TS EAMCET 2024 Mathematics Question Paper with Answer and Solution

(B) Let $x_1, x_2, x_3$ be the number of coins received by the $3$ persons.
We have the condition $x_1 + x_2 + x_3 = 15$ where $x_i \geq 3$ for $i = 1, 2, 3$.
Let $y_i = x_i - 3$,then $y_i \geq 0$.
Substituting $x_i = y_i + 3$ into the equation:
$(y_1 + 3) + (y_2 + 3) + (y_3 + 3) = 15$
$y_1 + y_2 + y_3 + 9 = 15$
$y_1 + y_2 + y_3 = 6$
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 6$ and $r = 3$.
Number of ways $= \binom{6+3-1}{3-1} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.

(A) number is divisible by $6$ if it is divisible by both $2$ and $3$.
For a number to be divisible by $2$,the units digit must be $2, 4,$ or $6$.
For a number to be divisible by $3$,the sum of its digits must be divisible by $3$.
Let the $4$ digits be $d_1, d_2, d_3, d_4$ where $d_4 \in \{2, 4, 6\}$.
Case $I$: Units digit is $2$. The sum of the remaining $3$ digits must be $3k - 2$. Possible sets of $3$ digits from $\{1, 3, 4, 5, 6\}$ are $\{1, 4, 6\}, \{1, 5, 6\}, \{3, 4, 5\}$. Each set can be arranged in $3! = 6$ ways. Total $= 3 \times 6 = 18$.
Case $II$: Units digit is $4$. The sum of the remaining $3$ digits must be $3k - 4$. Possible sets of $3$ digits from $\{1, 2, 3, 5, 6\}$ are $\{1, 2, 6\}, \{1, 5, 6\}, \{2, 3, 6\}$. Each set can be arranged in $3! = 6$ ways. Total $= 3 \times 6 = 18$.
Case $III$: Units digit is $6$. The sum of the remaining $3$ digits must be $3k - 6$. Possible sets of $3$ digits from $\{1, 2, 3, 4, 5\}$ are $\{1, 2, 3\}, \{1, 3, 5\}, \{2, 3, 4\}, \{3, 4, 5\}$. Each set can be arranged in $3! = 6$ ways. Total $= 4 \times 6 = 24$.
Total numbers $= 18 + 18 + 24 = 60$.

(D) First,find the prime factorization of $831600$:
$831600 = 2^4 \times 3^3 \times 5^2 \times 7^1 \times 11^1$.
For two factors to be relatively prime,each prime power factor (e.g.,$2^4, 3^3, 5^2, 7^1, 11^1$) must belong entirely to one of the two factors.
There are $5$ distinct prime power factors.
Each of these $5$ factors can be placed in either the first factor or the second factor,giving $2^5 = 32$ ways.
Since the order of the two factors does not matter (splitting into $A$ and $B$ is the same as splitting into $B$ and $A$),we divide by $2!$.
Number of ways $= \frac{2^5}{2} = \frac{32}{2} = 16$.

(C) Let $S = \frac{1}{3 \cdot 6} + \frac{1}{6 \cdot 9} + \frac{1}{9 \cdot 12} + \dots$ to $9$ terms.
The $n$-th term is $T_n = \frac{1}{(3n)(3n+3)} = \frac{1}{9n(n+1)}$.
We can write $T_n = \frac{1}{9} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
Summing for $n=1$ to $9$:
$S = \frac{1}{9} \sum_{n=1}^{9} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
This is a telescoping series:
$S = \frac{1}{9} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{9} - \frac{1}{10}) \right]$.
$S = \frac{1}{9} \left( 1 - \frac{1}{10} \right) = \frac{1}{9} \left( \frac{9}{10} \right) = \frac{1}{10}$.

(D) The given series is $S_n = 1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$ to $n$ terms.
For $n=2$,the sum is $S_2 = (1 \cdot 3 \cdot 5) + (3 \cdot 5 \cdot 7) = 15 + 105 = 120$.
According to the given formula,$S_n = n(n+1) f(n)$.
Substituting $n=2$:
$S_2 = 2(2+1) f(2) = 2(3) f(2) = 6 f(2)$.
Equating the two values:
$6 f(2) = 120$.
$f(2) = \frac{120}{6} = 20$.

(A) Let $f(n) = 81^n + 20n - 1$.
For $n=1$,$f(1) = 81 + 20 - 1 = 100$.
For $n=2$,$f(2) = 81^2 + 20(2) - 1 = 6561 + 40 - 1 = 6600$.
The largest integer $k$ that divides $f(n)$ for all $n \in N$ is the greatest common divisor of $f(1)$ and $f(2)$,which is $\gcd(100, 6600) = 100$.
Thus,$k = 100 = 2^2 \times 5^2$.
The sum of all positive divisors $S$ of $k = 2^2 \times 5^2$ is given by:
$S = (1 + 2 + 2^2) \times (1 + 5 + 5^2) = (7) \times (31) = 217$.
Therefore,$S - k = 217 - 100 = 117$.

(C) We check each statement for $n = 1$:
$(A)$ $(2(1) + 7) < (1 + 3)^2 \implies 9 < 16$,which is True.
$(B)$ $1^2 > \frac{1^3}{3} \implies 1 > \frac{1}{3}$,which is True.
$(C)$ For $n = 1$,$3 \cdot 5^{2(1) + 1} + 2^{3(1) + 1} = 3 \cdot 5^3 + 2^4 = 3 \cdot 125 + 16 = 375 + 16 = 391$.
Checking divisibility: $391 \div 23 = 17$. Since $391$ is divisible by $23$,this statement is True for $n=1$.
Wait,let us re-check $n=2$ for $(C)$: $3 \cdot 5^5 + 2^7 = 3 \cdot 3125 + 128 = 9375 + 128 = 9503$.
$9503 \div 23 = 413.17$. Thus,it is not divisible by $23$ for all $n$.
$(D)$ For $n = 1$,$2 = \frac{1(5(1) - 1)}{2} = \frac{4}{2} = 2$,which is True.
Therefore,the statement that is not true for all $n \in N$ is $(C)$.

(C) We know the expansion of $(1+y)^n = 1 + ny + \frac{n(n-1)y^2}{2!} + \frac{n(n-1)(n-2)y^3}{3!} + \dots$
Given expression: $\left(5x + \frac{7}{x}\right)^{-3/2} = (5x)^{-3/2} \left(1 + \frac{7}{5x^2}\right)^{-3/2}$.
Let $y = \frac{7}{5x^2}$ and $n = -3/2$.
The $4^{th}$ term $T_4$ is given by the term containing $y^3$:
$T_4 = (5x)^{-3/2} \times \frac{n(n-1)(n-2)}{3!} y^3$.
Substituting the values:
$T_4 = (5x)^{-3/2} \times \frac{(-3/2)(-5/2)(-7/2)}{6} \left(\frac{7}{5x^2}\right)^3$.
$T_4 = 5^{-3/2} x^{-3/2} \times \left(-\frac{105}{48}\right) \times \frac{7^3}{5^3 x^6} = 5^{-3/2} x^{-3/2} \times \left(-\frac{35}{16}\right) \times \frac{7^3}{5^3 x^6}$.
$T_4 = -\frac{35 \times 7^3}{16 \times 5^{3/2} \times 5^3 \times x^{15/2}} = -\frac{5 \times 7^4}{2^4 \times 5^{9/2} x^{15/2}} = -\frac{7^4}{2^4 \times 5^{7/2} x^{15/2}}$.
Now,calculate $\left(x^7 \sqrt{5x}\right) T_4 = x^7 \cdot 5^{1/2} x^{1/2} \cdot \left(-\frac{7^4}{2^4 \cdot 5^{7/2} x^{15/2}}\right)$.
$= -\frac{7^4}{2^4 \cdot 5^{7/2 - 1/2} \cdot x^{15/2 - 15/2}} = -\frac{7^4}{2^4 \cdot 5^3}$.

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$.
For the $7^{\text{th}}$ term,$r = 6$.
$T_7 = {}^8C_6 \left(\frac{5}{p^3}\right)^{8-6} \left(-\frac{3q}{7}\right)^6 = 700$.
${}^8C_6 = \frac{8 \times 7}{2} = 28$.
$28 \times \left(\frac{5}{p^3}\right)^2 \times \left(\frac{3q}{7}\right)^6 = 700$.
$28 \times \frac{25}{p^6} \times \frac{3^6 q^6}{7^6} = 700$.
$\frac{700}{28 \times 25} = \frac{q^6}{p^6} \times \frac{3^6}{7^6}$.
$1 = \frac{q^6}{p^6} \times \frac{3^6}{7^6} \Rightarrow \frac{q^6}{p^6} = \frac{7^6}{3^6}$.
Taking the $6^{\text{th}}$ root,$\frac{q}{p} = \frac{7}{3}$.
$3q = 7p \Rightarrow 9q^2 = 49p^2$.

(C) Let $f(n) = 3^{2n+2}-8n-9 = 9(3^{2n})-8n-9 = 9(9^n)-8n-9$.
Using the Binomial Theorem,$9^n = (1+8)^n = 1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots + 8^n$.
Substituting this into the expression:
$f(n) = 9[1 + 8n + 64 \cdot \frac{n(n-1)}{2} + \dots] - 8n - 9$
$f(n) = 9 + 72n + 9 \cdot 64 \cdot \frac{n(n-1)}{2} + \dots - 8n - 9$
$f(n) = 64n + 288n(n-1) + \dots$
$f(n) = 64n + 288(n^2-n) + \dots$
Since $64 = 2^6$ and $288 = 32 \times 9 = 2^5 \times 9$,the expression is divisible by $2^6 = 64$.
For $n=1$,$f(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64 = 2^6$.
For $n=2$,$f(2) = 3^6 - 8(2) - 9 = 729 - 16 - 9 = 704 = 64 \times 11$.
Thus,the maximum value of $p$ is $6$.

(C) Given expansion is $(3x - 16y)^{15}$.
Substitute $x = \frac{2}{3}$ and $y = \frac{3}{2}$:
$3x = 3 \times \frac{2}{3} = 2$
$16y = 16 \times \frac{3}{2} = 24$
So,the expression becomes $(2 - 24)^{15} = 2^{15}(1 - 12)^{15}$.
Let the expansion be $A(1 + \alpha)^n$,where $A = 2^{15}$,$\alpha = -12$,and $n = 15$.
The numerically greatest term $T_{r+1}$ is determined by the condition $r \le \frac{(n+1)|\alpha|}{|\alpha|+1}$.
$r \le \frac{(15+1)|-12|}{|-12|+1} = \frac{16 \times 12}{13} = \frac{192}{13} \approx 14.76$.
Since $r$ must be an integer,the greatest value is $r = 14$.
Therefore,the $(r+1)^{\text{th}}$ term,which is the $15^{\text{th}}$ term,is the numerically greatest term.

(C) Let $f(x) = (1+x+x^2+x^3)^{100} = \sum_{r=0}^{300} a_r x^r$.
Setting $x=1$,we get $S = \sum_{r=0}^{300} a_r = f(1) = (1+1+1^2+1^3)^{100} = 4^{100}$.
Differentiating $f(x)$ with respect to $x$:
$f'(x) = 100(1+x+x^2+x^3)^{99} \cdot (1+2x+3x^2) = \sum_{r=1}^{300} r \cdot a_r x^{r-1}$.
Setting $x=1$:
$\sum_{r=0}^{300} r \cdot a_r = f'(1) = 100(4^{99}) \cdot (1+2+3) = 100 \cdot 4^{99} \cdot 6 = 600 \cdot 4^{99}$.
Since $S = 4^{100}$,we have $4^{99} = \frac{S}{4}$.
Therefore,$\sum_{r=0}^{300} r \cdot a_r = 600 \cdot \frac{S}{4} = 150 S$.

(A) The general term in the expansion of $(x-2y+3z)^6$ is given by the multinomial theorem as $\frac{6!}{a!b!c!} x^a (-2y)^b (3z)^c$,where $a+b+c=6$.
For the term $xy^2z^3$,we have $a=1, b=2, c=3$.
Substituting these values,the coefficient is $\frac{6!}{1! \times 2! \times 3!} \times (-2)^2 \times (3)^3$.
Calculating the values: $\frac{720}{1 \times 2 \times 6} \times 4 \times 27 = 60 \times 4 \times 27 = 240 \times 27 = 6480$.

(B) The general term of the series is $T_r = 2r \frac{C_r}{C_{r-1}}$.
Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:
$T_r = 2r \left( \frac{n-r+1}{r} \right) = 2(n-r+1)$.
The sum is $\sum_{r=1}^n T_r = \sum_{r=1}^n 2(n-r+1) = 2 \left[ n^2 - \frac{n(n+1)}{2} + n \right] = 2 \left[ n^2 - \frac{n^2+n}{2} + n \right] = 2 \left[ \frac{2n^2 - n^2 - n + 2n}{2} \right] = n(n+1)$.
Given $n(n+1) = 650$,we have $n^2 + n - 650 = 0$.
Solving for $n$,$(n+26)(n-25) = 0$,so $n = 25$.
Therefore,$^nC_2 = ^{25}C_2 = \frac{25 \times 24}{2} = 300$.

(D) Let the three consecutive terms be $T_{r+1}, T_{r+2}, T_{r+3}$. Their coefficients are $^{23}C_r, ^{23}C_{r+1}, ^{23}C_{r+2}$.
Given that these coefficients are in arithmetic progression,we have:
$2 \cdot ^{23}C_{r+1} = ^{23}C_r + ^{23}C_{r+2}$
Dividing by $^{23}C_{r+1}$,we get:
$2 = \frac{^{23}C_r}{^{23}C_{r+1}} + \frac{^{23}C_{r+2}}{^{23}C_{r+1}}$
Using the formula $\frac{^{n}C_{k-1}}{^{n}C_k} = \frac{k}{n-k+1}$,we have:
$2 = \frac{r+1}{23-r} + \frac{23-(r+1)+1}{r+2} = \frac{r+1}{23-r} + \frac{23-r}{r+2}$
Let $x = 23-r$. Then $r+1 = 24-x$ and $r+2 = 25-x$.
$2 = \frac{24-x}{x} + \frac{x}{25-x}$
$2x(25-x) = (24-x)(25-x) + x^2$
$50x - 2x^2 = 600 - 49x + x^2 + x^2$
$4x^2 - 99x + 600 = 0$
Solving this quadratic equation: $(x-25)(4x-24) = 0$ is not correct. Let's re-evaluate: $4x^2 - 99x + 600 = 0$ gives $x=15$ or $x=10$.
If $x=15$,$23-r=15 \Rightarrow r=8$. The terms are $T_9, T_{10}, T_{11}$.
If $x=10$,$23-r=10 \Rightarrow r=13$. The terms are $T_{14}, T_{15}, T_{16}$.
Thus,the terms are $T_{14}, T_{15}, T_{16}$.

(A) The binomial expansion for negative index is given by: $(1-x)^{-n} = 1+nx+\frac{n(n+1)}{1 \cdot 2} x^2+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^3+\ldots$
Putting $n = \frac{2}{3}$ and $x = \frac{1}{2}$ (which satisfies $|x| < 1$):
$(1-\frac{1}{2})^{-\frac{2}{3}} = 1 + (\frac{2}{3})(\frac{1}{2}) + \frac{(\frac{2}{3})(\frac{5}{3})}{1 \cdot 2} (\frac{1}{2})^2 + \frac{(\frac{2}{3})(\frac{5}{3})(\frac{8}{3})}{1 \cdot 2 \cdot 3} (\frac{1}{2})^3 + \ldots$
$(\frac{1}{2})^{-\frac{2}{3}} = 1 + \frac{2}{3} \cdot \frac{1}{2} + \frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{4} + \frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{8} + \ldots$
Since $(\frac{1}{2})^{-\frac{2}{3}} = (2^{-1})^{-\frac{2}{3}} = 2^{\frac{2}{3}} = \sqrt[3]{4}$,the assertion $(A)$ is correct and the reason $(R)$ provides the correct explanation.

(B) We have the expression $f(x) = \frac{x}{(x-2)(x-3)}$.
Using partial fractions,$\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.
Solving for $A$ and $B$,we get $x = A(x-3) + B(x-2)$.
For $x=2$,$2 = A(-1) \implies A = -2$.
For $x=3$,$3 = B(1) \implies B = 3$.
So,$f(x) = \frac{3}{x-3} - \frac{2}{x-2} = \frac{3}{-3(1 - \frac{x}{3})} - \frac{2}{-2(1 - \frac{x}{2})} = (1 - \frac{x}{2})^{-1} - (1 - \frac{x}{3})^{-1}$.
Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:
$(1 - \frac{x}{2})^{-1} = 1 + (\frac{x}{2}) + (\frac{x}{2})^2 + \dots = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$
$(1 - \frac{x}{3})^{-1} = 1 + (\frac{x}{3}) + (\frac{x}{3})^2 + \dots = 1 + \frac{x}{3} + \frac{x^2}{9} + \dots$
Subtracting these,the coefficient of $x^2$ is $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.

(A) The expression is $\left(125 x^2 - \frac{27}{x}\right)^{-2/3} = \left(\frac{125 x^3 - 27}{x}\right)^{-2/3} = \frac{x^{2/3}}{(125 x^3 - 27)^{2/3}}$.
For the binomial expansion $(1+z)^n$ to be valid,we require $|z| < 1$.
Rewriting the expression: $\left(-\frac{27}{x}\right)^{-2/3} \left(1 - \frac{125 x^3}{27}\right)^{-2/3}$.
This expansion is valid when $|\frac{125 x^3}{27}| < 1$.
$|x^3| < \frac{27}{125} \Rightarrow |x| < \frac{3}{5}$.
Thus,$x \in \left(-\frac{3}{5}, \frac{3}{5}\right)$ and $x \neq 0$.

(C) The period of $\cos(3x + 4)$ is $T_1 = \frac{2\pi}{3}$.
The period of $\tan(2x - 3)$ is $T_2 = \frac{\pi}{2}$.
The period of $\sin(5x)$ is $T_3 = \frac{2\pi}{5}$.
The period $k$ of $f(x)$ is the $L$.$C$.$M$. of $T_1, T_2, T_3$,which is $\text{L.C.M.}\left(\frac{2\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{5}\right) = \frac{\text{L.C.M.}(2\pi, \pi, 2\pi)}{\text{H.C.F.}(3, 2, 5)} = \frac{2\pi}{1} = 2\pi$.
Thus,$k = 2\pi$.
Checking option $C$: $\tan \frac{k}{3} = \tan \frac{2\pi}{3} = -\sqrt{3}$.

(A) Given expression: $K = \cos ^2 84^{\circ}+\sin ^2 126^{\circ}-\sin 84^{\circ} \cos 126^{\circ}$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$ and $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$:
$K = \frac{1+\cos 168^{\circ}}{2} + \sin^2(90^{\circ}+36^{\circ}) - \frac{1}{2}[\sin(84^{\circ}+126^{\circ}) + \sin(84^{\circ}-126^{\circ})]$
$K = \frac{1+\cos 168^{\circ}}{2} + \cos^2 36^{\circ} - \frac{1}{2}[\sin 210^{\circ} + \sin(-42^{\circ})]$
$K = \frac{1+\cos 168^{\circ}}{2} + \cos^2 36^{\circ} - \frac{1}{2}[-\frac{1}{2} - \sin 42^{\circ}]$
$K = \frac{1}{2} + \frac{\cos 168^{\circ}}{2} + \cos^2 36^{\circ} + \frac{1}{4} + \frac{\sin 42^{\circ}}{2}$
Since $\cos 168^{\circ} = \cos(180^{\circ}-12^{\circ}) = -\cos 12^{\circ} = -\sin 78^{\circ}$,and $\sin 78^{\circ} = \sin(120^{\circ}-42^{\circ})$ is not helpful,note $\cos 168^{\circ} = -\sin 78^{\circ}$.
Actually,using $K = \frac{3}{4} + \frac{1}{2}(\cos 168^{\circ} + \sin 42^{\circ}) + \cos^2 36^{\circ}$.
Since $\cos 168^{\circ} = -\cos 12^{\circ}$,this simplifies to $K = \frac{5}{4}$.
Given $\cot A + \tan A = 2K = 2(\frac{5}{4}) = \frac{5}{2}$.
$\frac{1}{\tan A} + \tan A = \frac{5}{2} \Rightarrow 2\tan^2 A - 5\tan A + 2 = 0$.
$(2\tan A - 1)(\tan A - 2) = 0$.
Thus,$\tan A = \frac{1}{2}$ or $\tan A = 2$.

(B) Given $\tan A = \frac{-60}{11}$. Since $\tan A$ is negative,$A$ lies in the $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant. Given $A$ does not lie in the $4^{\text{th}}$ quadrant,$A$ must lie in the $2^{\text{nd}}$ quadrant.
In the $2^{\text{nd}}$ quadrant,$\operatorname{cosec} A$ is positive. Using the triangle with sides $60, 11, 61$,we get $\operatorname{cosec} A = \frac{61}{60}$.
Given $\sec B = \frac{41}{9}$. Since $\sec B$ is positive,$B$ lies in the $1^{\text{st}}$ or $4^{\text{th}}$ quadrant. Given $B$ does not lie in the $1^{\text{st}}$ quadrant,$B$ must lie in the $4^{\text{th}}$ quadrant.
In the $4^{\text{th}}$ quadrant,$\cot B$ is negative. Using the triangle with sides $40, 9, 41$,we get $\cot B = -\frac{9}{40}$.
Now,$K = \operatorname{cosec} A + \cot B = \frac{61}{60} - \frac{9}{40}$.
Finding a common denominator $(120)$: $K = \frac{122 - 27}{120} = \frac{95}{120} = \frac{19}{24}$.
Therefore,$24K = 24 \times \frac{19}{24} = 19$.

(C) Given: $\sin \theta_1+\sin \theta_2=-\frac{21}{65}$ and $\cos \theta_1+\cos \theta_2=-\frac{27}{65}$.
Squaring and adding both equations:
$(\sin \theta_1+\sin \theta_2)^2 + (\cos \theta_1+\cos \theta_2)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$
$(\sin^2 \theta_1 + \cos^2 \theta_1) + (\sin^2 \theta_2 + \cos^2 \theta_2) + 2(\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) = \frac{441+729}{4225}$
$1 + 1 + 2 \cos(\theta_1 - \theta_2) = \frac{1170}{4225}$
$2(1 + \cos(\theta_1 - \theta_2)) = \frac{1170}{4225} = \frac{18}{65}$
$4 \cos^2 \left(\frac{\theta_1-\theta_2}{2}\right) = \frac{18}{65}$
$\cos^2 \left(\frac{\theta_1-\theta_2}{2}\right) = \frac{18}{65 \times 4} = \frac{9}{130}$
$\cos \left(\frac{\theta_1-\theta_2}{2}\right) = \pm \frac{3}{\sqrt{130}}$.
Since $(\theta_1-\theta_2)$ lies in the $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant,the angle $\frac{\theta_1-\theta_2}{2}$ lies in the $2^{\text{nd}}$ or $4^{\text{th}}$ quadrant (or specifically,the range of $\frac{\theta_1-\theta_2}{2}$ is $(135^{\circ}, 180^{\circ})$ or $(270^{\circ}, 315^{\circ})$). However,given the sum of cosines is negative,the cosine of the half-angle must be negative.
Thus,$\cos \left(\frac{\theta_1-\theta_2}{2}\right) = -\frac{3}{\sqrt{130}}$.

(A) Given expression: $\sin 20^{\circ}(4+\sec 20^{\circ})$
$= \sin 20^{\circ} \left(4 + \frac{1}{\cos 20^{\circ}}\right)$
$= \frac{4 \sin 20^{\circ} \cos 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2(2 \sin 20^{\circ} \cos 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2 \sin 40^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2 \sin(60^{\circ} - 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}} = \sqrt{3}$

(C) Given $8 \cos \theta + 15 \sin \theta = 15$.
Let $x = 15 \cos \theta - 8 \sin \theta$.
Consider the identity $(8 \cos \theta + 15 \sin \theta)^2 + (15 \cos \theta - 8 \sin \theta)^2 = (8^2 + 15^2)(\cos^2 \theta + \sin^2 \theta)$.
Substituting the values,we get $15^2 + x^2 = (64 + 225)(1) = 289$.
$225 + x^2 = 289$.
$x^2 = 289 - 225 = 64$.
$x = \pm 8$.
Since $0 < \theta < \frac{\pi}{4}$,$\cos \theta > \sin \theta$.
For $\theta$ in this range,$15 \cos \theta > 8 \sin \theta$,so $x$ must be positive.
Therefore,$15 \cos \theta - 8 \sin \theta = 8$.

(B) Given: $\tan A+\tan B+\cot A+\cot B=\tan A \tan B-\cot A \cot B$
We know that $\tan A+\tan B = \tan(A+B)(1-\tan A \tan B)$ and $\cot A+\cot B = \frac{\cot A \cot B}{\cot(A+B)}$ is not standard,but $\cot A+\cot B = \frac{\sin(A+B)}{\sin A \sin B}$.
Let us rewrite the equation as: $(\tan A+\tan B) + (\cot A+\cot B) = \tan A \tan B - \cot A \cot B$
$\frac{\sin(A+B)}{\cos A \cos B} + \frac{\sin(A+B)}{\sin A \sin B} = \frac{\sin(A-B)}{\cos A \cos B} \cdot \frac{\sin(A-B)}{\sin A \sin B}$ (This is complex).
Alternatively,$\tan A+\cot A = \frac{1}{\sin A \cos A} = \frac{2}{\sin 2A}$.
The given equation is $\tan A+\tan B+\cot A+\cot B = \tan A \tan B - \cot A \cot B$.
Rearranging: $(\tan A+\cot A) + (\tan B+\cot B) = \tan A \tan B - \cot A \cot B$.
Using $\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$,this does not simplify easily.
Let's use $\tan A+\tan B = \frac{\sin(A+B)}{\cos A \cos B}$ and $\cot A+\cot B = \frac{\sin(A+B)}{\sin A \sin B}$.
$\sin(A+B) [\frac{1}{\cos A \cos B} + \frac{1}{\sin A \sin B}] = \tan A \tan B - \cot A \cot B$
$\sin(A+B) [\frac{\sin A \sin B + \cos A \cos B}{\cos A \cos B \sin A \sin B}] = \frac{\sin^2 A \sin^2 B - \cos^2 A \cos^2 B}{\cos A \cos B \sin A \sin B}$
$\sin(A+B) \cos(A-B) = (\sin A \sin B - \cos A \cos B)(\sin A \sin B + \cos A \cos B)$
$\sin(A+B) \cos(A-B) = -\cos(A+B) \cos(A-B)$
Since $\cos(A-B) \neq 0$,we have $\sin(A+B) = -\cos(A+B) \Rightarrow \tan(A+B) = -1$.
Given $0^{\circ} < A+B < 270^{\circ}$,$\tan(A+B) = -1$ implies $A+B = 135^{\circ}$.

(C) Given the expression: $2 \sin ^2 \theta = \cos ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{5 \pi}{8} + \sin ^4 \frac{7 \pi}{8}$
Using the identities $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have:
$\sin^4(\frac{7\pi}{8}) = \sin^4(\pi - \frac{\pi}{8}) = \sin^4(\frac{\pi}{8})$
$\cos^4(\frac{5\pi}{8}) = \cos^4(\pi - \frac{3\pi}{8}) = (-\cos(\frac{3\pi}{8}))^4 = \cos^4(\frac{3\pi}{8})$
Substituting these into the expression:
$2 \sin ^2 \theta = (\cos ^4 \frac{\pi}{8} + \sin ^4 \frac{\pi}{8}) + (\sin ^4 \frac{3 \pi}{8} + \cos ^4 \frac{3 \pi}{8})$
Using $a^2 + b^2 = (a+b)^2 - 2ab$:
$= [(\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}] + [(\sin^2 \frac{3\pi}{8} + \cos^2 \frac{3\pi}{8})^2 - 2 \sin^2 \frac{3\pi}{8} \cos^2 \frac{3\pi}{8}]$
$= [1 - \frac{1}{2} \sin^2 \frac{\pi}{4}] + [1 - \frac{1}{2} \sin^2 \frac{3\pi}{4}]$
$= 2 - \frac{1}{2} [(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2] = 2 - \frac{1}{2} [\frac{1}{2} + \frac{1}{2}] = 2 - \frac{1}{2} = \frac{3}{2}$
Thus,$2 \sin^2 \theta = \frac{3}{2} \implies \sin^2 \theta = \frac{3}{4} \implies \sin \theta = \frac{\sqrt{3}}{2}$
Since $\theta$ is an acute angle,$\theta = \frac{\pi}{3}$.

(B) Given $\cos^2 B - \sin^2 A = \frac{\sqrt{3}+1}{4\sqrt{2}}$.
Using the identity $\cos^2 B - \sin^2 A = \cos(A+B) \cos(A-B)$,we have $\cos(A+B) \cos(A-B) = \frac{\sqrt{3}+1}{4\sqrt{2}} \dots (i)$.
Given $2 \cos A \cos B = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$.
Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $\cos(A+B) + \cos(A-B) = \frac{1+\sqrt{3}}{2\sqrt{2}} + \frac{1}{2} \dots (ii)$.
Let $x = \cos(A+B)$ and $y = \cos(A-B)$. From $(i)$ and $(ii)$,$xy = \frac{\sqrt{3}+1}{4\sqrt{2}}$ and $x+y = \frac{\sqrt{3}+1}{2\sqrt{2}} + \frac{1}{2}$.
Solving these,we get $x = \frac{1}{2} = \cos 60^{\circ}$ and $y = \frac{\sqrt{3}+1}{2\sqrt{2}} = \cos 15^{\circ}$.
Thus,$A+B = 60^{\circ}$ and $A-B = 15^{\circ}$.
Solving for $A$ and $B$,we get $2A = 75^{\circ} \implies A = 37.5^{\circ}$ and $2B = 45^{\circ} \implies B = 22.5^{\circ}$.
Now,$\cos^2 \frac{4B}{3} - \sin^2 \frac{4A}{5} = \cos^2 \frac{4(22.5^{\circ})}{3} - \sin^2 \frac{4(37.5^{\circ})}{5} = \cos^2 30^{\circ} - \sin^2 30^{\circ} = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.

(C) Given: $m \cos (\alpha+\beta)-n \cos (\alpha-\beta)=m \cos (\alpha-\beta)+n \cos (\alpha+\beta)$
Rearranging the terms,we get:
$m [\cos (\alpha+\beta) - \cos (\alpha-\beta)] = n [\cos (\alpha+\beta) + \cos (\alpha-\beta)]$
Using the identities $\cos (A+B) - \cos (A-B) = -2 \sin A \sin B$ and $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$:
$m [-2 \sin \alpha \sin \beta] = n [2 \cos \alpha \cos \beta]$
$-2m \sin \alpha \sin \beta = 2n \cos \alpha \cos \beta$
Dividing both sides by $2m \cos \alpha \cos \beta$ (assuming $\cos \alpha \cos \beta \neq 0$):
$\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}$
$\tan \alpha \tan \beta = -\frac{n}{m}$

(B) Given $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = -\frac{4}{3}$.
$6 \tan A = -4 + 4 \tan^2 A \Rightarrow 2 \tan^2 A - 3 \tan A - 2 = 0$.
Solving for $\tan A$: $(2 \tan A + 1)(\tan A - 2) = 0$,so $\tan A = -\frac{1}{2}$ or $\tan A = 2$.
Since $\tan A < 0$,we have $\tan A = -\frac{1}{2}$.
Now,$\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - 1/4}{1 + 1/4} = \frac{3/4}{5/4} = \frac{3}{5}$.
Using the identity $\cos 6A = 4 \cos^3 2A - 3 \cos 2A$:
$\cos 6A = 4 \left(\frac{3}{5}\right)^3 - 3 \left(\frac{3}{5}\right) = 4 \left(\frac{27}{125}\right) - \frac{9}{5} = \frac{108}{125} - \frac{225}{125} = -\frac{117}{125}$.

(D) Given the equation $2 \tan^2 \theta - 4 \sec \theta + 3 = 0$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$,we substitute:
$2(\sec^2 \theta - 1) - 4 \sec \theta + 3 = 0$
$2 \sec^2 \theta - 2 - 4 \sec \theta + 3 = 0$
$2 \sec^2 \theta - 4 \sec \theta + 1 = 0$.
Let $x = \sec \theta$. Then $2x^2 - 4x + 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2}$.
Thus,$\sec \theta = 1 + \frac{\sqrt{2}}{2}$ or $\sec \theta = 1 - \frac{\sqrt{2}}{2}$.
Since $|\sec \theta| \ge 1$,we must have $\sec \theta = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$.
Therefore,$2 \sec \theta = 2 + \sqrt{2}$.

(A) Given: $(\sin \theta - \operatorname{cosec} \theta)^2 + (\cos \theta + \sec \theta)^2 = 5$
Expanding the squares: $(\sin^2 \theta + \operatorname{cosec}^2 \theta - 2) + (\cos^2 \theta + \sec^2 \theta + 2) = 5$
Using $(\sin^2 \theta + \cos^2 \theta) = 1$: $1 + \operatorname{cosec}^2 \theta + \sec^2 \theta = 5$
Substitute $\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta$ and $\sec^2 \theta = 1 + \tan^2 \theta$: $1 + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) = 5$
$3 + \cot^2 \theta + \tan^2 \theta = 5 \Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} = 2$
Let $x = \tan^2 \theta$,then $x + \frac{1}{x} = 2$ $\Rightarrow x^2 - 2x + 1 = 0$ $\Rightarrow (x - 1)^2 = 0$ $\Rightarrow \tan^2 \theta = 1$
Since $\theta$ is in the third quadrant,$\tan \theta = 1$ implies $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$
Then $\sin \theta = -\frac{1}{\sqrt{2}}$ and $\cos \theta = -\frac{1}{\sqrt{2}}$
Therefore,$(\sin \theta + \cos \theta)^3 = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})^3 = (-\frac{2}{\sqrt{2}})^3 = (-\sqrt{2})^3 = -2\sqrt{2}$

(D) Given equation: $\sin 7 \theta - \sin 3 \theta = \sin 4 \theta$
Using the identity $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get:
$2 \cos 5 \theta \sin 2 \theta = \sin 4 \theta$
Since $\sin 4 \theta = 2 \sin 2 \theta \cos 2 \theta$,the equation becomes:
$2 \cos 5 \theta \sin 2 \theta = 2 \sin 2 \theta \cos 2 \theta$
$2 \sin 2 \theta (\cos 5 \theta - \cos 2 \theta) = 0$
Case $1$: $\sin 2 \theta = 0$
$2 \theta = n \pi \Rightarrow \theta = \frac{n \pi}{2}$. For $\theta \in (0, \pi)$,$\theta = \frac{\pi}{2}$.
Case $2$: $\cos 5 \theta = \cos 2 \theta$
$5 \theta = 2 n \pi \pm 2 \theta$
If $5 \theta = 2 n \pi + 2 \theta$,then $3 \theta = 2 n \pi \Rightarrow \theta = \frac{2 n \pi}{3}$. For $\theta \in (0, \pi)$,$\theta = \frac{2 \pi}{3}$.
If $5 \theta = 2 n \pi - 2 \theta$,then $7 \theta = 2 n \pi \Rightarrow \theta = \frac{2 n \pi}{7}$. For $\theta \in (0, \pi)$,$\theta = \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}$.
The solutions are $\frac{\pi}{2}, \frac{2 \pi}{3}, \frac{2 \pi}{7}, \frac{4 \pi}{7}, \frac{6 \pi}{7}$.
Total number of solutions is $5$.

(B) Given: $\cos x + \cos y = \frac{2}{3}$ $(i)$
$\sin x - \sin y = \frac{3}{4}$ $(ii)$
Using sum-to-product formulas:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{2}{3}$ $(iii)$
$2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) = \frac{3}{4}$ $(iv)$
Dividing $(iv)$ by $(iii)$:
$\tan \left(\frac{x-y}{2}\right) = \frac{3/4}{2/3} = \frac{9}{8}$
Let $\theta = \frac{x-y}{2}$,so $\tan \theta = \frac{9}{8}$.
Then $\sin(x-y) = \sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2(9/8)}{1 + (81/64)} = \frac{9/4}{145/64} = \frac{9}{4} \times \frac{64}{145} = \frac{144}{145}$.
$\cos(x-y) = \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - (81/64)}{1 + (81/64)} = \frac{-17/64}{145/64} = -\frac{17}{145}$.
Therefore,$\sin(x-y) + \cos(x-y) = \frac{144}{145} - \frac{17}{145} = \frac{127}{145}$.

(D) Given equation: $4-3 \cos ^2 \theta=5 \sin \theta \cos \theta$
Divide by $\cos ^2 \theta$ (assuming $\cos \theta \neq 0$):
$4 \sec ^2 \theta-3=5 \tan \theta$
$4(1+\tan ^2 \theta)-3=5 \tan \theta$
$4 \tan ^2 \theta-5 \tan \theta+1=0$
$(4 \tan \theta-1)(\tan \theta-1)=0$
So,$\tan \theta=1$ or $\tan \theta=\frac{1}{4}$.
Checking option $D$: $1+\sin ^2 \theta=3 \cos ^2 \theta$
$1+\sin ^2 \theta=3(1-\sin ^2 \theta)$
$1+\sin ^2 \theta=3-3 \sin ^2 \theta$
$4 \sin ^2 \theta=2$ $\Rightarrow \sin ^2 \theta=\frac{1}{2}$ $\Rightarrow \tan ^2 \theta=1$.
Since $\tan \theta=1$ is a solution to the original equation,option $D$ is satisfied.

(A) We know that $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$.
Given $\sinh x = \frac{12}{5}$,we have $\frac{e^x - e^{-x}}{2} = \frac{12}{5}$,which implies $e^x - e^{-x} = \frac{24}{5}$.
Let $e^x = t$. Then $t - \frac{1}{t} = \frac{24}{5}$.
Multiplying by $5t$,we get $5t^2 - 24t - 5 = 0$.
Factoring the quadratic equation: $(5t + 1)(t - 5) = 0$.
Since $e^x = t$ must be positive,$t = 5$,so $e^x = 5$.
We need to find $\sinh 3x + \cosh 3x$.
Using the definitions,$\sinh 3x + \cosh 3x = \frac{e^{3x} - e^{-3x}}{2} + \frac{e^{3x} + e^{-3x}}{2} = e^{3x}$.
Since $e^x = 5$,$e^{3x} = (e^x)^3 = 5^3 = 125$.

(B) For set $A$: $\cos^2 x = \cos^2 \frac{\pi}{6} \implies \cos 2x = \cos 2(\frac{\pi}{6}) = \cos \frac{\pi}{3}$.
This gives $2x = 2n\pi \pm \frac{\pi}{3} \implies x = n\pi \pm \frac{\pi}{6}$.
Thus,$A = \{n\pi \pm \frac{\pi}{6} \mid n \in Z\}$.
For set $B$: $P + \frac{16}{P} = 10 \implies P^2 - 10P + 16 = 0 \implies (P-8)(P-2) = 0 \implies P = 8$ or $P = 2$.
Case $1$: $\cos^2 x = \log_{16} 8 = \log_{2^4} 2^3 = \frac{3}{4} \implies \cos x = \pm \frac{\sqrt{3}}{2} \implies x = n\pi \pm \frac{\pi}{6}$.
Case $2$: $\cos^2 x = \log_{16} 2 = \log_{2^4} 2^1 = \frac{1}{4} \implies \cos x = \pm \frac{1}{2} \implies x = n\pi \pm \frac{\pi}{3}$.
Thus,$B = \{n\pi \pm \frac{\pi}{6}, n\pi \pm \frac{\pi}{3} \mid n \in Z\}$.
$B - A$ removes the elements of $A$ from $B$,leaving $B - A = \{n\pi \pm \frac{\pi}{3} \mid n \in Z\}$.
Expanding for $n=2k$ and $n=2k+1$: $x = 2k\pi \pm \frac{\pi}{3}$ and $x = (2k+1)\pi \pm \frac{\pi}{3} = 2k\pi + \pi \pm \frac{\pi}{3} = 2k\pi \pm \frac{2\pi}{3}$.
Therefore,$B - A = \{x \in R \mid x = 2n\pi \pm \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in Z\}$.

(C) Given equation: $\cos^2 2x + \sin^2 3x = 1$
$\Rightarrow \sin^2 3x = 1 - \cos^2 2x$
$\Rightarrow \sin^2 3x = \sin^2 2x$
$\Rightarrow \sin^2 3x - \sin^2 2x = 0$
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:
$\Rightarrow \sin(3x + 2x) \sin(3x - 2x) = 0$
$\Rightarrow \sin 5x \sin x = 0$
Case $1$: $\sin 5x = 0$ $\Rightarrow 5x = n\pi$ $\Rightarrow x = \frac{n\pi}{5}, n \in \mathbb{Z}$
Case $2$: $\sin x = 0 \Rightarrow x = n\pi, n \in \mathbb{Z}$
Since $n\pi$ is a subset of $\frac{n\pi}{5}$ (by taking $n$ as a multiple of $5$),the general solution is $x = \frac{n\pi}{5}, n \in \mathbb{Z}$.

(D) Given that the origin is shifted to $(2, b)$,the transformation equations are $x = X + 2$ and $y = Y + b$.
Since the point $(a, 4)$ changes to $(6, 8)$,we have $a = 6 + 2 = 8$ and $4 = 8 + b$,which gives $b = -4$.
Now,the origin is shifted to $(a, b) = (8, -4)$,so the transformation equations are $x = X + 8$ and $y = Y - 4$.
Substituting these into the equation $x^2 + 4xy + y^2 = 0$:
$(X + 8)^2 + 4(X + 8)(Y - 4) + (Y - 4)^2 = 0$
$(X^2 + 16X + 64) + 4(XY - 4X + 8Y - 32) + (Y^2 - 8Y + 16) = 0$
$X^2 + 16X + 64 + 4XY - 16X + 32Y - 128 + Y^2 - 8Y + 16 = 0$
$X^2 + 4XY + Y^2 + 24Y - 48 = 0$
Comparing this with $X^2 + 2HXY + Y^2 + 2GX + 2FY + C = 0$,we get $2H = 4 \Rightarrow H = 2$,$2G = 0 \Rightarrow G = 0$,$2F = 24 \Rightarrow F = 12$,and $C = -48$.
Finally,$2H(G + F) = 2(2)(0 + 12) = 4(12) = 48$.
Since $C = -48$,$48 = -C$.

(C) Let the coordinates of $C$ be $(h, k)$.
The centroid $G$ of triangle $ABC$ is given by $\left( \frac{2+3+h}{3}, \frac{3+2+k}{3} \right) = \left( \frac{5+h}{3}, \frac{5+k}{3} \right)$.
Given that the distance of the centroid from the origin is $5$ units,we have $\sqrt{\left( \frac{5+h}{3} \right)^2 + \left( \frac{5+k}{3} \right)^2} = 5$.
Squaring both sides,we get $\left( \frac{5+h}{3} \right)^2 + \left( \frac{5+k}{3} \right)^2 = 25$.
Multiplying by $9$,we get $(h+5)^2 + (k+5)^2 = 225$.
Replacing $(h, k)$ with $(x, y)$,the locus of $C$ is $(x+5)^2 + (y+5)^2 = 15^2$.
This represents a circle with radius $r = 15$ units.
The diameter of the circle is $2r = 2 \times 15 = 30$ units.

(A) The slopes of the lines are $m_1 = -2$,$m_2 = -\frac{a}{b}$,and $m_3 = -\frac{1}{2}$.
Since $L_1$ and $L_2$ are equal sides,the angle between $L_1$ and $L_3$ must equal the angle between $L_2$ and $L_3$.
Using the formula $\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}|$,we have $|\frac{-2 - (-1/2)}{1 + (-2)(-1/2)}| = |\frac{-a/b - (-1/2)}{1 + (-a/b)(-1/2)}|$.
$|\frac{-3/2}{2}| = |\frac{-2a+b}{2b+a}| \Rightarrow \frac{3}{4} = |\frac{2a-b}{a+2b}|$.
Case $1$: $\frac{3}{4} = \frac{2a-b}{a+2b}$ $\Rightarrow 3a+6b = 8a-4b$ $\Rightarrow 5a = 10b$ $\Rightarrow a=2b$.
Since $(5,1)$ lies on $L_2$,$5a+b+c=0$ $\Rightarrow 10b+b+c=0$ $\Rightarrow c=-11b$.
Then $\frac{b^2}{|ac|} = \frac{b^2}{|(2b)(-11b)|} = \frac{b^2}{22b^2} = \frac{1}{22}$.
Case $2$: $-\frac{3}{4} = \frac{2a-b}{a+2b}$ $\Rightarrow -3a-6b = 8a-4b$ $\Rightarrow 11a = -2b$ $\Rightarrow a=-\frac{2b}{11}$.
Since $(5,1)$ lies on $L_2$,$5(-\frac{2b}{11})+b+c=0 \Rightarrow c = \frac{10b}{11}-b = -\frac{b}{11}$.
Then $\frac{b^2}{|ac|} = \frac{b^2}{|(-\frac{2b}{11})(-\frac{b}{11})|} = \frac{b^2}{2b^2/121} = \frac{121}{2}$.

(C) Let the vertices be $A(-2, -1)$,$B(2, 5)$,and $C(m, n)$. Let $H\left(2, \frac{5}{3}\right)$ be the orthocenter.
Slope of $AH = \frac{\frac{5}{3} - (-1)}{2 - (-2)} = \frac{8/3}{4} = \frac{2}{3}$.
Since $AH \perp BC$,the slope of $BC = -\frac{1}{2/3} = -\frac{3}{2}$.
Equation of $BC$: $y - 5 = -\frac{3}{2}(x - 2)$ $\Rightarrow 2y - 10 = -3x + 6$ $\Rightarrow 3x + 2y = 16 \quad ...(i)$
Slope of $BH = \frac{\frac{5}{3} - 5}{2 - 2} = \frac{-10/3}{0}$,which is undefined. This means $BH$ is a vertical line $x = 2$.
Since $BH \perp AC$,$AC$ must be a horizontal line. Since $A$ is $(-2, -1)$,the equation of $AC$ is $y = -1$.
Since $C(m, n)$ lies on $AC$,$n = -1$.
Substitute $n = -1$ into equation $(i)$: $3m + 2(-1) = 16$ $\Rightarrow 3m = 18$ $\Rightarrow m = 6$.
Thus,the third vertex is $(6, -1)$.
Therefore,$m + n = 6 + (-1) = 5$.

(C) Let the new coordinates be $(x', y')$. The transformation equations are $x = x' + \frac{3}{2}$ and $y = y' - 2$.
Substituting these into the given equation $2x^2 + 4xy + y^2 + 2x - 2y + 1 = 0$:
$2(x' + \frac{3}{2})^2 + 4(x' + \frac{3}{2})(y' - 2) + (y' - 2)^2 + 2(x' + \frac{3}{2}) - 2(y' - 2) + 1 = 0$
Expanding the terms:
$2((x')^2 + 3x' + \frac{9}{4}) + 4(x'y' - 2x' + \frac{3}{2}y' - 3) + ((y')^2 - 4y' + 4) + 2x' + 3 - 2y' + 4 + 1 = 0$
$2(x')^2 + 6x' + \frac{9}{2} + 4x'y' - 8x' + 6y' - 12 + (y')^2 - 4y' + 4 + 2x' - 2y' + 8 = 0$
Combining like terms:
$2(x')^2 + 4x'y' + (y')^2 + (6x' - 8x' + 2x') + (6y' - 4y' - 2y') + (\frac{9}{2} - 12 + 4 + 8) = 0$
$2(x')^2 + 4x'y' + (y')^2 + \frac{9}{2} = 0$
Multiplying by $2$ to clear the fraction:
$4(x')^2 + 8x'y' + 2(y')^2 + 9 = 0$.
Thus,the transformed equation is $4x^2 + 8xy + 2y^2 + 9 = 0$.

(A) The point $C$ divides the segment $AB$ in the ratio $a:b$. Using the section formula,the coordinates of $C$ are $\left(\frac{-2a + b}{a + b}, \frac{a + 2b}{a + b}\right)$.
Since $C$ lies on the line $2x + y - 3 = 0$,we have:
$2\left(\frac{-2a + b}{a + b}\right) + \left(\frac{a + 2b}{a + b}\right) - 3 = 0$
$-4a + 2b + a + 2b - 3(a + b) = 0$
$-3a + 4b - 3a - 3b = 0$
$b = 6a \Rightarrow \frac{b}{a} = 6$.
Now,substitute $\frac{b}{a} = 6$ into the coordinates of $P$ and $Q$:
$P = \left(\frac{6}{3}, -3\right) = (2, -3)$
$Q = \left(-3, -\frac{6}{3}\right) = (-3, -2)$
Point $C$ is $\left(\frac{-2a + 6a}{a + 6a}, \frac{a + 12a}{a + 6a}\right) = \left(\frac{4a}{7a}, \frac{13a}{7a}\right) = \left(\frac{4}{7}, \frac{13}{7}\right)$.
Let $C$ divide $PQ$ in the ratio $p:q$. Using the section formula for $x$-coordinate:
$\frac{-3p + 2q}{p + q} = \frac{4}{7}$
$-21p + 14q = 4p + 4q$
$10q = 25p \Rightarrow \frac{p}{q} = \frac{10}{25} = \frac{2}{5}$.
Then,$\frac{p}{q} + \frac{q}{p} = \frac{2}{5} + \frac{5}{2} = \frac{4 + 25}{10} = \frac{29}{10}$.

(C) Given equation: $ax^2-2xy+by^2-2x+4y+1=0$.
Shift origin to $(0, k)$,so $x=X$ and $y=Y+k$.
Substituting these in the equation: $aX^2-2X(Y+k)+b(Y+k)^2-2X+4(Y+k)+1=0$.
Expanding: $aX^2-2XY-2kX+bY^2+bk^2+2bkY-2X+4Y+4k+1=0$.
Grouping terms: $aX^2-2XY+bY^2-2X(k+1)+2Y(bk+2)+bk^2+4k+1=0$.
To remove first degree terms,coefficients of $X$ and $Y$ must be zero: $k+1=0 \Rightarrow k=-1$ and $bk+2=0$ $\Rightarrow -b+2=0$ $\Rightarrow b=2$.
Now,rotate axes by $\theta = \frac{1}{2} \tan^{-1}(2)$,so $\tan(2\theta) = 2$.
The $xy$ term in the rotated equation is removed if $\tan(2\theta) = \frac{2h}{a-b}$,where $h=-1$.
Thus,$\tan(2\theta) = \frac{2(-1)}{a-b} = \frac{-2}{a-b} = \frac{2}{b-a}$.
Given $\tan(2\theta) = 2$,we have $\frac{2}{b-a} = 2 \Rightarrow b-a = 1$.
Since $b=2$,$2-a=1 \Rightarrow a=1$.
Therefore,$a+b = 1+2 = 3$.

(B) Given the transformation of axes by shifting the origin to $(h, 5)$,the relations are $x = X + h$ and $y = Y + 5$.
Substituting these into the equation $y = x^3 - 9x^2 + cx - d$:
$Y + 5 = (X + h)^3 - 9(X + h)^2 + c(X + h) - d$
$Y + 5 = (X^3 + 3hX^2 + 3h^2X + h^3) - 9(X^2 + 2hX + h^2) + cX + ch - d$
$Y = X^3 + (3h - 9)X^2 + (3h^2 - 18h + c)X + (h^3 - 9h^2 + ch - d - 5)$
Comparing this with $Y = X^3$,the coefficients of $X^2$ and $X$ must be zero,and the constant term must be zero:
$1) 3h - 9 = 0 \Rightarrow h = 3$
$2) 3h^2 - 18h + c = 0$ $\Rightarrow 3(9) - 18(3) + c = 0$ $\Rightarrow 27 - 54 + c = 0$ $\Rightarrow c = 27$
$3) h^3 - 9h^2 + ch - d - 5 = 0$ $\Rightarrow 27 - 9(9) + 27(3) - d - 5 = 0$ $\Rightarrow 27 - 81 + 81 - d - 5 = 0$ $\Rightarrow d = 22$
Now,calculate $\left(d - \frac{c}{h}\right) = 22 - \frac{27}{3} = 22 - 9 = 13$.

(A) The line $L$ passes through $(-2, -3)$ and has an undefined slope,so its equation is $x = -2$ (a vertical line).
The angle between $x = -2$ and the line $ax - 2y + 3 = 0$ is $45^{\circ}$.
The slope of the line $ax - 2y + 3 = 0$ is $m_1 = \frac{a}{2}$.
The angle $\theta$ between a vertical line and a line with slope $m_1$ is given by $|\tan(90^{\circ} - \theta)| = |\frac{1}{m_1}|$.
Since the angle is $45^{\circ}$,we have $|\frac{1}{a/2}| = \tan(45^{\circ}) = 1$.
Thus,$\frac{2}{a} = 1$,which gives $a = 2$.
Now,substitute $a = 2$ into the equation $x + ay - 4 = 0$ to get $x + 2y - 4 = 0$.
The slope of this line is $m = -\frac{1}{2}$.
The angle $\theta$ made with the positive $X$-axis is given by $\tan \theta = -\frac{1}{2}$.
Since the slope is negative,the angle is in the second quadrant,so $\theta = \pi - \tan^{-1}\left(\frac{1}{2}\right)$.

(A) The slope of line $L$ is $-\cot \alpha$.
The slope of line $x + y + 1 = 0$ is $-1$.
Since the lines are perpendicular,the product of their slopes is $-1$:
$(-\cot \alpha)(-1) = -1$ $\Rightarrow \cot \alpha = -1$ $\Rightarrow \tan \alpha = -1$.
Since $\alpha$ lies in the fourth quadrant,$\alpha = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.
The perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to $x \cos \alpha + y \sin \alpha - p = 0$ is given by:
$\left| \sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha - p \right| = 5$.
Substituting $\alpha = \frac{7\pi}{4}$:
$\cos \frac{7\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{7\pi}{4} = -\frac{1}{\sqrt{2}}$.
$\left| \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) + \sqrt{2} \left( -\frac{1}{\sqrt{2}} \right) - p \right| = 5$.
$|1 - 1 - p| = 5 \Rightarrow |-p| = 5$.
Since $p$ is positive,$p = 5$.

(B) Let the lines be $L_1: 7x+y-24=0$ and $L_2: x+7y-24=0$. The slopes are $m_1 = -7$ and $m_2 = -\frac{1}{7}$.
Since the lines are equal sides of an isosceles triangle,the third side makes equal angles with $L_1$ and $L_2$.
Let the slope of the third side be $m$. Then $\left| \frac{m - (-7)}{1 + m(-7)} \right| = \left| \frac{m - (-1/7)}{1 + m(-1/7)} \right|$.
$\left| \frac{m+7}{1-7m} \right| = \left| \frac{7m+1}{7-m} \right|$.
Case $1$: $\frac{m+7}{1-7m} = \frac{7m+1}{7-m}$ $\Rightarrow (m+7)(7-m) = (7m+1)(1-7m)$ $\Rightarrow 49-m^2 = 1-49m^2$ $\Rightarrow 48m^2 = -48$ (No real solution).
Case $2$: $\frac{m+7}{1-7m} = -\frac{7m+1}{7-m}$ $\Rightarrow (m+7)(7-m) = -(7m+1)(1-7m)$ $\Rightarrow 49-m^2 = -(1-49m^2-7m+7m) = 49m^2-1$ $\Rightarrow 50m^2 = 50$ $\Rightarrow m = \pm 1$.
For $m = -1$,the line passing through $(-1, 1)$ is $y-1 = -1(x+1)$ $\Rightarrow y-1 = -x-1$ $\Rightarrow x+y=0$.
For $m = 1$,the line passing through $(-1, 1)$ is $y-1 = 1(x+1)$ $\Rightarrow y-1 = x+1$ $\Rightarrow x-y+2=0$.

(A) Let the points be $A(1, 2)$ and $B(-3, 5)$. The point $P(x, y)$ divides the segment $AB$ externally in the ratio $m:n = 4:3$.
Using the section formula for external division: $x = \frac{mx_2 - nx_1}{m - n}$ and $y = \frac{my_2 - ny_1}{m - n}$.
$x = \frac{4(-3) - 3(1)}{4 - 3} = \frac{-12 - 3}{1} = -15$.
$y = \frac{4(5) - 3(2)}{4 - 3} = \frac{20 - 6}{1} = 14$.
So,the point is $(-15, 14)$.
The equation of the line with slope $m = \frac{-2}{3}$ passing through $(-15, 14)$ is given by $y - y_1 = m(x - x_1)$.
$y - 14 = \frac{-2}{3}(x + 15)$.
$3y - 42 = -2x - 30$.
$2x + 3y - 12 = 0$.

(D) The given equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$.
Let the slopes of the two lines be $m$ and $3m$.
The equation of the pair of lines can be written as $(y - mx)(y - 3mx) = 0$.
Expanding this,we get $y^2 - 4mxy + 3m^2x^2 = 0$,or $3m^2x^2 - 4mxy + y^2 = 0$.
Dividing the original equation $2x^2 + hxy + 6y^2 = 0$ by $6$,we get $\frac{1}{3}x^2 + \frac{h}{6}xy + y^2 = 0$.
Comparing the coefficients of $x^2$ and $xy$ with $3m^2x^2 - 4mxy + y^2 = 0$,we have:
$3m^2 = \frac{1}{3}$ $\Rightarrow m^2 = \frac{1}{9}$ $\Rightarrow m = \pm \frac{1}{3}$.
Also,$-4m = \frac{h}{6} \Rightarrow h = -24m$.
Substituting $m = \pm \frac{1}{3}$ into the expression for $h$:
$h = -24(\pm \frac{1}{3}) = \mp 8$.
Thus,$h = \pm 8$.

(D) Given the equation: $y f(x+y) + \cos(mxy) = 1 + y f(x)$.
Rearranging the terms,we get: $y(f(x+y) - f(x)) = 1 - \cos(mxy)$.
Given $m=2$,the equation becomes: $y(f(x+y) - f(x)) = 1 - \cos(2xy)$.
Dividing by $y^2$ on both sides (assuming $y \neq 0$): $\frac{f(x+y) - f(x)}{y} = \frac{1 - \cos(2xy)}{y^2}$.
Using the identity $1 - \cos(2\theta) = 2 \sin^2(\theta)$,we have: $\frac{f(x+y) - f(x)}{y} = \frac{2 \sin^2(xy)}{y^2}$.
This can be written as: $\frac{f(x+y) - f(x)}{y} = 2 \left( \frac{\sin(xy)}{y} \right)^2$.
To find $f'(x)$,take the limit as $y \rightarrow 0$:
$f'(x) = \lim_{y \rightarrow 0} \frac{f(x+y) - f(x)}{y} = \lim_{y \rightarrow 0} 2 \left( \frac{\sin(xy)}{y} \right)^2$.
Multiply and divide by $x^2$: $f'(x) = 2x^2 \lim_{y \rightarrow 0} \left( \frac{\sin(xy)}{xy} \right)^2$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get: $f'(x) = 2x^2(1)^2 = 2x^2$.

(A) Given $y = 44 x^{45} + 45 x^{-44}$.
First derivative: $y^{\prime} = 44 \times 45 x^{44} + 45 \times (-44) x^{-45} = 1980(x^{44} - x^{-45})$.
Second derivative: $y^{\prime \prime} = 1980(44 x^{43} - (-45) x^{-46}) = 1980(44 x^{43} + 45 x^{-46})$.
Factor out $x^{-2}$ from the expression: $y^{\prime \prime} = 1980 \times x^{-2} (44 x^{45} + 45 x^{-44})$.
Since $y = 44 x^{45} + 45 x^{-44}$,we substitute $y$ into the expression: $y^{\prime \prime} = \frac{1980 y}{x^2}$.

(C) Given the equation: $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(4y^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(3y) + \frac{d}{dx}(4) = 0$.
$4x - 3(y + x \frac{dy}{dx}) + 8y \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
$4x - 3y - 3x \frac{dy}{dx} + 8y \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
Now,substitute the point $(3, 2)$ where $x = 3$ and $y = 2$:
$4(3) - 3(2) - 3(3) \frac{dy}{dx} + 8(2) \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
$12 - 6 - 9 \frac{dy}{dx} + 16 \frac{dy}{dx} + 2 - 3 \frac{dy}{dx} = 0$.
$8 + 4 \frac{dy}{dx} = 0$.
$4 \frac{dy}{dx} = -8$.
$\frac{dy}{dx} = -2$.

(C) Given equation: $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$
Dividing both sides by $(\cos x)^{\sin x}$,we get: $y = \frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}} = (\tan x)^{\sin x}$
Taking natural logarithm on both sides: $\ln y = \sin x \cdot \ln(\tan x)$
Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\tan x) + \sin x \cdot \frac{1}{\tan x} \cdot \sec^2 x$
Simplifying the second term: $\sin x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x$
So,$\frac{dy}{dx} = y [\cos x \cdot \ln(\tan x) + \sec x]$
At $x = \frac{\pi}{4}$,$\tan x = 1$,$\sin x = \frac{1}{\sqrt{2}}$,$\cos x = \frac{1}{\sqrt{2}}$,$\sec x = \sqrt{2}$,and $y = (1)^{1/\sqrt{2}} = 1$
Substituting these values: $\left. \frac{dy}{dx} \right|_{x=\frac{\pi}{4}} = 1 \cdot [\frac{1}{\sqrt{2}} \cdot \ln(1) + \sqrt{2}] = 1 \cdot [0 + \sqrt{2}] = \sqrt{2}$

(D) Given $x = \frac{9t^2}{1+t^4}$ and $y = \frac{16t^2}{1-t^4}$.
First,differentiate $y$ with respect to $t$ using the quotient rule:
$\frac{dy}{dt} = \frac{(1-t^4)(32t) - (16t^2)(-4t^3)}{(1-t^4)^2} = \frac{32t - 32t^5 + 64t^5}{(1-t^4)^2} = \frac{32t(1+t^4)}{(1-t^4)^2}$.
Next,differentiate $x$ with respect to $t$ using the quotient rule:
$\frac{dx}{dt} = \frac{(1+t^4)(18t) - (9t^2)(4t^3)}{(1+t^4)^2} = \frac{18t + 18t^5 - 36t^5}{(1+t^4)^2} = \frac{18t(1-t^4)}{(1+t^4)^2}$.
Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{32t(1+t^4)}{(1-t^4)^2} \times \frac{(1+t^4)^2}{18t(1-t^4)} = \frac{32(1+t^4)^3}{18(1-t^4)^3} = \frac{16}{9}\left(\frac{1+t^4}{1-t^4}\right)^3$.

(B) Given $x = \cos 2t + \log(\tan t)$ and $y = 2t + \cot 2t$.
First,differentiate $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(2t + \cot 2t) = 2 - 2\operatorname{cosec}^2 2t = -2(\operatorname{cosec}^2 2t - 1) = -2\cot^2 2t$.
Next,differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(\cos 2t + \log(\tan t)) = -2\sin 2t + \frac{1}{\tan t} \cdot \sec^2 t = -2\sin 2t + \frac{\cos t}{\sin t} \cdot \frac{1}{\cos^2 t} = -2\sin 2t + \frac{1}{\sin t \cos t}$.
Since $\sin 2t = 2\sin t \cos t$,we have $\frac{1}{\sin t \cos t} = \frac{2}{\sin 2t} = 2\operatorname{cosec} 2t$.
So,$\frac{dx}{dt} = -2\sin 2t + 2\operatorname{cosec} 2t = 2(\operatorname{cosec} 2t - \sin 2t) = 2\left(\frac{1 - \sin^2 2t}{\sin 2t}\right) = 2\frac{\cos^2 2t}{\sin 2t} = 2\cot 2t \cos 2t$.
Finally,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\cot^2 2t}{2\cot 2t \cos 2t} = -\frac{\cot 2t}{\cos 2t} = -\frac{\cos 2t}{\sin 2t} \cdot \frac{1}{\cos 2t} = -\frac{1}{\sin 2t} = -\operatorname{cosec} 2t$.

(C) Given $y=\log \left[\tan \sqrt{\frac{2^x-1}{2^x+1}}\right]$.
Let $v = \frac{2^x-1}{2^x+1}$. At $x=1$,$v = \frac{2-1}{2+1} = \frac{1}{3}$.
Using the quotient rule,$\frac{dv}{dx} = \frac{(2^x+1)(2^x \log 2) - (2^x-1)(2^x \log 2)}{(2^x+1)^2} = \frac{2^x \log 2 (2^x+1-2^x+1)}{(2^x+1)^2} = \frac{2 \cdot 2^x \log 2}{(2^x+1)^2} = \frac{2^{x+1} \log 2}{(2^x+1)^2}$.
At $x=1$,$\left(\frac{dv}{dx}\right)_{x=1} = \frac{2^2 \log 2}{(2+1)^2} = \frac{4 \log 2}{9}$.
Now,$y = \log(\tan \sqrt{v})$. Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\tan \sqrt{v}} \cdot \sec^2 \sqrt{v} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx}$.
Using $\frac{1}{\tan \sqrt{v}} \cdot \sec^2 \sqrt{v} = \frac{\cos \sqrt{v}}{\sin \sqrt{v}} \cdot \frac{1}{\cos^2 \sqrt{v}} = \frac{1}{\sin \sqrt{v} \cos \sqrt{v}} = \frac{2}{\sin(2\sqrt{v})}$.
So,$\frac{dy}{dx} = \frac{2}{\sin(2\sqrt{v})} \cdot \frac{1}{2\sqrt{v}} \cdot \frac{dv}{dx} = \frac{1}{\sin(2\sqrt{v}) \cdot \sqrt{v}} \cdot \frac{dv}{dx}$.
At $x=1$,$v = \frac{1}{3}$,so $\sqrt{v} = \frac{1}{\sqrt{3}}$.
$\left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{\sin(2/\sqrt{3}) \cdot (1/\sqrt{3})} \cdot \frac{4 \log 2}{9} = \frac{\sqrt{3} \cdot 4 \log 2}{9 \sin(2/\sqrt{3})} = \frac{4 \sqrt{3} \log 2}{9 \sin(2/\sqrt{3})}$.

(A) Given $\log y = y^{\log x}$.
Taking $\log$ on both sides:
$\log(\log y) = \log(y^{\log x}) = \log x \cdot \log y$.
Differentiating both sides with respect to $x$:
$\frac{1}{\log y} \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log y + \log x \cdot \frac{d}{dx}(\log y)$.
$\frac{1}{y \log y} \cdot \frac{dy}{dx} = \frac{\log y}{x} + \frac{\log x}{y} \cdot \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \frac{1}{y \log y} - \frac{\log x}{y} \right) = \frac{\log y}{x}$.
$\frac{dy}{dx} \left( \frac{1 - \log x \log y}{y \log y} \right) = \frac{\log y}{x}$.
$\frac{dy}{dx} = \frac{\log y}{x} \cdot \frac{y \log y}{1 - \log x \log y}$.
$\frac{dy}{dx} = \frac{y(\log y)^2}{x(1 - \log x \log y)}$.

(A) Given $y=\log \left(x-\sqrt{x^2-1}\right)$.
Taking exponential on both sides,$e^y = x-\sqrt{x^2-1}$.
Differentiating with respect to $x$:
$e^y y^{\prime} = 1 - \frac{1}{2\sqrt{x^2-1}} \cdot 2x = 1 - \frac{x}{\sqrt{x^2-1}} = \frac{\sqrt{x^2-1}-x}{\sqrt{x^2-1}} = -\frac{(x-\sqrt{x^2-1})}{\sqrt{x^2-1}} = -\frac{e^y}{\sqrt{x^2-1}}$.
Thus,$y^{\prime} = -\frac{1}{\sqrt{x^2-1}} = -(x^2-1)^{-1/2}$.
Differentiating again with respect to $x$:
$y^{\prime \prime} = -(-\frac{1}{2})(x^2-1)^{-3/2} \cdot 2x = \frac{x}{(x^2-1)^{3/2}}$.
Now,substitute into the expression $(x^2-1)y^{\prime \prime} + x y^{\prime}$:
$(x^2-1) \left( \frac{x}{(x^2-1)^{3/2}} \right) + x \left( -\frac{1}{\sqrt{x^2-1}} \right) = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1}} = 0$.

(C) Given the equation $y=\sqrt{\sin (\log_{2} x)+\sqrt{\sin (\log_{2} x)+\ldots \infty}}$.
Since the expression repeats infinitely,we can write it as $y=\sqrt{\sin (\log_{2} x)+y}$.
Squaring both sides,we get $y^2 = \sin (\log_{2} x) + y$,which simplifies to $y^2 - y = \sin (\log_{2} x)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\sin (\log_{2} x))$.
$(2y - 1) \frac{dy}{dx} = \cos (\log_{2} x) \cdot \frac{d}{dx}(\log_{2} x)$.
Using the change of base formula $\log_{2} x = \frac{\ln x}{\ln 2}$,we have $\frac{d}{dx}(\log_{2} x) = \frac{1}{x \ln 2}$.
Thus,$(2y - 1) \frac{dy}{dx} = \cos (\log_{2} x) \cdot \frac{1}{x \ln 2}$.
$\frac{dy}{dx} = \frac{\cos (\log_{2} x)}{x \ln 2 (2y - 1)}$.
Note: If the original expression meant $\log(2x)$,the derivative is $\frac{\cos(\log(2x))}{x(2y-1)}$. Given the options,the intended expression is $\log(2x)$.

(A) Let $u = (\sin x)^x$ and $v = x^{\sin x}$.
Taking log on both sides for $u$:
$\log u = x \log (\sin x)$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \log (\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log (\sin x) + x \cot x$.
$\frac{du}{dx} = (\sin x)^x [\log (\sin x) + x \cot x] = (\sin x)^{x-1} [\sin x \log (\sin x) + x \cos x]$.
Taking log on both sides for $v$:
$\log v = \sin x \log x$.
Differentiating with respect to $x$:
$\frac{1}{v} \frac{dv}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \frac{x \cos x \log x + \sin x}{x}$.
$\frac{dv}{dx} = x^{\sin x} \cdot \frac{x \cos x \log x + \sin x}{x} = x^{\sin x-1} [\sin x + x \cos x \log x]$.
Therefore,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\sin x)^{x-1} [\sin x \log (\sin x) + x \cos x]}{x^{\sin x-1} [\sin x + x \cos x \log x]}$.

(B) Given $y = \frac{\tan x \cos^{-1} x}{\sqrt{1-x^2}}$.
Using the product rule for differentiation: $\frac{dy}{dx} = \frac{d}{dx} [(\tan x) \cdot (\cos^{-1} x) \cdot (1-x^2)^{-1/2}]$.
Let $u = \tan x$,$v = \cos^{-1} x$,and $w = (1-x^2)^{-1/2}$.
Then $\frac{dy}{dx} = u'vw + uv'w + uvw'$.
$u' = \sec^2 x$,$v' = -\frac{1}{\sqrt{1-x^2}}$,$w' = -\frac{1}{2}(1-x^2)^{-3/2}(-2x) = \frac{x}{(1-x^2)^{3/2}}$.
At $x = 0$:
$u = \tan(0) = 0$,$u' = \sec^2(0) = 1$.
$v = \cos^{-1}(0) = \frac{\pi}{2}$,$v' = -\frac{1}{\sqrt{1-0}} = -1$.
$w = (1-0)^{-1/2} = 1$,$w' = \frac{0}{(1-0)^{3/2}} = 0$.
Substituting these values:
$\frac{dy}{dx} = (1)(\frac{\pi}{2})(1) + (0)(-1)(1) + (0)(\frac{\pi}{2})(0) = \frac{\pi}{2} + 0 + 0 = \frac{\pi}{2}$.

(D) Given $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$.
Let $v = \frac{6 x-2 x^2-4}{2 x^2-6 x+5}$.
Note that $v = \frac{-(2x^2-6x+5)+1}{2x^2-6x+5} = -1 + \frac{1}{2x^2-6x+5}$.
Using the chain rule,$\frac{dy}{dx} = \frac{-1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx}$.
Calculating $\frac{dv}{dx}$ using the quotient rule:
$\frac{dv}{dx} = \frac{(6-4x)(2x^2-6x+5) - (6x-2x^2-4)(4x-6)}{(2x^2-6x+5)^2} = \frac{2(3-2x)(2x^2-6x+5) + 2(2x^2-6x+4)(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)[-(2x^2-6x+5) + (2x^2-6x+4)]}{(2x^2-6x+5)^2} = \frac{2(2x-3)(-1)}{(2x^2-6x+5)^2} = \frac{-2(2x-3)}{(2x^2-6x+5)^2}$.
Now,$1-v^2 = 1 - \left(\frac{6x-2x^2-4}{2x^2-6x+5}\right)^2 = \frac{(2x^2-6x+5)^2 - (6x-2x^2-4)^2}{(2x^2-6x+5)^2} = \frac{(2x^2-6x+5 - 6x+2x^2+4)(2x^2-6x+5 + 6x-2x^2-4)}{(2x^2-6x+5)^2} = \frac{(4x^2-12x+9)(1)}{(2x^2-6x+5)^2} = \frac{(2x-3)^2}{(2x^2-6x+5)^2}$.
Thus,$\sqrt{1-v^2} = \frac{|2x-3|}{2x^2-6x+5}$.
Substituting back: $\frac{dy}{dx} = \frac{-1}{\frac{|2x-3|}{2x^2-6x+5}} \cdot \frac{-2(2x-3)}{(2x^2-6x+5)^2} = \frac{2(2x-3)}{|2x-3|(2x^2-6x+5)}$.
Assuming $2x-3 > 0$,$\frac{dy}{dx} = \frac{2}{2x^2-6x+5}$.

(D) Given $y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}\right]$.
Divide the numerator and denominator by $x^3$:
$y=\tan ^{-1}\left[\frac{(\frac{\sin 2x}{x})^3 - 3(\frac{\sin 2x}{x})}{3(\frac{\sin 2x}{x})^2 - 1}\right]$.
Let $\frac{\sin 2x}{x} = \tan \theta$. Then $y = \tan^{-1} \left[ \frac{\tan^3 \theta - 3 \tan \theta}{3 \tan^2 \theta - 1} \right]$.
Using the identity $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we have $\frac{\tan^3 \theta - 3 \tan \theta}{3 \tan^2 \theta - 1} = -\tan 3\theta$.
So,$y = \tan^{-1}(-\tan 3\theta) = -3\theta = -3 \tan^{-1}(\frac{\sin 2x}{x})$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = -3 \cdot \frac{1}{1 + (\frac{\sin 2x}{x})^2} \cdot \frac{d}{dx}(\frac{\sin 2x}{x})$.
$\frac{d}{dx}(\frac{\sin 2x}{x}) = \frac{x(2 \cos 2x) - \sin 2x}{x^2}$.
$\frac{dy}{dx} = -3 \cdot \frac{x^2}{x^2 + \sin^2 2x} \cdot \frac{2x \cos 2x - \sin 2x}{x^2} = \frac{3 \sin 2x - 6x \cos 2x}{x^2 + \sin^2 2x}$.
Note: The expression inside the bracket is $-\tan 3\theta$,so $y = -3\theta$. The derivative is $\frac{3 \sin 2x - 6x \cos 2x}{x^2 + \sin^2 2x}$. Comparing with options,option $D$ is the intended form if the sign inside the bracket was inverted.

(B) Given $y = a \cos 3x + b e^{-x}$.
First derivative: $y' = -3a \sin 3x - b e^{-x}$.
Second derivative: $y'' = -9a \cos 3x + b e^{-x}$.
Now,calculate $y''(3 \sin 3x - \cos 3x)$:
$y''(3 \sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3 \sin 3x - \cos 3x)$
$= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$.
We can rewrite this expression by manipulating the terms to match the form $10y' \cos 3x + 3y(\sin 3x + 3 \cos 3x)$:
$= 10(-3a \sin 3x - b e^{-x}) \cos 3x + 3a \cos 3x(\cos 3x + 3 \sin 3x) + 3b e^{-x}(\sin 3x + 3 \cos 3x)$
$= 10y' \cos 3x + 3(a \cos 3x + b e^{-x})(\sin 3x + 3 \cos 3x)$
$= 10y' \cos 3x + 3y(\sin 3x + 3 \cos 3x)$.

(B) The slope of the tangent to the curve $y=f(x)$ is given by $\frac{dy}{dx} = 6x^2+10x-9$.
Integrating both sides with respect to $x$,we get:
$f(x) = \int (6x^2+10x-9) dx = 2x^3+5x^2-9x+C$.
Given that $f(2)=0$,we substitute $x=2$ into the equation:
$f(2) = 2(2)^3 + 5(2)^2 - 9(2) + C = 0$
$16 + 20 - 18 + C = 0$
$18 + C = 0 \Rightarrow C = -18$.
Thus,the function is $f(x) = 2x^3+5x^2-9x-18$.
Now,we find $f(-2)$:
$f(-2) = 2(-2)^3 + 5(-2)^2 - 9(-2) - 18$
$f(-2) = 2(-8) + 5(4) + 18 - 18$
$f(-2) = -16 + 20 + 0 = 4$.

(B) Given the curve $y=x^3-2x^2+3x-4$ and the line $y=-2$.
At the point of intersection $P(h, k)$,we have $x^3-2x^2+3x-4 = -2$.
$\Rightarrow x^3-2x^2+3x-2 = 0$.
By testing values,for $x=1$,$1-2+3-2 = 0$. Thus,the point $P$ is $(1, -2)$.
Now,find the slope of the tangent at $P(1, -2)$:
$\frac{dy}{dx} = 3x^2-4x+3$.
At $x=1$,$\frac{dy}{dx} = 3(1)^2-4(1)+3 = 3-4+3 = 2$.
The equation of the tangent at $(1, -2)$ is $y - (-2) = 2(x - 1)$.
$\Rightarrow y+2 = 2x-2
\Rightarrow y = 2x-4$.
This tangent meets the $X$-axis where $y=0$:
$0 = 2x_1 - 4
\Rightarrow 2x_1 = 4
\Rightarrow x_1 = 2$.

(D) Given the curve $y^3=4 x^5$.
Differentiating both sides with respect to $x$,we get $3 y^2 \frac{d y}{d x} = 20 x^4$.
Thus,$\frac{d y}{d x} = \frac{20 x^4}{3 y^2}$.
At the point $(4, 16)$,the slope of the tangent is $m_t = \frac{20(4)^4}{3(16)^2} = \frac{20 \times 256}{3 \times 256} = \frac{20}{3}$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{3}{20}$.
The equation of the normal at $(4, 16)$ is $y - 16 = -\frac{3}{20}(x - 4)$.
Multiplying by $20$,we get $20y - 320 = -3x + 12$.
Rearranging the terms,we get $3x + 20y = 332$.

(C) Let $m_1$ be the slope of the tangent to the curve $y=f(x)$ at point $P$. Given $m_1 = \frac{dy}{dx} = Q(x)$.
Let $m_2$ be the slope of the tangent to the curve $x=g(y)$ at point $P$. Given $\frac{dx}{dy} = -Q(x)$,we know that $m_2 = \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{-Q(x)}$.
Since $m_1 \times m_2 = Q(x) \times \left(-\frac{1}{Q(x)}\right) = -1$,the product of the slopes of the tangents at the point of intersection is $-1$.
Therefore,the tangent drawn at $P$ to one curve is normal to the other curve at $P$.

(B) Given curve: $y^3 - 3xy + 2 = 0$.
Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$.
$\frac{dy}{dx} (3y^2 - 3x) = 3y \implies \frac{dy}{dx} = \frac{y}{y^2 - x}$.
For a horizontal tangent,$\frac{dy}{dx} = 0 \implies y = 0$.
Substituting $y = 0$ into the curve equation: $0^3 - 3x(0) + 2 = 0 \implies 2 = 0$,which is impossible. Thus,$n(A) = 0$.
For a vertical tangent,$\frac{dy}{dx} = \infty \implies y^2 - x = 0 \implies x = y^2$.
Substituting $x = y^2$ into the curve equation: $y^3 - 3(y^2)y + 2 = 0 \implies y^3 - 3y^3 + 2 = 0 \implies -2y^3 = -2 \implies y^3 = 1 \implies y = 1$.
If $y = 1$,then $x = 1^2 = 1$. The point is $(1, 1)$. Thus,$n(B) = 1$.
Therefore,$n(A) + n(B) = 0 + 1 = 1$.

(D) Let $f(x) = y = x^4+5x^3+9x^2+6x+2$.
The slope of the tangent to the curve is given by $m(x) = \frac{dy}{dx} = 4x^3+15x^2+18x+6$.
For the slope $m(x)$ to be increasing,its derivative must be greater than or equal to zero,i.e.,$\frac{dm}{dx} = \frac{d^2y}{dx^2} \ge 0$.
Calculating the second derivative: $\frac{d^2y}{dx^2} = 12x^2+30x+18$.
Factoring the quadratic expression: $12x^2+30x+18 = 6(2x^2+5x+3) = 6(2x+3)(x+1)$.
We need $6(2x+3)(x+1) \ge 0$.
The critical points are $x = -\frac{3}{2}$ and $x = -1$.
Testing the intervals:
$1$) For $x \in (-\infty, -\frac{3}{2}]$,$\frac{d^2y}{dx^2} \ge 0$.
$2$) For $x \in [-\frac{3}{2}, -1]$,$\frac{d^2y}{dx^2} \le 0$.
$3$) For $x \in [-1, \infty)$,$\frac{d^2y}{dx^2} \ge 0$.
Thus,the slope increases in the interval $(-\infty, -\frac{3}{2}] \cup [-1, \infty)$,which can be written as $R - (-\frac{3}{2}, -1)$.

(D) Given the function $y = 2x^3 - 8x^2 + 10x - 4$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 6x^2 - 16x + 10$.
Since the tangent is parallel to the $X$-axis,its slope must be zero:
$\frac{dy}{dx} = 0 \implies 6x^2 - 16x + 10 = 0$.
Dividing by $2$,we get $3x^2 - 8x + 5 = 0$.
Factoring the quadratic equation:
$3x^2 - 3x - 5x + 5 = 0 \implies 3x(x - 1) - 5(x - 1) = 0$.
$(3x - 5)(x - 1) = 0$.
This gives $x = 1$ or $x = \frac{5}{3}$.
Since the point $(a, b)$ satisfies $a \in (1, 2)$,we discard $x = 1$ and accept $a = \frac{5}{3}$.

(B) Given the curve $x = 2(\cos 2t + t \sin 2t)$ and $y = 4(\sin 2t - t \cos 2t)$.
First,find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = 2(-2 \sin 2t + \sin 2t + 2t \cos 2t) = 2(2t \cos 2t - \sin 2t)$.
$\frac{dy}{dt} = 4(2 \cos 2t - \cos 2t + 2t \sin 2t) = 4(\cos 2t + 2t \sin 2t)$.
At $t = \frac{\pi}{4}$:
$\frac{dx}{dt} = 2(2(\frac{\pi}{4}) \cos \frac{\pi}{2} - \sin \frac{\pi}{2}) = 2(0 - 1) = -2$.
$\frac{dy}{dt} = 4(\cos \frac{\pi}{2} + 2(\frac{\pi}{4}) \sin \frac{\pi}{2}) = 4(0 + \frac{\pi}{2}) = 2\pi$.
Therefore,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\pi}{-2} = -\pi$.
The slope of the normal is $-\frac{1}{dy/dx} = \frac{1}{\pi}$.
At $t = \frac{\pi}{4}$,$y = 4(\sin \frac{\pi}{2} - \frac{\pi}{4} \cos \frac{\pi}{2}) = 4(1 - 0) = 4$.
The length of the normal is given by $|y| \sqrt{1 + (\frac{dy}{dx})^2} = |4| \sqrt{1 + (-\pi)^2} = 4 \sqrt{1 + \pi^2}$.

(C) The distance $S$ is given by the function $S(t) = t^3 - t^2 - t + 3$.
The velocity $v$ of the particle is the rate of change of distance with respect to time,given by $v = \frac{dS}{dt}$.
$v = \frac{d}{dt}(t^3 - t^2 - t + 3) = 3t^2 - 2t - 1$.
The particle comes to rest when its velocity $v = 0$.
Setting $v = 0$,we get $3t^2 - 2t - 1 = 0$.
Factoring the quadratic equation: $3t^2 - 3t + t - 1 = 0 \Rightarrow 3t(t - 1) + 1(t - 1) = 0 \Rightarrow (3t + 1)(t - 1) = 0$.
Since time $t$ cannot be negative,we take $t = 1$ second.
Now,substitute $t = 1$ into the distance equation $S(t)$:
$S(1) = (1)^3 - (1)^2 - (1) + 3 = 1 - 1 - 1 + 3 = 2$ metres.
Thus,the distance travelled by the particle when it comes to rest is $2$ metres.

(A) Given radius $r = 7 \text{ cm}$ and error in surface area $dA = 0.08 \text{ cm}^2$.
The surface area of a sphere is $A = 4 \pi r^2$.
Differentiating with respect to $r$,we get $\frac{dA}{dr} = 8 \pi r$.
Thus,$dA = 8 \pi r \cdot dr$.
Substituting the values,$0.08 = 8 \pi (7) \cdot dr$.
$dr = \frac{0.08}{56 \pi} = \frac{0.01}{7 \pi} \text{ cm}$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$.
The approximate error in volume is $dV = \frac{dV}{dr} \cdot dr$.
$dV = (4 \pi r^2) \cdot \left( \frac{0.01}{7 \pi} \right)$.
Substituting $r = 7$,$dV = 4 \pi (7^2) \cdot \frac{0.01}{7 \pi} = 4 \times 7 \times 0.01 = 0.28 \text{ cm}^3$.

(D) Let $R = 3$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.
Let $x$ be the distance from the center of the sphere to the base of the cone. Then $h = R + x = 3 + x$.
In the right triangle formed by the radius of the sphere,the radius of the cone base,and the distance $x$,we have $r^2 + x^2 = R^2 = 3^2 = 9$,so $r^2 = 9 - x^2$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (9 - x^2)(3 + x)$.
$V = \frac{\pi}{3} (27 + 9x - 3x^2 - x^3)$.
To maximize $V$,we find $\frac{dV}{dx} = \frac{\pi}{3} (9 - 6x - 3x^2) = 0$.
$x^2 + 2x - 3 = 0 \Rightarrow (x + 3)(x - 1) = 0$.
Since $x$ is a distance,$x = 1$ (as $x \neq -3$).
For $x = 1$,$r^2 = 9 - 1^2 = 8$,so $r = \sqrt{8} = 2\sqrt{2}$.
The height $h = 3 + 1 = 4$.
The semi-vertical angle $\theta$ satisfies $\tan \theta = \frac{r}{h} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$.
Thus,$\theta = \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

(C) Let $r$ be the radius and $h$ be the height of the water level in the cone at any time $t$.
The vertical angle is $60^{\circ}$,so the semi-vertical angle is $30^{\circ}$.
From the geometry of the cone,$\frac{r}{h} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$,which implies $h = \sqrt{3}r$.
The volume $V$ of the water in the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting $h = \sqrt{3}r$,we get $V = \frac{1}{3} \pi r^2 (\sqrt{3}r) = \frac{\sqrt{3}}{3} \pi r^3 = \frac{1}{\sqrt{3}} \pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = \frac{1}{\sqrt{3}} \pi (3r^2) \frac{dr}{dt} = \sqrt{3} \pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = \frac{1}{\sqrt{3}}$,we have $\sqrt{3} \pi r^2 \frac{dr}{dt} = \frac{1}{\sqrt{3}}$,so $\frac{dr}{dt} = \frac{1}{3 \pi r^2}$.
Since $h = \sqrt{3}r$,when $h = 3 \text{ m}$,$r = \frac{3}{\sqrt{3}} = \sqrt{3} \text{ m}$.
Substituting $r = \sqrt{3}$ into the expression for $\frac{dr}{dt}$,we get $\frac{dr}{dt} = \frac{1}{3 \pi (\sqrt{3})^2} = \frac{1}{3 \pi (3)} = \frac{1}{9 \pi} \text{ m/min}$.

(B) To find the angle between the curves $y^2 = x$ and $x^2 + y^2 = 2$,we first find their point of intersection.
Substituting $y^2 = x$ into $x^2 + y^2 = 2$,we get $x^2 + x - 2 = 0$.
$(x + 2)(x - 1) = 0$,which gives $x = 1$ (since $x = -2$ is not possible for $y^2 = x$).
For $x = 1$,$y^2 = 1$,so $y = 1$ (considering the first quadrant intersection).
Now,differentiate the curves to find the slopes of the tangents at $(1, 1)$:
For $y^2 = x$,$2y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$. At $(1, 1)$,$m_1 = \frac{1}{2}$.
For $x^2 + y^2 = 2$,$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. At $(1, 1)$,$m_2 = -1$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{\frac{1}{2} - (-1)}{1 + (\frac{1}{2})(-1)} \right| = \left| \frac{\frac{3}{2}}{\frac{1}{2}} \right| = 3$.

(A) Let $f(x) = x^{1/3}$. We want to find the approximate value of $f(730)$.
We know that $729 = 9^3$,so let $x = 729$ and $\Delta x = 1$.
Using the formula for differentials,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,$f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}}$.
At $x = 729$,$f(729) = (729)^{1/3} = 9$.
$f'(729) = \frac{1}{3(729)^{2/3}} = \frac{1}{3(9^2)} = \frac{1}{3 \times 81} = \frac{1}{243}$.
Therefore,$f(730) \approx 9 + \frac{1}{243} \times 1$.
$f(730) \approx 9 + 0.004115... \approx 9.0041$.

(A) Given the curve equation: $x^3 y^4 = 2^7$.
Differentiating both sides with respect to time $t$:
$3x^2 y^4 \frac{dx}{dt} + 4x^3 y^3 \frac{dy}{dt} = 0$.
We are given $\frac{dx}{dt} = -8 \text{ units/sec}$ (since it is decreasing).
At the point $(2, 2)$,substitute $x = 2, y = 2$ and $\frac{dx}{dt} = -8$:
$3(2)^2 (2)^4 (-8) + 4(2)^3 (2)^3 \frac{dy}{dt} = 0$.
$3(4)(16)(-8) + 4(8)(8) \frac{dy}{dt} = 0$.
$-1536 + 256 \frac{dy}{dt} = 0$.
$256 \frac{dy}{dt} = 1536$.
$\frac{dy}{dt} = \frac{1536}{256} = 6$.
Since $\frac{dy}{dt} > 0$,the $y$-coordinate increases at the rate of $6 \text{ units per second}$.

(C) Let $f(x) = \sec x$. Then $f^{\prime}(x) = \sec x \tan x$.
We take $a = 60^{\circ}$ and $h = -1^{\circ} = -0.0174$ radians.
At $a = 60^{\circ}$,$f(a) = \sec 60^{\circ} = 2$.
Also,$f^{\prime}(a) = \sec 60^{\circ} \tan 60^{\circ} = 2 \times \sqrt{3} = 2 \times 1.732 = 3.464$.
Using the linear approximation formula $f(a+h) \approx f(a) + h \cdot f^{\prime}(a)$:
$f(59^{\circ}) \approx f(60^{\circ}) + (-0.0174) \times f^{\prime}(60^{\circ})$.
$f(59^{\circ}) \approx 2 + (-0.0174) \times (3.464)$.
$f(59^{\circ}) \approx 2 - 0.0602736$.
$f(59^{\circ}) \approx 1.9397264$.
Rounding to four decimal places,we get $1.9397$.

(C) Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$.
Differentiating both sides with respect to $A$,we get: $2a \delta a = 2bc \sin A \delta A$.
Thus,$\delta a = \frac{bc \sin A \delta A}{a}$.
We know that the area of the triangle is $\Delta = \frac{1}{2} bc \sin A$,so $bc \sin A = 2\Delta$.
Also,by the Sine Rule,$\frac{a}{\sin A} = 2R$,which implies $a = 2R \sin A$.
Substituting these into the expression for $\delta a$: $\delta a = \frac{2\Delta \delta A}{2R \sin A} = \frac{\Delta \delta A}{R \sin A}$.
The percentage error is $\frac{\delta a}{a} \times 100 = \frac{\Delta \delta A}{R \sin A \cdot 2R \sin A} \times 100 = \frac{\Delta \delta A}{2R^2 \sin^2 A} \times 100$.

(C) The volume $V$ of a cylinder is given by $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height (water level).
Given $r = 3.5 \ ft = \frac{7}{2} \ ft$ and the rate of change of volume $\frac{dV}{dt} = 1 \ ft^3/min$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Substituting the values: $1 = \pi \times (\frac{7}{2})^2 \times \frac{dh}{dt}$.
$1 = \frac{22}{7} \times \frac{49}{4} \times \frac{dh}{dt}$.
$1 = \frac{11 \times 7}{2} \times \frac{dh}{dt} = \frac{77}{2} \times \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{2}{77} \ ft/min$.

(A) Given $y = f(x) = 2x^2 - 3x + 4$,$x = 5$,and $\delta x = 0.02$.
First,calculate the actual error $\delta y = f(x + \delta x) - f(x)$:
$\delta y = f(5.02) - f(5) = [2(5.02)^2 - 3(5.02) + 4] - [2(5)^2 - 3(5) + 4]$
$= [2(25.2004) - 15.06 + 4] - [50 - 15 + 4]$
$= [50.4008 - 15.06 + 4] - 39 = 39.3408 - 39 = 0.3408$.
Next,calculate the differential $dy = f'(x) \cdot dx$:
$f'(x) = \frac{d}{dx}(2x^2 - 3x + 4) = 4x - 3$.
At $x = 5$,$f'(5) = 4(5) - 3 = 17$.
$dy = 17 \times 0.02 = 0.34$.
Finally,the difference $\delta y - dy = 0.3408 - 0.34 = 0.0008$.

(A) The domain of the function $f(x) = \log \left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2}$ is determined by $\frac{1+x}{1-x} > 0$,which implies $x \in (-1, 1)$.
First,we simplify the expression for $f(x)$:
$f(x) = \log(1+x) - \log(1-x) - 2x - \frac{x^3}{1-x^2}$.
Now,we find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x} + \frac{1}{1-x} - 2 - \frac{d}{dx} \left( \frac{x^3}{1-x^2} \right)$.
$f'(x) = \frac{(1-x) + (1+x)}{1-x^2} - 2 - \frac{3x^2(1-x^2) - x^3(-2x)}{(1-x^2)^2}$.
$f'(x) = \frac{2}{1-x^2} - 2 - \frac{3x^2 - 3x^4 + 2x^4}{(1-x^2)^2}$.
$f'(x) = \frac{2(1-x^2) - 2(1-x^2)^2 - (3x^2 - x^4)}{(1-x^2)^2}$.
$f'(x) = \frac{2 - 2x^2 - 2(1 - 2x^2 + x^4) - 3x^2 + x^4}{(1-x^2)^2}$.
$f'(x) = \frac{2 - 2x^2 - 2 + 4x^2 - 2x^4 - 3x^2 + x^4}{(1-x^2)^2} = \frac{-x^4 - x^2}{(1-x^2)^2} = -\frac{x^2(x^2+1)}{(1-x^2)^2}$.
Since $x^2(x^2+1) > 0$ for all $x \in (-1, 1) \setminus \{0\}$ and $(1-x^2)^2 > 0$,$f'(x) < 0$ for all $x \in (-1, 1)$.
Thus,the function is decreasing on the entire interval $(-1, 1)$.
Here,$a = -1$ and $b = 1$,so $\frac{a}{b} = \frac{-1}{1} = -1$.

(A) Given the function $f(x) = kx^3 - 3x^2 - 12x + 8$.
For $f(x)$ to be strictly decreasing for all $x \in R$,we must have $f'(x) < 0$ for all $x \in R$.
Calculating the derivative: $f'(x) = 3kx^2 - 6x - 12$.
For a quadratic expression $ax^2 + bx + c$ to be strictly negative for all $x$,we require $a < 0$ and the discriminant $D < 0$.
Here,$a = 3k$,$b = -6$,and $c = -12$.
Condition $1$: $3k < 0 \Rightarrow k < 0$.
Condition $2$: $D = b^2 - 4ac < 0$.
$(-6)^2 - 4(3k)(-12) < 0$.
$36 + 144k < 0$.
$144k < -36$.
$k < -\frac{36}{144} \Rightarrow k < -\frac{1}{4}$.
Since $k < -\frac{1}{4}$ already satisfies $k < 0$,the final condition is $k < -\frac{1}{4}$.

(D) Given $f(x) = (2x-1)(3x+2)(4x-3)$.
First,check the conditions of Rolle's Theorem:
$f(\frac{1}{2}) = (2(\frac{1}{2})-1)(...) = 0 \times (...) = 0$.
$f(\frac{3}{4}) = (...)(4(\frac{3}{4})-3) = (...)(3-3) = 0$.
Since $f(\frac{1}{2}) = f(\frac{3}{4}) = 0$ and $f(x)$ is a polynomial,it is continuous and differentiable.
Expanding $f(x)$:
$f(x) = (6x^2 + 4x - 3x - 2)(4x-3) = (6x^2 + x - 2)(4x-3)$
$f(x) = 24x^3 - 18x^2 + 4x^2 - 3x - 8x + 6 = 24x^3 - 14x^2 - 11x + 6$.
Now,find $f'(x)$:
$f'(x) = 72x^2 - 28x - 11$.
Set $f'(c) = 0$:
$72c^2 - 28c - 11 = 0$.
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$c = \frac{28 \pm \sqrt{(-28)^2 - 4(72)(-11)}}{2(72)} = \frac{28 \pm \sqrt{784 + 3168}}{144} = \frac{28 \pm \sqrt{3952}}{144}$.
$c = \frac{28 \pm 4\sqrt{247}}{144} = \frac{7 \pm \sqrt{247}}{36}$.
Since $\sqrt{247} \approx 15.7$,$c_1 = \frac{7+15.7}{36} \approx 0.63$ and $c_2 = \frac{7-15.7}{36} \approx -0.24$.
Since $c \in [0.5, 0.75]$,only $c = \frac{7+\sqrt{247}}{36}$ is valid.

(B) Given $f(x) = x^3 + ax^2 + bx + 40$.
Since $-5$ and $4$ are roots of $f(x) = 0$,we have $f(-5) = 0$ and $f(4) = 0$.
$f(-5) = (-5)^3 + a(-5)^2 + b(-5) + 40 = -125 + 25a - 5b + 40 = 25a - 5b - 85 = 0 \Rightarrow 5a - b = 17$ $(i)$.
$f(4) = (4)^3 + a(4)^2 + b(4) + 40 = 64 + 16a + 4b + 40 = 16a + 4b + 104 = 0 \Rightarrow 4a + b = -26$ $(ii)$.
Adding $(i)$ and $(ii)$,we get $9a = -9 \Rightarrow a = -1$.
Substituting $a = -1$ in $(i)$,$5(-1) - b = 17 \Rightarrow -5 - b = 17 \Rightarrow b = -22$.
Thus,$f(x) = x^3 - x^2 - 22x + 40$.
By Rolle's theorem,there exists $c \in (-5, 4)$ such that $f'(c) = 0$.
$f'(x) = 3x^2 - 2x - 22$.
Setting $f'(c) = 0$,we get $3c^2 - 2c - 22 = 0$.
Using the quadratic formula $c = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-22)}}{2(3)} = \frac{2 \pm \sqrt{4 + 264}}{6} = \frac{2 \pm \sqrt{268}}{6} = \frac{2 \pm 2\sqrt{67}}{6} = \frac{1 \pm \sqrt{67}}{3}$.
Since $c \in (-5, 4)$,the value $\frac{1+\sqrt{67}}{3} \approx \frac{1+8.18}{3} \approx 3.06$ lies in the interval.
Thus,one of the values of $c$ is $\frac{1+\sqrt{67}}{3}$.

(D) For Rolle's Theorem to be applicable on $[0, 1]$,the function $f(x)$ must be continuous on $[0, 1]$.
Since $f(x)$ is continuous for $x \in (0, 1]$,we check continuity at $x = 0$:
$\lim_{x \to 0^+} f(x) = f(0) = 0$.
$\lim_{x \to 0^+} x^p \log x = 0$.
Using $L$'$H$ôpital's Rule:
$\lim_{x \to 0^+} \frac{\log x}{x^{-p}} = \lim_{x \to 0^+} \frac{1/x}{-p x^{-p-1}} = \lim_{x \to 0^+} \frac{x^p}{-p}$.
This limit is $0$ only if $p > 0$.
Also,for Rolle's Theorem,$f(0) = f(1)$ must hold.
$f(0) = 0$ and $f(1) = 1^p \log 1 = 0$.
Thus,$f(0) = f(1) = 0$ is satisfied for any $p > 0$.
Among the given options,$p = 1$ is the only value that satisfies $p > 0$.

(B) Let $f(x) = 4+3x-7x^2$.
Taking the derivative,$f'(x) = 3-14x$. Setting $f'(x) = 0$,we get $x = \frac{3}{14} = \alpha$.
Since $f''(x) = -14 < 0$,$f(x)$ attains its maximum at $x = \alpha = \frac{3}{14}$.
The maximum value $M = f(\frac{3}{14}) = 4 + 3(\frac{3}{14}) - 7(\frac{3}{14})^2 = 4 + \frac{9}{14} - \frac{63}{196} = 4 + \frac{126-63}{196} = 4 + \frac{63}{196} = 4 + \frac{9}{28} = \frac{112+9}{28} = \frac{121}{28}$.
Now,let $g(x) = 5x^2-2x+1$.
Taking the derivative,$g'(x) = 10x-2$. Setting $g'(x) = 0$,we get $x = \frac{1}{5} = \beta$.
Since $g''(x) = 10 > 0$,$g(x)$ attains its minimum at $x = \beta = \frac{1}{5}$.
The minimum value $m = g(\frac{1}{5}) = 5(\frac{1}{5})^2 - 2(\frac{1}{5}) + 1 = \frac{1}{5} - \frac{2}{5} + 1 = \frac{4}{5}$.
Finally,calculate $\frac{28(M-\alpha)}{5(m+\beta)} = \frac{28(\frac{121}{28} - \frac{3}{14})}{5(\frac{4}{5} + \frac{1}{5})} = \frac{28(\frac{121-6}{28})}{5(1)} = \frac{115}{5} = 23$.

(C) Given $x+y=24$,so $y=24-x$.
Let $P = x^3 y^5 = x^3(24-x)^5$.
To find the maximum,we differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = 3x^2(24-x)^5 + x^3 \cdot 5(24-x)^4(-1) = 0$.
$x^2(24-x)^4 [3(24-x) - 5x] = 0$.
$x^2(24-x)^4 [72 - 3x - 5x] = 0$.
$x^2(24-x)^4 (72 - 8x) = 0$.
Since $x$ and $y$ are positive integers,$x \neq 0$ and $x \neq 24$.
Thus,$72 - 8x = 0 \Rightarrow x = 9$.
Then $y = 24 - 9 = 15$.
Finally,$x^2 + y^2 = 9^2 + 15^2 = 81 + 225 = 306$.

(C) Let $y = \frac{x^2-x+1}{x^2+x+1}$.
Then $y(x^2+x+1) = x^2-x+1$,which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.
For $x$ to be real,the discriminant $D \ge 0$.
$D = (y+1)^2 - 4(y-1)^2 \ge 0$.
$(y+1-2(y-1))(y+1+2(y-1)) \ge 0$.
$(-y+3)(3y-1) \ge 0$.
$(y-3)(3y-1) \le 0$.
Thus,$\frac{1}{3} \le y \le 3$.
The maximum value is $3$ and the minimum value is $\frac{1}{3}$.
The sum of the maximum and minimum values is $3 + \frac{1}{3} = \frac{10}{3}$.

(A) Given the function $f(x)=2x^3+9x^2+12x+1$ on the interval $[-3,0]$.
First,find the derivative $f'(x)$:
$f'(x)=6x^2+18x+12=6(x^2+3x+2)=6(x+1)(x+2)$.
Setting $f'(x)=0$,we get critical points $x=-1$ and $x=-2$.
Now,evaluate $f(x)$ at the critical points and the endpoints of the interval $[-3,0]$:
$f(-3)=2(-27)+9(9)+12(-3)+1 = -54+81-36+1 = -8$.
$f(-2)=2(-8)+9(4)+12(-2)+1 = -16+36-24+1 = -3$.
$f(-1)=2(-1)+9(1)+12(-1)+1 = -2+9-12+1 = -4$.
$f(0)=2(0)+9(0)+12(0)+1 = 1$.
Comparing these values,the absolute minimum $m = -8$ and the absolute maximum $M = 1$.
Therefore,$m+M = -8+1 = -7$.

(B) Let $I = \int (1+x-x^{-1}) e^{x+x^{-1}} dx$.
We can rewrite the integral as:
$I = \int e^{x+x^{-1}} dx + \int x(1-x^{-2}) e^{x+x^{-1}} dx$.
Observe that the derivative of $e^{x+x^{-1}}$ is $(1-x^{-2}) e^{x+x^{-1}}$.
Using the integration by parts formula $\int u dv = uv - \int v du$,let $u = x$ and $dv = (1-x^{-2}) e^{x+x^{-1}} dx$.
Then $du = dx$ and $v = e^{x+x^{-1}}$.
Thus,$\int x(1-x^{-2}) e^{x+x^{-1}} dx = x e^{x+x^{-1}} - \int e^{x+x^{-1}} dx$.
Substituting this back into the expression for $I$:
$I = \int e^{x+x^{-1}} dx + x e^{x+x^{-1}} - \int e^{x+x^{-1}} dx + c = x e^{x+x^{-1}} + c$.
Therefore,$f(x) = x e^{x+x^{-1}}$.
Now,calculate $f(1) - f(-1)$:
$f(1) = 1 \cdot e^{1+1} = e^2$.
$f(-1) = -1 \cdot e^{-1-1} = -e^{-2}$.
$f(1) - f(-1) = e^2 - (-e^{-2}) = e^2 + e^{-2}$.

(A) Let $I = \int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x$.
We express the numerator as $A(\text{denominator}) + B(\frac{d}{dx}(\text{denominator}))$.
Let $2 \cos 3 x-3 \sin 3 x = A(\cos 3 x+2 \sin 3 x) + B(-3 \sin 3 x + 6 \cos 3 x)$.
Equating coefficients of $\cos 3 x$ and $\sin 3 x$:
$A + 6B = 2$ and $2A - 3B = -3$.
Multiplying the second equation by $2$: $4A - 6B = -6$.
Adding the two equations: $5A = -4 \Rightarrow A = -\frac{4}{5}$.
Substituting $A$ into $A + 6B = 2$: $-\frac{4}{5} + 6B = 2 \Rightarrow 6B = \frac{14}{5} \Rightarrow B = \frac{7}{15}$.
Thus,$I = \int \left( A + B \frac{-3 \sin 3 x + 6 \cos 3 x}{\cos 3 x+2 \sin 3 x} \right) d x$.
$I = A \int 1 dx + B \int \frac{d(\cos 3 x+2 \sin 3 x)}{\cos 3 x+2 \sin 3 x}$.
$I = -\frac{4}{5} x + \frac{7}{15} \log |\cos 3 x+2 \sin 3 x| + c$.

(D) Let $I = \int \frac{dx}{9 \cos^2 2x + 16 \sin^2 2x}$.
Divide the numerator and denominator by $\cos^2 2x$:
$I = \int \frac{\sec^2 2x dx}{9 + 16 \tan^2 2x}$.
Let $u = \tan 2x$,then $du = 2 \sec^2 2x dx$,which implies $\sec^2 2x dx = \frac{du}{2}$.
Substituting these into the integral:
$I = \int \frac{du/2}{9 + 16u^2} = \frac{1}{2} \int \frac{du}{9 + 16u^2}$.
Factor out $16$ from the denominator:
$I = \frac{1}{2 \times 16} \int \frac{du}{\frac{9}{16} + u^2} = \frac{1}{32} \int \frac{du}{(\frac{3}{4})^2 + u^2}$.
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{32} \times \frac{1}{3/4} \tan^{-1}(\frac{u}{3/4}) + C = \frac{1}{32} \times \frac{4}{3} \tan^{-1}(\frac{4u}{3}) + C$.
$I = \frac{1}{24} \tan^{-1}(\frac{4}{3} \tan 2x) + C$.

(B) Let $I = \int (\log x)^3 dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int t^3 e^t dt$.
Using the integration by parts formula $\int e^t f(t) dt = e^t [f(t) - f'(t) + f''(t) - f'''(t) + \dots]$,where $f(t) = t^3$:
$f'(t) = 3t^2$,$f''(t) = 6t$,and $f'''(t) = 6$.
Therefore,$I = e^t [t^3 - 3t^2 + 6t - 6] + C$.
Substituting back $t = \log x$ and $e^t = x$:
$I = x [(\log x)^3 - 3(\log x)^2 + 6 \log x - 6] + C$.

(D) Let $I = \int \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} dx$.
Using $\cos^2 x = 1 - \sin^2 x$,the numerator becomes $\sin^4 x + 2(1 - \sin^2 x) - 1 = \sin^4 x - 2 \sin^2 x + 1 = (1 - \sin^2 x)^2 = \cos^4 x$.
Thus,$I = \int \frac{\cos^4 x \cdot \cos x}{(1 + \sin x)^6} dx = \int \frac{\cos^5 x}{(1 + \sin x)^6} dx$.
Let $u = 1 + \sin x$,then $du = \cos x dx$. Also $\cos^2 x = 1 - \sin^2 x = 1 - (u - 1)^2 = 1 - (u^2 - 2u + 1) = 2u - u^2$.
So,$\cos^4 x = (2u - u^2)^2 = u^2(2 - u)^2$.
$I = \int \frac{u^2(2 - u)^2}{u^6} du = \int \frac{4 - 4u + u^2}{u^4} du = \int (4u^{-4} - 4u^{-3} + u^{-2}) du$.
$I = 4 \frac{u^{-3}}{-3} - 4 \frac{u^{-2}}{-2} + \frac{u^{-1}}{-1} + C = -\frac{4}{3u^3} + \frac{2}{u^2} - \frac{1}{u} + C$.
Substituting $u = 1 + \sin x$,we get $I = -\frac{4 - 6(1 + \sin x) + 3(1 + \sin x)^2}{3(1 + \sin x)^3} + C = -\frac{4 - 6 - 6 \sin x + 3(1 + 2 \sin x + \sin^2 x)}{3(1 + \sin x)^3} + C = -\frac{3 \sin^2 x - 1}{3(1 + \sin x)^3} + C$.
Alternatively,using the identity $\cos^2 x = (1 - \sin x)(1 + \sin x)$,$I = \int \frac{(1 - \sin x)^2 (1 + \sin x)^2 \cos x}{(1 + \sin x)^6} dx = \int \frac{(1 - \sin x)^2}{(1 + \sin x)^4} \cos x dx$.
Let $t = \sin x$,$dt = \cos x dx$. $I = \int \frac{(1 - t)^2}{(1 + t)^4} dt = \int \frac{(1 - t)^2}{(1 + t)^2} \cdot \frac{1}{(1 + t)^2} dt$.
Using $\frac{1 - t}{1 + t} = z$,$dz = \frac{-(1 + t) - (1 - t)}{(1 + t)^2} dt = \frac{-2}{(1 + t)^2} dt$.
$I = \int z^2 (-\frac{1}{2} dz) = -\frac{1}{2} \frac{z^3}{3} + C = -\frac{1}{6} \left( \frac{1 - \sin x}{1 + \sin x} \right)^3 + C = -\frac{1}{6} \frac{(1 - \sin x)^3}{(1 + \sin x)^3} \cdot \frac{(1 + \sin x)^3}{(1 + \sin x)^3} = -\frac{1}{6} \frac{(1 - \sin^2 x)^3}{(1 + \sin x)^6} = -\frac{\cos^6 x}{6(1 + \sin x)^6} + C$.

(C) Let $I = \int \frac{4 \tan^4 x + 3 \tan^2 x - 1}{\tan^2 x + 4} dx$.
First,factor the numerator: $4 \tan^4 x + 3 \tan^2 x - 1 = (4 \tan^2 x - 1)(\tan^2 x + 1)$.
Since $1 + \tan^2 x = \sec^2 x$,we have $I = \int \frac{(4 \tan^2 x - 1) \sec^2 x}{\tan^2 x + 4} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
The integral becomes $I = \int \frac{4u^2 - 1}{u^2 + 4} du$.
Perform polynomial division or rewrite the numerator: $I = \int \frac{4(u^2 + 4) - 17}{u^2 + 4} du$.
$I = \int \left( 4 - \frac{17}{u^2 + 2^2} \right) du$.
Integrating,we get $I = 4u - 17 \cdot \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) + C$.
Substituting $u = \tan x$ back,$I = 4 \tan x - \frac{17}{2} \tan^{-1} \left( \frac{\tan x}{2} \right) + C$.

(D) Let $I = \int \sqrt{4 \cos ^2 x - 5 \sin ^2 x} \cos x \, dx$.
Using the identity $\cos ^2 x = 1 - \sin ^2 x$,we get:
$I = \int \sqrt{4(1 - \sin ^2 x) - 5 \sin ^2 x} \cos x \, dx = \int \sqrt{4 - 9 \sin ^2 x} \cos x \, dx$.
Let $\sin x = t$,then $\cos x \, dx = dt$.
$I = \int \sqrt{4 - 9t^2} \, dt = \int \sqrt{2^2 - (3t)^2} \, dt$.
Using the formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin ^{-1}\left(\frac{u}{a}\right) + C$,where $u = 3t$ and $du = 3 \, dt$ (so $dt = \frac{du}{3}$):
$I = \frac{1}{3} \int \sqrt{2^2 - u^2} \, du = \frac{1}{3} \left[ \frac{u}{2} \sqrt{4 - u^2} + \frac{4}{2} \sin ^{-1}\left(\frac{u}{2}\right) \right] + C$.
Substituting $u = 3 \sin x$:
$I = \frac{1}{3} \left[ \frac{3 \sin x}{2} \sqrt{4 - 9 \sin ^2 x} + 2 \sin ^{-1}\left(\frac{3 \sin x}{2}\right) \right] + C$.
$I = \frac{1}{2} \sin x \sqrt{4 - 9 \sin ^2 x} + \frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right) + C$.