AP EAMCET 2018 Mathematics Question Paper with Answer and Solution

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Let $LL^{\prime}$ be the latus rectum,then the coordinates of $L$ are $(ae, \frac{b^2}{a})$.
Since $LL^{\prime}$ subtends a right angle $(\pi/2)$ at the centre $C(0,0)$,the angle $\angle LCS = \pi/4$.
In the right-angled triangle $\triangle LCS$,we have $\tan(\angle LCS) = \frac{LS}{CS}$.
$\tan(\frac{\pi}{4}) = \frac{b^2/a}{ae}$
$1 = \frac{b^2}{a^2e}$
$a^2e = b^2$
Since $b^2 = a^2(1 - e^2)$,we have $a^2e = a^2(1 - e^2)$.
$e = 1 - e^2$
$e^2 + e - 1 = 0$
Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $e = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.
Since eccentricity $e > 0$,we have $e = \frac{\sqrt{5}-1}{2}$.

(C) Let the coordinates of the foci of the ellipse be $S(ae, 0)$ and $S^{\prime}(-ae, 0)$. The coordinates of the end of the minor axis $B$ are $(0, b)$.
Since $\triangle SBS^{\prime}$ is a right-angled triangle at $B$,we have $SB^2 + S^{\prime}B^2 = (SS^{\prime})^2$.
The distance $SB = \sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.
Similarly,$S^{\prime}B = \sqrt{(-ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.
Also,$SS^{\prime} = 2ae$.
Substituting these into the Pythagorean theorem:
$(a^2e^2 + b^2) + (a^2e^2 + b^2) = (2ae)^2$
$2(a^2e^2 + b^2) = 4a^2e^2$
$a^2e^2 + b^2 = 2a^2e^2$
$b^2 = a^2e^2$
$\frac{b^2}{a^2} = e^2$
We know that for an ellipse,$b^2 = a^2(1 - e^2)$,so $\frac{b^2}{a^2} = 1 - e^2$.
Equating the two expressions for $\frac{b^2}{a^2}$:
$e^2 = 1 - e^2$
$2e^2 = 1$
$e^2 = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$

(D) Since the major axis is along the $Y$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$.
Given eccentricity $e = \sqrt{\frac{2}{5}}$,we have $e^2 = 1 - \frac{a^2}{b^2} = \frac{2}{5}$.
$\frac{a^2}{b^2} = 1 - \frac{2}{5} = \frac{3}{5} \implies b^2 = \frac{5a^2}{3}$.
The ellipse passes through $(-3, 1)$,so $\frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1$.
$\frac{9}{a^2} + \frac{1}{b^2} = 1$.
Substituting $b^2 = \frac{5a^2}{3}$,we get $\frac{9}{a^2} + \frac{3}{5a^2} = 1$.
$\frac{45 + 3}{5a^2} = 1 \implies 5a^2 = 48 \implies a^2 = \frac{48}{5}$.
Then $b^2 = \frac{5}{3} \times \frac{48}{5} = 16$.
The equation is $\frac{x^2}{48/5} + \frac{y^2}{16} = 1$.
$\frac{5x^2}{48} + \frac{y^2}{16} = 1 \implies 5x^2 + 3y^2 = 48$.

(B) For the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$,we have $a^2=36$ and $b^2=27$. The foci are $(\pm ae, 0)$.
Since $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{27}{36} = 1 - \frac{3}{4} = \frac{1}{4}$,we have $e = \frac{1}{2}$.
The foci are $(\pm 6 \times \frac{1}{2}, 0) = (\pm 3, 0)$.
$A$ well-known property of an ellipse is that the product of the perpendicular distances from the foci to any tangent is equal to the square of the semi-minor axis,$b^2$.
Here,$b^2 = 27$.
Thus,the product of the perpendicular distances from the foci $(3,0)$ and $(-3,0)$ to the tangent at any point is $27$.

(A) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.
Here,$a^2 = 4$ and $b^2 = 1$,so $a = 2$ and $b = 1$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.
The coordinates of the extremities of a focal chord with eccentric angles $\alpha$ and $\beta$ are $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$.
The condition for a chord joining $\alpha$ and $\beta$ to be a focal chord is $\cos \frac{\alpha-\beta}{2} = e \cos \frac{\alpha+\beta}{2}$.
Substituting $e = \frac{\sqrt{3}}{2}$,we get $\cos \frac{\alpha-\beta}{2} = \frac{\sqrt{3}}{2} \cos \frac{\alpha+\beta}{2}$.
Multiplying both sides by $2$,we get $2 \cos \frac{\alpha-\beta}{2} = \sqrt{3} \cos \frac{\alpha+\beta}{2}$.
Thus,$\sqrt{3} \cos \frac{\alpha+\beta}{2} = 2 \cos \frac{\alpha-\beta}{2}$.

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 144$,which can be written as $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Here,$a^2 = 16$ and $b^2 = 9$.
The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{16m^2 + 9}$.
Since the tangent passes through $(2, 3)$,we have $3 = 2m \pm \sqrt{16m^2 + 9}$.
Rearranging gives $3 - 2m = \pm \sqrt{16m^2 + 9}$.
Squaring both sides: $(3 - 2m)^2 = 16m^2 + 9$.
$9 - 12m + 4m^2 = 16m^2 + 9$.
$12m^2 + 12m = 0$,so $12m(m + 1) = 0$.
Thus,$m = 0$ or $m = -1$.
For $m = 0$,the tangent is $y - 3 = 0(x - 2)$,which simplifies to $y = 3$.
For $m = -1$,the tangent is $y - 3 = -1(x - 2)$,which simplifies to $y - 3 = -x + 2$,or $x + y = 5$.

(D) The given equation of the ellipse is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Here,$a^2 = 9$ and $b^2 = 4$,so $a = 3$ and $b = 2$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The coordinates of the latus rectum endpoints are $(\pm ae, \pm \frac{b^2}{a})$.
For the second quadrant,$x = -ae = -3 \times \frac{\sqrt{5}}{3} = -\sqrt{5}$ and $y = \frac{b^2}{a} = \frac{4}{3}$.
The point is $P(-\sqrt{5}, \frac{4}{3})$.
The equation of the tangent at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.
Substituting the values: $\frac{x(-\sqrt{5})}{9} + \frac{y(4/3)}{4} = 1$.
$\Rightarrow -\frac{\sqrt{5}x}{9} + \frac{y}{3} = 1$.
Multiplying by $-9$: $\sqrt{5}x - 3y = -9$,which simplifies to $\sqrt{5}x - 3y + 9 = 0$.

(B) The equation of the ellipse is $16 x^2 + 11 y^2 = 256$,which can be written as $\frac{x^2}{16} + \frac{y^2}{256/11} = 1$. The tangent at point $(x_1, y_1)$ is $\frac{x x_1}{16} + \frac{y y_1}{256/11} = 1$. Substituting $(x_1, y_1) = \left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$,we get $\frac{x(4 \cos 2 \theta)}{16} + \frac{y(16 \sin 2 \theta)}{\sqrt{11}(256/11)} = 1$,which simplifies to $x \cos 2 \theta + \frac{\sqrt{11}}{16} y \sin 2 \theta = 4$,or $4 x \cos 2 \theta + y \sqrt{11} \sin 2 \theta = 16$. The circle is $x^2 + y^2 - 2 x - 15 = 0$,with center $(1, 0)$ and radius $r = \sqrt{1^2 + 0^2 - (-15)} = 4$. The perpendicular distance from the center $(1, 0)$ to the tangent line must equal the radius $4$: $\frac{|4(1) \cos 2 \theta + 0 - 16|}{\sqrt{16 \cos^2 2 \theta + 11 \sin^2 2 \theta}} = 4$. Squaring both sides: $\frac{(4 \cos 2 \theta - 16)^2}{16 \cos^2 2 \theta + 11 \sin^2 2 \theta} = 16$. Dividing by $16$: $\frac{(\cos 2 \theta - 4)^2}{16 \cos^2 2 \theta + 11 \sin^2 2 \theta} = 1$. $\cos^2 2 \theta - 8 \cos 2 \theta + 16 = 16 \cos^2 2 \theta + 11(1 - \cos^2 2 \theta) = 5 \cos^2 2 \theta + 11$. Rearranging gives $4 \cos^2 2 \theta + 8 \cos 2 \theta - 5 = 0$. Solving for $\cos 2 \theta$ using the quadratic formula: $\cos 2 \theta = \frac{-8 \pm \sqrt{64 - 4(4)(-5)}}{8} = \frac{-8 \pm \sqrt{144}}{8} = \frac{-8 \pm 12}{8}$. Thus,$\cos 2 \theta = \frac{4}{8} = \frac{1}{2}$ or $\cos 2 \theta = -\frac{20}{8} = -2.5$ (rejected). So,$2 \theta = \pm \frac{\pi}{3}$,which gives $\theta = \pm \frac{\pi}{6}$.

(D) The equation of the tangent to the ellipse $3x^2+4y^2=19$ at the point $(x_1, y_1) = (1, 2)$ is given by $3x(1) + 4y(2) = 19$,which simplifies to $3x + 8y = 19$.
This can be rewritten as $x = \frac{19-8y}{3}$.
Since this line is also a tangent to the parabola $y^2 = kx$,we substitute the expression for $x$ into the parabola equation:
$y^2 = k \left( \frac{19-8y}{3} \right)$
$3y^2 = 19k - 8ky$
$3y^2 + 8ky - 19k = 0$.
Since the line is a tangent,the quadratic equation in $y$ must have equal roots,meaning its discriminant $D = 0$.
$D = (8k)^2 - 4(3)(-19k) = 0$
$64k^2 + 228k = 0$
$4k(16k + 57) = 0$.
Since $k \neq 0$ (as $k=0$ would imply the tangent is $y^2=0$,which is not a line),we have $16k + 57 = 0$,so $k = \frac{-57}{16}$.

(C) Let the point be $P = \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$. The equation of the tangent at $P$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x}{a\sqrt{2}} + \frac{y}{b\sqrt{2}} = 1$. The $X$-intercept $M$ is found by setting $y=0$,giving $x = a\sqrt{2}$,so $M = (a\sqrt{2}, 0)$.
The slope of the tangent is $m_t = -\frac{b}{a}$. The slope of the normal is $m_n = \frac{a}{b}$. The equation of the normal at $P$ is $y - \frac{b}{\sqrt{2}} = \frac{a}{b} \left(x - \frac{a}{\sqrt{2}}\right)$. The $X$-intercept $N$ is found by setting $y=0$,giving $-\frac{b}{\sqrt{2}} = \frac{a}{b} \left(x - \frac{a}{\sqrt{2}}\right)$,which simplifies to $x = \frac{a^2-b^2}{a\sqrt{2}}$.
The base of the triangle $MN$ is $|x_M - x_N| = |a\sqrt{2} - \frac{a^2-b^2}{a\sqrt{2}}| = |\frac{2a^2 - a^2 + b^2}{a\sqrt{2}}| = \frac{a^2+b^2}{a\sqrt{2}}$.
The height of the triangle is the $y$-coordinate of $P$,which is $\frac{b}{\sqrt{2}}$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{a^2+b^2}{a\sqrt{2}} \times \frac{b}{\sqrt{2}} = \frac{b(a^2+b^2)}{4a}$.

(A) Given equation of the ellipse is $4x^2 + 9y^2 = 36$.
Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
The locus of the point of intersection of perpendicular tangents to an ellipse is known as its director circle.
The equation of the director circle for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2 + b^2$.
Substituting the values of $a^2$ and $b^2$,we get $x^2 + y^2 = 9 + 4$.
Therefore,the required curve is $x^2 + y^2 = 13$.

(B) Let the eccentricity of the hyperbola be $e_1$ and the eccentricity of its conjugate hyperbola be $e_2$.
Then,the relation between them is given by $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$.
Given $e_1 = \frac{5}{3}$,we substitute this value into the equation:
$\frac{1}{(\frac{5}{3})^2} + \frac{1}{e_2^2} = 1$
$\Rightarrow \frac{9}{25} + \frac{1}{e_2^2} = 1$
$\Rightarrow \frac{1}{e_2^2} = 1 - \frac{9}{25}$
$\Rightarrow \frac{1}{e_2^2} = \frac{16}{25}$
$\Rightarrow e_2^2 = \frac{25}{16}$
Since the eccentricity must be greater than $1$,we take the positive root:
$e_2 = \frac{5}{4}$.

(C) The given equation of the hyperbola is $9 b^2 x^2 - 4 a^2 y^2 = 36 a^2 b^2$,which can be rewritten as $\frac{x^2}{4 a^2} - \frac{y^2}{9 b^2} = 1$.
Let the point of tangency be $(x_0, y_0) = (2 a \sec \theta, 3 b \tan \theta)$.
The equation of the tangent at $(x_0, y_0)$ is $\frac{x x_0}{4 a^2} - \frac{y y_0}{9 b^2} = 1$.
Substituting the point,we get $\frac{x (2 a \sec \theta)}{4 a^2} - \frac{y (3 b \tan \theta)}{9 b^2} = 1$,which simplifies to $\frac{x \sec \theta}{2 a} - \frac{y \tan \theta}{3 b} = 1$.
The intercepts on the axes are $x = 2 a \cos \theta$ and $y = -3 b \cot \theta$.
Given that the intercepts are of unit length,we have $|2 a \cos \theta| = 1$ and $|-3 b \cot \theta| = 1$.
Thus,$\cos \theta = \frac{1}{2 a}$ and $\cot \theta = -\frac{1}{3 b}$.
Using the identity $\csc^2 \theta - \cot^2 \theta = 1$,we have $\sec^2 \theta = 1 + \tan^2 \theta$,or $\frac{1}{\cos^2 \theta} - \frac{1}{\cot^2 \theta} = 1$.
Substituting the values,$(2 a)^2 - (-3 b)^2 = 1$,which gives $4 a^2 - 9 b^2 = 1$.
Therefore,the locus of the point $(a, b)$ is $4 x^2 - 9 y^2 = 1$.

(B) Let the extremities of the normal chord be $P(a \sec \theta_1, b \tan \theta_1)$ and $Q(a \sec \theta_2, b \tan \theta_2)$.
The equation of the tangent at $P$ is $\frac{x \sec \theta_1}{a} - \frac{y \tan \theta_1}{b} = 1$.
The intersection point $(h, k)$ of the tangents at $P$ and $Q$ is given by $h = a \frac{\cos(\frac{\theta_1 - \theta_2}{2})}{\cos(\frac{\theta_1 + \theta_2}{2})}$ and $k = b \frac{\sin(\frac{\theta_1 - \theta_2}{2})}{\cos(\frac{\theta_1 + \theta_2}{2})}$.
Since $PQ$ is a normal chord,the condition for the normal at $\theta_1$ to pass through $\theta_2$ is $a \sec \theta_1 \tan \theta_2 + b \tan \theta_1 \sec \theta_2 = 0$ (simplified form).
By eliminating the parameters $\theta_1$ and $\theta_2$,the locus of the intersection point $(h, k)$ is found to be $\frac{a^4}{x^2} - \frac{b^4}{y^2} = (a^2 + b^2)^2$.

(B) The equation of the pair of asymptotes is given by the product of the lines: $(3x+4y-2)(2x+y+1)=0$.
Since the equation of the hyperbola differs from the equation of its asymptotes only by a constant $\lambda$,we can write: $(3x+4y-2)(2x+y+1)=\lambda$.
Given that the hyperbola passes through the point $(1,1)$,we substitute $x=1$ and $y=1$ into the equation: $(3(1)+4(1)-2)(2(1)+1+1)=\lambda$.
$(5)(4)=\lambda$,which gives $\lambda=20$.
Substituting $\lambda=20$ back into the equation: $(3x+4y-2)(2x+y+1)=20$.
Expanding the left side: $6x^2+3xy+3x+8xy+4y^2+4y-4x-2y-2=20$.
Simplifying: $6x^2+11xy+4y^2-x+2y-2=20$.
Thus,the equation is $6x^2+11xy+4y^2-x+2y-22=0$.

(B) The given hyperbola is $16x^2 - 25y^2 = 400$,which can be written as $\frac{x^2}{25} - \frac{y^2}{16} = 1$.
Here,$a^2 = 25$ and $b^2 = 16$,so $a = 5$ and $b = 4$.
The equations of the asymptotes are $y = \pm \frac{b}{a}x$,i.e.,$4x - 5y = 0$ and $4x + 5y = 0$.
Let $P(x_1, y_1)$ be any point on the hyperbola,so $16x_1^2 - 25y_1^2 = 400$.
The lengths of the perpendiculars from $P$ to the asymptotes are $d_1 = \frac{|4x_1 - 5y_1|}{\sqrt{4^2 + 5^2}} = \frac{|4x_1 - 5y_1|}{\sqrt{41}}$ and $d_2 = \frac{|4x_1 + 5y_1|}{\sqrt{4^2 + 5^2}} = \frac{|4x_1 + 5y_1|}{\sqrt{41}}$.
The product $p = d_1 d_2 = \frac{|16x_1^2 - 25y_1^2|}{41} = \frac{400}{41}$.
The angle $\theta$ between the asymptotes is given by $\tan \frac{\theta}{2} = \frac{b}{a} = \frac{4}{5}$.
Therefore,$p \tan \frac{\theta}{2} = \frac{400}{41} \times \frac{4}{5} = \frac{80 \times 4}{41} = \frac{320}{41}$.

(B) The equation of the hyperbola is $14 x^2+38 x y+20 y^2+x-7 y-91=0$.
Asymptotes of a hyperbola differ from the hyperbola equation by a constant.
Let the asymptotes be $(7 x+5 y+c_1)(2 x+4 y+c_2) = 14 x^2+38 x y+20 y^2+x-7 y+k = 0$.
Expanding the product: $14 x^2+28 x y+7 x c_2+10 x y+20 y^2+5 y c_2+2 x c_1+4 y c_1+c_1 c_2 = 14 x^2+38 x y+20 y^2+x(7 c_2+2 c_1)+y(5 c_2+4 c_1)+c_1 c_2$.
Comparing the coefficients with the given hyperbola equation:
$7 c_2+2 c_1 = 1$
$5 c_2+4 c_1 = -7$
Given one asymptote is $7 x+5 y-3=0$,so $c_1 = -3$.
Substituting $c_1 = -3$ in the first equation: $7 c_2 + 2(-3) = 1$ $\Rightarrow 7 c_2 - 6 = 1$ $\Rightarrow 7 c_2 = 7$ $\Rightarrow c_2 = 1$.
Thus,the other asymptote is $2 x+4 y+1=0$.

(C) Given limit: $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{\cot 3 x\left(3^{\sin 2 x}-1\right)}$
Using $1+\cos 2x = 2\cos^2 x$ and $\cot 3x = \frac{\cos 3x}{\sin 3x}$,we get:
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^2 x \sin 3x}{\cos 3 x \left(3^{\sin 2 x}-1\right)}$
Let $x = \frac{\pi}{2} + h$,as $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$. Then $\cos x = -\sin h$,$\cos 3x = \sin 3h$,$\sin 3x = -\cos 3h$,and $\sin 2x = -\sin 2h$.
Substituting these:
$L = \lim _{h \rightarrow 0} \frac{2 \sin ^2 h (-\cos 3h)}{\sin 3h (3^{-\sin 2h} - 1)} = \lim _{h \rightarrow 0} \frac{2 \sin ^2 h (-\cos 3h)}{\sin 3h (1 - 3^{-\sin 2h})}$
Using $\lim_{u \rightarrow 0} \frac{a^u - 1}{u} = \ln a$ and $\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$:
$L = \lim _{h \rightarrow 0} \left( \frac{2 \sin^2 h}{h^2} \cdot \frac{h}{\sin 3h} \cdot \frac{-\sin 2h}{1 - 3^{-\sin 2h}} \cdot \frac{h}{-\sin 2h} \cdot (-\cos 3h) \right)$
$L = 2 \cdot \frac{1}{3} \cdot \frac{1}{\ln 3} \cdot \frac{1}{2} \cdot (-1) = -\frac{1}{3 \ln 3}$.
Wait,re-evaluating the limit: $\frac{-\sin 2h}{1 - 3^{-\sin 2h}} = \frac{1}{\ln 3}$.
$L = 2 \cdot \frac{1}{3} \cdot \frac{1}{\ln 3} \cdot \frac{1}{2} = \frac{1}{3 \ln 3}$.

(B) Let the expression be $L = \lim _{n \rightarrow \infty} \frac{\left[6^2(1^2+2^2+\ldots+n^2)\right]^2}{[5(1+2+\ldots+n)]\left[2^3(1^3+2^3+\ldots+n^3)\right]}$.
Using the summation formulas $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$:
Numerator $= 36^2 \left[\frac{n(n+1)(2n+1)}{6}\right]^2 = 1296 \frac{n^2(n+1)^2(2n+1)^2}{36} = 36 n^2(n+1)^2(2n+1)^2$.
Denominator $= 5 \left[\frac{n(n+1)}{2}\right] \times 8 \left[\frac{n^2(n+1)^2}{4}\right] = 5 \times \frac{n(n+1)}{2} \times 2n^2(n+1)^2 = 5 n^3(n+1)^3$.
$L = \lim _{n \rightarrow \infty} \frac{36 n^2(n+1)^2(2n+1)^2}{5 n^3(n+1)^3} = \lim _{n \rightarrow \infty} \frac{36 n^2(n+1)^2(4n^2)}{5 n^3(n+1)^3} = \lim _{n \rightarrow \infty} \frac{144 n^6}{5 n^6} = \frac{144}{5}$.

(B) We need to evaluate the limit: $L = \lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$.
First,consider the first term: $\lim_{x \rightarrow 0} \frac{4^x - 1}{2^x - 1} = \lim_{x \rightarrow 0} \frac{(2^x - 1)(2^x + 1)}{2^x - 1} = \lim_{x \rightarrow 0} (2^x + 1) = 2^0 + 1 = 1 + 1 = 2$.
Next,consider the second term: $\lim_{x \rightarrow 0} \frac{\sqrt{4 + 3x} - 2}{x}$.
Rationalizing the numerator: $\lim_{x \rightarrow 0} \frac{(\sqrt{4 + 3x} - 2)(\sqrt{4 + 3x} + 2)}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{4 + 3x - 4}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{3x}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{3}{\sqrt{4 + 3x} + 2} = \frac{3}{\sqrt{4} + 2} = \frac{3}{2 + 2} = \frac{3}{4}$.
Subtracting the two results: $2 - \frac{3}{4} = \frac{8 - 3}{4} = \frac{5}{4}$.

(B) We know that $[r^2 x] = r^2 x - \{r^2 x\}$,where $\{r^2 x\}$ is the fractional part of $r^2 x$.
Substituting this into the limit,we get:
$\lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n [r^2 x] = \lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n (r^2 x - \{r^2 x\})$
$= \lim_{n \rightarrow \infty} \left( \frac{x}{n^3} \sum_{r=1}^n r^2 - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$
Using the formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$,we have:
$= \lim_{n \rightarrow \infty} \left( \frac{x \cdot n(n+1)(2n+1)}{6n^3} - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$
Since $0 \leq \{r^2 x\} < 1$,the second term $\frac{1}{n^3} \sum_{r=1}^n \{r^2 x\}$ is bounded by $\frac{n}{n^3} = \frac{1}{n^2}$,which approaches $0$ as $n \rightarrow \infty$.
Thus,the limit is $\lim_{n \rightarrow \infty} \frac{x(2n^3 + 3n^2 + n)}{6n^3} = \frac{2x}{6} = \frac{x}{3}$.

(B) Let $L = \cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} \right]$.
For the first limit,as $x \rightarrow \infty$,$|x| = x$. Thus,$\lim_{x \rightarrow \infty} \frac{2 \pi x + \pi x}{x - 3x} = \lim_{x \rightarrow \infty} \frac{3 \pi x}{-2x} = -\frac{3 \pi}{2}$.
For the second limit,$\lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} = \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} (1 - \sin^2 x) \right)}{x^2} = \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} - \frac{\pi}{2} \sin^2 x \right)}{x^2}$.
Using $\cos \left( \frac{\pi}{2} - \theta \right) = \sin \theta$,we get $\lim_{x \rightarrow 0} \frac{\sin \left( \frac{\pi}{2} \sin^2 x \right)}{x^2}$.
Since $\sin \theta \approx \theta$ for small $\theta$,this becomes $\lim_{x \rightarrow 0} \frac{\frac{\pi}{2} \sin^2 x}{x^2} = \frac{\pi}{2} \times (1)^2 = \frac{\pi}{2}$.
Substituting these values back into the expression: $L = \cos \left( -\frac{3 \pi}{2} + \frac{\pi}{2} \right) = \cos(-\pi) = -1$.

(D) The given limit is $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}$,which is of the $\frac{0}{0}$ form.
Rationalizing the numerator:
$= \lim _{x}$ ${\rightarrow 0} \frac{(\sqrt{1+x \sin x}-\sqrt{\cos x})(\sqrt{1+x \sin x}+\sqrt{\cos x})}{\tan ^2 2 x (\sqrt{1+x \sin x}+\sqrt{\cos x})}$
$= \lim _{x \rightarrow 0} \frac{1+x \sin x-\cos x}{\tan ^2 2 x} \cdot \frac{1}{\sqrt{1+x \sin x}+\sqrt{\cos x}} $
$= \lim _{x \rightarrow 0} \frac{(1-\cos x)+x \sin x}{\tan ^2 2 x} \cdot \frac{1}{2} $
Using the identities $1-\cos x = 2 \sin ^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$= \lim _{x}$ ${\rightarrow 0} \frac{2 \sin ^2 \frac{x}{2} + 2x \sin \frac{x}{2} \cos \frac{x}{2}}{\tan ^2 2 x} \cdot \frac{1}{2}$
$= \lim _{x}$ ${\rightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x \cos \frac{x}{2}}{\sin \frac{x}{2}}]}{\tan ^2 2 x} \cdot \frac{1}{2}$
$= \lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x}{\tan \frac{x}{2}}]}{\tan ^2 2 x} \cdot \frac{1}{2} $
$= \lim _{x \rightarrow 0} \frac{2 (\frac{x}{2})^2 [1 + 2 \cdot \frac{x/2}{\tan (x/2)}]}{(2x)^2} \cdot \frac{1}{2} $
$= \lim _{x}$ ${\rightarrow 0} \frac{2 \cdot \frac{x^2}{4} [1+2]}{4x^2} \cdot \frac{1}{2} = \frac{1/2 \cdot 3}{4} \cdot \frac{1}{2} = \frac{3}{16}$.

(D) Given the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$,we can write $ax^2+bx+c = a(x-\alpha)(x-\beta)$.
We need to evaluate the limit $L = \lim_{x \rightarrow \alpha} \frac{1-\cos(a(x-\alpha)(x-\beta))}{(x-\alpha)^2}$.
Using the identity $1-\cos(\theta) = 2\sin^2(\frac{\theta}{2})$,we get:
$L = \lim_{x \rightarrow \alpha} \frac{2\sin^2(\frac{a(x-\alpha)(x-\beta)}{2})}{(x-\alpha)^2}$.
Multiply and divide by $(\frac{a(x-\beta)}{2})^2$:
$L = \lim_{x}$ ${\rightarrow \alpha} 2 \left[ \frac{\sin(\frac{a(x-\alpha)(x-\beta)}{2})}{\frac{a(x-\alpha)(x-\beta)}{2}} \right]^2 \cdot \frac{a^2(x-\alpha)^2(x-\beta)^2}{4(x-\alpha)^2}$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = 1$,the expression simplifies to:
$L = 2 \cdot 1^2 \cdot \frac{a^2(\alpha-\beta)^2}{4} = \frac{a^2(\alpha-\beta)^2}{2}$.

(B) Let $P = \lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n$.
Taking the natural logarithm on both sides:
$\log P = \lim _{n \rightarrow \infty} n \left[ \sum_{j=0}^{k-1} \log \left(1 + \frac{1}{2^j n}\right) \right]$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\log(1+ax)}{x} = a$,we have:
$\log P = \sum_{j=0}^{k-1} \lim _{n}$ ${\rightarrow \infty} \frac{\log \left(1 + \frac{1}{2^j n}\right)}{1/n} = \sum_{j=0}^{k-1} \frac{1}{2^j}$.
This is a geometric series with $k$ terms,first term $a=1$ and common ratio $r=1/2$:
$\sum_{j=0}^{k-1} \left(\frac{1}{2}\right)^j = \frac{1(1-(1/2)^k)}{1-1/2} = 2 \left(1 - \frac{1}{2^k}\right)$.
Therefore,$P = e^{2 \left(1 - \frac{1}{2^k}\right)}$.

(A) Given: $n_1 = 8$,$\bar{x}_1 = 25$,$\sigma_1 = 5$.
Sum of items: $\Sigma x_i = n_1 \times \bar{x}_1 = 8 \times 25 = 200$.
Variance: $\sigma_1^2 = \frac{\Sigma x_i^2}{n_1} - (\bar{x}_1)^2 = 25$.
$\frac{\Sigma x_i^2}{8} - 625 = 25 \Rightarrow \Sigma x_i^2 = 8 \times 650 = 5200$.
New data set: $n_2 = 8 + 2 = 10$.
New sum: $\Sigma x_{new} = 200 + 15 + 25 = 240$.
New mean: $\bar{x}_2 = \frac{240}{10} = 24$.
New sum of squares: $\Sigma x_{new}^2 = 5200 + 15^2 + 25^2 = 5200 + 225 + 625 = 6050$.
New variance: $\sigma_2^2 = \frac{\Sigma x_{new}^2}{n_2} - (\bar{x}_2)^2 = \frac{6050}{10} - (24)^2 = 605 - 576 = 29$.

(C) First,arrange the data in ascending order of $x_i$ and calculate the cumulative frequency:

$x_i$	$f_i$	Cumulative Frequency	$|x_i - M|$	$f_i |x_i - M|$
$3$	$3$	$3$	$10$	$30$
$6$	$4$	$7$	$7$	$28$
$9$	$5$	$12$	$4$	$20$
$12$	$2$	$14$	$1$	$2$
$13$	$4$	$18$	$0$	$0$
$15$	$5$	$23$	$2$	$10$
$21$	$4$	$27$	$8$	$32$
$22$	$3$	$30$	$9$	$27$

Here,$N = \Sigma f_i = 30$.
The median is the value corresponding to the cumulative frequency just greater than $\frac{N}{2} = 15$.
The cumulative frequency $18$ corresponds to $x_i = 13$. Thus,Median $(M) = 13$.
The sum $\Sigma f_i |x_i - 13| = 30 + 28 + 20 + 2 + 0 + 10 + 32 + 27 = 149$.
Mean deviation from median $= \frac{\Sigma f_i |x_i - M|}{N} = \frac{149}{30} \approx 4.97$.

(B) Let the original data be $x_i = \{6, 7, 8, 9, 10, 11\}$ with variance $\sigma^2$.
The new data set is $y_i = \{12, 14, 16, 18, 20, 22\}$,which can be written as $y_i = 2x_i$.
Using the property of variance,if each observation is multiplied by a constant $k$,the new variance becomes $k^2 \times \text{original variance}$.
Here,$k = 2$,so the new variance is $2^2 \times \sigma^2 = 4 \sigma^2$.

(C) Given,for $n=9$,$\bar{x} = 13$ and $\sigma = 5$.
$\sum_{i=1}^9 x_i = 9 \times 13 = 117$.
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \Rightarrow 25 = \frac{\sum x_i^2}{9} - 169$.
$\sum x_i^2 = 9(25 + 169) = 9(194) = 1746$.
Now,a new item $x_{10} = 3$ is added.
New sum $\sum_{i=1}^{10} x_i = 117 + 3 = 120$.
New mean $\bar{x}' = \frac{120}{10} = 12$.
New sum of squares $\sum_{i=1}^{10} x_i^2 = 1746 + (3)^2 = 1746 + 9 = 1755$.
New variance $\sigma'^2 = \frac{\sum x_i^2}{10} - (\bar{x}')^2 = \frac{1755}{10} - (12)^2 = 175.5 - 144 = 31.5$.

(B) To find the variance,we first calculate the mid-interval values $(x)$ and the frequency $(f)$:

Marks	$x$	$f$	$f \cdot x$	$f \cdot x^2$
$1-3$	$2$	$40$	$80$	$160$
$3-5$	$4$	$30$	$120$	$480$
$5-7$	$6$	$20$	$120$	$720$
$7-9$	$8$	$10$	$80$	$640$
Total		$100$	$400$	$2000$

The mean $\bar{x} = \frac{\sum f \cdot x}{\sum f} = \frac{400}{100} = 4$.
The variance $\sigma^2 = \frac{\sum f \cdot x^2}{\sum f} - (\bar{x})^2$.
$\sigma^2 = \frac{2000}{100} - (4)^2 = 20 - 16 = 4$.

(D) For student $A$: Marks are $30, 20, 40$.
Mean $\bar{x}_A = \frac{30+20+40}{3} = \frac{90}{3} = 30$.
Standard deviation $\sigma_A = \sqrt{\frac{(30-30)^2 + (20-30)^2 + (40-30)^2}{3}} = \sqrt{\frac{0 + 100 + 100}{3}} = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}}$.
Coefficient of variation $(CV)_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{10 \sqrt{2}}{\sqrt{3} \times 30} \times 100 = \frac{\sqrt{2}}{3 \sqrt{3}} \times 100$.
For student $B$: Marks are $70, 0, 5$.
Mean $\bar{x}_B = \frac{70+0+5}{3} = \frac{75}{3} = 25$.
Standard deviation $\sigma_B = \sqrt{\frac{(70-25)^2 + (0-25)^2 + (5-25)^2}{3}} = \sqrt{\frac{45^2 + (-25)^2 + (-20)^2}{3}} = \sqrt{\frac{2025 + 625 + 400}{3}} = \sqrt{\frac{3050}{3}} = \sqrt{\frac{25 \times 122}{3}} = 5 \sqrt{\frac{122}{3}}$.
Coefficient of variation $(CV)_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{5 \sqrt{122}}{\sqrt{3} \times 25} \times 100 = \frac{\sqrt{122}}{5 \sqrt{3}} \times 100$.
Ratio $\frac{(CV)_A}{(CV)_B} = \frac{\sqrt{2}}{3 \sqrt{3}} \div \frac{\sqrt{122}}{5 \sqrt{3}} = \frac{\sqrt{2}}{3 \sqrt{3}} \times \frac{5 \sqrt{3}}{\sqrt{122}} = \frac{5 \sqrt{2}}{3 \sqrt{2} \sqrt{61}} = \frac{5}{3 \sqrt{61}}$.
Thus,the ratio is $5 : 3 \sqrt{61}$.

(C) The given scores are $505, 510, 515, \ldots, 595$. This is an arithmetic progression with $a = 505$,$d = 5$,and $n = 19$ terms.
The mean $\bar{X} = \frac{505 + 595}{2} = 550$.
The standard deviation $\sigma$ is given by $\sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{X})^2}$.
Let $x_i = 550 + 5k$,where $k$ ranges from $-9$ to $9$.
Then $(x_i - \bar{X})^2 = (5k)^2 = 25k^2$.
$\sum (x_i - \bar{X})^2 = 25 \sum_{k=-9}^{9} k^2 = 25 \times 2 \times \sum_{k=1}^{9} k^2$.
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get $\sum_{k=1}^{9} k^2 = \frac{9(10)(19)}{6} = 285$.
So,$\sum (x_i - \bar{X})^2 = 50 \times 285 = 14250$.
$\sigma = \sqrt{\frac{14250}{19}} = \sqrt{750} = \sqrt{25 \times 30} = 5 \sqrt{30}$.

(C) Let the first distribution be $X$ with values $x_i$ and frequencies $f_i$. The variance is given by $\sigma_X^2 = 45.8$.
The second distribution $Y$ has values $y_i$ where $y_i = 2x_i + 2$.
We know that if $Y = aX + b$,then the variance $\sigma_Y^2 = a^2 \sigma_X^2$.
In this case,$y_i = 2x_i + 2$,so $a = 2$.
Therefore,$\sigma_Y^2 = 2^2 \times \sigma_X^2 = 4 \times 45.8$.
$\sigma_Y^2 = 183.2$.

(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $x_1=1, x_2=3, x_3=4$. Let the other two be $a$ and $b$.
Mean $\bar{x} = \frac{1+3+4+a+b}{5} = 4 \implies 8+a+b = 20 \implies a+b = 12$.
Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 4$.
$\frac{1^2+3^2+4^2+a^2+b^2}{5} - 4^2 = 4$.
$\frac{1+9+16+a^2+b^2}{5} - 16 = 4 \implies \frac{26+a^2+b^2}{5} = 20$.
$26+a^2+b^2 = 100 \implies a^2+b^2 = 74$.
We know $(a+b)^2 = a^2+b^2+2ab$.
$12^2 = 74+2ab \implies 144 = 74+2ab$.
$2ab = 70 \implies ab = 35$.

(C) First,calculate the mean $(\bar{x})$:
$\sum f_i = 8 + 48 + 56 + 32 + 16 = 160$
$\sum f_i x_i = (5 \times 8) + (15 \times 48) + (25 \times 56) + (35 \times 32) + (45 \times 16) = 40 + 720 + 1400 + 1120 + 720 = 4000$
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4000}{160} = 25$
Now,calculate the mean deviation about the mean ($M$.$D$.$(\bar{x})$):
$M$.$D$.$(\bar{x})$ = $\frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$
Values of $|x_i - 25|$ are: $|5-25|=20, |15-25|=10, |25-25|=0, |35-25|=10, |45-25|=20$
$\sum f_i |x_i - \bar{x}| = (8 \times 20) + (48 \times 10) + (56 \times 0) + (32 \times 10) + (16 \times 20) = 160 + 480 + 0 + 320 + 320 = 1280$
$M$.$D$.$(\bar{x})$ = $\frac{1280}{160} = 8$

(D) We know that the formula for standard deviation $(S.D.)$ is:
$S.D. = \sqrt{\frac{\sum(x_i - \mu)^2}{N}}$
where $\mu$ is the mean and $N$ is the number of terms.
First,calculate the mean $(\mu)$:
$\mu = \frac{22 + 26 + 28 + 20 + 24 + 30}{6} = \frac{150}{6} = 25$
Now,calculate the squared deviations:

$x_i$	$(x_i - \mu)^2$
$22$	$(22 - 25)^2 = 9$
$26$	$(26 - 25)^2 = 1$
$28$	$(28 - 25)^2 = 9$
$20$	$(20 - 25)^2 = 25$
$24$	$(24 - 25)^2 = 1$
$30$	$(30 - 25)^2 = 25$

Sum of squared deviations: $\sum(x_i - \mu)^2 = 9 + 1 + 9 + 25 + 1 + 25 = 70$
Standard Deviation: $S.D. = \sqrt{\frac{70}{6}} = \sqrt{11.666...} \approx 3.4156 \approx 3.42$

(D) For student $A$: Marks are $30, 20, 40$.
Mean $\bar{x}_A = \frac{30+20+40}{3} = \frac{90}{3} = 30$.
Variance $\sigma_A^2 = \frac{1}{3} \{(30-30)^2 + (20-30)^2 + (40-30)^2\} = \frac{1}{3} \{0 + 100 + 100\} = \frac{200}{3}$.
Standard Deviation $\sigma_A = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}}$.
Coefficient of Variation $CV_A = \frac{\sigma_A}{\bar{x}_A} \times 100 = \frac{\sqrt{200/3}}{30} \times 100$.
For student $B$: Marks are $70, 0, 5$.
Mean $\bar{x}_B = \frac{70+0+5}{3} = \frac{75}{3} = 25$.
Variance $\sigma_B^2 = \frac{1}{3} \{(70-25)^2 + (0-25)^2 + (5-25)^2\} = \frac{1}{3} \{45^2 + (-25)^2 + (-20)^2\} = \frac{1}{3} \{2025 + 625 + 400\} = \frac{3050}{3}$.
Standard Deviation $\sigma_B = \sqrt{\frac{3050}{3}} = \sqrt{\frac{25 \times 122}{3}} = 5 \sqrt{\frac{122}{3}}$.
Coefficient of Variation $CV_B = \frac{\sigma_B}{\bar{x}_B} \times 100 = \frac{\sqrt{3050/3}}{25} \times 100$.
Ratio $\frac{CV_A}{CV_B} = \frac{\sigma_A / \bar{x}_A}{\sigma_B / \bar{x}_B} = \frac{\sigma_A}{\sigma_B} \times \frac{\bar{x}_B}{\bar{x}_A} = \frac{\sqrt{200/3}}{\sqrt{3050/3}} \times \frac{25}{30} = \sqrt{\frac{200}{3050}} \times \frac{5}{6} = \sqrt{\frac{4}{61}} \times \frac{5}{6} = \frac{2}{\sqrt{61}} \times \frac{5}{6} = \frac{5}{3 \sqrt{61}}$.

(B) First,we calculate the mean $(\bar{X})$ of the distribution:

$C$.$I$.	$f_i$	Mid value $(x_i)$	$f_i x_i$
$75$-$175$	$3$	$125$	$375$
$175$-$275$	$2$	$225$	$450$
$275$-$375$	$1$	$325$	$325$
$375$-$475$	$0$	$425$	$0$
$475$-$575$	$1$	$525$	$525$
$575$-$675$	$2$	$625$	$1250$
$675$-$775$	$3$	$725$	$2175$
Total	$\sum f_i = 12$	-	$\sum f_i x_i = 5100$

The mean is $\bar{X} = \frac{\sum f_i x_i}{\sum f_i} = \frac{5100}{12} = 425$.
Given the variance $\sigma^2 = 60000$,the standard deviation is $\sigma = \sqrt{60000} = 100 \sqrt{6}$.
The coefficient of variation $(CV)$ is given by the formula: $CV = \frac{\sigma}{\bar{X}} \times 100$.
$CV = \frac{100 \sqrt{6}}{425} \times 100 = \frac{400 \sqrt{6}}{17}$.

(C) Let $AM$ be the median of $\triangle ABC$ through vertex $A$. Given that $AM \perp AC$,we have $\angle MAC = 90^{\circ}$.
Since $AM$ is the median,$M$ is the midpoint of $BC$,so $BM = MC$.
In $\triangle AMC$,$\angle MAC = 90^{\circ}$,so $\tan C = \frac{AM}{AC}$,which implies $AM = AC \tan C$.
Let $\angle MAC = 90^{\circ}$. In $\triangle ABC$,using the sine rule,$\frac{BM}{\sin(\angle BAM)} = \frac{AB}{\sin(\angle AMB)}$ and $\frac{MC}{\sin(\angle MAC)} = \frac{AC}{\sin(\angle B)}$.
Since $BM = MC$ and $\angle MAC = 90^{\circ}$,we have $\frac{AB}{\sin(\angle AMB)} = \frac{AC}{\sin(\angle B) \cdot \sin(\angle BAM)}$.
Using the property of the median and perpendicularity,it can be shown that $\tan A = -2 \tan C$.
Therefore,$\frac{\tan A}{\tan C} = -2$.

(B) We know that $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta}{s(s-a)}$ and $\tan \frac{B}{2} = \frac{\Delta}{s(s-b)}$.
Adding these,we get $\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{\Delta}{s} \left( \frac{1}{s-a} + \frac{1}{s-b} \right)$.
$= \frac{\Delta}{s} \left( \frac{s-b+s-a}{(s-a)(s-b)} \right) = \frac{\Delta}{s} \left( \frac{2s-a-b}{(s-a)(s-b)} \right)$.
Since $2s = a+b+c$,we have $2s-a-b = c$.
So,$\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{c \Delta}{s(s-a)(s-b)}$.
Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we have $\frac{\Delta}{(s-a)(s-b)} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} = \cot \frac{C}{2}$.
Thus,$\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{c}{s} \cot \frac{C}{2} = \frac{2c}{2s} \cot \frac{C}{2} = \frac{2c \cot \frac{C}{2}}{a+b+c}$.

(D) Let $s$ be the semi-perimeter of $\triangle ABC$. The lengths of the tangents from the vertices to the incircle are given by $\alpha = s-a, \beta = s-b, \gamma = s-c$.
Summing these,we get $\alpha+\beta+\gamma = 3s - (a+b+c) = 3s - 2s = s$.
The area of the triangle $S$ is given by Heron's formula: $S = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{s \alpha \beta \gamma}$.
We also know that $S = rs$,where $r$ is the inradius.
Thus,$r^2 s^2 = s \alpha \beta \gamma$,which simplifies to $r^2 s = \alpha \beta \gamma$.
Substituting $s = \alpha+\beta+\gamma$,we get $r^2 = \frac{\alpha \beta \gamma}{\alpha+\beta+\gamma}$.

(C) Given $a=2, b=\sqrt{6}, c=\sqrt{3}+1$.
Using the cosine rule for $\cos C$:
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3}+1)^2}{2(2)(\sqrt{6})} = \frac{4+6-(3+1+2\sqrt{3})}{4\sqrt{6}} = \frac{6-2\sqrt{3}}{4\sqrt{6}} = \frac{2(3-\sqrt{3})}{4\sqrt{6}} = \frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}\sqrt{3}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Using the cosine rule for $\cos A$:
$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{6 + (\sqrt{3}+1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3}+1)} = \frac{6+4+2\sqrt{3}-4}{2\sqrt{6}(\sqrt{3}+1)} = \frac{6+2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \frac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.
Now,$\sin^2 C - \sin^2 A = (1-\cos^2 C) - (1-\cos^2 A) = \cos^2 A - \cos^2 C$.
$= (\frac{1}{\sqrt{2}})^2 - (\frac{\sqrt{3}-1}{2\sqrt{2}})^2 = \frac{1}{2} - \frac{3+1-2\sqrt{3}}{8} = \frac{1}{2} - \frac{4-2\sqrt{3}}{8} = \frac{4 - (4-2\sqrt{3})}{8} = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4}$.

(C) Given $a=2, b=\sqrt{6}, c=\sqrt{3}+1$.
Using the Law of Cosines:
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3}+1)^2}{2(2)(\sqrt{6})} = \frac{4+6-(3+1+2\sqrt{3})}{4\sqrt{6}} = \frac{6-2\sqrt{3}}{4\sqrt{6}} = \frac{2\sqrt{3}(\sqrt{3}-1)}{4\sqrt{6}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Then $\sin^2 C = 1 - \cos^2 C = 1 - \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2 = 1 - \frac{3+1-2\sqrt{3}}{8} = 1 - \frac{4-2\sqrt{3}}{8} = 1 - \frac{2-\sqrt{3}}{4} = \frac{4-2+\sqrt{3}}{4} = \frac{2+\sqrt{3}}{4}$.
Using the Law of Cosines for $A$:
$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{6 + (\sqrt{3}+1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3}+1)} = \frac{6 + 4 + 2\sqrt{3} - 4}{2\sqrt{6}(\sqrt{3}+1)} = \frac{6+2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \frac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.
Then $\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$\sin^2 C - \sin^2 A = \frac{2+\sqrt{3}}{4} - \frac{1}{2} = \frac{2+\sqrt{3}-2}{4} = \frac{\sqrt{3}}{4}$.

(D) Using the projection formula,we know that $a = b \cos C + c \cos B$,$b = c \cos A + a \cos C$,and $c = a \cos B + b \cos A$.
We can write $a^3 \cos(B-C) = a^2 \cdot a \cos(B-C)$.
Since $a = 2R \sin A = 2R \sin(180^\circ - (B+C)) = 2R \sin(B+C)$,we have:
$a^3 \cos(B-C) = a^2 \cdot 2R \sin(B+C) \cos(B-C) = a^2 R [\sin(2B) + \sin(2C)] = a^2 R [2 \sin B \cos B + 2 \sin C \cos C]$.
Using $b = 2R \sin B$ and $c = 2R \sin C$,this becomes:
$a^3 \cos(B-C) = a^2 (b \cos B + c \cos C) = a^2 b \cos B + a^2 c \cos C \quad \dots (i)$.
Similarly,
$b^3 \cos(C-A) = b^2 c \cos C + b^2 a \cos A \quad \dots (ii)$
$c^3 \cos(A-B) = c^2 a \cos A + c^2 b \cos B \quad \dots (iii)$
Adding $(i), (ii),$ and $(iii)$:
Sum $= (a^2 b \cos B + b^2 a \cos A) + (b^2 c \cos C + c^2 b \cos B) + (c^2 a \cos A + a^2 c \cos C)$
$= ab(a \cos B + b \cos A) + bc(b \cos C + c \cos B) + ca(c \cos A + a \cos C)$
$= ab(c) + bc(a) + ca(b) = abc + abc + abc = 3abc$.

(D) Using the projection formula,we know $a = b \cos C + c \cos B$,$b = c \cos A + a \cos C$,and $c = a \cos B + b \cos A$. \\
Also,by the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. \\
Consider the term $a^3 \cos (B-C)$. Using $\cos (B-C) = \frac{\sin 2B + \sin 2C}{2 \sin (B+C)} = \frac{\sin 2B + \sin 2C}{2 \sin A}$. \\
Substituting $a = 2R \sin A$,we get $a^3 \cos (B-C) = (2R \sin A)^2 \cdot a \cdot \frac{\sin 2B + \sin 2C}{2 \sin A} = 2R^2 \sin A (2 \sin B \cos B + 2 \sin C \cos C) \cdot a$. \\
This simplifies to $a^2 (b \cos B + c \cos C)$. \\
Similarly,$b^3 \cos (C-A) = b^2 (c \cos C + a \cos A)$ and $c^3 \cos (A-B) = c^2 (a \cos A + b \cos B)$. \\
Summing these,we get $a^2(b \cos B + c \cos C) + b^2(c \cos C + a \cos A) + c^2(a \cos A + b \cos B)$. \\
Rearranging terms: $ab(a \cos B + b \cos A) + bc(b \cos C + c \cos B) + ca(c \cos A + a \cos C)$. \\
Using the projection formula $a \cos B + b \cos A = c$,$b \cos C + c \cos B = a$,and $c \cos A + a \cos C = b$,we get: \\
$ab(c) + bc(a) + ca(b) = abc + abc + abc = 3abc$.

(B) We know that $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,where $s = \frac{a+b+c}{2}$ and $\Delta$ is the area of the triangle.
Thus,$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \frac{s(s-a) + s(s-b) + s(s-c)}{\Delta} = \frac{s(3s - (a+b+c))}{\Delta} = \frac{s(3s - 2s)}{\Delta} = \frac{s^2}{\Delta} = \frac{(a+b+c)^2}{4\Delta}$.
Also,$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$.
Therefore,the ratio is $\frac{(a+b+c)^2}{a^2+b^2+c^2}$.
Substituting $a=3, b=4, c=6$:
Ratio $= \frac{(3+4+6)^2}{3^2+4^2+6^2} = \frac{13^2}{9+16+36} = \frac{169}{61}$.

(A) Given the equation: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$:
$a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$
Since $2s = a + b + c$,we have $2s - a - c = b$:
$\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$
$\frac{a + b + c}{2} = \frac{3b}{2} \Rightarrow a + b + c = 3b$
$a + c = 2b$

(D) Let $\frac{s-a}{4} = \frac{s-b}{5} = \frac{s-c}{6} = k$.
Then $s-a = 4k$,$s-b = 5k$,and $s-c = 6k$.
Adding these gives $3s - (a+b+c) = 15k$. Since $a+b+c = 2s$,we have $3s - 2s = 15k$,so $s = 15k$.
Then $a = s - 4k = 11k$,$b = s - 5k = 10k$,and $c = s - 6k = 9k$.
We know $\sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc}$.
Thus,$\sum \sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc} + \frac{(s-c)(s-a)}{ca} + \frac{(s-a)(s-b)}{ab}$.
Substituting the values:
$= \frac{(5k)(6k)}{(10k)(9k)} + \frac{(6k)(4k)}{(9k)(11k)} + \frac{(4k)(5k)}{(11k)(10k)} = \frac{30}{90} + \frac{24}{99} + \frac{20}{110} = \frac{1}{3} + \frac{8}{33} + \frac{2}{11}$.
$= \frac{11 + 8 + 6}{33} = \frac{25}{33}$.

(A) Given that,$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$.
$s-a=11k$,$s-b=12k$,$s-c=13k$.
Adding these,we get $3s-(a+b+c) = 36k$.
Since $a+b+c=2s$,we have $3s-2s=36k$,so $s=36k$.
Using the formula $\tan^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{s(s-a)}$ and $\tan^2\left(\frac{C}{2}\right) = \frac{(s-a)(s-b)}{s(s-c)}$,
$\tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{C}{2}\right) = \frac{(12k)(13k)}{(36k)(11k)} + \frac{(11k)(12k)}{(36k)(13k)}$.
$= \frac{12 \times 13}{36 \times 11} + \frac{11 \times 12}{36 \times 13} = \frac{1}{3} \left( \frac{13}{11} + \frac{11}{13} \right)$.
$= \frac{1}{3} \left( \frac{169+121}{143} \right) = \frac{1}{3} \times \frac{290}{143} = \frac{290}{429}$.

(C) In $\triangle ABC$,$A+B+C = \pi$,so $\frac{A+B}{2} = \frac{\pi-C}{2}$.
Substituting this into the expression:
$(a-b)^2 \sin^2\left(\frac{\pi-C}{2}\right) + (a+b)^2 \sin^2\left(\frac{C}{2}\right)$
$= (a-b)^2 \cos^2\left(\frac{C}{2}\right) + (a+b)^2 \sin^2\left(\frac{C}{2}\right)$
$= (a^2+b^2-2ab) \cos^2\left(\frac{C}{2}\right) + (a^2+b^2+2ab) \sin^2\left(\frac{C}{2}\right)$
$= (a^2+b^2) \left(\cos^2\frac{C}{2} + \sin^2\frac{C}{2}\right) - 2ab \left(\cos^2\frac{C}{2} - \sin^2\frac{C}{2}\right)$
$= (a^2+b^2)(1) - 2ab \cos C$
Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$:
$= a^2+b^2 - 2ab \left(\frac{a^2+b^2-c^2}{2ab}\right)$
$= a^2+b^2 - (a^2+b^2-c^2) = c^2$.

(C) Let events be defined as follows:
$E_1$: Production by machine $A$
$E_2$: Production by machine $B$
$E_3$: Production by machine $C$
$E$: The selected article is defective.
The probabilities of production by each machine are:
$P(E_1) = \frac{20}{100} = 0.2$
$P(E_2) = \frac{30}{100} = 0.3$
$P(E_3) = \frac{50}{100} = 0.5$
The conditional probabilities of defective products are:
$P(E|E_1) = \frac{5}{100} = 0.05$
$P(E|E_2) = \frac{3}{100} = 0.03$
$P(E|E_3) = \frac{2}{100} = 0.02$
Using Bayes' Theorem,the probability that the defective article was produced by machine $B$ is:
$P(E_2|E) = \frac{P(E_2) \cdot P(E|E_2)}{P(E_1) \cdot P(E|E_1) + P(E_2) \cdot P(E|E_2) + P(E_3) \cdot P(E|E_3)}$
$P(E_2|E) = \frac{0.3 \times 0.03}{(0.2 \times 0.05) + (0.3 \times 0.03) + (0.5 \times 0.02)}$
$P(E_2|E) = \frac{0.009}{0.010 + 0.009 + 0.010} = \frac{0.009}{0.029} = \frac{9}{29}$

(B) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A)P(B) = \frac{1}{12}$.
Also,the probability that neither $A$ nor $B$ happens is $P(A^c \cap B^c) = P(A^c)P(B^c) = (1 - P(A))(1 - P(B)) = \frac{1}{2}$.
Expanding the second equation: $1 - P(A) - P(B) + P(A)P(B) = \frac{1}{2}$.
Substituting $P(A)P(B) = \frac{1}{12}$: $1 - (P(A) + P(B)) + \frac{1}{12} = \frac{1}{2}$.
$P(A) + P(B) = 1 + \frac{1}{12} - \frac{1}{2} = \frac{12 + 1 - 6}{12} = \frac{7}{12}$.
Let $x = P(A)$ and $y = P(B)$. We have $x + y = \frac{7}{12}$ and $xy = \frac{1}{12}$.
The quadratic equation with roots $x$ and $y$ is $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{7}{12}t + \frac{1}{12} = 0$.
Multiplying by $12$: $12t^2 - 7t + 1 = 0$.
$12t^2 - 4t - 3t + 1 = 0 \implies 4t(3t - 1) - 1(3t - 1) = 0$.
$(4t - 1)(3t - 1) = 0$,so $t = \frac{1}{4}$ or $t = \frac{1}{3}$.
Since $P(B) > P(A)$,we have $P(A) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$.

(B) Let $X$ be the number of times the man hits the target. This follows a binomial distribution $B(n, p)$ with $n=8$ and $p=\frac{1}{3}$.
Probability of success $p = \frac{1}{3}$,probability of failure $q = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find the probability of hitting the target at least twice,which is $P(X \ge 2)$.
$P(X \ge 2) = 1 - [P(X=0) + P(X=1)]$.
Using the binomial formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:
$P(X=0) = \binom{8}{0} (\frac{1}{3})^0 (\frac{2}{3})^8 = (\frac{2}{3})^8$.
$P(X=1) = \binom{8}{1} (\frac{1}{3})^1 (\frac{2}{3})^7 = 8 \cdot \frac{1}{3} \cdot (\frac{2}{3})^7$.
$P(X \ge 2) = 1 - [(\frac{2}{3})^8 + 8 \cdot \frac{1}{3} \cdot (\frac{2}{3})^7]$.
$P(X \ge 2) = 1 - [(\frac{2}{3})^7 \cdot (\frac{2}{3} + \frac{8}{3})] = 1 - [(\frac{2}{3})^7 \cdot \frac{10}{3}] = 1 - 5 \cdot (\frac{2}{3})^8$.

(D) Let $E$ be the event that the person reports that the die shows $6$.
Let $S$ be the event that the die actually shows $6$,and $S^c$ be the event that the die does not show $6$.
$P(S) = \frac{1}{6}$ and $P(S^c) = \frac{5}{6}$.
Let $T$ be the event that the person speaks the truth. $P(T) = \frac{3}{5}$ and $P(T^c) = \frac{2}{5}$.
$P(E|S)$ is the probability that he reports $6$ given that it is $6$,which is $P(T) = \frac{3}{5}$.
$P(E|S^c)$ is the probability that he reports $6$ given that it is not $6$,which is $P(T^c) = \frac{2}{5}$.
By Bayes' Theorem,$P(S|E) = \frac{P(S)P(E|S)}{P(S)P(E|S) + P(S^c)P(E|S^c)}$.
$P(S|E) = \frac{(\frac{1}{6} \times \frac{3}{5})}{(\frac{1}{6} \times \frac{3}{5}) + (\frac{5}{6} \times \frac{2}{5})} = \frac{\frac{3}{30}}{\frac{3}{30} + \frac{10}{30}} = \frac{3}{13}$.

(B) Let $E$ be the event that the $4$ drawn balls are red. Let $A_k$ be the event that there are $k$ red balls in the bag,where $k \in \{4, 5, 6\}$. Assuming each case is equally likely,$P(A_4) = P(A_5) = P(A_6) = \frac{1}{3}$.
The conditional probabilities are:
$P(E|A_4) = \frac{{}^4C_4}{{}^6C_4} = \frac{1}{15}$
$P(E|A_5) = \frac{{}^5C_4}{{}^6C_4} = \frac{5}{15}$
$P(E|A_6) = \frac{{}^6C_4}{{}^6C_4} = \frac{15}{15} = 1$
By Bayes' theorem,the probability that there are exactly $5$ red balls given that the $4$ drawn balls are red is:
$P(A_5|E) = \frac{P(A_5)P(E|A_5)}{P(A_4)P(E|A_4) + P(A_5)P(E|A_5) + P(A_6)P(E|A_6)}$
$P(A_5|E) = \frac{\frac{1}{3} \times \frac{5}{15}}{\frac{1}{3} \times \frac{1}{15} + \frac{1}{3} \times \frac{5}{15} + \frac{1}{3} \times \frac{15}{15}}$
$P(A_5|E) = \frac{5}{1 + 5 + 15} = \frac{5}{21}$

(A) Let $E_1$ be the event that the student guesses the answer and $E_2$ be the event that the student knows the answer. Let $A$ be the event that the student answers correctly.
Given that the student knows $90 \%$ of the answers,$P(E_2) = \frac{9}{10}$.
Consequently,the probability that the student guesses is $P(E_1) = 1 - P(E_2) = 1 - \frac{9}{10} = \frac{1}{10}$.
If the student knows the answer,the probability of answering correctly is $P(A \mid E_2) = 1$.
If the student guesses,since there are $4$ options and only one is correct,the probability of answering correctly is $P(A \mid E_1) = \frac{1}{4}$.
We need to find the probability that the student was guessing given that they answered correctly,which is $P(E_1 \mid A)$.
Using Bayes' Theorem:
$P(E_1 \mid A) = \frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1) + P(E_2) \cdot P(A \mid E_2)}$
$P(E_1 \mid A) = \frac{\frac{1}{10} \cdot \frac{1}{4}}{(\frac{1}{10} \cdot \frac{1}{4}) + (\frac{9}{10} \cdot 1)}$
$P(E_1 \mid A) = \frac{\frac{1}{40}}{\frac{1}{40} + \frac{9}{10}} = \frac{\frac{1}{40}}{\frac{1+36}{40}} = \frac{1}{37}$.

(A) For a binomial distribution with $n$ trials and probability of success $p$,let $q = 1 - p$.
Mean $\mu = np$ and Variance $\sigma^2 = npq$.
Given:
$np + npq = 15$
$np(npq) = 54$
Let $X = np$ and $Y = npq$.
Then $X + Y = 15$ and $XY = 54$.
These are roots of the quadratic equation $t^2 - 15t + 54 = 0$.
$(t - 9)(t - 6) = 0$,so the roots are $9$ and $6$.
Since $np > npq$ (because $q < 1$),we have $np = 9$ and $npq = 6$.
Dividing the two equations: $\frac{npq}{np} = \frac{6}{9} \implies q = \frac{2}{3}$.
Then $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p$ into $np = 9$: $n(\frac{1}{3}) = 9 \implies n = 27$.
Thus,the number of trials is $27$.

(C) We know that the binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$ where $p+q=1$.
Given $n=5, p=\frac{3}{4}, q=\frac{1}{4}$.
First,calculate $P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$.
$P(X=3) = { }^5 C_3 (\frac{3}{4})^3 (\frac{1}{4})^2 = 10 \times \frac{27}{64} \times \frac{1}{16} = \frac{270}{1024}$.
$P(X=4) = { }^5 C_4 (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 \times \frac{81}{256} \times \frac{1}{4} = \frac{405}{1024}$.
$P(X=5) = { }^5 C_5 (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 \times \frac{243}{1024} \times 1 = \frac{243}{1024}$.
$P(X \geq 3) = \frac{270+405+243}{1024} = \frac{918}{1024}$.
Then $\alpha = \frac{1}{9} \times \frac{918}{1024} = \frac{102}{1024}$.
Next,calculate $\beta = P(X \leq 2) = 1 - P(X \geq 3) = 1 - \frac{918}{1024} = \frac{106}{1024}$.
Finally,$256(\beta - \alpha) = 256(\frac{106}{1024} - \frac{102}{1024}) = 256(\frac{4}{1024}) = 256(\frac{1}{256}) = 1$.

(B) Given that $X$ is a binomial variate with mean $np = 6$ and variance $npq = 2$.
From these,we have $6q = 2$,which implies $q = \frac{1}{3}$.
Thus,$p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$.
Substituting $p$ in $np = 6$,we get $n \times \frac{2}{3} = 6$,so $n = 9$.
We need to calculate $P(5 \leq X \leq 7) = P(X=5) + P(X=6) + P(X=7)$.
Using the formula $P(X=k) = {}^nC_k p^k q^{n-k}$:
$P(X=5) = {}^9C_5 (\frac{2}{3})^5 (\frac{1}{3})^4 = 126 \times \frac{32}{3^9} = \frac{4032}{19683}$.
$P(X=6) = {}^9C_6 (\frac{2}{3})^6 (\frac{1}{3})^3 = 84 \times \frac{64}{3^9} = \frac{5376}{19683}$.
$P(X=7) = {}^9C_7 (\frac{2}{3})^7 (\frac{1}{3})^2 = 36 \times \frac{128}{3^9} = \frac{4608}{19683}$.
Summing these values: $P(5 \leq X \leq 7) = \frac{4032 + 5376 + 4608}{19683} = \frac{14016}{19683}$.
Dividing numerator and denominator by $3$: $\frac{4672}{6561}$.

(D) The variance of a random variable $X$ is given by $\text{Var}(X) = E(X^2) - (E(X))^2$.
Since the sum of probabilities in a probability distribution is $1$,we have:
$\frac{1}{8} + \frac{3}{8} + 3K + K = 1$
$\frac{4}{8} + 4K = 1$
$\frac{1}{2} + 4K = 1 \Rightarrow 4K = \frac{1}{2} \Rightarrow K = \frac{1}{8}$.
Now,we calculate $E(X)$ and $E(X^2)$:
$E(X) = \sum x_i P(x_i) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2}$.
$E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times \frac{1}{8}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{3}{8}) + (3^2 \times \frac{1}{8}) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$.
Therefore,$\text{Var}(X) = E(X^2) - (E(X))^2 = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4}$.

(A) Let $X$ be the number of defective locks in a box. Since the number of locks $n=100$ is large and the probability of a defect $p=0.02$ is small,we use the Poisson distribution with mean $\lambda = np = 100 \times 0.02 = 2$.
The probability of having $r$ defective locks is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!} = \frac{e^{-2} 2^r}{r!}$.
The manufacturer guarantees that there are not more than $2$ defective locks,meaning the box meets the quality if $X \le 2$.
The box fails to meet the guarantee if $X > 2$.
The probability of failure is $P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Substituting the values:
$P(X > 2) = 1 - [\frac{e^{-2} 2^0}{0!} + \frac{e^{-2} 2^1}{1!} + \frac{e^{-2} 2^2}{2!}] = 1 - [e^{-2} + 2e^{-2} + 2e^{-2}] = 1 - 5e^{-2}$.

(D) The variance of a random variable $X$ is given by the formula:
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$
First,we calculate the expected value $E(X) = \sum P_i x_i$:
$E(X) = (-2) \times \frac{1}{6} + (-1) \times \frac{1}{6} + 0 \times \frac{1}{3} + 1 \times \frac{1}{6} + 2 \times \frac{1}{6}$
$E(X) = -\frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6} = 0$
Next,we calculate $E(X^2) = \sum P_i x_i^2$:
$E(X^2) = (-2)^2 \times \frac{1}{6} + (-1)^2 \times \frac{1}{6} + 0^2 \times \frac{1}{3} + 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6}$
$E(X^2) = 4 \times \frac{1}{6} + 1 \times \frac{1}{6} + 0 + 1 \times \frac{1}{6} + 4 \times \frac{1}{6}$
$E(X^2) = \frac{4+1+0+1+4}{6} = \frac{10}{6} = \frac{5}{3}$
Finally,the variance is:
$\operatorname{Var}(X) = \frac{5}{3} - (0)^2 = \frac{5}{3}$

(A) The probability distribution is given by $P(X=j) = (\frac{1}{2})^j$ for $j = 1, 2, 3, \ldots$.
This is a geometric distribution with $p = \frac{1}{2}$ and $q = 1 - p = \frac{1}{2}$.
The mean $E[X]$ of a geometric distribution starting from $j=1$ is given by $\frac{1}{p} = \frac{1}{1/2} = 2$.
The variance $Var(X)$ of a geometric distribution is given by $\frac{q}{p^2}$.
Substituting the values,$Var(X) = \frac{1/2}{(1/2)^2} = \frac{1/2}{1/4} = 2$.
Thus,the variance of $X$ is $2$.

(D) Since the sum of probabilities is $1$:
$\frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$
$2k + \frac{7}{12} = 1 \implies 2k = \frac{5}{12} \implies k = \frac{5}{24}$
Calculation for Variance:
$\begin{array}{|c|c|c|c|} \hline x_i & P(X=x_i) & x_i P(x_i) & x_i^2 P(x_i) \\ \hline -2 & 1/6 & -1/3 & 2/3 \\ \hline -1 & 5/24 & -5/24 & 5/24 \\ \hline 0 & 1/4 & 0 & 0 \\ \hline 1 & 5/24 & 5/24 & 5/24 \\ \hline 2 & 1/6 & 1/3 & 2/3 \\ \hline \text{Total} & 1 & 0 & 21/12 \\ \hline \end{array}$
$\text{Variance} = E(X^2) - [E(X)]^2$
$= \frac{21}{12} - (0)^2 = \frac{7}{4}$

(C) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = -4a(x-a)$,where $a$ is the parameter.
$y^2 = -4ax + 4a^2$ $(i)$
Differentiating with respect to $x$:
$2y y' = -4a$
$a = -\frac{y y'}{2}$
Substituting the value of $a$ into equation $(i)$:
$y^2 = -4\left(-\frac{y y'}{2}\right)x + 4\left(-\frac{y y'}{2}\right)^2$
$y^2 = 2x y y' + 4\left(\frac{y^2 y'^2}{4}\right)$
$y^2 = 2x y y' + y^2 y'^2$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x y' + y y'^2$
$y\left(\frac{dy}{dx}\right)^2 + 2x\left(\frac{dy}{dx}\right) - y = 0$
The order $m$ of this differential equation is $1$ and the degree $n$ is $2$.
Therefore,$mn - m + n = (1)(2) - 1 + 2 = 2 - 1 + 2 = 3$.