallel to the given line,their slopes are equal:
$-\frac{1}{9} = -\frac{1}{3(x^2-1)} \Rightarrow 3(x^2-1) = 9 \Rightarrow x^2-1 = 3 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Case $1$: If $x=2$,then $y = (2)^3 - 3(2) = 8-6 = 2$. The point is $(2, 2)$.
The equation of the normal is $y-2 = -\frac{1}{9}(x-2) \Rightarrow 9y-18 = -x+2 \Rightarrow x+9y = 20$.
Case $2$: If $x=-2$,then $y = (-2)^3 - 3(-2) = -8+6 = -2$. The point is $(-2, -2)$.
The equation of the normal is $y-(-2) = -\frac{1}{9}(x-(-2)) \Rightarrow 9(y+2) = -(x+2) \Rightarrow 9y+18 = -x-2 \Rightarrow x+9y = -20$.
Combining both cases,the required equations are $x+9y = \pm 20$.

(C) Given the curve $y = e^{2x}$.
First,we find the derivative $\frac{dy}{dx} = 2e^{2x}$.
At the point $(0, 1)$,the slope of the tangent is $m = \left(\frac{dy}{dx}\right)_{(0, 1)} = 2e^{2(0)} = 2(1) = 2$.
The equation of the tangent line at $(x_1, y_1) = (0, 1)$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - 1 = 2(x - 0)$,which simplifies to $y = 2x + 1$.
To find where the tangent meets the $x$-axis,we set $y = 0$.
$0 = 2x + 1 \implies 2x = -1 \implies x = -\frac{1}{2}$.
Thus,the tangent meets the $x$-axis at the point $(-\frac{1}{2}, 0)$.

(C) Given the curve $y = x + \frac{2}{x}$.
At $x = 2$,$y = 2 + \frac{2}{2} = 3$. So,the point is $(2, 3)$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 - \frac{2}{x^2}$.
At $x = 2$,the slope of the tangent $m_t = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}$.
The slope of the normal $m_n = -\frac{1}{m_t} = -2$.
The equation of the normal at $(2, 3)$ is $y - 3 = -2(x - 2)$,which simplifies to $y - 3 = -2x + 4$,or $2x + y = 7$.
To find the points $A$ and $B$ where the normal meets the coordinate axes:
For the $x$-intercept $(y = 0)$,$2x = 7 \implies x = \frac{7}{2}$. So,$A = (\frac{7}{2}, 0)$.
For the $y$-intercept $(x = 0)$,$y = 7$. So,$B = (0, 7)$.
The length of $AB = \sqrt{(\frac{7}{2} - 0)^2 + (0 - 7)^2} = \sqrt{\frac{49}{4} + 49} = \sqrt{49(\frac{1}{4} + 1)} = 7 \sqrt{\frac{5}{4}} = \frac{7\sqrt{5}}{2}$.

(B) Let $N(t)$ be the number of bacteria at time $t$. The growth is given by $N(t) = t^3$.
The rate of growth of the bacteria is given by the derivative $\frac{dN}{dt} = \frac{d}{dt}(t^3) = 3t^2$.
We are given that the rate of growth is $1200 \ \text{per } s$.
Therefore,$3t^2 = 1200$.
Dividing by $3$,we get $t^2 = 400$.
Taking the square root,$t = \sqrt{400} = 20 \ s$.
Thus,the time taken is $20 \ s$.

(A) The area $A$ of an equilateral triangle with side $a$ is given by:
$A = \frac{\sqrt{3}}{4} a^2$
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2a \cdot \frac{da}{dt}$
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} a \cdot \frac{da}{dt}$
Given that the rate of increase of the side is $\frac{da}{dt} = 2 \text{ cm s}^{-1}$ and the side length is $a = 10 \text{ cm}$,we substitute these values into the equation:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2$
$\frac{dA}{dt} = 10 \sqrt{3} \text{ cm}^2 \text{ s}^{-1}$
Thus,the area of the triangle increases at the rate of $10 \sqrt{3} \text{ cm}^2 \text{ s}^{-1}$.

(D) Given, radius $(r) = 7 \text{ m}$ and error in radius $(dr) = 0.02 \text{ m}$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$, we get $\frac{dV}{dr} = 4 \pi r^2$.
The approximate error in volume $(dV)$ is given by $dV = \frac{dV}{dr} \times dr$.
Substituting the values, $dV = 4 \pi (7)^2 \times 0.02$.
$dV = 4 \pi (49) \times 0.02$.
$dV = 196 \pi \times 0.02 = 3.92 \pi \text{ m}^3$.
Thus, the approximate error in calculating the volume is $3.92 \pi \text{ m}^3$. Therefore, option $(D)$ is correct.

(A) Let the principal be $P$. Given that the principal increases continuously at a rate of $6 \%$ per year,we have the differential equation:
$\frac{dP}{dt} = \frac{6}{100} P = 0.06 P$
Separating the variables,we get:
$\frac{dP}{P} = 0.06 dt$
Integrating both sides:
$\int \frac{dP}{P} = \int 0.06 dt$
$\log P = 0.06 t + C$
Initially,at $t = 0$,$P = 6000$. So,$\log 6000 = C$.
Thus,$\log P = 0.06 t + \log 6000$,which implies $\log(\frac{P}{6000}) = 0.06 t$.
We want to find the time $t$ when the principal doubles,i.e.,$P = 12000$.
$\log(\frac{12000}{6000}) = 0.06 t$
$\log 2 = \frac{6}{100} t$
$t = \frac{100}{6} \log 2 = \frac{50}{3} \log 2$
Therefore,the time required is $\frac{50}{3} \log 2$ years. Option $(A)$ is correct.

(D) Given function is $y = \ln(\ln(x))$ for $x > 1$.
First,we find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot \frac{d}{dx}(\ln(x)) = \frac{1}{x \ln(x)}$.
For the function to be decreasing,we must have $\frac{dy}{dx} < 0$.
So,$\frac{1}{x \ln(x)} < 0$.
Since the domain is given as $x > 1$,we know that $\ln(x) > 0$ and $x > 1$.
Therefore,the product $x \ln(x)$ is always positive for all $x > 1$.
Since $\frac{1}{x \ln(x)}$ is always positive for $x > 1$,there is no interval where the function is decreasing.
Thus,the correct option is $D$.

(D) Let the initial population be $P_0$. The population after $5 \text{ yr}$ at an annual growth rate of $3 \%$ is given by $P = P_0(1 + \frac{3}{100})^5$.
The percentage increase is calculated as:
$\frac{P - P_0}{P_0} \times 100 = [ (1 + 0.03)^5 - 1 ] \times 100$
$= [ (1.03)^5 - 1 ] \times 100$
Using the approximation $(1.03)^5 \approx 1.15927$,we get:
$\approx [1.15927 - 1] \times 100 = 15.927 \% \approx 15.9 \%$.
Hence,option $D$ is correct.

(D) Let a function $f(x) = \sqrt{x}$.
We choose $x = 196$ and $\Delta x = 3$ because $196$ is the nearest perfect square to $199$.
Using the differential formula $\Delta y \approx \frac{dy}{dx} \times \Delta x$:
$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.
At $x = 196$,$\frac{dy}{dx} = \frac{1}{2\sqrt{196}} = \frac{1}{2 \times 14} = \frac{1}{28}$.
Now,$\Delta y \approx \frac{1}{28} \times 3 = \frac{3}{28} \approx 0.10714$.
Therefore,$\sqrt{199} = \sqrt{196} + \Delta y \approx 14 + 0.1071 = 14.1071$.

(C) Let $y = f(x) = x^{1/3}$.
We choose $x = 27$ such that $x + \Delta x = 26$.
Then $\Delta x = 26 - 27 = -1$.
We know that $y + \Delta y \approx f(x + \Delta x) = (x + \Delta x)^{1/3}$.
The differential $\Delta y$ is given by $\Delta y \approx \frac{dy}{dx} \Delta x$.
Since $y = x^{1/3}$,we have $\frac{dy}{dx} = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}}$.
At $x = 27$,$\frac{dy}{dx} = \frac{1}{3(27)^{2/3}} = \frac{1}{3(9)} = \frac{1}{27}$.
Thus,$\Delta y \approx \frac{1}{27} \times (-1) = -\frac{1}{27} \approx -0.037037$.
Therefore,$\sqrt[3]{26} = y + \Delta y = 27^{1/3} - 0.037037 = 3 - 0.037037 = 2.962963$.
Rounding to three decimal places,we get $2.963$. However,based on the provided options,$2.962$ is the closest value.
Hence,option $C$ is correct.

(C) Given the function $f(x) = x^3 + 3x^2 - 2$.
To find the stationary points,we calculate the first derivative of the function with respect to $x$:
$f'(x) = \frac{d}{dx}(x^3 + 3x^2 - 2) = 3x^2 + 6x$.
Stationary points occur where the first derivative is equal to zero:
$f'(x) = 0 \Rightarrow 3x^2 + 6x = 0$.
Factoring the expression,we get:
$3x(x + 2) = 0$.
This gives the values $x = 0$ and $x = -2$.
Therefore,the values of $x$ at the stationary points are $0$ and $-2$.

(A) Let $f(x) = (1/2)^x$ for all $x \in R$.
To determine the nature of the function,we find its derivative:
$f'(x) = \frac{d}{dx} \left( (1/2)^x \right) = (1/2)^x \ln(1/2)$.
Since $\ln(1/2) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2)$,we have:
$f'(x) = -(1/2)^x \ln(2)$.
Since $(1/2)^x > 0$ for all $x \in R$ and $\ln(2) > 0$,it follows that $f'(x) < 0$ for all $x \in R$.
Because the derivative $f'(x)$ is strictly less than $0$ for all $x$ in the domain,the function $f(x)$ is strictly decreasing on $R$.

(B) Given function is $f(x) = k(x + \sin x) + k$.
To find the interval where the function is increasing,we calculate the derivative $f'(x)$.
$f'(x) = \frac{d}{dx}[k(x + \sin x) + k] = k(1 + \cos x)$.
For a function to be increasing,we must have $f'(x) \geq 0$.
So,$k(1 + \cos x) \geq 0$.
We know that for all real $x$,$-1 \leq \cos x \leq 1$,which implies $0 \leq 1 + \cos x \leq 2$.
Since $(1 + \cos x)$ is always non-negative,for the product $k(1 + \cos x)$ to be greater than or equal to $0$,$k$ must be greater than $0$.
Therefore,$k > 0$.

(B) Given,$y = \frac{ax - b}{(x - 1)(x - 4)} . . . . . . (i)$ has a turning point $P(2, -1)$.
Since point $P$ lies on the curve,it must satisfy equation $(i)$:
$-1 = \frac{2a - b}{(2 - 1)(2 - 4)} = \frac{2a - b}{-2}$
$2a - b = 2 . . . . . . (ii)$
At a turning point,the derivative $\frac{dy}{dx} = 0$.
From equation $(i)$,$y(x^2 - 5x + 4) = ax - b$.
Differentiating with respect to $x$ using the product rule:
$y(2x - 5) + (x^2 - 5x + 4) \frac{dy}{dx} = a$.
At $x = 2$ and $y = -1$,$\frac{dy}{dx} = 0$:
$-1(2(2) - 5) + (2^2 - 5(2) + 4)(0) = a$
$-1(4 - 5) = a$
$a = 1$.
Substituting $a = 1$ into equation $(ii)$:
$2(1) - b = 2$
$b = 0$.
Thus,$a = 1$ and $b = 0$.

(B) Let $R = 22 \ cm$ be the radius of the sphere and $h$ be the height of the cylinder inscribed in it. Let $r$ be the radius of the cylinder.
From the geometry of the sphere,we have $r^2 + (h/2)^2 = R^2$,which implies $r^2 = R^2 - h^2/4$.
The curved surface area $A$ of the cylinder is given by $A = 2\pi rh = 2\pi h \sqrt{R^2 - h^2/4} = \pi h \sqrt{4R^2 - h^2}$.
To maximize $A$,we maximize $A^2 = \pi^2 h^2 (4R^2 - h^2) = \pi^2 (4R^2h^2 - h^4)$.
Let $f(h) = 4R^2h^2 - h^4$. Differentiating with respect to $h$,we get $f'(h) = 8R^2h - 4h^3$.
Setting $f'(h) = 0$,we get $4h(2R^2 - h^2) = 0$. Since $h > 0$,we have $h^2 = 2R^2$,so $h = R\sqrt{2}$.
Given $R = 22 \ cm$,the height $h = 22\sqrt{2} \ cm$.

(A) Given function is $f(x) = -x + \sin 2x$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
First,find the critical points by setting $f'(x) = 0$.
$f'(x) = -1 + 2 \cos 2x = 0$
$\cos 2x = \frac{1}{2}$
Since $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$2x \in [-\pi, \pi]$.
Thus,$2x = \pm \frac{\pi}{3}$,which gives $x = \pm \frac{\pi}{6}$.
Now,evaluate $f(x)$ at the critical points and the endpoints:
$f(-\frac{\pi}{2}) = -(-\frac{\pi}{2}) + \sin(-\pi) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \approx 1.57$
$f(-\frac{\pi}{6}) = -(-\frac{\pi}{6}) + \sin(-\frac{\pi}{3}) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} \approx 0.52 - 0.866 = -0.346$
$f(\frac{\pi}{6}) = -(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} \approx -0.52 + 0.866 = 0.346$
$f(\frac{\pi}{2}) = -(\frac{\pi}{2}) + \sin(\pi) = -\frac{\pi}{2} + 0 = -\frac{\pi}{2} \approx -1.57$
Comparing these values,the greatest value is $M = \frac{\pi}{2}$ and the least value is $m = -\frac{\pi}{2}$.
The difference is $M - m = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.