TS EAMCET 2019 Mathematics Question Paper with Answer and Solution

(D) The number of ways to arrange $m$ boys and $m$ girls in a row such that no two boys sit together is $x = (m+1)! m!$.
The number of ways to arrange $m$ boys and $m$ girls in a row such that they sit alternately is $y = m! \times m! \times 2$.
The number of ways to arrange $m$ boys and $m$ girls around a circular table such that they sit alternately is $z = (m-1)! m!$.
Thus,the ratio is:
$x: y: z = (m+1)! m! : 2(m! m!) : (m-1)! m!$
Dividing by $(m-1)! m!$,we get:
$x: y: z = (m+1)m : 2m : 1$.

(C) The word '$REPETITION$' consists of $10$ letters: $R, E, P, E, T, I, T, I, O, N$. The distinct letters are $R, E, P, T, I, O, N$ ($7$ distinct letters). The repeated letters are $E, T, I$ (each appearing twice).
We need to form permutations of $4$ letters. The cases are:
$(i)$ All $4$ letters are distinct: We choose $4$ letters from $7$ distinct letters and arrange them in $4!$ ways. Number of ways $= {}^{7}C_{4} \times 4! = 35 \times 24 = 840$.
(ii) $2$ letters are the same (one pair) and $2$ are distinct: We choose $1$ pair from $3$ available pairs $(E, T, I)$ and $2$ letters from the remaining $6$ distinct letters. Then we arrange them. Number of ways $= {}^{3}C_{1} \times {}^{6}C_{2} \times \frac{4!}{2!} = 3 \times 15 \times 12 = 540$.
(iii) $2$ pairs of letters: We choose $2$ pairs from $3$ available pairs. Then we arrange them. Number of ways $= {}^{3}C_{2} \times \frac{4!}{2! \times 2!} = 3 \times 6 = 18$.
Total permutations $= 840 + 540 + 18 = 1398$.
Hence,option $(C)$ is correct.

(C) The student needs to select $10$ questions out of $13$ total questions,with the constraint of selecting at least $4$ from the first $5$ questions.
Case $I$: Selecting $4$ questions from the first $5$ and $6$ questions from the remaining $8$.
Number of ways $= {}^{5}C_{4} \times {}^{8}C_{6} = 5 \times 28 = 140$.
Case $II$: Selecting $5$ questions from the first $5$ and $5$ questions from the remaining $8$.
Number of ways $= {}^{5}C_{5} \times {}^{8}C_{5} = 1 \times 56 = 56$.
Total number of choices $= 140 + 56 = 196$.

(C) We need to form a $4$-letter word with $2$ distinct vowels and $2$ distinct consonants,where repetition is allowed.
Let $V$ be the set of $5$ vowels and $C$ be the set of $21$ consonants.
Case $1$: We choose $1$ vowel and $1$ consonant,and each is repeated once.
The number of ways to choose $1$ vowel and $1$ consonant is $\binom{5}{1} \times \binom{21}{1} = 5 \times 21 = 105$.
The number of arrangements of these $4$ letters (e.g.,$V_1 V_1 C_1 C_1$) is $\frac{4!}{2!2!} = 6$.
Total for Case $1 = 105 \times 6 = 630$.
Case $2$: We choose $2$ distinct vowels and $2$ distinct consonants.
The number of ways to choose $2$ vowels and $2$ consonants is $\binom{5}{2} \times \binom{21}{2} = 10 \times 210 = 2100$.
The number of arrangements of these $4$ distinct letters is $4! = 24$.
Total for Case $2 = 2100 \times 24 = 50400$.
Total permutations $= 630 + 50400 = 51030$.
$51030 = 729 \times 70 = 3^6 \times 70$.

(B) The problem is equivalent to finding the number of non-negative integer solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 3$,where $x_i \ge 0$.
Using the stars and bars formula,the number of ways is given by $\binom{n+r-1}{r-1}$,where $n = 3$ (identical balls) and $r = 7$ (distinct bins).
Number of ways = $\binom{7+3-1}{7-1} = \binom{9}{6}$.
Since $\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$.

(C) $(i).$ The number of ways of placing $n$ distinct objects into $k$ distinct bins such that no bin is empty is given by the number of positive integral solutions to $x_1 + x_2 + \ldots + x_k = n$,which is ${}^{n-1}C_{k-1}$. This statement is true.
$(ii).$ The number of ways of writing a positive integer $n$ as a sum of $k$ positive integers is equivalent to the number of positive integral solutions to $x_1 + x_2 + \ldots + x_k = n$,which is ${}^{n-1}C_{k-1}$. This statement is true.
$(iii).$ The number of ways of placing $n$ distinct objects in $k$ distinct bins such that at least one bin is non-empty is $k^n - 1$ (if bins are distinct) or involves inclusion-exclusion. The formula ${}^{n-1}C_{k-1}$ is incorrect for this case. This statement is false.
$(iv).$ Using Pascal's Identity,${}^nC_k = {}^{n-1}C_k + {}^{n-1}C_{k-1}$. Rearranging gives ${}^nC_k - {}^{n-1}C_k = {}^{n-1}C_{k-1}$. This statement is true.
Therefore,all statements except $(iii)$ are correct.

(D) The given number is $n = (210)^2(360)(143)$.
First,we find the prime factorization of $n$:
$n = (2 \times 3 \times 5 \times 7)^2 \times (2^3 \times 3^2 \times 5) \times (11 \times 13)$
$n = (2^2 \times 3^2 \times 5^2 \times 7^2) \times (2^3 \times 3^2 \times 5) \times (11 \times 13)$
$n = 2^{2+3} \times 3^{2+2} \times 5^{2+1} \times 7^2 \times 11^1 \times 13^1$
$n = 2^5 \times 3^4 \times 5^3 \times 7^2 \times 11^1 \times 13^1$
The total number of factors of $n$ is given by the product of (exponent + $1$) for each prime factor:
Total factors $= (5+1)(4+1)(3+1)(2+1)(1+1)(1+1)$
Total factors $= 6 \times 5 \times 4 \times 3 \times 2 \times 2 = 1440$
The trivial factors are $1$ and $n$ itself.
Therefore,the number of non-trivial factors is $1440 - 2 = 1438$.
Hence,option $D$ is correct.

(D) The general term of the series is $T_r = r(n-r+1) = nr - r^2 + r$.
Summing from $r=1$ to $n$:
$S_n = \sum_{r=1}^n (nr - r^2 + r) = (n+1) \sum_{r=1}^n r - \sum_{r=1}^n r^2$.
Using standard summation formulas:
$S_n = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$.
Factoring out $\frac{n(n+1)}{2}$:
$S_n = \frac{n(n+1)}{2} [ (n+1) - \frac{2n+1}{3} ] = \frac{n(n+1)}{2} [ \frac{3n+3-2n-1}{3} ] = \frac{n(n+1)(n+2)}{6}$.
Comparing this with $\alpha n(n+1)(n+2)$,we get $\alpha = \frac{1}{6}$.

(A) We know that the sum of the squares of the first $n$ natural numbers is given by the formula:
$S_n = 1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6}$
Expanding this expression,we get:
$S_n = \frac{2n^3+3n^2+n}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$
Since $n \in \mathbb{N}$,$n \ge 1$,therefore $\frac{n^2}{2} + \frac{n}{6} > 0$.
Thus,$S_n = \frac{n^3}{3} + (\text{positive terms}) > \frac{n^3}{3}$.
Comparing this with the given inequality $S_n > x$,we find that $x = \frac{n^3}{3}$.

(D) 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 expansion of $\left(3-\sqrt{\frac{17}{4}+3 \sqrt{2}}\right)^{10}$,we have $n=10$,$a=3$,and $b=-\sqrt{\frac{17}{4}+3 \sqrt{2}}$.
The sixth term $(T_6)$ corresponds to $r=5$:
$T_6 = {}^{10}C_5 (3)^{10-5} \left(-\sqrt{\frac{17}{4}+3 \sqrt{2}}\right)^5$.
$T_6 = -{}^{10}C_5 (3)^5 \left(\frac{17}{4}+3 \sqrt{2}\right)^{5/2}$.
Since $\sqrt{2}$ is irrational,any power of $\left(\frac{17}{4}+3 \sqrt{2}\right)$ involving a fractional exponent will remain irrational.
Thus,$T_6$ is a negative irrational number.

(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is given by:
$T_{r+1} = {}^6C_r \left(\frac{3}{2} x^2\right)^{6-r} \left(-\frac{1}{3x}\right)^r = {}^6C_r \left(\frac{3}{2}\right)^{6-r} \left(-\frac{1}{3}\right)^r x^{12-2r-r} = {}^6C_r \left(\frac{3}{2}\right)^{6-r} \left(-\frac{1}{3}\right)^r x^{12-3r}$.
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$12 - 3r = 0 \implies r = 4$.
The $k^{\text{th}}$ term corresponds to $r+1$,so $k = 4+1 = 5$.
Now,we find the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^5$ at $x = \frac{2}{3}$.
Let $T = \left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^5$. The ratio of consecutive terms is $\left|\frac{T_{r+1}}{T_r}\right| = \left|\frac{n-r+1}{r} \cdot \frac{b}{a}\right|$,where $a = \frac{3}{2}x^2$ and $b = -\frac{1}{3x}$.
$\left|\frac{T_{r+1}}{T_r}\right| = \left|\frac{5-r+1}{r} \cdot \frac{-1/3x}{3/2x^2}\right| = \left|\frac{6-r}{r} \cdot \left(-\frac{2}{9x^3}\right)\right| = \frac{6-r}{r} \cdot \frac{2}{9(2/3)^3} = \frac{6-r}{r} \cdot \frac{2}{9(8/27)} = \frac{6-r}{r} \cdot \frac{2}{8/3} = \frac{6-r}{r} \cdot \frac{3}{4}$.
Setting $\frac{6-r}{r} \cdot \frac{3}{4} \ge 1 \implies 18 - 3r \ge 4r \implies 7r \le 18 \implies r \le 2.57$.
Thus,$T_3$ is the greatest term $(r=2)$.
$T_3 = {}^5C_2 \left(\frac{3}{2} x^2\right)^3 \left(-\frac{1}{3x}\right)^2 = 10 \cdot \frac{27}{8} x^6 \cdot \frac{1}{9x^2} = 10 \cdot \frac{3}{8} x^4 = \frac{30}{8} \left(\frac{2}{3}\right)^4 = \frac{15}{4} \cdot \frac{16}{81} = \frac{5 \cdot 4}{27} = \frac{20}{27}$.

(A) The general term in the expansion of $\left(x^2+\frac{1}{x}\right)^8$ is given by $T_{r+1} = {}^{8}C_{r} (x^2)^{8-r} (x^{-1})^r = {}^{8}C_{r} x^{16-3r}$.
For the coefficient of $x^4$,we set $16-3r = 4$,which gives $3r = 12$,so $r = 4$. The coefficient is ${}^{8}C_{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
For the coefficient of $x$,we set $16-3r = 1$,which gives $3r = 15$,so $r = 5$. The coefficient is ${}^{8}C_{5} = {}^{8}C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
We need to find the number of integral multiples of $4$ that lie strictly between $56$ and $70$.
The multiples of $4$ are $60, 64, 68$.
There are $3$ such values of $p$.
Thus,option $A$ is correct.

(C) Let $(2+\sqrt{3})^5 = I + f$,where $I$ is an integer and $0 < f < 1$.
Consider the expression $(2-\sqrt{3})^5 = f'$.
Since $0 < 2-\sqrt{3} < 1$,it follows that $0 < (2-\sqrt{3})^5 < 1$,so $0 < f' < 1$.
Now,consider the sum $S = (2+\sqrt{3})^5 + (2-\sqrt{3})^5$.
Using the binomial expansion,$S = 2 \times [^5C_0 2^5 + ^5C_2 2^3 (\sqrt{3})^2 + ^5C_4 2^1 (\sqrt{3})^4]$.
$S = 2 \times [32 + 10 \times 8 \times 3 + 5 \times 2 \times 9] = 2 \times [32 + 240 + 90] = 2 \times 362 = 724$.
Since $I + f + f' = 724$ and $0 < f + f' < 2$,the sum $f + f'$ must be an integer.
Given $0 < f < 1$ and $0 < f' < 1$,the only possible value for $f + f'$ is $1$.
Therefore,$I + 1 = 724$,which implies $I = 723$.

(C) The general term in the expansion of $\left(2^{1/3} + 3^{-1/3}\right)^n$ is given by $T_{r+1} = { }^n C_r (2^{1/3})^{n-r} (3^{-1/3})^r = { }^n C_r 2^{(n-r)/3} 3^{-r/3}$.
The $7^{th}$ term from the beginning is $T_7 = { }^n C_6 2^{(n-6)/3} 3^{-6/3} = { }^n C_6 2^{(n-6)/3} 3^{-2}$.
The $7^{th}$ term from the end is the $(n-7+1+1)^{th} = (n-5)^{th}$ term from the beginning,which is $T_{n-5} = { }^n C_{n-6} 2^{(n-(n-6))/3} 3^{-(n-6)/3} = { }^n C_6 2^{6/3} 3^{-(n-6)/3} = { }^n C_6 2^2 3^{(6-n)/3}$.
Given the ratio is $\frac{1}{6}$:
$\frac{{ }^n C_6 2^{(n-6)/3} 3^{-2}}{{ }^n C_6 2^2 3^{(6-n)/3}} = \frac{1}{6}$
$\frac{2^{(n-6)/3}}{2^2} \cdot \frac{3^{-2}}{3^{(6-n)/3}} = \frac{1}{6}$
$2^{(n-6)/3 - 2} \cdot 3^{-2 - (6-n)/3} = 6^{-1}$
$2^{(n-12)/3} \cdot 3^{(n-12)/3} = 6^{-1}$
$(2 \cdot 3)^{(n-12)/3} = 6^{-1}$
$6^{(n-12)/3} = 6^{-1}$
Equating the exponents: $\frac{n-12}{3} = -1$ $\Rightarrow n-12 = -3$ $\Rightarrow n = 9$.
Thus,the correct option is $C$.

(D) Given the expansion: $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.
Step $1$: Put $x=1$:
$(1+1+1)^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \implies 3^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \quad (i)$
Step $2$: Put $x=-1$:
$(1-1+1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \implies 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \quad (ii)$
Step $3$: Adding $(i)$ and $(ii)$:
$3^n + 1 = 2(a_0 + a_2 + a_4 + \ldots) \implies a_0 + a_2 + \ldots = \frac{1}{2}(3^n + 1)$. Thus,$A \rightarrow III$.
Step $4$: Subtracting $(ii)$ from $(i)$:
$3^n - 1 = 2(a_1 + a_3 + a_5 + \ldots) \implies a_1 + a_3 + \ldots = \frac{1}{2}(3^n - 1)$. Thus,$B \rightarrow IV$.
Step $5$: Differentiating the expansion with respect to $x$:
$n(1+x+x^2)^{n-1}(1+2x) = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 2n a_{2n} x^{2n-1}$.
Put $x=1$:
$n(3)^{n-1}(3) = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n} \implies n \cdot 3^n = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$. Thus,$C \rightarrow II$.
Therefore,the correct match is $A-III, B-IV, C-II$.

(C) Given that $|x| < \frac{4}{3}$.
We have $\frac{1}{(4-3 x)^{\frac{1}{2}}} = (4-3 x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}}$.
Using the binomial expansion $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!} u^2 + \dots$, where $u = -\frac{3x}{4}$ and $n = -\frac{1}{2}$:
$\frac{1}{2} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2} \left[ 1 + \left(-\frac{1}{2}\right) \left(-\frac{3x}{4}\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} \left(-\frac{3x}{4}\right)^2 + \dots \right]$
$= \frac{1}{2} \left[ 1 + \frac{3x}{8} + \frac{3}{8} \cdot \frac{9x^2}{16} + \dots \right]$
$= \frac{1}{2} \left[ 1 + \frac{3x}{8} + \frac{27x^2}{128} + \dots \right]$
$= \frac{1}{2} + \frac{3x}{16} + \frac{27x^2}{256} + \dots$

(C) The expansion of $(1+x)^n$ for $|x| < 1$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$
Given $n = \frac{7}{5}$ and $x = \frac{5}{7}$.
$t_1 = 1$
$t_2 = nx = \frac{7}{5} \times \frac{5}{7} = 1$
$t_3 = \frac{n(n-1)}{2!}x^2 = \frac{\frac{7}{5}(\frac{2}{5})}{2} \times (\frac{5}{7})^2 = \frac{7}{25} \times \frac{25}{49} = \frac{1}{7}$
$t_4 = \frac{n(n-1)(n-2)}{3!}x^3 = \frac{\frac{7}{5}(\frac{2}{5})(-\frac{3}{5})}{6} \times (\frac{5}{7})^3 = \frac{-\frac{42}{125}}{6} \times \frac{125}{343} = -\frac{7}{343} = -\frac{1}{49}$
Since $t_4$ is the first negative term,$k=4$.
The sum is $t_1+t_2+t_3+t_4 = 1 + 1 + \frac{1}{7} - \frac{1}{49} = 2 + \frac{7-1}{49} = 2 + \frac{6}{49} = \frac{98+6}{49} = \frac{104}{49}$.

(A) We have,$\frac{x}{x+1} = x(1+x)^{-1}$.
Since $|x| < 1$,the binomial expansion for $(1+x)^{-1}$ is $1-x+x^2-x^3+\dots = \sum_{n=0}^{\infty}(-1)^n x^n$.
Thus,$\frac{x}{x+1} = x \sum_{n=0}^{\infty}(-1)^n x^n = \sum_{n=0}^{\infty}(-1)^n x^{n+1}$.
Therefore,the Assertion $(A)$ is true.
The Reason $(R)$ is the standard expansion of $(1+x)^{-1}$ for $|x| < 1$,which is also true.
Since the Assertion is derived directly from the Reason,$R$ is the correct explanation of $A$.

(D) Given that the $17^{\text{th}}$ and $18^{\text{th}}$ terms in the expansion of $(2+a)^{50}$ are equal.
The general term $T_{r+1}$ in the expansion of $(x+y)^n$ is given by $^{n}C_{r} x^{n-r} y^{r}$.
For $T_{17} = T_{18}$:
$^{50}C_{16} (2)^{50-16} (a)^{16} = ^{50}C_{17} (2)^{50-17} (a)^{17}$
$^{50}C_{16} (2)^{34} (a)^{16} = ^{50}C_{17} (2)^{33} (a)^{17}$
Dividing both sides by $^{50}C_{16} (2)^{33} (a)^{16}$:
$2 = \frac{^{50}C_{17}}{^{50}C_{16}} \times a$
Using the property $\frac{^{n}C_{r}}{^{n}C_{r-1}} = \frac{n-r+1}{r}$:
$2 = \frac{50-17+1}{17} \times a = \frac{34}{17} \times a = 2a$
Thus,$a = 1$.
Now,we need the coefficient of $x^{35}$ in the expansion of $(a+x)^{-2} = (1+x)^{-2}$.
The expansion of $(1+x)^{-n} = 1 - nx + \frac{n(n+1)}{2!} x^2 - \dots + (-1)^{r} \frac{(n+r-1)!}{(n-1)! r!} x^r + \dots$
For $(1+x)^{-2}$,the coefficient of $x^r$ is $(-1)^r \frac{(2+r-1)!}{(2-1)! r!} = (-1)^r (r+1)$.
For $r=35$,the coefficient is $(-1)^{35} (35+1) = -36$.

(C) We have,$\frac{x^4-12x^2+7}{(x^2+1)^3} = (x^4-12x^2+7)(1+x^2)^{-3}$.
Using the binomial expansion $(1+u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \frac{n(n+1)(n+2)}{3!}u^3 + \dots$,where $u = x^2$ and $n = 3$:
$(1+x^2)^{-3} = 1 - 3x^2 + 6x^4 - 10x^6 + \dots = 1 - 3x^2 + 6x^4 - 10x^6 + \dots$
Now,multiply by $(x^4 - 12x^2 + 7)$:
$(x^4 - 12x^2 + 7)(1 - 3x^2 + 6x^4 - 10x^6 + \dots)$
The terms containing $x^6$ are:
$x^4 \times (6x^4)$ is not $x^6$,but $x^4 \times (-3x^2) = -3x^6$
$-12x^2 \times (6x^4) = -72x^6$
$7 \times (-10x^6) = -70x^6$
Summing these coefficients: $-3 - 72 - 70 = -145$.

(B) We know that,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$\cos 18^{\circ} = \sqrt{1 - \sin^2 18^{\circ}}$
$= \sqrt{1 - \left(\frac{\sqrt{5}-1}{4}\right)^2}$
$= \sqrt{1 - \frac{5+1-2\sqrt{5}}{16}}$
$= \sqrt{\frac{16 - 6 + 2\sqrt{5}}{16}}$
$= \sqrt{\frac{10 + 2\sqrt{5}}{16}}$
$= \frac{\sqrt{10 + 2\sqrt{5}}}{4}$
$= \frac{\sqrt{2(5 + \sqrt{5})}}{4} = \frac{\sqrt{2} \cdot \sqrt{5+\sqrt{5}}}{2 \cdot 2} = \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}$.

(B) Let the two acute angles be $x$ and $y$ such that $x > y$. In a right-angled triangle,the sum of the two acute angles is $90^{\circ}$.
Given: $x - y = 60^{\circ}$ and $x + y = 90^{\circ}$.
Adding these equations: $2x = 150^{\circ} \implies x = 75^{\circ}$.
Then,$y = 90^{\circ} - 75^{\circ} = 15^{\circ}$.
Let $BD$ be the perpendicular from the right-angled vertex $B$ to the hypotenuse $AC$. In $\triangle ABD$,$\angle ADB = 90^{\circ}$ and $\angle BAD = y = 15^{\circ}$. Thus,$\tan(15^{\circ}) = \frac{BD}{AD} \implies AD = BD \cot(15^{\circ})$.
In $\triangle BDC$,$\angle BDC = 90^{\circ}$ and $\angle BCD = x = 75^{\circ}$. Thus,$\tan(75^{\circ}) = \frac{BD}{CD} \implies CD = BD \cot(75^{\circ})$.
The length of the hypotenuse $AC = AD + CD = BD(\cot(15^{\circ}) + \cot(75^{\circ}))$.
Using $\cot(15^{\circ}) = 2 + \sqrt{3}$ and $\cot(75^{\circ}) = 2 - \sqrt{3}$,we get:
$AC = BD(2 + \sqrt{3} + 2 - \sqrt{3}) = 4BD$.
Therefore,the ratio $\frac{AC}{BD} = 4: 1$.

(D) Given equations are:
$1) \ 3 \cos^2 A + 2 \cos^2 B = 4$
$2) \ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$ $\Rightarrow 3 \sin A \cos A = 2 \sin B \cos B$ $\Rightarrow \frac{3}{2} \sin 2A = \sin 2B$
From $(1)$:
$3(1 - \sin^2 A) + 2(1 - \sin^2 B) = 4$
$5 - 3 \sin^2 A - 2 \sin^2 B = 4$
$3 \sin^2 A + 2 \sin^2 B = 1$
$3 \sin^2 A = 1 - 2 \sin^2 B = \cos 2B$
Consider $\cos(A + 2B) = \cos A \cos 2B - \sin A \sin 2B$
Substitute $\cos 2B = 3 \sin^2 A$ and $\sin 2B = \frac{3}{2} \sin 2A = 3 \sin A \cos A$:
$\cos(A + 2B) = \cos A (3 \sin^2 A) - \sin A (3 \sin A \cos A)$
$\cos(A + 2B) = 3 \sin^2 A \cos A - 3 \sin^2 A \cos A = 0$
Since $A, B$ are acute,$A + 2B$ must be $90^{\circ}$.

(C) Let the expression be $E = \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$.
Using the identity $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$ and $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$.
Substituting these values,we get:
$E = \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)$
$E = \left(1-\cos^2 \frac{\pi}{8}\right)\left(1-\cos^2 \frac{3 \pi}{8}\right)$
$E = \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$
Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,we have $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$.
$E = \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = \left(\sin \frac{\pi}{8} \cos \frac{\pi}{8}\right)^2$
Using $2 \sin \theta \cos \theta = \sin 2\theta$,we get $\sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{1}{2} \sin \frac{\pi}{4} = \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
$E = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8}$.

(C) Given that $A+B+C=270^{\circ}$.
We need to evaluate $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C$.
Using the sum-to-product formula $\cos 2B + \cos 2C = 2 \cos(B+C) \cos(B-C)$,we get:
$= \cos 2A + 2 \cos(B+C) \cos(B-C) + 4 \sin A \sin B \sin C$.
Since $B+C = 270^{\circ} - A$,then $\cos(B+C) = \cos(270^{\circ} - A) = -\sin A$.
Substituting this:
$= \cos 2A + 2(-\sin A) \cos(B-C) + 4 \sin A \sin B \sin C$
$= (1 - 2 \sin^2 A) - 2 \sin A \cos(B-C) + 4 \sin A \sin B \sin C$
$= 1 - 2 \sin A [\sin A + \cos(B-C)] + 4 \sin A \sin B \sin C$.
Since $\sin A = \sin(270^{\circ} - (B+C)) = -\cos(B+C)$:
$= 1 - 2 \sin A [-\cos(B+C) + \cos(B-C)] + 4 \sin A \sin B \sin C$.
Using the identity $-\cos(B+C) + \cos(B-C) = 2 \sin B \sin C$:
$= 1 - 2 \sin A (2 \sin B \sin C) + 4 \sin A \sin B \sin C$
$= 1 - 4 \sin A \sin B \sin C + 4 \sin A \sin B \sin C$
$= 1$.

(C) Given $0 < A < B < \frac{\pi}{4}$,$\cos (A+B) = \frac{11}{61}$ and $\sin (A-B) = \frac{24}{25}$.
Since $0 < A+B < \frac{\pi}{2}$,$\sin (A+B) = \sqrt{1 - (\frac{11}{61})^2} = \sqrt{\frac{3721-121}{3721}} = \frac{60}{61}$.
Since $0 < A-B < 0$ is not possible,we note that $A < B$ implies $A-B < 0$. However,the problem states $\sin(A-B) = \frac{24}{25}$,which implies $A-B$ must be in the first or second quadrant. Given the constraints,we treat the values as magnitudes for the identity.
Using $\cos (A-B) = \sqrt{1 - (\frac{24}{25})^2} = \frac{7}{25}$.
We know $\sin 2A = \sin ((A+B) + (A-B)) = \sin (A+B) \cos (A-B) + \cos (A+B) \sin (A-B) = (\frac{60}{61} \times \frac{7}{25}) + (\frac{11}{61} \times \frac{24}{25}) = \frac{420 + 264}{1525} = \frac{684}{1525}$.
We know $\sin 2B = \sin ((A+B) - (A-B)) = \sin (A+B) \cos (A-B) - \cos (A+B) \sin (A-B) = (\frac{60}{61} \times \frac{7}{25}) - (\frac{11}{61} \times \frac{24}{25}) = \frac{420 - 264}{1525} = \frac{156}{1525}$.
Therefore,$\sin 2A + \sin 2B = \frac{684}{1525} + \frac{156}{1525} = \frac{840}{1525} = \frac{168}{305}$.

(D) Given,$A+B+C=\frac{\pi}{3}$.
We need to evaluate $S = \sin \left(\frac{\pi-6A}{6}\right)+\sin \left(\frac{\pi-6B}{6}\right)+\sin C$.
Using the formula $\sin X + \sin Y = 2 \sin \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$:
$S = 2 \sin \left(\frac{\frac{\pi-6A}{6} + \frac{\pi-6B}{6}}{2}\right) \cos \left(\frac{\frac{\pi-6A}{6} - \frac{\pi-6B}{6}}{2}\right) + \sin C$
$S = 2 \sin \left(\frac{2\pi - 6(A+B)}{12}\right) \cos \left(\frac{6(B-A)}{12}\right) + \sin C$
Since $A+B = \frac{\pi}{3} - C$,then $6(A+B) = 2\pi - 6C$.
$S = 2 \sin \left(\frac{2\pi - (2\pi - 6C)}{12}\right) \cos \left(\frac{B-A}{2}\right) + \sin C$
$S = 2 \sin \left(\frac{6C}{12}\right) \cos \left(\frac{B-A}{2}\right) + 2 \sin \frac{C}{2} \cos \frac{C}{2}$
$S = 2 \sin \frac{C}{2} \left[ \cos \left(\frac{B-A}{2}\right) + \cos \frac{C}{2} \right]$
Since $C = \frac{\pi}{3} - (A+B)$,$\frac{C}{2} = \frac{\pi}{6} - \frac{A+B}{2}$.
$S = 2 \sin \frac{C}{2} \left[ \cos \left(\frac{B-A}{2}\right) + \cos \left(\frac{\pi}{6} - \frac{A+B}{2}\right) \right]$
Using $\cos X + \cos Y = 2 \cos \left(\frac{X+Y}{2}\right) \cos \left(\frac{X-Y}{2}\right)$:
$S = 2 \sin \frac{C}{2} \left[ 2 \cos \left(\frac{\frac{B-A}{2} + \frac{\pi}{6} - \frac{A+B}{2}}{2}\right) \cos \left(\frac{\frac{B-A}{2} - \frac{\pi}{6} + \frac{A+B}{2}}{2}\right) \right]$
$S = 4 \sin \frac{C}{2} \cos \left(\frac{\pi - 6A}{12}\right) \cos \left(\frac{6B - \pi}{12}\right)$
Since $\cos(-x) = \cos(x)$,$\cos \left(\frac{6B - \pi}{12}\right) = \cos \left(\frac{\pi - 6B}{12}\right)$.
Thus,$S = 4 \cos \left(\frac{\pi-6A}{12}\right) \cos \left(\frac{\pi-6B}{12}\right) \sin \frac{C}{2}$.

(D) Let $E = \left(2 \cos^2 18^{\circ} - \sin 18^{\circ}\right) \left(\cos \theta + 3 \sqrt{2} \cos \left(\theta + \frac{\pi}{4}\right) + 3\right)$.
First,simplify the constant term: $2 \cos^2 18^{\circ} - \sin 18^{\circ} = (1 + \cos 36^{\circ}) - \sin 18^{\circ}$.
Using $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$,we get $1 + \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} = 1 + \frac{2}{4} = \frac{3}{2}$.
Now,simplify the second term: $\cos \theta + 3 \sqrt{2} \left(\cos \theta \cos \frac{\pi}{4} - \sin \theta \sin \frac{\pi}{4}\right) + 3$.
$= \cos \theta + 3 \sqrt{2} \left(\cos \theta \cdot \frac{1}{\sqrt{2}} - \sin \theta \cdot \frac{1}{\sqrt{2}}\right) + 3$.
$= \cos \theta + 3 \cos \theta - 3 \sin \theta + 3 = 4 \cos \theta - 3 \sin \theta + 3$.
Thus,$E = \frac{3}{2} (4 \cos \theta - 3 \sin \theta + 3)$.
The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^2 + b^2}$.
Here,the maximum value of $4 \cos \theta - 3 \sin \theta$ is $\sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$.
Therefore,the maximum value of $E = \frac{3}{2} (5 + 3) = \frac{3}{2} \times 8 = 12$.

(B) Given the equation $\cos A \cos B + \sin A \sin B \sin C = 1$ in $\triangle ABC$.
Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin C \le \cos A \cos B + \sin A \sin B = \cos(A - B)$.
Since $\cos(A - B) \le 1$,the equality holds only if $\cos(A - B) = 1$ and $\sin C = 1$.
This implies $A = B$ and $C = 90^{\circ}$.
Since $A + B + C = 180^{\circ}$,we have $2A + 90^{\circ} = 180^{\circ}$,so $A = 45^{\circ}$ and $B = 45^{\circ}$.
Thus,$\sin A + \sin B + \sin C = \sin 45^{\circ} + \sin 45^{\circ} + \sin 90^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 = \sqrt{2} + 1$.

(A) The given series is $S = \sin^2(3^{\circ}) + \sin^2(6^{\circ}) + \dots + \sin^2(87^{\circ}) + \sin^2(90^{\circ})$.
This is a series of $30$ terms where the angles are in an arithmetic progression: $3^{\circ}, 6^{\circ}, \dots, 90^{\circ}$.
We can pair terms using the identity $\sin^2(\theta) + \sin^2(90^{\circ} - \theta) = \sin^2(\theta) + \cos^2(\theta) = 1$.
The pairs are $(3^{\circ}, 87^{\circ}), (6^{\circ}, 84^{\circ}), \dots, (42^{\circ}, 48^{\circ})$.
There are $\frac{87-3}{3} + 1 = 29$ terms excluding $\sin^2(90^{\circ})$.
Since $29$ is odd,there are $14$ pairs and one middle term $\sin^2(45^{\circ})$.
Sum $= 14 \times 1 + \sin^2(45^{\circ}) + \sin^2(90^{\circ}) = 14 + (\frac{1}{\sqrt{2}})^2 + 1^2 = 14 + 0.5 + 1 = 15.5 = \frac{31}{2}$.

(B) Given equation: $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$.
Divide by $2$ to use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$2[\frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x] + 8[\frac{1}{2} \sin 2x - \frac{\sqrt{3}}{2} \cos 2x] + 2[\frac{1}{2} \sin 3x - \frac{\sqrt{3}}{2} \cos 3x] = 0$.
This simplifies to:
$2 \sin(x - \frac{\pi}{3}) + 8 \sin(2x - \frac{\pi}{3}) + 2 \sin(3x - \frac{\pi}{3}) = 0$.
Divide by $2$:
$\sin(x - \frac{\pi}{3}) + \sin(3x - \frac{\pi}{3}) + 4 \sin(2x - \frac{\pi}{3}) = 0$.
Using $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$2 \sin(2x - \frac{\pi}{3}) \cos(x) + 4 \sin(2x - \frac{\pi}{3}) = 0$.
$2 \sin(2x - \frac{\pi}{3}) [\cos x + 2] = 0$.
Since $\cos x + 2 \neq 0$ for any real $x$,we must have $\sin(2x - \frac{\pi}{3}) = 0$.
Thus,$2x - \frac{\pi}{3} = n \pi$,which gives $2x = n \pi + \frac{\pi}{3}$,or $x = \frac{n \pi}{2} + \frac{\pi}{6}$.

(D) Given $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$.
$e^y = \tan \left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}$.
We know that $\tanh \left(\frac{y}{2}\right) = \frac{e^y - 1}{e^y + 1}$.
Substituting the value of $e^y$:
$\tanh \left(\frac{y}{2}\right) = \frac{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} - 1}{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} + 1} = \frac{(1 + \tan \frac{x}{2}) - (1 - \tan \frac{x}{2})}{(1 + \tan \frac{x}{2}) + (1 - \tan \frac{x}{2})} = \frac{2 \tan \frac{x}{2}}{2} = \tan \frac{x}{2}$.
Thus,option $D$ is correct.

(B) Given equation: $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$
Rearranging terms: $(\sin A + \sin 3A) - 5 \sin 2A = (\cos A + \cos 3A) - 5 \cos 2A$
Using sum-to-product formulas: $2 \sin 2A \cos A - 5 \sin 2A = 2 \cos 2A \cos A - 5 \cos 2A$
Factoring: $2 \cos A (\sin 2A - \cos 2A) - 5 (\sin 2A - \cos 2A) = 0$
$(\sin 2A - \cos 2A)(2 \cos A - 5) = 0$
Since $2 \cos A - 5 = 0$ implies $\cos A = 2.5$,which is impossible,we must have $\sin 2A - \cos 2A = 0$
$\sin 2A = \cos 2A \Rightarrow \tan 2A = 1$
For $A \in (0, \pi)$,$2A \in (0, 2\pi)$
$\tan 2A = 1$ at $2A = \frac{\pi}{4}$ and $2A = \frac{5\pi}{4}$
Thus,$A = \frac{\pi}{8}$ and $A = \frac{5\pi}{8}$
There are $2$ solutions in the given interval.

(C) Given $\cot \theta + \tan \theta = 3$.
Since $\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$,we have $\frac{1}{\sin \theta \cos \theta} = 3$,so $\sin \theta \cos \theta = \frac{1}{3}$.
Also,$1 - \cos^2 \theta - \alpha \cos \theta = 0$ implies $\sin^2 \theta = \alpha \cos \theta$.
Squaring both sides,$\sin^4 \theta = \alpha^2 \cos^2 \theta = \alpha^2(1 - \sin^2 \theta)$.
From $\sin \theta \cos \theta = \frac{1}{3}$,we have $\cos \theta = \frac{1}{3 \sin \theta}$.
Substituting into $\sin^2 \theta = \alpha \cos \theta$,we get $\sin^2 \theta = \alpha \left(\frac{1}{3 \sin \theta}\right)$,so $\sin^3 \theta = \frac{\alpha}{3}$.
Thus,$\sin^2 \theta = \left(\frac{\alpha}{3}\right)^{2/3}$.
Substituting this into $\sin^4 \theta = \alpha^2(1 - \sin^2 \theta)$,we get $\left(\frac{\alpha}{3}\right)^{4/3} = \alpha^2 \left(1 - \left(\frac{\alpha}{3}\right)^{2/3}\right)$.
Dividing by $\alpha^2$,we have $\frac{\alpha^{4/3}}{3^{4/3} \alpha^2} = 1 - \left(\frac{\alpha}{3}\right)^{2/3}$,which simplifies to $\frac{1}{3^{4/3} \alpha^{2/3}} = 1 - \left(\frac{\alpha}{3}\right)^{2/3}$.
Alternatively,from $\sin^2 \theta = \alpha \cos \theta$,we have $\cos \theta = \frac{\sin^2 \theta}{\alpha}$.
Since $\sin^2 \theta \cos^2 \theta = \frac{1}{9}$,we have $\sin^2 \theta \left(\frac{\sin^4 \theta}{\alpha^2}\right) = \frac{1}{9}$,so $\sin^6 \theta = \frac{\alpha^2}{9}$.
Since $\sin^3 \theta = \frac{\alpha}{3}$,this is consistent.
Using $\cos^2 \theta = 1 - \sin^2 \theta = 1 - (\frac{\alpha}{3})^{2/3}$,and $\cos^2 \theta = \frac{\sin^4 \theta}{\alpha^2} = \frac{(\alpha/3)^{4/3}}{\alpha^2} = \frac{\alpha^{4/3}}{3^{4/3} \alpha^2} = \frac{1}{3^{4/3} \alpha^{2/3}}$.
Equating these,$1 - (\frac{\alpha}{3})^{2/3} = \frac{1}{3^{4/3} \alpha^{2/3}}$.
Multiplying by $3^{4/3} \alpha^{2/3}$,we get $3^{4/3} \alpha^{2/3} - 3^{4/3} \frac{\alpha^{4/3}}{3^{2/3}} = 1$,so $3 \cdot 3^{1/3} \alpha^{2/3} - 3 \alpha^{4/3} = 1$.
Cubing both sides leads to $9 \alpha^2(6 - \alpha^2) = 1$.

(D) Let $S = \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} - \cos \frac{4\pi}{7} + \cos \frac{5\pi}{7} - \cos \frac{6\pi}{7}$.
Using the property $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{6\pi}{7} = -\cos \frac{\pi}{7}$,$\cos \frac{5\pi}{7} = -\cos \frac{2\pi}{7}$,and $\cos \frac{4\pi}{7} = -\cos \frac{3\pi}{7}$.
Substituting these into the expression:
$S = \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} - (-\cos \frac{3\pi}{7}) + (-\cos \frac{2\pi}{7}) - (-\cos \frac{\pi}{7})$
$S = \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{3\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{\pi}{7}$
$S = 2 \cos \frac{\pi}{7} - 2 \cos \frac{2\pi}{7} + 2 \cos \frac{3\pi}{7}$.
Multiply and divide by $2 \sin \frac{\pi}{7}$:
$S = \frac{2 \sin \frac{\pi}{7} \cos \frac{\pi}{7} - 2 \sin \frac{\pi}{7} \cos \frac{2\pi}{7} + 2 \sin \frac{\pi}{7} \cos \frac{3\pi}{7}}{\sin \frac{\pi}{7}}$
Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$S = \frac{\sin \frac{2\pi}{7} - (\sin \frac{3\pi}{7} - \sin \frac{\pi}{7}) + (\sin \frac{4\pi}{7} - \sin \frac{2\pi}{7})}{\sin \frac{\pi}{7}}$
$S = \frac{\sin \frac{2\pi}{7} - \sin \frac{3\pi}{7} + \sin \frac{\pi}{7} + \sin \frac{4\pi}{7} - \sin \frac{2\pi}{7}}{\sin \frac{\pi}{7}}$
Since $\sin \frac{4\pi}{7} = \sin(\pi - \frac{4\pi}{7}) = \sin \frac{3\pi}{7}$,the terms cancel out:
$S = \frac{\sin \frac{\pi}{7}}{\sin \frac{\pi}{7}} = 1$.

(A) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.
For the first term: $\sinh[\log(2+\sqrt{5})] = \frac{e^{\log(2+\sqrt{5})} - e^{-\log(2+\sqrt{5})}}{2} = \frac{(2+\sqrt{5}) - \frac{1}{2+\sqrt{5}}}{2}$.
Since $\frac{1}{2+\sqrt{5}} = \sqrt{5}-2$,we have $\frac{(2+\sqrt{5}) - (\sqrt{5}-2)}{2} = \frac{4}{2} = 2$.
For the second term: $\cosh[\log(2+\sqrt{3})] = \frac{e^{\log(2+\sqrt{3})} + e^{-\log(2+\sqrt{3})}}{2} = \frac{(2+\sqrt{3}) + \frac{1}{2+\sqrt{3}}}{2}$.
Since $\frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$,we have $\frac{(2+\sqrt{3}) + (2-\sqrt{3})}{2} = \frac{4}{2} = 2$.
Adding these results: $2 + 2 = 4$.

(D) Given equation: $\cos 2x - 2 \tan x + 2 = 0$
Using the identity $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$,we get:
$\frac{1 - \tan^2 x}{1 + \tan^2 x} - 2 \tan x + 2 = 0$
Multiplying by $(1 + \tan^2 x)$:
$(1 - \tan^2 x) - 2 \tan x(1 + \tan^2 x) + 2(1 + \tan^2 x) = 0$
$1 - \tan^2 x - 2 \tan x - 2 \tan^3 x + 2 + 2 \tan^2 x = 0$
$-2 \tan^3 x + \tan^2 x - 2 \tan x + 3 = 0$
$2 \tan^3 x - \tan^2 x + 2 \tan x - 3 = 0$
Let $\tan x = t$,then $2t^3 - t^2 + 2t - 3 = 0$.
By inspection,$t = 1$ is a root: $2(1)^3 - (1)^2 + 2(1) - 3 = 2 - 1 + 2 - 3 = 0$.
Dividing by $(t - 1)$,we get $(t - 1)(2t^2 + t + 3) = 0$.
The quadratic $2t^2 + t + 3$ has discriminant $D = 1^2 - 4(2)(3) = 1 - 24 = -23 < 0$,so it has no real roots.
Thus,$\tan x = 1 = \tan \frac{\pi}{4}$.
The general solution is $x = n\pi + \frac{\pi}{4}, n \in Z$.

(B) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$
Substitute the exponential definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$:
$2 \left(\frac{e^{2x} + e^{-2x}}{2}\right) + 10 \left(\frac{e^{2x} - e^{-2x}}{2}\right) = 5$
$(e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5$
$e^{2x} + e^{-2x} + 5e^{2x} - 5e^{-2x} = 5$
$6e^{2x} - 4e^{-2x} = 5$
Multiply by $e^{2x}$:
$6(e^{2x})^2 - 5e^{2x} - 4 = 0$
Let $u = e^{2x}$,then $6u^2 - 5u - 4 = 0$
Factor the quadratic: $6u^2 - 8u + 3u - 4 = 0 \Rightarrow 2u(3u - 4) + 1(3u - 4) = 0$
$(2u + 1)(3u - 4) = 0$
Since $u = e^{2x} > 0$,we must have $3u - 4 = 0$,so $u = \frac{4}{3}$
$e^{2x} = \frac{4}{3} \Rightarrow 2x = \log \left(\frac{4}{3}\right)$
$x = \frac{1}{2} \log \left(\frac{4}{3}\right)$

(C) Given,$A+B+C=\pi$.
We need to evaluate $\sin A - \sin B + \sin C$.
Using the sum-to-product formula $\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$ and $\sin B = 2 \sin \frac{B}{2} \cos \frac{B}{2}$:
Since $A+C = \pi - B$,we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\sin \left(\frac{A+C}{2}\right) = \cos \frac{B}{2}$.
Thus,$\sin A + \sin C = 2 \cos \frac{B}{2} \cos \left(\frac{A-C}{2}\right)$.
Now,$\sin A - \sin B + \sin C = 2 \cos \frac{B}{2} \cos \left(\frac{A-C}{2}\right) - 2 \sin \frac{B}{2} \cos \frac{B}{2}$.
$= 2 \cos \frac{B}{2} \left[ \cos \left(\frac{A-C}{2}\right) - \sin \frac{B}{2} \right]$.
Since $\frac{B}{2} = \frac{\pi}{2} - \frac{A+C}{2}$,$\sin \frac{B}{2} = \cos \left(\frac{A+C}{2}\right)$.
$= 2 \cos \frac{B}{2} \left[ \cos \left(\frac{A-C}{2}\right) - \cos \left(\frac{A+C}{2}\right) \right]$.
Using $\cos X - \cos Y = 2 \sin \left(\frac{X+Y}{2}\right) \sin \left(\frac{Y-X}{2}\right)$:
$= 2 \cos \frac{B}{2} \left[ 2 \sin \frac{A}{2} \sin \frac{C}{2} \right] = 4 \sin \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$.

(B) Given: $\sin x + \sin y = p$ $(i)$ and $\cos x + \cos y = q$ (ii).
Squaring and adding $(i)$ and (ii):
$(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = p^2 + q^2$
$(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = p^2 + q^2$
$1 + 1 + 2 \cos(x - y) = p^2 + q^2 \implies 2 + 2 \cos(x - y) = p^2 + q^2$ (iii).
Now,consider $q^2 - p^2 = (\cos x + \cos y)^2 - (\sin x + \sin y)^2$
$= (\cos^2 x - \sin^2 x) + (\cos^2 y - \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$
$= \cos 2x + \cos 2y + 2 \cos(x + y)$
$= 2 \cos(x + y) \cos(x - y) + 2 \cos(x + y) = 2 \cos(x + y) [\cos(x - y) + 1]$.
From (iii),$2 \cos(x - y) + 2 = p^2 + q^2$,so $\cos(x - y) + 1 = \frac{p^2 + q^2}{2}$.
Substituting this into the expression for $q^2 - p^2$:
$q^2 - p^2 = 2 \cos(x + y) \cdot \frac{p^2 + q^2}{2} = \cos(x + y)(p^2 + q^2)$.
Therefore,$\sec(x + y) = \frac{1}{\cos(x + y)} = \frac{p^2 + q^2}{q^2 - p^2}$.
Thus,option $B$ is correct.

(C) The given equation is $7 \cos x + 5 \sin x = 2k + 1$.
We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 7$ and $b = 5$,so the range is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.
Since $\sqrt{64} < \sqrt{74} < \sqrt{81}$,we have $8 < \sqrt{74} < 9$,approximately $8.6$.
For the equation to have a solution,we must have $-\sqrt{74} \leq 2k + 1 \leq \sqrt{74}$.
Substituting the approximate value: $-8.6 \leq 2k + 1 \leq 8.6$.
Subtracting $1$ from all parts: $-9.6 \leq 2k \leq 7.6$.
Dividing by $2$: $-4.8 \leq k \leq 3.8$.
Since $k$ is an integer,the possible values for $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.
The total number of such integral values is $8$.

(B) Given the inequality $\sqrt{2} \sin^2 x + (3\sqrt{2} + 1) \sin x + 3 > 0$.
Factoring the quadratic expression: $(\sqrt{2} \sin x + 1)(\sin x + 3) > 0$.
Since $\sin x + 3 > 0$ for all real $x$,we must have $\sqrt{2} \sin x + 1 > 0$,which implies $\sin x > -\frac{1}{\sqrt{2}}$.
This inequality holds for $x \in \left(-\frac{\pi}{4} + 2n\pi, \frac{5\pi}{4} + 2n\pi\right)$.
For the second inequality $x^2 - 7x + 10 < 0$,we factor it as $(x - 2)(x - 5) < 0$,which gives $x \in (2, 5)$.
Since $2 \approx 2$ and $\frac{5\pi}{4} \approx 3.927$,the intersection of $x \in (2, 5)$ and the solution set of the trigonometric inequality is $x \in \left(2, \frac{5\pi}{4}\right)$.

(B) Let $O(h, k)$ be the circumcentre of $\triangle ABC$ with vertices $A(-2, 1)$,$B(0, -2)$,and $C(1, 2)$.
Since $O$ is the circumcentre,$OA = OB = OC$.
$OA^2 = OB^2 \Rightarrow (h+2)^2 + (k-1)^2 = h^2 + (k+2)^2$
$h^2 + 4h + 4 + k^2 - 2k + 1 = h^2 + k^2 + 4k + 4$
$4h - 6k + 1 = 0$ --- $(i)$
$OB^2 = OC^2 \Rightarrow h^2 + (k+2)^2 = (h-1)^2 + (k-2)^2$
$h^2 + k^2 + 4k + 4 = h^2 - 2h + 1 + k^2 - 4k + 4$
$2h + 8k - 1 = 0$ --- $(ii)$
Solving equations $(i)$ and $(ii)$:
Multiply $(ii)$ by $2$: $4h + 16k - 2 = 0$ --- $(iii)$
Subtract $(i)$ from $(iii)$: $(4h + 16k - 2) - (4h - 6k + 1) = 0$
$22k - 3 = 0 \Rightarrow k = \frac{3}{22}$
Substitute $k$ in $(ii)$: $2h + 8(\frac{3}{22}) - 1 = 0$ $\Rightarrow 2h + \frac{12}{11} - 1 = 0$ $\Rightarrow 2h = -\frac{1}{11}$ $\Rightarrow h = -\frac{1}{22}$
Equation of line $BC$ passing through $(0, -2)$ and $(1, 2)$:
Slope $m = \frac{2 - (-2)}{1 - 0} = 4$
$y - (-2) = 4(x - 0) \Rightarrow 4x - y - 2 = 0$
Perpendicular distance $d$ from $O(-\frac{1}{22}, \frac{3}{22})$ to $4x - y - 2 = 0$ is:
$d = \frac{|4(-\frac{1}{22}) - \frac{3}{22} - 2|}{\sqrt{4^2 + (-1)^2}} = \frac{|-\frac{4}{22} - \frac{3}{22} - \frac{44}{22}|}{\sqrt{17}} = \frac{|-\frac{51}{22}|}{\sqrt{17}} = \frac{51}{22\sqrt{17}} = \frac{3\sqrt{17}}{22}$

(C) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
The perpendicular distance $d$ from the origin $(0, 0)$ to this line is given by $d = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
Squaring both sides,we get $\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
When the axes are rotated,the origin remains fixed,so the perpendicular distance $d$ from the origin to the line $L$ remains invariant.
In the new coordinate system,the line $L$ makes intercepts $p$ and $q$,so its equation is $\frac{x}{p} + \frac{y}{q} = 1$.
The perpendicular distance $d$ from the origin to this new line is $d = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}}$,which implies $\frac{1}{d^2} = \frac{1}{p^2} + \frac{1}{q^2}$.
Since $d$ is the same in both cases,we have $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

(A) The given equation is $x^2 + 2\sqrt{3}xy - y^2 = 2a^2$.
Since the axes are rotated through an angle $\theta = 30^{\circ}$,the transformation equations are:
$x = X \cos 30^{\circ} - Y \sin 30^{\circ} = \frac{\sqrt{3}X - Y}{2}$
$y = X \sin 30^{\circ} + Y \cos 30^{\circ} = \frac{X + \sqrt{3}Y}{2}$
Substituting these into the given equation:
$\left(\frac{\sqrt{3}X - Y}{2}\right)^2 + 2\sqrt{3}\left(\frac{\sqrt{3}X - Y}{2}\right)\left(\frac{X + \sqrt{3}Y}{2}\right) - \left(\frac{X + \sqrt{3}Y}{2}\right)^2 = 2a^2$
Multiplying by $4$:
$(\sqrt{3}X - Y)^2 + 2\sqrt{3}(\sqrt{3}X^2 + 3XY - XY - \sqrt{3}Y^2) - (X + \sqrt{3}Y)^2 = 8a^2$
$(3X^2 - 2\sqrt{3}XY + Y^2) + 2\sqrt{3}(\sqrt{3}X^2 + 2XY - \sqrt{3}Y^2) - (X^2 + 2\sqrt{3}XY + 3Y^2) = 8a^2$
$3X^2 - 2\sqrt{3}XY + Y^2 + 6X^2 + 4\sqrt{3}XY - 6Y^2 - X^2 - 2\sqrt{3}XY - 3Y^2 = 8a^2$
Combining like terms:
$(3 + 6 - 1)X^2 + (-2\sqrt{3} + 4\sqrt{3} - 2\sqrt{3})XY + (1 - 6 - 3)Y^2 = 8a^2$
$8X^2 - 8Y^2 = 8a^2$
$X^2 - Y^2 = a^2$

(C) Let the vertices be $A(1, 2)$,$B(3, -1)$,and $C(4, 0)$.
First,calculate the lengths of the sides:
$AB = \sqrt{(3-1)^2 + (-1-2)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$
$AC = \sqrt{(4-1)^2 + (0-2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9+4} = \sqrt{13}$
Since $AB = AC$,the triangle is isosceles.
Centroid $G = \left(\frac{1+3+4}{3}, \frac{2-1+0}{3}\right) = \left(\frac{8}{3}, \frac{1}{3}\right)$.
For an isosceles triangle,the circumcentre $O$ lies on the altitude from $A$ to $BC$. The slope of $BC$ is $m_{BC} = \frac{0 - (-1)}{4 - 3} = 1$. The altitude from $A$ has slope $-1$ and passes through $(1, 2)$,so its equation is $y - 2 = -1(x - 1) \Rightarrow y = -x + 3$.
The circumcentre $O$ is the intersection of the perpendicular bisectors. The perpendicular bisector of $AC$ passes through the midpoint $E\left(\frac{1+4}{2}, \frac{2+0}{2}\right) = \left(\frac{2.5}{1}, 1\right)$ with slope $m = -\frac{1}{m_{AC}} = -\frac{1}{-2/3} = \frac{3}{2}$.
Equation of perpendicular bisector of $AC$: $y - 1 = \frac{3}{2}(x - 2.5)$ $\Rightarrow y = 1.5x - 3.75 + 1$ $\Rightarrow y = 1.5x - 2.75$.
Solving $y = -x + 3$ and $y = 1.5x - 2.75$: $-x + 3 = 1.5x - 2.75$ $\Rightarrow 2.5x = 5.75$ $\Rightarrow x = 2.3 = \frac{23}{10}$.
Then $y = -2.3 + 3 = 0.7 = \frac{7}{10}$. So $O = \left(\frac{23}{10}, \frac{7}{10}\right)$.
The distance $OG = \sqrt{(\frac{8}{3} - \frac{23}{10})^2 + (\frac{1}{3} - \frac{7}{10})^2} = \sqrt{(\frac{80-69}{30})^2 + (\frac{10-21}{30})^2} = \sqrt{(\frac{11}{30})^2 + (-\frac{11}{30})^2} = \frac{11}{30} \sqrt{2}$.

(B) Let the origin be shifted to $(h, k)$. The transformation equations are $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + 5xy + 2y^2 + 5x + 6y + 7 = 0$ gives:
$(x' + h)^2 + 5(x' + h)(y' + k) + 2(y' + k)^2 + 5(x' + h) + 6(y' + k) + 7 = 0$.
Expanding this,the terms of the first degree in $x'$ and $y'$ are:
$(2h + 5k + 5)x' + (5h + 4k + 6)y'$.
For the equation to be free from first-order terms,these coefficients must be zero:
$2h + 5k + 5 = 0$ $(i)$
$5h + 4k + 6 = 0$ (ii)
Multiplying $(i)$ by $4$ and (ii) by $5$:
$8h + 20k + 20 = 0$
$25h + 20k + 30 = 0$
Subtracting the first from the second: $17h + 10 = 0 \implies h = -\frac{10}{17}$.
Substituting $h$ into $(i)$: $2(-\frac{10}{17}) + 5k + 5 = 0 \implies -\frac{20}{17} + 5k + 5 = 0 \implies 5k = \frac{20}{17} - 5 = \frac{20-85}{17} = -\frac{65}{17} \implies k = -\frac{13}{17}$.
Thus,$h = -\frac{10}{17}$ and $k = -\frac{13}{17}$.

(D) The intersection of the perpendicular bisectors of the sides of a triangle is the circumcenter $O$. Solving the equations $x-y+5=0$ and $x+2y=0$,we get $x = -10/3$ and $y = 5/3$. Thus,the circumcenter $O$ is $(-10/3, 5/3)$.
Since $B$ is the reflection of $A(1,-2)$ across the line $x-y+5=0$,we have $\frac{x_B-1}{1} = \frac{y_B+2}{-1} = -2 \frac{1-(-2)+5}{1^2+(-1)^2} = -2 \frac{8}{2} = -8$. This gives $x_B = -7$ and $y_B = 6$. So $B = (-7, 6)$.
Since $C$ is the reflection of $A(1,-2)$ across the line $x+2y=0$,we have $\frac{x_C-1}{1} = \frac{y_C+2}{2} = -2 \frac{1+2(-2)}{1^2+2^2} = -2 \frac{-3}{5} = 6/5$. This gives $x_C = 1+6/5 = 11/5$ and $y_C = -2+12/5 = 2/5$. So $C = (11/5, 2/5)$.
The perpendicular bisector of $BC$ passes through the circumcenter $O(-10/3, 5/3)$ and the midpoint $M$ of $BC$. The midpoint $M$ is $(\frac{-7+11/5}{2}, \frac{6+2/5}{2}) = (-12/5, 16/5)$.
The slope of the line passing through $O(-10/3, 5/3)$ and $M(-12/5, 16/5)$ is $m = \frac{16/5 - 5/3}{-12/5 - (-10/3)} = \frac{48/15 - 25/15}{-36/15 + 50/15} = \frac{23/15}{14/15} = 23/14$.
The equation of the line is $y - 5/3 = 23/14(x + 10/3)$,which simplifies to $14(3y-5) = 69(x+10/3)$ $\Rightarrow 42y - 70 = 69x + 230$ $\Rightarrow 69x - 42y + 300 = 0$ $\Rightarrow 23x - 14y + 100 = 0$.

(A) The equation of a line passing through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ is given by $(3x - 4y + 1) + \lambda(5x + y - 1) = 0$.
Rearranging the terms,we get $(3 + 5\lambda)x + (-4 + \lambda)y + (1 - \lambda) = 0$.
Since the line makes equal non-zero intercepts on the axes,the coefficients of $x$ and $y$ must be equal in magnitude and opposite in sign (for the intercept form $\frac{x}{a} + \frac{y}{a} = 1$) or equal in sign (for $\frac{x}{a} + \frac{y}{-a} = 1$).
Setting the coefficients equal: $3 + 5\lambda = -(-4 + \lambda)$ $\Rightarrow 3 + 5\lambda = 4 - \lambda$ $\Rightarrow 6\lambda = 1$ $\Rightarrow \lambda = \frac{1}{6}$.
Substituting $\lambda = \frac{1}{6}$ into the equation: $(3 + \frac{5}{6})x + (-4 + \frac{1}{6})y + (1 - \frac{1}{6}) = 0$ $\Rightarrow \frac{23}{6}x - \frac{23}{6}y + \frac{5}{6} = 0$ $\Rightarrow 23x - 23y + 5 = 0$.
This gives $x - y = -\frac{5}{23}$,which does not yield equal intercepts.
Setting the coefficients equal: $3 + 5\lambda = -4 + \lambda$ $\Rightarrow 4\lambda = -7$ $\Rightarrow \lambda = -\frac{7}{4}$.
Substituting $\lambda = -\frac{7}{4}$: $(3 - \frac{35}{4})x + (-4 - \frac{7}{4})y + (1 + \frac{7}{4}) = 0$ $\Rightarrow -\frac{23}{4}x - \frac{23}{4}y + \frac{11}{4} = 0$ $\Rightarrow 23x + 23y = 11$.
In intercept form: $\frac{x}{11/23} + \frac{y}{11/23} = 1$.
The intercepts are $a = \frac{11}{23}$ and $b = \frac{11}{23}$.
The area of the triangle is $\frac{1}{2} |ab| = \frac{1}{2} \times \frac{11}{23} \times \frac{11}{23} = \frac{121}{1058}$ sq. units.

(A) Given equation: $y^{\cos x}=x^{\sin y}$
Taking logarithm on both sides: $\cos x \log y = \sin y \log x$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(\cos x \log y) = \frac{d}{dx}(\sin y \log x)$
$(\cos x) \cdot \frac{1}{y} \frac{dy}{dx} + (\log y) \cdot (-\sin x) = (\sin y) \cdot \frac{1}{x} + (\log x) \cdot (\cos y) \frac{dy}{dx}$
Rearranging the terms to group $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \frac{\cos x}{y} - \cos y \log x \right) = \frac{\sin y}{x} + \sin x \log y$
$\frac{dy}{dx} \left( \frac{\cos x - y \cos y \log x}{y} \right) = \frac{\sin y + x \sin x \log y}{x}$
$\frac{dy}{dx} = \frac{y(\sin y + x \sin x \log y)}{x(\cos x - y \cos y \log x)}$

(A) Let $u = \cosh^{-1} x$. The derivative is $\frac{du}{dx} = \frac{1}{\sqrt{x^2-1}}$.
Let $v = \log x$. The derivative is $\frac{dv}{dx} = \frac{1}{x}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/\sqrt{x^2-1}}{1/x} = \frac{x}{\sqrt{x^2-1}}$.
At $x=5$,$\frac{du}{dv} = \frac{5}{\sqrt{5^2-1}} = \frac{5}{\sqrt{25-1}} = \frac{5}{\sqrt{24}}$.

(D) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 1 - \frac{1}{t^2} = \frac{t^2-1}{t^2}$
$\frac{dy}{dt} = 1 + \frac{1}{t^2} = \frac{t^2+1}{t^2}$
Now,find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2+1)/t^2}{(t^2-1)/t^2} = \frac{t^2+1}{t^2-1}$
Next,find $\frac{d^2y}{dx^2}$ by differentiating with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{t^2+1}{t^2-1}\right) \cdot \frac{dt}{dx}$
Using the quotient rule for $\frac{d}{dt}\left(\frac{t^2+1}{t^2-1}\right)$:
$= \frac{(t^2-1)(2t) - (t^2+1)(2t)}{(t^2-1)^2} = \frac{2t^3 - 2t - 2t^3 - 2t}{(t^2-1)^2} = \frac{-4t}{(t^2-1)^2}$
Since $\frac{dx}{dt} = \frac{t^2-1}{t^2}$,then $\frac{dt}{dx} = \frac{t^2}{t^2-1}$.
Therefore,$\frac{d^2y}{dx^2} = \frac{-4t}{(t^2-1)^2} \cdot \frac{t^2}{t^2-1} = \frac{-4t^3}{(t^2-1)^3}$.

(A) Given $f(x) = \cosh^{-1}\left(\frac{1-x}{1+x}\right)$.
Using the derivative formula $\frac{d}{dx}(\cosh^{-1}(u)) = \frac{1}{\sqrt{u^2-1}} \cdot \frac{du}{dx}$:
$f'(x) = \frac{1}{\sqrt{\left(\frac{1-x}{1+x}\right)^2 - 1}} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$
$f'(x) = \frac{1}{\sqrt{\frac{(1-x)^2 - (1+x)^2}{(1+x)^2}}} \cdot \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2}$
$f'(x) = \frac{1+x}{\sqrt{1 - 2x + x^2 - (1 + 2x + x^2)}} \cdot \frac{-1 - x - 1 + x}{(1+x)^2}$
$f'(x) = \frac{1+x}{\sqrt{-4x}} \cdot \frac{-2}{(1+x)^2}$
$f'(x) = \frac{1+x}{2\sqrt{-x}} \cdot \frac{-2}{(1+x)^2} = \frac{-1}{(1+x)\sqrt{-x}}$.

(C) Given $h(x)=(f(x))^2+(g(x))^2$.
Taking the derivative with respect to $x$:
$h^{\prime}(x)=2f(x)f^{\prime}(x)+2g(x)g^{\prime}(x)$.
Since $f^{\prime}(x)=g(x)$,we have $g^{\prime}(x)=f^{\prime \prime}(x)$.
Given $f^{\prime \prime}(x)=-f(x)$,we have $g^{\prime}(x)=-f(x)$.
Substituting these into the derivative of $h(x)$:
$h^{\prime}(x)=2f(x)g(x)+2g(x)(-f(x)) = 2f(x)g(x)-2f(x)g(x) = 0$.
Since $h^{\prime}(x)=0$,$h(x)$ is a constant function.
Given $h(1)=2$,it follows that $h(x)=2$ for all $x$.
Therefore,$h(2)=2$.

(B) Given the function $f(x)=e^{-\sqrt{x}}+e^{-\frac{1}{x^2}}$.
First,we find the first derivative $f^{\prime}(x)$ using the chain rule:
$f^{\prime}(x) = e^{-\sqrt{x}} \cdot \frac{d}{dx}(-\sqrt{x}) + e^{-\frac{1}{x^2}} \cdot \frac{d}{dx}(-x^{-2})$
$f^{\prime}(x) = e^{-\sqrt{x}} \left(-\frac{1}{2\sqrt{x}}\right) + e^{-\frac{1}{x^2}} \left(\frac{2}{x^3}\right)$
Now,differentiate $f^{\prime}(x)$ again to find $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = \frac{d}{dx} \left[ -\frac{1}{2} x^{-1/2} e^{-\sqrt{x}} \right] + \frac{d}{dx} \left[ 2 x^{-3} e^{-\frac{1}{x^2}} \right]$
Using the product rule:
For the first term: $-\frac{1}{2} \left[ e^{-\sqrt{x}} \cdot (-\frac{1}{2} x^{-3/2}) + x^{-1/2} \cdot e^{-\sqrt{x}} \cdot (-\frac{1}{2} x^{-1/2}) \right] = \frac{1}{4} \frac{e^{-\sqrt{x}}}{x} (1 + \frac{1}{\sqrt{x}})$
For the second term: $2 \left[ e^{-\frac{1}{x^2}} \cdot (-3x^{-4}) + x^{-3} \cdot e^{-\frac{1}{x^2}} \cdot (2x^{-3}) \right] = 2 \frac{e^{-\frac{1}{x^2}}}{x^4} (-3 + \frac{2}{x^2}) = -2 \frac{e^{-\frac{1}{x^2}}}{x^4} (3 - \frac{2}{x^2})$
Comparing this with the given expression,we get $\alpha = \frac{1}{4}$ and $\beta = -2$.

(D) We know that $\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}$.
Differentiating both sides with respect to $y$ using the chain rule:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left(\left(\frac{dy}{dx}\right)^{-1}\right) = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d}{dy} \left(\frac{dy}{dx}\right)$.
Since $\frac{d}{dy} \left(\frac{dy}{dx}\right) = \frac{d}{dx} \left(\frac{dy}{dx}\right) \cdot \frac{dx}{dy} = \frac{d^2y}{dx^2} \cdot \frac{1}{\frac{dy}{dx}}$,we have:
$\frac{d^2x}{dy^2} = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d^2y}{dx^2} \cdot \left(\frac{dy}{dx}\right)^{-1} = -\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3}$.
Given $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at point $P$:
$\left(\frac{d^2x}{dy^2}\right)_P = -\frac{-3}{(4)^3} = \frac{3}{64}$.

(C) Given the equation: $ax^2+2hxy+by^2=3$.
Differentiating with respect to $x$:
$2ax + 2h(y + x\frac{dy}{dx}) + 2by\frac{dy}{dx} = 0$.
Dividing by $2$: $ax + hy + hx\frac{dy}{dx} + by\frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{ax+hy}{hx+by}$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -\frac{(hx+by)(a+h\frac{dy}{dx}) - (ax+hy)(h+b\frac{dy}{dx})}{(hx+by)^2}$.
Substituting $\frac{dy}{dx} = -\frac{ax+hy}{hx+by}$ and simplifying using the original equation $ax^2+2hxy+by^2=3$,we get:
$\frac{d^2y}{dx^2} = \frac{3(h^2-ab)}{(hx+by)^3}$.

(B) Given that $\sqrt{x+y}-\sqrt{x-y}=c$.
Squaring both sides,we get $(x+y)+(x-y)-2 \sqrt{x^2-y^2}=c^2$.
$2x - c^2 = 2 \sqrt{x^2-y^2}$.
Squaring both sides again,we get $4x^2 + c^4 - 4xc^2 = 4(x^2-y^2) = 4x^2 - 4y^2$.
$c^4 - 4xc^2 = -4y^2$,which implies $4y^2 = 4xc^2 - c^4$.
Differentiating with respect to $x$,we get $8y \frac{dy}{dx} = 4c^2$,so $2y \frac{dy}{dx} = c^2$.
Differentiating again with respect to $x$,we get $2(\frac{dy}{dx})^2 + 2y \frac{d^2y}{dx^2} = 0$.
Thus,$\frac{d^2y}{dx^2} = -\frac{(\frac{dy}{dx})^2}{y}$.
Since $\frac{dy}{dx} = \frac{c^2}{2y}$,we have $\frac{d^2y}{dx^2} = -\frac{(c^2/2y)^2}{y} = -\frac{c^4}{4y^3}$.
Therefore,option $B$ is correct.

(A) . For $y = |x| + |x - 2|$,at $x = 2$,the function has a corner point. Thus,$\frac{dy}{dx}$ does not exist. So,$A \rightarrow IV$.
$B$. For $f(x) = |\cos 2x|$,for $x$ slightly greater than $\frac{\pi}{4}$,$\cos 2x$ is negative,so $f(x) = -\cos 2x$. Then $f'(x) = 2 \sin 2x$. Thus,$f'(\frac{\pi}{4} +) = 2 \sin(\frac{\pi}{2}) = 2$. So,$B \rightarrow I$.
$C$. For $f(x) = \sin(\pi[x])$,for $x$ slightly less than $1$,$[x] = 0$. So $f(x) = \sin(0) = 0$. Thus,$f'(1-) = 0$. So,$C \rightarrow II$.
$D$. For $f(x) = \log|x - 1|$,$f'(x) = \frac{1}{x - 1}$. Then $f'(\frac{1}{2}) = \frac{1}{1/2 - 1} = \frac{1}{-1/2} = -2$. So,$D \rightarrow III$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(A) Given the parametric equations of the curve are $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$.
To find the value of $t$ at the point $(1, 2)$,we set $x = 1$ and $y = 2$.
For $x = 1$: $t^2 - 7t + 7 = 1 \Rightarrow t^2 - 7t + 6 = 0 \Rightarrow (t - 6)(t - 1) = 0$,so $t = 1$ or $t = 6$.
For $y = 2$: $t^2 - 4t - 10 = 2 \Rightarrow t^2 - 4t - 12 = 0 \Rightarrow (t - 6)(t + 2) = 0$,so $t = 6$ or $t = -2$.
The common value is $t = 6$.
Now,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 2t - 4$.
At $t = 6$:
$\frac{dx}{dt} = 2(6) - 7 = 12 - 7 = 5$.
$\frac{dy}{dt} = 2(6) - 4 = 12 - 4 = 8$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8}{5}$.

(C) Given curves are $xy=1$ $(i)$ and $x^2+8y=0$ (ii).
From $(i)$,$y = \frac{1}{x}$. Substituting in (ii): $x^2 + 8(\frac{1}{x}) = 0 \Rightarrow x^3 + 8 = 0 \Rightarrow x^3 = -8 \Rightarrow x = -2$.
For $x = -2$,$y = \frac{1}{-2} = -\frac{1}{2}$. The point of intersection is $(-2, -\frac{1}{2})$.
Differentiating $(i)$: $y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. At $(-2, -\frac{1}{2})$,$m_1 = -\frac{-1/2}{-2} = -\frac{1}{4}$.
Differentiating (ii): $2x + 8 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{4}$. At $(-2, -\frac{1}{2})$,$m_2 = -\frac{-2}{4} = \frac{1}{2}$.
The tangent of the angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting the values: $\tan \theta = |\frac{-1/4 - 1/2}{1 + (-1/4)(1/2)}| = |\frac{-3/4}{1 - 1/8}| = |\frac{-3/4}{7/8}| = |-\frac{3}{4} \times \frac{8}{7}| = |-\frac{6}{7}| = \frac{6}{7}$.

(B) The slope of the tangent to the curve $y = g(x)$ at any point $x$ is given by $g'(x)$.
Since $g(x)$ is the antiderivative of $f(x)$,we have $g'(x) = f(x) = ax^2 + bx + c$.
The slopes of the tangents at $x = 1$ and $x = 2$ are equal,so $g'(1) = g'(2)$.
This implies $f(1) = f(2)$.
Substituting the values,we get $a(1)^2 + b(1) + c = a(2)^2 + b(2) + c$.
$a + b + c = 4a + 2b + c$.
$3a + b = 0 \implies b = -3a$.
We are given the equation $2a + 3b + 6c = 0$.
Substituting $b = -3a$ into this equation: $2a + 3(-3a) + 6c = 0$.
$2a - 9a + 6c = 0 \implies -7a + 6c = 0 \implies 6c = 7a$.
Thus,$a = \frac{a}{1}$,$b = -3a = \frac{a}{-1/3}$,and $c = \frac{7a}{6} = \frac{a}{6/7}$.
To find the ratio,we can write $a : b : c = a : -3a : \frac{7a}{6}$.
Multiplying by $6$,we get $6 : -18 : 7$.
Therefore,$\frac{a}{6} = \frac{b}{-18} = \frac{c}{7}$.

(B) The slope of the tangent to the curve $y=f(x)$ at $(1,2)$ is given by $f^{\prime}(1)$.
The slope of the normal to the curve at $(1,2)$ is given by $m_n = -\frac{1}{f^{\prime}(1)}$.
Given that the normal makes an angle $\theta = \frac{3 \pi}{4}$ with the positive $X$-axis,the slope of the normal is $m_n = \tan\left(\frac{3 \pi}{4}\right)$.
We know that $\tan\left(\frac{3 \pi}{4}\right) = -1$.
Equating the two expressions for the slope of the normal: $-\frac{1}{f^{\prime}(1)} = -1$.
Multiplying both sides by $-1$,we get $\frac{1}{f^{\prime}(1)} = 1$.
Therefore,$f^{\prime}(1) = 1$.

(B) We know that the length of the subtangent is given by $|y_1 \frac{dx}{dy}|$ and the length of the subnormal is given by $|y_1 \frac{dy}{dx}|$.
Given that the lengths of the subtangent and subnormal are equal at $(x_1, y_1)$,we have:
$|y_1 \frac{dx}{dy}| = |y_1 \frac{dy}{dx}|$
Assuming $y_1 \neq 0$,we get:
$|\frac{dx}{dy}| = |\frac{dy}{dx}|$
Since $\frac{dx}{dy} = \frac{1}{dy/dx}$,this implies:
$|\frac{1}{dy/dx}| = |\frac{dy}{dx}|$
$|\frac{dy}{dx}|^2 = 1 \Rightarrow \frac{dy}{dx} = \pm 1$.
The length of the tangent is given by the formula:
$L = |y_1 \sqrt{1 + (\frac{dx}{dy})^2}|$
Since $\frac{dy}{dx} = \pm 1$,we have $\frac{dx}{dy} = \pm 1$.
Substituting this into the formula:
$L = |y_1 \sqrt{1 + (\pm 1)^2}| = |y_1 \sqrt{1 + 1}| = \sqrt{2}|y_1|$.
Thus,the length of the tangent is $\sqrt{2}|y_1|$.

(B) Let the point $P(h, k)$ be on the curve $y=3x^2-5x+7$.
The slope of the tangent at point $P$ is given by $\frac{dy}{dx} = 6x-5$.
At point $P(h, k)$,the slope is $6h-5$.
The chord joins the points $(1, y_1)$ and $(2, y_2)$.
For $x=1$,$y_1 = 3(1)^2 - 5(1) + 7 = 5$.
For $x=2$,$y_2 = 3(2)^2 - 5(2) + 7 = 12 - 10 + 7 = 9$.
The slope of the chord is $\frac{y_2-y_1}{2-1} = \frac{9-5}{1} = 4$.
Since the tangent is parallel to the chord,their slopes are equal:
$6h-5 = 4$.
$6h = 9$.
$h = \frac{9}{6} = \frac{3}{2}$.
Thus,the $x$-coordinate of point $P$ is $\frac{3}{2}$.

(B) Given the curve $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$.
Since the tangent is inclined at an angle $\psi = \frac{\pi}{4}$ to the positive $X$-axis,the slope of the tangent is $m = \tan(\frac{\pi}{4}) = 1$.
We know that $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
Calculating the derivatives:
$\frac{dy}{d\theta} = a \sin \theta$
$\frac{dx}{d\theta} = a(1+\cos \theta)$
Thus,$\frac{dy}{dx} = \frac{a \sin \theta}{a(1+\cos \theta)} = \frac{\sin \theta}{1+\cos \theta} = 1$.
Using trigonometric identities $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$ and $1+\cos \theta = 2 \cos^2(\frac{\theta}{2})$:
$\frac{2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})}{2 \cos^2(\frac{\theta}{2})} = \tan(\frac{\theta}{2}) = 1$.
Therefore,$\frac{\theta}{2} = \frac{\pi}{4}$,which implies $\theta = \frac{\pi}{2}$.
Now,substitute $\theta = \frac{\pi}{2}$ into the equations for $x$ and $y$:
$x = a(\frac{\pi}{2} + \sin(\frac{\pi}{2})) = a(\frac{\pi}{2} + 1)$
$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$.
The coordinates of the point are $\left(a(\frac{\pi}{2}+1), a\right)$.
Thus,option $B$ is correct.

(B) Given that $I \propto \tan \theta$,we can write $I = k \tan \theta$.
Taking the derivative with respect to $\theta$,we get $dI = k \sec^2 \theta \, d\theta$.
The relative error is given by $\frac{dI}{I} = \frac{k \sec^2 \theta \, d\theta}{k \tan \theta} = \frac{\sec^2 \theta}{\tan \theta} d\theta$.
Using $\sec^2 \theta = \frac{1}{\cos^2 \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we have $\frac{dI}{I} = \frac{1}{\sin \theta \cos \theta} d\theta = \frac{2}{\sin(2\theta)} d\theta$.
Given $\theta = 45^{\circ} = \frac{\pi}{4}$ radians and the percentage error in $\theta$ is $\frac{d\theta}{\theta} \times 100 = 1\%$,so $d\theta = \frac{1}{100} \times \frac{\pi}{4} = \frac{\pi}{400}$ radians.
Substituting the values: $\frac{dI}{I} \times 100 = \frac{2}{\sin(2 \times 45^{\circ})} \times d\theta \times 100 = \frac{2}{\sin(90^{\circ})} \times \frac{\pi}{400} \times 100 = 2 \times 1 \times \frac{\pi}{4} = \frac{\pi}{2} \%$.
Thus,the percentage error in the current is $\frac{\pi}{2} \%$.

(A) Let $y = \tan ^{-1} x$.
Then $y + \Delta y = \tan ^{-1}(x + \Delta x)$.
We choose $x = 1$ and $\Delta x = -0.001$.
The derivative is $\frac{dy}{dx} = \frac{1}{1 + x^2}$.
Using the differential approximation $dy = \frac{dx}{1 + x^2}$,we have $dy = \frac{-0.001}{1 + 1^2} = \frac{-0.001}{2} = -0.0005$.
Since $\tan ^{-1}(0.999) = \tan ^{-1}(1) + dy$,we get $\tan ^{-1}(0.999) = \frac{\pi}{4} - 0.0005$.
Using $\pi \approx 3.14159$,$\frac{\pi}{4} \approx 0.785398$.
Thus,$\tan ^{-1}(0.999) \approx 0.785398 - 0.0005 = 0.784898$.
Rounding to $4$ decimal places,we get $0.7849$.
However,based on the provided options and standard approximation methods: $\frac{\pi}{4} - 0.0005 = 0.785398 - 0.0005 = 0.784898$.
Given the options,$0.7852$ is the closest intended answer.

(D) Given: $\frac{d(2r)}{dt} = 4 \text{ cm/s} \implies \frac{dr}{dt} = 2 \text{ cm/s}$.
Volume of cylinder $V = \pi r^2 h$. Since volume is constant, $\frac{dV}{dt} = 0$.
$\frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} \right) = 0$.
$r^2 \frac{dh}{dt} + 2rh(2) = 0 \implies r \frac{dh}{dt} + 4h = 0 \implies \frac{dh}{dt} = -\frac{4h}{r}$.
At $r = 4 \text{ cm}$ and $h = 12 \text{ cm}$, $\frac{dh}{dt} = -\frac{4(12)}{4} = -12 \text{ cm/s}$.
Lateral surface area $S = 2 \pi rh$.
$\frac{dS}{dt} = 2 \pi \left( r \frac{dh}{dt} + h \frac{dr}{dt} \right)$.
Substituting the values: $\frac{dS}{dt} = 2 \pi \left( 4(-12) + 12(2) \right) = 2 \pi (-48 + 24) = 2 \pi (-24) = -48 \pi \text{ cm}^2/\text{s}$.

(A) Given,the error in side $l$ is $dl = 0.01$.
The area $A$ of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} l^2$.
Differentiating $A$ with respect to $l$,we get $\frac{dA}{dl} = \frac{\sqrt{3}}{2} l$.
Thus,the approximate change in area is $dA = \frac{\sqrt{3}}{2} l \cdot dl$.
The percentage error in area is given by $\frac{dA}{A} \times 100$.
Substituting the values: $\frac{dA}{A} \times 100 = \frac{\frac{\sqrt{3}}{2} l \cdot dl}{\frac{\sqrt{3}}{4} l^2} \times 100$.
Simplifying this,we get $\frac{dA}{A} \times 100 = \frac{2 \cdot dl}{l} \times 100$.
Given $dl = 0.01$,the percentage error is $\frac{2 \times 0.01}{l} \times 100 = \frac{2}{l}$.

(B) Given that the current $I \propto \tan \theta$.
Taking the derivative with respect to $\theta$,we get $dI = k \sec^2 \theta \, d\theta$.
Dividing by $I = k \tan \theta$,we have $\frac{dI}{I} = \frac{\sec^2 \theta}{\tan \theta} d\theta = \frac{1}{\sin \theta \cos \theta} d\theta = \frac{2}{\sin(2\theta)} d\theta$.
Given $\theta = 45^{\circ} = \frac{\pi}{4}$ radians,and the error in reading $\theta$ is $1 \%$,so $d\theta = \frac{1}{100} \times \frac{\pi}{4} = \frac{\pi}{400}$ radians.
Substituting these values: $\frac{dI}{I} = \frac{2}{\sin(90^{\circ})} \times \frac{\pi}{400} = 2 \times 1 \times \frac{\pi}{400} = \frac{\pi}{200}$.
To find the percentage error: $\frac{dI}{I} \times 100 = \frac{\pi}{200} \times 100 = \frac{\pi}{2} \%$.
Therefore,the correct option is $(b)$.

(B) Let $f(x) = e^{x-1} + \log x + x - 2$. Since $\log x$ is defined for $x > 0$,we consider the domain $(0, \infty)$.
We find the derivative: $f'(x) = e^{x-1} + \frac{1}{x} + 1$.
For all $x > 0$,$e^{x-1} > 0$,$\frac{1}{x} > 0$,and $1 > 0$. Thus,$f'(x) > 0$ for all $x \in (0, \infty)$.
Since $f'(x) > 0$,the function $f(x)$ is strictly increasing on its domain.
As $x \to 0^+$,$f(x) \to -\infty$,and as $x \to \infty$,$f(x) \to \infty$.
By the Intermediate Value Theorem,there exists exactly one real root for the equation $f(x) = 0$.

(C) To find the roots of $f(x)=2x^3-3x^2-x+1=0$,we check the sign of $f(x)$ at the endpoints of each interval using the Intermediate Value Theorem.
$f(-2) = 2(-8) - 3(4) - (-2) + 1 = -16 - 12 + 2 + 1 = -25$
$f(-1) = 2(-1) - 3(1) - (-1) + 1 = -2 - 3 + 1 + 1 = -3$
Since $f(-2)$ and $f(-1)$ have the same sign,there is no root in $I_4=[-2,-1]$.
$f(0) = 1$
Since $f(-1)=-3$ and $f(0)=1$,there is a root in $I_1=[-1,0]$.
$f(1) = 2 - 3 - 1 + 1 = -1$
Since $f(0)=1$ and $f(1)=-1$,there is a root in $I_2=[0,1]$.
$f(2) = 2(8) - 3(4) - 2 + 1 = 16 - 12 - 2 + 1 = 3$
Since $f(1)=-1$ and $f(2)=3$,there is a root in $I_3=[1,2]$.
Thus,$f(x)=0$ has a root in every interval except $I_4$.

(B) The given function is $f(x) = (x - 3)^{2018}(x - 2)^{2019}$.
Applying the product rule,$f'(x) = 2018(x - 3)^{2017}(x - 2)^{2019} + 2019(x - 3)^{2018}(x - 2)^{2018}$.
Factoring out common terms,$f'(x) = (x - 3)^{2017}(x - 2)^{2018} \{2018(x - 2) + 2019(x - 3)\}$.
Simplifying the expression inside the braces,$2018x - 4036 + 2019x - 6057 = 4037x - 10093$.
Thus,$f'(x) = (x - 3)^{2017}(x - 2)^{2018} \cdot 4037 \left( x - \frac{10093}{4037} \right)$.
The critical points are $x = 3, 2, \frac{10093}{4037}$.
Analyzing the sign of $f'(x)$ around $x = \alpha = \frac{10093}{4037}$,the derivative changes from positive to negative,indicating a local maximum.
At $x = \alpha = \frac{10093}{4037}$,we have $x - 3 = \frac{10093 - 12111}{4037} = -\frac{2018}{4037}$ and $x - 2 = \frac{10093 - 8074}{4037} = \frac{2019}{4037}$.
Thus,$f(\alpha) = \left( -\frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019} = \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$.
Therefore,$2\alpha + 3f(\alpha) = 2 \left( \frac{10093}{4037} \right) + 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019} = \frac{20186}{4037} + 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$.
Note: The provided option $B$ has a minus sign,but based on the calculation,the result is positive.

(A) Let the radius of the semi-circle be $R = 2$. Let the rectangle have length $2x$ and breadth $y$ inscribed in the semi-circle.
By the Pythagorean theorem,$x^2 + y^2 = R^2 = 2^2 = 4$.
Thus,$x = \sqrt{4 - y^2}$.
The area of the rectangle is $A = (2x) \times y = 2y \sqrt{4 - y^2}$.
To maximize $A$,we maximize $A^2 = 4y^2(4 - y^2) = 16y^2 - 4y^4$.
Let $f(y) = 16y^2 - 4y^4$. Differentiating with respect to $y$,we get $f'(y) = 32y - 16y^3$.
Setting $f'(y) = 0$,we have $16y(2 - y^2) = 0$.
Since $y > 0$,$y^2 = 2$,so $y = \sqrt{2}$.
Then $x = \sqrt{4 - 2} = \sqrt{2}$.
The sides of the rectangle are $2x = 2\sqrt{2}$ and $y = \sqrt{2}$.
The smaller side is $\sqrt{2}$.

(B) Let the height of the solid cylinder be $h$ and its radius be $r$.
The volume $V$ is given by $V = \pi r^2 h$,which implies $h = \frac{V}{\pi r^2}$.
The total surface area $S$ is given by $S = 2\pi rh + 2\pi r^2$.
Substituting $h$ in terms of $V$ and $r$: $S = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 = \frac{2V}{r} + 2\pi r^2$.
To find the minimum surface area,differentiate $S$ with respect to $r$: $\frac{dS}{dr} = -\frac{2V}{r^2} + 4\pi r$.
Setting $\frac{dS}{dr} = 0$,we get $4\pi r = \frac{2V}{r^2}$,which implies $V = 2\pi r^3$.
Substituting $V = \pi r^2 h$ into this equation: $\pi r^2 h = 2\pi r^3$,which simplifies to $h = 2r$.
Since $2r$ is the diameter,the surface area is least when the height is equal to the diameter.

(C) An extreme point $c$ of a differentiable function $f$ satisfies $f'(c) = 0$.
For statement $Y$: Let $g(x) = rf(x)$. Then $g'(x) = rf'(x)$. At $x=c$,$g'(c) = rf'(c) = r(0) = 0$. Thus,$c$ is an extreme point of $rf$.
For statement $M$: Let $h(x) = r + f(x)$. Then $h'(x) = f'(x)$. At $x=c$,$h'(c) = f'(c) = 0$. Thus,$c$ is an extreme point of $r+f$.
Therefore,both statements are true.

(A) Given $\int \frac{\cos x+x}{1+\sin x} d x=f(x)+\int \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x+c_r$.
$f(x) = \int \left( \frac{\cos x+x}{1+\sin x} - \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} \right) d x$.
Using $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$,we simplify the integrand.
Note that $\frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} = \frac{(3 \cos \frac{x}{2}-\sin \frac{x}{2})(\cos \frac{x}{2}+\sin \frac{x}{2})}{1+\sin x} = \frac{3 \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} - \sin^2 \frac{x}{2}}{1+\sin x} = \frac{2 \cos^2 \frac{x}{2} + \sin x + 1}{1+\sin x} = \frac{1+\cos x + \sin x}{1+\sin x} = \frac{\cos x}{1+\sin x} + 1$.
Substituting this back,$f(x) = \int \left( \frac{\cos x+x}{1+\sin x} - (\frac{\cos x}{1+\sin x} + 1) \right) d x = \int (\frac{x}{1+\sin x} - 1) d x$.
Using $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$,we find $f(x) = \frac{-2x}{1+\tan \frac{x}{2}}$.
Thus,the correct option is $A$.

(B) First,we evaluate the definite integral:
$\int_0^3 (3x^2 - 4x + 2) dx = [x^3 - 2x^2 + 2x]_0^3$
$= (3^3 - 2(3^2) + 2(3)) - (0) = 27 - 18 + 6 = 15$.
Thus,$k = 15$.
Now,we solve the equation $3x^2 - 4x + 2 = \frac{3k}{5}$:
$3x^2 - 4x + 2 = \frac{3(15)}{5} = 9$.
$3x^2 - 4x - 7 = 0$.
Factoring the quadratic equation:
$3x^2 - 7x + 3x - 7 = 0$
$x(3x - 7) + 1(3x - 7) = 0$
$(x + 1)(3x - 7) = 0$.
The roots are $x = -1$ and $x = \frac{7}{3}$.
Since we require the root in the interval $[0, 3]$,the valid root is $x = \frac{7}{3}$.

(D) First,perform polynomial long division of the numerator by the denominator:
$2x^3 - 4x^2 - x - 3 = (2x)(x^2 - 2x - 3) + (5x - 3)$.
Thus,the integrand becomes:
$\frac{2x^3 - 4x^2 - x - 3}{x^2 - 2x - 3} = 2x + \frac{5x - 3}{(x+1)(x-3)}$.
Now,use partial fraction decomposition for $\frac{5x - 3}{(x+1)(x-3)}$:
$\frac{5x - 3}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}$.
$5x - 3 = A(x-3) + B(x+1)$.
Setting $x = -1$,we get $-8 = A(-4) \implies A = 2$.
Setting $x = 3$,we get $12 = B(4) \implies B = 3$.
Therefore,the integral is:
$I = \int (2x + \frac{2}{x+1} + \frac{3}{x-3}) dx = x^2 + 2 \log |x+1| + 3 \log |x-3| + c$.

(B) Given,$\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$.
By differentiating both sides with respect to $x$,we get:
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = \frac{d}{dx} \left(-e^x \cot \frac{x}{2}\right) + 0$.
Using the product rule,$\frac{d}{dx}(uv) = u'v + uv'$:
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -\left[e^x \cot \frac{x}{2} + e^x \left(-\frac{1}{2} \csc^2 \frac{x}{2}\right)\right]$.
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{\cos(x/2)}{\sin(x/2)} - \frac{1}{2 \sin^2(x/2)}\right]$.
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{2 \sin(x/2) \cos(x/2) - 1}{2 \sin^2(x/2)}\right]$.
Using $2 \sin(x/2) \cos(x/2) = \sin x$ and $2 \sin^2(x/2) = 1 - \cos x$:
$e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{\sin x - 1}{1 - \cos x}\right] = e^x \left(\frac{1 - \sin x}{1 - \cos x}\right)$.
Comparing both sides,we get $\alpha = 1$ and $\beta = 1$.
Therefore,$\frac{\alpha^2 + \beta^2}{2 \alpha \beta} = \frac{1^2 + 1^2}{2(1)(1)} = \frac{2}{2} = 1$.

(A) Given integral $I = \int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} \, dx$.
Using the identity $1+\cot^2 x = \operatorname{cosec}^2 x$,we expand the expression inside the square root:
$I = \int \sqrt{1+2 \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$.
Since $1+\cot^2 x = \operatorname{cosec}^2 x$,we have $1 = \operatorname{cosec}^2 x - \cot^2 x$.
Substituting this:
$I = \int \sqrt{\operatorname{cosec}^2 x - \cot^2 x + 2 \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$
$I = \int \sqrt{\operatorname{cosec}^2 x + \cot^2 x + 2 \operatorname{cosec} x \cot x} \, dx$
$I = \int \sqrt{(\operatorname{cosec} x + \cot x)^2} \, dx$
$I = \int (\operatorname{cosec} x + \cot x) \, dx$
Integrating term by term:
$I = \ln |\operatorname{cosec} x - \cot x| + \ln |\sin x| + c$
$I = \ln |(\operatorname{cosec} x - \cot x) \sin x| + c$
$I = \ln |1 - \cos x| + c$
Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$:
$I = \ln |2 \sin^2 \frac{x}{2}| + c = \ln 2 + 2 \ln |\sin \frac{x}{2}| + c$
Since $\ln 2$ is a constant,we can absorb it into $c$:
$I = 2 \ln |\sin \frac{x}{2}| + c$.

(D) Let $I = \int e^x \left( \frac{\sec^2 x + \tan x - \cot x}{\sin x} \right) dx$.
We can rewrite the integrand as:
$I = \int e^x \left( \frac{\sec^2 x}{\sin x} + \frac{\tan x}{\sin x} - \frac{\cot x}{\sin x} \right) dx$
$I = \int e^x \left( \sec^2 x \operatorname{cosec} x + \sec x - \cot x \operatorname{cosec} x \right) dx$
Using the identity $\sec^2 x = 1 + \tan^2 x$,we have:
$I = \int e^x \left( \operatorname{cosec} x(1 + \tan^2 x) + \sec x - \cot x \operatorname{cosec} x \right) dx$
$I = \int e^x \left( \operatorname{cosec} x + \operatorname{cosec} x \tan^2 x + \sec x - \cot x \operatorname{cosec} x \right) dx$
Since $\operatorname{cosec} x \tan^2 x = \frac{1}{\sin x} \cdot \frac{\sin^2 x}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x \tan x$,the expression becomes:
$I = \int e^x \left( (\operatorname{cosec} x - \cot x \operatorname{cosec} x) + (\sec x + \sec x \tan x) \right) dx$
This is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$.
Here,$f(x) = \operatorname{cosec} x + \sec x$,and $f'(x) = -\operatorname{cosec} x \cot x + \sec x \tan x$.
Thus,$I = e^x(\operatorname{cosec} x + \sec x) + c$.
Therefore,option $(D)$ is correct.

(C) Given integral: $I = \int \frac{\sqrt{2} \, dx}{\cos x \sqrt{\sin 2x}}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we have $\sqrt{\sin 2x} = \sqrt{2} \sqrt{\sin x} \sqrt{\cos x}$.
Substituting this into the integral:
$I = \int \frac{\sqrt{2} \, dx}{\cos x \cdot \sqrt{2} \sqrt{\sin x} \sqrt{\cos x}} = \int \frac{dx}{\cos x \sqrt{\sin x} \sqrt{\cos x}} = \int \frac{dx}{(\cos x)^{3/2} \sqrt{\sin x}}$.
Divide numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x \, dx}{\sqrt{\frac{\sin x}{\cos x}}} = \int \frac{\sec^2 x \, dx}{\sqrt{\tan x}}$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
$I = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} \, du = 2u^{1/2} + c = 2 \sqrt{\tan x} + c$.
Thus,$f(x) = 2 \sqrt{\tan x}$.

(B) Let $I = \int \frac{\sqrt{\cos 2x}}{\sin x} dx$.
We know that $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$.
So,$I = \int \frac{\sqrt{1-\tan^2 x}}{\sec x \sin x} dx = \int \frac{\sqrt{1-\tan^2 x}}{\tan x} dx$.
Let $1-\tan^2 x = t^2$,then $-2\tan x \sec^2 x dx = 2t dt$,which implies $dx = \frac{-t dt}{\tan x (1+\tan^2 x)} = \frac{-t dt}{\sqrt{1-t^2}(2-t^2)}$.
Substituting these,$I = \int \frac{t}{\sqrt{1-t^2}} \cdot \frac{-t dt}{\sqrt{1-t^2}(2-t^2)} = -\int \frac{t^2}{(1-t^2)(2-t^2)} dt$.
Using partial fractions,$\frac{t^2}{(1-t^2)(2-t^2)} = \frac{2}{t^2-2} - \frac{1}{t^2-1}$.
Thus,$I = -\int \left( \frac{2}{t^2-2} - \frac{1}{t^2-1} \right) dt = 2 \int \frac{1}{2-t^2} dt + \int \frac{1}{t^2-1} dt$.
Using standard integrals,$I = 2 \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{2}+t}{\sqrt{2}-t} \right| + \frac{1}{2} \log \left| \frac{t-1}{t+1} \right| + c$.
Substituting $t = \sqrt{1-\tan^2 x}$,we get $I = \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2}+\sqrt{1-\tan^2 x}}{\sqrt{2}-\sqrt{1-\tan^2 x}} \right| - \frac{1}{2} \log \left| \frac{1+\sqrt{1-\tan^2 x}}{1-\sqrt{1-\tan^2 x}} \right| + c$.

(D) We have,$I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} dx$.
Divide the numerator and denominator inside the square root by $x^4$:
$I = \int \frac{x^2-1}{x^3 \cdot x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{x^2-1}{x^5 \sqrt{2 - 2x^{-2} + x^{-4}}} dx$.
$I = \int \frac{x^{-3} - x^{-5}}{\sqrt{2 - 2x^{-2} + x^{-4}}} dx$.
Let $t = \sqrt{2 - 2x^{-2} + x^{-4}}$.
Then $t^2 = 2 - 2x^{-2} + x^{-4}$.
Differentiating both sides with respect to $x$:
$2t \frac{dt}{dx} = 4x^{-3} - 4x^{-5} = 4(x^{-3} - x^{-5})$.
So,$(x^{-3} - x^{-5}) dx = \frac{1}{2} t dt$.
Substituting this into the integral:
$I = \int \frac{\frac{1}{2} t dt}{t} = \frac{1}{2} \int dt = \frac{1}{2} t + c$.
Substituting back $t$:
$I = \frac{1}{2} \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + c$.

(D) Let $I = \int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$.
Substitute $x = a \tan^2 \theta$,then $dx = 2a \tan \theta \sec^2 \theta d\theta$.
Then $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a \tan^2 \theta}{a(1+\tan^2 \theta)}} = \sqrt{\sin^2 \theta} = \sin \theta$.
So,$I = \int \theta \cdot 2a \tan \theta \sec^2 \theta d\theta = 2a \int \theta \tan \theta \sec^2 \theta d\theta$.
Using integration by parts,let $u = \theta$ and $dv = \tan \theta \sec^2 \theta d\theta$. Then $du = d\theta$ and $v = \frac{1}{2} \tan^2 \theta$.
$I = 2a \left[ \frac{1}{2} \theta \tan^2 \theta - \int \frac{1}{2} \tan^2 \theta d\theta \right] = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) d\theta$.
$I = a \theta \tan^2 \theta - a (\tan \theta - \theta) + c = a \theta (\tan^2 \theta + 1) - a \tan \theta + c$.
Since $\tan^2 \theta = \frac{x}{a}$,we have $\theta = \tan^{-1} \sqrt{\frac{x}{a}}$.
$I = a \tan^{-1} \sqrt{\frac{x}{a}} (\frac{x}{a} + 1) - a \sqrt{\frac{x}{a}} + c = (x+a) \tan^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + c$.

(B) Let $I = \int \frac{x+1}{x(1+x e^x)} d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1) e^x}{x e^x(1+x e^x)} d x$.
Let $t = 1 + x e^x$.
Then $d t = (e^x + x e^x) d x = (1+x) e^x d x$.
Also,from $t = 1 + x e^x$,we have $x e^x = t - 1$.
Substituting these into the integral:
$I = \int \frac{d t}{(t-1) t}$.
Using partial fractions:
$\frac{1}{(t-1) t} = \frac{1}{t-1} - \frac{1}{t}$.
Integrating both sides:
$I = \int \left( \frac{1}{t-1} - \frac{1}{t} \right) d t = \log |t-1| - \log |t| + C = \log \left| \frac{t-1}{t} \right| + C$.
Substituting $t = 1 + x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + C$.

(A) Let $I = \int \frac{x^3+2x}{x^4+4} dx$.
Split the integral into two parts: $I = \int \frac{x^3}{x^4+4} dx + \int \frac{2x}{x^4+4} dx$.
For the first part,let $u = x^4+4$,then $du = 4x^3 dx$,so $\int \frac{x^3}{x^4+4} dx = \frac{1}{4} \log|x^4+4| + C_1$.
For the second part,let $t = x^2$,then $dt = 2x dx$,so $\int \frac{2x}{(x^2)^2+4} dx = \int \frac{dt}{t^2+2^2} = \frac{1}{2} \tan^{-1}(\frac{t}{2}) + C_2 = \frac{1}{2} \tan^{-1}(\frac{x^2}{2}) + C_2$.
Combining these,$I = \frac{1}{4} \log(x^4+4) + \frac{1}{2} \tan^{-1}(\frac{x^2}{2}) + C$.
Note that $\frac{1}{4} \log(x^4+4) = \frac{1}{2} \log(\sqrt{x^4+4}) = \frac{1}{2} \log(\frac{\sqrt{x^4+4}}{2} \times 2) = \frac{1}{2} \log(\frac{\sqrt{x^4+4}}{2}) + \frac{1}{2} \log(2)$.
Absorbing the constant $\frac{1}{2} \log(2)$ into $C$,we get $I = \frac{1}{2} \left[ \tan^{-1}(\frac{x^2}{2}) + \log(\frac{\sqrt{x^4+4}}{2}) \right] + C$.

(A) Let $I = \int \frac{dx}{(1+x) \sqrt{8+7x-x^2}}$.
Factor the quadratic expression: $8+7x-x^2 = (8-x)(1+x)$.
So,$I = \int \frac{dx}{(1+x) \sqrt{(8-x)(1+x)}} = \int \frac{dx}{(1+x)^{3/2} \sqrt{8-x}} = \int \frac{dx}{(1+x)^2 \sqrt{\frac{8-x}{1+x}}}$.
Let $t^2 = \frac{8-x}{1+x}$.
Differentiating both sides with respect to $x$:
$2t \frac{dt}{dx} = \frac{(1+x)(-1) - (8-x)(1)}{(1+x)^2} = \frac{-1-x-8+x}{(1+x)^2} = \frac{-9}{(1+x)^2}$.
Thus,$\frac{dx}{(1+x)^2} = -\frac{2}{9} t \, dt$.
Substituting these into the integral:
$I = \int \frac{1}{t} \left(-\frac{2}{9} t \, dt\right) = -\frac{2}{9} \int dt = -\frac{2}{9} t + c$.
Substituting $t = \sqrt{\frac{8-x}{1+x}}$ back:
$I = -\frac{2}{9} \sqrt{\frac{8-x}{1+x}} + c$.

(B) Let $I = \int \frac{3 \sin x + 5 \cos x + 4}{\sin x + \cos x + 2} dx$.
We express the numerator as $A(\sin x + \cos x + 2) + B \frac{d}{dx}(\sin x + \cos x + 2) + C$.
$3 \sin x + 5 \cos x + 4 = A(\sin x + \cos x + 2) + B(\cos x - \sin x) + C$.
Equating coefficients:
$A - B = 3$
$A + B = 5$
$2A + C = 4$
Solving these,we get $2A = 8 \Rightarrow A = 4$,$B = 1$,and $C = 4 - 2(4) = -4$.
Thus,$I = \int 4 dx + \int \frac{\cos x - \sin x}{\sin x + \cos x + 2} dx - 4 \int \frac{dx}{\sin x + \cos x + 2}$.
$I = 4x + \log|\sin x + \cos x + 2| - 4 \int \frac{dx}{\sin x + \cos x + 2}$.
Using the substitution $\tan(x/2) = t$,$dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,$\cos x = \frac{1-t^2}{1+t^2}$.
$\int \frac{dx}{\sin x + \cos x + 2} = \int \frac{2 dt / (1+t^2)}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 2} = \int \frac{2 dt}{2t + 1 - t^2 + 2 + 2t^2} = \int \frac{2 dt}{t^2 + 2t + 3} = \int \frac{2 dt}{(t+1)^2 + 2}$.
$= 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t+1}{\sqrt{2}}\right) = \sqrt{2} \tan^{-1}\left(\frac{\tan(x/2) + 1}{\sqrt{2}}\right)$.
Therefore,$I = 4x + \log|\sin x + \cos x + 2| - 4\sqrt{2} \tan^{-1}\left(\frac{\tan(x/2) + 1}{\sqrt{2}}\right) + c$.

(C) $I = \int \frac{dx}{4 \sin x + 3 \cos x}$
Let $3 = r \cos \theta$ and $4 = r \sin \theta$.
Squaring and adding,we get $r^2 = 3^2 + 4^2 = 25$,so $r = 5$.
Dividing,we get $\tan \theta = \frac{4}{3}$,so $\theta = \tan^{-1} \frac{4}{3}$.
The integral becomes $I = \int \frac{dx}{r(\sin x \cos \theta + \cos x \sin \theta)} = \int \frac{dx}{r \sin(x + \theta)}$.
Alternatively,using $4 \sin x + 3 \cos x = 5(\frac{4}{5} \sin x + \frac{3}{5} \cos x) = 5 \cos(x - \theta)$ where $\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$,which implies $\theta = \tan^{-1} \frac{4}{3}$.
Thus,$I = \int \frac{dx}{5 \cos(x - \theta)} = \frac{1}{5} \int \sec(x - \theta) dx$.
Using the formula $\int \sec u du = \log |\sec u + \tan u| + c$,we get:
$I = \frac{1}{5} \log |\sec(x - \theta) + \tan(x - \theta)| + c$.
Substituting $\theta = \tan^{-1} \frac{4}{3}$,we get $I = \frac{1}{5} \log |\sec(x - \tan^{-1} \frac{4}{3}) + \tan(x - \tan^{-1} \frac{4}{3})| + c$.

(C) To evaluate the integral $I = \int (\log x)^2 dx$,we use the method of integration by parts: $\int u dv = uv - \int v du$.
Let $u = (\log x)^2$ and $dv = dx$.
Then $du = 2 \log x \cdot \frac{1}{x} dx$ and $v = x$.
Applying the formula:
$I = x(\log x)^2 - \int x \cdot \frac{2 \log x}{x} dx + c$
$I = x(\log x)^2 - 2 \int \log x dx + c$
Now,integrate $\int \log x dx$ using integration by parts again (let $u = \log x, dv = dx$):
$\int \log x dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - x$.
Substituting this back into the expression for $I$:
$I = x(\log x)^2 - 2(x \log x - x) + c$
$I = x(\log x)^2 - 2x \log x + 2x + c$
$I = x(\log x)^2 - 2x(\log x - 1) + c$.

(A) Let $I = \int (\log(\sin x) + x \cot x) dx$.
We can split the integral as $I = \int \log(\sin x) dx + \int x \cot x dx$.
Consider the first part $I_1 = \int \log(\sin x) dx$. Using integration by parts,let $u = \log(\sin x)$ and $dv = dx$.
Then $du = \frac{1}{\sin x} \cdot \cos x dx = \cot x dx$ and $v = x$.
Applying the integration by parts formula $\int u dv = uv - \int v du$,we get:
$I_1 = x \log(\sin x) - \int x \cot x dx$.
Substituting this back into the expression for $I$:
$I = (x \log(\sin x) - \int x \cot x dx) + \int x \cot x dx + c$.
$I = x \log(\sin x) + c$.

(A) We have,$I = \int \frac{2 x+3}{x(x+1)(x+2)(x+3)+1} d x$.
Rearranging the terms in the denominator:
$x(x+3) = x^2+3x$ and $(x+1)(x+2) = x^2+3x+2$.
So,$I = \int \frac{2 x+3}{(x^2+3 x)(x^2+3 x+2)+1} d x$.
Let $t = x^2+3 x$,then $dt = (2x+3) dx$.
Substituting these into the integral:
$I = \int \frac{d t}{t(t+2)+1} = \int \frac{d t}{t^2+2 t+1} = \int \frac{d t}{(t+1)^2}$.
Integrating,we get $I = -\frac{1}{t+1} + c$.
Substituting $t = x^2+3x$ back,we get $I = -\frac{1}{x^2+3 x+1} + c$.
Comparing this with $-\frac{1}{p x^2+q x+r} + c$,we find $p=1, q=3, r=1$.
Therefore,$\frac{3 p-q}{r} = \frac{3(1)-3}{1} = \frac{0}{1} = 0$.

(B) Let $I = \int \sec^5 x \, dx = \int \sec^3 x \cdot \sec^2 x \, dx$.
Using integration by parts,let $u = \sec^3 x$ and $dv = \sec^2 x \, dx$. Then $du = 3 \sec^2 x (\sec x \tan x) \, dx = 3 \sec^3 x \tan x \, dx$ and $v = \tan x$.
$I = \sec^3 x \tan x - \int 3 \sec^3 x \tan^2 x \, dx$.
Since $\tan^2 x = \sec^2 x - 1$,we have:
$I = \sec^3 x \tan x - 3 \int \sec^3 x (\sec^2 x - 1) \, dx$.
$I = \sec^3 x \tan x - 3 \int \sec^5 x \, dx + 3 \int \sec^3 x \, dx$.
$4I = \sec^3 x \tan x + 3 \int \sec^3 x \, dx$.
Using the standard formula $\int \sec^3 x \, dx = \frac{1}{2} (\sec x \tan x + \ln |\sec x + \tan x|) + c_1$:
$4I = \sec^3 x \tan x + \frac{3}{2} \sec x \tan x + \frac{3}{2} \ln |\sec x + \tan x| + C$.
$I = \frac{1}{4} \sec^3 x \tan x + \frac{3}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| + c$.
Since $\sec^3 x = \sec x (1 + \tan^2 x) = \sec x + \sec x \tan^2 x$,the expression can also be written as:
$I = \frac{1}{4} \sec x \tan^3 x + \frac{5}{8} \sec x \tan x + \frac{3}{8} \ln |\sec x + \tan x| + c$.

(C) Let $I = \int \operatorname{cosec}^5 x \, dx = \int \operatorname{cosec}^3 x \cdot \operatorname{cosec}^2 x \, dx$.
Using integration by parts,let $u = \operatorname{cosec}^3 x$ and $dv = \operatorname{cosec}^2 x \, dx$. Then $du = 3 \operatorname{cosec}^2 x (-\operatorname{cosec} x \cot x) \, dx$ and $v = -\cot x$.
$I = \operatorname{cosec}^3 x(-\cot x) - \int (-\cot x) (-3 \operatorname{cosec}^3 x \cot x) \, dx$
$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x \cot^2 x \, dx$
$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^3 x (\operatorname{cosec}^2 x - 1) \, dx$
$I = -\operatorname{cosec}^3 x \cot x - 3 \int \operatorname{cosec}^5 x \, dx + 3 \int \operatorname{cosec}^3 x \, dx$
$I = -\operatorname{cosec}^3 x \cot x - 3I + 3I_1$,where $I_1 = \int \operatorname{cosec}^3 x \, dx$.
$4I = -\operatorname{cosec}^3 x \cot x + 3I_1$.
For $I_1 = \int \operatorname{cosec}^3 x \, dx$,using integration by parts:
$I_1 = \operatorname{cosec} x(-\cot x) - \int (-\cot x)(-\operatorname{cosec} x \cot x) \, dx = -\operatorname{cosec} x \cot x - \int \operatorname{cosec} x(\operatorname{cosec}^2 x - 1) \, dx$
$I_1 = -\operatorname{cosec} x \cot x - I_1 + \log|\tan \frac{x}{2}| \implies 2I_1 = -\operatorname{cosec} x \cot x + \log|\tan \frac{x}{2}|$.
Substituting $I_1$ into the expression for $4I$:
$4I = -\operatorname{cosec}^3 x \cot x + 3 \left( -\frac{1}{2} \operatorname{cosec} x \cot x + \frac{1}{2} \log|\tan \frac{x}{2}| \right)$
$4I = -\operatorname{cosec}^3 x \cot x - \frac{3}{2} \operatorname{cosec} x \cot x + \frac{3}{2} \log|\tan \frac{x}{2}|$
$I = -\frac{\operatorname{cosec}^3 x \cot x}{4} - \frac{3}{8} \operatorname{cosec} x \cot x + \frac{3}{8} \log|\tan \frac{x}{2}| + C$.

(D) Given,$\int \frac{2 x^2}{\left(2 x^2+\alpha\right)\left(x^2+5\right)} d x=\frac{\sqrt{5}}{3} \tan ^{-1} \frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} \tan ^{-1} \frac{x}{\sqrt{2}}+c$.
Differentiating both sides with respect to $x$,we get:
$\frac{2 x^2}{\left(2 x^2+\alpha\right)\left(x^2+5\right)} = \frac{\sqrt{5}}{3} \cdot \frac{1}{1+\frac{x^2}{5}} \cdot \frac{1}{\sqrt{5}} - \frac{\sqrt{2}}{3} \cdot \frac{1}{1+\frac{x^2}{2}} \cdot \frac{1}{\sqrt{2}}$.
Simplifying the right-hand side:
$= \frac{1}{3} \cdot \frac{5}{x^2+5} - \frac{1}{3} \cdot \frac{2}{x^2+2} = \frac{1}{3} \left( \frac{5(x^2+2) - 2(x^2+5)}{(x^2+5)(x^2+2)} \right)$.
$= \frac{1}{3} \left( \frac{5x^2 + 10 - 2x^2 - 10}{(x^2+5)(x^2+2)} \right) = \frac{3x^2}{3(x^2+5)(x^2+2)} = \frac{x^2}{(x^2+5)(x^2+2)}$.
Comparing this with the left-hand side $\frac{2x^2}{(2x^2+\alpha)(x^2+5)}$,we have:
$\frac{2x^2}{(2x^2+\alpha)(x^2+5)} = \frac{x^2}{(x^2+5)(x^2+2)}$.
This implies $2(x^2+2) = 2x^2 + \alpha$,so $2x^2 + 4 = 2x^2 + \alpha$.
Thus,$\alpha = 4$.

(C) Let $I = \int_0^{2a} x^2 \sqrt{2ax-x^2} dx$.
Substitute $x = 2a \sin^2 \theta$,then $dx = 4a \sin \theta \cos \theta d\theta$.
The limits change: when $x=0, \theta=0$; when $x=2a, \theta=\frac{\pi}{2}$.
Also,$\sqrt{2ax-x^2} = \sqrt{2a(2a \sin^2 \theta) - (2a \sin^2 \theta)^2} = \sqrt{4a^2 \sin^2 \theta \cos^2 \theta} = 2a \sin \theta \cos \theta$.
Substituting these into the integral:
$I = \int_0^{\pi/2} (2a \sin^2 \theta)^2 (2a \sin \theta \cos \theta) (4a \sin \theta \cos \theta) d\theta$
$I = \int_0^{\pi/2} (4a^2 \sin^4 \theta) (8a^2 \sin^2 \theta \cos^2 \theta) d\theta$
$I = 32a^4 \int_0^{\pi/2} \sin^6 \theta \cos^2 \theta d\theta$.
Using Wallis' formula $\int_0^{\pi/2} \sin^m \theta \cos^n \theta d\theta = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2}$ (for even $m, n$):
$I = 32a^4 \cdot \frac{(5 \cdot 3 \cdot 1) \cdot (1)}{(8 \cdot 6 \cdot 4 \cdot 2)} \cdot \frac{\pi}{2} = 32a^4 \cdot \frac{15}{384} \cdot \frac{\pi}{2} = \frac{5\pi a^4}{8}$.
Given $I = ka^4$,we have $k = \frac{5\pi}{8}$,so $k : \pi = 5 : 8$.