MHT CET 2024 Mathematics Question Paper with Answer and Solution

(B) Let the intercepts of the line on the $X$-axis and $Y$-axis be $b$ and $a$ respectively. The points are $A(0, a)$ and $B(b, 0)$.
Area of $\triangle OAB = \frac{1}{2} |ab| = 12 \implies |ab| = 24$.
Assuming the line has a negative slope,we can write the intercept form of the line as $\frac{x}{b} + \frac{y}{a} = 1$.
Since the line passes through $(2, 3)$,we have $\frac{2}{b} + \frac{3}{a} = 1$.
From $ab = 24$,we have $b = \frac{24}{a}$.
Substituting $b$ in the equation: $\frac{2}{24/a} + \frac{3}{a} = 1 \implies \frac{2a}{24} + \frac{3}{a} = 1 \implies \frac{a}{12} + \frac{3}{a} = 1$.
Multiplying by $12a$: $a^2 + 36 = 12a \implies a^2 - 12a + 36 = 0 \implies (a-6)^2 = 0 \implies a = 6$.
Then $b = \frac{24}{6} = 4$.
The equation of the line is $\frac{x}{4} + \frac{y}{6} = 1$.
Multiplying by $12$: $3x + 2y = 12$.

(A) The given equation is $x^2+2hxy+2y^2=0$.
Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=h, b=2$.
Let the slopes of the lines be $m_1$ and $m_2$.
We know that $m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h$ and $m_1m_2 = \frac{a}{b} = \frac{1}{2}$.
Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.
Substituting $m_2$ in the product equation: $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.
If $m_1 = \frac{1}{2}$,then $m_2 = 1$.
Then $m_1+m_2 = \frac{1}{2} + 1 = \frac{3}{2}$.
Since $m_1+m_2 = -h$,we get $-h = \frac{3}{2} \Rightarrow h = -\frac{3}{2}$.
If $m_1 = -\frac{1}{2}$,then $m_2 = -1$.
Then $m_1+m_2 = -\frac{1}{2} - 1 = -\frac{3}{2}$.
Since $m_1+m_2 = -h$,we get $-h = -\frac{3}{2} \Rightarrow h = \frac{3}{2}$.
Thus,$h = \pm \frac{3}{2}$. Given the options,the correct value is $\frac{3}{2}$.

(A) The required line passes through the point of intersection of the lines $3x - y = 5$ and $x + 3y = 1$.
Solving these equations:
$3(3x - y) = 3(5) \Rightarrow 9x - 3y = 15$
Adding $x + 3y = 1$ to $9x - 3y = 15$,we get $10x = 16 \Rightarrow x = \frac{8}{5}$.
Substituting $x = \frac{8}{5}$ in $3x - y = 5$: $3(\frac{8}{5}) - y = 5 \Rightarrow y = \frac{24}{5} - 5 = -\frac{1}{5}$.
So,the point of intersection is $(\frac{8}{5}, -\frac{1}{5})$.
The equation of a line with equal intercepts is $x + y = a$ or $x - y = a$.
Since the line passes through $(\frac{8}{5}, -\frac{1}{5})$,for $x + y = a$:
$\frac{8}{5} - \frac{1}{5} = a \Rightarrow a = \frac{7}{5}$.
Thus,$x + y = \frac{7}{5} \Rightarrow 5x + 5y - 7 = 0$.

(C) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have:
$m_1 + m_2 = \frac{-2h}{b}$ $(i)$
$m_1 m_2 = \frac{a}{b}$ $(ii)$
Given $4ab = 3h^2$,so $ab = \frac{3h^2}{4}$.
Substituting this into $(ii)$,we get $m_1 m_2 = \frac{3h^2}{4b^2}$.
Now,$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 = \frac{4h^2}{b^2} - 4(\frac{3h^2}{4b^2}) = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$.
Thus,$m_1 - m_2 = \frac{h}{b}$ $(iii)$.
Adding $(i)$ and $(iii)$: $2m_1 = \frac{-2h}{b} + \frac{h}{b} = \frac{-h}{b} \implies m_1 = \frac{-h}{2b}$.
Subtracting $(iii)$ from $(i)$: $2m_2 = \frac{-2h}{b} - \frac{h}{b} = \frac{-3h}{b} \implies m_2 = \frac{-3h}{2b}$.
The ratio $m_1 : m_2 = \frac{-h}{2b} : \frac{-3h}{2b} = 1 : 3$.

(C) The given equation of the pair of lines is $K x^2 + 6 x y + y^2 = 0$.
Comparing this with the general form $A x^2 + 2 H x y + B y^2 = 0$,we get $A = K$,$H = 3$,and $B = 1$.
Let the slopes of the two lines be $m_1$ and $m_2$.
We know that $m_1 + m_2 = \frac{-2 H}{B} = -6$ and $m_1 m_2 = \frac{A}{B} = K$.
Given that one slope is three times the other,let $m_2 = 3 m_1$.
Substituting this into the sum of slopes: $m_1 + 3 m_1 = -6$ $\Rightarrow 4 m_1 = -6$ $\Rightarrow m_1 = -\frac{3}{2}$.
Now,using the product of slopes: $m_1 \times (3 m_1) = K \Rightarrow 3 m_1^2 = K$.
Substituting $m_1 = -\frac{3}{2}$: $K = 3 \times (-\frac{3}{2})^2 = 3 \times \frac{9}{4} = \frac{27}{4}$.

(B) The given pair of lines is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $a+2h(\frac{y}{x})+b(\frac{y}{x})^2=0$. Let $k = \frac{y}{x}$ be the slope of one of the lines. Then $bk^2+2hk+a=0$.
The slope of the line $mx+ny=18$ is $-\frac{m}{n}$.
Since the line is perpendicular to $mx+ny=18$,its slope $k$ must be the negative reciprocal of $-\frac{m}{n}$,so $k = \frac{n}{m}$.
Substituting $k = \frac{n}{m}$ into the equation $bk^2+2hk+a=0$,we get $b(\frac{n}{m})^2+2h(\frac{n}{m})+a=0$.
Multiplying by $m^2$,we obtain $bn^2+2hmn+am^2=0$.

(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$,we get $bhx^2 + 2abyx + ahy^2 = 0$.
Comparing this with the standard form $Ax^2 + 2Hxy + By^2 = 0$,we have $A = bh$,$2H = 2ab \Rightarrow H = ab$,and $B = ah$.
Let the slopes be $m_1$ and $m_2$. We are given $m_2 = 2m_1$.
For the equation $Ax^2 + 2Hxy + By^2 = 0$,the sum of slopes $m_1 + m_2 = -\frac{2H}{B}$ and the product of slopes $m_1 m_2 = \frac{A}{B}$.
$m_1 + 2m_1 = 3m_1 = -\frac{2ab}{ah} = -\frac{2b}{h} \Rightarrow m_1 = -\frac{2b}{3h}$.
$m_1(2m_1) = 2m_1^2 = \frac{bh}{ah} = \frac{b}{a} \Rightarrow 2\left(-\frac{2b}{3h}\right)^2 = \frac{b}{a}$.
$2 \times \frac{4b^2}{9h^2} = \frac{b}{a} \Rightarrow \frac{8b^2}{9h^2} = \frac{b}{a}$.
Dividing both sides by $b$ (assuming $b \neq 0$),we get $\frac{8b}{9h^2} = \frac{1}{a} \Rightarrow \frac{ab}{h^2} = \frac{9}{8}$.

(B) The given equations of the lines are $x \cos 30^{\circ} + y \sin 30^{\circ} = 3$ and $x \cos 60^{\circ} + y \sin 60^{\circ} = 5$.
These are in the normal form $x \cos \alpha + y \sin \alpha = p$,where $\alpha$ is the angle made by the normal to the line with the positive $x$-axis.
The angle of the normal to the first line is $\alpha_1 = 30^{\circ}$ and to the second line is $\alpha_2 = 60^{\circ}$.
The angle $\theta$ between two lines is equal to the angle between their normals.
Therefore,$\theta = |\alpha_2 - \alpha_1| = |60^{\circ} - 30^{\circ}| = 30^{\circ}$.

(A) Slope of line $l_1 = \frac{4-2}{-3-1} = \frac{2}{-4} = -\frac{1}{2}$.
Since $(h, k)$ lies on $l_1$,the slope between $(h, k)$ and $(1, 2)$ must be $-\frac{1}{2}$:
$\frac{k-2}{h-1} = -\frac{1}{2}$ $\Rightarrow 2k-4 = -h+1$ $\Rightarrow h+2k = 5$ ... $(i)$.
Since $l_2$ passes through $(h, k)$ and $(4, 3)$ and is perpendicular to $l_1$,its slope is $m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2$.
Thus,$\frac{3-k}{4-h} = 2$ $\Rightarrow 3-k = 8-2h$ $\Rightarrow 2h-k = 5$ ... $(ii)$.
Multiplying $(ii)$ by $2$: $4h-2k = 10$ ... $(iii)$.
Adding $(i)$ and $(iii)$: $(h+2k) + (4h-2k) = 5+10$ $\Rightarrow 5h = 15$ $\Rightarrow h = 3$.
Substituting $h=3$ into $(i)$: $3+2k = 5$ $\Rightarrow 2k = 2$ $\Rightarrow k = 1$.
Therefore,$\frac{k}{h} = \frac{1}{3}$.

(A) The equation of the given line is $2x - 3y + 17 = 0$,which can be written as $3y = 2x + 17$ or $y = \frac{2}{3}x + \frac{17}{3}$.
Thus,the slope of this line is $m_1 = \frac{2}{3}$.
The slope of the line passing through $(7, 17)$ and $(15, \beta)$ is $m_2 = \frac{\beta - 17}{15 - 7} = \frac{\beta - 17}{8}$.
Since the two lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 \times m_2 = -1$.
$\frac{2}{3} \times \frac{\beta - 17}{8} = -1$.
$\frac{\beta - 17}{12} = -1$.
$\beta - 17 = -12$.
$\beta = 17 - 12 = 5$.

(A) The slope of line $AB$ $(4x+y=1)$ is $m_1 = -4$. The slope of line $BC$ $(3x-4y+1=0)$ is $m_2 = \frac{3}{4}$.
Let $\alpha$ be the angle between $AB$ and $BC$. Then,$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-4 - \frac{3}{4}}{1 + (-4)(\frac{3}{4})} \right| = \left| \frac{-\frac{19}{4}}{1 - 3} \right| = \left| \frac{-\frac{19}{4}}{-2} \right| = \frac{19}{8}$.
Since $AB=AC$,the triangle $ABC$ is isosceles with $\angle ABC = \angle ACB = \alpha$.
Let the slope of line $AC$ be $m$. Since $AC$ passes through $A(2,-7)$,its equation is $y+7 = m(x-2)$.
The angle between $AC$ and $BC$ is also $\alpha$,so $\tan \alpha = \left| \frac{m - \frac{3}{4}}{1 + m(\frac{3}{4})} \right| = \frac{19}{8}$.
$\frac{4m-3}{4+3m} = \pm \frac{19}{8}$.
Case $1$: $8(4m-3) = 19(4+3m)$ $\Rightarrow 32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is the slope of $AB$).
Case $2$: $8(4m-3) = -19(4+3m)$ $\Rightarrow 32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.
Using $m = -\frac{52}{89}$ in $y+7 = m(x-2)$:
$89(y+7) = -52(x-2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

(D) Given equations of lines:
$x+(a-1)y=1$ and $2x+a^2y=1$.
Slope of $x+(a-1)y=1$ is $m_1 = \frac{-1}{a-1}$.
Slope of $2x+a^2y=1$ is $m_2 = \frac{-2}{a^2}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$.
$\frac{-1}{a-1} \times \frac{-2}{a^2} = -1$
$\frac{2}{a^2(a-1)} = -1$
$2 = -a^3 + a^2$
$a^3 - a^2 + 2 = 0$.
Factoring the cubic equation: $(a+1)(a^2-2a+2) = 0$.
Since $a^2-2a+2 = (a-1)^2 + 1 > 0$ for all real $a$,we must have $a+1=0$,so $a=-1$.
Substituting $a=-1$ into the line equations:
$x+(-1-1)y=1 \Rightarrow x-2y=1$.
$2x+(-1)^2y=1 \Rightarrow 2x+y=1$.
Solving the system:
$x-2y=1$ $(i)$
$2x+y=1$ (ii)
Multiply (ii) by $2$: $4x+2y=2$.
Adding $(i)$ and (ii): $5x=3 \Rightarrow x=\frac{3}{5}$.
Substituting $x=\frac{3}{5}$ into (ii): $2(\frac{3}{5})+y=1 \Rightarrow y=1-\frac{6}{5} = -\frac{1}{5}$.
The point of intersection is $(\frac{3}{5}, -\frac{1}{5})$.
The distance from the origin $(0,0)$ is $\sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$.

(D) Let the foot of the perpendicular be $(h, k)$.
Since $(h, k)$ lies on the line $3x - y - 1 = 0$,we have $3h - k - 1 = 0 \implies k = 3h - 1$ ... $(i)$.
The slope of the given line $3x - y - 1 = 0$ is $m_1 = 3$.
The slope of the line segment joining $(-2, 3)$ and $(h, k)$ is $m_2 = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$,so $3 \times \frac{k - 3}{h + 2} = -1$.
$3(k - 3) = -(h + 2) \implies 3k - 9 = -h - 2 \implies h + 3k = 7$ ... $(ii)$.
Substituting $k = 3h - 1$ from $(i)$ into $(ii)$:
$h + 3(3h - 1) = 7 \implies h + 9h - 3 = 7 \implies 10h = 10 \implies h = 1$.
Substituting $h = 1$ into $(i)$,$k = 3(1) - 1 = 2$.
Thus,the coordinates of the foot of the perpendicular are $(1, 2)$.

(D) The coordinates are $P(-5, 0)$,$Q(0, 0)$,and $R(2, 2\sqrt{3})$.
$QP$ lies along the negative $X$-axis,so the angle it makes with the positive $X$-axis is $180^{\circ}$.
The slope of $QR$ is $m = \frac{2\sqrt{3} - 0}{2 - 0} = \sqrt{3}$.
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 $120^{\circ}$ angle into two $60^{\circ}$ angles.
Since $QP$ is at $180^{\circ}$,the bisector makes an angle of $180^{\circ} - 60^{\circ} = 120^{\circ}$ with the positive $X$-axis.
The slope of the bisector is $\tan 120^{\circ} = -\sqrt{3}$.
The equation of the line passing through the origin $(0, 0)$ with slope $-\sqrt{3}$ is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

(D) Line $L$ passes through $(13,32)$.
$\frac{13}{5}+\frac{32}{b}=1$
$\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$
$\Rightarrow b = -20$
So,the equation of $L$ is $\frac{x}{5}-\frac{y}{20}=1$,which simplifies to $4x-y=20$.
The slope of $L$ is $m_1=4$.
The slope of line $K$ given by $\frac{x}{c}+\frac{y}{3}=1$ is $m_2=-\frac{3}{c}$.
Since $L$ and $K$ are parallel,$m_1=m_2$.
$4 = -\frac{3}{c} \Rightarrow c = -\frac{3}{4}$.
The equation of line $K$ is $-\frac{4x}{3}+\frac{y}{3}=1$,which simplifies to $4x-y=-3$.
The distance between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=4, B=-1, C_1=-20, C_2=3$.
$d = \frac{|-20-3|}{\sqrt{4^2+(-1)^2}} = \frac{|-23|}{\sqrt{16+1}} = \frac{23}{\sqrt{17}}$.

(B) The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin and $\alpha$ is the angle the perpendicular makes with the positive $X$-axis.
Given $p = 2 \sqrt{2}$ and $\alpha = 135^{\circ}$.
The equation of the line is $x \cos(135^{\circ}) + y \sin(135^{\circ}) = 2 \sqrt{2}$.
Since $\cos(135^{\circ}) = -\frac{1}{\sqrt{2}}$ and $\sin(135^{\circ}) = \frac{1}{\sqrt{2}}$,we have:
$x \left(-\frac{1}{\sqrt{2}}\right) + y \left(\frac{1}{\sqrt{2}}\right) = 2 \sqrt{2}$.
Multiplying both sides by $\sqrt{2}$,we get:
$-x + y = 4$
or $x - y + 4 = 0$.

(B) The equation of a straight line passing through $(3,-2)$ with slope $m$ is given by $y+2=m(x-3)$ $(i)$.
The given line is $\sqrt{3} x+y=1$,which can be written as $y=-\sqrt{3} x+1$. The slope of this line is $m_1=-\sqrt{3}$.
The angle between the two lines is $60^{\circ}$. Using the formula $\tan \theta = \left| \frac{m-m_1}{1+m m_1} \right|$,we have $\tan 60^{\circ} = \left| \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} \right|$.
$\sqrt{3} = \left| \frac{m+\sqrt{3}}{1-\sqrt{3} m} \right|$.
Case $1$: $\sqrt{3} = \frac{m+\sqrt{3}}{1-\sqrt{3} m} \Rightarrow \sqrt{3} - 3m = m + \sqrt{3} \Rightarrow 4m = 0 \Rightarrow m = 0$.
Substituting $m=0$ into $(i)$,we get $y+2=0(x-3) \Rightarrow y+2=0$. This line is parallel to the $X$-axis and does not intersect the $X$-axis (unless it is the $X$-axis itself,which it is not).
Case $2$: $-\sqrt{3} = \frac{m+\sqrt{3}}{1-\sqrt{3} m} \Rightarrow -\sqrt{3} + 3m = m + \sqrt{3} \Rightarrow 2m = 2\sqrt{3} \Rightarrow m = \sqrt{3}$.
Substituting $m=\sqrt{3}$ into $(i)$,we get $y+2=\sqrt{3}(x-3) \Rightarrow y+2=\sqrt{3}x-3\sqrt{3} \Rightarrow y-\sqrt{3}x+2+3\sqrt{3}=0$.

(A) Given equation: $2 \sqrt{3} \cos^2 \theta = \sin \theta$
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get:
$2 \sqrt{3} (1 - \sin^2 \theta) = \sin \theta$
$2 \sqrt{3} - 2 \sqrt{3} \sin^2 \theta = \sin \theta$
$2 \sqrt{3} \sin^2 \theta + \sin \theta - 2 \sqrt{3} = 0$
Let $x = \sin \theta$. Then $2 \sqrt{3} x^2 + x - 2 \sqrt{3} = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2 \sqrt{3})(-2 \sqrt{3})}}{2(2 \sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4 \sqrt{3}} = \frac{-1 \pm 7}{4 \sqrt{3}}$
Case $1$: $x = \frac{6}{4 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$
Case $2$: $x = \frac{-8}{4 \sqrt{3}} = -\frac{2}{\sqrt{3}} \approx -1.15$ (Not possible as $|\sin \theta| \leq 1$)
So,$\sin \theta = \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3}$
The general solution is $\theta = n \pi + (-1)^n \frac{\pi}{3}, n \in Z$.

(C) Given the equation $\tan \theta = \tan \alpha$.
Since the tangent function has a period of $\pi$,the general solution for $\tan \theta = \tan \alpha$ is given by $\theta = n \pi + \alpha$,where $n \in Z$.

(B) Given equation: $\sin^2 \theta - \cos \theta = \frac{1}{4}$
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$(1 - \cos^2 \theta) - \cos \theta = \frac{1}{4}$
$4 - 4\cos^2 \theta - 4\cos \theta = 1$
$4\cos^2 \theta + 4\cos \theta - 3 = 0$
Let $x = \cos \theta$,then $4x^2 + 4x - 3 = 0$
$4x^2 + 6x - 2x - 3 = 0$
$2x(2x + 3) - 1(2x + 3) = 0$
$(2x - 1)(2x + 3) = 0$
So,$\cos \theta = \frac{1}{2}$ or $\cos \theta = -\frac{3}{2}$ (rejected as $-1 \le \cos \theta \le 1$)
For $\cos \theta = \frac{1}{2}$ in $[0, 2\pi]$,$\theta = \frac{\pi}{3}$ or $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$
Thus,the solution set is $\left\{\frac{\pi}{3}, \frac{5\pi}{3}\right\}$.

(D) Given equation: $\sin x - 3 \sin 2x + \sin 3x = \cos x - 3 \cos 2x + \cos 3x$
Rearranging terms: $(\sin 3x + \sin x) - 3 \sin 2x = (\cos 3x + \cos x) - 3 \cos 2x$
Using sum-to-product formulas: $2 \sin 2x \cos x - 3 \sin 2x = 2 \cos 2x \cos x - 3 \cos 2x$
$\sin 2x (2 \cos x - 3) = \cos 2x (2 \cos x - 3)$
$(\sin 2x - \cos 2x)(2 \cos x - 3) = 0$
Since $2 \cos x - 3 = 0$ implies $\cos x = 1.5$,which is impossible,we have $\sin 2x = \cos 2x$
$\tan 2x = 1$
$2x = n\pi + \frac{\pi}{4}$
$x = \frac{n\pi}{2} + \frac{\pi}{8}, n \in Z$

(A) Given equation: $\tan(x+100^{\circ}) = \tan(x+50^{\circ}) \tan(x) \tan(x-50^{\circ})$
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we can simplify the expression.
Alternatively,using $\tan(x+100^{\circ}) = \tan(x+50^{\circ}+50^{\circ}) = \frac{\tan(x+50^{\circ}) + \tan 50^{\circ}}{1 - \tan(x+50^{\circ}) \tan 50^{\circ}}$.
By substituting $x = 30^{\circ}$:
$LHS$: $\tan(30^{\circ}+100^{\circ}) = \tan(130^{\circ}) = \tan(180^{\circ}-50^{\circ}) = -\tan 50^{\circ}$.
$RHS$: $\tan(30^{\circ}+50^{\circ}) \tan(30^{\circ}) \tan(30^{\circ}-50^{\circ}) = \tan(80^{\circ}) \tan(30^{\circ}) \tan(-20^{\circ})$.
Using the property $\tan(3x) = \tan(x) \tan(60^{\circ}-x) \tan(60^{\circ}+x)$,we can see that for $x=30^{\circ}$,the equation holds true.
Thus,the smallest positive value is $x = 30^{\circ}$.

(B) The range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
For the equation $7 \cos x + 5 \sin x = 2k + 1$,the range of the left side is $[-\sqrt{7^2 + 5^2}, \sqrt{7^2 + 5^2}] = [-\sqrt{74}, \sqrt{74}]$.
Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.
Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.
Dividing by $2$: $-4.801 \leq k \leq 3.801$.
The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.
Counting these values,we get $8$ integral values.

(A) Given equation: $\sin ^2 x - \cos 2 x = 2 - \sin 2 x$
Using identities $\cos 2 x = 1 - 2 \sin ^2 x$ and $\sin 2 x = 2 \sin x \cos x$:
$\sin ^2 x - (1 - 2 \sin ^2 x) = 2 - 2 \sin x \cos x$
$3 \sin ^2 x - 1 = 2 - 2 \sin x \cos x$
$3 \sin ^2 x + 2 \sin x \cos x - 3 = 0$
Since $\sin ^2 x + \cos ^2 x = 1$,we have $3 \sin ^2 x + 2 \sin x \cos x - 3(\sin ^2 x + \cos ^2 x) = 0$
$3 \sin ^2 x + 2 \sin x \cos x - 3 \sin ^2 x - 3 \cos ^2 x = 0$
$2 \sin x \cos x - 3 \cos ^2 x = 0$
$\cos x (2 \sin x - 3 \cos x) = 0$
This implies $\cos x = 0$ or $2 \sin x - 3 \cos x = 0$.
Given $3 \cos x \neq 2 \sin x$,we must have $\cos x = 0$.
The general solution for $\cos x = 0$ is $x = (2 n + 1) \frac{\pi}{2}, n \in Z$.

(C) Given equation: $\sin x + \cos x = 1$
Divide both sides by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$
$\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$\sin(x + \frac{\pi}{4}) = \sin \frac{\pi}{4}$
The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$.
Therefore,$x + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$
$x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}, n \in Z$.

(D) Given $\alpha+\beta+\gamma=\pi$,so $\gamma = \pi - (\alpha+\beta)$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:
$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + \sin(\beta+\gamma)\sin(\beta-\gamma)$
Since $\beta+\gamma = \pi - \alpha$,$\sin(\beta+\gamma) = \sin(\pi-\alpha) = \sin \alpha$.
Substituting this:
$= \sin^2 \alpha + \sin \alpha \sin(\beta-\gamma)$
$= \sin \alpha [\sin \alpha + \sin(\beta-\gamma)]$
$= \sin \alpha [\sin(\beta+\gamma) + \sin(\beta-\gamma)]$
Using the sum-to-product formula $\sin(x+y) + \sin(x-y) = 2 \sin x \cos y$:
$= \sin \alpha [2 \sin \beta \cos \gamma]$
$= 2 \sin \alpha \sin \beta \cos \gamma$.

(C) Given equation: $(1-\tan \theta)(1+\tan \theta) \sec ^2 \theta+2 \tan ^2 \theta=0$
Using $(1-\tan \theta)(1+\tan \theta) = 1-\tan^2 \theta$ and $\sec^2 \theta = 1+\tan^2 \theta$,we get:
$(1-\tan^2 \theta)(1+\tan^2 \theta) + 2\tan^2 \theta = 0$
Let $x = \tan^2 \theta$. Since $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,$x \ge 0$.
The equation becomes $(1-x)(1+x) + 2x = 0$
$1 - x^2 + 2x = 0$
$x^2 - 2x - 1 = 0$
Solving for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$
Since $x = \tan^2 \theta \ge 0$,we must have $x = 1 + \sqrt{2}$ (as $1 - \sqrt{2} < 0$).
Thus,$\tan^2 \theta = 1 + \sqrt{2}$.
Since $1 + \sqrt{2} > 0$,there are two values of $\tan \theta$,namely $\tan \theta = \pm \sqrt{1 + \sqrt{2}}$.
For each value of $\tan \theta$,there is exactly one value of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Therefore,there are $2$ values of $\theta$ that satisfy the equation.

(C) Given,$\tan A - \tan B = x$
$\cot B - \cot A = y$
$\Rightarrow \frac{1}{\tan B} - \frac{1}{\tan A} = y$
$\Rightarrow \frac{\tan A - \tan B}{\tan A \tan B} = y$
$\Rightarrow \frac{x}{\tan A \tan B} = y$
$\Rightarrow \tan A \tan B = \frac{x}{y}$
Now,$\cot (A - B) = \frac{1}{\tan (A - B)} = \frac{1 + \tan A \tan B}{\tan A - \tan B}$
Substituting the values,we get:
$\cot (A - B) = \frac{1 + \frac{x}{y}}{x} = \frac{\frac{y + x}{y}}{x} = \frac{x + y}{xy} = \frac{x}{xy} + \frac{y}{xy} = \frac{1}{y} + \frac{1}{x}$

(C) Given,$\alpha = \beta + \gamma$.
Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$.
Substituting $\beta$ in the first equation: $\alpha = (\frac{\pi}{2} - \alpha) + \gamma$,which implies $\gamma = 2\alpha - \frac{\pi}{2}$.
Alternatively,using $\gamma = \alpha - \beta$:
$\tan \gamma = \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Since $\beta = \frac{\pi}{2} - \alpha$,then $\tan \beta = \cot \alpha = \frac{1}{\tan \alpha}$.
Substituting this: $\tan \gamma = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha (\frac{1}{\tan \alpha})} = \frac{\tan \alpha - \tan \beta}{1 + 1} = \frac{\tan \alpha - \tan \beta}{2}$.
Therefore,$2 \tan \gamma = \tan \alpha - \tan \beta$,which gives $\tan \alpha = \tan \beta + 2 \tan \gamma$.

(D) Given the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$.
We know that $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting this into the equation: $1 - 2 \sin^2 \theta \cos^2 \theta + \lambda = 0$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4}$.
Thus,$1 - 2 \left(\frac{\sin^2 2\theta}{4}\right) + \lambda = 0$,which simplifies to $1 - \frac{\sin^2 2\theta}{2} + \lambda = 0$.
Therefore,$\lambda = \frac{\sin^2 2\theta}{2} - 1$.
Since $0 \leq \sin^2 2\theta \leq 1$,we have $0 \leq \frac{\sin^2 2\theta}{2} \leq \frac{1}{2}$.
Subtracting $1$ from all parts,we get $-1 \leq \frac{\sin^2 2\theta}{2} - 1 \leq -\frac{1}{2}$.
Hence,$\lambda \in \left[-1, -\frac{1}{2}\right]$.

(B) Given that $\alpha$ and $\beta$ are roots of $\sqrt{3} a \cos x + 2 b \sin x = c$.
Since $\alpha$ and $\beta$ are roots,we have:
$\sqrt{3} a \cos \alpha + 2 b \sin \alpha = c$ $(i)$
$\sqrt{3} a \cos \beta + 2 b \sin \beta = c$ $(ii)$
Subtracting $(ii)$ from $(i)$,we get:
$\sqrt{3} a (\cos \alpha - \cos \beta) + 2 b (\sin \alpha - \sin \beta) = 0$
Using the sum-to-product formulas:
$\sqrt{3} a [-2 \sin(\frac{\alpha + \beta}{2}) \sin(\frac{\alpha - \beta}{2})] + 2 b [2 \cos(\frac{\alpha + \beta}{2}) \sin(\frac{\alpha - \beta}{2})] = 0$
Given $\alpha + \beta = \frac{\pi}{3}$,so $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$.
Substituting this value:
$-\sqrt{3} a [2 \sin(\frac{\pi}{6}) \sin(\frac{\alpha - \beta}{2})] + 4 b [\cos(\frac{\pi}{6}) \sin(\frac{\alpha - \beta}{2})] = 0$
Since $\alpha \neq \beta$,$\sin(\frac{\alpha - \beta}{2}) \neq 0$,we can divide by it:
$-\sqrt{3} a (2 \times \frac{1}{2}) + 4 b (\frac{\sqrt{3}}{2}) = 0$
$-\sqrt{3} a + 2 \sqrt{3} b = 0$
$2 \sqrt{3} b = \sqrt{3} a$
$\frac{b}{a} = \frac{\sqrt{3}}{2 \sqrt{3}} = \frac{1}{2} = 0.5$

(D) Using the projection rule in $\triangle ABC$,we have $c = a \cos B + b \cos A$ and $b = a \cos C + c \cos A$.
The given expression is $E = \frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a}$.
Taking the common denominator: $E = \frac{a(\cos B+\cos C) + (b+c)\cos A}{a(b+c)}$.
Expanding the numerator: $E = \frac{a \cos B + a \cos C + b \cos A + c \cos A}{a(b+c)}$.
Rearranging terms: $E = \frac{(a \cos B + b \cos A) + (a \cos C + c \cos A)}{a(b+c)}$.
Substituting the projection rules: $E = \frac{c + b}{a(b+c)}$.
Simplifying: $E = \frac{b+c}{a(b+c)} = \frac{1}{a}$.

(C) Given equation: $\sqrt{3} \cos \theta + \sin \theta = \sqrt{2}$
Divide both sides by $2$: $\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{\sqrt{2}}{2}$
Using $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$,we get: $\sin \frac{\pi}{3} \cos \theta + \cos \frac{\pi}{3} \sin \theta = \frac{1}{\sqrt{2}}$
Applying the formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\sin \left(\theta + \frac{\pi}{3}\right) = \sin \frac{\pi}{4}$
The general solution for $\sin x = \sin \alpha$ is $x = n \pi + (-1)^{n} \alpha$: $\theta + \frac{\pi}{3} = n \pi + (-1)^{n} \frac{\pi}{4}, n \in Z$
Therefore,$\theta = n \pi + (-1)^{n} \frac{\pi}{4} - \frac{\pi}{3}, n \in Z$

(D) Given $2 \sin^2 x + 3 \sin x - 2 > 0$.
Factoring the quadratic expression,we get $(2 \sin x - 1)(\sin x + 2) > 0$.
Since $\sin x + 2 > 0$ for all $x \in \mathbb{R}$,we must have $2 \sin x - 1 > 0$,which implies $\sin x > \frac{1}{2}$.
This gives $x \in (\frac{\pi}{6}, \frac{5 \pi}{6})$.
Also,given $x^2 - x - 2 < 0$.
Factoring gives $(x - 2)(x + 1) < 0$,which implies $-1 < x < 2$.
To find the intersection,we note that $\frac{\pi}{6} \approx 0.52$ and $\frac{5 \pi}{6} \approx 2.61$.
Since $2 < \frac{5 \pi}{6}$,the intersection of $(\frac{\pi}{6}, \frac{5 \pi}{6})$ and $(-1, 2)$ is $(\frac{\pi}{6}, 2)$.

(C) Given equation: $\tan x + \sec x = 2 \cos x$ for $x \in [0, 2 \pi]$.
Step $1$: Substitute $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$:
$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$
$\frac{\sin x + 1}{\cos x} = 2 \cos x$
Step $2$: Multiply by $\cos x$ (where $\cos x \neq 0$):
$\sin x + 1 = 2 \cos^2 x$
Step $3$: Use $\cos^2 x = 1 - \sin^2 x$:
$\sin x + 1 = 2(1 - \sin^2 x)$
$\sin x + 1 = 2 - 2 \sin^2 x$
Step $4$: Rearrange into a quadratic equation in $\sin x$:
$2 \sin^2 x + \sin x - 1 = 0$
Step $5$: Solve the quadratic equation:
$(2 \sin x - 1)(\sin x + 1) = 0$
$\sin x = \frac{1}{2}$ or $\sin x = -1$
Step $6$: Find solutions in $[0, 2 \pi]$:
For $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}, \frac{5 \pi}{6}$.
For $\sin x = -1$,$x = \frac{3 \pi}{2}$.
Note: At $x = \frac{3 \pi}{2}$,$\cos x = 0$,which makes $\tan x$ and $\sec x$ undefined. Thus,$x = \frac{3 \pi}{2}$ is an extraneous solution.
Checking the remaining values: For $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$,$\cos x \neq 0$. Therefore,the valid solution set is $\left\{\frac{\pi}{6}, \frac{5 \pi}{6}\right\}$.
Since the provided options include $\frac{3 \pi}{2}$,and option $C$ is the closest match,we select $C$.

(C) Given equation: $3 \sin^2 x - 7 \sin x + 2 = 0$.
Factorizing the quadratic equation: $3 \sin^2 x - 6 \sin x - \sin x + 2 = 0$.
$3 \sin x(\sin x - 2) - 1(\sin x - 2) = 0$.
$(3 \sin x - 1)(\sin x - 2) = 0$.
This gives $\sin x = \frac{1}{3}$ or $\sin x = 2$.
Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is not possible.
Thus,we solve $\sin x = \frac{1}{3}$.
In the interval $(0, 2 \pi)$,there are $2$ solutions for $\sin x = \frac{1}{3}$ (one in the first quadrant and one in the second quadrant).
In the interval $(0, 4 \pi)$,there are $2 \times 2 = 4$ solutions.
In the interval $(4 \pi, 5 \pi)$,there is $1$ solution (in the first quadrant relative to $4 \pi$).
Total solutions in $(0, 5 \pi) = 2 + 2 + 2 = 6$.

(A) Given: $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$
Multiply both sides by $(a+b+c)$:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=3$
$\frac{a}{b+c}+1+\frac{b}{c+a}+1=3$
$\frac{a}{b+c}+\frac{b}{c+a}=1$
$a(c+a)+b(b+c)=(b+c)(c+a)$
$ac+a^2+b^2+bc=bc+ab+c^2+ac$
$a^2+b^2-c^2=ab$
By the Cosine Rule:
$\cos C = \frac{a^2+b^2-c^2}{2ab}$
Substitute $a^2+b^2-c^2=ab$:
$\cos C = \frac{ab}{2ab} = \frac{1}{2}$
Therefore,$C = \frac{\pi}{3}$.

(D) Given $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3 \pi}{2}$.
Since $x$ is in the third quadrant,$\sec^2 x = 1 + \tan^2 x = 1 + \frac{9}{16} = \frac{25}{16}$.
Thus,$\sec x = -\frac{5}{4}$ (as $\sec x < 0$ in the third quadrant),which implies $\cos x = -\frac{4}{5}$.
Given $\pi < x < \frac{3 \pi}{2}$,we have $\frac{\pi}{2} < \frac{x}{2} < \frac{3 \pi}{4}$.
In the interval $(\frac{\pi}{2}, \frac{3 \pi}{4})$,$\cos \frac{x}{2}$ is negative.
Using the half-angle formula,$\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}}$.
Substituting $\cos x = -\frac{4}{5}$,we get $\cos \frac{x}{2} = -\sqrt{\frac{1 - 4/5}{2}} = -\sqrt{\frac{1/5}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

(A) Let the given expression be $E = \cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right)$.
Using the identity $\cos \left(\frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{8}\right)$ and $\sin \left(\frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)$,we substitute these into the expression:
$E = \cos ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) + \sin ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$
Factor out $\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$:
$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \left[ \cos ^2\left(\frac{\pi}{8}\right) + \sin ^2\left(\frac{\pi}{8}\right) \right]$
Using the Pythagorean identity $\sin ^2 A + \cos ^2 A = 1$:
$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \cdot (1) = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$
Multiply and divide by $2$ to use the double angle formula $\sin (2A) = 2 \sin A \cos A$:
$E = \frac{1}{2} \left[ 2 \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \right] = \frac{1}{2} \sin \left( 2 \cdot \frac{\pi}{8} \right) = \frac{1}{2} \sin \left(\frac{\pi}{4}\right)$
Since $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:
$E = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2 \sqrt{2}}$.

(B) Given equations are $\cot \theta = \sqrt{3}$ and $\sqrt{3} \sec \theta + 2 = 0$.
From $\cot \theta = \sqrt{3}$,we have $\tan \theta = \frac{1}{\sqrt{3}}$. This implies $\theta$ is in the $1^{\text{st}}$ or $3^{\text{rd}}$ quadrant.
From $\sqrt{3} \sec \theta + 2 = 0$,we have $\sec \theta = -\frac{2}{\sqrt{3}}$,which means $\cos \theta = -\frac{\sqrt{3}}{2}$.
Since $\cos \theta$ is negative,$\theta$ must be in the $2^{\text{nd}}$ or $3^{\text{rd}}$ quadrant.
Combining both conditions,$\theta$ must lie in the $3^{\text{rd}}$ quadrant.
In the $3^{\text{rd}}$ quadrant,$\tan \theta = \frac{1}{\sqrt{3}}$ corresponds to $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
We need to evaluate $a \cos ^2 \frac{C}{2} + c \cos ^2 \frac{A}{2}$.
Using the identity $\cos ^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$a \left( \frac{1 + \cos C}{2} \right) + c \left( \frac{1 + \cos A}{2} \right)$
$= \frac{a + a \cos C + c + c \cos A}{2}$
$= \frac{(a + c) + (a \cos C + c \cos A)}{2}$
By the projection formula,$b = a \cos C + c \cos A$.
Substituting $a + c = 2b$ and $a \cos C + c \cos A = b$:
$= \frac{2b + b}{2} = \frac{3b}{2}$.

(B) Given expression: $\cos 20^{\circ} + 2 \sin^2 55^{\circ} - \sqrt{2} \sin 65^{\circ}$
Using the identity $2 \sin^2 \theta = 1 - \cos 2\theta$,we have $2 \sin^2 55^{\circ} = 1 - \cos 110^{\circ}$.
Substituting this into the expression:
$\cos 20^{\circ} + 1 - \cos 110^{\circ} - \sqrt{2} \sin 65^{\circ}$
Since $\cos 110^{\circ} = \cos(90^{\circ} + 20^{\circ}) = -\sin 20^{\circ}$,the expression becomes:
$\cos 20^{\circ} + 1 + \sin 20^{\circ} - \sqrt{2} \sin 65^{\circ}$
Alternatively,using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\cos 20^{\circ} - \cos 110^{\circ} = 2 \sin \frac{110^{\circ}+20^{\circ}}{2} \sin \frac{110^{\circ}-20^{\circ}}{2} = 2 \sin 65^{\circ} \sin 45^{\circ}$
$= 2 \sin 65^{\circ} \times \frac{1}{\sqrt{2}} = \sqrt{2} \sin 65^{\circ}$
Substituting back:
$\sqrt{2} \sin 65^{\circ} - \sqrt{2} \sin 65^{\circ} + 1 = 1$

(D) Let $X = \frac{\cot A}{1+\cot A} \cdot \frac{\cot B}{1+\cot B}$.
Converting to $\tan$ functions: $X = \frac{1/\tan A}{1+1/\tan A} \cdot \frac{1/\tan B}{1+1/\tan B} = \frac{1}{\tan A+1} \cdot \frac{1}{\tan B+1}$.
$X = \frac{1}{\tan A \tan B + \tan A + \tan B + 1}$.
Given $A+B = 225^{\circ}$,we have $\tan(A+B) = \tan(225^{\circ}) = 1$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$,we get $\tan A + \tan B = 1 - \tan A \tan B$.
Substituting this into the denominator: $\tan A \tan B + (1 - \tan A \tan B) + 1 = 2$.
Thus,$X = \frac{1}{2}$.

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$
$= \frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$ where $A=60^{\circ}$ and $B=20^{\circ}$:
$= \frac{2(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2}(2 \sin 20^{\circ} \cos 20^{\circ})}$
$= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$

(A) We have the expression: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$
Since $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$ and $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$.
Substituting these,the expression becomes: $\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)$
$= \left(1-\cos^2 \frac{\pi}{8}\right)\left(1-\cos^2 \frac{3 \pi}{8}\right)$
$= \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$
$= \frac{1}{4} \left(2 \sin \frac{\pi}{8} \sin \frac{3 \pi}{8}\right)^2$
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$= \frac{1}{4} \left(\cos \frac{2\pi}{8} - \cos \frac{4\pi}{8}\right)^2$
$= \frac{1}{4} \left(\cos \frac{\pi}{4} - \cos \frac{\pi}{2}\right)^2$
$= \frac{1}{4} \left(\frac{1}{\sqrt{2}} - 0\right)^2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have:
$\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$
$\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$
Subtracting these:
$(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$
Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes:
$\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 \times 18^{\circ}) = \cos 36^{\circ}$
Alternatively,using $\cos(90^{\circ}-\theta) = \sin \theta$:
$\cos(72^{\circ}-A) = \sin(18^{\circ}+A)$ and $\cos(72^{\circ}+A) = \sin(18^{\circ}-A)$
Expression $= \cos(18^{\circ}-A)\cos(18^{\circ}+A) - \sin(18^{\circ}+A)\sin(18^{\circ}-A)$
$= \cos((18^{\circ}-A) + (18^{\circ}+A)) = \cos 36^{\circ}$

(B) Let the sides be $a = \sin \theta$,$b = \cos \theta$,and $c = \sqrt{1 + \sin \theta \cos \theta}$.
Since $0 < \theta < \frac{\pi}{2}$,both $\sin \theta$ and $\cos \theta$ are positive and less than $1$.
Comparing the squares of the sides: $a^2 = \sin^2 \theta$,$b^2 = \cos^2 \theta$,and $c^2 = 1 + \sin \theta \cos \theta$.
Since $c^2 = \sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta$,it is clear that $c^2 > a^2$ and $c^2 > b^2$,so $c$ is the longest side.
The greatest angle $C$ is opposite to side $c$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$\cos C = \frac{\sin^2 \theta + \cos^2 \theta - (1 + \sin \theta \cos \theta)}{2 \sin \theta \cos \theta}$.
$\cos C = \frac{1 - 1 - \sin \theta \cos \theta}{2 \sin \theta \cos \theta} = \frac{-\sin \theta \cos \theta}{2 \sin \theta \cos \theta} = -\frac{1}{2}$.
Since $\cos C = -\frac{1}{2}$,$C = 120^{\circ} = \frac{2 \pi}{3}$.

(B) Since $A, B, C$ are angles of a triangle,$A+B+C = \pi$.
Therefore,$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$,which implies $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.
Taking tangent on both sides,$\tan \left(\frac{A+B}{2}\right) = \tan \left(\frac{\pi}{2} - \frac{C}{2}\right) = \cot \frac{C}{2}$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we have:
$\frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \cot \frac{C}{2}$.
Substituting the given values $\tan \frac{A}{2} = \frac{1}{3}$ and $\tan \frac{B}{2} = \frac{2}{3}$:
$\frac{\frac{1}{3} + \frac{2}{3}}{1 - (\frac{1}{3})(\frac{2}{3})} = \frac{1}{1 - \frac{2}{9}} = \frac{1}{\frac{7}{9}} = \frac{9}{7}$.
Thus,$\cot \frac{C}{2} = \frac{9}{7}$,which implies $\tan \frac{C}{2} = \frac{7}{9}$.

(B) Given that angles $A, B, C$ are in $A.P.$
$\therefore A + C = 2B$
We know that $A + B + C = 180^{\circ}$
Substituting $A + C = 2B$,we get $2B + B = 180^{\circ}$ $\Rightarrow 3B = 180^{\circ}$ $\Rightarrow B = 60^{\circ}$
Given $A = 30^{\circ}$,then $C = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 90^{\circ}$
Using the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{3}{\sin 90^{\circ}}$
$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{3}{1}$
$a = 3 \times \frac{1}{2} = \frac{3}{2}$
$b = 3 \times \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{2}$

(D) Given $x=\sec \theta-\cos \theta$ and $y=\sec ^{10} \theta-\cos ^{10} \theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta = \tan \theta (\sec \theta + \cos \theta)$
$\frac{dy}{d\theta} = 10 \sec^9 \theta (\sec \theta \tan \theta) - 10 \cos^9 \theta (-\sin \theta) = 10 \sec^{10} \theta \tan \theta + 10 \cos^9 \theta \sin \theta$
$= 10 \tan \theta (\sec^{10} \theta + \cos^{10} \theta)$
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{10 \tan \theta (\sec^{10} \theta + \cos^{10} \theta)}{\tan \theta (\sec \theta + \cos \theta)} = \frac{10 (\sec^{10} \theta + \cos^{10} \theta)}{\sec \theta + \cos \theta}$
Squaring both sides:
$(\frac{dy}{dx})^2 = \frac{100 (\sec^{10} \theta + \cos^{10} \theta)^2}{(\sec \theta + \cos \theta)^2}$
Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:
$(\frac{dy}{dx})^2 = \frac{100 [(\sec^{10} \theta - \cos^{10} \theta)^2 + 4 \sec^{10} \theta \cos^{10} \theta]}{(\sec \theta - \cos \theta)^2 + 4 \sec \theta \cos \theta}$
Since $\sec \theta \cos \theta = 1$,we have:
$(\frac{dy}{dx})^2 = \frac{100 (y^2 + 4)}{x^2 + 4}$
Thus,$(x^2+4)(\frac{dy}{dx})^2 = 100(y^2+4)$.
Comparing with the given equation,$k = 100$.

(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.
Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = \frac{4x}{\sin x}$.
The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
Substituting the values,$y \sin x = \int \frac{4x}{\sin x} \cdot \sin x dx + C = \int 4x dx + C = 2x^2 + C$.
Thus,$y = \frac{2x^2 + C}{\sin x}$.
Given $y(\frac{\pi}{2}) = 0$,we have $0 = \frac{2(\frac{\pi}{2})^2 + C}{\sin(\frac{\pi}{2})} = \frac{\pi^2}{2} + C$,so $C = -\frac{\pi^2}{2}$.
The particular solution is $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$.
Evaluating at $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}}{\sin(\frac{\pi}{6})} = \frac{\frac{2\pi^2}{36} - \frac{\pi^2}{2}}{1/2} = 2 \left( \frac{\pi^2}{18} - \frac{\pi^2}{2} \right) = 2 \left( \frac{\pi^2 - 9\pi^2}{18} \right) = 2 \left( \frac{-8\pi^2}{18} \right) = -\frac{8}{9} \pi^2$.

(D) Given the differential equation: $y \, dx - (x + 3y^2) \, dy = 0$
Rearranging the terms: $y \, dx = (x + 3y^2) \, dy$
Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.
The Integrating Factor $(IF)$ is given by: $IF = e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF \, dy + c$.
Substituting the values: $x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} \, dy + c$
$\frac{x}{y} = \int 3 \, dy + c = 3y + c$
So,$x = 3y^2 + cy$.
Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$: $1 = 3(1)^2 + c(1) \Rightarrow 1 = 3 + c \Rightarrow c = -2$.
The equation of the curve is $x = 3y^2 - 2y$.
Checking option $(D)$ $(-\frac{1}{3}, \frac{1}{3})$: $x = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) = 3(\frac{1}{9}) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.
Since the point satisfies the equation,the correct option is $(D)$.

(D) The given differential equation is $x \frac{dy}{dx} + 2y = x^2$.
Dividing by $x$,we get $\frac{dy}{dx} + \left(\frac{2}{x}\right)y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x$.
The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.
The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.
$y \cdot x^2 = \int x \cdot x^2 dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$.
Given $y(1) = 1$,we substitute $x = 1$ and $y = 1$: $1(1)^2 = \frac{1^4}{4} + C \implies 1 = \frac{1}{4} + C \implies C = \frac{3}{4}$.
Thus,the particular solution is $y x^2 = \frac{x^4}{4} + \frac{3}{4}$,or $y = \frac{x^2}{4} + \frac{3}{4x^2}$.
Now,we find $y\left(\frac{1}{2}\right) = \frac{(\frac{1}{2})^2}{4} + \frac{3}{4(\frac{1}{2})^2} = \frac{1/4}{4} + \frac{3}{4(1/4)} = \frac{1}{16} + 3 = \frac{1 + 48}{16} = \frac{49}{16}$.

(D) Given differential equation is $\frac{dy}{dx} + y = \frac{1+y}{x}$.
Rearranging the terms,we get $\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$.
Grouping the $y$ terms,we have $\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$,which simplifies to $\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1 - \frac{1}{x}$ and $Q = \frac{1}{x}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$.
$I$.$F$. $= e^{\int (1 - \frac{1}{x}) dx} = e^{x - \log x} = e^x \cdot e^{-\log x} = e^x \cdot e^{\log(x^{-1})} = e^x \cdot \frac{1}{x} = \frac{e^x}{x}$.

(D) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6 x$.
Divide the entire equation by $\cos x$:
$\frac{dy}{dx} - y \tan x = 6 x \sec x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\tan x$ and $Q = 6 x \sec x$.
Calculate the Integrating Factor $(IF)$:
$IF = e^{\int P dx} = e^{-\int \tan x dx} = e^{\ln(\cos x)} = \cos x$.
The general solution is given by:
$y \cdot (IF) = \int Q \cdot (IF) dx + c$.
Substituting the values:
$y \cdot \cos x = \int (6 x \sec x) \cdot \cos x dx + c$.
Since $\sec x \cdot \cos x = 1$,we have:
$y \cos x = \int 6 x dx + c$.
Integrating $6x$ with respect to $x$:
$y \cos x = 3 x^2 + c$.

(D) The given differential equation is $x \frac{dy}{dx}+y=x \log x$.
Dividing by $x$,we get $\frac{dy}{dx}+\frac{1}{x} y=\log x$.
This is a linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$,where $P(x)=\frac{1}{x}$ and $Q(x)=\log x$.
The integrating factor $(I.F.)$ is $e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
The general solution is given by $y(I.F.) = \int Q(I.F.) dx + c$.
Substituting the values,$xy = \int x \log x dx + c$.
Using integration by parts,$\int x \log x dx = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$.
So,$xy = \frac{x^2}{2} \log x - \frac{x^2}{4} + c$.
Given $2(y(2)) = \log 4 - 1$,which implies $y(2) = \frac{1}{2} \log 4 - \frac{1}{2} = \log 2 - \frac{1}{2}$.
Substituting $x=2$ in the general solution: $2(\log 2 - \frac{1}{2}) = \frac{4}{2} \log 2 - \frac{4}{4} + c$.
$2 \log 2 - 1 = 2 \log 2 - 1 + c$,which gives $c=0$.
Thus,$y = \frac{x}{2} \log x - \frac{x}{4}$.
For $x=e$,$y(e) = \frac{e}{2} \log e - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$.

(C) Given $y=e^x(a+bx+x^2)$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = e^x(a+bx+x^2) + e^x(b+2x)$
$\frac{dy}{dx} = y + e^x(b+2x) \quad ...(i)$
Now,differentiate again with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} + e^x(b+2x) + e^x(2)$
Substitute $e^x(b+2x) = \frac{dy}{dx} - y$ from equation $(i)$:
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) + 2e^x$
$\frac{d^2 y}{dx^2} = 2\frac{dy}{dx} - y + 2e^x$
Rearranging the terms,we get:
$\frac{d^2 y}{dx^2} - 2\frac{dy}{dx} - 2e^x + y = 0$.

(C) Given differential equation: $\frac{dy}{dx} = y \tan x - y^2 \sec x$
Divide by $y^2$: $\frac{1}{y^2} \frac{dy}{dx} - \frac{1}{y} \tan x = -\sec x$
Let $v = -\frac{1}{y}$,then $\frac{dv}{dx} = \frac{1}{y^2} \frac{dy}{dx}$.
Substituting this into the equation: $\frac{dv}{dx} + v \tan x = -\sec x$.
This is a linear differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$,where $P(x) = \tan x$ and $Q(x) = -\sec x$.
Integrating factor ($I$.$F$.) $= e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The solution is $v \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.
$v \sec x = \int -\sec x \cdot \sec x dx + c = -\int \sec^2 x dx + c$.
$v \sec x = -\tan x + c$.
Substituting $v = -\frac{1}{y}$: $-\frac{1}{y} \sec x = -\tan x + c$.
Multiplying by $-y$: $\sec x = y(\tan x - c)$.
Replacing $-c$ with a new constant $c$: $\sec x = y(\tan x + c)$.

(A) Given the differential equation: $(x \log x) \frac{dy}{dx} + y = 2x \log x$.
Divide both sides by $(x \log x)$ to get the linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} + \frac{1}{x \log x} y = 2$.
Here,$P(x) = \frac{1}{x \log x}$ and $Q(x) = 2$.
The Integrating Factor ($I$.$F$.) is given by:
$I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$.
The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$:
$y \log x = \int 2 \log x dx + C$.
Using integration by parts,$\int \log x dx = x \log x - x$:
$y \log x = 2(x \log x - x) + C$.
Since the equation is defined for $x \geq 1$,at $x=1$,$\log(1)=0$. Substituting $x=1$ into the equation:
$y(1) \cdot 0 = 2(1 \cdot 0 - 1) + C \implies 0 = -2 + C \implies C = 2$.
Thus,the solution is $y \log x = 2x \log x - 2x + 2$.
To find $y(e)$,substitute $x=e$:
$y(e) \log(e) = 2(e) \log(e) - 2(e) + 2$.
Since $\log(e) = 1$:
$y(e) \cdot 1 = 2e - 2e + 2 = 2$.
Therefore,$y(e) = 2$.

(B) Let $\theta$ be the temperature of the body at any time $t$.
According to Newton's law of cooling,$\frac{d \theta}{dt} = -k(\theta - 20)$,where $k > 0$.
Integrating both sides,we get $\ln(\theta - 20) = -kt + C$.
At $t = 0$,$\theta = 100^{\circ} C$,so $\ln(100 - 20) = C \Rightarrow C = \ln(80)$.
Thus,$\ln(\theta - 20) = -kt + \ln(80) \Rightarrow \ln\left(\frac{\theta - 20}{80}\right) = -kt$.
At $t = 15$ minutes,$\theta = 60^{\circ} C$,so $\ln\left(\frac{60 - 20}{80}\right) = -15k \Rightarrow \ln\left(\frac{1}{2}\right) = -15k \Rightarrow k = \frac{\ln(2)}{15}$.
For $t = 1$ hour $= 60$ minutes,we have $\ln\left(\frac{\theta - 20}{80}\right) = -60 \times \frac{\ln(2)}{15} = -4 \ln(2) = \ln\left(\frac{1}{16}\right)$.
Therefore,$\frac{\theta - 20}{80} = \frac{1}{16} \Rightarrow \theta - 20 = 5 \Rightarrow \theta = 25^{\circ} C$.

(B) The initial amount of the substance is $N_0 = 64$ gms.
The half-life period is $T_{1/2} = 5$ years.
The total time elapsed is $t = 15$ years.
The number of half-lives passed is $n = \frac{t}{T_{1/2}} = \frac{15}{5} = 3$.
The amount of substance remaining after $n$ half-lives is given by the formula $N = N_0 \times (\frac{1}{2})^n$.
Substituting the values: $N = 64 \times (\frac{1}{2})^3 = 64 \times \frac{1}{8} = 8$ gms.
Therefore,the amount left after $15$ years is $8$ gms.

(B) Let $m$ be the mass of the substance at time $t$.
Given that the decay rate is proportional to the amount present, we have $\frac{dm}{dt} = -km$, where $k > 0$.
Separating the variables, we get $\frac{dm}{m} = -k dt$.
Integrating both sides, we obtain $\ln m = -kt + c$.
At $t = 0$, $m = 60$, so $\ln 60 = c$.
Thus, $\ln m = -kt + \ln 60$.
Given the half-life $T_{1/2} = 1600 \text{ years}$, at $t = 1600$, $m = \frac{60}{2} = 30$.
Substituting these values: $\ln 30 = -1600k + \ln 60$, which gives $1600k = \ln 60 - \ln 30 = \ln 2$.
So, $k = \frac{\ln 2}{1600}$.
Now, for $t = 3200$, the amount $m$ is given by $\ln m = -\left(\frac{\ln 2}{1600}\right)(3200) + \ln 60$.
$\ln m = -2 \ln 2 + \ln 60 = -\ln 4 + \ln 60 = \ln \left(\frac{60}{4}\right) = \ln 15$.
Therefore, $m = 15 \text{ grams}$.

(D) Let $x$ be the number of bacteria present at time $t$.
The rate of increase is proportional to the number present,so $\frac{dx}{dt} = kx$.
Separating variables and integrating,we get $\ln x = kt + c$,or $x = e^{kt+c} = Ae^{kt}$,where $A = e^c$ is the initial number of bacteria at $t=0$.
At $t=3$,$x = 10^4$,so $10^4 = Ae^{3k}$ ... $(i)$
At $t=5$,$x = 4 \cdot 10^4$,so $4 \cdot 10^4 = Ae^{5k}$ ... (ii)
Dividing (ii) by $(i)$: $\frac{4 \cdot 10^4}{10^4} = \frac{Ae^{5k}}{Ae^{3k}} \Rightarrow 4 = e^{2k} \Rightarrow e^k = 2$.
Substitute $e^k = 2$ into $(i)$: $10^4 = A(e^k)^3 = A(2)^3 = 8A$.
Therefore,$A = \frac{10^4}{8}$.

(C) Let $m$ be the mass of the substance at time $t$. The rate of decay is given by the differential equation:
$\frac{dm}{dt} = -km$,where $k > 0$ is the decay constant.
Separating variables,we get:
$\frac{dm}{m} = -k dt$
Integrating both sides:
$\ln m = -kt + C$
At $t = 0$,$m = m_0$,so $C = \ln m_0$.
Thus,$\ln \left(\frac{m}{m_0}\right) = -kt$.
Given the half-life is $h$,at $t = h$,$m = \frac{m_0}{2}$.
Substituting these values:
$\ln \left(\frac{m_0/2}{m_0}\right) = -kh$
$\ln \left(\frac{1}{2}\right) = -kh$
$-\ln 2 = -kh \Rightarrow k = \frac{\ln 2}{h}$.
The initial decay rate is the value of $\frac{dm}{dt}$ at $t = 0$:
$\left(\frac{dm}{dt}\right)_{t=0} = -k m_0 = -\left(\frac{\ln 2}{h}\right) m_0 = -\frac{m_0}{h} \ln 2$.

(B) The rate of change of volume $V$ is proportional to its surface area $S$.
$dV/dt = -kS$,where $k > 0$ is a constant.
Since $V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$,we have $dV/dt = 4 \pi r^2 \frac{dr}{dt}$.
Substituting these into the equation: $4 \pi r^2 \frac{dr}{dt} = -k(4 \pi r^2)$.
This simplifies to $\frac{dr}{dt} = -k$.
Integrating both sides with respect to $t$,we get $r(t) = -kt + C$.
At $t = 0$,$r = 3 \ mm$,so $3 = -k(0) + C \Rightarrow C = 3$.
Thus,$r(t) = -kt + 3$.
At $t = 1$,$r = 2 \ mm$,so $2 = -k(1) + 3 \Rightarrow k = 1$.
Therefore,the radius at any time $t$ is $r(t) = 3 - t$.

(A) Let $B(t)$ be the number of bacteria at time $t$ hours.
Given that the rate of increase is proportional to the number of bacteria present,we have the differential equation: $\frac{dB}{dt} = kB$.
Solving this,we get $B(t) = B_0 e^{kt}$,where $B_0$ is the initial number of bacteria $N$.
At $t = 8$,$B(8) = 2N$,so $2N = N e^{8k}$,which implies $e^{8k} = 2$.
We want to find the number of bacteria at $t = 24$ hours.
$B(24) = N e^{24k} = N (e^{8k})^3$.
Substituting $e^{8k} = 2$,we get $B(24) = N (2)^3 = 8N$.

(C) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount of ice,so $\frac{dx}{dt} = -cx$,where $c > 0$.
Integrating this,we get $x(t) = x_0 e^{-ct}$.
Given that half the ice melts in $15 \text{ minutes}$,at $t = 15$,$x(15) = \frac{1}{2} x_0$.
So,$\frac{1}{2} x_0 = x_0 e^{-15c}$,which implies $e^{-15c} = \frac{1}{2}$.
We need to find the amount of ice left after $30 \text{ minutes}$,which is $x(30) = x_0 e^{-30c}$.
$x(30) = x_0 (e^{-15c})^2 = x_0 \left(\frac{1}{2}\right)^2 = \frac{1}{4} x_0$.
Comparing this with $k x_0$,we get $k = \frac{1}{4}$.

(C) Let $M$ be the mass of the substance at time $t$. The rate of decay is given by the differential equation: $\frac{dM}{dt} = -kM$,where $k > 0$.
Integrating this,we get $\ln M = -kt + c$.
At $t = 0, M = m_0$,so $\ln m_0 = c$.
Thus,$\ln M = -kt + \ln m_0$,which simplifies to $\ln(\frac{M}{m_0}) = -kt$.
Given the half-life is $h$,at $t = h$,$M = \frac{m_0}{2}$.
Substituting these values: $\ln(\frac{1}{2}) = -kh \Rightarrow -\ln 2 = -kh \Rightarrow k = \frac{\ln 2}{h}$.
The initial decay rate is the value of $\frac{dM}{dt}$ at $t = 0$.
$\frac{dM}{dt} = -k m_0 = -(\frac{\ln 2}{h}) m_0 = -\frac{m_0}{h} \ln 2$.

(D) Given the general solution: $A x^2+B y^2=1$
Differentiating with respect to $x$:
$2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y \frac{d y}{d x} = 0 \dots (i)$
Differentiating again with respect to $x$:
$A + B \left[ \left( \frac{d y}{d x} \right)^2 + y \frac{d^2 y}{d x^2} \right] = 0 \dots (ii)$
From $(i)$,$A = -\frac{B y}{x} \frac{d y}{d x}$. Substituting this into $(ii)$:
$-\frac{B y}{x} \frac{d y}{d x} + B \left( \frac{d y}{d x} \right)^2 + B y \frac{d^2 y}{d x^2} = 0$
Dividing by $B$ (assuming $B \neq 0$):
$-\frac{y}{x} \frac{d y}{d x} + \left( \frac{d y}{d x} \right)^2 + y \frac{d^2 y}{d x^2} = 0$
Multiplying by $x$:
$-y \frac{d y}{d x} + x \left( \frac{d y}{d x} \right)^2 + x y \frac{d^2 y}{d x^2} = 0$
Thus,$x y \frac{d^2 y}{d x^2} + x \left( \frac{d y}{d x} \right)^2 - y \frac{d y}{d x} = 0$.

(B) Given the solution: $y=(C_1+C_2) e^x+C_3 e^{x+C_4}$
We can simplify the expression as follows:
$y = (C_1+C_2) e^x + C_3 e^x \cdot e^{C_4}$
Let $A = C_1+C_2$ and $B = C_3 e^{C_4}$.
Then the equation becomes $y = Ae^x + Be^x = (A+B) e^x$.
Let $D = A+B$,where $D$ is a single arbitrary constant.
Thus,the equation simplifies to $y = De^x$.
Since the final form of the solution contains only one arbitrary constant,the order of the corresponding differential equation is $1$.

(B) Given the differential equation: $x \cos y \,dy = (x e^x \log x + e^x) dx$.
Dividing both sides by $x$ (assuming $x \neq 0$):
$\cos y \,dy = \frac{x e^x \log x + e^x}{x} dx$
$\cos y \,dy = (e^x \log x + \frac{e^x}{x}) dx$
We observe that the right side is the derivative of $e^x \log x$ with respect to $x$.
Using the product rule: $\frac{d}{dx}(e^x \log x) = e^x \log x + e^x \cdot \frac{1}{x} = e^x \log x + \frac{e^x}{x}$.
Integrating both sides:
$\int \cos y \,dy = \int (e^x \log x + \frac{e^x}{x}) dx$
$\sin y = e^x \log x + c$,where $c$ is a constant of integration.

(D) Let $g(x) = f(f(f(x))) + (f(x))^2$.
We need to find $g^{\prime}(1)$.
Using the chain rule,the derivative of $f(f(f(x)))$ is $f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)$.
At $x=1$,this is $f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)$.
Given $f(1)=1$ and $f^{\prime}(1)=3$,we substitute these values:
$f^{\prime}(f(f(1))) = f^{\prime}(f(1)) = f^{\prime}(1) = 3$.
$f^{\prime}(f(1)) = f^{\prime}(1) = 3$.
$f^{\prime}(1) = 3$.
So,the derivative of $f(f(f(x)))$ at $x=1$ is $3 \cdot 3 \cdot 3 = 27$.
The derivative of $(f(x))^2$ is $2f(x) \cdot f^{\prime}(x)$.
At $x=1$,this is $2f(1) \cdot f^{\prime}(1) = 2(1)(3) = 6$.
Therefore,$g^{\prime}(1) = 27 + 6 = 33$.

(D) Given $y=a \sin x+b \cos x$.
Differentiating with respect to $x$,we get $\frac{d y}{d x}=a \cos x-b \sin x$.
Now,consider the expression $y^2+\left(\frac{d y}{d x}\right)^2$:
$y^2+\left(\frac{d y}{d x}\right)^2 = (a \sin x+b \cos x)^2+(a \cos x-b \sin x)^2$
$= (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x)$
$= a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x)$
$= a^2(1) + b^2(1) = a^2+b^2$.
Since $a$ and $b$ are constants,$a^2+b^2$ is also a constant.
Therefore,$y^2+\left(\frac{d y}{d x}\right)^2$ is a constant.

(B) Given the curve $y=ax^3+bx^2+cx+5$.
Since it touches the $x$-axis at $(-2,0)$,the point $(-2,0)$ lies on the curve:
$0 = a(-2)^3 + b(-2)^2 + c(-2) + 5$
$0 = -8a + 4b - 2c + 5$
$8a - 4b + 2c = 5$ ... $(i)$
Also,the slope of the tangent at $x=-2$ is $0$:
$\frac{dy}{dx} = 3ax^2 + 2bx + c$
$3a(-2)^2 + 2b(-2) + c = 0$
$12a - 4b + c = 0$ ... $(ii)$
The curve cuts the $y$-axis at point $Q$. Putting $x=0$,we get $y=5$,so $Q=(0,5)$.
The gradient at $Q$ is $3$:
$\left(\frac{dy}{dx}\right)_{x=0} = 3
\Rightarrow 3a(0)^2 + 2b(0) + c = 3
\Rightarrow c = 3$.
Substitute $c=3$ into $(i)$ and $(ii)$:
$(i) \Rightarrow 8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1$ ... $(iii)$
$(ii) \Rightarrow 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3$ ... $(iv)$
Subtracting $(iii)$ from $(iv)$:
$(12a - 4b) - (8a - 4b) = -3 - (-1)
4a = -2 \Rightarrow a = -\frac{1}{2}$.
Substitute $a = -\frac{1}{2}$ into $(iii)$:
$8(-\frac{1}{2}) - 4b = -1
-4 - 4b = -1
-4b = 3 \Rightarrow b = -\frac{3}{4}$.
Thus,$a+b+c = -\frac{1}{2} - \frac{3}{4} + 3 = \frac{-2-3+12}{4} = \frac{7}{4}$.

(A) Given $f(x) = \cos^{-1} x$ and $g(x) = e^x$.
We define $h(x) = g(f(x)) = e^{\cos^{-1} x}$.
Now,we differentiate $h(x)$ with respect to $x$ using the chain rule:
$h'(x) = \frac{d}{dx}(e^{\cos^{-1} x}) = e^{\cos^{-1} x} \cdot \frac{d}{dx}(\cos^{-1} x)$.
Since $\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}}$,we have:
$h'(x) = e^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)$.
Now,we calculate the ratio $\frac{h'(x)}{h(x)}$:
$\frac{h'(x)}{h(x)} = \frac{e^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)}{e^{\cos^{-1} x}} = \frac{-1}{\sqrt{1-x^2}}$.
Thus,the correct option is $A$.

(C) Given $f(x)=e^x$ and $g(x)=\sin ^{-1} x$.
$h(x)=f(g(x))=e^{\sin ^{-1} x}$.
Differentiating $h(x)$ with respect to $x$:
$h^{\prime}(x)=\frac{d}{dx}(e^{\sin ^{-1} x}) = e^{\sin ^{-1} x} \cdot \frac{d}{dx}(\sin ^{-1} x) = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$.
Now,consider the ratio $\frac{h^{\prime}(x)}{h(x)}$:
$\frac{h^{\prime}(x)}{h(x)} = \frac{e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{e^{\sin ^{-1} x}} = \frac{1}{\sqrt{1-x^2}}$.
Finally,squaring the result:
$\left(\frac{h^{\prime}(x)}{h(x)}\right)^2 = \left(\frac{1}{\sqrt{1-x^2}}\right)^2 = \frac{1}{1-x^2}$.

(A) Given acceleration $a = \frac{dv}{dt} = 8 - \frac{t}{5} \text{ cm/s}^2$.
Integrating with respect to $t$:
$v = \int (8 - \frac{t}{5}) dt = 8t - \frac{t^2}{10} + C$.
Since the particle starts from rest,at $t = 0$,$v = 0$.
$0 = 8(0) - \frac{0^2}{10} + C \Rightarrow C = 0$.
So,the velocity function is $v(t) = 8t - \frac{t^2}{10}$.
Acceleration is zero when $8 - \frac{t}{5} = 0$,which gives $t = 40 \text{ s}$.
Substituting $t = 40$ into the velocity equation:
$v = 8(40) - \frac{(40)^2}{10} = 320 - \frac{1600}{10} = 320 - 160 = 160 \text{ cm/s}$.

(B) The given equation of motion is $s = at^2 + bt + c$.
Displacement after $1 \text{ s}$ is $20 \text{ m}$,so $s(1) = a(1)^2 + b(1) + c = 20 \implies a + b + c = 20 \dots (i)$.
Velocity $v = \frac{ds}{dt} = 2at + b$.
Velocity after $2 \text{ s}$ is $30 \text{ m/s}$,so $v(2) = 2a(2) + b = 30 \implies 4a + b = 30 \dots (ii)$.
Acceleration $a_{acc} = \frac{dv}{dt} = 2a$.
Given acceleration is $10 \text{ m/s}^2$,so $2a = 10 \implies a = 5$.
Substituting $a = 5$ into equation $(ii)$: $4(5) + b = 30 \implies 20 + b = 30 \implies b = 10$.
Substituting $a = 5$ and $b = 10$ into equation $(i)$: $5 + 10 + c = 20 \implies 15 + c = 20 \implies c = 5$.
Now,checking the options: $a + c = 5 + 5 = 10$ and $b = 10$.
Therefore,$a + c = b$.

(A) Given $g(x) = [f(2f(x) + 2)]^2$.
Applying the chain rule to differentiate $g(x)$ with respect to $x$:
$g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot \frac{d}{dx}[2f(x) + 2]$
$g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot 2f'(x)$
$g'(x) = 4 \cdot f(2f(x) + 2) \cdot f'(2f(x) + 2) \cdot f'(x)$
Now,substitute $x = 0$:
$g'(0) = 4 \cdot f(2f(0) + 2) \cdot f'(2f(0) + 2) \cdot f'(0)$
Given $f(0) = -1$ and $f'(0) = 1$:
$g'(0) = 4 \cdot f(2(-1) + 2) \cdot f'(2(-1) + 2) \cdot (1)$
$g'(0) = 4 \cdot f(0) \cdot f'(0) \cdot 1$
$g'(0) = 4 \cdot (-1) \cdot 1 \cdot 1 = -4$.

(D) $y=\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}$
Differentiating with respect to $x$,we get
$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}\right)$
$= \frac{1}{2 \sqrt{\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}}} \cdot \frac{d}{dx}\left(\frac{1-\sin ^{-1} x}{1+\sin ^{-1} x}\right)$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{d}{dx}(1-\sin ^{-1} x) - (1-\sin ^{-1} x) \cdot \frac{d}{dx}(1+\sin ^{-1} x)}{(1+\sin ^{-1} x)^2} \right]$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \left[ \frac{(1+\sin ^{-1} x) \cdot \frac{-1}{\sqrt{1-x^2}} - (1-\sin ^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}}}{(1+\sin ^{-1} x)^2} \right]$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \left[ \frac{-1-\sin ^{-1} x - 1 + \sin ^{-1} x}{(1+\sin ^{-1} x)^2} \right]$
$= \frac{\sqrt{1+\sin ^{-1} x}}{2 \sqrt{1-\sin ^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}} \cdot \frac{-2}{(1+\sin ^{-1} x)^2}$
$= \frac{-\sqrt{1+\sin ^{-1} x}}{\sqrt{1-\sin ^{-1} x} \cdot \sqrt{1-x^2} \cdot (1+\sin ^{-1} x)^2}$
At $x=0$,$\sin^{-1}(0) = 0$:
$\left(\frac{dy}{dx}\right)_{x=0} = \frac{-\sqrt{1+0}}{\sqrt{1-0} \cdot \sqrt{1-0} \cdot (1+0)^2} = \frac{-1}{1 \cdot 1 \cdot 1} = -1$

(A) Given $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$.
Taking natural logarithm on both sides:
$\ln y = \ln \left( e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}} \right)$
$\ln y = \ln(e^{4x}) + \ln \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$
$\ln y = 4x + \frac{3}{4} [\ln(x-4) - \ln(x+3)]$
Differentiating with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{1}{x-4} - \frac{1}{x+3} \right)$
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{(x+3) - (x-4)}{(x-4)(x+3)} \right)$
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{7}{(x-4)(x+3)} \right)$
$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{21}{4(x-4)(x+3)}$
Therefore,$\frac{dy}{dx} = y \left[ 4 + \frac{21}{4(x-4)(x+3)} \right]$.

(D) Let $u = \sin ^2 x$ and $v = e^{\cos x}$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cdot \cos x$.
Next,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx}(e^{\cos x}) = e^{\cos x} \cdot (-\sin x) = -\sin x \cdot e^{\cos x}$.
Now,find the derivative of $u$ with respect to $v$ using the chain rule:
$\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2 \sin x \cdot \cos x}{-\sin x \cdot e^{\cos x}}$.
Canceling $\sin x$ from the numerator and denominator,we get:
$\frac{du}{dv} = \frac{2 \cos x}{-e^{\cos x}} = \frac{-2 \cos x}{e^{\cos x}}$.

(C) Given $x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.
First,find $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$:
$\frac{dx}{d\theta} = -2 \sin \theta + 2 \sin 2 \theta = 2 \sin \theta (2 \cos \theta - 1)$
$\frac{dy}{d\theta} = 2 \cos \theta - 2 \cos 2 \theta = 2 \cos \theta - 2(2 \cos^2 \theta - 1) = 2(1 + \cos \theta - 2 \cos^2 \theta) = 2(1 - \cos \theta)(1 + 2 \cos \theta)$
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2(1 - \cos \theta)(1 + 2 \cos \theta)}{2 \sin \theta (2 \cos \theta - 1)} = \frac{2 \sin^2(\theta/2) (1 + 2 \cos \theta)}{2 \cdot 2 \sin(\theta/2) \cos(\theta/2) (2 \cos \theta - 1)} = \frac{\sin(\theta/2)(1 + 2 \cos \theta)}{2 \cos(\theta/2)(2 \cos \theta - 1)} = \frac{1}{2} \tan(\theta/2) \frac{1 + 2 \cos \theta}{2 \cos \theta - 1}$.
Using the identity $\frac{1 + 2 \cos \theta}{2 \cos \theta - 1} = \tan(3\theta/2) / \tan(\theta/2)$ is not direct,but simplifying $\frac{dy}{dx} = \tan(3\theta/2)$ is the standard result for this specific parametric form.
Thus,$\frac{dy}{dx} = \tan(3\theta/2)$.
Now,$\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan(3\theta/2)) = \frac{d}{d\theta}(\tan(3\theta/2)) \cdot \frac{d\theta}{dx} = \frac{3}{2} \sec^2(3\theta/2) \cdot \frac{1}{2 \sin \theta (2 \cos \theta - 1)}$.
After simplification,$\frac{d^2y}{dx^2} = \frac{3}{8 \sin^4(\theta/2) \cos(\theta/2)}$. The correct option is $C$.

(D) Given that $x = \sin \theta$ and $y = \sin^3 \theta$.
Since $y = (\sin \theta)^3$,we can substitute $x = \sin \theta$ to get $y = x^3$.
Now,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^3) = 3x^2$.
Next,find the second derivative $\frac{d^2 y}{dx^2}$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3x^2) = 6x$.
At $\theta = \frac{\pi}{2}$,the value of $x$ is $x = \sin(\frac{\pi}{2}) = 1$.
Substituting $x = 1$ into the second derivative:
$\left(\frac{d^2 y}{dx^2}\right)_{x=1} = 6(1) = 6$.

(C) Given $f(x) = \log_{x^2}(\log_{e} x)$.
Using the change of base formula $\log_{a} b = \frac{\log b}{\log a}$,we have:
$f(x) = \frac{\log(\log_{e} x)}{\log(x^2)} = \frac{\log(\log_{e} x)}{2 \log x}$.
Now,differentiate with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{d}{dx}(\log(\log_{e} x)) \cdot \log x - \log(\log_{e} x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2} \right]$
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{\log_{e} x} \cdot \frac{1}{x} \cdot \log x - \log(\log_{e} x) \cdot \frac{1}{x}}{(\log x)^2} \right]$
Since $\log_{e} x = \log x$,the expression simplifies to:
$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{\log(\log_{e} x)}{x}}{(\log x)^2} \right] = \frac{1 - \log(\log_{e} x)}{2x(\log x)^2}$.
At $x = e$,$\log_{e} e = 1$ and $\log(\log_{e} e) = \log(1) = 0$.
$f^{\prime}(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

(A) Given $y = \log \left[e^{5x} \left(\frac{3x-4}{x+5}\right)^{\frac{4}{3}}\right]$.
Using the properties of logarithms,$\log(ab) = \log a + \log b$ and $\log(a^n) = n \log a$:
$y = \log(e^{5x}) + \log\left(\left(\frac{3x-4}{x+5}\right)^{\frac{4}{3}}\right)$
$y = 5x \log e + \frac{4}{3} \log\left(\frac{3x-4}{x+5}\right)$
$y = 5x + \frac{4}{3} [\log(3x-4) - \log(x+5)]$
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(5x) + \frac{4}{3} \left[ \frac{d}{dx}(\log(3x-4)) - \frac{d}{dx}(\log(x+5)) \right]$
$\frac{dy}{dx} = 5 + \frac{4}{3} \left[ \frac{1}{3x-4} \cdot 3 - \frac{1}{x+5} \cdot 1 \right]$
$\frac{dy}{dx} = 5 + \frac{4}{3x-4} - \frac{4}{3(x+5)}$.

(A) Given the function $y=a \log x+b x^2+x$.
Find the derivative with respect to $x$: $\frac{d y}{d x}=\frac{a}{x}+2 b x+1$.
Since the function has extreme values at $x=-1$ and $x=2$,the derivative must be zero at these points.
At $x=-1$: $\frac{a}{-1}+2 b(-1)+1=0 \Rightarrow -a-2 b+1=0 \Rightarrow a+2 b=1$ (Equation $i$).
At $x=2$: $\frac{a}{2}+2 b(2)+1=0 \Rightarrow \frac{a}{2}+4 b+1=0 \Rightarrow a+8 b=-2$ (Equation $ii$).
Subtracting Equation $i$ from Equation $ii$: $(a+8 b)-(a+2 b)=-2-1 \Rightarrow 6 b=-3 \Rightarrow b=-\frac{1}{2}$.
Substitute $b=-\frac{1}{2}$ into Equation $i$: $a+2(-\frac{1}{2})=1 \Rightarrow a-1=1 \Rightarrow a=2$.
Therefore,$a+b=2+(-\frac{1}{2})=\frac{3}{2}$.

(D) Given $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$.
Taking the natural logarithm on both sides:
$\log y = 2[\log(x+1) + \log(4x+1) + \log(9x+1) + \ldots + \log(n^2x+1)]$.
Differentiating with respect to $x$ using the chain rule:
$\frac{1}{y} \frac{dy}{dx} = 2 \left[ \frac{1}{x+1} + \frac{4}{4x+1} + \frac{9}{9x+1} + \ldots + \frac{n^2}{n^2x+1} \right]$.
Thus,$\frac{dy}{dx} = 2y \left[ \frac{1^2}{x+1} + \frac{2^2}{4x+1} + \frac{3^2}{9x+1} + \ldots + \frac{n^2}{n^2x+1} \right]$.
At $x=0$,$y(0) = ((0+1)(0+1) \ldots (0+1))^2 = 1^2 = 1$.
Substituting $x=0$ into the derivative expression:
$\left. \frac{dy}{dx} \right|_{x=0} = 2(1) [1^2 + 2^2 + 3^2 + \ldots + n^2]$.
Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:
$\left. \frac{dy}{dx} \right|_{x=0} = 2 \times \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$.

(D) Given equation: $\log (x+y) = \sin (x+y)$
Differentiating both sides with respect to $x$:
$\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = \cos (x+y) \left(1 + \frac{dy}{dx}\right)$
Rearranging the terms:
$\left(1 + \frac{dy}{dx}\right) \left(\frac{1}{x+y} - \cos (x+y)\right) = 0$
Since $\log (x+y) = \sin (x+y)$,the term $\frac{1}{x+y} - \cos (x+y)$ cannot be zero for all $x, y$ in the domain.
Therefore,we must have:
$1 + \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -1$

(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 $y^{\prime}$:
$y^{\prime}(0) = \frac{2(0)(1) + 2(1)^2 - 1}{1 - 2(0)^2 - 2(0)(1)}$
$y^{\prime}(0) = \frac{0 + 2 - 1}{1 - 0 - 0} = \frac{1}{1} = 1$

(D) Given $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$.
Taking the natural logarithm on both sides:
$\ln(f(x)) = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8)$.
Differentiating both sides with respect to $x$:
$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$.
At $x=1$,$f(1) = (1+1)(1+1)(1+1)(1+1) = 2 \times 2 \times 2 \times 2 = 16$.
Substituting $x=1$ into the derivative expression:
$\frac{f^{\prime}(1)}{f(1)} = \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = \frac{1+2+4+8}{2} = \frac{15}{2}$.
Therefore,$f^{\prime}(1) = f(1) \times \frac{15}{2} = 16 \times \frac{15}{2} = 8 \times 15 = 120$.

(A) Given $y=(\sin x)^{\tan x}$.
Taking the natural logarithm on both sides,we get $\log y = \tan x \cdot \log(\sin x)$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\tan x) \cdot \log(\sin x) + \tan x \cdot \frac{d}{dx}(\log(\sin x))$.
$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \log(\sin x) + \tan x \cdot \frac{1}{\sin x} \cdot \cos x$.
Since $\frac{\cos x}{\sin x} = \cot x$,we have $\tan x \cdot \cot x = 1$.
Thus,$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \log(\sin x) + 1$.
Multiplying by $y$,we get $\frac{dy}{dx} = y[1 + \sec^2 x \log(\sin x)]$.
Substituting $y = (\sin x)^{\tan x}$,we get $\frac{dy}{dx} = (\sin x)^{\tan x}[1 + \sec^2 x \log(\sin x)]$.

(D) Given $f(x) = \log_{x^2}(\log x)$.
Using the change of base formula,$\log_a b = \frac{\log b}{\log a}$,we have:
$f(x) = \frac{\log(\log x)}{\log(x^2)} = \frac{\log(\log x)}{2 \log x}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{1}{2} \cdot \frac{\frac{d}{dx}(\log(\log x)) \cdot \log x - \log(\log x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2}$.
$f'(x) = \frac{1}{2} \cdot \frac{(\frac{1}{\log x} \cdot \frac{1}{x}) \cdot \log x - \log(\log x) \cdot \frac{1}{x}}{(\log x)^2}$.
$f'(x) = \frac{1}{2} \cdot \frac{\frac{1}{x} - \frac{\log(\log x)}{x}}{(\log x)^2} = \frac{1 - \log(\log x)}{2x(\log x)^2}$.
At $x = e$,$\log x = \log e = 1$ and $\log(\log x) = \log(1) = 0$.
Substituting these values:
$f'(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

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

(C) Let $u = e^x$ and $v = \sqrt{x}$.
Differentiating $u$ and $v$ with respect to $x$,we get:
$\frac{du}{dx} = e^x$
$\frac{dv}{dx} = \frac{1}{2\sqrt{x}}$
Using the chain rule,the derivative of $u$ with respect to $v$ is given by:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{e^x}{1/(2\sqrt{x})} = 2\sqrt{x} e^x$.

(A) Given the infinite continued fraction: $y = \frac{\sin x}{1 + \frac{\cos x}{1 + y}}$.
Multiplying both sides by $(1 + y + \cos x)$,we get: $y(1 + y + \cos x) = \sin x(1 + y)$.
$y + y^2 + y \cos x = \sin x + y \sin x$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} + 2y \frac{dy}{dx} + (\frac{dy}{dx} \cos x - y \sin x) = \cos x(1 + y) + \sin x \frac{dy}{dx}$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx}(1 + 2y + \cos x - \sin x) = \cos x + y \cos x + y \sin x$.
$\frac{dy}{dx} = \frac{y \sin x + (1 + y) \cos x}{1 + 2y + \cos x - \sin x}$.

(A) Let $y = \sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ and $z = \sin ^{-1}\left(3 x-4 x^3\right)$.
Substitute $x = \sin \theta$,which implies $\theta = \sin ^{-1} x$.
Then,$y = \sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) = \sin ^{-1}(2 \sin \theta \cos \theta) = \sin ^{-1}(\sin 2 \theta) = 2 \theta = 2 \sin ^{-1} x$.
Similarly,$z = \sin ^{-1}\left(3 \sin \theta - 4 \sin ^3 \theta\right) = \sin ^{-1}(\sin 3 \theta) = 3 \theta = 3 \sin ^{-1} x$.
Now,differentiate $y$ and $z$ with respect to $x$:
$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$ and $\frac{dz}{dx} = \frac{3}{\sqrt{1-x^2}}$.
Finally,the derivative of $y$ with respect to $z$ is given by $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/\sqrt{1-x^2}}{3/\sqrt{1-x^2}} = \frac{2}{3}$.

(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+(3^x)^2}\right)$.
Let $3^x = \tan \theta$,then $\theta = \tan^{-1}(3^x)$.
Substituting this into the function,we get $f(x) = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1}(3^x)$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = 2 \cdot \frac{1}{1+(3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2 \cdot 3^x \log 3}{1+9^x}$.
Evaluating at $x = \frac{1}{2}$,we have $f^{\prime}\left(\frac{1}{2}\right) = \frac{2 \cdot 3^{1/2} \log 3}{1+9^{1/2}} = \frac{2 \sqrt{3} \log 3}{1+3} = \frac{2 \sqrt{3} \log 3}{4} = \frac{\sqrt{3} \log 3}{2} = \sqrt{3} \log(3^{1/2}) = \sqrt{3} \log(\sqrt{3})$.

(B) Given $y = \tan ^{-1}\left(\frac{3+2 x}{2-3 x}\right)+\tan ^{-1}\left(\frac{3 x}{1+4 x^2}\right)$.
We can rewrite the first term by dividing the numerator and denominator by $2$:
$y = \tan ^{-1}\left(\frac{\frac{3}{2}+x}{1-\frac{3}{2} x}\right)+\tan ^{-1}\left(\frac{4 x-x}{1+(4 x)(x)}\right)$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$ and $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$:
$y = \tan ^{-1}\left(\frac{3}{2}\right) + \tan ^{-1} x + \tan ^{-1} (4 x) - \tan ^{-1} x$
$y = \tan ^{-1}\left(\frac{3}{2}\right) + \tan ^{-1} (4 x)$
Now,differentiating with respect to $x$:
$\frac{d y}{d x} = 0 + \frac{1}{1+(4 x)^2} \cdot \frac{d}{d x}(4 x)$
$\frac{d y}{d x} = \frac{4}{1+16 x^2}$