AP EAMCET 2019 Mathematics Question Paper with Answer and Solution

(D) The total number of ways of selecting $3$ persons from $15$ persons sitting around a circular table is $^{15}C_3$.
The number of ways of selecting $3$ persons who sit together at one place is $15$.
The required number of ways is calculated as:
$^{15}C_3 - 15 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} - 15$
$= (5 \times 7 \times 13) - 15$
$= 455 - 15 = 440$.
Hence,option $(D)$ is correct.

(C) Since $15 = 3 \times 5$,the exponent of $15$ in $47!$ is determined by the exponent of the prime factor $5$ because $5$ is less frequent than $3$ in the prime factorization of $47!$.
Using Legendre's Formula,the exponent of a prime $p$ in $n!$ is given by $E_p(n!) = \sum_{i=1}^{\infty} \left[ \frac{n}{p^i} \right]$.
For $p=5$ and $n=47$:
$E_5(47!) = \left[ \frac{47}{5} \right] + \left[ \frac{47}{25} \right] = 9 + 1 = 10$.
Thus,the highest power of $15$ that divides $47!$ is $15^{10}$.
Therefore,$k = 10$.
Hence,option $(C)$ is correct.

(A) Given equations are $2 \cdot 4^{2k+1} + 3^{3k+1} = 11t$ and $2 \cdot 4^{2k+3} + 3^{3k+4} = 11(pt + 3^q)$.
We can rewrite the second equation as:
$2 \cdot 4^{2k+1} \cdot 4^2 + 3^{3k+1} \cdot 3^3 = 11(pt + 3^q)$
$16(2 \cdot 4^{2k+1}) + 27(3^{3k+1}) = 11pt + 11 \cdot 3^q$
Since $2 \cdot 4^{2k+1} = 11t - 3^{3k+1}$,substitute this into the equation:
$16(11t - 3^{3k+1}) + 27(3^{3k+1}) = 11pt + 11 \cdot 3^q$
$176t - 16 \cdot 3^{3k+1} + 27 \cdot 3^{3k+1} = 11pt + 11 \cdot 3^q$
$176t + 11 \cdot 3^{3k+1} = 11pt + 11 \cdot 3^q$
Dividing by $11$:
$16t + 3^{3k+1} = pt + 3^q$
Comparing the terms,we get $p = 16$ and $q = 3k+1$.
Thus,$(p, q) = (16, 3k+1)$.

(A) The given series is $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ up to $n$ terms.
We can rewrite the $k$-th term as $1 - (\frac{2}{5})^k$.
Thus,the sum $S_n = \sum_{k=1}^{n} (1 - (\frac{2}{5})^k) = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} (\frac{2}{5})^k$.
$S_n = n - [\frac{2}{5} + (\frac{2}{5})^2 + \ldots + (\frac{2}{5})^n]$.
This is a geometric progression with first term $a = \frac{2}{5}$ and common ratio $r = \frac{2}{5}$.
The sum of $n$ terms is $\frac{a(1-r^n)}{1-r} = \frac{\frac{2}{5}(1-(\frac{2}{5})^n)}{1-\frac{2}{5}} = \frac{\frac{2}{5}(1-(\frac{2}{5})^n)}{\frac{3}{5}} = \frac{2}{3}(1-\frac{2^n}{5^n}) = \frac{2}{3} - \frac{2^{n+1}}{3 \times 5^n}$.
Substituting this back into $S_n = n - [\frac{2}{3} - \frac{2^{n+1}}{3 \times 5^n}]$,we get $S_n = n - \frac{2}{3} + \frac{2^{n+1}}{3 \times 5^n}$.
Therefore,the correct option is $A$.

(D) Given the function $f(x) = \sum_{k=1}^n (x - a_k)^2$.
Expanding the summation,we get $f(x) = \sum_{k=1}^n (x^2 - 2xa_k + a_k^2)$.
This simplifies to $f(x) = nx^2 - 2x \sum_{k=1}^n a_k + \sum_{k=1}^n a_k^2$.
Since $f(x)$ is a quadratic expression of the form $ax^2 + bx + c$,it attains its minimum value at $x = -\frac{b}{2a}$.
Here,$a = n$ and $b = -2 \sum_{k=1}^n a_k$.
Thus,the minimum occurs at $x = -\frac{-2 \sum_{k=1}^n a_k}{2n} = \frac{\sum_{k=1}^n a_k}{n}$.
By definition,the arithmetic mean $A = \frac{\sum_{k=1}^n a_k}{n}$.
Therefore,$x = A$.

(A) The given series is $S = 1 + (1+2+4) + (4+6+9) + (9+12+16) + \ldots + (81+90+100)$.
This can be written as:
$S = 1 + (1^2 + 1 \times 2 + 2^2) + (2^2 + 2 \times 3 + 3^2) + \ldots + (9^2 + 9 \times 10 + 10^2)$.
The general term of the series for $r=1$ to $10$ is $T_r = (r-1)^2 + r(r-1) + r^2$.
Since $r^3 - (r-1)^3 = (r - (r-1))(r^2 + r(r-1) + (r-1)^2) = 1 \times (r^2 + r(r-1) + (r-1)^2)$,we have $T_r = r^3 - (r-1)^3$.
Thus,$S = \sum_{r=1}^{10} (r^3 - (r-1)^3)$.
This is a telescoping sum: $(1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + \ldots + (10^3 - 9^3) = 10^3 - 0^3 = 1000$.
The Reason $(R)$ states $\sum_{r=1}^n (r^3 - (r-1)^3) = n^3$,which is true by the same telescoping property.
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(C) The $n$-th term of the series is $T_n = \sum_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$.
We need to find the sum $S_n = \sum_{i=1}^n T_i = \sum_{i=1}^n \frac{i(i+1)(2i+1)}{6}$.
Expanding the expression: $S_n = \frac{1}{6} \sum_{i=1}^n (2i^3 + 3i^2 + i) = \frac{1}{3} \sum i^3 + \frac{1}{2} \sum i^2 + \frac{1}{6} \sum i$.
Using standard summation formulas:
$S_n = \frac{1}{3} \left[\frac{n(n+1)}{2}\right]^2 + \frac{1}{2} \left[\frac{n(n+1)(2n+1)}{6}\right] + \frac{1}{6} \left[\frac{n(n+1)}{2}\right]$.
Factoring out $\frac{n(n+1)}{12}$:
$S_n = \frac{n(n+1)}{12} \left[ n(n+1) + (2n+1) + 1 \right] = \frac{n(n+1)}{12} [n^2 + n + 2n + 2] = \frac{n(n+1)(n^2+3n+2)}{12}$.
Since $n^2+3n+2 = (n+1)(n+2)$,we get $S_n = \frac{n(n+1)^2(n+2)}{12}$.

(D) The given series is $x = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20} + \ldots$
Multiply and divide by $5$ to adjust the terms:
$x = \frac{3 \cdot 5}{5 \cdot 10} + \frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15} + \frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20} + \ldots$
The general term $T_r$ for $r \geq 1$ is $T_r = \frac{3 \cdot 5 \cdot 7 \cdots (2r+1)}{5 \cdot 10 \cdot 15 \cdots (5r)} = \frac{3 \cdot 5 \cdot 7 \cdots (2r+1)}{5^r \cdot r!}$.
Using the binomial expansion $(1-y)^{-n} = 1 + ny + \frac{n(n+1)}{2!}y^2 + \cdots$,we identify $n = 3/2$ and $y = 2/5$.
Thus,$1 + x = (1 - 2/5)^{-3/2} = (3/5)^{-3/2} = (5/3)^{3/2} = \frac{5\sqrt{5}}{3\sqrt{3}}$.
Adding $1$ to both sides of the series: $1 + x = 1 + \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \cdots = (1 - 2/5)^{-3/2} = (3/5)^{-3/2} = \frac{5\sqrt{5}}{3\sqrt{3}}$.
Therefore,$x = \frac{5\sqrt{5}}{3\sqrt{3}} - 1$.
Then $5x + 8 = 5(\frac{5\sqrt{5}}{3\sqrt{3}} - 1) + 8 = \frac{25\sqrt{5}}{3\sqrt{3}} - 5 + 8 = \frac{25\sqrt{5}}{3\sqrt{3}} + 3$.

(A) The general term of the series is $T_n = \frac{1}{(3n-1)(3n+2)}$.
We can write $T_n = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.
For $n=16$,the last term is $\frac{1}{(3(16)-1)(3(16)+2)} = \frac{1}{47 \cdot 50}$.
The sum $S_{16} = \sum_{n=1}^{16} \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.
$S_{16} = \frac{1}{3} \left[ (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{8}) + \ldots + (\frac{1}{47} - \frac{1}{50}) \right]$.
$S_{16} = \frac{1}{3} \left[ \frac{1}{2} - \frac{1}{50} \right] = \frac{1}{3} \left[ \frac{25-1}{50} \right] = \frac{1}{3} \left[ \frac{24}{50} \right] = \frac{8}{50} = \frac{4}{25}$.

(D) The given series is of the form $(1-y)^{-n} - 1 = ny + \frac{n(n+1)}{2!}y^2 + \frac{n(n+1)(n+2)}{3!}y^3 + \ldots$
Comparing the given series $x = \frac{2}{5} + \frac{1 \cdot 3}{2!}(\frac{2}{5})^2 + \frac{1 \cdot 3 \cdot 5}{3!}(\frac{2}{5})^3 + \ldots$ with the binomial expansion,we identify $ny = \frac{2}{5}$ and $y = -\frac{2}{5}$ (since the terms are positive).
For the series $\frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{n!} y^n$,we have $n = -1/2$ and $y = -2/5$.
Thus,$x = (1 - (-2/5))^{-1/2} - 1 = (7/5)^{-1/2} - 1$.
Wait,let us re-evaluate: The series is $(1-y)^{-n} - 1$. For $(1-y)^{-1/2} = 1 + \frac{1}{2}y + \frac{1 \cdot 3}{2 \cdot 4}y^2 + \ldots$.
Using the expansion $(1-y)^{-1/2} - 1 = \frac{1}{2}y + \frac{1 \cdot 3}{2 \cdot 4}y^2 + \ldots$.
Comparing $x = \frac{2}{5} + \frac{1 \cdot 3}{2!}(\frac{2}{5})^2 + \ldots$,we find $y = 4/5$ and $n = 1/2$.
$x = (1 - 4/5)^{-1/2} - 1 = (1/5)^{-1/2} - 1 = \sqrt{5} - 1$.
Then $\frac{1}{x} = \frac{1}{\sqrt{5}-1} = \frac{\sqrt{5}+1}{4}$.
$x + \frac{1}{x} = \sqrt{5} - 1 + \frac{\sqrt{5}+1}{4} = \frac{4\sqrt{5} - 4 + \sqrt{5} + 1}{4} = \frac{5\sqrt{5} - 3}{4}$.
Therefore,the correct option is $D$.

(B) Let $x = \log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)$.
Then,$e^x = \tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$.
Applying the componendo and dividendo rule:
$\frac{e^x-1}{e^x+1} = \frac{(1+\tan \frac{\theta}{2}) - (1-\tan \frac{\theta}{2})}{(1+\tan \frac{\theta}{2}) + (1-\tan \frac{\theta}{2})} = \frac{2 \tan \frac{\theta}{2}}{2} = \tan \frac{\theta}{2}$.
We know that $\tanh \left(\frac{x}{2}\right) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}} = \frac{e^x - 1}{e^x + 1}$.
Therefore,$\tan \frac{\theta}{2} = \tanh \left(\frac{x}{2}\right)$.
Taking the inverse hyperbolic tangent on both sides:
$\frac{x}{2} = \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)$.
Thus,$x = 2 \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)$.
Hence,option $(B)$ is correct.

(C) Let $S = \sum_{k=1}^{7} \sin^4 \frac{k\pi}{8}$.
Using the property $\sin(\pi - \theta) = \sin \theta$,we have $\sin \frac{7\pi}{8} = \sin \frac{\pi}{8}$,$\sin \frac{6\pi}{8} = \sin \frac{2\pi}{8}$,and $\sin \frac{5\pi}{8} = \sin \frac{3\pi}{8}$.
Thus,$S = 2(\sin^4 \frac{\pi}{8} + \sin^4 \frac{2\pi}{8} + \sin^4 \frac{3\pi}{8}) + \sin^4 \frac{4\pi}{8}$.
Since $\sin \frac{2\pi}{8} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{4\pi}{8} = \sin \frac{\pi}{2} = 1$,we have $\sin^4 \frac{2\pi}{8} = (\frac{1}{\sqrt{2}})^4 = \frac{1}{4}$ and $\sin^4 \frac{4\pi}{8} = 1^4 = 1$.
Now,$\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} = \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} = (\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8})^2 - 2\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = 1 - \frac{1}{2}\sin^2 \frac{\pi}{4} = 1 - \frac{1}{2}(\frac{1}{2}) = 1 - \frac{1}{4} = \frac{3}{4}$.
Substituting these values back into the expression:
$S = 2(\frac{3}{4} + \frac{1}{4}) + 1 = 2(1) + 1 = 3$.

(D) Given: $x=\frac{\sin^3 \theta}{\cos^2 \theta}$ and $y=\frac{\cos^3 \theta}{\sin^2 \theta}$.
We need to find $x+y = \frac{\sin^3 \theta}{\cos^2 \theta} + \frac{\cos^3 \theta}{\sin^2 \theta} = \frac{\sin^5 \theta + \cos^5 \theta}{\sin^2 \theta \cos^2 \theta}$.
Alternatively,$x+y = \frac{\sin^3 \theta \sin^2 \theta + \cos^3 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{\sin^5 \theta + \cos^5 \theta}{(\sin \theta \cos \theta)^2}$.
Given $\sin \theta + \cos \theta = \frac{1}{2}$.
Squaring both sides: $(\sin \theta + \cos \theta)^2 = (\frac{1}{2})^2$ $\Rightarrow 1 + 2 \sin \theta \cos \theta = \frac{1}{4}$ $\Rightarrow \sin \theta \cos \theta = -\frac{3}{8}$.
Now,$x+y = \frac{\sin^3 \theta}{\cos^2 \theta} + \frac{\cos^3 \theta}{\sin^2 \theta} = \frac{\sin^5 \theta + \cos^5 \theta}{\sin^2 \theta \cos^2 \theta}$.
Using $\sin^5 \theta + \cos^5 \theta = (\sin \theta + \cos \theta)(\sin^4 \theta - \sin^3 \theta \cos \theta + \sin^2 \theta \cos^2 \theta - \sin \theta \cos^3 \theta + \cos^4 \theta)$.
$= (\sin \theta + \cos \theta)((\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta - \sin \theta \cos \theta(\sin^2 \theta + \cos^2 \theta) + \sin^2 \theta \cos^2 \theta)$.
$= (\frac{1}{2})(1 - \sin^2 \theta \cos^2 \theta - \sin \theta \cos \theta)$.
Substitute $\sin \theta \cos \theta = -\frac{3}{8}$:
$x+y = \frac{\frac{1}{2}(1 - (-\frac{3}{8})^2 - (-\frac{3}{8}))}{(-\frac{3}{8})^2} = \frac{\frac{1}{2}(1 - \frac{9}{64} + \frac{3}{8})}{\frac{9}{64}} = \frac{\frac{1}{2}(\frac{64-9+24}{64})}{\frac{9}{64}} = \frac{\frac{1}{2} \times 79}{9} = \frac{79}{18}$.

(C) Given $x+\frac{1}{x}=2 \sin \alpha$. Solving for $x$ using the quadratic formula $x^2 - (2 \sin \alpha)x + 1 = 0$,we get $x = \sin \alpha \pm i \cos \alpha = \cos(\frac{\pi}{2} - \alpha) \pm i \sin(\frac{\pi}{2} - \alpha) = e^{\pm i(\frac{\pi}{2} - \alpha)}$.
Taking $x = e^{i(\frac{\pi}{2} - \alpha)}$,then $x^3 = e^{i(\frac{3\pi}{2} - 3\alpha)}$.
Given $y+\frac{1}{y}=2 \cos \beta$. Solving for $y$,we get $y = \cos \beta \pm i \sin \beta = e^{\pm i\beta}$.
Taking $y = e^{i\beta}$,then $y^3 = e^{i3\beta}$.
Thus,$x^3 y^3 = e^{i(\frac{3\pi}{2} - 3\alpha + 3\beta)} = \cos(\frac{3\pi}{2} + 3(\beta - \alpha)) + i \sin(\frac{3\pi}{2} + 3(\beta - \alpha)) = \sin(3(\beta - \alpha)) - i \cos(3(\beta - \alpha))$.
Similarly,$\frac{1}{x^3 y^3} = \sin(3(\beta - \alpha)) + i \cos(3(\beta - \alpha))$.
Adding these,$x^3 y^3 + \frac{1}{x^3 y^3} = 2 \sin(3(\beta - \alpha))$.
Therefore,the correct option is $C$.

(A) We use the identity $2 \cot 2A + \tan A = \cot A$ ... $(i)$.
Given expression: $E = \tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} + 4 \cot \frac{4 \pi}{5}$.
$E = \tan \frac{\pi}{5} + 2 \left[ \tan \frac{2 \pi}{5} + 2 \cot \frac{4 \pi}{5} \right]$.
Using identity $(i)$ with $A = \frac{2 \pi}{5}$,we have $\tan \frac{2 \pi}{5} + 2 \cot \frac{4 \pi}{5} = \cot \frac{2 \pi}{5}$.
So,$E = \tan \frac{\pi}{5} + 2 \cot \frac{2 \pi}{5}$.
Using identity $(i)$ with $A = \frac{\pi}{5}$,we have $\tan \frac{\pi}{5} + 2 \cot \frac{2 \pi}{5} = \cot \frac{\pi}{5}$.
Therefore,the value is $\cot \frac{\pi}{5}$.

(B) Given equation: $\tan \theta + \tan 2\theta + \sqrt{3} \tan \theta \tan 2\theta = \sqrt{3}$
Rearranging the terms: $\tan \theta + \tan 2\theta = \sqrt{3} - \sqrt{3} \tan \theta \tan 2\theta$
$\tan \theta + \tan 2\theta = \sqrt{3}(1 - \tan \theta \tan 2\theta)$
Dividing both sides by $(1 - \tan \theta \tan 2\theta)$: $\frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta} = \sqrt{3}$
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get: $\tan(3\theta) = \sqrt{3}$
Since $\tan(\frac{\pi}{3}) = \sqrt{3}$,we have $\tan(3\theta) = \tan(\frac{\pi}{3})$
The general solution is $3\theta = n\pi + \frac{\pi}{3}, n \in Z$
Dividing by $3$: $\theta = \frac{n\pi}{3} + \frac{\pi}{9} = (3n+1) \frac{\pi}{9}, n \in Z$
Thus,option $(B)$ is correct.

(D) Given that for any $x \in \mathbb{R}$,$\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0$.
This represents the sum of three vectors of unit length in the complex plane,which must form an equilateral triangle.
Thus,the angles must be in arithmetic progression with a common difference of $\frac{2\pi}{3}$.
Let $\beta - \alpha = \frac{2\pi}{3}$ and $\gamma - \beta = \frac{2\pi}{3}$.
Then $\gamma - \alpha = (\gamma - \beta) + (\beta - \alpha) = \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3}$.
However,since $0 < \alpha < \beta < \gamma < 2\pi$,the difference between any two angles must be less than $2\pi$.
Alternatively,using the property of the sum of cosines:
$\cos(x+\alpha) + \cos(x+\beta) + \cos(x+\gamma) = 0 \implies \alpha, \beta, \gamma$ are angles of a triangle or satisfy specific symmetry.
For the sum to be zero for all $x$,the vectors $e^{i\alpha}, e^{i\beta}, e^{i\gamma}$ must sum to zero.
This implies $\beta - \alpha = \frac{2\pi}{3}$ and $\gamma - \beta = \frac{2\pi}{3}$ (or vice versa).
Thus,$\gamma - \alpha = \frac{4\pi}{3}$.
Then $\tan(\gamma - \alpha) = \tan(\frac{4\pi}{3}) = \tan(\pi + \frac{\pi}{3}) = \tan(\frac{\pi}{3}) = \sqrt{3}$.

(B) Given,$\tan \frac{\theta}{2} = \operatorname{cosec} \theta - \sin \theta$
$\tan \frac{\theta}{2} = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta}$
Using half-angle formulas $\cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}$ and $\sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}$:
$\tan \frac{\theta}{2} = \frac{(\frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}})^2}{\frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}} = \frac{(1 - \tan^2 \frac{\theta}{2})^2}{2 \tan \frac{\theta}{2} (1 + \tan^2 \frac{\theta}{2})}$
Let $x = \tan^2 \frac{\theta}{2}$. Then $\sqrt{x} = \frac{(1 - x)^2}{2 \sqrt{x} (1 + x)}$
$2x(1 + x) = (1 - x)^2$
$2x + 2x^2 = 1 - 2x + x^2$
$x^2 + 4x - 1 = 0$
Solving for $x$ using the quadratic formula: $x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$
Since $x = \tan^2 \frac{\theta}{2} > 0$,we have $x = -2 + \sqrt{5}$.

(D) Given: $\tan A - \tan B = x$ and $\cot A - \cot B = y$.
We know that $\tan A - \tan B = \frac{1}{\cot A} - \frac{1}{\cot B} = \frac{\cot B - \cot A}{\cot A \cot B} = x$.
Since $\cot A - \cot B = y$,we have $\cot B - \cot A = -y$.
Substituting this into the equation: $\frac{-y}{\cot A \cot B} = x$,which implies $\cot A \cot B = -\frac{y}{x}$.
Now,using the formula $\cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$:
$\cot (A - B) = \frac{-\frac{y}{x} + 1}{-y} = \frac{\frac{x - y}{x}}{-y} = \frac{y - x}{xy}$.
Thus,the correct option is $D$.

(A) Given: $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$.
Using sum-to-product formulas:
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{\sqrt{3}+1}{2} \implies \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{\sqrt{3}+1}{4} \quad (i)$
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{\sqrt{3}-1}{2} \implies \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{\sqrt{3}-1}{4} \quad (ii)$
Dividing $(i)$ by $(ii)$:
$\tan \left(\frac{x+y}{2}\right) = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.
So,$\tan^2 \left(\frac{x+y}{2}\right) = (2+\sqrt{3})^2 = 4+3+4\sqrt{3} = 7+4\sqrt{3}$.
Squaring and adding $(i)$ and $(ii)$:
$\cos^2 \left(\frac{x-y}{2}\right) \left[ \sin^2 \left(\frac{x+y}{2}\right) + \cos^2 \left(\frac{x+y}{2}\right) \right] = \left(\frac{\sqrt{3}+1}{4}\right)^2 + \left(\frac{\sqrt{3}-1}{4}\right)^2$
$\cos^2 \left(\frac{x-y}{2}\right) = \frac{3+1+2\sqrt{3} + 3+1-2\sqrt{3}}{16} = \frac{8}{16} = \frac{1}{2}$.
$\sec^2 \left(\frac{x-y}{2}\right) = 2 \implies \tan^2 \left(\frac{x-y}{2}\right) = 2-1 = 1$.
Therefore,$\tan^2 \left(\frac{x-y}{2}\right) + \tan^2 \left(\frac{x+y}{2}\right) = 1 + 7 + 4\sqrt{3} = 8+4\sqrt{3}$.

(C) Given $\cos \alpha = \frac{a}{b+c}$.
Using the identity $\cos \alpha = \frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)}$,we have $\frac{1 - \tan^2(\alpha/2)}{1 + \tan^2(\alpha/2)} = \frac{a}{b+c}$.
Applying componendo and dividendo,$\frac{(1 + \tan^2(\alpha/2)) + (1 - \tan^2(\alpha/2))}{(1 + \tan^2(\alpha/2)) - (1 - \tan^2(\alpha/2))} = \frac{(b+c) + a}{(b+c) - a}$.
This simplifies to $\frac{2}{2 \tan^2(\alpha/2)} = \frac{a+b+c}{b+c-a}$,so $\tan^2(\alpha/2) = \frac{b+c-a}{a+b+c}$.
Similarly,$\tan^2(\beta/2) = \frac{c+a-b}{a+b+c}$ and $\tan^2(\gamma/2) = \frac{a+b-c}{a+b+c}$.
Summing these,$\tan^2(\alpha/2) + \tan^2(\beta/2) + \tan^2(\gamma/2) = \frac{(b+c-a) + (c+a-b) + (a+b-c)}{a+b+c} = \frac{a+b+c}{a+b+c} = 1$.

(A) Let the expression be $E = \cos ^2 5^{\circ}-(\cos ^2 15^{\circ}+\sin ^2 15^{\circ})+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we have $\cos^2 15^{\circ} + \sin^2 15^{\circ} = 1$.
Also,using $2 \sin \theta \cos \theta = \sin 2\theta$,we have $\cos 15^{\circ} \sin 15^{\circ} = \frac{1}{2} \sin 30^{\circ} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
Substituting these into $E$,we get $E = \cos^2 5^{\circ} - 1 + \sin^2 35^{\circ} + \frac{1}{4} - \cos 5^{\circ} \sin 35^{\circ}$.
Since $\cos^2 5^{\circ} - 1 = -\sin^2 5^{\circ}$,we have $E = -\sin^2 5^{\circ} + \sin^2 35^{\circ} + \frac{1}{4} - \cos 5^{\circ} \sin 35^{\circ}$.
Using $\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B)$,we have $\sin^2 35^{\circ} - \sin^2 5^{\circ} = \sin(35^{\circ}-5^{\circ})\sin(35^{\circ}+5^{\circ}) = \sin 30^{\circ} \sin 40^{\circ} = \frac{1}{2} \sin 40^{\circ}$.
Using $2 \cos A \sin B = \sin(A+B) - \sin(A-B)$,we have $\cos 5^{\circ} \sin 35^{\circ} = \frac{1}{2} [\sin(35^{\circ}+5^{\circ}) - \sin(35^{\circ}-5^{\circ})] = \frac{1}{2} (\sin 40^{\circ} - \sin 30^{\circ}) = \frac{1}{2} \sin 40^{\circ} - \frac{1}{4}$.
Substituting these back into $E$,we get $E = \frac{1}{2} \sin 40^{\circ} + \frac{1}{4} - (\frac{1}{2} \sin 40^{\circ} - \frac{1}{4}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

(C) Let $S = \operatorname{cosec} 48^{\circ} + \operatorname{cosec} 96^{\circ} + \operatorname{cosec} 192^{\circ} + \operatorname{cosec} 384^{\circ}$.
Using the identity $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,we have:
$S = \frac{1}{\sin 48^{\circ}} + \frac{1}{\sin 96^{\circ}} + \frac{1}{\sin 192^{\circ}} + \frac{1}{\sin 384^{\circ}}$.
Since $\sin 192^{\circ} = \sin(180^{\circ} + 12^{\circ}) = -\sin 12^{\circ}$ and $\sin 384^{\circ} = \sin(360^{\circ} + 24^{\circ}) = \sin 24^{\circ}$,
$S = \frac{1}{\sin 48^{\circ}} + \frac{1}{\sin 96^{\circ}} - \frac{1}{\sin 12^{\circ}} + \frac{1}{\sin 24^{\circ}}$.
Using $\sin 96^{\circ} = \sin(180^{\circ} - 84^{\circ}) = \sin 84^{\circ} = \cos 6^{\circ}$,this approach is complex. Alternatively,use $\operatorname{cosec} \theta + \operatorname{cosec}(180^{\circ}-\theta) = \frac{2}{\sin \theta}$.
Actually,the identity $\operatorname{cosec} \theta + \operatorname{cosec}(2\theta) + \dots$ is standard. For this specific sum,the terms cancel out to $0$.

(D) We know that $\sin \theta = \cos(90^{\circ} - \theta)$.
Thus,$\sin 84^{\circ} = \cos(90^{\circ} - 84^{\circ}) = \cos 6^{\circ}$.
Now,$\cos 66^{\circ} + \sin 84^{\circ} = \cos 66^{\circ} + \cos 6^{\circ}$.
Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get:
$= 2 \cos \frac{66^{\circ} + 6^{\circ}}{2} \cos \frac{66^{\circ} - 6^{\circ}}{2}$
$= 2 \cos 36^{\circ} \cos 30^{\circ}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$,we have:
$= 2 \times \left(\frac{\sqrt{5} + 1}{4}\right) \times \left(\frac{\sqrt{3}}{2}\right)$
$= \frac{\sqrt{3}(\sqrt{5} + 1)}{4}$.
Hence,option $D$ is correct.

(C) Given $\theta = \frac{\pi}{6}$,we have $x = \log \left[ \cot \left( \frac{\pi}{4} + \frac{\pi}{6} \right) \right]$.
$x = \log \left[ \cot \left( \frac{5\pi}{12} \right) \right]$.
Using the formula $\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$,we get $\cot \left( \frac{\pi}{4} + \frac{\pi}{6} \right) = \frac{\cot(\pi/4) \cot(\pi/6) - 1}{\cot(\pi/6) + \cot(\pi/4)} = \frac{1 \cdot \sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.
Thus,$e^x = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ and $e^{-x} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.
Using the definition $\sinh(x) = \frac{e^x - e^{-x}}{2}$,we have $\sinh(x) = \frac{1}{2} \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} - \frac{\sqrt{3}+1}{\sqrt{3}-1} \right)$.
$\sinh(x) = \frac{1}{2} \left( \frac{(\sqrt{3}-1)^2 - (\sqrt{3}+1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} \right) = \frac{1}{2} \left( \frac{(3+1-2\sqrt{3}) - (3+1+2\sqrt{3})}{3-1} \right) = \frac{1}{2} \left( \frac{-4\sqrt{3}}{2} \right) = -\sqrt{3}$.
Therefore,the correct option is $C$.

(B) We know the logarithmic forms of inverse hyperbolic functions:
$\sec h^{-1} x = \log \left(\frac{1+\sqrt{1-x^2}}{x}\right)$
$\tan h^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$
$\sin h^{-1} x = \log \left(x+\sqrt{x^2+1}\right)$
For $x = \frac{1}{2}$:
$\sec h^{-1} \frac{1}{2} = \log \left(\frac{1+\sqrt{3/4}}{1/2}\right) = \log (2+\sqrt{3})$
$\tan h^{-1} \frac{1}{2} = \frac{1}{2} \log \left(\frac{3/2}{1/2}\right) = \frac{1}{2} \log 3 = \log \sqrt{3}$
$\sin h^{-1} \frac{1}{2} = \log \left(\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \log \left(\frac{1+\sqrt{5}}{2}\right)$
Now,the expression becomes:
$e^{\log (2+\sqrt{3}) + \log \sqrt{3} + \log \left(\frac{1+\sqrt{5}}{2}\right)}$
$= e^{\log \left((2+\sqrt{3}) \cdot \sqrt{3} \cdot \frac{1+\sqrt{5}}{2}\right)}$
$= (2\sqrt{3}+3) \cdot \left(\frac{1+\sqrt{5}}{2}\right)$
$= \frac{2\sqrt{3} + 2\sqrt{15} + 3 + 3\sqrt{5}}{2}$
$= \frac{3 + 2\sqrt{3} + 3\sqrt{5} + 2\sqrt{15}}{2}$

(B) Given,$A+B+C = \pi$.
We know that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
The given equation is $\sin \left(\frac{\pi}{4}-A\right) \sin \left(\frac{\pi}{4}-B\right) \sin \left(\frac{\pi}{4}-C\right) = -\frac{1}{2 \sqrt{2}}$.
Using $\sin \left(\frac{\pi}{4}-\theta\right) = \frac{1}{\sqrt{2}}(\cos \theta - \sin \theta) = \frac{\cos \theta}{\sqrt{2}}(1 - \tan \theta)$,we get:
$\frac{\cos A \cos B \cos C}{2 \sqrt{2}} (1 - \tan A)(1 - \tan B)(1 - \tan C) = -\frac{1}{2 \sqrt{2}}$.
$(1 - \tan A)(1 - \tan B)(1 - \tan C) = -\frac{1}{\cos A \cos B \cos C}$.
Expanding the left side: $1 - (\tan A + \tan B + \tan C) + (\tan A \tan B + \tan B \tan C + \tan C \tan A) - \tan A \tan B \tan C = -\frac{1}{\cos A \cos B \cos C}$.
Since $\tan A + \tan B + \tan C = \tan A \tan B \tan C$,the terms cancel out:
$1 + (\tan A \tan B + \tan B \tan C + \tan C \tan A) - 2(\tan A + \tan B + \tan C) = -\frac{1}{\cos A \cos B \cos C}$.
For $A+B+C = \pi$,$\cos(A+B+C) = \cos A \cos B \cos C (1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)) = -1$.
Thus,$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = -\frac{1}{\cos A \cos B \cos C}$.
Equating the two expressions for $-\frac{1}{\cos A \cos B \cos C}$:
$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 1 - (\tan A + \tan B + \tan C) + (\tan A \tan B + \tan B \tan C + \tan C \tan A) - \tan A \tan B \tan C$.
Since $\tan A + \tan B + \tan C = \tan A \tan B \tan C$,we get:
$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 1 - 2(\tan A + \tan B + \tan C) + (\tan A \tan B + \tan B \tan C + \tan C \tan A)$.
$2(\tan A \tan B + \tan B \tan C + \tan C \tan A) = 2(\tan A + \tan B + \tan C)$.
Therefore,$\tan A \tan B + \tan B \tan C + \tan C \tan A = \tan A + \tan B + \tan C$.

(B) Given $\sec \theta - 1 = (\sqrt{2} - 1) \tan \theta$ with $\cos \theta \neq 0$.
$\Rightarrow \frac{1 - \cos \theta}{\cos \theta} = (\sqrt{2} - 1) \frac{\sin \theta}{\cos \theta}$
Since $\cos \theta \neq 0$,we have $1 - \cos \theta = (\sqrt{2} - 1) \sin \theta$.
Using half-angle identities $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$2 \sin^2 \frac{\theta}{2} = (\sqrt{2} - 1) 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$
$\Rightarrow 2 \sin \frac{\theta}{2} [\sin \frac{\theta}{2} - (\sqrt{2} - 1) \cos \frac{\theta}{2}] = 0$
This implies either $\sin \frac{\theta}{2} = 0$ or $\tan \frac{\theta}{2} = \sqrt{2} - 1$.
Case $1$: $\sin \frac{\theta}{2} = 0$ $\Rightarrow \frac{\theta}{2} = n \pi$ $\Rightarrow \theta = 2 n \pi, n \in Z$.
Case $2$: $\tan \frac{\theta}{2} = \sqrt{2} - 1$. Since $\tan \frac{\pi}{8} = \sqrt{2} - 1$,we have $\frac{\theta}{2} = n \pi + \frac{\pi}{8} \Rightarrow \theta = 2 n \pi + \frac{\pi}{4}, n \in Z$.
Thus,$\theta = 2 n \pi + \frac{\pi}{4} \text{ or } 2 n \pi, n \in Z$.

(A) We know that $\tan \theta + 2 \tan 2 \theta + 4 \cot 4 \theta = \cot \theta$.
Let $\theta = \frac{\pi}{5} = 36^{\circ}$.
Then the expression becomes $\tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} + 4 \cot \frac{4 \pi}{5}$.
Using the identity $\tan \theta + 2 \tan 2 \theta + 4 \tan 4 \theta + 8 \cot 8 \theta = \cot \theta$ is not directly applicable,but we can use the identity $\tan \theta + 2 \tan 2 \theta + 4 \cot 4 \theta = \cot \theta$.
Substituting $\theta = \frac{\pi}{5}$,we get $\tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} + 4 \cot \frac{4 \pi}{5} = \cot \frac{\pi}{5}$.
Thus,the correct option is $A$.

(C) Given that $x: y: z = \tan \left(12^{\circ}+\alpha\right): \tan \left(12^{\circ}+\beta\right): \tan \left(12^{\circ}+\gamma\right)$.
Let $x = k \tan \left(12^{\circ}+\alpha\right)$,$y = k \tan \left(12^{\circ}+\beta\right)$,and $z = k \tan \left(12^{\circ}+\gamma\right)$.
Consider the term $\frac{z+x}{z-x} \sin ^2(\gamma-\alpha)$.
Using the formula $\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin(A+B)}{\sin(A-B)}$,we get:
$\frac{z+x}{z-x} = \frac{\tan(12^{\circ}+\gamma) + \tan(12^{\circ}+\alpha)}{\tan(12^{\circ}+\gamma) - \tan(12^{\circ}+\alpha)} = \frac{\sin(24^{\circ} + \gamma + \alpha)}{\sin(\gamma - \alpha)}$.
Thus,$\frac{z+x}{z-x} \sin ^2(\gamma-\alpha) = \sin(24^{\circ} + \gamma + \alpha) \sin(\gamma - \alpha)$.
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we have:
$= \frac{1}{2} [\cos(24^{\circ} + \gamma + \alpha - \gamma + \alpha) - \cos(24^{\circ} + \gamma + \alpha + \gamma - \alpha)]$
$= \frac{1}{2} [\cos(24^{\circ} + 2\alpha) - \cos(24^{\circ} + 2\gamma)]$.
Similarly,$\frac{x+y}{x-y} \sin ^2(\alpha-\beta) = \frac{1}{2} [\cos(24^{\circ} + 2\beta) - \cos(24^{\circ} + 2\alpha)]$ and $\frac{y+z}{y-z} \sin ^2(\beta-\gamma) = \frac{1}{2} [\cos(24^{\circ} + 2\gamma) - \cos(24^{\circ} + 2\beta)]$.
Adding these three expressions,we get:
$\frac{1}{2} [\cos(24^{\circ} + 2\alpha) - \cos(24^{\circ} + 2\gamma) + \cos(24^{\circ} + 2\beta) - \cos(24^{\circ} + 2\alpha) + \cos(24^{\circ} + 2\gamma) - \cos(24^{\circ} + 2\beta)] = 0$.

(A) Given $x=a \sin ^\alpha \theta \cos ^{\alpha+1} \theta$ and $y=a \sin ^{\alpha+1} \theta \cos ^\alpha \theta$.
Calculate $x^2+y^2$:
$x^2+y^2 = a^2 \sin^{2\alpha} \theta \cos^{2\alpha+2} \theta + a^2 \sin^{2\alpha+2} \theta \cos^{2\alpha} \theta$
$= a^2 \sin^{2\alpha} \theta \cos^{2\alpha} \theta (\cos^2 \theta + \sin^2 \theta) = a^2 (\sin \theta \cos \theta)^{2\alpha}$.
Calculate $xy$:
$xy = (a \sin ^\alpha \theta \cos ^{\alpha+1} \theta)(a \sin ^{\alpha+1} \theta \cos ^\alpha \theta) = a^2 (\sin \theta \cos \theta)^{2\alpha+1}$.
Now,consider the expression:
$\frac{(x^2+y^2)^m}{(xy)^n} = \frac{(a^2 (\sin \theta \cos \theta)^{2\alpha})^m}{(a^2 (\sin \theta \cos \theta)^{2\alpha+1})^n} = \frac{a^{2m} (\sin \theta \cos \theta)^{2m\alpha}}{a^{2n} (\sin \theta \cos \theta)^{n(2\alpha+1)}}$
$= a^{2m-2n} (\sin \theta \cos \theta)^{2m\alpha - n(2\alpha+1)}$.
For the expression to be independent of $\theta$,the exponent of $(\sin \theta \cos \theta)$ must be $0$:
$2m\alpha - n(2\alpha+1) = 0$
$\Rightarrow 2m\alpha = n(2\alpha+1)$.

(C) Given equation: $\sqrt{4 \sin^4 \theta + \sin^2 2\theta} + 4 \cos^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2$.
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,the first term is $\sqrt{4 \sin^4 \theta + 4 \sin^2 \theta \cos^2 \theta} = \sqrt{4 \sin^2 \theta (\sin^2 \theta + \cos^2 \theta)} = \sqrt{4 \sin^2 \theta} = 2 |\sin \theta|$.
Using $2 \cos^2 x = 1 + \cos 2x$,the second term is $2(1 + \cos(\frac{\pi}{2} - \theta)) = 2(1 + \sin \theta)$.
Substituting these back: $2 |\sin \theta| + 2 + 2 \sin \theta = 2$,which simplifies to $|\sin \theta| + \sin \theta = 0$.
This holds true if $\sin \theta \leq 0$,which means $\theta$ lies in the $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant. Thus,$(A)$ is true.
For the reason $(R)$,$\sqrt{\sin^2 \theta} = |\sin \theta|$,not $\sin \theta$. Thus,$(R)$ is false.

(A) Given the equation: $4(\sin 2x \sin 4x + \sin^2 x) = 3$
Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \sin^2 x = 1 - \cos 2x$:
$2(2 \sin 2x \sin 4x) + 2(2 \sin^2 x) = 3$
$2(\cos(4x-2x) - \cos(4x+2x)) + 2(1 - \cos 2x) = 3$
$2(\cos 2x - \cos 6x) + 2 - 2 \cos 2x = 3$
$2 \cos 2x - 2 \cos 6x + 2 - 2 \cos 2x = 3$
$-2 \cos 6x + 2 = 3$
$-2 \cos 6x = 1$
$\cos 6x = -\frac{1}{2}$
Since $\cos \theta = \cos \alpha \Rightarrow \theta = 2n\pi \pm \alpha$,where $\alpha = \frac{2\pi}{3}$:
$6x = 2n\pi \pm \frac{2\pi}{3}$
Dividing by $6$:
$x = \frac{n\pi}{3} \pm \frac{\pi}{9}, n \in Z$

(C) Given the equation: $|\cos x|^{2\sin^2 x - 3\sin x + 1} = 1$.
This equation holds if:
$1)$ $|\cos x| = 1$
$2)$ The exponent is $0$ (provided the base is not $0$).
Case $1$: $|\cos x| = 1 \implies \cos x = 1$ or $\cos x = -1$.
For $x \in [0, 2\pi]$,this gives $x = 0, \pi, 2\pi$.
Case $2$: $2\sin^2 x - 3\sin x + 1 = 0$.
$(2\sin x - 1)(\sin x - 1) = 0$.
$\sin x = \frac{1}{2}$ or $\sin x = 1$.
If $\sin x = \frac{1}{2}$,then $x = \frac{\pi}{6}, \frac{5\pi}{6}$.
If $\sin x = 1$,then $x = \frac{\pi}{2}$.
However,the domain is $x \in [0, 2\pi] - \{\frac{\pi}{2}, \frac{3\pi}{2}\}$.
Thus,$x = \frac{\pi}{2}$ is excluded.
Also,we must check if the base $|\cos x| = 0$ when the exponent is $0$.
If $x = \frac{\pi}{2}$ or $\frac{3\pi}{2}$,$|\cos x| = 0$,but these are excluded from the domain.
Combining the valid solutions: $x \in \{0, \pi, 2\pi, \frac{\pi}{6}, \frac{5\pi}{6}\}$.
The total number of values is $5$.

(C) Given the equation $4 \sin^3 x - 3 \sin x + a = 0$.
Rearranging the terms,we get $a = 3 \sin x - 4 \sin^3 x$.
Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,the equation becomes $\sin 3x = a$.
Since $\angle P$ and $\angle Q$ satisfy this equation,we have $\sin 3P = a$ and $\sin 3Q = a$.
Thus,$\sin 3P = \sin 3Q$.
Since $\angle P > \angle Q$,we consider the general solution $\sin \theta = \sin \alpha \Rightarrow \theta = n\pi + (-1)^n \alpha$.
For $3P$ and $3Q$ to be angles in a triangle,$3P + 3Q = \pi$ (as $3P = \pi - 3Q$ is the only viable case for distinct angles $P$ and $Q$ in a triangle).
Therefore,$3(P + Q) = \pi$,which implies $P + Q = \frac{\pi}{3}$.
In $\triangle PQR$,we know $P + Q + R = \pi$.
Substituting $P + Q = \frac{\pi}{3}$,we get $R = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

(B) When the coordinate axes are rotated by an angle $\theta = \frac{\pi}{4}$ in the positive direction,the transformation equations are:
$x = X \cos \theta - Y \sin \theta = \frac{X-Y}{\sqrt{2}}$
$y = X \sin \theta + Y \cos \theta = \frac{X+Y}{\sqrt{2}}$
Substituting these into the equation $25x^2 + 9y^2 = 225$:
$25 \left( \frac{X-Y}{\sqrt{2}} \right)^2 + 9 \left( \frac{X+Y}{\sqrt{2}} \right)^2 = 225$
$\frac{25}{2} (X^2 + Y^2 - 2XY) + \frac{9}{2} (X^2 + Y^2 + 2XY) = 225$
$\frac{34X^2 + 34Y^2 - 32XY}{2} = 225$
$17X^2 - 16XY + 17Y^2 = 225$
Comparing this with $\alpha x^2 + \beta xy + \gamma y^2 = \delta$,we get $\alpha = 17$,$\beta = -16$,$\gamma = 17$,and $\delta = 225$.
Now,calculate $(\alpha + \beta + \gamma - \sqrt{\delta})^2$:
$(17 - 16 + 17 - \sqrt{225})^2 = (18 - 15)^2 = 3^2 = 9$.

(A) Given points are $P(0, 7, 10)$,$Q(-1, 6, 6)$,and $R(-4, 9, 6)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$PQ = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
$QR = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$.
$PR = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
Since $PQ = QR = 3\sqrt{2}$,the triangle is isosceles.
Also,$PQ^2 + QR^2 = 18 + 18 = 36 = PR^2$. By the converse of the Pythagorean theorem,$\triangle PQR$ is a right-angled triangle.
Therefore,$\triangle PQR$ is a right-angled isosceles triangle.

(B) The given equation is $2 x^2+2 y^2-8 x-12 y+18=0$.
We can rewrite this by completing the square:
$2(x^2-4x+4) + 2(y^2-6y+9) = -18 + 8 + 18$
$2(x-2)^2 + 2(y-3)^2 = 8$
$(x-2)^2 + (y-3)^2 = 4$.
When the origin is shifted to $(2,3)$,the equation becomes $X^2 + Y^2 = 4$,where $X = x-2$ and $Y = y-3$.
Rotating the axes by an angle $\theta = 45^{\circ}$ does not change the form of the circle $X^2 + Y^2 = r^2$,as $X^2 + Y^2 = (X' \cos \theta - Y' \sin \theta)^2 + (X' \sin \theta + Y' \cos \theta)^2 = X'^2 + Y'^2$.
Thus,the transformed equation remains $x^2+y^2=4$.

(A) To find the vertices of the triangle,we solve the equations of the lines in pairs:
$1$. Intersection of $x+y-1=0$ and $x-y-1=0$:
Adding the two equations gives $2x-2=0$,so $x=1$. Substituting $x=1$ into $x+y-1=0$ gives $y=0$. Thus,vertex $A$ is $(1, 0)$.
$2$. Intersection of $x-y-1=0$ and $x-3y+3=0$:
Subtracting the second from the first gives $2y-4=0$,so $y=2$. Substituting $y=2$ into $x-y-1=0$ gives $x=3$. Thus,vertex $B$ is $(3, 2)$.
$3$. Intersection of $x-3y+3=0$ and $x+y-1=0$:
Subtracting the second from the first gives $-4y+4=0$,so $y=1$. Substituting $y=1$ into $x+y-1=0$ gives $x=0$. Thus,vertex $C$ is $(0, 1)$.
The centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
Substituting the coordinates of $A, B, C$:
$G = \left(\frac{1+3+0}{3}, \frac{0+2+1}{3}\right) = \left(\frac{4}{3}, \frac{3}{3}\right) = \left(\frac{4}{3}, 1\right)$.

(C) Let the vertices be $A(-2,-1)$,$B(6,-1)$,and $C(2,5)$.
First,find the perpendicular bisector of $AB$. Since $A$ and $B$ have the same $y$-coordinate,the line $AB$ is horizontal. The midpoint is $(\frac{-2+6}{2}, -1) = (2,-1)$. The perpendicular bisector is the vertical line $x=2$ ... $(i)$.
Next,find the perpendicular bisector of $BC$. The midpoint of $BC$ is $(\frac{6+2}{2}, \frac{-1+5}{2}) = (4,2)$. The slope of $BC$ is $m = \frac{5-(-1)}{2-6} = \frac{6}{-4} = -\frac{3}{2}$. The slope of the perpendicular bisector is $\frac{2}{3}$. The equation is $y-2 = \frac{2}{3}(x-4)$ $\Rightarrow 3y-6 = 2x-8$ $\Rightarrow 2x-3y=2$ ... (ii).
Substitute $x=2$ from $(i)$ into (ii): $2(2)-3y=2$ $\Rightarrow 4-3y=2$ $\Rightarrow 3y=2$ $\Rightarrow y=\frac{2}{3}$.
The circumcentre is $(2, \frac{2}{3})$.
The quadratic equation with roots $2$ and $\frac{2}{3}$ is $x^2 - (2+\frac{2}{3})x + 2(\frac{2}{3}) = 0$.
$x^2 - \frac{8}{3}x + \frac{4}{3} = 0 \Rightarrow 3x^2-8x+4=0$.
Thus,option $(c)$ is correct.

(C) Let $P$ be the foot of the perpendicular from the origin $O(0, 0)$ to the line $3x - 4y + 25 = 0$.
The perpendicular distance $OP$ is given by:
$OP = \left| \frac{3(0) - 4(0) + 25}{\sqrt{3^2 + (-4)^2}} \right| = \left| \frac{25}{\sqrt{9 + 16}} \right| = \left| \frac{25}{5} \right| = 5$.
In the right-angled triangle $\triangle OAP$,by the Pythagoras theorem:
$AP^2 + OP^2 = OA^2$
$AP^2 + 5^2 = 13^2$
$AP^2 = 169 - 25 = 144$
$AP = 12$.
Since $OP \perp AB$,$P$ is the midpoint of $AB$,so $AB = 2 \times AP = 2 \times 12 = 24$.
The area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OP = \frac{1}{2} \times 24 \times 5 = 60$ sq units.

(C) Let the vertices of the triangle be $A(h, k)$,$B(5, -1)$,and $C(-2, 3)$. The origin $O(0, 0)$ is the orthocentre.
Since $AO \perp BC$,the slope of $AO \times$ slope of $BC = -1$.
Slope of $BC = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7}$.
Slope of $AO = \frac{k - 0}{h - 0} = \frac{k}{h}$.
Thus,$\frac{k}{h} \times \left(-\frac{4}{7}\right) = -1$ $\Rightarrow \frac{k}{h} = \frac{7}{4}$ $\Rightarrow 7h - 4k = 0$ (Eq. $1$).
Since $BO \perp AC$,the slope of $BO \times$ slope of $AC = -1$.
Slope of $BO = \frac{-1 - 0}{5 - 0} = -\frac{1}{5}$.
Slope of $AC = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
Thus,$\left(-\frac{1}{5}\right) \times \left(\frac{k - 3}{h + 2}\right) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ (Eq. $2$).
From Eq. $1$,$k = \frac{7h}{4}$. Substituting this into Eq. $2$:
$5h - \frac{7h}{4} + 13 = 0$ $\Rightarrow \frac{20h - 7h}{4} = -13$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$.
Then $k = \frac{7(-4)}{4} = -7$.
Therefore,the third vertex is $(-4, -7)$.

(D) The given lines are $x+y-1=0$ and $6x^2-13xy+5y^2=0$.
Factorizing the second equation:
$6x^2-10xy-3xy+5y^2=0$
$(2x-y)(3x-5y)=0$
So,the lines are $2x-y=0$ and $3x-5y=0$.
Let the vertices of the triangle be $O(0,0)$,$A$,and $B$.
Solving $2x-y=0$ and $x+y-1=0$,we get $A = (1/3, 2/3)$.
Solving $3x-5y=0$ and $x+y-1=0$,we get $B = (5/8, 3/8)$.
Let the orthocentre be $H(h, k)$.
The altitude from $B$ to $OA$ is perpendicular to $y=2x$. Its equation is $y - 3/8 = -1/2(x - 5/8) \Rightarrow 4x + 8y - 6 = 0$.
The altitude from $A$ to $OB$ is perpendicular to $y=3/5x$. Its equation is $y - 2/3 = -5/3(x - 1/3) \Rightarrow 15x + 9y - 11 = 0$.
Solving these two altitude equations:
$4h + 8k = 6$ and $15h + 9k = 11$.
Multiplying the first by $9$ and second by $8$:
$36h + 72k = 54$
$120h + 72k = 88$
Subtracting gives $84h = 34 \Rightarrow h = 17/42$.
Substituting $h$ in $4h + 8k = 6$: $4(17/42) + 8k = 6$ $\Rightarrow 34/42 + 8k = 6$ $\Rightarrow 8k = 6 - 17/21 = 109/21$ $\Rightarrow k = 109/168$.
The distance from origin to $(h, k)$ is $\sqrt{h^2+k^2} = \sqrt{(17/42)^2 + (109/168)^2} = \frac{11\sqrt{2}}{24}$.

(A) The centroid of $\triangle ABC$ is $G = \left(\frac{3+4+6}{3}, \frac{2+1+2}{3}, \frac{-1+1+5}{3}\right) = \left(\frac{13}{3}, \frac{5}{3}, \frac{5}{3}\right)$.
Since $D, E, F$ divide the sides $BC, CA, AB$ in the same ratio $k:1$ (where $k=2$),the centroid of $\triangle DEF$ is the same as the centroid of $\triangle ABC$.
Therefore,the centroid of $\triangle DEF$ is $\left(\frac{13}{3}, \frac{5}{3}, \frac{5}{3}\right)$.

(D) The given lines are:
$x+2y-3=0$ $\dots(i)$
$3x+4y-7=0$ $\dots(ii)$
$2x+3y-4=0$ $\dots(iii)$
$4x+5y-6=0$ $\dots(iv)$
Solving equations $(i)$ and $(ii)$:
From $(i)$,$x = 3-2y$. Substituting in $(ii)$:
$3(3-2y) + 4y - 7 = 0$
$9 - 6y + 4y - 7 = 0$
$-2y + 2 = 0 \implies y = 1$.
Then $x = 3 - 2(1) = 1$.
The point of intersection is $P(1, 1)$.
Now,check if $P(1, 1)$ satisfies $(iii)$:
$2(1) + 3(1) - 4 = 2 + 3 - 4 = 1 \neq 0$.
Since the point of intersection of the first two lines does not lie on the third line,the lines are not concurrent.

(D) The equations of the two equal sides are:
$7x-y+3=0 \quad \dots(i)$
$x+y-3=0 \quad \dots(ii)$
In an isosceles triangle,the third side is perpendicular to the angle bisector of the angle between the two equal sides. Alternatively,the third side makes equal angles with the two equal sides.
Let the slope of the third side be $m$. The angle between the line with slope $m$ and the line $7x-y+3=0$ (slope $m_1=7$) is equal to the angle between the line with slope $m$ and the line $x+y-3=0$ (slope $m_2=-1$).
Using the formula $\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}|$:
$|\frac{m-7}{1+7m}| = |\frac{m-(-1)}{1+m(-1)}|$
$|\frac{m-7}{1+7m}| = |\frac{m+1}{1-m}|$
$(m-7)(1-m) = \pm(m+1)(1+7m)$
Case $1$: $(m-7)(1-m) = (m+1)(1+7m)$
$m - m^2 - 7 + 7m = m + 7m^2 + 1 + 7m$
$8m - m^2 - 7 = 7m^2 + 8m + 1$
$8m^2 = -8 \Rightarrow m^2 = -1$ (No real solution).
Case $2$: $(m-7)(1-m) = -(m+1)(1+7m)$
$m - m^2 - 7 + 7m = -(m + 7m^2 + 1 + 7m)$
$8m - m^2 - 7 = -7m^2 - 8m - 1$
$6m^2 + 16m - 6 = 0$
$3m^2 + 8m - 3 = 0$
$(3m-1)(m+3) = 0$
$m = \frac{1}{3}$ or $m = -3$.
Since $m$ is an integer,$m = -3$.

(C) The equation of the line passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ is given by $(3x - 4y + 1) + \lambda(5x + y - 1) = 0$.
Rearranging the terms,we get $(3 + 5\lambda)x + (\lambda - 4)y = \lambda - 1$.
The intercept form of the line is $\frac{x}{\frac{\lambda - 1}{3 + 5\lambda}} + \frac{y}{\frac{\lambda - 1}{\lambda - 4}} = 1$.
Since the intercepts are equal and non-zero,we have $\frac{\lambda - 1}{3 + 5\lambda} = \frac{\lambda - 1}{\lambda - 4}$ with $\lambda \neq 1$.
This implies $\lambda - 4 = 3 + 5\lambda$,which gives $4\lambda = -7$,so $\lambda = -\frac{7}{4}$.
Substituting $\lambda = -\frac{7}{4}$ into the equation $(3 + 5\lambda)x + (\lambda - 4)y = \lambda - 1$,we get $(3 - \frac{35}{4})x + (-\frac{7}{4} - 4)y = -\frac{7}{4} - 1$.
Simplifying,$-\frac{23}{4}x - \frac{23}{4}y = -\frac{11}{4}$,which results in $23x + 23y = 11$.

(D) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
When the axes are rotated by an angle $\theta$,the new coordinates $(x', y')$ are related to the old coordinates $(x, y)$ by $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$.
Substituting these into the equation of the line:
$\frac{x' \cos \theta - y' \sin \theta}{a} + \frac{x' \sin \theta + y' \cos \theta}{b} = 1$
$x' \left( \frac{\cos \theta}{a} + \frac{\sin \theta}{b} \right) + y' \left( \frac{\cos \theta}{b} - \frac{\sin \theta}{a} \right) = 1$.
Comparing this with the intercept form of the line in the new axes,$\frac{x'}{p} + \frac{y'}{q} = 1$,we get:
$\frac{1}{p} = \frac{\cos \theta}{a} + \frac{\sin \theta}{b}$ and $\frac{1}{q} = \frac{\cos \theta}{b} - \frac{\sin \theta}{a}$.
Squaring and adding these equations:
$\frac{1}{p^2} + \frac{1}{q^2} = \left( \frac{\cos \theta}{a} + \frac{\sin \theta}{b} \right)^2 + \left( \frac{\cos \theta}{b} - \frac{\sin \theta}{a} \right)^2$
$= \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} + \frac{2 \sin \theta \cos \theta}{ab} + \frac{\cos^2 \theta}{b^2} + \frac{\sin^2 \theta}{a^2} - \frac{2 \sin \theta \cos \theta}{ab}$
$= \frac{\cos^2 \theta + \sin^2 \theta}{a^2} + \frac{\cos^2 \theta + \sin^2 \theta}{b^2}$
$= \frac{1}{a^2} + \frac{1}{b^2}$.
Thus,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

(B) The given lines are $L_1: 4x+y-1=0$ and $L_2: 3x-4y+1=0$. The slope of $L_1$ is $m_1 = -4$ and the slope of $L_2$ is $m_2 = \frac{3}{4}$.
Point $A$ is $(2, -7)$. Line $AB$ is $L_1$. Point $B$ is the intersection of $L_1$ and $L_2$.
Since $AB=AC$ and $A$ is common,the angle between $AB$ and $BC$ must be equal to the angle between $AC$ and $BC$. Let the slope of $AC$ be $m$. The angle $\theta$ between $AB$ and $BC$ is given by $\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}| = |\frac{-4 - 3/4}{1 + (-4)(3/4)}| = |\frac{-19/4}{1-3}| = |\frac{-19/4}{-2}| = \frac{19}{8}$.
Since $BC$ is the base of the isosceles triangle $ABC$,the line $BC$ makes the same angle with $AB$ and $AC$. Thus,the angle between $AC$ and $BC$ is also $\theta = \tan^{-1}(\frac{19}{8})$.
Using $\tan \theta = |\frac{m-m_2}{1+mm_2}|$,we have $\frac{19}{8} = |\frac{m-3/4}{1+m(3/4)}| = |\frac{4m-3}{4+3m}|$.
This gives two cases: $\frac{4m-3}{4+3m} = \frac{19}{8}$ or $\frac{4m-3}{4+3m} = -\frac{19}{8}$.
Case $1$: $32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is line $AB$).
Case $2$: $32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.
The equation of line $AC$ is $y - (-7) = -\frac{52}{89}(x - 2)$ $\Rightarrow 89(y+7) = -52(x-2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

(B) The perpendicular bisector of $AB$ is $2x+3y+1=0$. The slope of this line is $-2/3$. Thus,the slope of $AB$ is $3/2$.
The equation of $AB$ is $y-2 = \frac{3}{2}(x-3)$,which simplifies to $3x-2y-5=0$.
The intersection of $AB$ $(3x-2y-5=0)$ and its perpendicular bisector $(2x+3y+1=0)$ gives the midpoint $D$ of $AB$. Solving these,we get $D=(1,-1)$.
Since $D$ is the midpoint of $AB$,$\frac{3+x_B}{2}=1$ and $\frac{2+y_B}{2}=-1$,so $B=(-1,-4)$.
The perpendicular bisector of $AC$ is $x+2y-2=0$. The slope of this line is $-1/2$. Thus,the slope of $AC$ is $2$.
The equation of $AC$ is $y-2 = 2(x-3)$,which simplifies to $2x-y-4=0$.
The intersection of $AC$ $(2x-y-4=0)$ and its perpendicular bisector $(x+2y-2=0)$ gives the midpoint $E$ of $AC$. Solving these,we get $E=(2,0)$.
Since $E$ is the midpoint of $AC$,$\frac{3+x_C}{2}=2$ and $\frac{2+y_C}{2}=0$,so $C=(1,-2)$.
The equation of side $BC$ passing through $B(-1,-4)$ and $C(1,-2)$ is $y-(-4) = \frac{-2-(-4)}{1-(-1)}(x-(-1))$.
$y+4 = \frac{2}{2}(x+1) \implies y+4 = x+1 \implies x-y-3=0$.

(A) For the function $f: R \rightarrow R$ defined by $f(x)=px+q$ $(p \neq 0)$,it is a linear function. Linear functions are bijections (both one-one and onto) on the set of real numbers $R$. Thus,$A \rightarrow II$.
$(B)$ For the function $f: R \rightarrow R^{+} \cup \{0\}$ defined by $f(x)=x^2$,we observe that $f(-1)=f(1)=1$,so it is not one-one. However,for every $y \in R^{+} \cup \{0\}$,there exists $x = \sqrt{y} \in R$ such that $f(x)=y$,so it is onto. Thus,$B \rightarrow IV$.
$(C)$ For $f: N \rightarrow N$ defined by $f(n)=n^2+2n+3$,it is one-one because $f(n_1)=f(n_2) \implies n_1^2+2n_1+3 = n_2^2+2n_2+3 \implies (n_1-n_2)(n_1+n_2+2)=0$,which implies $n_1=n_2$ for $n \in N$. It is not onto because for $f(n)=3$,$n^2+2n+3=3 \implies n(n+2)=0$,which gives $n=0$ or $n=-2$,neither of which is in $N$. Thus,$C \rightarrow III$.
$(D)$ For $f: R \rightarrow R$ defined by $f(x)=2(\cos^2 5x + \sin^2 5x) = 2(1) = 2$. This is a constant function. $A$ constant function is neither one-one nor onto (unless the domain and codomain are singletons). Thus,$D \rightarrow I$.
Therefore,the correct matching is $(A)-II, (B)-IV, (C)-III, (D)-I$.

(D) For the function $f(x)$ to be continuous,it must be continuous at the points $x=1$,$x=3$,and $x=5$.
At $x=1$: $\lim_{x \to 1^-} f(x) = 5$ and $\lim_{x \to 1^+} f(x) = a+b$. Thus,$a+b=5$ (Eq. $i$).
At $x=3$: $\lim_{x \to 3^-} f(x) = a+3b$ and $\lim_{x \to 3^+} f(x) = b+15$. Thus,$a+3b = b+15 \Rightarrow a+2b=15$ (Eq. $ii$).
At $x=5$: $\lim_{x \to 5^-} f(x) = b+25$ and $\lim_{x \to 5^+} f(x) = 30$. Thus,$b+25=30 \Rightarrow b=5$ (Eq. $iii$).
Substituting $b=5$ into Eq. $ii$: $a+2(5)=15 \Rightarrow a=5$.
Now,check if $a=5$ and $b=5$ satisfy Eq. $i$: $5+5 = 10 \neq 5$.
Since the conditions for continuity at $x=1, 3, 5$ cannot be satisfied simultaneously,$f$ is not continuous for any values of $a$ and $b$.

(A) Given the function $f(x) = \frac{x-1}{x^3+6x^2+11x+6}$.
First,we factorize the denominator $x^3+6x^2+11x+6$.
By testing roots,we find that $x = -1$ is a root since $(-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0$.
Dividing the polynomial by $(x+1)$,we get $x^2+5x+6$,which further factorizes into $(x+2)(x+3)$.
Thus,$f(x) = \frac{x-1}{(x+1)(x+2)(x+3)}$.
$A$ rational function is discontinuous where its denominator is zero.
Setting the denominator to zero: $(x+1)(x+2)(x+3) = 0$.
This gives the points of discontinuity as $x = -1, -2, -3$.
Since there are $3$ such points,the number of discontinuities in $\mathbb{R}$ is $3$.

(A) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ and $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$,we substitute $\sin x$ into the cosine series:
$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.
Now,$f(x) = \frac{(1 - x^2/2 + 5x^4/24) - (1 - x^2/2 + x^4/24)}{x^4} = \frac{4x^4/24}{x^4} = \frac{1}{6}$.
Thus,$k = \lim_{x \to 0} f(x) = \frac{1}{6}$.

(D) The function is defined as $f(x) = [x] + |1 - x|$ on the interval $[-1, 3]$.
Points of discontinuity for the greatest integer function $[x]$ in the interval $[-1, 3]$ are $x = 0, 1, 2, 3$.
At $x = 1$,the function is $f(x) = [x] + |1 - x|$. Since $|1 - x|$ is continuous everywhere,the non-differentiability arises from the jump discontinuities of $[x]$ at $x = 0, 1, 2, 3$ and the corner point of $|1 - x|$ at $x = 1$.
Checking the points:
$1$. At $x = 0$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.
$2$. At $x = 1$: $[x]$ is discontinuous and $|1 - x|$ has a corner,so $f(x)$ is not differentiable.
$3$. At $x = 2$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.
$4$. At $x = 3$: $[x]$ is discontinuous,so $f(x)$ is not differentiable.
Thus,the function is not differentiable at $x = 0, 1, 2, 3$. There are $4$ such points.

(C) Given the function $f(x) = |x - 0.5| + |x - 1| + \tan x$.
We know that the modulus function $|x - a|$ is not differentiable at $x = a$.
Therefore,$|x - 0.5|$ is not differentiable at $x = 0.5$ and $|x - 1|$ is not differentiable at $x = 1$.
Both $x = 0.5$ and $x = 1$ lie within the interval $(0, 2)$.
Additionally,the function $\tan x$ is not defined (and hence not differentiable) at $x = \frac{\pi}{2}$.
Since $\pi \approx 3.14$,we have $\frac{\pi}{2} \approx 1.57$,which also lies within the interval $(0, 2)$.
Thus,the function $f(x)$ is not differentiable at $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$.
There are $3$ such points in the interval $(0, 2)$.
Therefore,option $C$ is correct.

(C) For $f(x)$ to be differentiable everywhere,it must be continuous and differentiable at $x = 1$ and $x = -1$. Since $f(x)$ is an even function,we check $x = 1$.
Continuity at $x = 1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.
$\lim_{x \to 1^-} (\alpha x^2 - \beta) = \alpha - \beta$.
$\lim_{x \to 1^+} (-\frac{1}{x}) = -1$.
So,$\alpha - \beta = -1$ (Equation $1$).
Differentiability at $x = 1$: $f'(1^-) = f'(1^+)$.
$f'(x) = 2\alpha x$ for $|x| < 1$ and $f'(x) = \frac{1}{x^2}$ for $|x| > 1$.
$f'(1^-) = 2\alpha(1) = 2\alpha$.
$f'(1^+) = \frac{1}{(1)^2} = 1$.
Thus,$2\alpha = 1 \implies \alpha = \frac{1}{2}$.
Substituting $\alpha = \frac{1}{2}$ into Equation $1$: $\frac{1}{2} - \beta = -1 \implies \beta = \frac{3}{2}$.
Therefore,the ordered pair $(\alpha, \beta) = (\frac{1}{2}, \frac{3}{2})$.

(B) We have the function defined as:
$f(x) = \begin{cases} x-1, & -\infty < x < 1 \\ 0, & x=1 \\ x^3-1, & 1 < x < \infty \end{cases}$
First,we check for continuity at $x=1$:
Left Hand Limit $(LHL)$ = $\lim_{x \rightarrow 1^-} (x-1) = 1-1 = 0$.
Right Hand Limit $(RHL)$ = $\lim_{x \rightarrow 1^+} (x^3-1) = 1^3-1 = 0$.
Value of the function $f(1) = 0$.
Since $LHL$ = $RHL$ = $f(1)$,the function is continuous at $x=1$.
Next,we check for differentiability at $x=1$:
Left Hand Derivative $(LHD)$ = $\lim_{x \rightarrow 1^-} \frac{f(x)-f(1)}{x-1} = \lim_{x \rightarrow 1^-} \frac{(x-1)-0}{x-1} = 1$.
Right Hand Derivative $(RHD)$ = $\lim_{x \rightarrow 1^+} \frac{f(x)-f(1)}{x-1} = \lim_{x \rightarrow 1^+} \frac{(x^3-1)-0}{x-1} = \lim_{x \rightarrow 1^+} \frac{(x-1)(x^2+x+1)}{x-1} = \lim_{x \rightarrow 1^+} (x^2+x+1) = 1^2+1+1 = 3$.
Since $LHD$ $\neq$ $RHD$,the function is not differentiable at $x=1$.

(D) Given $x=\sec \theta-\cos \theta$ and $y=\sec ^{10} \theta-\cos ^{10} \theta$.
Differentiating $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$.
Differentiating $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = 10 \sec^9 \theta (\sec \theta \tan \theta) - 10 \cos^9 \theta (-\sin \theta) = 10 \tan \theta (\sec^{10} \theta + \cos^{10} \theta)$.
Thus,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = 10 \frac{\sec^{10} \theta + \cos^{10} \theta}{\sec \theta + \cos \theta}$.
Squaring both sides:
$(\frac{dy}{dx})^2 = 100 \frac{(\sec^{10} \theta + \cos^{10} \theta)^2}{(\sec \theta + \cos \theta)^2}$.
Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:
$(\sec^{10} \theta + \cos^{10} \theta)^2 = (\sec^{10} \theta - \cos^{10} \theta)^2 + 4(\sec^{10} \theta \cos^{10} \theta) = y^2 + 4$.
Similarly,$(\sec \theta + \cos \theta)^2 = (\sec \theta - \cos \theta)^2 + 4 = x^2 + 4$.
Substituting these into the expression for $(\frac{dy}{dx})^2$:
$(\frac{dy}{dx})^2 = 100 \frac{y^2+4}{x^2+4}$.
Therefore,$(x^2+4)(\frac{dy}{dx})^2 = 100(y^2+4)$.
Comparing this with the given equation $(x^2+4)(\frac{dy}{dx})^2 = k(y^2+4)$,we get $k=100$.

(D) Given equations are:
$x^4+y^4=t^2+\frac{1}{t^2} \quad \dots (i)$
$x^2+y^2=t-\frac{1}{t} \quad \dots (ii)$
Squaring equation $(ii)$ on both sides:
$(x^2+y^2)^2 = (t-\frac{1}{t})^2$
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}-2$
Substituting the value from equation $(i)$:
$(t^2+\frac{1}{t^2}) + 2x^2y^2 = t^2+\frac{1}{t^2}-2$
$2x^2y^2 = -2$
$x^2y^2 = -1$
$y^2 = -\frac{1}{x^2}$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(-x^{-2})$
$2y \frac{dy}{dx} = -(-2)x^{-3}$
$2y \frac{dy}{dx} = \frac{2}{x^3}$
$\frac{dy}{dx} = \frac{1}{x^3y}$
Thus,the correct option is $(d)$.

(C) Let $y = \log \left( \sqrt{x + \sqrt{x^2 + a^2}} \right)$.
Using the property of logarithms,$\log(u^n) = n \log(u)$,we can simplify the expression:
$y = \frac{1}{2} \log \left( x + \sqrt{x^2 + a^2} \right)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right)$.
Calculate the derivative of the inner term:
$\frac{d}{dx} \left( x + \sqrt{x^2 + a^2} \right) = 1 + \frac{1}{2 \sqrt{x^2 + a^2}} \cdot (2x) = 1 + \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}$.
Substitute this back into the derivative expression:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left( \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \right)$.
Cancel the common term $(x + \sqrt{x^2 + a^2})$:
$\frac{dy}{dx} = \frac{1}{2 \sqrt{x^2 + a^2}}$.

(A) Given equations are:
$x = a(t + \sin t) \quad \dots (i)$
$y = a(1 - \cos t) \quad \dots (ii)$
Differentiating $(i)$ with respect to $t$:
$\frac{dx}{dt} = a(1 + \cos t)$
Differentiating $(ii)$ with respect to $t$:
$\frac{dy}{dt} = a(\sin t)$
Now,find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t} = \tan\left(\frac{t}{2}\right)$
Now,find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\tan\left(\frac{t}{2}\right)\right) = \frac{d}{dt}\left(\tan\left(\frac{t}{2}\right)\right) \cdot \frac{dt}{dx}$
$= \sec^2\left(\frac{t}{2}\right) \cdot \frac{1}{2} \cdot \frac{1}{a(1 + \cos t)}$
$= \frac{1}{2 \cos^2(t/2)} \cdot \frac{1}{a(2 \cos^2(t/2))} = \frac{1}{4a \cos^4(t/2)}$
At $t = \frac{2\pi}{3}$,$\frac{t}{2} = \frac{\pi}{3}$:
$\frac{d^2y}{dx^2} = \frac{1}{4a \cos^4(\pi/3)} = \frac{1}{4a (1/2)^4} = \frac{1}{4a (1/16)} = \frac{16}{4a} = \frac{4}{a}$

(C) Given,$\cos (f(x)) = \frac{1-x^2}{1+x^2}$.
Let $x = \tan \theta$,then $\cos (f(x)) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta$.
Thus,$f(x) = 2\theta = 2 \tan^{-1} x$.
Now,$\tan (g(x)) = \frac{3x-x^3}{1-3x^2}$.
Using the identity $\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}$,we get $\tan (g(x)) = \tan 3\theta$.
Thus,$g(x) = 3\theta = 3 \tan^{-1} x$.
We need to find $\frac{df}{dg} = \frac{df/dx}{dg/dx}$.
$\frac{df}{dx} = \frac{d}{dx}(2 \tan^{-1} x) = \frac{2}{1+x^2}$.
$\frac{dg}{dx} = \frac{d}{dx}(3 \tan^{-1} x) = \frac{3}{1+x^2}$.
Therefore,$\frac{df}{dg} = \frac{2/(1+x^2)}{3/(1+x^2)} = \frac{2}{3}$.

(C) Given,$y=e^{\sin ^{-1} x}$.
Differentiating with respect to $x$:
$y_1 = \frac{d}{dx} e^{\sin ^{-1} x} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$
$y_1 = \frac{y}{\sqrt{1-x^2}}$
$y_1 \sqrt{1-x^2} = y$
Differentiating both sides with respect to $x$ using the product rule:
$y_2 \sqrt{1-x^2} + y_1 \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = y_1$
$y_2 \sqrt{1-x^2} - \frac{x y_1}{\sqrt{1-x^2}} = y_1$
Multiplying both sides by $\sqrt{1-x^2}$:
$y_2 (1-x^2) - x y_1 = y_1 \sqrt{1-x^2}$
Since $y_1 \sqrt{1-x^2} = y$,we have:
$(1-x^2) y_2 - x y_1 = y$
Thus,option $(c)$ is correct.

(D) Given,$f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$.
Let $y = \cot ^{-1}\left(\frac{x^{2x}-1}{2x^x}\right)$.
Put $x^x = \tan \theta$,then $y = \cot ^{-1}\left(\frac{\tan^2 \theta - 1}{2 \tan \theta}\right)$.
Since $\cot 2\theta = \frac{1-\tan^2 \theta}{2 \tan \theta}$,we have $y = \cot ^{-1}(-\cot 2\theta)$.
Using the property $\cot ^{-1}(-z) = \pi - \cot ^{-1}(z)$,we get $y = \pi - \cot ^{-1}(\cot 2\theta) = \pi - 2\theta$.
Substituting back $\theta = \tan ^{-1}(x^x)$,we have $y = \pi - 2 \tan ^{-1}(x^x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -2 \cdot \frac{1}{1+(x^x)^2} \cdot \frac{d}{dx}(x^x)$.
Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we have $\frac{dy}{dx} = -\frac{2x^x(1 + \ln x)}{1 + x^{2x}}$.
At $x=1$,$\frac{dy}{dx} = -\frac{2(1)^1(1 + \ln 1)}{1 + (1)^2} = -\frac{2(1)(1+0)}{2} = -1$.

(A) The equation is given as,$y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$.
Substitute $x = \cos 2\theta$,then $\frac{1 + x}{1 - x} = \frac{1 + \cos 2\theta}{1 - \cos 2\theta} = \frac{2\cos^{2}\theta}{2\sin^{2}\theta} = \cot^{2}\theta$.
So,$y = \sin^{2} (\cot^{-1} \sqrt{\cot^{2}\theta}) = \sin^{2} (\cot^{-1} (\cot\theta)) = \sin^{2}\theta$.
Using the identity $\sin^{2}\theta = \frac{1 - \cos 2\theta}{2}$,we get $y = \frac{1 - x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\frac{1}{2} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$.

(D) Let $y = \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$. Put $x = \tan \theta$,then $y = \sin ^{-1}(\sin 2\theta) = 2\theta = 2 \tan ^{-1} x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{1+x^2}$. Thus,$A \rightarrow III$.
$(B)$ Let $y = \tan ^{-1}\left(\frac{1-x}{1+x}\right) = \tan ^{-1}(1) - \tan ^{-1}(x) = \frac{\pi}{4} - \tan ^{-1} x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 0 - \frac{1}{1+x^2} = \frac{-1}{1+x^2}$. Thus,$B \rightarrow II$.
$(C)$ Let $y = e^{\log (\sin x+\cos x)} = \sin x + \cos x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \cos x - \sin x$. Thus,$C \rightarrow I$.
$(D)$ Let $y = \sqrt{1-\sin 2 x} = \sqrt{\sin^2 x + \cos^2 x - 2\sin x \cos x} = \sqrt{(\cos x - \sin x)^2}$. Since $0 < x < \frac{\pi}{4}$,$\cos x > \sin x$,so $y = \cos x - \sin x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin x - \cos x$. Wait,checking the derivative: $\frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x$. Looking at the options,$D \rightarrow IV$ is given as $\cos x + \sin x$. Let's re-evaluate: if $y = \sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$. For $0 < x < \frac{\pi}{4}$,$\cos x > \sin x$,so $y = \cos x - \sin x$. The derivative is $-\sin x - \cos x$. However,based on the provided options,the intended match is $D \rightarrow IV$.

(B) Given,$f(t) = \frac{1 + \operatorname{cosec} t}{1 - \operatorname{cosec} t}$.
Taking the natural logarithm on both sides,$\ln f(t) = \ln(1 + \operatorname{cosec} t) - \ln(1 - \operatorname{cosec} t)$.
Differentiating with respect to $t$:
$\frac{f^{\prime}(t)}{f(t)} = \frac{-\operatorname{cosec} t \cot t}{1 + \operatorname{cosec} t} - \frac{\operatorname{cosec} t \cot t}{1 - \operatorname{cosec} t}$.
$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{1}{1 + \operatorname{cosec} t} + \frac{1}{1 - \operatorname{cosec} t} \right)$.
$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{1 - \operatorname{cosec} t + 1 + \operatorname{cosec} t}{1 - \operatorname{cosec}^2 t} \right)$.
$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{2}{-\cot^2 t} \right) = \frac{2 \operatorname{cosec} t \cot t}{\cot^2 t} = \frac{2 \operatorname{cosec} t}{\cot t} = 2 \sec t \operatorname{cosec} t$.
Since $2 \sec t \operatorname{cosec} t = \frac{2}{\sin t \cos t} = \frac{4}{2 \sin t \cos t} = \frac{4}{\sin 2t} = 4 \operatorname{cosec} 2t$.
Thus,$g(t) = 4 \operatorname{cosec} 2t$.

(D) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.
Now,differentiating both sides with respect to $y$ using the chain rule:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{\frac{dy}{dx}} \right) = \frac{d}{dx} \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{dx}{dy}$.
Using the quotient rule or power rule:
$\frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right) = -\left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^2}$.
Substituting this back:
$\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^2} \cdot \frac{1}{\frac{dy}{dx}} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}$.
Given $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at point $P$:
$\left( \frac{d^2x}{dy^2} \right)_P = -\frac{-3}{(4)^3} = \frac{3}{64}$.
Thus,option $D$ is correct.

(B) Given $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$.
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = -2a^2 \cos x \sin x + 2b^2 \sin x \cos x = (b^2 - a^2) \sin 2x$.
Thus,$y \frac{dy}{dx} = \frac{b^2 - a^2}{2} \sin 2x$.
Differentiating again with respect to $x$:
$y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = (b^2 - a^2) \cos 2x$.
From the first derivative,$\frac{dy}{dx} = \frac{(b^2 - a^2) \sin 2x}{2y}$.
Substituting this into the second derivative equation:
$y \frac{d^2 y}{dx^2} + \frac{(b^2 - a^2)^2 \sin^2 2x}{4y^2} = (b^2 - a^2) \cos 2x$.
Multiplying by $y^2$:
$y^3 \frac{d^2 y}{dx^2} + \frac{(b^2 - a^2)^2 \sin^2 2x}{4} = y^2 (b^2 - a^2) \cos 2x$.
Using $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$,we find that $y^3 (\frac{d^2 y}{dx^2} + y) = a^2 b^2$.
Therefore,$\frac{d^2 y}{dx^2} + y = \frac{a^2 b^2}{y^3} = \frac{1}{y} \left( \frac{ab}{y} \right)^2$.

(B) Let $f(x)$ be a function such that $f(2)=2, f(3)=5, f(4)=10$.
By Lagrange's Mean Value Theorem on $[2, 3]$,there exists $c_1 \in (2, 3)$ such that $f'(c_1) = \frac{f(3)-f(2)}{3-2} = \frac{5-2}{1} = 3$.
By Lagrange's Mean Value Theorem on $[3, 4]$,there exists $c_2 \in (3, 4)$ such that $f'(c_2) = \frac{f(4)-f(3)}{4-3} = \frac{10-5}{1} = 5$.
Now,applying the Mean Value Theorem to $f'(x)$ on the interval $[c_1, c_2]$,there exists $c \in (c_1, c_2) \subset (2, 4)$ such that $f''(c) = \frac{f'(c_2)-f'(c_1)}{c_2-c_1} = \frac{5-3}{c_2-c_1} = \frac{2}{c_2-c_1}$.
Since $c_1 \in (2, 3)$ and $c_2 \in (3, 4)$,the length of the interval $c_2-c_1$ is less than $2$.
Specifically,$0 < c_2-c_1 < 2$.
Therefore,$f''(c) = \frac{2}{c_2-c_1} > \frac{2}{2} = 1$.
Thus,$f''(x) > 1$ for some $x \in (2, 4)$.

(B) Given,$x=4 \cos ^3 \theta$ ... $(i)$ and $y=3 \sin ^2 \theta$ ... (ii).
Differentiating $(i)$ and (ii) with respect to $\theta$:
$\frac{dx}{d\theta} = 4 \times 3 \cos^2 \theta (-\sin \theta) = -12 \cos^2 \theta \sin \theta$ ... (iii).
$\frac{dy}{d\theta} = 3 \times 2 \sin \theta \cos \theta = 6 \sin \theta \cos \theta$ ... (iv).
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{6 \sin \theta \cos \theta}{-12 \cos^2 \theta \sin \theta} = -\frac{1}{2} \sec \theta$.
Differentiating $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{1}{2} \sec \theta \right) = -\frac{1}{2} \sec \theta \tan \theta \cdot \frac{d\theta}{dx}$.
Substituting $\frac{d\theta}{dx} = \frac{1}{-12 \cos^2 \theta \sin \theta}$:
$\frac{d^2y}{dx^2} = -\frac{1}{2} \sec \theta \tan \theta \cdot \left( \frac{1}{-12 \cos^2 \theta \sin \theta} \right) = \frac{1}{24} \cdot \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos^2 \theta \sin \theta} = \frac{1}{24 \cos^4 \theta}$.
At $\theta = \frac{\pi}{4}$,$\cos \theta = \frac{1}{\sqrt{2}}$,so $\cos^4 \theta = \left( \frac{1}{\sqrt{2}} \right)^4 = \frac{1}{4}$.
Thus,$\frac{d^2y}{dx^2} = \frac{1}{24 \times (1/4)} = \frac{1}{6}$.

(A) Given,$x = \sin \theta$ and $y = \cos(p \theta)$.
$\frac{dx}{d\theta} = \cos \theta$ and $\frac{dy}{d\theta} = -p \sin(p \theta)$.
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-p \sin(p \theta)}{\cos \theta}$.
Since $\sin(p \theta) = \sqrt{1 - y^2}$ and $\cos \theta = \sqrt{1 - x^2}$,we have $\frac{dy}{dx} = -p \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}$.
$\sqrt{1 - x^2} y_1 = -p \sqrt{1 - y^2}$.
Squaring both sides: $(1 - x^2) y_1^2 = p^2 (1 - y^2)$.
Differentiating with respect to $x$:
$(1 - x^2) \cdot 2 y_1 y_2 + y_1^2 (-2x) = p^2 (-2y y_1)$.
Dividing by $2 y_1$ (assuming $y_1 \neq 0$):
$(1 - x^2) y_2 - x y_1 = -p^2 y$.
Therefore,$(1 - x^2) y_2 = x y_1 - p^2 y$.
Thus,option $A$ is correct.

(C) Let $x^3 = \sin \theta$ and $y^3 = \sin \phi$. Then the equation becomes $\cos \theta + \cos \phi = a(\sin \theta - \sin \phi)$.
Using trigonometric identities,$2 \cos \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2} = a \cdot 2 \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2}$.
This implies $\cot \frac{\theta-\phi}{2} = a$,which is a constant.
Thus,$\frac{\theta-\phi}{2} = \text{constant} \Rightarrow \theta - \phi = C$.
Substituting back,$\arcsin(x^3) - \arcsin(y^3) = C$.
Differentiating with respect to $x$:
$\frac{3x^2}{\sqrt{1-x^6}} - \frac{3y^2}{\sqrt{1-y^6}} \frac{dy}{dx} = 0$.
Rearranging the terms,we get $\frac{3y^2}{\sqrt{1-y^6}} \frac{dy}{dx} = \frac{3x^2}{\sqrt{1-x^6}}$.
Therefore,$y^2 \frac{dy}{dx} = x^2 \sqrt{\frac{1-y^6}{1-x^6}}$.

(A) Given $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$.
Since $f(0)=f(2)$:
$f(0) = 0^3 + p(0)^2 + q(0) = 0$
$f(2) = 2^3 + p(2)^2 + q(2) = 8 + 4p + 2q$
Setting $f(0)=f(2)$ gives $8 + 4p + 2q = 0$,which simplifies to $2p + q + 4 = 0$ (Equation $i$).
Now,find the derivative $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^2 + 2px + q$
Given $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$:
$3\left(1+\frac{1}{\sqrt{3}}\right)^2 + 2p\left(1+\frac{1}{\sqrt{3}}\right) + q = 0$
$3\left(1 + \frac{1}{3} + \frac{2}{\sqrt{3}}\right) + 2p + \frac{2p}{\sqrt{3}} + q = 0$
$3\left(\frac{4}{3} + \frac{2}{\sqrt{3}}\right) + 2p + \frac{2p}{\sqrt{3}} + q = 0$
$4 + 2\sqrt{3} + 2p + \frac{2p}{\sqrt{3}} + q = 0$ (Equation $ii$).
Subtracting Equation $i$ $(2p + q = -4)$ from Equation $ii$:
$(4 + 2\sqrt{3} + 2p + \frac{2p}{\sqrt{3}} + q) - (2p + q) = 0 - (-4)$
$4 + 2\sqrt{3} + \frac{2p}{\sqrt{3}} = 4$
$2\sqrt{3} + \frac{2p}{\sqrt{3}} = 0$
$2\sqrt{3} = -\frac{2p}{\sqrt{3}}$
$2p = -2(3) = -6 \Rightarrow p = -3$.
Substitute $p = -3$ into Equation $i$:
$2(-3) + q + 4 = 0$
$-6 + q + 4 = 0 \Rightarrow q = 2$.
Therefore,$p^2 + q^2 = (-3)^2 + (2)^2 = 9 + 4 = 13$.
Thus,option $A$ is correct.

(D) The given curve is $y^2(x-a)=x^2(x+a)$.
For a tangent to be parallel to the $X$-axis,the slope $\frac{dy}{dx}$ must be $0$.
Differentiating $y^2(x-a)=x^2(x+a)$ with respect to $x$ using implicit differentiation:
$2y \frac{dy}{dx}(x-a) + y^2(1) = 2x(x+a) + x^2(1)$
Setting $\frac{dy}{dx} = 0$,we get:
$y^2 = 2x^2 + 2ax + x^2 = 3x^2 + 2ax$
Substitute $y^2 = \frac{x^2(x+a)}{x-a}$ into the equation:
$\frac{x^2(x+a)}{x-a} = 3x^2 + 2ax$
Since $x=0$ is a solution (giving $y=0$),we check for other solutions by dividing by $x$ $(x \neq 0)$:
$\frac{x(x+a)}{x-a} = 3x + 2a$
$x^2 + ax = (3x + 2a)(x - a)$
$x^2 + ax = 3x^2 - 3ax + 2ax - 2a^2$
$x^2 + ax = 3x^2 - ax - 2a^2$
$2x^2 - 2ax - 2a^2 = 0$
$x^2 - ax - a^2 = 0$
The roots are $x = \frac{a \pm \sqrt{a^2 - 4(1)(-a^2)}}{2} = \frac{a \pm a\sqrt{5}}{2}$.
For these values of $x$,$y^2 = 3x^2 + 2ax$. Since $x^2 - ax - a^2 = 0$,$x^2 = ax + a^2$.
$y^2 = 3(ax + a^2) + 2ax = 5ax + 3a^2$.
For $x = \frac{a(1+\sqrt{5})}{2}$,$y^2 = 5a(\frac{a(1+\sqrt{5})}{2}) + 3a^2 > 0$,giving two values of $y$.
For $x = \frac{a(1-\sqrt{5})}{2}$,$y^2 = 5a(\frac{a(1-\sqrt{5})}{2}) + 3a^2 = \frac{5a^2 - 5a^2\sqrt{5} + 6a^2}{2} = \frac{11-5\sqrt{5}}{2}a^2 < 0$,which is impossible.
Thus,there are $2$ tangents.

(A) The equation of the tangent is $2y = 3x - 1$,which can be written as $y = \frac{3}{2}x - \frac{1}{2}$.
Comparing this with $y = mx + c$,the slope of the tangent is $m = \frac{3}{2}$.
Given the curve $y^2 = ax^3 + b$,we differentiate with respect to $x$:
$2y \frac{dy}{dx} = 3ax^2 \implies \frac{dy}{dx} = \frac{3ax^2}{2y}$.
At the point $(1, 1)$,the slope is $\frac{dy}{dx} = \frac{3a(1)^2}{2(1)} = \frac{3a}{2}$.
Equating the slopes: $\frac{3a}{2} = \frac{3}{2} \implies a = 1$.
Since the point $(1, 1)$ lies on the curve $y^2 = ax^3 + b$,we substitute $x=1, y=1, a=1$:
$1^2 = 1(1)^3 + b \implies 1 = 1 + b \implies b = 0$.
Thus,$(a, b) = (1, 0)$.

(B) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y = \sin x$,we have $k = \sin h$.
The slope of the tangent at $(h, k)$ is $\frac{dy}{dx} = \cos x$. At $(h, k)$,the slope is $\cos h$.
The equation of the tangent at $(h, k)$ is $y - k = \cos h(x - h)$.
Since the tangent passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$:
$0 - k = \cos h(0 - h) \Rightarrow -k = -h \cos h \Rightarrow k = h \cos h$.
From $k = \sin h$,we have $\cos h = \frac{k}{h}$.
Using the identity $\sin^2 h + \cos^2 h = 1$,we substitute $\sin h = k$ and $\cos h = \frac{k}{h}$:
$k^2 + \left(\frac{k}{h}\right)^2 = 1
k^2 + \frac{k^2}{h^2} = 1
h^2 k^2 + k^2 = h^2
h^2 - k^2 = h^2 k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 = x^2 y^2$.

(C) The equation of the given curve is $3 y^2 = 4 x^3$ ...$(i)$
Let $P(h, k)$ be a point on the curve.
Differentiating equation $(i)$ with respect to $x$,we get $6 y \frac{dy}{dx} = 12 x^2$,which implies $\frac{dy}{dx} = \frac{2 x^2}{y}$.
At point $(h, k)$,the slope $m = \frac{2 h^2}{k}$.
The length of the subtangent $T = \left| \frac{k}{m} \right| = \left| \frac{k}{2 h^2 / k} \right| = \frac{k^2}{2 h^2}$.
Since $3 k^2 = 4 h^3$,we have $k^2 = \frac{4}{3} h^3$.
Substituting this into $T$,we get $T = \frac{4 h^3}{3(2 h^2)} = \frac{2}{3} h$.
The length of the subnormal $N = |k m| = \left| k \cdot \frac{2 h^2}{k} \right| = 2 h^2$.
We need to find the relation between $T$ and $N$.
From $N = 2 h^2$,we have $h^2 = \frac{N}{2}$,so $h = \sqrt{\frac{N}{2}}$.
Then $T = \frac{2}{3} \sqrt{\frac{N}{2}}$.
Squaring both sides,$T^2 = \frac{4}{9} \cdot \frac{N}{2} = \frac{2 N}{9}$.
Thus,$9 T^2 = 2 N$,which can be written as $(3 T)^2 = 2 N$.
Comparing this with $(\beta T)^2 = 2 N$,we find $\beta = 3$.
Therefore,the correct option is $C$.

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g$.
Differentiating both sides,we get $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.
Here,the clock loses $2$ minutes per day,so $\Delta T = -2$ minutes.
The total time in a day is $24 \times 60 = 1440$ minutes.
Thus,the fractional change in time is $\frac{\Delta T}{T} = \frac{-2}{1440} = -\frac{1}{720}$.
Using the relation $\frac{\Delta L}{L} = 2 \frac{\Delta T}{T}$,we have $\frac{\Delta L}{L} = 2 \times \left( -\frac{1}{720} \right) = -\frac{1}{360}$.
To express this as a percentage change,we multiply by $100$:
$\frac{\Delta L}{L} \% = -\frac{1}{360} \times 100 = -\frac{10}{36} = -\frac{5}{18} \%$.
Therefore,the length must be changed by $-\frac{5}{18} \%$.

(D) Let $r$ be the radius of the sphere.
Given,the rate of change in radius is $\frac{dr}{dt} = 0.04 \text{ cm/sec}$.
The volume of the sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$,we get:
$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \cdot \frac{dr}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \quad \dots(i)$
The surface area of the sphere is $S = 4 \pi r^2$.
Differentiating with respect to $t$,we get:
$\frac{dS}{dt} = 4 \pi (2r) \cdot \frac{dr}{dt} = 8 \pi r \cdot \frac{dr}{dt} \quad \dots(ii)$
Dividing equation $(i)$ by equation $(ii)$,we get:
$\frac{dV/dt}{dS/dt} = \frac{dV}{dS} = \frac{4 \pi r^2 \cdot \frac{dr}{dt}}{8 \pi r \cdot \frac{dr}{dt}}$
$\frac{dV}{dS} = \frac{r}{2}$
Given $r = 10 \text{ cm}$,we substitute the value:
$\frac{dV}{dS} = \frac{10}{2} = 5 \text{ cm}$.
Thus,the rate of increase in volume with respect to surface area is $5 \text{ cm}$.
Hence,option $(D)$ is correct.

(C) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the upper end of the ladder from the ground.
Given the length of the ladder is $5 \ m$,by the Pythagorean theorem,we have $x^2 + y^2 = 5^2 = 25$.
Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
We are given $\frac{dx}{dt} = 3 \ m/sec$ and the ladder is descending at $\frac{dy}{dt} = -4 \ m/sec$.
Substituting these values into the differentiated equation: $x(3) + y(-4) = 0$,which gives $3x = 4y$,or $x = \frac{4}{3}y$.
Substitute $x = \frac{4}{3}y$ into the original equation $x^2 + y^2 = 25$:
$(\frac{4}{3}y)^2 + y^2 = 25$
$\frac{16}{9}y^2 + y^2 = 25$
$\frac{25}{9}y^2 = 25$
$y^2 = 9$
$y = 3 \ m$ (since height must be positive).
Thus,the height of the upper end is $3 \ m$.

(B) Given,$f(x)=(2 k+1) x-3-k e^{-x}+2 e^x$.
Since $f(x)$ is monotonically increasing for all $x \in R$,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x \in R$.
$f'(x) = (2k+1) + k e^{-x} + 2 e^x \geq 0$.
Multiplying by $e^x$ (which is always positive),we get:
$(2k+1)e^x + k + 2e^{2x} \geq 0$.
Rearranging the terms:
$2e^{2x} + (2k+1)e^x + k \geq 0$.
Factoring the quadratic expression in terms of $e^x$:
$2e^{2x} + 2ke^x + e^x + k \geq 0$.
$2e^x(e^x + k) + 1(e^x + k) \geq 0$.
$(2e^x + 1)(e^x + k) \geq 0$.
Since $2e^x + 1 > 0$ for all $x \in R$,we must have $e^x + k \geq 0$ for all $x \in R$.
This implies $k \geq -e^x$ for all $x \in R$.
As $x \to -\infty$,$e^x \to 0$,so $k \geq 0$.
Thus,the least value of $k$ is $0$.

(D) Let $f(x) = \frac{\log(7+x)}{\log(3+x)}$. For the function to be decreasing,we require $f'(x) < 0$.
Using the quotient rule,$f'(x) = \frac{\frac{1}{7+x} \log(3+x) - \frac{1}{3+x} \log(7+x)}{(\log(3+x))^2}$.
For $f'(x) < 0$,we need $\frac{\log(3+x)}{7+x} < \frac{\log(7+x)}{3+x}$.
This is equivalent to $\frac{\log(3+x)}{3+x} < \frac{\log(7+x)}{7+x}$.
Consider the function $g(t) = \frac{\log t}{t}$ for $t > 3$. The derivative $g'(t) = \frac{1 - \log t}{t^2}$,which is negative for $t > e \approx 2.718$.
Since $3+x > 3 > e$,the function $g(t)$ is strictly decreasing for $t > 3$.
Thus,$3+x < 7+x$ implies $g(3+x) > g(7+x)$,which means $\frac{\log(3+x)}{3+x} > \frac{\log(7+x)}{7+x}$.
Therefore,$f'(x) = \frac{1}{3+x} \cdot \frac{1}{7+x} \left( \frac{\log(3+x)}{3+x} \cdot (3+x) - \frac{\log(7+x)}{7+x} \cdot (7+x) \right)$ is not correct; let's re-evaluate: $f'(x) = \frac{(3+x)\log(3+x) - (7+x)\log(7+x)}{(7+x)(3+x)(\log(3+x))^2}$.
Since $g(t) = \frac{\log t}{t}$ is decreasing for $t > e$,for $x > 0$,$3+x < 7+x$ implies $g(3+x) > g(7+x)$,so $(3+x)\log(3+x) < (7+x)\log(7+x)$ is false; actually,$\frac{\log(3+x)}{3+x} > \frac{\log(7+x)}{7+x}$ implies $(3+x)\log(3+x) > (7+x)\log(7+x)$ is false. Re-calculating: $f'(x) < 0$ implies $(3+x)\log(3+x) < (7+x)\log(7+x)$ is false. The function $f(x)$ is actually decreasing for all $x > 0$ because the numerator grows slower than the denominator relative to the log base change. Thus,the interval is $(0, \infty)$.

(D) The function $f(x)$ is defined on the interval $[-1, 1]$.
For $x \in [-1, 0)$,$f(x) = 2^x + 1$. As $x \to 0^-$,$f(x) \to 2^0 + 1 = 2$. Since $2^x$ is strictly increasing,$f(x)$ increases from $f(-1) = 2^{-1} + 1 = 1.5$ to $2$.
At $x = 0$,$f(0) = 1$.
For $x \in (0, 1]$,$f(x) = 2^x - 1$. As $x \to 0^+$,$f(x) \to 2^0 - 1 = 0$. Since $2^x$ is strictly increasing,$f(x)$ increases from $0$ to $f(1) = 2^1 - 1 = 1$.
The range of the function is $[1.5, 2) \cup \{1\} \cup (0, 1]$.
Combining these,the range is $(0, 1] \cup [1.5, 2)$.
Since the range is not a closed interval and the function values do not attain a supremum or infimum within the domain (the values approach $2$ but never reach it,and approach $0$ but never reach it),the function has neither a maximum nor a minimum value on $[-1, 1]$.

(B) Given,$f(x)=a \log |x|+b x^2+x$.
The derivative is $f^{\prime}(x) = \frac{a}{x} + 2bx + 1$.
Since the function has extreme values at $x=-1$ and $x=2$,we have $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$.
For $x=-1$: $f^{\prime}(-1) = \frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a + 2b = 1$ $(i)$.
For $x=2$: $f^{\prime}(2) = \frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b = -2$ (ii).
Subtracting $(i)$ from (ii): $(a + 8b) - (a + 2b) = -2 - 1 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$.
Substituting $b = -\frac{1}{2}$ into $(i)$: $a + 2(-\frac{1}{2}) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$.
Thus,the ordered pair $(a, b) = \left(2, -\frac{1}{2}\right)$.

(B) Given $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$.
Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Let $t = \sin ^{-1} x$. Since $x \in [-1, 1]$,$t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Then $f(t) = t^2 + (\frac{\pi}{2} - t)^2 = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$.
This is a parabola opening upwards. The vertex is at $t = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$.
Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $t = \frac{\pi}{4}$.
$\alpha = f(\frac{\pi}{4}) = 2(\frac{\pi^2}{16}) - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$.
The maximum value occurs at the endpoints of the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
At $t = -\frac{\pi}{2}$,$f(-\frac{\pi}{2}) = 2(-\frac{\pi}{2})^2 - \pi(-\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$.
At $t = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 - \pi(\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}$.
Thus,$\beta = \frac{5\pi^2}{4}$.
Finally,$8(\alpha + \beta) = 8(\frac{\pi^2}{8} + \frac{5\pi^2}{4}) = 8(\frac{\pi^2 + 10\pi^2}{8}) = 11\pi^2$.

(C) Given,$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
To find the critical points,we calculate the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.
Setting $f'(x) = 0$,we get $6(x^2 - 3ax + 2a^2) = 0$,which factors as $6(x - a)(x - 2a) = 0$.
Thus,the critical points are $x = a$ and $x = 2a$.
We find the second derivative: $f''(x) = 12x - 18a$.
For $x = a$,$f''(a) = 12a - 18a = -6a$. Since $a > 0$,$f''(a) < 0$,so $f(x)$ has a maximum at $p = a$.
For $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a$. Since $a > 0$,$f''(2a) > 0$,so $f(x)$ has a minimum at $q = 2a$.
Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$.
Since $a > 0$,we can divide by $a$ to get $a = 2$.

(C) Given the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$.
First,we find the critical points by setting the derivative $f'(x) = 0$.
$f'(x) = 2x e^{2x} + x^2 (2 e^{2x}) = 2x e^{2x} (1 + x)$.
Setting $f'(x) = 0$,we get $x = 0$ or $x = -1$.
Now,we evaluate the function at the critical points and the endpoints of the interval $[-2, 2]$:
$f(-2) = (-2)^2 e^{2(-2)} = 4 e^{-4}$.
$f(-1) = (-1)^2 e^{2(-1)} = 1 e^{-2} = e^{-2}$.
$f(0) = (0)^2 e^{2(0)} = 0$.
$f(2) = (2)^2 e^{2(2)} = 4 e^4$.
Comparing these values,the global maximum $p = 4 e^4$ and the global minimum $q = 0$.
Finally,we calculate $p e^{-4} + q e^4 = (4 e^4) e^{-4} + (0) e^4 = 4 e^0 + 0 = 4(1) = 4$.

(A) Let $h$ be the height and $r$ be the radius of the cylinder inscribed in a sphere of radius $R = 3$ units.
From the geometry of the inscribed cylinder,we have the relation $r^2 + (h/2)^2 = R^2$.
Substituting $R = 3$,we get $r^2 + h^2/4 = 9$,which implies $r^2 = 9 - h^2/4$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
Substituting $r^2$,we get $V = \pi (9 - h^2/4) h = \pi (9h - h^3/4)$.
To find the maximum volume,we differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \pi (9 - 3h^2/4)$.
Setting $\frac{dV}{dh} = 0$,we get $9 - 3h^2/4 = 0$,which implies $3h^2/4 = 9$,so $h^2 = 12$ and $h = 2\sqrt{3}$ (since height must be positive).
Then $r^2 = 9 - (12/4) = 9 - 3 = 6$.
The maximum volume is $V = \pi (6) (2\sqrt{3}) = 12\sqrt{3}\pi$ cubic units.

(D) Since $f$ is a polynomial function,it is continuous on $[2,7]$ and differentiable on $(2,7)$.
According to the Lagrange's Mean Value Theorem,there exists some $c \in (2,7)$ such that:
$\frac{f(7)-f(2)}{7-2} = f^{\prime}(c)$
Given that $f^{\prime}(x) \leq 5$ for all $x \in (2,7)$,we have $f^{\prime}(c) \leq 5$.
Substituting the values,we get:
$\frac{f(7)-3}{5} \leq 5$
$f(7)-3 \leq 25$
$f(7) \leq 28$
Thus,the maximum possible value of $f(7)$ is $28$.

(A) Given the function $f(x) = 2x^3 - 3x^2 - 12x + 5$.
To find the critical points,we calculate the derivative $f'(x) = 6x^2 - 6x - 12$.
Setting $f'(x) = 0$,we get $6(x^2 - x - 2) = 0$,which factors as $6(x + 1)(x - 2) = 0$.
Thus,the critical points are $x = -1$ and $x = 2$,both of which lie within the interval $[-2, 4]$.
Now,we evaluate the function at the critical points and the endpoints of the interval:
$f(-2) = 2(-8) - 3(4) - 12(-2) + 5 = -16 - 12 + 24 + 5 = 1$.
$f(-1) = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$.
$f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$.
$f(4) = 2(64) - 3(16) - 12(4) + 5 = 128 - 48 - 48 + 5 = 37$.
Comparing these values,the absolute maximum is $37$,which occurs at $x = 4$.
Therefore,the correct option is $A$.

(C) Given that Rolle's theorem is applicable for the function $f(x) = x^{2m-1}(a-x)^{2n}$ in $(0, a)$.
We find the derivative $f'(x)$ using the product rule:
$f'(x) = (2m-1)x^{2m-2}(a-x)^{2n} - 2n(a-x)^{2n-1}x^{2m-1}$.
According to Rolle's theorem,there exists a $c \in (0, a)$ such that $f'(c) = 0$.
$(2m-1)c^{2m-2}(a-c)^{2n} - 2nc^{2m-1}(a-c)^{2n-1} = 0$.
Dividing by $c^{2m-2}(a-c)^{2n-1}$ (since $c \neq 0$ and $c \neq a$):
$(2m-1)(a-c) = 2nc$.
$(2m-1)a - (2m-1)c = 2nc$.
$(2m-1)a = (2m-1+2n)c$.
$c = \frac{a(2m-1)}{2m+2n-1}$.
Thus,option $C$ is correct.

(A) Given,$f(x)=a x^3+b x^2+11 x-6$ satisfies Rolle's theorem in $[1,3]$.
Therefore,$f(1)=f(3)$.
$f(1) = a(1)^3 + b(1)^2 + 11(1) - 6 = a + b + 5$
$f(3) = a(3)^3 + b(3)^2 + 11(3) - 6 = 27a + 9b + 33 - 6 = 27a + 9b + 27$
Since $f(1)=f(3)$,we have $a+b+5 = 27a+9b+27$,which simplifies to $26a+8b = -22$,or $13a+4b = -11$ ... $(i)$.
Also,$f'(x) = 3ax^2 + 2bx + 11$.
Given $f'(2 + \frac{1}{\sqrt{3}}) = 0$.
$f'(x) = 0$ at $x = 2 \pm \frac{1}{\sqrt{3}}$ for a cubic polynomial satisfying Rolle's theorem on $[1,3]$.
Using the sum of roots of $f'(x)=0$,$x_1 + x_2 = -\frac{2b}{3a}$.
Here,$(2 - \frac{1}{\sqrt{3}}) + (2 + \frac{1}{\sqrt{3}}) = 4 = -\frac{2b}{3a} \Rightarrow 4 = -\frac{2b}{3a} \Rightarrow b = -6a$.
Substitute $b = -6a$ into $(i)$:
$13a + 4(-6a) = -11$
$13a - 24a = -11$
$-11a = -11 \Rightarrow a = 1$.
Then $b = -6(1) = -6$.
Thus,$a+b = 1 + (-6) = -5$.

(A) In $\triangle AOB$,the semi-vertical angle $\theta = 45^{\circ}$.
$\tan 45^{\circ} = \frac{r}{h} \implies 1 = \frac{r}{h} \implies r = h$.
Using the slant height formula $l = \sqrt{r^2 + h^2}$,we get $l = \sqrt{h^2 + h^2} = \sqrt{2h^2} = h\sqrt{2}$.
The lateral surface area $S = \pi r l$.
Substituting $r = h$ and $l = h\sqrt{2}$,we get $S = \pi (h)(h\sqrt{2}) = \sqrt{2} \pi h^2$.
Given $h = 20.025 \ cm$,we have $h^2 = (20.025)^2 = 401.000625 \approx 401$.
Thus,$S \approx \sqrt{2} \pi (401) = 401 \sqrt{2} \pi \ cm^2$.
Therefore,option $A$ is correct.

(A) Let $AB$ be the lamp-post and $PQ$ be the man. Let $C$ be the tip of the shadow. Let $AP = x$ be the distance of the man from the lamp-post and $PC = y$ be the length of his shadow.
Given: $AB = 9 \ m$,$PQ = 2 \ m$,and $\frac{dx}{dt} = 7 \ m/min$.
Since $\triangle CAB$ and $\triangle CPQ$ are similar triangles,we have:
$\frac{PC}{AC} = \frac{PQ}{AB}$
$\frac{y}{x+y} = \frac{2}{9}$
$9y = 2x + 2y$
$7y = 2x$
$x = \frac{7}{2}y$
Differentiating both sides with respect to time $t$:
$\frac{dx}{dt} = \frac{7}{2} \frac{dy}{dt}$
Substituting $\frac{dx}{dt} = 7$:
$7 = \frac{7}{2} \frac{dy}{dt}$
$\frac{dy}{dt} = 2 \ m/min$.
Thus,the length of his shadow increases at a rate of $2 \ m/min$.

(D) Given integral,$I = \int \frac{d x}{x+\sqrt{x-1}}$
Put $x-1 = t^2 \Rightarrow dx = 2t \, dt$
Then $I = \int \frac{2t}{(t^2+1)+t} \, dt = \int \frac{(2t+1)-1}{t^2+t+1} \, dt$
$= \int \frac{2t+1}{t^2+t+1} \, dt - \int \frac{dt}{t^2+t+1}$
$= \log _e|t^2+t+1| - \int \frac{dt}{(t+\frac{1}{2})^2 + \frac{3}{4}}$
$= \log _e|t^2+t+1| - \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + c$
$= \log _e|t^2+t+1| - \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2t+1}{\sqrt{3}}\right) + c$
On substituting $t = \sqrt{x-1}$,we get
$I = \log _e|x+\sqrt{x-1}| - \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}\right) + c$
Hence,option $D$ is correct.

(C) Given the integral $I = \int \frac{\cos 4x + 1}{\cot x - \tan x} dx$.
Using the identity $\cos 4x + 1 = 2 \cos^2 2x$,we have:
$I = \int \frac{2 \cos^2 2x}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} dx$
$I = \int \frac{2 \cos^2 2x}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} dx$
Since $\cos^2 x - \sin^2 x = \cos 2x$ and $\sin x \cos x = \frac{1}{2} \sin 2x$,we get:
$I = \int \frac{2 \cos^2 2x}{\frac{\cos 2x}{\frac{1}{2} \sin 2x}} dx = \int \frac{2 \cos^2 2x \cdot \frac{1}{2} \sin 2x}{\cos 2x} dx$
$I = \int \cos 2x \sin 2x dx = \frac{1}{2} \int \sin 4x dx$
$I = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) + c = -\frac{1}{8} \cos 4x + c$.
Comparing with $k \cos 4x + c$,we find $k = -\frac{1}{8}$.
Thus,option $(C)$ is correct.

(A) We need to evaluate the integral $I = \int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x$.
Split the integral into two parts: $I = \int e^{2 x} \cos (3 x+4) d x + 5 \int e^{2 x} x^2 d x$.
Using the formula $\int e^{a x} \cos (b x+c) d x = \frac{e^{a x}}{a^2+b^2}(a \cos (b x+c)+b \sin (b x+c))$,we get:
$\int e^{2 x} \cos (3 x+4) d x = \frac{e^{2 x}}{2^2+3^2}(2 \cos (3 x+4)+3 \sin (3 x+4)) = \frac{e^{2 x}}{13}(2 \cos (3 x+4)+3 \sin (3 x+4))$.
Now,evaluate $5 \int e^{2 x} x^2 d x$ using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$:
$5 \int x^2 e^{2 x} d x = 5 \left[ x^2 \frac{e^{2 x}}{2} - \int 2x \frac{e^{2 x}}{2} d x \right] = \frac{5 x^2 e^{2 x}}{2} - 5 \int x e^{2 x} d x$.
Applying integration by parts again:
$5 \int x e^{2 x} d x = 5 \left[ x \frac{e^{2 x}}{2} - \int 1 \cdot \frac{e^{2 x}}{2} d x \right] = \frac{5 x e^{2 x}}{2} - \frac{5}{2} \cdot \frac{e^{2 x}}{2} = \frac{5 x e^{2 x}}{2} - \frac{5 e^{2 x}}{4}$.
Combining all parts:
$I = e^{2 x} \left[ \frac{2}{13} \cos (3 x+4) + \frac{3}{13} \sin (3 x+4) \right] + \frac{5 x^2 e^{2 x}}{2} - \left( \frac{5 x e^{2 x}}{2} - \frac{5 e^{2 x}}{4} \right) + c$.
$I = e^{2 x} \left[ \frac{2}{13} \cos (3 x+4) + \frac{3}{13} \sin (3 x+4) + \frac{5 x^2}{2} - \frac{5 x}{2} + \frac{5}{4} \right] + c$.
Thus,option $(A)$ is correct.

(C) Given $I_{m, n} = \int e^{mx} \cdot x^n \, dx$.
Using integration by parts,let $u = x^n$ and $dv = e^{mx} \, dx$.
Then $du = nx^{n-1} \, dx$ and $v = \frac{e^{mx}}{m}$.
$I_{m, n} = \int u \, dv = uv - \int v \, du$.
$I_{m, n} = x^n \left( \frac{e^{mx}}{m} \right) - \int \left( \frac{e^{mx}}{m} \right) (nx^{n-1}) \, dx$.
$I_{m, n} = \frac{x^n e^{mx}}{m} - \frac{n}{m} \int e^{mx} x^{n-1} \, dx$.
Since $I_{m, n-1} = \int e^{mx} x^{n-1} \, dx$,we have:
$I_{m, n} = \frac{x^n e^{mx}}{m} - \frac{n}{m} I_{m, n-1} + c$.
Rearranging the terms,we get:
$I_{m, n} + \frac{n}{m} I_{m, n-1} = \frac{x^n e^{mx}}{m} + c$.
Hence,option $C$ is correct.