MHT CET 2024 Mathematics Question Paper with Answer and Solution

(A) The word $MANAMA$ contains $6$ letters: $M$ appears $2$ times and $A$ appears $3$ times,and $N$ appears $1$ time.
First,we arrange the letters other than $M$,which are $A, A, A, N$.
The number of ways to arrange these $4$ letters is $\frac{4!}{3!1!} = 4$.
These $4$ letters create $5$ possible gaps (including the ends) where the $2$ $M$'s can be placed so that they are not adjacent.
The number of ways to choose $2$ gaps out of $5$ is $\binom{5}{2} = 10$.
Since the $2$ $M$'s are identical,the number of ways to place them is $\binom{5}{2} = 10$.
Therefore,the total number of arrangements is $4 \times 10 = 40$.

(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. The sum of all given digits is $0+1+2+3+4+5 = 15$. Since we need a $5$-digit number,we must exclude one digit such that the sum of the remaining $5$ digits is divisible by $3$.
Case $1$: Exclude $0$. The remaining digits are ${1, 2, 3, 4, 5}$. The sum is $15$,which is divisible by $3$. The number of $5$-digit numbers formed is $5! = 120$.
Case $2$: Exclude $3$. The remaining digits are ${0, 1, 2, 4, 5}$. The sum is $12$,which is divisible by $3$. The number of $5$-digit numbers formed is $5! - 4! = 120 - 24 = 96$.
Note: If we exclude any other digit (like $1, 2, 4, 5$),the sum of the remaining digits will not be divisible by $3$.
Total ways $= 120 + 96 = 216$.

(A) Let the colours be $R, B, G$. The number of ways to colour the vertices of a cycle graph $C_n$ with $k$ colours such that no two adjacent vertices have the same colour is given by the chromatic polynomial $P(C_n, k) = (k-1)^n + (-1)^n(k-1)$.
Here,$n = 5$ (number of persons) and $k = 3$ (number of colours).
Substituting the values:
$P(C_5, 3) = (3-1)^5 + (-1)^5(3-1)$
$= 2^5 - 2$
$= 32 - 2$
$= 30$
Thus,there are $30$ ways to distribute the hats.

(A) Each of the $3$ persons can choose any of the $3$ houses independently.
Total number of ways for $3$ persons to choose houses $= 3 \times 3 \times 3 = 27$.
For all $3$ persons to apply for the same house,they must all choose house $1$,or all choose house $2$,or all choose house $3$.
Thus,the number of favourable cases $= 3$.
Therefore,the required probability $= \frac{3}{27} = \frac{1}{9}$.

(A) Given that,the probability that exactly one of them occurs is $P(A \cup B) - P(A \cap B) = \frac{2}{5} \dots (i)$
The probability that $A$ or $B$ occurs is $P(A \cup B) = \frac{1}{2} \dots (ii)$
We know that the probability of exactly one event occurring is given by $P(A \cup B) - P(A \cap B)$.
Substituting the values from $(i)$ and $(ii)$:
$\frac{1}{2} - P(A \cap B) = \frac{2}{5}$
$P(A \cap B) = \frac{1}{2} - \frac{2}{5}$
$P(A \cap B) = \frac{5 - 4}{10} = \frac{1}{10} = 0.1$

(C) Let $N$ be the number of newspapers.
Total number of reading sessions performed by students $= 300 \times 5 = 1500$.
Since each newspaper is read by $60$ students,the total number of reading sessions is also equal to $60 \times N$.
Equating the two,we get:
$60 \times N = 1500$
$N = \frac{1500}{60} = 25$.
Thus,the number of newspapers is $25$.

(B) The total number of balls is $3 + 4 + 2 = 9$.
The number of ways to choose $3$ balls out of $9$ is given by $^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
To have three balls of different colours,we must select one red,one blue,and one green ball.
The number of ways to select one ball of each colour is $^{3}C_{1} \times ^{4}C_{1} \times ^{2}C_{1} = 3 \times 4 \times 2 = 24$.
The required probability is $\frac{24}{84} = \frac{2}{7}$.

(C) When two fair dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
We want the sum of the numbers on the upper faces to be at least $9$,which means the sum can be $9, 10, 11,$ or $12$.
The favorable outcomes are:
Sum $= 9: (3, 6), (4, 5), (5, 4), (6, 3)$
Sum $= 10: (4, 6), (5, 5), (6, 4)$
Sum $= 11: (5, 6), (6, 5)$
Sum $= 12: (6, 6)$
The total number of favorable outcomes is $4 + 3 + 2 + 1 = 10$.
Therefore,the required probability is $\frac{10}{36} = \frac{5}{18}$.

(B) Given that events $A, B, C$ are mutually exclusive and exhaustive,we have $P(A) + P(B) + P(C) = 1$.
Odds against $A$ are $8:3$,so $P(A) = \frac{3}{8+3} = \frac{3}{11}$.
Odds against $B$ are $5:2$,so $P(B) = \frac{2}{5+2} = \frac{2}{7}$.
Since $P(A) + P(B) + P(C) = 1$,we have $P(C) = 1 - (\frac{3}{11} + \frac{2}{7}) = 1 - (\frac{21+22}{77}) = 1 - \frac{43}{77} = \frac{34}{77}$.
The odds against $C$ are given by $\frac{P(C^c)}{P(C)} = \frac{1 - P(C)}{P(C)} = \frac{1 - 34/77}{34/77} = \frac{43/77}{34/77} = \frac{43}{34}$.
Given the odds against $C$ are $43:17k$,we equate: $\frac{43}{17k} = \frac{43}{34}$.
Thus,$17k = 34$,which implies $k = 2$.

(B) Let $H$ be the event that the husband is selected and $W$ be the event that the wife is selected.
Given $P(H) = \frac{1}{7}$ and $P(W) = \frac{1}{5}$.
The probability that the husband is not selected is $P(H') = 1 - \frac{1}{7} = \frac{6}{7}$.
The probability that the wife is not selected is $P(W') = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability that only one of them is selected is given by $P(\text{only } H) + P(\text{only } W)$.
$P(\text{only } H) = P(H) \times P(W') = \frac{1}{7} \times \frac{4}{5} = \frac{4}{35}$.
$P(\text{only } W) = P(W) \times P(H') = \frac{1}{5} \times \frac{6}{7} = \frac{6}{35}$.
Therefore,the required probability is $\frac{4}{35} + \frac{6}{35} = \frac{10}{35} = \frac{2}{7}$.

(D) Let $A, B, C,$ and $D$ be the events that the four persons hit the target respectively.
Given probabilities are $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}, P(D) = \frac{1}{5}$.
The probability that the target is not hit by any of them is the probability that all four miss the target.
$P(A') = 1 - \frac{1}{2} = \frac{1}{2}$
$P(B') = 1 - \frac{1}{3} = \frac{2}{3}$
$P(C') = 1 - \frac{1}{4} = \frac{3}{4}$
$P(D') = 1 - \frac{1}{5} = \frac{4}{5}$
Since the events are independent,the probability that none hit the target is:
$P(\text{None hit}) = P(A') \times P(B') \times P(C') \times P(D') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{1}{5}$.
The probability that the target is hit at least once is:
$P(\text{Target is hit}) = 1 - P(\text{None hit}) = 1 - \frac{1}{5} = \frac{4}{5}$.

(B) Given: $P(A) = \frac{2}{5}$ and $P(B) = \frac{4}{7}$.
We need to find the probability that exactly one of them is selected.
This is given by the formula: $P(\text{exactly one}) = P(A \cap B') + P(A' \cap B)$.
Since the events are independent,$P(A \cap B') = P(A) \times P(B')$ and $P(A' \cap B) = P(A') \times P(B)$.
Here,$P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$ and $P(B') = 1 - P(B) = 1 - \frac{4}{7} = \frac{3}{7}$.
Substituting the values:
$P(\text{exactly one}) = \left(\frac{2}{5} \times \frac{3}{7}\right) + \left(\frac{3}{5} \times \frac{4}{7}\right)$
$= \frac{6}{35} + \frac{12}{35}$
$= \frac{18}{35}$.

(B) The angles of triangle $ABC$ are in the ratio $2:3:7$. Let the common multiple be $x$.
$\angle A = 2x, \angle B = 3x, \angle C = 7x$.
Since the sum of the angles of a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ} \implies 12x = 180^{\circ} \implies x = 15^{\circ}$.
Thus,$\angle A = 30^{\circ}, \angle B = 45^{\circ}, \angle C = 105^{\circ}$.
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Substituting the values: $\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 105^{\circ}}$.
We know $\sin 30^{\circ} = \frac{1}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,and $\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Therefore,$\frac{a}{1/2} = \frac{b}{1/\sqrt{2}} = \frac{c}{(\sqrt{3}+1)/(2\sqrt{2})}$.
Multiplying by $\frac{1}{2}$,we get $\frac{a}{\sqrt{2}} = \frac{b}{2} = \frac{c}{\sqrt{3}+1}$.
Thus,the ratio $a:b:c = \sqrt{2}: 2: (\sqrt{3}+1)$.

(C) Given that: $m \angle A = 45^{\circ}, m \angle B = 75^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} - (45^{\circ} + 75^{\circ}) = 60^{\circ}$.
Using the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin 45^{\circ} = \frac{k}{\sqrt{2}}$,$b = k \sin 75^{\circ} = k \sin(45^{\circ} + 30^{\circ}) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = \frac{k(\sqrt{3} + 1)}{2\sqrt{2}}$,and $c = k \sin 60^{\circ} = \frac{k\sqrt{3}}{2}$.
We need to find $a + c\sqrt{2}$:
$a + c\sqrt{2} = \frac{k}{\sqrt{2}} + \frac{k\sqrt{3}}{2} \cdot \sqrt{2} = \frac{k}{\sqrt{2}} + \frac{k\sqrt{3}}{\sqrt{2}} = \frac{k(1 + \sqrt{3})}{\sqrt{2}}$.
From the expression for $b$,we have $b = \frac{k(\sqrt{3} + 1)}{2\sqrt{2}}$,which implies $2b = \frac{k(\sqrt{3} + 1)}{\sqrt{2}}$.
Therefore,$a + c\sqrt{2} = 2b$.

(C) Given that the angles $A, B, C$ are in $A$.$P$.,we have $A+C = 2B$. Since $A+B+C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,we have $\frac{b}{c} = \frac{\sin B}{\sin C}$.
Substituting the given values,$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}/2}{\sin C}$.
This simplifies to $\sin C = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Thus,$C = 45^{\circ}$.
Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}$.

(D) Let the angles of the triangle be $5x, x, 6x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $5x + x + 6x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.
The angles are $75^{\circ}, 15^{\circ}, 90^{\circ}$.
Let the sides opposite to these angles be $a, b, c$ respectively. By the Sine Rule,$\frac{a}{\sin 75^{\circ}} = \frac{b}{\sin 15^{\circ}} = \frac{c}{\sin 90^{\circ}} = k$.
The smallest side is $b$ (opposite to $15^{\circ}$) and the greatest side is $c$ (opposite to $90^{\circ}$).
The ratio of the smallest side to the greatest side is $\frac{b}{c} = \frac{\sin 15^{\circ}}{\sin 90^{\circ}}$.
We know $\sin 15^{\circ} = \sin(45^{\circ}-30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Since $\sin 90^{\circ} = 1$,the ratio is $\frac{\sqrt{3}-1}{2\sqrt{2}} : 1 = \sqrt{3}-1 : 2\sqrt{2}$.

(C) We know that in any triangle $ABC$,$A+B+C = \pi$.
Therefore,$A+C = \pi - B$.
Substituting this into the expression:
$2 ac \sin \left(\frac{A-B+C}{2}\right) = 2 ac \sin \left(\frac{(A+C)-B}{2}\right)$
$= 2 ac \sin \left(\frac{(\pi-B)-B}{2}\right)$
$= 2 ac \sin \left(\frac{\pi-2B}{2}\right)$
$= 2 ac \sin \left(\frac{\pi}{2} - B\right)$
$= 2 ac \cos B$
Using the cosine rule,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.
Substituting this value:
$= 2 ac \left(\frac{a^2+c^2-b^2}{2ac}\right)$
$= a^2+c^2-b^2$.

(B) Let $a, b, c$ be the side lengths opposite to vertices $A, B, C$ respectively. Here,$a = 3$,$b = 4$,and $c = \sqrt{23}$.
Using the cotangent rule,$\cot A = \frac{b^2+c^2-a^2}{4\Delta}$,$\cot B = \frac{a^2+c^2-b^2}{4\Delta}$,and $\cot C = \frac{a^2+b^2-c^2}{4\Delta}$,where $\Delta$ is the area of the triangle.
Then,$\frac{\cot A+\cot C}{\cot B} = \frac{\frac{b^2+c^2-a^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta}}{\frac{a^2+c^2-b^2}{4\Delta}} = \frac{b^2+c^2-a^2+a^2+b^2-c^2}{a^2+c^2-b^2} = \frac{2b^2}{a^2+c^2-b^2}$.
Substituting the values: $a^2 = 9$,$b^2 = 16$,$c^2 = 23$.
$\frac{2(16)}{9+23-16} = \frac{32}{16} = 2$.

(C) Let the sides of the triangle be $a=3$,$b=5$,and $c=7$.
Since the largest side is $c=7$,the largest angle is $\angle C$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5} = \frac{9 + 25 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$.
Therefore,$\angle C = \arccos(-\frac{1}{2}) = \frac{2\pi}{3}$.

(C) Using the cosine rule,we find the side $a$:
$a^2 = b^2 + c^2 - 2bc \cos A$
$a^2 = 3^2 + 8^2 - 2(3)(8) \cos 60^{\circ}$
$a^2 = 9 + 64 - 48 \times \frac{1}{2}$
$a^2 = 73 - 24 = 49$
$a = 7$
Using the sine rule,the circumradius $R$ is given by:
$R = \frac{a}{2 \sin A}$
$R = \frac{7}{2 \sin 60^{\circ}}$
$R = \frac{7}{2 \times \frac{\sqrt{3}}{2}}$
$R = \frac{7}{\sqrt{3}}$

(B) Given the equation: $(a+b) \cos C + (b+c) \cos A + (c+a) \cos B = 72$
Expanding the terms: $a \cos C + b \cos C + b \cos A + c \cos A + c \cos B + a \cos B = 72$
Rearranging the terms: $(a \cos C + c \cos A) + (b \cos A + a \cos B) + (b \cos C + c \cos B) = 72$
Using the projection formula: $b = c \cos A + a \cos C$,$c = a \cos B + b \cos A$,and $a = b \cos C + c \cos B$
Substituting these into the equation: $b + c + a = 72$
Given $a = 18$ and $b = 24$: $18 + 24 + c = 72$ $\Rightarrow 42 + c = 72$ $\Rightarrow c = 30$
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{72}{2} = 36$
Using Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{36(36-18)(36-24)(36-30)} = \sqrt{36 \times 18 \times 12 \times 6}$
$\text{Area} = \sqrt{36 \times 1296} = 6 \times 36 = 216 \text{ sq. units}$

(B) Let the sides be $a = 7$,$b = 4\sqrt{3}$,and $c = \sqrt{13}$.
Since $c$ is the smallest side,the smallest angle is $C$.
Using the law of cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos C = \frac{7^2 + (4\sqrt{3})^2 - (\sqrt{13})^2}{2 \times 7 \times 4\sqrt{3}}$.
$\cos C = \frac{49 + 48 - 13}{56\sqrt{3}} = \frac{84}{56\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Therefore,$C = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.

(D) Given the quadratic equation $(\cos p - 1) x^2 + (\cos p) x + \sin p = 0$.
For the equation to have real roots,the discriminant $D = b^2 - 4ac \geq 0$.
Here,$a = \cos p - 1$,$b = \cos p$,and $c = \sin p$.
Substituting these into the discriminant condition:
$(\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0$
$\cos^2 p - 4\sin p \cos p + 4\sin p \geq 0$
Since $\cos p - 1 \neq 0$,we have $\cos p \neq 1$,which implies $p \neq 2n\pi$.
For $p \in (0, \pi)$,$\sin p > 0$ and $\cos p - 1 < 0$.
The condition simplifies to $p \in (0, \pi)$ as it satisfies the discriminant inequality for real roots.

(D) Given that $\alpha$ and $\beta$ are the roots of $x^2-px+r=0$.
From the relation between roots and coefficients,we have $\alpha+\beta=p$ $(i)$ and $\alpha\beta=r$.
Given that $\frac{\alpha}{2}$ and $2\beta$ are the roots of $x^2-qx+r=0$.
From the relation between roots and coefficients,we have $\frac{\alpha}{2}+2\beta=q$,which implies $\alpha+4\beta=2q$ $(ii)$.
Also,the product of roots is $\frac{\alpha}{2} \times 2\beta = r$,so $\alpha\beta=r$.
Subtracting $(i)$ from $(ii)$,we get $(\alpha+4\beta)-(\alpha+\beta)=2q-p$,which simplifies to $3\beta=2q-p$,so $\beta=\frac{2q-p}{3}$.
Substituting $\beta$ into $(i)$,we get $\alpha=p-\frac{2q-p}{3} = \frac{3p-2q+p}{3} = \frac{4p-2q}{3} = \frac{2(2p-q)}{3}$.
Since $r=\alpha\beta$,we have $r = \left(\frac{2(2p-q)}{3}\right) \left(\frac{2q-p}{3}\right) = \frac{2}{9}(2p-q)(2q-p)$.

(A) In $\triangle PQR$,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Since $\angle R = \frac{\pi}{2} = 90^{\circ}$,we have $\angle P + \angle Q = 90^{\circ}$.
Dividing by $2$,we get $\frac{P}{2} + \frac{Q}{2} = 45^{\circ} = \frac{\pi}{4}$.
Given that $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are roots of $ax^2 + bx + c = 0$,by Vieta's formulas:
Sum of roots: $\tan \left(\frac{P}{2}\right) + \tan \left(\frac{Q}{2}\right) = -\frac{b}{a}$.
Product of roots: $\tan \left(\frac{P}{2}\right) \tan \left(\frac{Q}{2}\right) = \frac{c}{a}$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\tan \left(\frac{P}{2} + \frac{Q}{2}\right) = \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}}$.
Substituting the values: $\tan \left(\frac{\pi}{4}\right) = \frac{-b/a}{1 - c/a}$.
$1 = \frac{-b/a}{(a-c)/a} = \frac{-b}{a-c}$.
$a - c = -b$,which implies $a + b = c$.

(D) Let $e^{\sin x} = y$. Since $e^{\sin x} > 0$,we have $y > 0$.
The equation becomes $y - \frac{1}{y} = 4$,which simplifies to $y^2 - 4y - 1 = 0$.
Solving for $y$ using the quadratic formula,$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.
Since $y > 0$,we must have $y = 2 + \sqrt{5}$.
Thus,$e^{\sin x} = 2 + \sqrt{5}$.
Taking the natural logarithm on both sides,$\sin x = \ln(2 + \sqrt{5})$.
Since $\sqrt{5} \approx 2.236$,$2 + \sqrt{5} \approx 4.236$.
We know that $e \approx 2.718$,so $\ln(4.236) > \ln(e) = 1$.
Therefore,$\sin x > 1$,which is impossible for any real $x$.
Hence,the equation has no real solutions.

(A) Given that $\sin (\theta-\alpha), \sin \theta, \sin (\theta+\alpha)$ are in $H.P.$
$\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}$ are in $A.P.$
$\therefore \frac{2}{\sin \theta} = \frac{1}{\sin (\theta-\alpha)} + \frac{1}{\sin (\theta+\alpha)}$
$\Rightarrow \frac{2}{\sin \theta} = \frac{\sin (\theta+\alpha) + \sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)}$
Using the formula $\sin (A+B) + \sin (A-B) = 2 \sin A \cos B$ and $\sin (A-B) \sin (A+B) = \sin^2 A - \sin^2 B$:
$\Rightarrow \frac{2}{\sin \theta} = \frac{2 \sin \theta \cos \alpha}{\sin^2 \theta - \sin^2 \alpha}$
$\Rightarrow \sin^2 \theta - \sin^2 \alpha = \sin^2 \theta \cos \alpha$
$\Rightarrow \sin^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$\Rightarrow \sin^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$
$\Rightarrow \sin^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
Since $\sin^2 \theta = 1 - \cos^2 \theta$:
$1 - \cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\Rightarrow \cos^2 \theta = 1 - 2 \cos^2 \frac{\alpha}{2}$

(C) Given $n(A)=4$ and $n(B)=2$.
The number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = 4 \times 2 = 8$.
We need to find the number of subsets of $A \times B$ having at least $3$ elements.
The total number of subsets of a set with $8$ elements is $2^8 = 256$.
The number of subsets with $0, 1,$ or $2$ elements are given by $\binom{8}{0} + \binom{8}{1} + \binom{8}{2}$.
$\binom{8}{0} = 1$,$\binom{8}{1} = 8$,and $\binom{8}{2} = \frac{8 \times 7}{2} = 28$.
Sum of these subsets $= 1 + 8 + 28 = 37$.
The number of subsets with at least $3$ elements $= 2^8 - (\binom{8}{0} + \binom{8}{1} + \binom{8}{2}) = 256 - 37 = 219$.

(C) The given statement $p$ is: $\exists x \in S$ such that $x > 0$.
To find the negation $\sim p$,we apply the rule $\sim(\exists x, P(x)) \equiv \forall x, \sim P(x)$.
$\sim p : \forall x \in S, x \leq 0$.
This means that every rational number $x \in S$ satisfies $x \leq 0$.

(B) Let the sixth test score be $x$.
Given the mean of six tests is $48$,we have:
$\frac{54+45+41+43+57+x}{6} = 48$
$240 + x = 288$
$x = 48$.
The marks are $54, 45, 41, 43, 57, 48$.
The mean $\bar{x} = 48$.
The standard deviation $\sigma = \sqrt{\frac{1}{n} \sum (x_i - \bar{x})^2}$.
$\sigma = \sqrt{\frac{(54-48)^2 + (45-48)^2 + (41-48)^2 + (43-48)^2 + (57-48)^2 + (48-48)^2}{6}}$
$\sigma = \sqrt{\frac{6^2 + (-3)^2 + (-7)^2 + (-5)^2 + 9^2 + 0^2}{6}}$
$\sigma = \sqrt{\frac{36 + 9 + 49 + 25 + 81 + 0}{6}}$
$\sigma = \sqrt{\frac{200}{6}} = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}$.

(A) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$.
The mean $\bar{x} = \frac{2+4+\dots+100}{50} = \frac{2(1+2+\dots+50)}{50} = \frac{2 \times \frac{50 \times 51}{2}}{50} = 51$.
The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 4^2 + \dots + 100^2 = 4(1^2 + 2^2 + \dots + 50^2)$.
Using the sum of squares formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$:
$\sum x_i^2 = 4 \times \frac{50 \times 51 \times 101}{6} = 2 \times 50 \times 17 \times 101 = 171700$.
$\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

(C) Let the sum of $n$ observations be $S_n = n \bar{x}$.
When three observations $n+1, n-1, 2n-1$ are added,the new sum becomes $S_{new} = n \bar{x} + (n+1) + (n-1) + (2n-1) = n \bar{x} + 4n - 1$.
The new number of observations is $n+3$.
Given that the mean remains the same,we have:
$\bar{x} = \frac{n \bar{x} + 4n - 1}{n+3}$
Multiplying both sides by $(n+3)$:
$\bar{x}(n+3) = n \bar{x} + 4n - 1$
$n \bar{x} + 3 \bar{x} = n \bar{x} + 4n - 1$
Subtracting $n \bar{x}$ from both sides:
$3 \bar{x} = 4n - 1$
$4n = 3 \bar{x} + 1$
$n = \frac{3 \bar{x} + 1}{4}$

(A) Given the mean $\bar{x} = 6$ for the numbers $a, b, 8, 5, 10$:
$\frac{a+b+8+5+10}{5} = 6$
$a+b+23 = 30$ $\Rightarrow a+b = 7$ $\Rightarrow a = 7-b$ $(i)$
Given the variance $\sigma^2 = 6.8$:
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$6.8 = \frac{a^2+b^2+8^2+5^2+10^2}{5} - 6^2$
$6.8 = \frac{a^2+b^2+64+25+100}{5} - 36$
$6.8 + 36 = \frac{a^2+b^2+189}{5}$
$42.8 \times 5 = a^2+b^2+189$
$214 = a^2+b^2+189 \Rightarrow a^2+b^2 = 25$ (ii)
Substitute $(i)$ into (ii):
$(7-b)^2 + b^2 = 25$
$49 - 14b + b^2 + b^2 = 25$
$2b^2 - 14b + 24 = 0$
$b^2 - 7b + 12 = 0$
$(b-3)(b-4) = 0$
So,$b=3$ or $b=4$.
If $b=3$,$a=4$. If $b=4$,$a=3$. Thus,the pair $(a, b)$ is $(3, 4)$ or $(4, 3)$.

(D) Let the observations be $x_1, x_2, x_3, \ldots, x_{50}$.
Given that the sum of deviations from $30$ is $50$,we have:
$\sum_{i=1}^{50} (x_i - 30) = 50$
Expanding the summation:
$\sum_{i=1}^{50} x_i - \sum_{i=1}^{50} 30 = 50$
$\sum_{i=1}^{50} x_i - (50 \times 30) = 50$
$\sum_{i=1}^{50} x_i - 1500 = 50$
$\sum_{i=1}^{50} x_i = 1550$
Now,the mean $\bar{x}$ is given by:
$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{n} = \frac{1550}{50} = 31$

(C) Let the unknown observations be $x$ and $y$.
Given the mean of $7$ observations is $8$:
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42+x+y = 56$
$x+y = 14$ ... $(i)$
Given the variance is $16$:
$\frac{\sum x_i^2}{n} - (\text{mean})^2 = 16$
$\frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2 = 16$
$\frac{4+16+100+144+196+x^2+y^2}{7} = 16+64$
$460+x^2+y^2 = 7 \times 80 = 560$
$x^2+y^2 = 100$ ... $(ii)$
From $(i)$,$y = 14-x$. Substituting into $(ii)$:
$x^2 + (14-x)^2 = 100$
$x^2 + 196 - 28x + x^2 = 100$
$2x^2 - 28x + 96 = 0$
$x^2 - 14x + 48 = 0$
$(x-6)(x-8) = 0$
So,$x=6$ and $y=8$ (or vice versa).
The product is $6 \times 8 = 48$.

(A) Given: $n = 100$,$\bar{x} = 50$,and $\sigma = 5$.
We know that the mean $\bar{x} = \frac{\sum x_i}{n} = 50$.
Therefore,$\sum x_i = 50 \times 100 = 5000$.
The formula for standard deviation is $\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2}$.
Substituting the values: $5 = \sqrt{\frac{\sum x_i^2}{100} - (50)^2}$.
Squaring both sides: $25 = \frac{\sum x_i^2}{100} - 2500$.
$\frac{\sum x_i^2}{100} = 2500 + 25 = 2525$.
$\sum x_i^2 = 2525 \times 100 = 252500$.

(C) The mean of $a, b, c$ is $\bar{x} = \frac{a+b+c}{3}$.
Since $b = a + c$,we have $\bar{x} = \frac{b+b}{3} = \frac{2b}{3}$.
The standard deviation of $a+2, b+2, c+2$ is the same as the standard deviation of $a, b, c$,which is $d$.
Thus,$d^2 = \frac{a^2+b^2+c^2}{3} - (\bar{x})^2$.
$d^2 = \frac{a^2+b^2+c^2}{3} - \left(\frac{2b}{3}\right)^2$.
$d^2 = \frac{a^2+b^2+c^2}{3} - \frac{4b^2}{9}$.
$d^2 = \frac{3(a^2+b^2+c^2) - 4b^2}{9}$.
$9d^2 = 3(a^2+c^2+b^2) - 4b^2$.
$9d^2 = 3(a^2+c^2) + 3b^2 - 4b^2$.
$9d^2 = 3(a^2+c^2) - b^2$.
Therefore,$b^2 = 3(a^2+c^2) - 9d^2$.

(A) Let the unknown observations be $x$ and $y$.
Given mean $\bar{x} = 8$ for $n = 7$ observations.
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42 + x + y = 56 \Rightarrow x + y = 14 \dots (i)$
Given variance $\sigma^2 = 16$.
$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$
$16 = \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2$
$16 + 64 = \frac{4+16+100+144+196+x^2+y^2}{7}$
$80 \times 7 = 460 + x^2 + y^2$
$560 = 460 + x^2 + y^2 \Rightarrow x^2 + y^2 = 100 \dots (ii)$
From $(i)$, $y = 14 - x$. Substituting in $(ii)$:
$x^2 + (14 - x)^2 = 100$
$x^2 + 196 - 28x + x^2 = 100$
$2x^2 - 28x + 96 = 0 \Rightarrow x^2 - 14x + 48 = 0$
$(x - 6)(x - 8) = 0$. Thus, $x = 6$ or $x = 8$.
If $x = 6$, $y = 8$. If $x = 8$, $y = 6$.
The product $xy = 48$.
The square root of the product is $\sqrt{48} = \sqrt{16 \times 3} = 4 \sqrt{3}$.

(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$.
If each observation $x_i$ is transformed to $y_i = p x_i - q$,then the new mean $\bar{y} = p \bar{x} - q$ and the new standard deviation $\sigma_y = |p| \sigma$.
Given that the new mean is half of the original mean: $\bar{y} = \frac{20}{2} = 10$.
So,$p(20) - q = 10 \implies 20p - q = 10$ $(i)$.
Given that the new standard deviation is half of the original standard deviation: $\sigma_y = \frac{2}{2} = 1$.
So,$|p| \times 2 = 1 \implies |p| = \frac{1}{2} \implies p = \pm \frac{1}{2}$.
Case $1$: If $p = \frac{1}{2}$,then $20(\frac{1}{2}) - q = 10 \implies 10 - q = 10 \implies q = 0$. This contradicts the condition $q \neq 0$.
Case $2$: If $p = -\frac{1}{2}$,then $20(-\frac{1}{2}) - q = 10 \implies -10 - q = 10 \implies q = -20$.
Thus,$q = -20$.

(D) Let the observations be $x_1, x_2, \dots, x_{20}$ with variance $\sigma^2 = 5$.
If each observation is multiplied by a constant $a$,the new variance becomes $a^2 \sigma^2$.
Here,$a = 3$,so the new variance is $3^2 \times 5 = 9 \times 5 = 45$.
Adding a constant $b$ to each observation does not change the variance.
Therefore,the final variance remains $45$.

(D) Let the original observations be $x_1, x_2, \dots, x_6$ with variance $\sigma^2 = 12$.
When each observation is multiplied by a constant $k$,the new variance $\sigma'^2$ is given by $\sigma'^2 = k^2 \sigma^2$.
Here,$k = 3$ and $\sigma^2 = 12$.
Therefore,the new variance is:
$\sigma'^2 = 3^2 \times 12$
$\sigma'^2 = 9 \times 12$
$\sigma'^2 = 108$

(B) Let the unknown numbers be $x$ and $y$.
Given mean $\bar{x} = 8$.
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42 + x + y = 56$
$x + y = 14$ $... (i)$
Given variance $\sigma^2 = 16$.
We know $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$16 = \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2$
$16 + 64 = \frac{4+16+100+144+196+x^2+y^2}{7}$
$80 \times 7 = 460 + x^2 + y^2$
$560 = 460 + x^2 + y^2$
$x^2 + y^2 = 100$ $... (ii)$
From $(x+y)^2 = x^2 + y^2 + 2xy$,we have $14^2 = 100 + 2xy$.
$196 - 100 = 2xy$ $\Rightarrow 2xy = 96$ $\Rightarrow xy = 48$.
Now,$(x-y)^2 = (x+y)^2 - 4xy = 14^2 - 4(48) = 196 - 192 = 4$.
$|x-y| = \sqrt{4} = 2$.

(A) Given: $n = 15$,$\sum x = 170$,and $\sum x^2 = 2830$.
Corrected sum of observations: $\sum x = 170 - 20 + 30 = 180$.
Corrected sum of squares: $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
Corrected variance is given by the formula: $\sigma^2 = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$.
Substituting the values: $\sigma^2 = \frac{3330}{15} - \left(\frac{180}{15}\right)^2$.
$\sigma^2 = 222 - (12)^2 = 222 - 144 = 78$.

(B) The total number of students is $N = 20$.
Sum of frequencies $\Sigma f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.
Expanding the terms: $(x^2+2x+1) + 2x - 5 + x^2 - 3x + x = 20$.
$2x^2 + 2x - 4 = 20$ $\Rightarrow 2x^2 + 2x - 24 = 0$ $\Rightarrow x^2 + x - 12 = 0$.
Factoring the quadratic: $(x+4)(x-3) = 0$.
Since $x \in R^{+} \cup \{0\}$,we have $x = 3$.
Now,calculate the frequencies for $x=3$:
Marks $2$: $(3+1)^2 = 16$.
Marks $3$: $2(3)-5 = 1$.
Marks $5$: $3^2-3(3) = 0$.
Marks $7$: $3$.
Check sum: $16+1+0+3 = 20$. Correct.
Mean $\bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{2(16) + 3(1) + 5(0) + 7(3)}{20} = \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$.

(D) The variance $\sigma^2$ of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - (\bar{x})^2$.
Here,$\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\bar{x} = \frac{n+1}{2}$.
Substituting these values:
$\sigma^2 = \frac{1}{n} \left( \frac{n(n+1)(2n+1)}{6} \right) - \left( \frac{n+1}{2} \right)^2$
$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$
Taking $\frac{n+1}{2}$ as a common factor:
$\sigma^2 = \frac{n+1}{2} \left( \frac{2n+1}{3} - \frac{n+1}{2} \right)$
$\sigma^2 = \frac{n+1}{2} \left( \frac{4n+2 - 3n-3}{6} \right)$
$\sigma^2 = \frac{n+1}{2} \left( \frac{n-1}{6} \right)$
$\sigma^2 = \frac{n^2-1}{12}$

(C) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$.
Since the line passes through the point $(2,3)$,we have $\frac{2}{a} + \frac{3}{b} = 1$,which implies $2b + 3a = ab$.
Given the area of the triangle is $12$ sq. units,we have $\frac{1}{2}|ab| = 12$,so $ab = \pm 24$.
Case $I$: $ab = 24$. Substituting $b = \frac{24}{a}$ into $2b + 3a = ab$,we get $2(\frac{24}{a}) + 3a = 24$,which simplifies to $3a^2 - 24a + 48 = 0$,or $a^2 - 8a + 16 = 0$. This gives $(a-4)^2 = 0$,so $a = 4$ and $b = 6$. This provides $1$ line.
Case $II$: $ab = -24$. Substituting $b = \frac{-24}{a}$ into $2b + 3a = ab$,we get $2(\frac{-24}{a}) + 3a = -24$,which simplifies to $3a^2 + 24a - 48 = 0$,or $a^2 + 8a - 16 = 0$. The roots are $a = \frac{-8 \pm \sqrt{64 - 4(1)(-16)}}{2} = -4 \pm 4\sqrt{2}$. Since $b = \frac{-24}{a}$,each value of $a$ gives a distinct value of $b$. This provides $2$ additional lines.
Thus,the total number of possible lines is $1 + 2 = 3$.

(A) Given vertices are $A(-4, 5, p)$,$B(3, 1, 4)$,and $C(-2, 0, q)$.
The centroid $G(r, q, 1)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Equating the coordinates,we have:
$r = \frac{-4+3-2}{3} = \frac{-3}{3} = -1$
$q = \frac{5+1+0}{3} = \frac{6}{3} = 2$
$1 = \frac{p+4+q}{3} \Rightarrow p+4+q = 3$
Substituting $q=2$ into the equation: $p+4+2 = 3$ $\Rightarrow p+6 = 3$ $\Rightarrow p = -3$.
Now,calculating $2p + q - r$:
$2(-3) + 2 - (-1) = -6 + 2 + 1 = -3$.

(D) Line $L$ passes through $(13, 32)$.
$\frac{13}{5} + \frac{32}{b} = 1$
$\frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$
$b = -20$
So,the equation of $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y - 20 = 0$.
The slope of $L$ is $m_1 = 4$.
Since line $K$ is parallel to $L$,its slope $m_2$ must be $4$.
The equation of $K$ is $\frac{x}{c} + \frac{y}{3} = 1$,which can be written as $y = -\frac{3}{c}x + 3$.
Thus,$-\frac{3}{c} = -4 \Rightarrow c = \frac{3}{4}$.
The equation of $K$ is $\frac{x}{3/4} + \frac{y}{3} = 1$ $\Rightarrow \frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow 4x + y - 3 = 0$.
Wait,re-evaluating the slope: The equation $\frac{x}{c} + \frac{y}{3} = 1$ gives $y = -\frac{3}{c}x + 3$. For this to be parallel to $4x - y - 20 = 0$ (i.e.,$y = 4x - 20$),we need $-\frac{3}{c} = 4$,so $c = -\frac{3}{4}$.
The equation of $K$ becomes $\frac{x}{-3/4} + \frac{y}{3} = 1$ $\Rightarrow -\frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow -4x + y = 3$ $\Rightarrow 4x - y + 3 = 0$.
The distance between $4x - y - 20 = 0$ and $4x - y + 3 = 0$ is $\frac{|-20 - 3|}{\sqrt{4^2 + (-1)^2}} = \frac{23}{\sqrt{17}}$.

(D) Let $D$ be the midpoint of $AB$. The coordinates of $D$ are $\left(\frac{1+3}{2}, \frac{2+4}{2}\right) = (2,3)$.
The median $OD$ passes through $(0,0)$ and $(2,3)$. Its equation is $y = \frac{3}{2}x$,which simplifies to $3x-2y=0$.
The slope of line $AB$ is $m_{AB} = \frac{4-2}{3-1} = \frac{2}{2} = 1$.
The altitude $OE$ is perpendicular to $AB$,so its slope is $m_{OE} = -\frac{1}{m_{AB}} = -1$.
The equation of the altitude $OE$ is $y = -x$,which simplifies to $x+y=0$.
The joint equation of the median and altitude is $(3x-2y)(x+y) = 0$.
Expanding this,we get $3x^2 + 3xy - 2xy - 2y^2 = 0$,which is $3x^2 + xy - 2y^2 = 0$.

(B) Let $D$ be the midpoint of line $AB$.
Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$.
Then $D = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
The combined equation of sides $OA$ and $OB$ is $x^2-4xy+y^2=0$.
The equation of line $AB$ is $2x+3y-1=0$,so $x = \frac{1-3y}{2}$.
Substituting this into the combined equation:
$(\frac{1-3y}{2})^2 - 4(\frac{1-3y}{2})y + y^2 = 0$
$(1-3y)^2 - 8y(1-3y) + 4y^2 = 0$
$1 - 6y + 9y^2 - 8y + 24y^2 + 4y^2 = 0$
$37y^2 - 14y + 1 = 0$.
Sum of roots $y_1+y_2 = \frac{14}{37}$.
$y$-coordinate of $D = \frac{y_1+y_2}{2} = \frac{7}{37}$.
Since $D$ lies on $2x+3y-1=0$:
$2x + 3(\frac{7}{37}) - 1 = 0$
$2x + \frac{21}{37} - 1 = 0$
$2x = 1 - \frac{21}{37} = \frac{16}{37} \Rightarrow x = \frac{8}{37}$.
Thus,$D = (\frac{8}{37}, \frac{7}{37})$.
The equation of the median $OD$ passing through $(0,0)$ and $(\frac{8}{37}, \frac{7}{37})$ is:
$\frac{y-0}{x-0} = \frac{7/37}{8/37} = \frac{7}{8}$
$8y = 7x \Rightarrow 7x-8y=0$.

(D) Since the function is continuous at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$,we have:
At $x = \frac{\pi}{4}$:
$\lim_{x \to \frac{\pi}{4}^-} (x + a \sqrt{2} \sin x) = \lim_{x \to \frac{\pi}{4}^+} (2x \cot x + b)$
$\frac{\pi}{4} + a \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{\pi}{4} \cdot 1 + b$
$\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4}$ . . . $(i)$
At $x = \frac{\pi}{2}$:
$\lim_{x \to \frac{\pi}{2}^-} (2x \cot x + b) = \lim_{x \to \frac{\pi}{2}^+} (a \cos 2x - b \sin x)$
$2 \cdot \frac{\pi}{2} \cdot 0 + b = a \cos \pi - b \sin \frac{\pi}{2}$
$b = -a - b \implies a + 2b = 0$ . . . $(ii)$
Solving $(i)$ and $(ii)$:
From $(ii)$,$a = -2b$. Substituting in $(i)$:
$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$
Then $a = -2(-\frac{\pi}{12}) = \frac{\pi}{6}$.
Thus,$a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$.

(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \to 0^{-}} f(x) = f(0) = \lim_{x \to 0^{+}} f(x)$.
First,consider the left-hand limit: $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{a}{2}(x - |x|)$.
Since $x < 0$,$|x| = -x$,so $\lim_{x \to 0^{-}} \frac{a}{2}(x - (-x)) = \lim_{x \to 0^{-}} \frac{a}{2}(2x) = \lim_{x \to 0^{-}} ax = 0$.
This holds true for any real value of $a$.
Next,consider the right-hand limit: $\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} bx^2 \sin \left( \frac{1}{x} \right)$.
We know that $-1 \leq \sin \left( \frac{1}{x} \right) \leq 1$.
Multiplying by $bx^2$,we get $-|bx^2| \leq bx^2 \sin \left( \frac{1}{x} \right) \leq |bx^2|$.
By the Squeeze Theorem,as $x \to 0$,$bx^2 \sin \left( \frac{1}{x} \right) \to 0$.
This holds true for any real value of $b$.
Thus,$f(x)$ is continuous at $x = 0$ for any real values of $a$ and $b$.

(A) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the function value at $x = 0$.
$\lim_{x \rightarrow 0^-} f(x) = f(0)$
Given $f(0) = a$.
$\lim_{x \rightarrow 0^-} \frac{1 - \cos 4x}{x^2} = a$
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$,we have $1 - \cos 4x = 2 \sin^2 2x$.
$\lim_{x \rightarrow 0^-} \frac{2 \sin^2 2x}{x^2} = a$
Multiply and divide by $4$ to use the standard limit $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$.
$\lim_{x \rightarrow 0^-} 2 \times \left( \frac{\sin 2x}{2x} \right)^2 \times 4 = a$
$2 \times (1)^2 \times 4 = a$
$a = 8$.

(A) Since $f(x)$ is continuous everywhere,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.
At $x = -\frac{\pi}{2}$:
$\lim_{x \rightarrow -\frac{\pi}{2}^-} f(x) = \lim_{x \rightarrow -\frac{\pi}{2}^+} f(x)$
$-2 \sin(-\frac{\pi}{2}) = A \sin(-\frac{\pi}{2}) + B$
$-2(-1) = A(-1) + B$
$2 = -A + B$ ... $(i)$
At $x = \frac{\pi}{2}$:
$\lim_{x \rightarrow \frac{\pi}{2}^-} f(x) = \lim_{x \rightarrow \frac{\pi}{2}^+} f(x)$
$A \sin(\frac{\pi}{2}) + B = \cos(\frac{\pi}{2})$
$A(1) + B = 0$
$A + B = 0$ ... $(ii)$
Adding $(i)$ and $(ii)$:
$(-A + B) + (A + B) = 2 + 0$
$2B = 2 \Rightarrow B = 1$
Substituting $B = 1$ in $(ii)$:
$A + 1 = 0 \Rightarrow A = -1$
Thus,the values are $A = -1$ and $B = 1$.

(A) Since $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,it must be continuous at $x = \frac{\pi}{4}$.
Therefore,$f(\frac{\pi}{4}) = \lim_{x \to \frac{\pi}{4}} f(x) = \lim_{x \to \frac{\pi}{4}} \frac{1-\tan x}{4x-\pi}$.
This is a $\frac{0}{0}$ indeterminate form,so we apply $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$f(\frac{\pi}{4}) = \lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\tan x)}{\frac{d}{dx}(4x-\pi)} = \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4}$.
Substituting $x = \frac{\pi}{4}$:
$f(\frac{\pi}{4}) = \frac{-\sec^2(\frac{\pi}{4})}{4} = \frac{-(\sqrt{2})^2}{4} = \frac{-2}{4} = -\frac{1}{2}$.

(A) Since $f(x)$ is continuous on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$,it must be continuous at $x=\frac{\pi}{4}$.
Therefore,$f\left(\frac{\pi}{4}\right) = \lim_{x \rightarrow \frac{\pi}{4}} f(x) = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$.
This is a $\frac{0}{0}$ indeterminate form.
Applying $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$k = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x} = \lim_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \sin x}{\operatorname{cosec}^2 x}$.
Substituting $x = \frac{\pi}{4}$:
$k = \frac{\sqrt{2} \sin(\frac{\pi}{4})}{\operatorname{cosec}^2(\frac{\pi}{4})} = \frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{(\sqrt{2})^2} = \frac{1}{2}$.

(B) Given $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$.
Since $|x|$ is not differentiable at $x = 0$,we check the differentiability of $f(x)$ at $x = 0$.
For $x \ge 0$,$f(x) = \sin x - x + 2(x - \pi) \cos x$.
For $x < 0$,$f(x) = \sin(-x) - (-x) + 2(x - \pi) \cos(-x) = -\sin x + x + 2(x - \pi) \cos x$.
Now,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ at $x = 0$.
$f'(x)$ for $x > 0$ is $\cos x - 1 + 2 \cos x - 2(x - \pi) \sin x = 3 \cos x - 1 - 2(x - \pi) \sin x$.
$f'(0^+) = 3(1) - 1 - 2(-\pi)(0) = 2$.
$f'(x)$ for $x < 0$ is $-\cos x + 1 + 2 \cos x - 2(x - \pi) \sin x = \cos x + 1 - 2(x - \pi) \sin x$.
$f'(0^-) = 1 + 1 - 2(-\pi)(0) = 2$.
Since $f'(0^+) = f'(0^-) = 2$,the function is differentiable at $x = 0$.
Since the function is composed of differentiable functions everywhere else,$f(x)$ is differentiable for all $x \in \mathbb{R}$.
Therefore,the set $K$ of points where $f(x)$ is not differentiable is an empty set.

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[x^2 + \log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x \, dx$.
We can split the integral as $I = I_1 + I_2$,where $I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx$ and $I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \, dx$.
Since $f(x) = \log \left(\frac{\pi-x}{\pi+x}\right) \cos x$ is an odd function because $f(-x) = \log \left(\frac{\pi+x}{\pi-x}\right) \cos(-x) = -\log \left(\frac{\pi-x}{\pi+x}\right) \cos x = -f(x)$,we have $I_2 = 0$.
Now,$I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx$ (since $x^2 \cos x$ is an even function).
Using integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2 \sin x - 2 \left[x(-\cos x) - \int 1(-\cos x) \, dx\right] = x^2 \sin x + 2x \cos x - 2 \sin x$.
Evaluating from $0$ to $\frac{\pi}{2}$: $2 \left[x^2 \sin x + 2x \cos x - 2 \sin x\right]_{0}^{\frac{\pi}{2}} = 2 \left[\left(\frac{\pi^2}{4} \cdot 1 + 2 \cdot \frac{\pi}{2} \cdot 0 - 2 \cdot 1\right) - (0 + 0 - 0)\right] = 2 \left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^2}{2} - 4$.
Thus,$I = \frac{\pi^2}{2} - 4$.

(B) Let $I = \int \frac{dx}{3-2 \cos 2x}$.
Using the identity $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we have:
$I = \int \frac{dx}{3-2(\frac{1-\tan^2 x}{1+\tan^2 x})} = \int \frac{(1+\tan^2 x) dx}{3(1+\tan^2 x) - 2(1-\tan^2 x)}$.
$I = \int \frac{\sec^2 x dx}{3+3\tan^2 x - 2 + 2\tan^2 x} = \int \frac{\sec^2 x dx}{1+5\tan^2 x}$.
Let $t = \tan x$,then $dt = \sec^2 x dx$.
$I = \int \frac{dt}{1+(\sqrt{5}t)^2} = \frac{1}{\sqrt{5}} \tan^{-1}(\sqrt{5}t) + c$.
Substituting $t = \tan x$,we get $I = \frac{1}{\sqrt{5}} \tan^{-1}(\sqrt{5} \tan x) + c$.
Comparing this with $\frac{\tan^{-1}(f(x))}{\sqrt{5}} + c$,we find $f(x) = \sqrt{5} \tan x$.
Therefore,$f(\pi/4) = \sqrt{5} \tan(\pi/4) = \sqrt{5} \times 1 = \sqrt{5}$.

(A) Let $I = \int_{0}^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta$.
$= \frac{1}{\sqrt{2 k}} \int_{0}^{\frac{\pi}{3}} \frac{\sin \theta}{\cos \theta} \times \sqrt{\cos \theta} d \theta$.
$= \frac{1}{\sqrt{2 k}} \int_{0}^{\frac{\pi}{3}} \frac{\sin \theta}{\sqrt{\cos \theta}} d \theta$.
Let $\cos \theta = t$,then $-\sin \theta d \theta = dt$,so $\sin \theta d \theta = -dt$.
When $\theta = 0$,$t = 1$. When $\theta = \frac{\pi}{3}$,$t = \frac{1}{2}$.
$I = \frac{-1}{\sqrt{2 k}} \int_{1}^{\frac{1}{2}} t^{-\frac{1}{2}} dt = \frac{1}{\sqrt{2 k}} \int_{\frac{1}{2}}^{1} t^{-\frac{1}{2}} dt$.
$I = \frac{1}{\sqrt{2 k}} [2\sqrt{t}]_{\frac{1}{2}}^{1} = \frac{2}{\sqrt{2 k}} (1 - \frac{1}{\sqrt{2}}) = \frac{\sqrt{2}}{\sqrt{k}} (1 - \frac{1}{\sqrt{2}})$.
Given $I = 1 - \frac{1}{\sqrt{2}}$,we have $\frac{\sqrt{2}}{\sqrt{k}} = 1$,which implies $\sqrt{k} = \sqrt{2}$,so $k = 2$.

(C) $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x(\tan ^5 x+\cot ^5 x)} dx$
Using $\sin 2x = \frac{2 \tan x}{1+\tan ^2 x}$ and $\cot x = \frac{1}{\tan x}$,we get:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1+\tan ^2 x}{2 \tan x(\tan ^5 x + \frac{1}{\tan ^5 x})} dx$
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec ^2 x}{2 \tan x(\frac{\tan ^{10} x + 1}{\tan ^5 x})} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan ^4 x \sec ^2 x}{2(\tan ^{10} x + 1)} dx$
Let $\tan x = t$,then $\sec ^2 x dx = dt$.
When $x = \frac{\pi}{6}, t = \frac{1}{\sqrt{3}}$. When $x = \frac{\pi}{4}, t = 1$.
$I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^4}{2(t^{10} + 1)} dt$
Let $t^5 = u$,then $5t^4 dt = du$,so $t^4 dt = \frac{du}{5}$.
When $t = \frac{1}{\sqrt{3}}, u = (\frac{1}{\sqrt{3}})^5 = \frac{1}{9 \sqrt{3}}$. When $t = 1, u = 1$.
$I = \frac{1}{10} \int_{\frac{1}{9 \sqrt{3}}}^{1} \frac{du}{u^2 + 1} = \frac{1}{10} [\tan ^{-1} u]_{\frac{1}{9 \sqrt{3}}}^{1}$
$I = \frac{1}{10} (\tan ^{-1} 1 - \tan ^{-1}(\frac{1}{9 \sqrt{3}})) = \frac{1}{10} (\frac{\pi}{4} - \tan ^{-1}(\frac{1}{9 \sqrt{3}}))$

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x(\tan ^5 x+\cot ^5 x)}$
Using $\sin 2x = 2 \sin x \cos x$ and $\cot x = \frac{1}{\tan x}$,we have:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{2 \sin x \cos x(\tan ^5 x+\frac{1}{\tan ^5 x})}$
Multiply numerator and denominator by $\sec^2 x$:
$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec^2 x}{\tan x(\frac{\tan^{10} x+1}{\tan^5 x})} dx = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\tan^4 x \sec^2 x}{\tan^{10} x+1} dx$
Let $t = \tan^5 x$,then $dt = 5 \tan^4 x \sec^2 x dx$,so $\tan^4 x \sec^2 x dx = \frac{dt}{5}$.
When $x = \frac{\pi}{6}$,$t = (\frac{1}{\sqrt{3}})^5 = \frac{1}{9 \sqrt{3}}$. When $x = \frac{\pi}{4}$,$t = 1^5 = 1$.
$I = \frac{1}{2} \int_{\frac{1}{9 \sqrt{3}}}^{1} \frac{dt/5}{t^2+1} = \frac{1}{10} [\tan^{-1} t]_{\frac{1}{9 \sqrt{3}}}^{1}$
$I = \frac{1}{10} (\tan^{-1}(1) - \tan^{-1}(\frac{1}{9 \sqrt{3}})) = \frac{1}{10} (\frac{\pi}{4} - \tan^{-1}(\frac{1}{9 \sqrt{3}}))$

(A) Let $I = \int_0^a \frac{x-a}{x+a} dx$.
Substitute $t = x + a$,so $x = t - a$ and $dx = dt$.
When $x = 0$,$t = a$.
When $x = a$,$t = 2a$.
Thus,$I = \int_a^{2a} \frac{(t-a)-a}{t} dt = \int_a^{2a} \frac{t-2a}{t} dt$.
$I = \int_a^{2a} (1 - \frac{2a}{t}) dt$.
$I = [t]_a^{2a} - 2a [\log |t|]_a^{2a}$.
$I = (2a - a) - 2a (\log 2a - \log a)$.
$I = a - 2a \log(\frac{2a}{a})$.
$I = a - 2a \log 2$.

(D) Let $I = \int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$.
We can rewrite the expression as:
$I = \int_{-2}^0 ((x + 1)^3 + 4 + (x + 1) \cos(x + 1)) \, dx$.
Substitute $t = x + 1$,then $dt = dx$.
When $x = -2$,$t = -1$. When $x = 0$,$t = 1$.
So,$I = \int_{-1}^1 (t^3 + 4 + t \cos t) \, dt$.
We know that $t^3$ and $t \cos t$ are odd functions,and the integral of an odd function over a symmetric interval $[-a, a]$ is $0$.
Therefore,$I = \int_{-1}^1 t^3 \, dt + \int_{-1}^1 4 \, dt + \int_{-1}^1 t \cos t \, dt$.
$I = 0 + \int_{-1}^1 4 \, dt + 0$.
$I = 4 [t]_{-1}^1 = 4(1 - (-1)) = 4(2) = 8$.
The correct value is $8$.

(C) Let $I = \int_0^a f(x) g(x) dx$.
Using the property $\int_0^a h(x) dx = \int_0^a h(a-x) dx$,we have:
$I = \int_0^a f(a-x) g(a-x) dx$.
Since $f(x) = f(a-x)$ and $g(a-x) = 4 - g(x)$,we substitute these into the integral:
$I = \int_0^a f(x) (4 - g(x)) dx$.
$I = 4 \int_0^a f(x) dx - \int_0^a f(x) g(x) dx$.
$I = 4 \int_0^a f(x) dx - I$.
$2I = 4 \int_0^a f(x) dx$.
$I = 2 \int_0^a f(x) dx$.

(A) Let $I=\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin (\log 6-x^2)} d x$.
Put $x^2=t$,then $2x \, dx = dt$,so $x \, dx = \frac{1}{2} dt$.
The limits change as follows: when $x = \sqrt{\log 2}$,$t = \log 2$; when $x = \sqrt{\log 3}$,$t = \log 3$.
Thus,$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t}{\sin t + \sin (\log 6 - t)} dt \quad \dots (i)$
Using the property $\int_a^b f(t) dt = \int_a^b f(a+b-t) dt$,we get:
$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2 + \log 3 - t)}{\sin (\log 2 + \log 3 - t) + \sin (\log 6 - (\log 2 + \log 3 - t))} dt$
$I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6 - t)}{\sin (\log 6 - t) + \sin t} dt \quad \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6 - t)}{\sin t + \sin (\log 6 - t)} dt$
$2I = \frac{1}{2} \int_{\log 2}^{\log 3} 1 \, dt = \frac{1}{2} [t]_{\log 2}^{\log 3} = \frac{1}{2} (\log 3 - \log 2) = \frac{1}{2} \log \left(\frac{3}{2}\right)$
$I = \frac{1}{4} \log \left(\frac{3}{2}\right)$.

(A) Let $I = \int_{\frac{-1}{2}}^{\frac{1}{2}} \left([x] + \log_{e}\left(\frac{1+x}{1-x}\right)\right) dx$.
We can split the integral as:
$I = \int_{\frac{-1}{2}}^{0} [x] dx + \int_{0}^{\frac{1}{2}} [x] dx + \int_{\frac{-1}{2}}^{\frac{1}{2}} \log_{e}\left(\frac{1+x}{1-x}\right) dx$.
Let $g(x) = \log_{e}\left(\frac{1+x}{1-x}\right)$.
Then $g(-x) = \log_{e}\left(\frac{1-x}{1+x}\right) = \log_{e}\left(\left(\frac{1+x}{1-x}\right)^{-1}\right) = -\log_{e}\left(\frac{1+x}{1-x}\right) = -g(x)$.
Since $g(-x) = -g(x)$,$g(x)$ is an odd function.
Therefore,$\int_{\frac{-1}{2}}^{\frac{1}{2}} g(x) dx = 0$.
Now,for the greatest integer function $[x]$:
In the interval $[-\frac{1}{2}, 0)$,$[x] = -1$.
In the interval $[0, \frac{1}{2}]$,$[x] = 0$.
Thus,$I = \int_{\frac{-1}{2}}^{0} (-1) dx + \int_{0}^{\frac{1}{2}} (0) dx + 0$.
$I = [-x]_{\frac{-1}{2}}^{0} = -(0 - (-\frac{1}{2})) = -\frac{1}{2}$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx \quad ...(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we get:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-(-x)}} \,dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} \,dx \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \left( \frac{1}{1+e^{-x}} + \frac{1}{1+e^x} \right) dx$
Since $\frac{1}{1+e^{-x}} + \frac{1}{1+e^x} = \frac{e^x}{e^x+1} + \frac{1}{1+e^x} = 1$, we have:
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \,dx$
Since $f(x) = x^2 \cos x$ is an even function, $2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \,dx$, so $I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x \,dx$
Using integration by parts:
$I = [x^2 \sin x]_0^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x \,dx$
$I = [x^2 \sin x]_0^{\frac{\pi}{2}} - 2([-x \cos x]_0^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x \,dx)$
$I = [x^2 \sin x + 2x \cos x - 2 \sin x]_0^{\frac{\pi}{2}}$
$I = ((\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) + 2(\frac{\pi}{2}) \cos(\frac{\pi}{2}) - 2 \sin(\frac{\pi}{2})) - (0 + 0 - 0)$
$I = \frac{\pi^2}{4}(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$

(C) Let $I = \int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$.
The function $|\sin x - \cos x|$ changes sign at $x = \frac{\pi}{4}$ in the interval $[0, \frac{\pi}{2}]$.
For $0 \leq x \leq \frac{\pi}{4}$,$\cos x \geq \sin x$,so $|\sin x - \cos x| = \cos x - \sin x$.
For $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,$\sin x \geq \cos x$,so $|\sin x - \cos x| = \sin x - \cos x$.
Thus,$I = \int_0^{\frac{\pi}{4}}(\cos x - \sin x) d x + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x - \cos x) d x$.
Evaluating the integrals:
$I = [\sin x + \cos x]_0^{\frac{\pi}{4}} + [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$.
$I = [(\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) - (\sin 0 + \cos 0)] + [(-\cos \frac{\pi}{2} - \sin \frac{\pi}{2}) - (-\cos \frac{\pi}{4} - \sin \frac{\pi}{4})]$.
$I = [(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)] + [(0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})]$.
$I = [\frac{2}{\sqrt{2}} - 1] + [-1 + \frac{2}{\sqrt{2}}]$.
$I = \sqrt{2} - 1 - 1 + \sqrt{2} = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$.

(C) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x)^{-4} \,dx$.
Since $f(x) = (\sin x)^{-4} = \frac{1}{\sin^4 x}$ is an even function,we have:
$I = 2 \int_{0}^{\frac{\pi}{4}} \csc^4 x \,dx$.
Using the identity $\csc^2 x = 1 + \cot^2 x$,we get:
$I = 2 \int_{0}^{\frac{\pi}{4}} (1 + \cot^2 x) \csc^2 x \,dx$.
Let $u = \cot x$,then $du = -\csc^2 x \,dx$.
As $x \to 0^+$,$u \to \infty$. When $x = \frac{\pi}{4}$,$u = 1$.
$I = 2 \int_{\infty}^{1} (1 + u^2) (-du) = 2 \int_{1}^{\infty} (1 + u^2) \,du$.
Evaluating this integral: $2 [u + \frac{u^3}{3}]_{1}^{\infty} = \infty$.
The integral is divergent.

(D) Let $I = \int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x + \cos x}{\cos x}\right) dx$.
Simplifying the integrand,we get $I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^{\frac{\pi}{4}} \log \left[1 + \tan \left(\frac{\pi}{4} - x\right)\right] dx$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$I = \int_0^{\frac{\pi}{4}} \log \left[1 + \frac{1 - \tan x}{1 + \tan x}\right] dx$.
$I = \int_0^{\frac{\pi}{4}} \log \left(\frac{1 + \tan x + 1 - \tan x}{1 + \tan x}\right) dx$.
$I = \int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1 + \tan x}\right) dx$.
$I = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1 + \tan x)) dx$.
$I = \int_0^{\frac{\pi}{4}} \log 2 dx - \int_0^{\frac{\pi}{4}} \log(1 + \tan x) dx$.
$I = \log 2 [x]_0^{\frac{\pi}{4}} - I$.
$2I = \frac{\pi}{4} \log 2$.
$I = \frac{\pi}{8} \log 2$.

(C) We evaluate the definite integral by splitting the interval based on the jumps of the greatest integer function $[x]$:
$\int_{0.2}^{3.5} [x] \, dx = \int_{0.2}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx + \int_{3}^{3.5} [x] \, dx$
Since $[x] = 0$ for $x \in [0.2, 1)$,$[x] = 1$ for $x \in [1, 2)$,$[x] = 2$ for $x \in [2, 3)$,and $[x] = 3$ for $x \in [3, 3.5)$:
$= \int_{0.2}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{3.5} 3 \, dx$
$= 0 + [x]_{1}^{2} + 2[x]_{2}^{3} + 3[x]_{3}^{3.5}$
$= 0 + (2 - 1) + 2(3 - 2) + 3(3.5 - 3)$
$= 0 + 1 + 2(1) + 3(0.5)$
$= 1 + 2 + 1.5 = 4.5$

(C) Let $I = \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx$,we get:
$I = \int_0^{\frac{\pi}{4}} \log \left[ 1 + \tan \left( \frac{\pi}{4} - x \right) \right] \, dx$
Since $\tan \left( \frac{\pi}{4} - x \right) = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \frac{1 - \tan x}{1 + \tan x}$,we have:
$I = \int_0^{\frac{\pi}{4}} \log \left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) \, dx$
$I = \int_0^{\frac{\pi}{4}} \log \left( \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \right) \, dx$
$I = \int_0^{\frac{\pi}{4}} \log \left( \frac{2}{1 + \tan x} \right) \, dx$
$I = \int_0^{\frac{\pi}{4}} (\log 2 - \log (1 + \tan x)) \, dx$
$I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$
$I = \log 2 [x]_0^{\frac{\pi}{4}} - I$
$2I = \frac{\pi}{4} \log 2$
$I = \frac{\pi}{8} \log 2$

(D) We split the integral based on the definition of the greatest integer function $[x]$ in the interval $[0, 5]$:
$\int_0^5 x^2[x] d x = \int_0^1 x^2(0) d x + \int_1^2 x^2(1) d x + \int_2^3 x^2(2) d x + \int_3^4 x^2(3) d x + \int_4^5 x^2(4) d x$
$= 0 + \left[\frac{x^3}{3}\right]_1^2 + 2\left[\frac{x^3}{3}\right]_2^3 + 3\left[\frac{x^3}{3}\right]_3^4 + 4\left[\frac{x^3}{3}\right]_4^5$
$= \frac{1}{3}(8 - 1) + \frac{2}{3}(27 - 8) + \frac{3}{3}(64 - 27) + \frac{4}{3}(125 - 64)$
$= \frac{7}{3} + \frac{38}{3} + 37 + \frac{244}{3}$
$= \frac{7 + 38 + 111 + 244}{3} = \frac{400}{3}$

(C) Let $1+\tan x = t$. Then $\sec^2 x \, dx = dt$.
When $x = 0$,$t = 1+\tan(0) = 1$.
When $x = \frac{\pi}{4}$,$t = 1+\tan(\frac{\pi}{4}) = 2$.
The integral becomes:
$\int_1^2 \frac{dt}{t(t+1)}$.
Using partial fractions: $\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
So,the integral is $\int_1^2 \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$.
$= [\log|t| - \log|t+1|]_1^2 = [\log|\frac{t}{t+1}|]_1^2$.
$= \log(\frac{2}{3}) - \log(\frac{1}{2}) = \log(\frac{2/3}{1/2}) = \log(\frac{4}{3})$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \,d x$ ... $(i)$
Using the property $\int_{a}^{b} f(x) \,d x = \int_{a}^{b} f(a+b-x) \,d x$, we get:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 (-x)}{1+2^{-x}} \,d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+\frac{1}{2^x}} \,d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^x \sin ^2 x}{2^x+1} \,d x$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x (1+2^x)}{1+2^x} \,d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \,d x$
Since $\sin ^2 x$ is an even function, $2I = 2 \int_{0}^{\frac{\pi}{2}} \sin ^2 x \,d x$, so $I = \int_{0}^{\frac{\pi}{2}} \sin ^2 x \,d x$.
Using $\sin ^2 x = \frac{1-\cos 2x}{2}$:
$I = \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2x}{2} \,d x = \frac{1}{2} [x - \frac{\sin 2x}{2}]_{0}^{\frac{\pi}{2}} = \frac{1}{2} [(\frac{\pi}{2} - 0) - (0 - 0)] = \frac{\pi}{4}$.

(C) Let $I = \int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x$.
We know that $\sqrt{1+\cos x} = \sqrt{2\cos^2(x/2)} = \sqrt{2}\cos(x/2)$ and $\sqrt{1-\cos x} = \sqrt{2\sin^2(x/2)} = \sqrt{2}\sin(x/2)$.
Alternatively,multiply numerator and denominator by $\sqrt{1-\cos x}$:
$I = \int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1-\cos^2 x}}{(1-\cos x)^3} d x = \int_{\pi / 3}^{\pi / 2} \frac{\sin x}{(1-\cos x)^3} d x$.
Let $t = 1 - \cos x$,then $dt = \sin x \, dx$.
When $x = \pi/3$,$t = 1 - \cos(\pi/3) = 1 - 1/2 = 1/2$.
When $x = \pi/2$,$t = 1 - \cos(\pi/2) = 1 - 0 = 1$.
Thus,$I = \int_{1/2}^{1} t^{-3} dt = \left[ \frac{t^{-2}}{-2} \right]_{1/2}^{1} = -\frac{1}{2} [1 - (1/2)^{-2}] = -\frac{1}{2} [1 - 4] = -\frac{1}{2} (-3) = \frac{3}{2}$.

(B) To find the matrix of cofactors,we calculate the cofactor $A_{ij}$ for each element $a_{ij}$ using the formula $A_{ij} = (-1)^{i+j} M_{ij}$,where $M_{ij}$ is the minor of the element $a_{ij}$.
$A_{11} = (-1)^{1+1} \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1)(2-2) = 0$
$A_{12} = (-1)^{1+2} \begin{vmatrix} 3 & 2 \\ -1 & 2 \end{vmatrix} = (-1)(6 - (-2)) = -8$
$A_{13} = (-1)^{1+3} \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = (1)(3 - (-1)) = 4$
$A_{21} = (-1)^{2+1} \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (-1)(0 - (-1)) = -1$
$A_{22} = (-1)^{2+2} \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = (1)(4 - 1) = 3$
$A_{23} = (-1)^{2+3} \begin{vmatrix} 2 & 0 \\ -1 & 1 \end{vmatrix} = (-1)(2 - 0) = -2$
$A_{31} = (-1)^{3+1} \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (1)(0 - (-1)) = 1$
$A_{32} = (-1)^{3+2} \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} = (-1)(4 - (-3)) = -7$
$A_{33} = (-1)^{3+3} \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} = (1)(2 - 0) = 2$
Thus,the matrix of cofactors is $\left[\begin{array}{ccc}0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2\end{array}\right]$.

(B) Given $w = \frac{-1-i \sqrt{3}}{2}$,which is the complex cube root of unity,denoted as $\omega$.
We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
Let $\Delta = \left|\begin{array}{ccc}1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega\end{array}\right|$.
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left|\begin{array}{ccc}1 + \omega + \omega^2 & \omega & \omega^2 \\ \omega + \omega^2 + 1 & \omega^2 & 1 \\ \omega^2 + 1 + \omega & 1 & \omega\end{array}\right|$
Since $1 + \omega + \omega^2 = 0$,we get:
$\Delta = \left|\begin{array}{ccc}0 & \omega & \omega^2 \\ 0 & \omega^2 & 1 \\ 0 & 1 & \omega\end{array}\right|$
Since all elements of the first column are $0$,the value of the determinant is $0$.

(D) Given the matrix equation $AX = B$:
$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}$
This gives the system of linear equations:
$a - b + 2c = 3$ $(i)$
$2a + c = 1$ $(ii)$
$3a + 2b + c = 4$ $(iii)$
From $(ii)$,we have $c = 1 - 2a$.
Substitute $c$ into $(i)$: $a - b + 2(1 - 2a) = 3 \implies a - b + 2 - 4a = 3 \implies -3a - b = 1 \implies b = -3a - 1$.
Substitute $a, b, c$ into $(iii)$: $3a + 2(-3a - 1) + (1 - 2a) = 4$
$3a - 6a - 2 + 1 - 2a = 4$
$-5a - 1 = 4$
$-5a = 5 \implies a = -1$.
Now find $b$ and $c$:
$b = -3(-1) - 1 = 3 - 1 = 2$.
$c = 1 - 2(-1) = 1 + 2 = 3$.
Finally,calculate $2a - 3b + 4c$:
$2(-1) - 3(2) + 4(3) = -2 - 6 + 12 = 4$.

(B) Given the matrix equation $AX = B$,we have:
$\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$
Applying row operations $R_2 \rightarrow R_2 - 2R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\begin{bmatrix} 1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4 \\ -8 \\ -2 \end{bmatrix}$
From the third row,$2x_2 = -2$,which gives $x_2 = -1$.
Substituting $x_2 = -1$ into the second row equation $3x_2 - 5x_3 = -8$:
$3(-1) - 5x_3 = -8 \Rightarrow -3 - 5x_3 = -8 \Rightarrow -5x_3 = -5 \Rightarrow x_3 = 1$.
Substituting $x_2 = -1$ and $x_3 = 1$ into the first row equation $x_1 - x_2 + x_3 = 4$:
$x_1 - (-1) + 1 = 4 \Rightarrow x_1 + 1 + 1 = 4 \Rightarrow x_1 = 2$.
Thus,$X = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$.

(A) Given the differential equation: $\left[\frac{1+\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^{\frac{3}{2}}}\right]^2 = kx$
First,simplify the expression by removing the fractional exponent:
$\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^2}{\left(\frac{d^2y}{dx^2}\right)^3} = kx$
Now,multiply both sides by $\left(\frac{d^2y}{dx^2}\right)^3$:
$\left[1+\left(\frac{dy}{dx}\right)^2\right]^2 = kx \left(\frac{d^2y}{dx^2}\right)^3$
The order of a differential equation is the highest derivative present,which is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree is the power of the highest derivative when the equation is expressed as a polynomial in derivatives,which is $3$.
Therefore,the order is $2$ and the degree is $3$.

(B) Given the differential equation: $\left(\frac{d^2 y}{dx^2}\right)^5 + 4 \frac{\left(\frac{d^2 y}{dx^2}\right)^5}{\frac{d^3 y}{dx^3}} + \frac{d^3 y}{dx^3} = \sin x$.
Multiply the entire equation by $\frac{d^3 y}{dx^3}$ to eliminate the fraction:
$\left(\frac{d^2 y}{dx^2}\right)^5 \cdot \frac{d^3 y}{dx^3} + 4\left(\frac{d^2 y}{dx^2}\right)^5 + \left(\frac{d^3 y}{dx^3}\right)^2 = \sin x \cdot \frac{d^3 y}{dx^3}$.
The highest order derivative present is $\frac{d^3 y}{dx^3}$,so the order $m = 3$.
The highest power of the highest order derivative is $2$,so the degree $n = 2$.
Therefore,$m^2 + n^2 = 3^2 + 2^2 = 9 + 4 = 13$.

(B) Given the general solution: $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$.
We can simplify the expression by grouping the constants:
Let $A = c_1 + c_2$ and $B = c_4 e^{c_5}$.
Then the equation becomes $y = A \cos (x + c_3) - B e^x$.
Expanding the cosine term: $y = A (\cos x \cos c_3 - \sin x \sin c_3) - B e^x$.
$y = (A \cos c_3) \cos x - (A \sin c_3) \sin x - B e^x$.
Let $K_1 = A \cos c_3$,$K_2 = -A \sin c_3$,and $K_3 = -B$.
Thus,$y = K_1 \cos x + K_2 \sin x + K_3 e^x$.
There are $3$ independent arbitrary constants $(K_1, K_2, K_3)$.
The order of a differential equation is equal to the number of independent arbitrary constants in its general solution.
Therefore,the order of the differential equation is $3$.

(C) Given equation: $y^2 = (x + c)^3$
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 3(x + c)^2$
From this,we have $(x + c)^2 = \frac{2y}{3} \frac{dy}{dx}$.
Cubing both sides:
$(x + c)^6 = \left(\frac{2y}{3} \frac{dy}{dx}\right)^3$
Since $(x + c)^3 = y^2$,we have $(x + c)^6 = (y^2)^2 = y^4$.
Substituting this into the equation:
$y^4 = \frac{8y^3}{27} \left(\frac{dy}{dx}\right)^3$
Dividing both sides by $y^3$ (assuming $y \neq 0$):
$y = \frac{8}{27} \left(\frac{dy}{dx}\right)^3$
$27y = 8 \left(\frac{dy}{dx}\right)^3$

(D) Given the differential equation: $x \frac{dy}{dx} = y(\log y - \log x + 1)$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} (\log(\frac{y}{x}) + 1)$.
This is a homogeneous differential equation. Let $v = \frac{y}{x}$,so $y = vx$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v(\log v + 1)$.
$v + x \frac{dv}{dx} = v \log v + v$.
$x \frac{dv}{dx} = v \log v$.
Separating the variables: $\frac{dv}{v \log v} = \frac{dx}{x}$.
Integrating both sides: $\int \frac{1}{v \log v} dv = \int \frac{1}{x} dx$.
Let $u = \log v$,then $du = \frac{1}{v} dv$. The integral becomes $\int \frac{1}{u} du = \int \frac{1}{x} dx$.
$\log(u) = \log(x) + \log(c)$,where $\log(c)$ is the constant of integration.
$\log(\log v) = \log(cx)$.
Taking the exponential of both sides: $\log v = cx$.
Substituting $v = \frac{y}{x}$ back: $\log(\frac{y}{x}) = cx$.

(D) Given differential equation is $\frac{dy}{dx}=(x-y)^2$ $(i)$.
Let $x-y=t$. Then $1-\frac{dy}{dx}=\frac{dt}{dx}$,which implies $\frac{dy}{dx}=1-\frac{dt}{dx}$.
Substituting into $(i)$,we get $1-\frac{dt}{dx}=t^2$,so $\frac{dt}{dx}=1-t^2$.
Separating variables,$dx = \frac{1}{1-t^2} dt$.
Integrating both sides,$x = \int \frac{1}{1-t^2} dt = \frac{1}{2} \log \left|\frac{1+t}{1-t}\right| + c$.
Substituting $t=x-y$,we get $x = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| + c$.
Given $y(1)=1$,at $x=1$,$y=1$: $1 = \frac{1}{2} \log \left|\frac{1+1-1}{1-1+1}\right| + c \Rightarrow 1 = \frac{1}{2} \log(1) + c \Rightarrow c=1$.
Thus,$x = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| + 1$.
$x-1 = \frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| \Rightarrow 2(x-1) = \log \left|\frac{1+x-y}{1-x+y}\right|$.
Using $\log(a/b) = -\log(b/a)$,we get $2(x-1) = -\log \left|\frac{1-x+y}{1+x-y}\right|$.

(D) Given the differential equation: $(2+\sin x) \frac{dy}{dx} + (y+1) \cos x = 0$.
Rearranging the terms to separate the variables: $\frac{1}{y+1} dy = -\frac{\cos x}{2+\sin x} dx$.
Integrating both sides: $\int \frac{1}{y+1} dy = -\int \frac{\cos x}{2+\sin x} dx$.
This gives: $\ln(y+1) = -\ln(2+\sin x) + C$.
Using the initial condition $y(0)=1$: $\ln(1+1) = -\ln(2+\sin 0) + C \implies \ln 2 = -\ln 2 + C \implies C = 2\ln 2 = \ln 4$.
Substituting $C$ back into the equation: $\ln(y+1) = -\ln(2+\sin x) + \ln 4 = \ln\left(\frac{4}{2+\sin x}\right)$.
Taking the exponential of both sides: $y+1 = \frac{4}{2+\sin x} \implies y = \frac{4}{2+\sin x} - 1$.
Now,evaluating at $x = \frac{\pi}{2}$: $y\left(\frac{\pi}{2}\right) = \frac{4}{2+\sin(\pi/2)} - 1 = \frac{4}{2+1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

(A) Given the differential equation: $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx} + e^x = 0$.
Separating the variables,we get: $\frac{dy}{2+y} = -\frac{e^x}{5+e^x} dx$.
Integrating both sides: $\int \frac{dy}{2+y} = -\int \frac{e^x}{5+e^x} dx$.
This gives: $\log |2+y| = -\log |5+e^x| + C$.
Using the initial condition $y(0)=1$: $\log |2+1| = -\log |5+e^0| + C \Rightarrow \log 3 = -\log 6 + C \Rightarrow C = \log 3 + \log 6 = \log 18$.
Thus,$\log |2+y| = \log \left|\frac{18}{5+e^x}\right|$,which implies $2+y = \frac{18}{5+e^x}$.
So,$y(x) = \frac{18}{5+e^x} - 2$.
For $x = \log 13$,$y(\log 13) = \frac{18}{5+e^{\log 13}} - 2 = \frac{18}{5+13} - 2 = \frac{18}{18} - 2 = 1 - 2 = -1$.

(C) The equation of the given circle is $x^2+y^2-2x-2y=0$.
Completing the square,we get $(x-1)^2+(y-1)^2=2$.
Thus,the centre of the circle is $(1, 1)$.
The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{2y}{x^2}$.
Separating the variables,we have $\int \frac{1}{y} dy = \int \frac{2}{x^2} dx$.
Integrating both sides,we get $\log |y| = -\frac{2}{x} + c$.
Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$:
$\log |1| = -\frac{2}{1} + c \implies 0 = -2 + c \implies c = 2$.
Substituting $c=2$ into the general solution,we get $\log |y| = -\frac{2}{x} + 2$.
Multiplying by $x$,we get $x \log |y| = -2 + 2x = 2(x-1)$.

(B) Given differential equation: $(1+y^2)(1+\log x) dx + x dy = 0$
Rearranging the terms: $(1+y^2)(1+\log x) dx = -x dy$
Separating the variables: $\frac{1+\log x}{x} dx = -\frac{1}{1+y^2} dy$
Integrating both sides: $\int \frac{1+\log x}{x} dx = -\int \frac{1}{1+y^2} dy$
Let $1+\log x = t$,then $\frac{1}{x} dx = dt$.
Substituting this: $\int t dt = -\tan^{-1} y + C$
$\frac{t^2}{2} = -\tan^{-1} y + C$
$\frac{(1+\log x)^2}{2} = -\tan^{-1} y + C$
At $x=1, y=1$: $\frac{(1+\log 1)^2}{2} = -\tan^{-1}(1) + C$
$\frac{1}{2} = -\frac{\pi}{4} + C \implies C = \frac{1}{2} + \frac{\pi}{4}$
Substituting $C$ back: $\frac{(1+\log x)^2}{2} = -\tan^{-1} y + \frac{1}{2} + \frac{\pi}{4}$
$\frac{1 + 2\log x + (\log x)^2}{2} = -\tan^{-1} y + \frac{1}{2} + \frac{\pi}{4}$
$\frac{1}{2} + \log x + \frac{(\log x)^2}{2} = -\tan^{-1} y + \frac{1}{2} + \frac{\pi}{4}$
$\log x + \frac{(\log x)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$

(D) Given the differential equation: $\frac{1}{x} \frac{dy}{dx} = \tan^{-1} x$
Separating the variables,we get: $dy = x \tan^{-1} x dx$
Integrating both sides: $y = \int x \tan^{-1} x dx$
Using integration by parts,$\int u dv = uv - \int v du$,where $u = \tan^{-1} x$ and $dv = x dx$:
$y = \tan^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} dx$
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2} dx$
Adding and subtracting $1$ in the numerator:
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \int \frac{x^2+1-1}{1+x^2} dx$
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} \left( \int 1 dx - \int \frac{1}{1+x^2} dx \right)$
$y = \frac{x^2 \tan^{-1} x}{2} - \frac{1}{2} (x - \tan^{-1} x) + c$

(D) Given the differential equation: $\frac{dy}{dx} = \frac{y + \sqrt{x^2 - y^2}}{x} \dots (i)$
Since this is a homogeneous differential equation,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv}{dx} \dots (ii)$
Substituting $(ii)$ into $(i)$:
$v + x\frac{dv}{dx} = \frac{vx + \sqrt{x^2 - v^2x^2}}{x}$
$v + x\frac{dv}{dx} = v + \sqrt{1 - v^2}$
$x\frac{dv}{dx} = \sqrt{1 - v^2}$
Separating the variables:
$\int \frac{dv}{\sqrt{1 - v^2}} = \int \frac{dx}{x}$
Integrating both sides:
$\sin^{-1}(v) = \log|x| + c$
Substituting $v = \frac{y}{x}$ back:
$\sin^{-1}\left(\frac{y}{x}\right) = \log|x| + c$

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1} \dots (i)$
Let $x+y = v$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1 \dots (ii)$
Substituting $(ii)$ into $(i)$:
$\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$
$\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$
Separating the variables:
$\frac{v-1}{2v} dv = dx$
$\frac{1}{2} (1 - \frac{1}{v}) dv = dx$
Integrating both sides:
$\frac{1}{2} (v - \log|v|) = x + c_1$
$v - \log|v| = 2x + 2c_1$
Substituting $v = x+y$:
$(x+y) - \log|x+y| = 2x + c$
$y - \log|x+y| = x + c$
$y = x + \log|x+y| + c$,where $c = -2c_1$.

(A) Given differential equation: $e^{y-x} \frac{dy}{dx} = y \left( \frac{\sin x + \cos x}{1 + y \log y} \right)$
Rearranging the terms: $\frac{e^y}{e^x} \frac{dy}{dx} = \frac{y}{1 + y \log y} (\sin x + \cos x)$
Separating the variables: $\frac{e^y (1 + y \log y)}{y} dy = e^x (\sin x + \cos x) dx$
Simplifying the left side: $e^y \left( \log y + \frac{1}{y} \right) dy = e^x (\sin x + \cos x) dx$
Integrating both sides: $\int e^y \left( \log y + \frac{1}{y} \right) dy = \int e^x (\sin x + \cos x) dx$
Using the identity $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,we get: $e^y \log y = e^x \sin x + c$.

(C) Given differential equation is $x y \frac{dy}{dx} = x^2 + 2y^2$.
Dividing by $xy$,we get $\frac{dy}{dx} = \frac{x^2+2y^2}{xy} = \frac{x}{y} + \frac{2y}{x} \dots(i)$
This is a homogeneous differential equation. Put $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx} \dots(ii)$
Substituting $(ii)$ in $(i)$,we get $v + x \frac{dv}{dx} = \frac{1}{v} + 2v$.
Subtracting $v$ from both sides,$x \frac{dv}{dx} = \frac{1}{v} + v = \frac{1+v^2}{v}$.
Separating variables,$\frac{v}{1+v^2} dv = \frac{1}{x} dx$.
Integrating both sides,$\frac{1}{2} \ln(1+v^2) = \ln|x| + C$.
Multiplying by $2$,$\ln(1+v^2) = 2\ln|x| + 2C = \ln(x^2) + K$.
Thus,$1+v^2 = c x^2$.
Substituting $v = \frac{y}{x}$,we get $1 + \frac{y^2}{x^2} = c x^2$,which simplifies to $x^2 + y^2 = c x^4$.
Given $y(1) = 0$,we have $1^2 + 0^2 = c(1)^4$,so $c = 1$.
Therefore,the particular solution is $x^2 + y^2 = x^4$.

(B) Given the differential equation: $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$
Rearranging the terms: $\frac{dy}{dx} = \sin \left(\frac{x-y}{2}\right) - \sin \left(\frac{x+y}{2}\right)$
Using the trigonometric identity $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$,we get:
$\frac{dy}{dx} = 2 \cos \left(\frac{x}{2}\right) \sin \left(-\frac{y}{2}\right) = -2 \sin \left(\frac{y}{2}\right) \cos \left(\frac{x}{2}\right)$
Separating the variables: $\int \operatorname{cosec} \left(\frac{y}{2}\right) dy = -\int 2 \cos \left(\frac{x}{2}\right) dx$
Integrating both sides: $2 \log \tan \left(\frac{y}{4}\right) = -4 \sin \left(\frac{x}{2}\right) + c_1$
Dividing by $2$: $\log \tan \left(\frac{y}{4}\right) = -2 \sin \left(\frac{x}{2}\right) + C$,where $C = \frac{c_1}{2}$.

(A) Given the differential equation: $(x+2) \frac{dy}{dx} = x^2+4x-9$.
We can rewrite the right side as: $x^2+4x-9 = (x^2+4x+4) - 13 = (x+2)^2 - 13$.
So,$\frac{dy}{dx} = \frac{(x+2)^2 - 13}{x+2} = (x+2) - \frac{13}{x+2}$.
Integrating both sides with respect to $x$:
$\int dy = \int (x+2) dx - 13 \int \frac{1}{x+2} dx$.
$y = \frac{(x+2)^2}{2} - 13 \ln|x+2| + C$.
Given $y(0) = 0$,substitute $x=0$ and $y=0$:
$0 = \frac{(0+2)^2}{2} - 13 \ln|0+2| + C$.
$0 = 2 - 13 \ln(2) + C \implies C = 13 \ln(2) - 2$.
Thus,the solution is $y(x) = \frac{(x+2)^2}{2} - 13 \ln|x+2| + 13 \ln(2) - 2$.
Now,find $y(-4)$:
$y(-4) = \frac{(-4+2)^2}{2} - 13 \ln|-4+2| + 13 \ln(2) - 2$.
$y(-4) = \frac{(-2)^2}{2} - 13 \ln(2) + 13 \ln(2) - 2$.
$y(-4) = \frac{4}{2} - 2 = 2 - 2 = 0$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$
Simplify the numerator by factoring out $3e^{2x}$: $3e^{2x}(1 + e^{2x})$
Simplify the denominator: $e^x + e^{-x} = e^x + \frac{1}{e^x} = \frac{e^{2x} + 1}{e^x}$
Substitute these back into the equation: $\frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{\frac{e^{2x} + 1}{e^x}}$
Cancel the common term $(1 + e^{2x})$: $\frac{dy}{dx} = 3e^{2x} \cdot e^x = 3e^{3x}$
Integrate both sides with respect to $x$: $\int dy = \int 3e^{3x} dx$
$y = 3 \cdot \frac{e^{3x}}{3} + c = e^{3x} + c$

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \frac{y}{x} \dots (i)$
Substitute $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx} \dots (ii)$
Substituting $(ii)$ into $(i)$: $v + x \frac{dv}{dx} = v - \cos^2 v \Rightarrow x \frac{dv}{dx} = -\cos^2 v$
Separating variables: $\sec^2 v \, dv = -\frac{1}{x} \, dx$
Integrating both sides: $\int \sec^2 v \, dv = -\int \frac{1}{x} \, dx + C \Rightarrow \tan v = -\log |x| + C$
Since $v = \frac{y}{x}$,we have $\tan \frac{y}{x} = -\log x + C \dots (iii)$
The curve passes through $(1, \frac{\pi}{4})$,so $\tan \frac{\pi}{4} = -\log 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$
Thus,$\tan \frac{y}{x} = 1 - \log x = \log e - \log x = \log \frac{e}{x}$
Therefore,$y = x \tan^{-1} \left( \log \frac{e}{x} \right)$.