MHT CET 2023 Mathematics Question Paper with Answer and Solution

(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$,which implies $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$,so the triangle is equilateral.
Given $a = \frac{1}{\sqrt{6}}$,the area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2$.
Area $= \frac{\sqrt{3}}{4} \left( \frac{1}{\sqrt{6}} \right)^2 = \frac{\sqrt{3}}{4} \times \frac{1}{6} = \frac{\sqrt{3}}{24} = \frac{1}{8 \sqrt{3}}$ sq. units.

(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$.
$b+c=11k$ $(i)$,$c+a=12k$ $(ii)$,$a+b=13k$ $(iii)$.
Adding $(i), (ii), (iii)$,we get $2(a+b+c)=36k$,so $a+b+c=18k$ $(iv)$.
From $(iv)$,$a=(a+b+c)-(b+c)=18k-11k=7k$.
$b=(a+b+c)-(c+a)=18k-12k=6k$.
$c=(a+b+c)-(a+b)=18k-13k=5k$.
Using the cosine rule:
$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(6k)^2+(5k)^2-(7k)^2}{2(6k)(5k)} = \frac{36k^2+25k^2-49k^2}{60k^2} = \frac{12k^2}{60k^2} = \frac{1}{5}$.
$\cos B = \frac{c^2+a^2-b^2}{2ca} = \frac{(5k)^2+(7k)^2-(6k)^2}{2(5k)(7k)} = \frac{25k^2+49k^2-36k^2}{70k^2} = \frac{38k^2}{70k^2} = \frac{19}{35}$.
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{(7k)^2+(6k)^2-(5k)^2}{2(7k)(6k)} = \frac{49k^2+36k^2-25k^2}{84k^2} = \frac{60k^2}{84k^2} = \frac{5}{7}$.
Therefore,$\cos A+\cos B+\cos C = \frac{1}{5}+\frac{19}{35}+\frac{5}{7} = \frac{7+19+25}{35} = \frac{51}{35}$.

(D) Given that $A, B, C$ are in an Arithmetic Progression ($A$.$P$.).
Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
By the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$,so $\sin A = ak, \sin B = bk, \sin C = ck$.
The expression is $E = \frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$.
$E = \frac{a}{c} (2 \sin C \cos C) + \frac{c}{a} (2 \sin A \cos A)$.
Substituting $\sin C = ck$ and $\sin A = ak$:
$E = \frac{a}{c} (2 ck \cos C) + \frac{c}{a} (2 ak \cos A) = 2ak \cos C + 2ck \cos A$.
$E = 2k (a \cos C + c \cos A)$.
Using the projection formula $b = a \cos C + c \cos A$,we get:
$E = 2kb = 2 \sin B$.
Since $B = 60^{\circ}$,$E = 2 \sin 60^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.

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

(B) Let $h_1, h_2, h_3$ be the altitudes of $\triangle PQR$ corresponding to sides $a, b, c$ respectively.
Area of $\triangle PQR = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$.
Thus,$h_1 = \frac{2 \times \text{Area}}{a}$,$h_2 = \frac{2 \times \text{Area}}{b}$,and $h_3 = \frac{2 \times \text{Area}}{c}$.
By the sine rule,$\frac{a}{\sin P} = \frac{b}{\sin Q} = \frac{c}{\sin R} = 2R$ (where $R$ is the circumradius).
So,$a = 2R \sin P$,$b = 2R \sin Q$,and $c = 2R \sin R$.
Since $\sin P, \sin Q, \sin R$ are in $A$.$P$.,it follows that $a, b, c$ are in $A$.$P$.
Therefore,$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H$.$P$.
Substituting $a, b, c$ in terms of $h_1, h_2, h_3$,we get $\frac{h_1}{2 \times \text{Area}}, \frac{h_2}{2 \times \text{Area}}, \frac{h_3}{2 \times \text{Area}}$ are in $H$.$P$.
Hence,the altitudes $h_1, h_2, h_3$ are in $H$.$P$.

(A) Let the angles of the triangle be $4x, x$,and $x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $4x + x + x = 180^{\circ}$,which implies $6x = 180^{\circ}$,so $x = 30^{\circ}$.
The angles are $120^{\circ}, 30^{\circ}$,and $30^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,where $a$ is the longest side opposite to $120^{\circ}$.
Then $a = k \sin 120^{\circ}$,$b = k \sin 30^{\circ}$,and $c = k \sin 30^{\circ}$.
The ratio of the longest side to the perimeter is $\frac{a}{a+b+c} = \frac{\sin 120^{\circ}}{\sin 120^{\circ} + \sin 30^{\circ} + \sin 30^{\circ}}$.
Substituting the values: $\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}+2}{2}} = \frac{\sqrt{3}}{2+\sqrt{3}}$.

(A) In $\triangle ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.
Since $\angle C = \frac{\pi}{2} = 90^{\circ}$,we have $\angle A + \angle B = 90^{\circ}$.
Therefore,$\frac{A}{2} + \frac{B}{2} = 45^{\circ} = \frac{\pi}{4}$.
Given that $\tan \left(\frac{A}{2}\right)$ and $\tan \left(\frac{B}{2}\right)$ are the roots of $a_1 x^2 + b_1 x + c_1 = 0$,by the relation between roots and coefficients:
Sum of roots: $\tan \left(\frac{A}{2}\right) + \tan \left(\frac{B}{2}\right) = -\frac{b_1}{a_1}$.
Product of roots: $\tan \left(\frac{A}{2}\right) \cdot \tan \left(\frac{B}{2}\right) = \frac{c_1}{a_1}$.
Using the formula $\tan \left(\frac{A}{2} + \frac{B}{2}\right) = \frac{\tan \left(\frac{A}{2}\right) + \tan \left(\frac{B}{2}\right)}{1 - \tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)}$:
$\tan \left(\frac{\pi}{4}\right) = \frac{-\frac{b_1}{a_1}}{1 - \frac{c_1}{a_1}}$.
$1 = \frac{-b_1}{a_1 - c_1}$.
$a_1 - c_1 = -b_1$,which implies $a_1 + b_1 = c_1$.

(B) Let the sides of the triangle be $a, a+1, a+2$ where $a \in \mathbb{N}$. Let the angles opposite to these sides be $A, B, C$ respectively,such that $A < B < C$. Given $C = 2A$.
Using the Sine Rule: $\frac{\sin A}{a} = \frac{\sin C}{a+2} = k$.
Thus,$\sin C = k(a+2)$ and $\sin A = ka$.
Since $C = 2A$,$\sin C = 2 \sin A \cos A$,which implies $k(a+2) = 2(ka) \cos A$,so $\cos A = \frac{a+2}{2a}$.
Using the Cosine Rule: $\cos A = \frac{(a+1)^2 + (a+2)^2 - a^2}{2(a+1)(a+2)} = \frac{a^2+6a+5}{2(a^2+3a+2)}$.
Equating the two expressions for $\cos A$: $\frac{a+2}{2a} = \frac{(a+1)(a+5)}{2(a+1)(a+2)} = \frac{a+5}{2(a+2)}$.
$(a+2)^2 = a(a+5) \Rightarrow a^2+4a+4 = a^2+5a$.
$a = 4$.
The sides are $4, 5, 6$.

(B) Let the angles of the triangle be $4x, x,$ and $x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $4x + x + x = 180^{\circ}$,which implies $6x = 180^{\circ}$,so $x = 30^{\circ}$.
The angles are $120^{\circ}, 30^{\circ},$ and $30^{\circ}$.
By the sine rule,$\frac{a}{\sin 120^{\circ}} = \frac{b}{\sin 30^{\circ}} = \frac{c}{\sin 30^{\circ}} = k$.
Thus,$a = k \sin 120^{\circ} = k \frac{\sqrt{3}}{2}$,$b = k \sin 30^{\circ} = k \frac{1}{2}$,and $c = k \sin 30^{\circ} = k \frac{1}{2}$.
The greatest side is $a$ (opposite to $120^{\circ}$).
The perimeter is $a + b + c = k(\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{2}) = k(\frac{\sqrt{3}+2}{2})$.
The ratio of the greatest side to the perimeter is $\frac{a}{a+b+c} = \frac{k \frac{\sqrt{3}}{2}}{k \frac{\sqrt{3}+2}{2}} = \frac{\sqrt{3}}{2+\sqrt{3}}$.

(B) Let the sides be $n, n+1, n+2$. The smallest side is $n$ and the largest side is $n+2$. Let the angles opposite to these sides be $B$ and $A$ respectively. Thus,$A = 2B$.
By the sine rule,$\frac{\sin A}{n+2} = \frac{\sin B}{n} = \frac{\sin C}{n+1}$.
Since $A = 2B$,we have $\frac{\sin 2B}{n+2} = \frac{\sin B}{n}$ $\Rightarrow \frac{2 \sin B \cos B}{n+2} = \frac{\sin B}{n}$ $\Rightarrow \cos B = \frac{n+2}{2n}$.
Using the cosine rule,$\cos B = \frac{(n+1)^2 + n^2 - (n+2)^2}{2n(n+1)} = \frac{n^2+2n+1+n^2-n^2-4n-4}{2n(n+1)} = \frac{n^2-2n-3}{2n(n+1)} = \frac{(n-3)(n+1)}{2n(n+1)} = \frac{n-3}{2n}$.
Equating the two expressions for $\cos B$: $\frac{n+2}{2n} = \frac{n-3}{2n} \Rightarrow n+2 = n-3$,which is impossible.
Re-evaluating: Let sides be $a=n, b=n+1, c=n+2$. Smallest angle is $A$ (opposite to $n$),largest is $C$ (opposite to $n+2$). Given $C = 2A$.
By sine rule,$\frac{\sin A}{n} = \frac{\sin 2A}{n+2}$ $\Rightarrow \frac{\sin A}{n} = \frac{2 \sin A \cos A}{n+2}$ $\Rightarrow \cos A = \frac{n+2}{2n}$.
By cosine rule,$\cos A = \frac{(n+1)^2 + (n+2)^2 - n^2}{2(n+1)(n+2)} = \frac{n^2+2n+1+n^2+4n+4-n^2}{2(n+1)(n+2)} = \frac{n^2+6n+5}{2(n+1)(n+2)} = \frac{(n+1)(n+5)}{2(n+1)(n+2)} = \frac{n+5}{2(n+2)}$.
Equating: $\frac{n+2}{2n} = \frac{n+5}{2(n+2)}$ $\Rightarrow (n+2)^2 = n(n+5)$ $\Rightarrow n^2+4n+4 = n^2+5n$ $\Rightarrow n=4$.
The sides are $4, 5, 6$.

(D) The symbolic form of the given circuit is $(p \vee \sim q \vee \sim r) \wedge (p \vee (q \wedge r))$.
Applying the distributive law,we get $p \vee [(\sim q \vee \sim r) \wedge (q \wedge r)]$.
Using De Morgan's law,this becomes $p \vee [\sim (q \wedge r) \wedge (q \wedge r)]$.
Applying the complement law,we get $p \vee F$.
Finally,by the identity law,this simplifies to $p$.
Thus,the circuit is equivalent to a circuit with only switch $S_1$.

(C) Given that the mean $\bar{x} = 16$ and variance $\sigma^2 = 256$ for $n = 50$ observations.
$\text{Mean} = \frac{\sum x_i}{n}$ $\Rightarrow 16 = \frac{\sum x_i}{50}$ $\Rightarrow \sum x_i = 800$.
$\text{Variance} = \frac{\sum x_i^2}{n} - (\bar{x})^2 \Rightarrow 256 = \frac{\sum x_i^2}{50} - (16)^2$.
$256 = \frac{\sum x_i^2}{50} - 256$ $\Rightarrow \frac{\sum x_i^2}{50} = 512$ $\Rightarrow \sum x_i^2 = 25600$.
We need to find the mean of $(x_i - 5)^2$ for $i = 1, 2, \ldots, 50$.
$\text{New Mean} = \frac{\sum (x_i - 5)^2}{50} = \frac{\sum (x_i^2 - 10x_i + 25)}{50}$.
$= \frac{\sum x_i^2 - 10 \sum x_i + \sum 25}{50} = \frac{25600 - 10(800) + 25(50)}{50}$.
$= \frac{25600 - 8000 + 1250}{50} = \frac{18850}{50} = 377$.

(C) First,we find the midpoints $(x_i)$ for each class interval ($C$.$I$.):
For $0-6$,$x_1 = \frac{0+6}{2} = 3$
For $6-12$,$x_2 = \frac{6+12}{2} = 9$
For $12-18$,$x_3 = \frac{12+18}{2} = 15$
Now,calculate $\sum f_i$,$\sum f_i x_i$,and $\sum f_i x_i^2$:
$\sum f_i = 2 + 4 + 6 = 12$
$\sum f_i x_i = (2 \times 3) + (4 \times 9) + (6 \times 15) = 6 + 36 + 90 = 132$
$\sum f_i x_i^2 = (2 \times 3^2) + (4 \times 9^2) + (6 \times 15^2) = (2 \times 9) + (4 \times 81) + (6 \times 225) = 18 + 324 + 1350 = 1692$
The variance $V(X)$ is given by:
$V(X) = \frac{\sum f_i x_i^2}{\sum f_i} - \left( \frac{\sum f_i x_i}{\sum f_i} \right)^2$
$V(X) = \frac{1692}{12} - \left( \frac{132}{12} \right)^2$
$V(X) = 141 - (11)^2$
$V(X) = 141 - 121 = 20$
Standard deviation $\sigma = \sqrt{V(X)} = \sqrt{20} = 2 \sqrt{5}$.

(C) Given,$\text{Mean} = 5$ and $\text{S.D.} = 2$.
For the data $3, 5, 7, a, b$ with $n=5$:
$\text{Mean} = \frac{3+5+7+a+b}{5} = 5$
$\Rightarrow 15+a+b = 25$
$\Rightarrow a+b = 10$ ... $(i)$
Using the formula for variance,$\text{Var} = \text{S.D.}^2 = 2^2 = 4$:
$\text{Var} = \frac{\sum x_i^2}{n} - (\text{Mean})^2$
$4 = \frac{3^2+5^2+7^2+a^2+b^2}{5} - 5^2$
$4 = \frac{9+25+49+a^2+b^2}{5} - 25$
$29 = \frac{83+a^2+b^2}{5}$
$145 = 83 + a^2 + b^2$
$a^2 + b^2 = 62$ ... $(ii)$
We know that $(a+b)^2 = a^2 + b^2 + 2ab$.
$10^2 = 62 + 2ab$
$100 - 62 = 2ab$
$38 = 2ab \Rightarrow ab = 19$.
The quadratic equation with roots $a$ and $b$ is given by $x^2 - (a+b)x + ab = 0$.
Substituting the values,we get $x^2 - 10x + 19 = 0$.

(D) The variance of a set of observations $x_i$ is defined as $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$,where $\bar{x}$ is the mean.
When each observation is multiplied by a constant $\lambda$,the new mean becomes $\bar{x}' = \lambda \bar{x}$.
The new variance is $\sigma_{new}^2 = \frac{1}{n} \sum_{i=1}^{n} (\lambda x_i - \lambda \bar{x})^2$.
$\sigma_{new}^2 = \frac{1}{n} \sum_{i=1}^{n} \lambda^2 (x_i - \bar{x})^2$.
$\sigma_{new}^2 = \lambda^2 \left[ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right] = \lambda^2 \sigma_x^2$.
Therefore,the new variance is $\lambda^2 \sigma_x^2$.

(B) The first six prime numbers greater than $5$ are $7, 11, 13, 17, 19, 23$.
The mean $\bar{x} = \frac{7+11+13+17+19+23}{6} = \frac{90}{6} = 15$.
The variance $\sigma^2$ is given by $\frac{1}{n} \sum (x_i - \bar{x})^2$.
$\sigma^2 = \frac{(7-15)^2 + (11-15)^2 + (13-15)^2 + (17-15)^2 + (19-15)^2 + (23-15)^2}{6}$
$\sigma^2 = \frac{(-8)^2 + (-4)^2 + (-2)^2 + (2)^2 + (4)^2 + (8)^2}{6}$
$\sigma^2 = \frac{64 + 16 + 4 + 4 + 16 + 64}{6} = \frac{168}{6} = 28$.

(D) Let the original observations be $x_1, x_2, \dots, x_6$ with variance $\sigma^2 = 16$.
When each observation is multiplied by a constant $\lambda$,the new variance $\sigma'^2$ is given by $\sigma'^2 = \lambda^2 \sigma^2$.
Here,$\lambda = 3$ and $\sigma^2 = 16$.
Therefore,the new variance is:
$\sigma'^2 = 3^2 \times 16$
$\sigma'^2 = 9 \times 16$
$\sigma'^2 = 144$

(C) Given $n=50$,$\bar{x}=16$,and $\sigma_x^2=256$.
Since $\sigma_x^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$,we have $256 = \frac{1}{50} \sum x_i^2 - 16^2$.
$256 = \frac{1}{50} \sum x_i^2 - 256 \implies \frac{1}{50} \sum x_i^2 = 512 \implies \sum x_i^2 = 25600$.
We need the mean of $(x_i-5)^2$,which is $\frac{1}{50} \sum_{i=1}^{50} (x_i-5)^2$.
Expanding the sum: $\sum (x_i^2 - 10x_i + 25) = \sum x_i^2 - 10 \sum x_i + \sum 25$.
Since $\bar{x} = \frac{1}{50} \sum x_i = 16$,then $\sum x_i = 50 \times 16 = 800$.
Thus,$\sum (x_i-5)^2 = 25600 - 10(800) + 50(25) = 25600 - 8000 + 1250 = 18850$.
Required mean $= \frac{18850}{50} = 377$.

(B) The formula for variance is $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \bar{x}^2$.
Here,$n = 4$ and variance $\sigma^2 = 5$.
The mean $\bar{x} = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4}$.
Substituting the values into the formula:
$5 = \frac{1}{4} [(-1)^2 + 0^2 + 1^2 + k^2] - (\frac{k}{4})^2$.
$5 = \frac{2 + k^2}{4} - \frac{k^2}{16}$.
Multiply the entire equation by $16$ to clear the denominators:
$80 = 4(2 + k^2) - k^2$.
$80 = 8 + 4k^2 - k^2$.
$80 - 8 = 3k^2$.
$72 = 3k^2$.
$k^2 = 24$.
Since $k > 0$,$k = \sqrt{24} = 2 \sqrt{6}$.

(A) The standard deviation is independent of the change of origin.
$\therefore$ $S$.$D$. of $x_{i} = \text{S.D. of } (x_{i}-2)$.
Let $y_{i} = x_{i}-2$. Then $\sum y_{i} = 20$ and $\sum y_{i}^2 = 100$ for $n = 20$.
$\text{S.D. of } y = \sqrt{\frac{\sum y_{i}^2}{n} - \left(\frac{\sum y_{i}}{n}\right)^2}$.
$\text{S.D. of } y = \sqrt{\frac{100}{20} - \left(\frac{20}{20}\right)^2}$.
$\text{S.D. of } y = \sqrt{5 - 1^2} = \sqrt{4} = 2$.
Thus,the standard deviation of $x$ is $2$.

(D) The terms are $x_i = (i-1)d$ for $i=1, 2, \ldots, n$.
$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (i-1)d = \frac{d}{n} \frac{(n-1)n}{2} = \frac{(n-1)d}{2}$.
$\sum x_i^2 = \sum_{i=1}^{n} (i-1)^2 d^2 = d^2 \sum_{k=0}^{n-1} k^2 = d^2 \frac{(n-1)n(2n-1)}{6}$.
$\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = \frac{d^2(n-1)(2n-1)}{6} - \frac{(n-1)^2 d^2}{4}$.
$\sigma^2 = \frac{d^2(n-1)}{2} \left[ \frac{2n-1}{3} - \frac{n-1}{2} \right] = \frac{d^2(n-1)}{2} \left[ \frac{4n-2-3n+3}{6} \right] = \frac{d^2(n-1)(n+1)}{12} = \frac{(n^2-1)d^2}{12}$.

(D) The variance $\sigma^2$ of the first $N$ natural numbers is given by $\sigma^2 = \frac{N^2-1}{12}$.
Here,$N = 2n$.
Substituting $N = 2n$ into the formula:
$\sigma^2 = \frac{(2n)^2-1}{12}$
$\sigma^2 = \frac{4n^2-1}{12}$

(B) In $\triangle ABC$,by the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$.
Given $a = 2b$ and $A - B = 60^{\circ}$,so $A = 60^{\circ} + B$.
Substituting these into the sine rule:
$\frac{\sin(60^{\circ} + B)}{2b} = \frac{\sin B}{b}$
$\sin(60^{\circ} + B) = 2 \sin B$
$\sin 60^{\circ} \cos B + \cos 60^{\circ} \sin B = 2 \sin B$
$\frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B = 2 \sin B$
$\frac{\sqrt{3}}{2} \cos B = \frac{3}{2} \sin B$
$\tan B = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Thus,$B = 30^{\circ}$.
Then $A = 60^{\circ} + 30^{\circ} = 90^{\circ}$.
Since one angle is $90^{\circ}$,the triangle is right angled.

(A) The given equations of the lines are $5x - 12y + 39 = 0$ and $5x - 12y + 78 = 0$.
Since the coefficients of $x$ and $y$ are proportional,the lines are parallel.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left| \frac{c_1 - c_2}{\sqrt{a^2 + b^2}} \right|$.
Here,$a = 5$,$b = -12$,$c_1 = 39$,and $c_2 = 78$.
$d = \left| \frac{39 - 78}{\sqrt{5^2 + (-12)^2}} \right| = \left| \frac{-39}{\sqrt{25 + 144}} \right| = \left| \frac{-39}{\sqrt{169}} \right| = \frac{39}{13} = 3$ units.
Since the distance between two parallel sides of a square is equal to its side length,the side of the square is $s = 3$ units.
The area of the square is $s^2 = 3^2 = 9$ sq. units.

(D) Let the vertices of the triangle be $A(-2,3)$,$B(6,-1)$,and $C(4,3)$. Let $F(x,y)$ be the circumcentre. The circumcentre is equidistant from the vertices,so $FA^2 = FB^2 = FC^2$.
$FA^2 = (x+2)^2 + (y-3)^2$
$FB^2 = (x-6)^2 + (y+1)^2$
$FC^2 = (x-4)^2 + (y-3)^2$
Equating $FA^2 = FC^2$:
$(x+2)^2 + (y-3)^2 = (x-4)^2 + (y-3)^2$
$(x+2)^2 = (x-4)^2$
$x^2 + 4x + 4 = x^2 - 8x + 16$
$12x = 12 \implies x = 1$
Equating $FA^2 = FB^2$:
$(1+2)^2 + (y-3)^2 = (1-6)^2 + (y+1)^2$
$9 + y^2 - 6y + 9 = 25 + y^2 + 2y + 1$
$18 - 6y = 26 + 2y$
$-8 = 8y \implies y = -1$
Thus,the coordinates of the circumcentre are $(1,-1)$.

(A) The equation of line $PQ$ passing through $(1, 2)$ is $y - 2 = m(x - 1)$.
The $x$-intercept (Point $P$) is $(1 - \frac{2}{m}, 0)$ and the $y$-intercept (Point $Q$) is $(0, 2 - m)$.
Since the area must be positive,we consider the magnitude of the intercepts. For the triangle to exist in the first quadrant,$m$ must be negative. Let $m = -k$ where $k > 0$.
The intercepts are $P = (1 + \frac{2}{k}, 0)$ and $Q = (0, 2 + k)$.
Area $A = \frac{1}{2} \times (1 + \frac{2}{k}) \times (2 + k) = \frac{1}{2} (2 + k + \frac{4}{k} + 2) = \frac{1}{2} (4 + k + \frac{4}{k}) = 2 + \frac{k}{2} + \frac{2}{k}$.
By $AM$-$GM$ inequality,$\frac{k}{2} + \frac{2}{k} \ge 2 \sqrt{\frac{k}{2} \times \frac{2}{k}} = 2$.
Equality holds when $\frac{k}{2} = \frac{2}{k} \implies k^2 = 4 \implies k = 2$.
Since $m = -k$,the slope $m = -2$.

(B) Let the vertex be $A(1, 2)$ and the base $BC$ lie on the line $2x - y - 1 = 0$.
The altitude $AD$ from vertex $A$ to the base $BC$ is the perpendicular distance from $(1, 2)$ to the line $2x - y - 1 = 0$.
$AD = \left| \frac{2(1) - (2) - 1}{\sqrt{2^2 + (-1)^2}} \right| = \left| \frac{2 - 2 - 1}{\sqrt{5}} \right| = \frac{1}{\sqrt{5}}$.
In an equilateral triangle,the altitude $AD$ is related to the side length $s$ by $AD = s \sin 60^{\circ} = s \frac{\sqrt{3}}{2}$.
Therefore,$\frac{1}{\sqrt{5}} = s \frac{\sqrt{3}}{2}$.
$s = \frac{2}{\sqrt{5} \cdot \sqrt{3}} = \frac{2}{\sqrt{15}}$.

(C) $S$ is the midpoint of $QR = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.
Slope of $PS = \frac{2-1}{2-\frac{13}{2}} = \frac{1}{-\frac{9}{2}} = -\frac{2}{9}$.
The line is parallel to $PS$,so its slope is $m = -\frac{2}{9}$.
The equation of the line passing through $(1, -1)$ with slope $m = -\frac{2}{9}$ is $y - (-1) = -\frac{2}{9}(x - 1)$.
$9(y + 1) = -2(x - 1)$ $\Rightarrow 9y + 9 = -2x + 2$ $\Rightarrow 2x + 9y + 7 = 0$.
To find the $X$-intercept,set $y = 0$: $2x + 7 = 0 \Rightarrow x = -\frac{7}{2}$.
To find the $Y$-intercept,set $x = 0$: $9y + 7 = 0 \Rightarrow y = -\frac{7}{9}$.
Thus,the intercepts are $-\frac{7}{2}$ and $-\frac{7}{9}$.

(D) Given that $\angle A, \angle B, \angle C$ are in $A$.$P$.,we have $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Let the sides opposite to $A, B, C$ be $a, b, c$ respectively. Given $a = 10$ and $b = 9$ (the two larger sides).
Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting the values: $\cos 60^{\circ} = \frac{10^2 + c^2 - 9^2}{2(10)c}$.
$\frac{1}{2} = \frac{100 + c^2 - 81}{20c} \Rightarrow 10c = c^2 + 19$.
$c^2 - 10c + 19 = 0$.
Using the quadratic formula: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = \frac{10 \pm \sqrt{24}}{2} = 5 \pm \sqrt{6}$.
Since $b=9$ is the second largest side,$c$ must be the smallest side. Thus,$c = 5 - \sqrt{6}$.

(C) The coordinates of the points are $P(-3, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$.
$QP$ lies along the negative $X$-axis,so the angle it makes with the positive $X$-axis is $180^{\circ}$.
The slope of $QR$ is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,the angle $\theta$ that $QR$ makes with the positive $X$-axis is $60^{\circ}$.
The angle $\angle PQR$ is the angle between $QP$ and $QR$,which is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
The angle bisector of $\angle PQR$ divides this angle into two equal parts of $60^{\circ}$ each.
The angle that the bisector makes with the positive $X$-axis is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
The slope of the bisector is $\tan 120^{\circ} = -\sqrt{3}$.
Since the bisector passes through the origin $Q(0, 0)$,its equation is $y - 0 = -\sqrt{3}(x - 0)$,which simplifies to $y = -\sqrt{3}x$ or $\sqrt{3}x + y = 0$.

(C) Diagonals of a rectangle bisect each other.
Midpoint of $(1,3)$ and $(5,1)$ is $(\frac{1+5}{2}, \frac{3+1}{2}) = (3,2)$.
Since the other two vertices lie on the line $y=2x+c$,the midpoint $(3,2)$ must also lie on this line.
Substituting $(3,2)$ into $y=2x+c$ gives $2 = 2(3) + c$,so $c = -4$.
The equation of the line containing the other two vertices is $y=2x-4$.
Let the coordinates of a vertex be $(x, 2x-4)$.
Since the angle between adjacent sides of a rectangle is $90^{\circ}$,the product of the slopes of the sides meeting at a vertex is $-1$.
Let the given vertices be $A(1,3)$ and $C(5,1)$. Let the other vertices be $B(x, 2x-4)$ and $D$. The slope of $AB$ is $\frac{2x-4-3}{x-1} = \frac{2x-7}{x-1}$.
The slope of $BC$ is $\frac{2x-4-1}{x-5} = \frac{2x-5}{x-5}$.
Since $AB \perp BC$,we have $(\frac{2x-7}{x-1})(\frac{2x-5}{x-5}) = -1$.
$(2x-7)(2x-5) = -(x-1)(x-5)$
$4x^2 - 24x + 35 = -(x^2 - 6x + 5)$
$4x^2 - 24x + 35 = -x^2 + 6x - 5$
$5x^2 - 30x + 40 = 0$
$x^2 - 6x + 8 = 0$
$(x-4)(x-2) = 0$
So $x=4$ or $x=2$.
If $x=4$,$y=2(4)-4=4$. If $x=2$,$y=2(2)-4=0$.
The vertices are $(4,4)$ and $(2,0)$.
Thus,one of the vertices is $(2,0)$.

(B) The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the line passes through $(1, 3)$,we substitute these coordinates into the equation:
$\frac{1}{a} + \frac{3}{b} = 1$.
We are also given $b = 3a$. Substituting this into the equation:
$\frac{1}{a} + \frac{3}{3a} = 1$
$\frac{1}{a} + \frac{1}{a} = 1$
$\frac{2}{a} = 1 \Rightarrow a = 2$.
Since $b = 3a$,we have $b = 3(2) = 6$.
Substituting $a = 2$ and $b = 6$ into the intercept form:
$\frac{x}{2} + \frac{y}{6} = 1$.
Multiplying by $6$ to clear the denominators:
$3x + y = 6$.

(C) The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $\frac{1}{a}x + \frac{1}{b}y - 1 = 0$.
The length of the perpendicular $p$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \left| \frac{C}{\sqrt{A^2 + B^2}} \right|$.
Substituting the values $A = \frac{1}{a}$,$B = \frac{1}{b}$,and $C = -1$,we get $p = \left| \frac{-1}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} \right|$.
Squaring both sides,we have $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$.
Therefore,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$.

(D) Let the given line be $L: 3x + y + 4 = 0$. The slope of $L$ is $m = -3$.
Let the slopes of lines $L_1$ and $L_2$ be $m_1 = \frac{3}{4}$ and $m_2$ respectively.
Since the distances measured along $L_1$ and $L_2$ from point $A(2, 5)$ to the line $L$ are equal,the angles $\theta_1$ and $\theta_2$ that $L_1$ and $L_2$ make with the line $L$ must satisfy $\tan \theta_1 = \tan \theta_2$ or $\tan \theta_1 = -\tan \theta_2$.
Using the formula for the angle between two lines with slopes $m$ and $m'$,$\tan \theta = \left| \frac{m - m'}{1 + mm'} \right|$.
For $L_1$: $\tan \theta = \left| \frac{-3 - 3/4}{1 + (-3)(3/4)} \right| = \left| \frac{-15/4}{1 - 9/4} \right| = \left| \frac{-15/4}{-5/4} \right| = 3$.
For $L_2$: $\left| \frac{-3 - m_2}{1 - 3m_2} \right| = 3$.
Case $1$: $\frac{-3 - m_2}{1 - 3m_2} = 3$ $\Rightarrow -3 - m_2 = 3 - 9m_2$ $\Rightarrow 8m_2 = 6$ $\Rightarrow m_2 = \frac{3}{4}$ (This is $L_1$ itself).
Case $2$: $\frac{-3 - m_2}{1 - 3m_2} = -3$ $\Rightarrow -3 - m_2 = -3 + 9m_2$ $\Rightarrow 10m_2 = 0$ $\Rightarrow m_2 = 0$.
Thus,the slope of $L_2$ is $0$.

(A) The given equation is $x^2+4xy+4y^2+3x+6y-4=0$.
This can be written as $(x+2y)^2+3(x+2y)-4=0$.
Let $t = x+2y$,then $t^2+3t-4=0$.
$(t+4)(t-1)=0$,so $(x+2y+4)(x+2y-1)=0$.
The two parallel lines are $L_1: x+2y+4=0$ and $L_2: x+2y-1=0$.
The distance $\lambda$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $\lambda = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.
$\lambda = \frac{|4-(-1)|}{\sqrt{1^2+2^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$.
Therefore,$\lambda^2 = (\sqrt{5})^2 = 5$.

(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$.
Here,the perpendicular distance from the origin is $p = 7$ and the angle $\alpha = 120^{\circ}$.
Substituting these values into the equation:
$x \cos 120^{\circ} + y \sin 120^{\circ} = 7$
Since $\cos 120^{\circ} = -\frac{1}{2}$ and $\sin 120^{\circ} = \frac{\sqrt{3}}{2}$,we get:
$x(-\frac{1}{2}) + y(\frac{\sqrt{3}}{2}) = 7$
Multiplying both sides by $2$:
$-x + \sqrt{3}y = 14$
Rearranging the terms to the standard form:
$x - \sqrt{3}y + 14 = 0$

(A) Let the point be $P(x_1, y_1)$. Since $P$ lies on $2x - y = 5$,we have $y_1 = 2x_1 - 5$.
The distance of $P(x_1, y_1)$ from the line $3x + 4y - 5 = 0$ is given by $d = \left|\frac{3x_1 + 4y_1 - 5}{\sqrt{3^2 + 4^2}}\right| = 1$.
Substituting $y_1 = 2x_1 - 5$ into the distance formula:
$\left|\frac{3x_1 + 4(2x_1 - 5) - 5}{5}\right| = 1$
$\left|\frac{3x_1 + 8x_1 - 20 - 5}{5}\right| = 1$
$\left|\frac{11x_1 - 25}{5}\right| = 1$
$11x_1 - 25 = 5$ or $11x_1 - 25 = -5$
Case $1$: $11x_1 = 30 \implies x_1 = \frac{30}{11}$. Then $y_1 = 2(\frac{30}{11}) - 5 = \frac{60-55}{11} = \frac{5}{11}$.
Case $2$: $11x_1 = 20 \implies x_1 = \frac{20}{11}$. Then $y_1 = 2(\frac{20}{11}) - 5 = \frac{40-55}{11} = \frac{-15}{11}$.
Thus,the points are $\left(\frac{30}{11}, \frac{5}{11}\right)$ and $\left(\frac{20}{11}, \frac{-15}{11}\right)$.

(A) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. For this to represent a pair of lines,the determinant must be zero:
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ with the general form,we have $a=2, h=2, b=-p, g=2, f=q/2, c=1$.
Substituting these into the determinant condition:
$2(-p)(1) + 2(q/2)(2)(2) - 2(q/2)^2 - (-p)(2)^2 - 1(2)^2 = 0$.
$-2p + 4q - q^2/2 + 4p - 4 = 0$.
Multiplying by $-2$: $q^2 - 8q - 4p + 8 = 0$.
For $q$ to be an integer,the discriminant of this quadratic in $q$ must be a perfect square:
$D = (-8)^2 - 4(1)(8 - 4p) = 64 - 32 + 16p = 32 + 16p = 16(2 + p)$.
For $D$ to be a perfect square,$2 + p$ must be a perfect square $k^2$.
Given $p \in [-5, 5]$,$2 + p$ can take values from $-3$ to $7$.
The perfect squares in this range are $0, 1, 4$.
$2 + p = 0 \Rightarrow p = -2$.
$2 + p = 1 \Rightarrow p = -1$.
$2 + p = 4 \Rightarrow p = 2$.
Thus,the possible integral values of $p$ are $\{-2, -1, 2\}$.
The number of such values is $3$.

(A) The slope of the line $3x + 2y - 8 = 0$ is $m_1 = -\frac{3}{2}$.
Let $m$ be the slope of the lines passing through the origin that make an angle of $\frac{\pi}{4}$ with the given line.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + mm_1} \right|$,we have:
$\tan \frac{\pi}{4} = \left| \frac{m - (-3/2)}{1 + m(-3/2)} \right|$
$1 = \left| \frac{2m + 3}{2 - 3m} \right|$
Squaring both sides:
$(2 - 3m)^2 = (2m + 3)^2$
$4 - 12m + 9m^2 = 4m^2 + 12m + 9$
$5m^2 - 24m - 5 = 0$
Since the lines pass through the origin,their equations are $y = mx$,so $m = \frac{y}{x}$.
Substituting $m = \frac{y}{x}$ into the auxiliary equation:
$5(\frac{y}{x})^2 - 24(\frac{y}{x}) - 5 = 0$
Multiplying by $x^2$,we get:
$5y^2 - 24xy - 5x^2 = 0$
Rearranging,we get $5x^2 + 24xy - 5y^2 = 0$.

(A) For three lines $a_1 x + b_1 y + c_1 = 0$,$a_2 x + b_2 y + c_2 = 0$,and $a_3 x + b_3 y + c_3 = 0$ to be concurrent,the determinant of their coefficients must be zero:
$\begin{vmatrix} k & 2 & 2 \\ 2 & k & 3 \\ 3 & 3 & k \end{vmatrix} = 0$
Expanding the determinant:
$k(k^2 - 9) - 2(2k - 9) + 2(6 - 3k) = 0$
$k^3 - 9k - 4k + 18 + 12 - 6k = 0$
$k^3 - 19k + 30 = 0$
By testing roots,we find $(k - 2)$ is a factor:
$(k - 2)(k^2 + 2k - 15) = 0$
$(k - 2)(k + 5)(k - 3) = 0$
The possible values are $k_1 = 2$,$k_2 = -5$,and $k_3 = 3$.
The sum $\sum k_i = 2 + (-5) + 3 = 0$.

(B) Let the coordinates of the third vertex $C$ be $(x, y, z)$.
Since $G$ is the centroid of $\triangle ABC$,the coordinates of $G$ are given by the formula:
$G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}, \frac{z_A + z_B + z_C}{3} \right)$
Given $A(3, 1, 4)$,$B(-4, 5, -3)$,and $G(-1, 2, 1)$:
$-1 = \frac{3 - 4 + x}{3} \implies -3 = -1 + x \implies x = -2$
$2 = \frac{1 + 5 + y}{3} \implies 6 = 6 + y \implies y = 0$
$1 = \frac{4 - 3 + z}{3} \implies 3 = 1 + z \implies z = 2$
Therefore,the third vertex $C$ is $(-2, 0, 2)$.

(D) Let the coordinates of vertex $B$ be $(x_1, y_1, z_1)$.
The centroid $G$ of a triangle with vertices $A(x_A, y_A, z_A)$,$B(x_B, y_B, z_B)$,and $C(x_C, y_C, z_C)$ is given by $\left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3}\right)$.
Given $A(1,4,2)$,$C(5,-7,1)$,and $G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$,we have:
$\frac{1+x_1+5}{3} = \frac{4}{3} \Rightarrow 6+x_1 = 4 \Rightarrow x_1 = -2$
$\frac{4+y_1-7}{3} = 0 \Rightarrow y_1-3 = 0 \Rightarrow y_1 = 3$
$\frac{2+z_1+1}{3} = \frac{-2}{3} \Rightarrow 3+z_1 = -2 \Rightarrow z_1 = -5$
So,$B = (-2, 3, -5)$.
The midpoint of side $BC$ is $\left(\frac{x_B+x_C}{2}, \frac{y_B+y_C}{2}, \frac{z_B+z_C}{2}\right)$.
Midpoint $= \left(\frac{-2+5}{2}, \frac{3-7}{2}, \frac{-5+1}{2}\right) = \left(\frac{3}{2}, -2, -2\right)$.

(B) Given the equation $\frac{\tan 3x - 1}{\tan 3x + 1} = \sqrt{3}$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we can rewrite the equation as:
$\frac{\tan 3x - \tan(\frac{\pi}{4})}{1 + \tan 3x \tan(\frac{\pi}{4})} = \sqrt{3}$.
This simplifies to $\tan(3x - \frac{\pi}{4}) = \tan(\frac{\pi}{3})$.
The general solution for $\tan \theta = \tan \alpha$ is $\theta = n\pi + \alpha$.
So,$3x - \frac{\pi}{4} = n\pi + \frac{\pi}{3}$.
$3x = n\pi + \frac{\pi}{3} + \frac{\pi}{4} = n\pi + \frac{7\pi}{12}$.
$x = \frac{n\pi}{3} + \frac{7\pi}{36}$.
Comparing this with $x = \frac{n\pi}{p} + \frac{7\pi}{q}$,we get $p = 3$ and $q = 36$.
Therefore,$\frac{p}{q} = \frac{3}{36} = \frac{1}{12}$.

(A) Given equation: $\cos ^2 \theta - 2 \sin \theta + \frac{1}{4} = 0$
Substitute $\cos ^2 \theta = 1 - \sin ^2 \theta$:
$(1 - \sin ^2 \theta) - 2 \sin \theta + \frac{1}{4} = 0$
$\sin ^2 \theta + 2 \sin \theta - \frac{5}{4} = 0$
Multiply by $4$:
$4 \sin ^2 \theta + 8 \sin \theta - 5 = 0$
Factorize the quadratic:
$4 \sin ^2 \theta + 10 \sin \theta - 2 \sin \theta - 5 = 0$
$2 \sin \theta(2 \sin \theta + 5) - 1(2 \sin \theta + 5) = 0$
$(2 \sin \theta - 1)(2 \sin \theta + 5) = 0$
So,$\sin \theta = \frac{1}{2}$ or $\sin \theta = -\frac{5}{2}$.
Since $-1 \le \sin \theta \le 1$,$\sin \theta = -\frac{5}{2}$ is impossible.
Thus,$\sin \theta = \frac{1}{2} = \sin \frac{\pi}{6}$.
The general solution is $\theta = n \pi + (-1)^n \frac{\pi}{6}$.
Comparing with $\theta = \frac{n \pi}{A} + (-1)^n \frac{\pi}{B}$,we get $A = 1$ and $B = 6$.
Therefore,$A + B = 1 + 6 = 7$.

(C) Given equation: $3 \sec^2 \theta = 2 \operatorname{cosec} \theta$
$\Rightarrow \frac{3}{\cos^2 \theta} = \frac{2}{\sin \theta}$
$\Rightarrow \frac{3}{1 - \sin^2 \theta} = \frac{2}{\sin \theta}$
$\Rightarrow 3 \sin \theta = 2 - 2 \sin^2 \theta$
$\Rightarrow 2 \sin^2 \theta + 3 \sin \theta - 2 = 0$
$\Rightarrow (2 \sin \theta - 1)(\sin \theta + 2) = 0$
Since $\sin \theta$ cannot be $-2$,we have $\sin \theta = \frac{1}{2} = \sin \frac{\pi}{6}$.
The general solution for $\sin \theta = \sin \alpha$ is $\theta = n \pi + (-1)^n \alpha$.
Therefore,$\theta = n \pi + (-1)^n \frac{\pi}{6}, n \in Z$.

(A) The given equation is defined for $x \neq \frac{\pi}{2}, \frac{3 \pi}{2}$.
Given equation: $\sec x + \tan x = 2 \cos x$
$\Rightarrow \frac{1}{\cos x} + \frac{\sin x}{\cos x} = 2 \cos x$
$\Rightarrow 1 + \sin x = 2 \cos^2 x$
$\Rightarrow 1 + \sin x = 2(1 - \sin^2 x)$
$\Rightarrow 1 + \sin x = 2(1 - \sin x)(1 + \sin x)$
Since $\sin x = -1$ makes $\cos x = 0$,which is undefined,we have $1 + \sin x \neq 0$.
Dividing both sides by $(1 + \sin x)$,we get:
$1 = 2(1 - \sin x)$
$\Rightarrow 1 = 2 - 2 \sin x$
$\Rightarrow 2 \sin x = 1$
$\Rightarrow \sin x = \frac{1}{2}$
In the interval $[0, 2 \pi)$,the solutions are $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.

(D) The given equation is defined for $x \neq \frac{\pi}{2}, \frac{3 \pi}{2}$.
Given: $\tan x + \sec x = 2 \cos x$
$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$
$\Rightarrow \sin x + 1 = 2 \cos^2 x$
$\Rightarrow \sin x + 1 = 2(1 - \sin^2 x)$
$\Rightarrow \sin x + 1 = 2(1 - \sin x)(1 + \sin x)$
$\Rightarrow (1 + \sin x)[2(1 - \sin x) - 1] = 0$
Since $\sin x = -1$ implies $\cos x = 0$,which makes $\tan x$ and $\sec x$ undefined,we have $1 + \sin x \neq 0$.
Therefore,$2(1 - \sin x) - 1 = 0$
$\Rightarrow 2 - 2 \sin x - 1 = 0$
$\Rightarrow 2 \sin x = 1$
$\Rightarrow \sin x = \frac{1}{2}$
In the interval $[0, 2 \pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$.
Thus,the number of solutions is $2$.

(C) Given equation: $\sin 7 \theta + \sin \theta + \sin 4 \theta = 0$
Using the sum-to-product formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \sin 4 \theta \cos 3 \theta + \sin 4 \theta = 0$
$\sin 4 \theta (2 \cos 3 \theta + 1) = 0$
Case $1$: $\sin 4 \theta = 0 \implies 4 \theta = n \pi \implies \theta = \frac{n \pi}{4}$.
For $\theta \in (0, \pi)$,$n = 1, 2, 3$,so $\theta = \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}$.
Case $2$: $\cos 3 \theta = -\frac{1}{2} = \cos \frac{2 \pi}{3} \implies 3 \theta = 2 n \pi \pm \frac{2 \pi}{3} \implies \theta = \frac{2 n \pi}{3} \pm \frac{2 \pi}{9}$.
For $\theta \in (0, \pi)$:
If $n=0$,$\theta = \frac{2 \pi}{9}$ (since $-\frac{2 \pi}{9}$ is outside range).
If $n=1$,$\theta = \frac{2 \pi}{3} \pm \frac{2 \pi}{9} = \frac{8 \pi}{9}, \frac{4 \pi}{9}$.
If $n=2$,$\theta = \frac{4 \pi}{3} \pm \frac{2 \pi}{9}$ (both outside range).
The solutions are $\theta \in \{\frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{2 \pi}{9}, \frac{4 \pi}{9}, \frac{8 \pi}{9}\}$.
Total number of solutions $= 6$.

(C) Given: $\cos x + \cos y - \cos (x + y) = \frac{3}{2}$
Using the formula $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$ and $\cos (x+y) = 2 \cos^2 \left(\frac{x+y}{2}\right) - 1$,we get:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) - (2 \cos^2 \left(\frac{x+y}{2}\right) - 1) = \frac{3}{2}$
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) - 2 \cos^2 \left(\frac{x+y}{2}\right) = \frac{1}{2}$
Multiplying by $2$:
$4 \cos^2 \left(\frac{x+y}{2}\right) - 4 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) + 1 = 0$
Let $t = \cos \left(\frac{x+y}{2}\right)$. Then $4t^2 - 4t \cos \left(\frac{x-y}{2}\right) + 1 = 0$.
Since $t$ is real,the discriminant $D \geq 0$:
$(-4 \cos \left(\frac{x-y}{2}\right))^2 - 4(4)(1) \geq 0$
$16 \cos^2 \left(\frac{x-y}{2}\right) - 16 \geq 0$
$\cos^2 \left(\frac{x-y}{2}\right) \geq 1$.
Since the maximum value of $\cos^2 \theta$ is $1$,we must have $\cos^2 \left(\frac{x-y}{2}\right) = 1$.
This implies $\frac{x-y}{2} = 0$,so $x = y$.

(C) Given equation: $8 \cos^2 \theta + 14 \cos \theta + 5 = 0$
Factorizing the quadratic equation: $8 \cos^2 \theta + 10 \cos \theta + 4 \cos \theta + 5 = 0$
$\therefore 2 \cos \theta (4 \cos \theta + 5) + 1 (4 \cos \theta + 5) = 0$
$\therefore (2 \cos \theta + 1)(4 \cos \theta + 5) = 0$
$\therefore \cos \theta = -\frac{1}{2}$ or $\cos \theta = -\frac{5}{4}$
Since the range of $\cos \theta$ is $[-1, 1]$,the value $\cos \theta = -\frac{5}{4}$ is impossible.
$\therefore \cos \theta = -\frac{1}{2}$
In the interval $[0, 2\pi]$,$\cos \theta = -\frac{1}{2}$ at $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$.
Thus,the solution set is $\left\{\frac{2\pi}{3}, \frac{4\pi}{3}\right\}$.

(C) Given $f^{\prime}(x)=x-\frac{5}{x^5}$.
Integrating both sides with respect to $x$,we get:
$f(x) = \int \left(x - 5x^{-5}\right) dx$
$f(x) = \frac{x^2}{2} - 5 \left( \frac{x^{-4}}{-4} \right) + c$
$f(x) = \frac{x^2}{2} + \frac{5}{4x^4} + c$
Given $f(1) = 4$,substitute $x=1$:
$4 = \frac{1^2}{2} + \frac{5}{4(1)^4} + c$
$4 = \frac{1}{2} + \frac{5}{4} + c$
$4 = \frac{2+5}{4} + c$
$4 = \frac{7}{4} + c$
$c = 4 - \frac{7}{4} = \frac{16-7}{4} = \frac{9}{4}$
Thus,$f(x) = \frac{x^2}{2} + \frac{5}{4x^4} + \frac{9}{4}$.

(C) Given differential equation is $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$.
Rearranging the terms to separate the variables,we get:
$\cos x(1+\cos y) dx = \sin y(1+\sin x) dy$
$\frac{\cos x}{1+\sin x} dx = \frac{\sin y}{1+\cos y} dy$
Integrating both sides:
$\int \frac{\cos x}{1+\sin x} dx = \int \frac{\sin y}{1+\cos y} dy$
Let $u = 1+\sin x$,then $du = \cos x dx$.
Let $v = 1+\cos y$,then $dv = -\sin y dy$,so $\sin y dy = -dv$.
Substituting these into the integral:
$\int \frac{1}{u} du = \int -\frac{1}{v} dv$
$\ln|u| = -\ln|v| + \ln|c|$
$\ln|1+\sin x| = -\ln|1+\cos y| + \ln|c|$
$\ln|1+\sin x| + \ln|1+\cos y| = \ln|c|$
$\ln|(1+\sin x)(1+\cos y)| = \ln|c|$
Taking the exponential of both sides:
$(1+\sin x)(1+\cos y) = c$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$
Separate the variables: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$
Let $u = 1-y^2$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$.
The integral becomes: $-\frac{1}{2} \int u^{-1/2} du = x + C$
$-\frac{1}{2} (2u^{1/2}) = x + C$
$-\sqrt{1-y^2} = x + C$
Square both sides: $1-y^2 = (x+C)^2$
$(x+C)^2 + y^2 = 1$
This is the equation of a circle $(x-h)^2 + (y-k)^2 = r^2$ where $h = -C$,$k = 0$,and $r = 1$.
Thus,the radius is fixed at $1$ unit and the centre $(-C, 0)$ varies along the $X$-axis.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.
By separating the variables,we get: $\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$.
Integrating both sides: $\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$.
This yields: $\tan^{-1}(y) = \tan^{-1}(x) + C$,where $C$ is a constant of integration.
Rearranging the terms: $\tan^{-1}(y) - \tan^{-1}(x) = C$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}(\frac{A-B}{1+AB})$,we get: $\tan^{-1}(\frac{y-x}{1+xy}) = C$.
Taking the tangent of both sides: $\frac{y-x}{1+xy} = \tan(C)$.
Let $\tan(C) = c$,where $c$ is a new constant.
Thus,$y-x = c(1+xy)$.

(A) Given equation: $\log(x+y) = 2xy$ $(i)$
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$ $(ii)$
At $x=0$,from $(i)$: $\log(0+y) = 2(0)y \implies \log(y) = 0 \implies y = e^0 = 1$.
Substituting $x=0$ and $y=1$ into $(ii)$:
$\frac{1}{0+1} \left(1 + \frac{dy}{dx}\right) = 2(1) + 2(0) \frac{dy}{dx}$
$1 + \frac{dy}{dx} = 2$
$\frac{dy}{dx} = 2 - 1 = 1$
Thus,the value of $\frac{dy}{dx}$ at $x=0$ is $1$.

(B) Given the differential equation $\frac{dy}{dx} = y + 3$.
Separating the variables,we get $\frac{dy}{y + 3} = dx$.
Integrating both sides,we have $\int \frac{dy}{y + 3} = \int dx + C$.
This gives $\log|y + 3| = x + C$.
Given the initial condition $y(0) = 2$,substituting $x = 0$ and $y = 2$ into the equation:
$\log|2 + 3| = 0 + C \implies C = \log 5$.
Thus,the equation becomes $\log(y + 3) = x + \log 5$.
Taking the exponential of both sides,$y + 3 = e^{x + \log 5} = 5e^x$.
So,$y = 5e^x - 3$.
Now,calculating $y(\log 2)$:
$y(\log 2) = 5e^{\log 2} - 3$.
Since $e^{\log 2} = 2$,we get $y(\log 2) = 5(2) - 3 = 10 - 3 = 7$.

(C) Given the differential equation: $\cos (x+y) dy = dx$.
Dividing by $dy$,we get $\frac{dx}{dy} = \cos (x+y)$.
Let $x+y = u$. Differentiating with respect to $y$,we get $\frac{dx}{dy} + 1 = \frac{du}{dy}$,which implies $\frac{dx}{dy} = \frac{du}{dy} - 1$.
Substituting this into the differential equation: $\frac{du}{dy} - 1 = \cos u$.
Rearranging the terms: $\frac{du}{dy} = 1 + \cos u$.
Separating the variables: $\frac{du}{1 + \cos u} = dy$.
Using the identity $1 + \cos u = 2 \cos^2 \left(\frac{u}{2}\right)$,we get $\frac{du}{2 \cos^2 \left(\frac{u}{2}\right)} = dy$,which simplifies to $\frac{1}{2} \sec^2 \left(\frac{u}{2}\right) du = dy$.
Integrating both sides: $\int \frac{1}{2} \sec^2 \left(\frac{u}{2}\right) du = \int dy$.
This gives $\tan \left(\frac{u}{2}\right) = y + c$.
Substituting $u = x+y$ back,we get $\tan \left(\frac{x+y}{2}\right) = y + c$,or $y = \tan \left(\frac{x+y}{2}\right) - c$. Since $c$ is an arbitrary constant,we can write the general solution as $y = \tan \left(\frac{x+y}{2}\right) + c$.

(C) Given the differential equation: $e^{y-x} \frac{dy}{dx} = \frac{y(\sin x + \cos x)}{1 + y \log y}$
Separate the variables: $e^y \frac{1 + y \log y}{y} dy = e^x (\sin x + \cos x) dx$
Simplify the left side: $e^y (\log y + \frac{1}{y}) dy = e^x (\sin x + \cos x) dx$
Integrate both sides: $\int e^y (\log y + \frac{1}{y}) dy = \int e^x (\sin x + \cos x) dx$
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(y) = \log y$ and $f(x) = \sin x$:
$e^y \log y = e^x \sin x + c$
Thus,the correct option is $C$.

(B) Given the differential equation: $e^{\frac{dy}{dx}} = (x+1)$.
Taking the natural logarithm on both sides,we get: $\frac{dy}{dx} = \log(x+1)$.
Integrating both sides with respect to $x$: $\int dy = \int \log(x+1) dx + C$.
Using integration by parts for $\int \log(x+1) dx$ (let $u = \log(x+1)$ and $dv = dx$):
$y = (x+1) \log(x+1) - \int (x+1) \cdot \frac{1}{x+1} dx + C$.
$y = (x+1) \log(x+1) - \int 1 dx + C$.
$y = (x+1) \log(x+1) - x + C$.
Given the condition $y(0) = 3$,substitute $x = 0$ and $y = 3$:
$3 = (0+1) \log(0+1) - 0 + C$.
$3 = 1 \cdot \log(1) + C$.
Since $\log(1) = 0$,we have $3 = 0 + C$,so $C = 3$.
Therefore,the particular solution is $y = (x+1) \log(x+1) - x + 3$.

(D) Given differential equation: $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$
Dividing by $y e^{-x}$,we get: $\frac{y+1}{y} dy + e^x(\cos^2 x - \sin 2x) dx = 0$
Integrating both sides: $\int (1 + \frac{1}{y}) dy + \int e^x(\cos^2 x - 2 \sin x \cos x) dx = C$
Using the integral formula $\int e^x(f(x) + f'(x)) dx = e^x f(x) + C$,where $f(x) = \cos^2 x$ and $f'(x) = -2 \cos x \sin x = -\sin 2x$:
$y + \log |y| + e^x \cos^2 x = C$
Given $x=0, y=1$:
$1 + \log(1) + e^0 \cos^2(0) = C$
$1 + 0 + 1(1) = C \Rightarrow C = 2$
Thus,the solution is $y + \log |y| + e^x \cos^2 x = 2$.

(B) Given differential equation: $(1+xy)y \, dx + (1-xy)x \, dy = 0$
Expanding the terms: $y \, dx + xy^2 \, dx + x \, dy - x^2y \, dy = 0$
Rearranging: $(y \, dx + x \, dy) + xy(y \, dx - x \, dy) = 0$
Dividing by $x^2y^2$: $\frac{y \, dx + x \, dy}{x^2y^2} + \frac{y \, dx - x \, dy}{xy} = 0$
This can be written as: $\frac{d(xy)}{(xy)^2} - \left(\frac{x \, dy - y \, dx}{xy}\right) = 0$
Integrating both sides: $\int \frac{d(xy)}{(xy)^2} - \int d\left(\log \frac{x}{y}\right) = \int 0$
$-\frac{1}{xy} - \log \left(\frac{x}{y}\right) = -k$
$\log \left(\frac{x}{y}\right) = \frac{1}{xy} + k$

(C) Given differential equation is $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$.
Rearranging the terms to separate the variables,we get:
$\frac{\cos x}{1+\sin x} dx = \frac{\sin y}{1+\cos y} dy$.
Integrating both sides:
$\int \frac{\cos x}{1+\sin x} dx = \int \frac{\sin y}{1+\cos y} dy$.
Let $u = 1+\sin x$,then $du = \cos x dx$.
Let $v = 1+\cos y$,then $dv = -\sin y dy$.
Substituting these into the integral:
$\int \frac{1}{u} du = -\int \frac{1}{v} dv$.
$\ln |u| = -\ln |v| + \ln |c|$.
$\ln |u| + \ln |v| = \ln |c|$.
$\ln |uv| = \ln |c|$.
$uv = c$.
Substituting back the values of $u$ and $v$:
$(1+\sin x)(1+\cos y) = c$.

(B) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{3x^2}{1+x^3}$ and $Q = \frac{1}{1+x^3}$.
First,we find the Integrating Factor ($I$.$F$.):
$\text{I.F.} = e^{\int P dx} = e^{\int \frac{3x^2}{1+x^3} dx} = e^{\ln(1+x^3)} = 1+x^3$.
The general solution is given by $y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx + c$.
Substituting the values:
$y(1+x^3) = \int \left(\frac{1}{1+x^3}\right) \cdot (1+x^3) dx + c$
$y(1+x^3) = \int 1 dx + c$
$y(1+x^3) = x + c$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = \sin x$.
First,we find the Integrating Factor ($I$.$F$.):
$I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values:
$yx = \int (\sin x \cdot x) dx + c$.
Using integration by parts $(\int u v dx = u \int v dx - \int (u' \int v dx) dx)$:
Let $u = x$ and $v = \sin x$.
$yx = x(-\cos x) - \int (1 \cdot -\cos x) dx + c$.
$yx = -x \cos x + \int \cos x dx + c$.
$yx = -x \cos x + \sin x + c$.
Rearranging the terms:
$yx + x \cos x = \sin x + c$.
$x(y + \cos x) = \sin x + c$.

(B) Given equation: $\frac{dx}{dy} + \frac{x}{y} = x^2$
Divide both sides by $x^2$: $\frac{1}{x^2} \frac{dx}{dy} + \frac{1}{xy} = 1 \dots (i)$
Let $\frac{1}{x} = t$. Then $-\frac{1}{x^2} \frac{dx}{dy} = \frac{dt}{dy}$,so $\frac{1}{x^2} \frac{dx}{dy} = -\frac{dt}{dy}$.
Substituting into $(i)$: $-\frac{dt}{dy} + \frac{t}{y} = 1$,which simplifies to $\frac{dt}{dy} - \frac{t}{y} = -1$.
This is a linear differential equation in $t$. The Integrating Factor ($I$.$F$.) is $e^{\int -\frac{1}{y} dy} = e^{-\log y} = \frac{1}{y}$.
The solution is $t(I.F.) = \int (-1)(I.F.) dy + c$.
$t(\frac{1}{y}) = \int -\frac{1}{y} dy + c = -\log y + c$.
Substituting $t = \frac{1}{x}$: $\frac{1}{xy} = -\log y + c$.
Therefore,$\frac{1}{x} = cy - y \log y$.

(C) Given equation: $x \, dy = y(dx + y \, dy)$
Rearranging the terms: $x \, dy = y \, dx + y^2 \, dy$
$y \, dx = (x - y^2) \, dy$
Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x}{y} - y$
$\frac{dx}{dy} - \frac{1}{y} x = -y$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -y$.
Integrating Factor ($I$.$F$.) $= e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$.
The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) \, dy + C$.
$x \cdot \frac{1}{y} = \int (-y) \cdot \frac{1}{y} \, dy + C$
$\frac{x}{y} = \int -1 \, dy + C$
$\frac{x}{y} = -y + C$
Given $y(1) = 1$,substitute $x = 1$ and $y = 1$:
$\frac{1}{1} = -1 + C \Rightarrow C = 2$.
So,$\frac{x}{y} = -y + 2$.
To find $y(-3)$,substitute $x = -3$:
$\frac{-3}{y} = -y + 2$
$-3 = -y^2 + 2y$
$y^2 - 2y - 3 = 0$
$(y - 3)(y + 1) = 0$
Since $y(x) > 0$,we have $y = 3$.

(C) The general equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + (y-0)^2 = a^2$,where $(a, 0)$ is the center and $a$ is the radius.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0 ... (i)$.
Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $2a = 2x + 2y \frac{dy}{dx} ... (ii)$.
Substituting the value of $2a$ from equation $(ii)$ into equation $(i)$,we get $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) = 0$.
Simplifying this,$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.
Thus,$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$,which gives $y^2 = x^2 + 2xy \frac{dy}{dx}$.

(B) The rate of evaporation is proportional to the surface area $S = 4\pi R^2$. Since the volume $V = \frac{4}{3}\pi R^3$,the rate of change of volume is $\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$.
Given $\frac{dV}{dt} = -k(4\pi R^2)$,we have $4\pi R^2 \frac{dR}{dt} = -k(4\pi R^2)$,which simplifies to $\frac{dR}{dt} = -k$.
Integrating this,we get $R(t) = -kt + C$.
At $t = 0$,$R = 3$,so $C = 3$. Thus $R(t) = -kt + 3$.
At $t = 1$,$R = 2$,so $2 = -k(1) + 3$,which gives $k = 1$.
Therefore,$R(t) = -t + 3$.
Checking the options,we see that for $R(t) = 3 - t$,the expression $R(t + 2) = (3 - t)(t + 2)$ does not yield a constant. However,re-evaluating the provided options against the boundary conditions $R(0)=3$ and $R(1)=2$:
Option $(B)$ $R(t + 2) = 6 \implies R = \frac{6}{t + 2}$. At $t=0, R=3$. At $t=1, R=2$. This matches the given conditions.

(A) Let the semi-vertical angle be $\alpha = \tan^{-1}\left(\frac{1}{2}\right)$.
Then,$\tan \alpha = \frac{r}{h} = \frac{1}{2}$,which implies $r = \frac{h}{2}$.
The volume of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting $r = \frac{h}{2}$,we get $V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{12} \pi h^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{1}{12} \pi (3h^2) \frac{dh}{dt} = \frac{1}{4} \pi h^2 \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 5 \text{ m}^3/\text{min}$,we have $5 = \frac{1}{4} \pi h^2 \frac{dh}{dt}$,so $\frac{dh}{dt} = \frac{20}{\pi h^2}$.
At $h = 10 \text{ m}$,the rate of change of the water level is $\frac{dh}{dt} = \frac{20}{\pi (10)^2} = \frac{20}{100 \pi} = \frac{1}{5 \pi} \text{ m/min}$.

(A) Given the differential equation: $\frac{dP}{dt} = 0.5 P - 450 = \frac{1}{2}(P - 900)$.
Separating the variables,we get: $\frac{2 dP}{P - 900} = dt$.
Integrating both sides: $2 \int \frac{dP}{P - 900} = \int dt \Rightarrow 2 \ln |P - 900| = t + C$.
Using the initial condition $P(0) = 850$: $2 \ln |850 - 900| = 0 + C \Rightarrow C = 2 \ln 50$.
Thus,the equation is $2 \ln |P - 900| = t + 2 \ln 50$.
To find the time $t$ when the population $P$ becomes $0$:
$2 \ln |0 - 900| = t + 2 \ln 50$
$2 \ln 900 = t + 2 \ln 50$
$t = 2 \ln 900 - 2 \ln 50 = 2 \ln \left( \frac{900}{50} \right) = 2 \ln 18$.

(A) Let $x$ be the number of bacteria present at time $t$.
The rate of increase is proportional to the number of bacteria,so $\frac{dx}{dt} = kx$.
Separating variables,we get $\frac{dx}{x} = k dt$.
Integrating both sides,we get $\log x = kt + c$ ... $(i)$.
Given that at $t = 3$,$x = 10,000$,we have $\log(10,000) = 3k + c$ ... $(ii)$.
Given that at $t = 5$,$x = 40,000$,we have $\log(40,000) = 5k + c$ ... $(iii)$.
Subtracting $(ii)$ from $(iii)$,we get $\log(40,000) - \log(10,000) = 2k$,which simplifies to $\log(4) = 2k$,so $k = \log(2)$.
Substituting $k = \log(2)$ into $(ii)$,we get $\log(10,000) = 3\log(2) + c$.
Thus,$c = \log(10,000) - \log(8) = \log(\frac{10,000}{8}) = \log(1250)$.
Initially,at $t = 0$,$\log x = k(0) + c$,so $\log x = \log(1250)$.
Therefore,the initial number of bacteria is $x = 1250$.

(D) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x)f^{\prime}(x)$.
At $x = 1$,we have $f(1) = 1$ and $f^{\prime}(1) = 3$.
Substituting these values:
$f(f(1)) = f(1) = 1$.
$f(f(f(1))) = f(1) = 1$.
Therefore,$\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1)f^{\prime}(1)$.
$= f^{\prime}(1) \cdot f^{\prime}(1) \cdot f^{\prime}(1) + 2(1)(3)$.
$= 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

(C) Given $y = \log_{\sin x} \tan x$. Using the change of base formula,we can write $y = \frac{\log \tan x}{\log \sin x}$.
Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \log \tan x$ and $v = \log \sin x$:
$u' = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$.
$v' = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Thus,$\frac{dy}{dx} = \frac{(\log \sin x)(2 \csc 2x) - (\log \tan x)(\cot x)}{(\log \sin x)^2}$.
At $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$,$\tan x = 1$,$\log \tan x = 0$,$\csc 2x = \csc \frac{\pi}{2} = 1$,and $\cot x = 1$.
Substituting these values:
$\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = \frac{(\log \frac{1}{\sqrt{2}})(2 \cdot 1) - (0)(1)}{(\log \frac{1}{\sqrt{2}})^2} = \frac{2 \log(2^{-1/2})}{(\log 2^{-1/2})^2} = \frac{2 \cdot (-\frac{1}{2} \log 2)}{(-\frac{1}{2} \log 2)^2} = \frac{-\log 2}{\frac{1}{4} (\log 2)^2} = \frac{-4}{\log 2}$.

(A) $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^{\frac{3}{2}}$
Taking $\log$ on both sides,we get
$\log y = \frac{3}{2} [\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$
Differentiating with respect to $x$,we get
$\frac{1}{y} \frac{dy}{dx} = \frac{3}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$
$\therefore \frac{dy}{dx} = \frac{3y}{2} \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$
Now at $x=0$,$y = [(1)(1)(1) \ldots (1)]^{\frac{3}{2}} = 1$
$\left. \frac{dy}{dx} \right|_{x=0} = \frac{3(1)}{2} \left[ \frac{1}{0+1} + \frac{2}{0+1} + \frac{3}{0+1} + \ldots + \frac{n}{0+1} \right]$
$= \frac{3}{2} (1 + 2 + 3 + \ldots + n)$
$= \frac{3}{2} \times \frac{n(n+1)}{2}$
$= \frac{3n(n+1)}{4}$

(C) Given $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2n})$.
Taking $\log$ on both sides,we get $\log y = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \dots + \log(1+x^{2n})$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \dots + \frac{2n \cdot x^{2n-1}}{1+x^{2n}}$.
At $x=0$,the value of $y = (1+0)(1+0) \dots (1+0) = 1$.
Substituting $x=0$ and $y=1$ in the derivative equation,we get $\frac{1}{1} \left. \frac{dy}{dx} \right|_{x=0} = \frac{1}{1+0} + 0 + 0 + \dots + 0$.
Therefore,$\left. \frac{dy}{dx} \right|_{x=0} = 1$.

(C) Given $\tan y = \frac{x \sin \alpha}{1-x \cos \alpha}$.
Taking the derivative with respect to $x$ on both sides:
$\frac{d}{dx}(\tan y) = \frac{d}{dx}\left(\frac{x \sin \alpha}{1-x \cos \alpha}\right)$
$\sec^2 y \cdot \frac{dy}{dx} = \frac{(1-x \cos \alpha)(\sin \alpha) - (x \sin \alpha)(-\cos \alpha)}{(1-x \cos \alpha)^2}$
$\sec^2 y \cdot \frac{dy}{dx} = \frac{\sin \alpha - x \sin \alpha \cos \alpha + x \sin \alpha \cos \alpha}{(1-x \cos \alpha)^2} = \frac{\sin \alpha}{(1-x \cos \alpha)^2}$
Since $\tan y = \frac{x \sin \alpha}{1-x \cos \alpha}$,we have $\sec^2 y = 1 + \tan^2 y = 1 + \frac{x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{(1-x \cos \alpha)^2 + x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{1 - 2x \cos \alpha + x^2 \cos^2 \alpha + x^2 \sin^2 \alpha}{(1-x \cos \alpha)^2} = \frac{1 - 2x \cos \alpha + x^2}{(1-x \cos \alpha)^2}$.
Substituting this back:
$\frac{1 - 2x \cos \alpha + x^2}{(1-x \cos \alpha)^2} \cdot \frac{dy}{dx} = \frac{\sin \alpha}{(1-x \cos \alpha)^2}$
$\frac{dy}{dx} = \frac{\sin \alpha}{x^2 - 2x \cos \alpha + 1}$.
Comparing this with $\frac{dy}{dx} = \frac{m}{x^2+2nx+1}$,we get $m = \sin \alpha$ and $n = -\cos \alpha$.
Therefore,$m^2 + n^2 = \sin^2 \alpha + (-\cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha = 1$.

(C) Given $y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}$.
Taking logarithm on both sides:
$\log y = \frac{1}{3} \log (1+3 x) + \frac{1}{4} \log (1+4 x) + \frac{1}{5} \log (1+5 x) - \frac{1}{7} \log (1+7 x) - \frac{1}{8} \log (1+8 x)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{3} \cdot \frac{3}{1+3 x} + \frac{1}{4} \cdot \frac{4}{1+4 x} + \frac{1}{5} \cdot \frac{5}{1+5 x} - \frac{1}{7} \cdot \frac{7}{1+7 x} - \frac{1}{8} \cdot \frac{8}{1+8 x}$.
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1+3 x} + \frac{1}{1+4 x} + \frac{1}{1+5 x} - \frac{1}{1+7 x} - \frac{1}{1+8 x}$.
At $x=0$,$y = \frac{\sqrt[3]{1} \sqrt[4]{1} \sqrt[5]{1}}{\sqrt[7]{1} \sqrt[8]{1}} = 1$.
Substituting $x=0$ and $y=1$ into the derivative equation:
$\frac{1}{1} \left(\frac{d y}{d x}\right)_{x=0} = \frac{1}{1+0} + \frac{1}{1+0} + \frac{1}{1+0} - \frac{1}{1+0} - \frac{1}{1+0} = 1 + 1 + 1 - 1 - 1 = 1$.
Thus,$\left(\frac{d y}{d x}\right)_{x=0} = 1$.

(B) Given $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^n$.
Taking the natural logarithm on both sides:
$\log y = n [\log(x+1) + \log(2x+1) + \log(3x+1) + \ldots + \log(nx+1)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = n \left[ \frac{1}{x+1} + \frac{2}{2x+1} + \frac{3}{3x+1} + \ldots + \frac{n}{nx+1} \right]$.
At $x=0$,$y = [(1)(1)(1) \ldots (1)]^n = 1^n = 1$.
Substituting $x=0$ and $y=1$ into the derivative expression:
$\frac{1}{1} \left( \frac{dy}{dx} \right)_{x=0} = n [1 + 2 + 3 + \ldots + n]$.
Using the sum formula $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:
$\left( \frac{dy}{dx} \right)_{x=0} = n \left[ \frac{n(n+1)}{2} \right] = \frac{n^2(n+1)}{2}$.

(A) Given the expression $y=\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\ldots}}}$.
We can rewrite this as $y=\sqrt{(x-\sin x)+y}$.
Squaring both sides,we get $y^2 = x - \sin x + y$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 1 - \cos x + \frac{dy}{dx}$.
Rearranging the terms to solve for $\frac{dy}{dx}$,we have $2y \frac{dy}{dx} - \frac{dy}{dx} = 1 - \cos x$.
Factoring out $\frac{dy}{dx}$,we get $\frac{dy}{dx}(2y-1) = 1 - \cos x$.
Therefore,$\frac{dy}{dx} = \frac{1-\cos x}{2y-1}$.

(C) Given $y=f\left(\frac{2 x+3}{3-2 x}\right)$.
Applying the chain rule,we have $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$.
Using the quotient rule for $\frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$:
$\frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right) = \frac{(3-2 x)(2) - (2 x+3)(-2)}{(3-2 x)^2} = \frac{6-4 x+4 x+6}{(3-2 x)^2} = \frac{12}{(3-2 x)^2}$.
Since $f^{\prime}(x)=\sin (\log x)$,then $f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) = \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)$.
Thus,$\frac{d y}{d x} = \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right) \cdot \frac{12}{(3-2 x)^2}$.
At $x=1$,the argument of the function is $\frac{2(1)+3}{3-2(1)} = \frac{5}{1} = 5$.
The derivative at $x=1$ is $\sin (\log 5) \cdot \frac{12}{(3-2)^2} = 12 \sin (\log 5)$.

(C) Given equation: $\log(x+y)=2xy$ ... $(i)$
At $x=0$,$\log(0+y)=2(0)y$,which implies $\log(y)=0$,so $y=e^0=1$.
Differentiating both sides of $(i)$ with respect to $x$:
$\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$
Substitute $x=0$ and $y=1$ into the differentiated equation:
$\frac{1}{0+1} \left(1 + \frac{dy}{dx}\right) = 2(1) + 2(0) \frac{dy}{dx}$
$1 + \frac{dy}{dx} = 2$
$\frac{dy}{dx} = 2 - 1 = 1$
Thus,the value of $\frac{dy}{dx}$ at $x=0$ is $1$.

(B) Given the equation: $(2x)^{2y} = 4e^{2x-2y}$.
Taking the natural logarithm $(\log_e)$ on both sides:
$2y \log_e(2x) = \log_e 4 + (2x - 2y) \log_e e$
Since $\log_e 4 = 2 \log_e 2$ and $\log_e e = 1$,we have:
$2y \log_e(2x) = 2 \log_e 2 + 2x - 2y$
Dividing by $2$:
$y \log_e(2x) = \log_e 2 + x - y$
Rearranging to solve for $y$:
$y(1 + \log_e 2x) = x + \log_e 2$
$y = \frac{x + \log_e 2}{1 + \log_e 2x} \quad \dots(i)$
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} (1 + \log_e 2x) + y \left( \frac{1}{2x} \cdot 2 \right) = 1 - \frac{dy}{dx}$
$\frac{dy}{dx} (1 + \log_e 2x) + \frac{y}{x} = 1 - \frac{dy}{dx}$
$\frac{dy}{dx} (1 + \log_e 2x + 1) = 1 - \frac{y}{x}$
$\frac{dy}{dx} (2 + \log_e 2x) = \frac{x - y}{x}$
Wait,let us re-evaluate the derivative step:
From $y(1 + \log_e 2x) = x + \log_e 2$,differentiate:
$\frac{dy}{dx} (1 + \log_e 2x) + y \left( \frac{1}{x} \right) = 1$
$\frac{dy}{dx} (1 + \log_e 2x) = 1 - \frac{y}{x} = \frac{x - y}{x}$
Substitute $y$ from $(i)$:
$\frac{dy}{dx} (1 + \log_e 2x) = \frac{x - \frac{x + \log_e 2}{1 + \log_e 2x}}{x} = \frac{x(1 + \log_e 2x) - (x + \log_e 2)}{x(1 + \log_e 2x)}$
$= \frac{x + x \log_e 2x - x - \log_e 2}{x(1 + \log_e 2x)} = \frac{x \log_e 2x - \log_e 2}{x(1 + \log_e 2x)}$
Multiplying both sides by $(1 + \log_e 2x)$:
$(1 + \log_e 2x)^2 \frac{dy}{dx} = \frac{x \log_e 2x - \log_e 2}{x}$

(B) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.
Squaring the first equation on both sides,we get:
$(x^2+y^2)^2 = (t+\frac{1}{t})^2$
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$.
Since $x^4+y^4=t^2+\frac{1}{t^2}$,we substitute this into the equation:
$(t^2+\frac{1}{t^2})+2x^2y^2 = t^2+\frac{1}{t^2}+2$.
This simplifies to $2x^2y^2 = 2$,which implies $x^2y^2 = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2y^2) = \frac{d}{dx}(1)$
$x^2(2y\frac{dy}{dx}) + y^2(2x) = 0$.
$2x^2y\frac{dy}{dx} = -2xy^2$.
$\frac{dy}{dx} = \frac{-2xy^2}{2x^2y} = \frac{-y}{x}$.

(A) Given the equation: $\log (x+y)=2xy \quad ...(i)$
Differentiating both sides with respect to $x$ using the chain rule and product rule:
$\frac{1}{x+y} \cdot (1 + y^{\prime}) = 2(x y^{\prime} + y)$
Rearranging to solve for $y^{\prime}$:
$1 + y^{\prime} = 2(x+y)(x y^{\prime} + y)$
$1 + y^{\prime} = 2x^2 y^{\prime} + 2xy + 2xy y^{\prime} + 2y^2$
$y^{\prime}(1 - 2x^2 - 2xy) = 2xy + 2y^2 - 1$
$y^{\prime} = \frac{2xy + 2y^2 - 1}{1 - 2x^2 - 2xy}$
Now,find the value of $y$ when $x=0$ from equation $(i)$:
$\log(0+y) = 2(0)y$
$\log(y) = 0$
$y = e^0 = 1$
Substitute $x=0$ and $y=1$ into the expression for $y^{\prime}$:
$y^{\prime}(0) = \frac{2(0)(1) + 2(1)^2 - 1}{1 - 2(0)^2 - 2(0)(1)}$
$y^{\prime}(0) = \frac{0 + 2 - 1}{1 - 0 - 0} = \frac{1}{1} = 1$

(A) Given $f(x) = 3^x$,the derivative is $f^{\prime}(x) = 3^x \log 3$. Evaluating at $x = 0$,we get $f^{\prime}(0) = 3^0 \log 3 = \log 3$.
Given $g(x) = 4^x$,the derivative is $g^{\prime}(x) = 4^x \log 4$. Evaluating at $x = 0$,we get $g^{\prime}(0) = 4^0 \log 4 = \log 4$.
Substituting these values into the expression,we have $\frac{f^{\prime}(0)-g^{\prime}(0)}{1+f^{\prime}(0) g^{\prime}(0)} = \frac{\log 3 - \log 4}{1 + (\log 3)(\log 4)}$.
Using the logarithmic property $\log a - \log b = \log \left(\frac{a}{b}\right)$,the numerator becomes $\log \left(\frac{3}{4}\right)$.
Thus,the expression is $\frac{\log \left(\frac{3}{4}\right)}{1 + (\log 3)(\log 4)}$.

(C) Let $y = f(\sec x)$ and $z = g(\tan x)$.
Using the chain rule,we find the derivatives with respect to $x$:
$\frac{dy}{dx} = f^{\prime}(\sec x) \cdot \sec x \tan x$
$\frac{dz}{dx} = g^{\prime}(\tan x) \cdot \sec^2 x$
Now,the derivative of $y$ with respect to $z$ is given by:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{f^{\prime}(\sec x) \cdot \sec x \tan x}{g^{\prime}(\tan x) \cdot \sec^2 x} = \frac{f^{\prime}(\sec x) \tan x}{g^{\prime}(\tan x) \sec x}$
At $x = \frac{\pi}{4}$,we have $\sec(\frac{\pi}{4}) = \sqrt{2}$ and $\tan(\frac{\pi}{4}) = 1$.
Substituting these values:
$\left. \frac{dy}{dz} \right|_{x=\frac{\pi}{4}} = \frac{f^{\prime}(\sqrt{2}) \cdot 1}{g^{\prime}(1) \cdot \sqrt{2}} = \frac{4 \cdot 1}{2 \cdot \sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) Let $y = \sqrt{x^2+16}$.
Then,$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2+16}} \cdot (2x) = \frac{x}{\sqrt{x^2+16}}$.
Let $z = \frac{x}{x-1}$.
Using the quotient rule,$\frac{dz}{dx} = \frac{(x-1)(1) - x(1)}{(x-1)^2} = \frac{-1}{(x-1)^2}$.
We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.
$\frac{dy}{dz} = \frac{x}{\sqrt{x^2+16}} \div \left( \frac{-1}{(x-1)^2} \right) = -\frac{x(x-1)^2}{\sqrt{x^2+16}}$.
At $x=5$:
$\frac{dy}{dz} = -\frac{5(5-1)^2}{\sqrt{5^2+16}} = -\frac{5(16)}{\sqrt{25+16}} = -\frac{80}{\sqrt{41}}$.

(B) Given $x=3 \tan t$ and $y=3 \sec t$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 3 \sec^2 t$
$\frac{dy}{dt} = 3 \sec t \tan t$
Now,find the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \sin t$
Now,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \frac{d}{dt}(\sin t) \cdot \frac{dt}{dx} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos^3 t}{3}$
Finally,evaluate at $t = \frac{\pi}{4}$:
$\left(\frac{d^2y}{dx^2}\right)_{t=\frac{\pi}{4}} = \frac{(\cos(\pi/4))^3}{3} = \frac{(1/\sqrt{2})^3}{3} = \frac{1/(2\sqrt{2})}{3} = \frac{1}{6\sqrt{2}}$

(A) Let $y = \tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$ and $z = \cos ^{-1}(x^2)$.
Substitute $x^2 = \cos 2\theta$,which implies $\theta = \frac{1}{2} \cos ^{-1}(x^2)$.
Then,$y = \tan ^{-1}\left(\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right)$.
Using the identities $1+\cos 2\theta = 2\cos^2 \theta$ and $1-\cos 2\theta = 2\sin^2 \theta$,we get:
$y = \tan ^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right) = \tan ^{-1}\left(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\right)$.
Dividing numerator and denominator by $\cos \theta$,we have:
$y = \tan ^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right) = \tan ^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right) = \frac{\pi}{4} - \theta$.
Substituting $\theta = \frac{1}{2} \cos ^{-1}(x^2)$ back into the equation:
$y = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1}(x^2) = \frac{\pi}{4} - \frac{1}{2} z$.
Now,differentiating $y$ with respect to $z$:
$\frac{dy}{dz} = \frac{d}{dz}\left(\frac{\pi}{4} - \frac{1}{2} z\right) = -\frac{1}{2}$.

(A) Let $p = f(\tan x)$ and $q = g(\sec x)$.
Using the chain rule,we find the derivatives with respect to $x$:
$\frac{dp}{dx} = f^{\prime}(\tan x) \cdot \sec^2 x$
$\frac{dq}{dx} = g^{\prime}(\sec x) \cdot \sec x \tan x$
At $x = \frac{\pi}{4}$,we have $\tan x = 1$ and $\sec x = \sqrt{2}$.
$\left. \frac{dp}{dx} \right|_{x=\frac{\pi}{4}} = f^{\prime}(1) \cdot (\sqrt{2})^2 = 2 \cdot 2 = 4$.
$\left. \frac{dq}{dx} \right|_{x=\frac{\pi}{4}} = g^{\prime}(\sqrt{2}) \cdot (\sqrt{2} \cdot 1) = 4 \cdot \sqrt{2} = 4\sqrt{2}$.
The required derivative is $\frac{dp}{dq} = \frac{dp/dx}{dq/dx} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(B) Let $u = \tan ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$ and $v = \cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right]$.
Substitute $x = \tan \theta$,then $\theta = \tan ^{-1} x$.
For $u$:
$u = \tan ^{-1}\left[\frac{\sec \theta - 1}{\tan \theta}\right] = \tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] = \tan ^{-1}\left[\frac{2 \sin ^2 (\theta/2)}{2 \sin (\theta/2) \cos (\theta/2)}\right] = \tan ^{-1}(\tan (\theta/2)) = \frac{\theta}{2} = \frac{\tan ^{-1} x}{2}$.
For $v$:
$v = \cos ^{-1}\left[\sqrt{\frac{1+\sec \theta}{2 \sec \theta}}\right] = \cos ^{-1}\left[\sqrt{\frac{\cos \theta + 1}{2}}\right] = \cos ^{-1}\left[\sqrt{\frac{2 \cos ^2 (\theta/2)}{2}}\right] = \cos ^{-1}(\cos (\theta/2)) = \frac{\theta}{2} = \frac{\tan ^{-1} x}{2}$.
Since $u = v$,we have $\frac{du}{dv} = \frac{d}{dv}(v) = 1$.

(B) Given $x=\log _{e}\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$.
Dividing numerator and denominator by $\cos \frac{y}{2}$,we get $e^x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}} = \tan \left(\frac{\pi}{4}-\frac{y}{2}\right) \quad \dots(i)$.
Differentiating both sides with respect to $x$,we get $e^x = \sec^2 \left(\frac{\pi}{4}-\frac{y}{2}\right) \cdot \left(-\frac{1}{2}\right) \frac{dy}{dx}$.
Thus,$\frac{dy}{dx} = -2e^x \cos^2 \left(\frac{\pi}{4}-\frac{y}{2}\right)$.
When $t=\frac{1}{2}$,$\tan \frac{y}{2} = \sqrt{\frac{1-1/2}{1+1/2}} = \sqrt{\frac{1/2}{3/2}} = \frac{1}{\sqrt{3}}$.
So,$\frac{y}{2} = \frac{\pi}{6}$.
Substituting $\frac{y}{2} = \frac{\pi}{6}$ into equation $(i)$,$e^x = \tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right) = \tan \frac{\pi}{12} = 2-\sqrt{3}$.
Now,$\cos^2 \left(\frac{\pi}{4}-\frac{\pi}{6}\right) = \cos^2 \frac{\pi}{12} = \left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^2 = \frac{3+1+2\sqrt{3}}{8} = \frac{4+2\sqrt{3}}{8} = \frac{2+\sqrt{3}}{4}$.
Therefore,$\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}} = -2(2-\sqrt{3}) \cdot \frac{2+\sqrt{3}}{4} = -\frac{1}{2}(4-3) = -\frac{1}{2}$.

(A) Given $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}\right)$.
Let $x^2=a^2 \tan \theta$,then $\theta=\tan ^{-1}\left(\frac{x^2}{a^2}\right)$.
Substituting $x^2$ in the expression for $y$:
$y=\cos ^{-1}\left(\frac{a^2}{\sqrt{(a^2 \tan \theta)^2+a^4}}\right) = \cos ^{-1}\left(\frac{a^2}{\sqrt{a^4 \tan ^2 \theta+a^4}}\right)$
$y=\cos ^{-1}\left(\frac{a^2}{a^2 \sqrt{\tan ^2 \theta+1}}\right) = \cos ^{-1}\left(\frac{1}{\sec \theta}\right)$
$y=\cos ^{-1}(\cos \theta) = \theta$
Thus,$y=\tan ^{-1}\left(\frac{x^2}{a^2}\right)$.
Differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x} = \frac{1}{1+\left(\frac{x^2}{a^2}\right)^2} \cdot \frac{d}{d x}\left(\frac{x^2}{a^2}\right)$
$\frac{d y}{d x} = \frac{1}{1+\frac{x^4}{a^4}} \cdot \frac{2x}{a^2} = \frac{a^4}{a^4+x^4} \cdot \frac{2x}{a^2}$
$\frac{d y}{d x} = \frac{2 a^2 x}{x^4+a^4}$.

(A) Given $y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$.
Using properties of logarithms,$\log(\frac{e}{x^2}) = \log e - \log x^2 = 1 - 2\log x$ and $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.
So,$y = \tan^{-1}\left(\frac{1-2\log x}{1+2\log x}\right) + \tan^{-1}\left(\frac{4+2\log x}{1-8\log x}\right)$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we have $\tan^{-1}(1) - \tan^{-1}(2\log x) = \tan^{-1}\left(\frac{1-2\log x}{1+2\log x}\right)$.
Also,$\tan^{-1}(4) + \tan^{-1}(2\log x) = \tan^{-1}\left(\frac{4+2\log x}{1-4(2\log x)}\right) = \tan^{-1}\left(\frac{4+2\log x}{1-8\log x}\right)$.
Substituting these back,$y = \tan^{-1}(1) - \tan^{-1}(2\log x) + \tan^{-1}(4) + \tan^{-1}(2\log x)$.
$y = \tan^{-1}(1) + \tan^{-1}(4)$.
Since $y$ is a constant,$\frac{dy}{dx} = 0$.

(B) Given $y=\sqrt{\frac{1-\sin ^{-1}(x)}{1+\sin ^{-1}(x)}}$.
Taking the natural logarithm on both sides:
$\log y = \frac{1}{2} [\log(1-\sin^{-1}x) - \log(1+\sin^{-1}x)]$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1-\sin^{-1}x} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) - \frac{1}{1+\sin^{-1}x} \cdot \left(\frac{1}{\sqrt{1-x^2}}\right) \right]$.
Simplifying the expression:
$\frac{dy}{dx} = -\frac{y}{2\sqrt{1-x^2}} \left( \frac{1}{1-\sin^{-1}x} + \frac{1}{1+\sin^{-1}x} \right)$.
At $x=0$,$\sin^{-1}(0)=0$ and $y=1$:
$\left(\frac{dy}{dx}\right)_{(0,1)} = -\frac{1}{2\sqrt{1-0}} \left( \frac{1}{1-0} + \frac{1}{1+0} \right) = -\frac{1}{2} (1+1) = -1$.

(D) Let $u = \log x$. Then $f(x) = \sin^{-1}\left(\frac{2u}{1+u^2}\right)$.
Using the substitution $u = \tan \theta$,we have $\frac{2u}{1+u^2} = \sin(2\theta)$.
Thus,$f(x) = \sin^{-1}(\sin(2\theta)) = 2\theta = 2 \tan^{-1}(u) = 2 \tan^{-1}(\log x)$.
Now,differentiate with respect to $x$:
$f^{\prime}(x) = 2 \times \frac{1}{1+(\log x)^2} \times \frac{d}{dx}(\log x)$
$f^{\prime}(x) = \frac{2}{1+(\log x)^2} \times \frac{1}{x} = \frac{2}{x(1+(\log x)^2)}$.
Substitute $x = e$:
$f^{\prime}(e) = \frac{2}{e(1+(\log e)^2)} = \frac{2}{e(1+1^2)} = \frac{2}{e(2)} = \frac{1}{e}$.

(C) Given $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$.
Using the identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$:
$y = \tan^{-1} \left( \frac{8 \sin x \cos x}{\cos^2 x - \sin^2 x - 6 \sin^2 x} \right) = \tan^{-1} \left( \frac{8 \sin x \cos x}{\cos^2 x - 7 \sin^2 x} \right)$.
Divide numerator and denominator by $\cos^2 x$:
$y = \tan^{-1} \left( \frac{8 \tan x}{1 - 7 \tan^2 x} \right)$.
We can write $8 \tan x = 7 \tan x + \tan x$:
$y = \tan^{-1} \left( \frac{7 \tan x + \tan x}{1 - (7 \tan x)(\tan x)} \right)$.
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$:
$y = \tan^{-1}(7 \tan x) + \tan^{-1}(\tan x) = \tan^{-1}(7 \tan x) + x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1 + (7 \tan x)^2} \cdot 7 \sec^2 x + 1 = \frac{7 \sec^2 x}{1 + 49 \tan^2 x} + 1$.
At $x=0$,$\tan 0 = 0$ and $\sec 0 = 1$:
$\left(\frac{dy}{dx}\right)_{x=0} = \frac{7(1)^2}{1 + 49(0)^2} + 1 = \frac{7}{1} + 1 = 8$.

(A) Given $f^{\prime}(x) = \tan ^{-1}(\sec x + \tan x)$.
Simplifying the expression inside the inverse tangent:
$f^{\prime}(x) = \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$
Using half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$:
$f^{\prime}(x) = \tan ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}\right]$
$f^{\prime}(x) = \tan ^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}\right)$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$f^{\prime}(x) = \tan ^{-1}\left(\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}\right) = \tan ^{-1}\left[\tan \left(\frac{\pi}{4} + \frac{x}{2}\right)\right]$
Thus,$f^{\prime}(x) = \frac{\pi}{4} + \frac{x}{2}$.
Integrating with respect to $x$:
$f(x) = \int \left(\frac{\pi}{4} + \frac{x}{2}\right) dx = \frac{\pi x}{4} + \frac{x^2}{4} + C$.
Given $f(0) = 0$,we get $C = 0$.
So,$f(x) = \frac{\pi x + x^2}{4}$.
Therefore,$f(1) = \frac{\pi(1) + (1)^2}{4} = \frac{\pi+1}{4}$.

(B) Given $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$.
We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.
Substituting this,$y = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} + 2(\frac{\pi}{2} - \sin^{-1} x) \cdot (-\frac{1}{\sqrt{1-x^2}})$.
$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} (\sin^{-1} x - \frac{\pi}{2} + \sin^{-1} x) = \frac{2(2\sin^{-1} x - \frac{\pi}{2})}{\sqrt{1-x^2}}$.
Multiplying by $\sqrt{1-x^2}$: $\sqrt{1-x^2} y_1 = 4\sin^{-1} x - \pi$.
Differentiating again with respect to $x$:
$\sqrt{1-x^2} y_2 + y_1 \cdot (\frac{-2x}{2\sqrt{1-x^2}}) = \frac{4}{\sqrt{1-x^2}}$.
Multiplying throughout by $\sqrt{1-x^2}$: $(1-x^2) y_2 - x y_1 = 4$.

(C) Given $x = \sqrt{e^{\sin^{-1} t}}$ and $y = \sqrt{e^{\cos^{-1} t}}$.
Multiplying $x$ and $y$,we get:
$xy = \sqrt{e^{\sin^{-1} t}} \cdot \sqrt{e^{\cos^{-1} t}} = \sqrt{e^{\sin^{-1} t + \cos^{-1} t}}$.
Since $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have $xy = \sqrt{e^{\pi/2}}$,which is a constant.
Differentiating both sides with respect to $x$:
$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \quad ... (i)$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{dx^2} = -\frac{d}{dx} \left( \frac{y}{x} \right) = -\left( \frac{x \frac{dy}{dx} - y}{x^2} \right)$.
Substituting $\frac{dy}{dx} = -\frac{y}{x}$ from $(i)$:
$\frac{d^2 y}{dx^2} = -\left( \frac{x(-\frac{y}{x}) - y}{x^2} \right) = -\left( \frac{-y - y}{x^2} \right) = -\left( \frac{-2y}{x^2} \right) = \frac{2y}{x^2}$.