TS EAMCET 2019 Mathematics Question Paper with Answer and Solution

(A) Let the equation of the variable circle $S=0$ passing through the origin $(0,0)$ be $x^2+y^2+2gx+2fy=0$.
Since the circle touches the line $x-y=0$,the perpendicular distance from the center $(-g, -f)$ to the line must equal the radius $\sqrt{g^2+f^2}$.
$\frac{|-g+f|}{\sqrt{1^2+(-1)^2}} = \sqrt{g^2+f^2} \Rightarrow \frac{(g-f)^2}{2} = g^2+f^2$.
$g^2+f^2-2gf = 2g^2+2f^2$ $\Rightarrow g^2+f^2+2gf = 0$ $\Rightarrow (g+f)^2 = 0$ $\Rightarrow f = -g$.
Substituting $f = -g$ into the circle equation,we get $S = x^2+y^2+2gx-2gy = 0$.
The common chord of $S=0$ and $x^2+y^2+6x+8y-7=0$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2gx-2gy) - (x^2+y^2+6x+8y-7) = 0$.
$(2g-6)x - (2g+8)y + 7 = 0$.
$2g(x-y) - (6x+8y-7) = 0$.
This represents a family of lines passing through the intersection of $x-y=0$ and $6x+8y-7=0$.
Solving these: $x=y$,so $6x+8x-7=0$ $\Rightarrow 14x=7$ $\Rightarrow x=\frac{1}{2}$.
Thus,the fixed point is $\left(\frac{1}{2}, \frac{1}{2}\right)$.

(A) The given circle is $S^{\prime} \equiv x^2+y^2-2x+2y-7=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=1, c=-7$.
The centre is $C_1(-g, -f) = (1, -1)$ and the radius is $r_1 = \sqrt{g^2+f^2-c} = \sqrt{1+1+7} = 3$.
The pair of lines $5x^2-xy-5x+y=0$ are normals to the circle $S=0$.
Factoring the equation: $x(5x-y) - 1(5x-y) = 0 \Rightarrow (x-1)(5x-y) = 0$.
The lines are $x=1$ and $y=5x$.
The intersection of these normals is the centre $C_2$ of circle $S=0$.
Solving $x=1$ and $y=5x$,we get $x=1, y=5$. So,$C_2(1, 5)$.
Since the circles touch externally,the distance between centres $C_1C_2 = r_1 + r_2$.
$C_1C_2 = \sqrt{(1-1)^2 + (5 - (-1))^2} = \sqrt{0^2 + 6^2} = 6$.
Thus,$6 = 3 + r_2 \Rightarrow r_2 = 3$.
The equation of circle $S=0$ with centre $(1, 5)$ and radius $3$ is $(x-1)^2 + (y-5)^2 = 3^2$.
$x^2 - 2x + 1 + y^2 - 10y + 25 = 9 \Rightarrow x^2 + y^2 - 2x - 10y + 17 = 0$.
The chord of contact of point $C_1(1, -1)$ with respect to $S=0$ is given by $T=0$.
$x(1) + y(-1) - 1(x+1) - 5(y-1) + 17 = 0$.
$x - y - x - 1 - 5y + 5 + 17 = 0$.
$-6y + 21 = 0$ $\Rightarrow 6y = 21$ $\Rightarrow 2y = 7$ $\Rightarrow 2y-7=0$.

(D) The given circle is $x^2+y^2-4x+2y+3=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=1, c=3$.
The center is $(2, -1)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+1-3} = \sqrt{2}$.
If the angle between the tangents is $\frac{\pi}{2}$,the locus of the point $P$ is the director circle.
The equation of the director circle is $(x-h)^2+(y-k)^2 = 2r^2$,where $(h, k)$ is the center.
Substituting the values: $(x-2)^2+(y+1)^2 = 2(\sqrt{2})^2 = 4$.
Expanding this: $x^2-4x+4+y^2+2y+1 = 4$.
$x^2+y^2-4x+2y+1 = 0$.
Thus,option $D$ is correct.

(C) For statement $I$: $x = ly^2 + my + n \Rightarrow x - n = l(y^2 + \frac{m}{l}y) \Rightarrow x - n + \frac{m^2}{4l} = l(y + \frac{m}{2l})^2 \Rightarrow (y + \frac{m}{2l})^2 = \frac{1}{l}(x - (n - \frac{m^2}{4l}))$. The vertex is $(n - \frac{m^2}{4l}, -\frac{m}{2l})$. Thus,statement $I$ is true.
For statement $II$: $y = lx^2 + mx + n \Rightarrow y - n = l(x^2 + \frac{m}{l}x) \Rightarrow y - n + \frac{m^2}{4l} = l(x + \frac{m}{2l})^2 \Rightarrow (x + \frac{m}{2l})^2 = \frac{1}{l}(y - (n - \frac{m^2}{4l}))$. Here $4a = \frac{1}{l} \Rightarrow a = \frac{1}{4l}$. The focus is $(h, k+a) = (-\frac{m}{2l}, n - \frac{m^2}{4l} + \frac{1}{4l}) = (-\frac{m}{2l}, n - \frac{m^2-1}{4l})$. The given focus in the question is incorrect. Thus,statement $II$ is false.
For statement $III$: The pole of $lx + my + n = 0$ with respect to $x^2 = 4ay$ is found by comparing with the chord of contact $xx_1 = 2a(y + y_1) \Rightarrow xx_1 - 2ay - 2ay_1 = 0$. Comparing coefficients: $\frac{x_1}{l} = \frac{-2a}{m} = \frac{-2ay_1}{n}$. Thus $x_1 = -\frac{2al}{m}$ and $y_1 = -\frac{n}{m}$. The given pole is $(\frac{-2al}{m}, \frac{n}{m})$,which is incorrect as the $y$-coordinate is $-\frac{n}{m}$. Thus,statement $III$ is false.

(A) For $(A)$: Let $x = \frac{p}{2}\left(t+\frac{1}{t}\right)$ and $y = \frac{q}{2}\left(t-\frac{1}{t}\right)$.
$\Rightarrow \frac{2x}{p} = t+\frac{1}{t}$ and $\frac{2y}{q} = t-\frac{1}{t}$.
Squaring and subtracting: $\left(\frac{2x}{p}\right)^2 - \left(\frac{2y}{q}\right)^2 = \left(t+\frac{1}{t}\right)^2 - \left(t-\frac{1}{t}\right)^2 = 4$.
$\Rightarrow \frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$,which represents a hyperbola. Thus,$(A \rightarrow IV)$.
For $(B)$: Let $x = p+q \cos \theta$ and $y = r+q \sin \theta$.
$\Rightarrow (x-p) = q \cos \theta$ and $(y-r) = q \sin \theta$.
Squaring and adding: $(x-p)^2 + (y-r)^2 = q^2(\cos^2 \theta + \sin^2 \theta) = q^2$.
This represents a circle. Thus,$(B \rightarrow II)$.
For $(C)$: Let $x = p+\lambda^2$ and $y = q-\lambda$.
$\Rightarrow \lambda = q-y$.
Substituting $\lambda$ in $x$: $x = p + (q-y)^2$.
$\Rightarrow (y-q)^2 = x-p$,which represents a parabola. Thus,$(C \rightarrow I)$.
Therefore,the correct match is $(A$ $\rightarrow IV, B$ $\rightarrow II, C$ $\rightarrow I)$.

(A) The equation of the parabola is $y^2=4ax$,where $4a=4$,so $a=1$.
The condition for the line $y=mx+c$ to be a tangent to the parabola $y^2=4ax$ is $c = \frac{a}{m}$.
Given the line $y=mx+1$,we have $c=1$.
Substituting the values,we get $1 = \frac{1}{m}$,which implies $m=1$.

(B) The equation of the given parabola is $y^2=4ax$,which has its vertex at $V(0,0)$.
Let a point on the parabola be $P(at^2, 2at)$.
The slope of the line segment joining $V(0,0)$ and $P(at^2, 2at)$ is given by $\frac{2at-0}{at^2-0} = \tan \theta$.
Simplifying this,we get $\frac{2}{t} = \tan \theta$,which implies $t = 2 \cot \theta$.
The length of the line segment $VP$ is given by $\sqrt{(at^2-0)^2 + (2at-0)^2} = \sqrt{a^2t^4 + 4a^2t^2} = a|t|\sqrt{t^2+4}$.
Substituting $t = 2 \cot \theta$:
$VP = a(2 \cot \theta) \sqrt{4 \cot^2 \theta + 4} = 2a \cot \theta \cdot 2 \sqrt{\cot^2 \theta + 1} = 4a \cot \theta \operatorname{cosec} \theta$.
$VP = 4a \left(\frac{\cos \theta}{\sin \theta}\right) \left(\frac{1}{\sin \theta}\right) = \frac{4a \cos \theta}{\sin^2 \theta}$.
Thus,option $B$ is correct.

(A) Given: $C_1: y^2=4x$,where $a=1$.
$C_2: x^2+y^2-6x+1=0$,which has center $(3,0)$ and radius $r = \sqrt{3^2+0^2-1} = \sqrt{8} = 2\sqrt{2}$.
Equation of tangent to $C_1$ in slope form is $y = mx + \frac{1}{m}$,or $m^2x - my + 1 = 0$.
Since this is also a tangent to $C_2$,the perpendicular distance from the center $(3,0)$ to this line must equal the radius $2\sqrt{2}$.
$\frac{|m^2(3) - m(0) + 1|}{\sqrt{(m^2)^2 + (-m)^2}} = 2\sqrt{2}$
$\frac{|3m^2+1|}{\sqrt{m^4+m^2}} = 2\sqrt{2}$
Squaring both sides: $(3m^2+1)^2 = 8(m^4+m^2)$
$9m^4 + 6m^2 + 1 = 8m^4 + 8m^2$
$m^4 - 2m^2 + 1 = 0$
$(m^2-1)^2 = 0 \Rightarrow m = \pm 1$.
For $m=1$,the tangent is $y = x + 1 \Rightarrow x - y + 1 = 0$.
For $m=-1$,the tangent is $y = -x - 1 \Rightarrow x + y + 1 = 0$.
Since the slopes are $m_1=1$ and $m_2=-1$,their product $m_1 \cdot m_2 = -1$,which means the tangents are orthogonal.
Thus,both Assertion and Reason are true,and the Reason correctly explains the Assertion.

(A) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,so $a = 3$.
Point $P(3, 6)$ corresponds to $at^2 = 3$ and $2at = 6$. Substituting $a = 3$,we get $3t^2 = 3 \implies t = 1$.
The length of the normal chord at $t$ is given by $l_1 = 2a(t^2+2) \sqrt{t^2+1}$. For $t = 1$,$l_1 = 2(3)(1+2) \sqrt{1+1} = 6(3) \sqrt{2} = 18 \sqrt{2}$.
The length of the focal chord passing through $P(at^2, 2at)$ is $l_2 = a(t + \frac{1}{t})^2$. For $t = 1$,$l_2 = 3(1 + \frac{1}{1})^2 = 3(2)^2 = 12$.
Therefore,$\frac{l_1}{l_2} = \frac{18 \sqrt{2}}{12} = \frac{3 \sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.
Wait,re-evaluating the normal chord length formula: The length of the normal chord at $t$ is $l_1 = 2a(t^2+2) \sqrt{1+t^2}$. For $t=1$,$l_1 = 2(3)(3)\sqrt{2} = 18\sqrt{2}$.
Re-evaluating the focal chord length: The length of a focal chord with parameter $t$ is $a(t + 1/t)^2$. For $t=1$,$l_2 = 3(1+1)^2 = 12$.
Thus,$\frac{l_1}{l_2} = \frac{18\sqrt{2}}{12} = \frac{3\sqrt{2}}{2}$.

(B) Given the parabola $y^2=16x$, we have $4a=16$, so $a=4$.
The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$, which are $(4, 8)$ and $(4, -8)$.
Let us consider the point $(4, 8)$. The slope of the tangent at $(4, 8)$ is given by $2y \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{8}{y}$. At $(4, 8)$, the slope of the tangent is $m_t = \frac{8}{8} = 1$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -1$.
The equation of the normal at $(4, 8)$ is $y - 8 = -1(x - 4) \implies y = -x + 12$.
This normal meets the $X$-axis $(y=0)$ at $x=12$, so point $P$ is $(12, 0)$.
The chord passes through $P(12, 0)$ and is perpendicular to the normal. Since the normal has slope $-1$, the chord has slope $m_c = 1$.
The equation of the chord is $y - 0 = 1(x - 12) \implies y = x - 12$.
Substituting $y = x - 12$ into $y^2 = 16x$:
$(x - 12)^2 = 16x \implies x^2 - 24x + 144 = 16x \implies x^2 - 40x + 144 = 0$.
$(x - 36)(x - 4) = 0$, so $x = 4$ or $x = 36$.
For $x = 4$, $y = 4 - 12 = -8$. For $x = 36$, $y = 36 - 12 = 24$.
The endpoints of the chord are $(4, -8)$ and $(36, 24)$.
The length of the chord is $\sqrt{(36 - 4)^2 + (24 - (-8))^2} = \sqrt{32^2 + 32^2} = \sqrt{2 \times 32^2} = 32 \sqrt{2}$.

(D) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $a = 4$.
The equation of the normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.
Substituting $a = 4$,the normal is $y = mx - 8m - 4m^3$.
Given that the line $y = -x + k$ is a normal,we compare the slopes: $m = -1$.
Substituting $m = -1$ into the equation of the normal:
$y = (-1)x - 8(-1) - 4(-1)^3$
$y = -x + 8 + 4$
$y = -x + 12$
Comparing this with $y = -x + k$,we get $k = 12$.

(A) Given parabola is $y^2=4x$,which is of the form $y^2=4ax$. Comparing,we get $a=1$.
The equation of a tangent with slope $m$ to the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$.
Since the tangent passes through $(1,4)$,we have $4=m(1)+\frac{1}{m}$.
Multiplying by $m$,we get $m^2-4m+1=0$.
Let $m_1$ and $m_2$ be the slopes of the two tangents. Then $m_1+m_2=4$ and $m_1m_2=1$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1-m_2}{1+m_1m_2} \right|$.
Using the identity $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2}$,we get $\tan \theta = \frac{\sqrt{4^2-4(1)}}{1+1} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Thus,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

(D) Let $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$ be the endpoints of the chord.
Let the point $R(h, k)$ divide the chord $PQ$ internally in the ratio $1:3$.
Using the section formula,we have:
$h = \frac{1 \cdot t_2^2 + 3 \cdot t_1^2}{1+3} = \frac{t_2^2 + 3t_1^2}{4}$
$k = \frac{1 \cdot 2t_2 + 3 \cdot 2t_1}{1+3} = \frac{2t_2 + 6t_1}{4} = \frac{t_2 + 3t_1}{2}$
The slope of the chord $PQ$ is given as $2$:
$\frac{2t_2 - 2t_1}{t_2^2 - t_1^2} = 2$
$\frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = 2$
$\frac{2}{t_2 + t_1} = 2 \Rightarrow t_1 + t_2 = 1 \Rightarrow t_2 = 1 - t_1$
Substitute $t_2 = 1 - t_1$ into the expressions for $h$ and $k$:
$k = \frac{(1 - t_1) + 3t_1}{2} = \frac{2t_1 + 1}{2} = t_1 + \frac{1}{2} \Rightarrow t_1 = k - \frac{1}{2}$
$4h = 3t_1^2 + (1 - t_1)^2 = 3t_1^2 + 1 - 2t_1 + t_1^2 = 4t_1^2 - 2t_1 + 1$
Substitute $t_1 = k - \frac{1}{2}$ into the equation for $4h$:
$4h = 4(k - \frac{1}{2})^2 - 2(k - \frac{1}{2}) + 1$
$4h = 4(k^2 - k + \frac{1}{4}) - 2k + 1 + 1$
$4h = 4k^2 - 4k + 1 - 2k + 2 = 4k^2 - 6k + 3$
$h = k^2 - \frac{3}{2}k + \frac{3}{4}$
$k^2 - \frac{3}{2}k = h - \frac{3}{4}$
$(k - \frac{3}{4})^2 = h - \frac{3}{4} + \frac{9}{16} = h - \frac{3}{16}$
Comparing with $(y - k_0)^2 = 4a(x - h_0)$,the vertex is $\left(\frac{3}{16}, \frac{3}{4}\right)$.

(C) The definition of an ellipse is the locus of a point $P(x, y)$ such that the distance from the focus $S(1, -2)$ is $e$ times the distance from the directrix $x+y-2=0$.
$(x-1)^2 + (y+2)^2 = e^2 \frac{(x+y-2)^2}{1^2+1^2}$
$2(x^2 - 2x + 1 + y^2 + 4y + 4) = e^2(x^2 + y^2 + 4 + 2xy - 4x - 4y)$
$(2-e^2)x^2 - 2e^2xy + (2-e^2)y^2 + (4e^2-4)x + (8+4e^2)y + (10-4e^2) = 0$
Comparing this with the given equation $17x^2 - 2xy + 17y^2 - 32x + 76y + 86 = 0$:
$\frac{2-e^2}{17} = \frac{-2e^2}{-2} = \frac{4e^2-4}{-32}$
From $\frac{2-e^2}{17} = e^2$:
$2 - e^2 = 17e^2$ $\Rightarrow 18e^2 = 2$ $\Rightarrow e^2 = \frac{1}{9}$ $\Rightarrow e = \frac{1}{3}$.

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
The coordinates of the foci are $A(-ae, 0)$ and $B(ae, 0)$,and the endpoint of the semi-minor axis is $T(0, b)$.
The slope of $AT$ is $m_1 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.
The slope of $BT$ is $m_2 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.
Since $\angle ATB = 90^{\circ}$,the product of the slopes is $-1$:
$\left(\frac{b}{ae}\right) \left(-\frac{b}{ae}\right) = -1$.
$\Rightarrow \frac{b^2}{a^2 e^2} = 1$ $\Rightarrow b^2 = a^2 e^2$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$.
Substituting $b^2 = a^2 e^2$ into the equation:
$a^2 e^2 = a^2(1 - e^2)$.
$e^2 = 1 - e^2
$ $\Rightarrow 2e^2 = 1
$ $\Rightarrow e^2 = \frac{1}{2}
$ $\Rightarrow e = \frac{1}{\sqrt{2}}$.

(B) Let the rectangle be $ABCD$ with dimensions $2a$ and $2b$. The center of the ellipse is the origin $O(0,0)$. The vertices of the rectangle are $(a, b), (-a, b), (-a, -b),$ and $(a, -b)$.
Let the angle between the diagonals be $2\theta$. From the geometry of the rectangle,$\tan \theta = \frac{b}{a}$.
The angle between the diagonals is given as $\tan(2\theta) = 4\sqrt{3}$.
Using the formula $\tan(2\theta) = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have:
$4\sqrt{3} = \frac{2(b/a)}{1-(b/a)^2}$
$2\sqrt{3} = \frac{b/a}{1-(b/a)^2}$
Let $k = b/a$. Then $2\sqrt{3}(1-k^2) = k$,which implies $2\sqrt{3}k^2 + k - 2\sqrt{3} = 0$.
Solving the quadratic equation for $k$:
$k = \frac{-1 \pm \sqrt{1 - 4(2\sqrt{3})(-2\sqrt{3})}}{2(2\sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4\sqrt{3}} = \frac{-1 \pm 7}{4\sqrt{3}}$.
Since $k = b/a > 0$,we take $k = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - k^2}$.
$e = \sqrt{1 - (\frac{\sqrt{3}}{2})^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the correct option is $(b)$.

(B) Let the line perpendicular to the $X$-axis be $x = h = 3 \cos \alpha$. The point $A$ on the circle $x^2+y^2=9$ is $(3 \cos \alpha, 3 \sin \alpha)$. The tangent at $A$ is $x(3 \cos \alpha) + y(3 \sin \alpha) = 9$,which simplifies to $x \cos \alpha + y \sin \alpha = 3$. The slope $m_1 = -\cot \alpha$.
The point $B$ on the ellipse $4x^2+9y^2=36$ is $(3 \cos \alpha, 2 \sin \alpha)$. The tangent at $B$ is $4x(3 \cos \alpha) + 9y(2 \sin \alpha) = 36$,which simplifies to $2x \cos \alpha + 3y \sin \alpha = 6$. The slope $m_2 = -\frac{2}{3} \cot \alpha$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-\cot \alpha + \frac{2}{3} \cot \alpha}{1 + \frac{2}{3} \cot^2 \alpha} \right| = \frac{\frac{1}{3} \cot \alpha}{1 + \frac{2}{3} \cot^2 \alpha} = \frac{\cot \alpha}{3 + 2 \cot^2 \alpha}$.
To maximize $f(\cot \alpha) = \frac{\cot \alpha}{3 + 2 \cot^2 \alpha}$,let $u = \cot \alpha$. Then $f(u) = \frac{u}{3 + 2u^2}$.
Setting $f'(u) = 0$,we get $\frac{(3 + 2u^2)(1) - u(4u)}{(3 + 2u^2)^2} = 0$,which implies $3 - 2u^2 = 0$,so $u^2 = \frac{3}{2}$ or $u = \sqrt{\frac{3}{2}}$.
Substituting $u = \sqrt{\frac{3}{2}}$ into $f(u)$,we get $\tan \theta = \frac{\sqrt{3/2}}{3 + 2(3/2)} = \frac{\sqrt{3/2}}{6} = \frac{\sqrt{3}}{6 \sqrt{2}} = \frac{\sqrt{3}}{6 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{12} = \frac{1}{2 \sqrt{6}}$.

(B) The equation of the ellipse is $x^2+3y^2=3$,which can be written as $\frac{x^2}{3}+y^2=1$. Here $a^2=3$ and $b^2=1$.
The condition for $lx+my+n=0$ to be a normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2)^2}{n^2}$.
Substituting $l=1, m=1, a^2=3, b^2=1$,we get $\frac{3}{1^2}+\frac{1}{1^2}=\frac{(3-1)^2}{n^2} \Rightarrow 4=\frac{4}{n^2} \Rightarrow n^2=1$.
Since $n>0$,$n=1$. The normal is $x+y+1=0$.
The equation of the ellipse is $x^2+5y^2=5$,or $\frac{x^2}{5}+y^2=1$.
The condition for $lx+my+n=0$ to be a tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $n^2=a^2l^2+b^2m^2$.
Substituting $l=1, n=3, a^2=5, b^2=1$,we get $3^2=5(1)^2+1(m)^2 \Rightarrow 9=5+m^2 \Rightarrow m^2=4$.
Since $m < 0$,$m=-2$. The tangent is $x-2y+3=0$.
Solving $x+y+1=0$ and $x-2y+3=0$: Subtracting the equations gives $3y-2=0 \Rightarrow y=\frac{2}{3}$.
Then $x=-1-y=-1-\frac{2}{3}=-\frac{5}{3}$.
Checking the point $(-\frac{5}{3}, \frac{2}{3})$ in the options: For $x-5y+5=0$,we have $-\frac{5}{3}-5(\frac{2}{3})+5 = \frac{-5-10+15}{3} = 0$.
Thus,the point satisfies $x-5y+5=0$.

(A) The equation of the tangent at $(3 \sqrt{3} \cos \theta, \sin \theta)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ is given by $\frac{x(3 \sqrt{3} \cos \theta)}{27} + \frac{y \sin \theta}{1} = 1$,which simplifies to $\frac{x \cos \theta}{3 \sqrt{3}} + y \sin \theta = 1$.
The $x$-intercept is $a = 3 \sqrt{3} \sec \theta$ and the $y$-intercept is $b = \operatorname{cosec} \theta$.
The sum of the intercepts is $L(\theta) = 3 \sqrt{3} \sec \theta + \operatorname{cosec} \theta$.
To find the minimum,we differentiate with respect to $\theta$: $\frac{dL}{d\theta} = 3 \sqrt{3} \sec \theta \tan \theta - \operatorname{cosec} \theta \cot \theta$.
Setting $\frac{dL}{d\theta} = 0$,we get $3 \sqrt{3} \frac{\sin \theta}{\cos^2 \theta} = \frac{\cos \theta}{\sin^2 \theta}$,which implies $\tan^3 \theta = \frac{1}{3 \sqrt{3}} = \left(\frac{1}{\sqrt{3}}\right)^3$.
Thus,$\tan \theta = \frac{1}{\sqrt{3}}$,which gives $\theta = \frac{\pi}{6}$.

(A) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,so $a^2 = 16$ and $b^2 = 9$.
The tangent at $P(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$. It meets the major axis $(y=0)$ at $Q(\frac{a}{\cos \theta}, 0)$.
The normal at $P$ is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$. It meets the major axis at $R(\frac{(a^2 - b^2) \cos \theta}{a}, 0)$.
The distance $QR = \left| \frac{a}{\cos \theta} - \frac{(a^2 - b^2) \cos \theta}{a} \right| = \left| \frac{a^2 - (a^2 - b^2) \cos^2 \theta}{a \cos \theta} \right| = \left| \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a \cos \theta} \right|$.
Thus,Reason $(R)$ is true.
For $a=4, b=3, \theta = \frac{\pi}{3}$,$QR = \left| \frac{16 \cdot \frac{3}{4} + 9 \cdot \frac{1}{4}}{4 \cdot \frac{1}{2}} \right| = \left| \frac{12 + 2.25}{2} \right| = \frac{14.25}{2} = \frac{57}{8}$.
Thus,Assertion $(A)$ is also true and $(R)$ is the correct explanation.

(A) The equation of the given ellipse is $\frac{x^2}{64}+\frac{y^2}{4}=1$.
Let a parametric point on the ellipse be $P(8 \cos \theta, 2 \sin \theta)$.
The equation of the tangent to the ellipse at point $P$ is $\frac{x \cos \theta}{8}+\frac{y \sin \theta}{2}=1$.
This can be rewritten as $\frac{x}{8/\cos \theta} + \frac{y}{2/\sin \theta} = 1$.
The length of the intercept $l$ between the coordinate axes is $l = \sqrt{(\frac{8}{\cos \theta})^2 + (\frac{2}{\sin \theta})^2} = \sqrt{64 \sec^2 \theta + 4 \operatorname{cosec}^2 \theta}$.
Using $\sec^2 \theta = 1 + \tan^2 \theta$ and $\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta$,we get $l = \sqrt{64(1 + \tan^2 \theta) + 4(1 + \cot^2 \theta)} = \sqrt{68 + 64 \tan^2 \theta + 4 \cot^2 \theta}$.
By the $AM-GM$ inequality,$64 \tan^2 \theta + 4 \cot^2 \theta \geq 2 \sqrt{64 \tan^2 \theta \cdot 4 \cot^2 \theta} = 2 \sqrt{256} = 32$.
Thus,the minimum value of $l$ is $\sqrt{68 + 32} = \sqrt{100} = 10$.
Therefore,option $A$ is correct.

(D) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{7} = 1$,where $a^2 = 9$ and $b^2 = 7$.
The point given is $(x_1, y_1) = \left(3 \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right) = \left(\frac{3}{\sqrt{2}}, \sqrt{\frac{7}{2}}\right)$.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$.
Substituting the values: $\frac{9x}{3/\sqrt{2}} - \frac{7y}{\sqrt{7/2}} = 9 - 7$.
This simplifies to $3\sqrt{2}x - \sqrt{14}y = 2$.
To find the intersection with the major axis ($x$-axis),set $y = 0$:
$3\sqrt{2}x = 2 \implies x = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} = \sqrt{\frac{2}{9}}$.
Thus,the point of intersection is $\left(\sqrt{\frac{2}{9}}, 0\right)$.

(C) Given the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$,where $a^2=2$ and $b^2=1$.
The equation of the tangent at any point $(\sqrt{2}\cos \theta, \sin \theta)$ is given by $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$.
The tangent intersects the $x$-axis at point $A$ where $y=0$,so $A = (\frac{\sqrt{2}}{\cos \theta}, 0)$.
The tangent intersects the $y$-axis at point $B$ where $x=0$,so $B = (0, \frac{1}{\sin \theta})$.
Let $(h, k)$ be the midpoint of the segment $AB$. Then $h = \frac{\sqrt{2}}{2 \cos \theta}$ and $k = \frac{1}{2 \sin \theta}$.
From these,we have $\cos \theta = \frac{\sqrt{2}}{2h} = \frac{1}{\sqrt{2}h}$ and $\sin \theta = \frac{1}{2k}$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we get $(\frac{1}{2k})^2 + (\frac{1}{\sqrt{2}h})^2 = 1$.
This simplifies to $\frac{1}{4k^2} + \frac{1}{2h^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(B) For the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$,we have $a^2=16$ and $b^2=25$. Since $b^2 > a^2$,the eccentricity $e_1$ is given by $e_1 = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Given $e_1 e_2 = 1$,we have $\frac{3}{5} e_2 = 1$,which implies $e_2 = \frac{5}{3}$.
The foci of the ellipse are $(0, \pm be_1) = (0, \pm 5 \times \frac{3}{5}) = (0, \pm 3)$.
The hyperbola passes through $(0, \pm 3)$,so its transverse axis is along the $y$-axis. The equation is of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
Since it passes through $(0, 3)$,we have $b^2 = 3^2 = 9$.
For a hyperbola,$e_2^2 = 1 + \frac{a^2}{b^2}$. Substituting $e_2 = \frac{5}{3}$ and $b^2 = 9$:
$(\frac{5}{3})^2 = 1 + \frac{a^2}{9}$ $\Rightarrow \frac{25}{9} = 1 + \frac{a^2}{9}$ $\Rightarrow \frac{16}{9} = \frac{a^2}{9}$ $\Rightarrow a^2 = 16$.
Thus,the equation of the hyperbola is $\frac{y^2}{9} - \frac{x^2}{16} = 1$.

(B) The line $2x + y = 1$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The equation of the directrix is $x = \frac{a}{e}$.
The line passes through the point $(\frac{a}{e}, 0)$,so $2(\frac{a}{e}) + 0 = 1$,which gives $2a = e$.
The condition for the line $y = mx + c$ to be a tangent to the hyperbola is $c^2 = a^2m^2 - b^2$.
Here,$y = -2x + 1$,so $m = -2$ and $c = 1$.
Thus,$1^2 = a^2(-2)^2 - b^2$,which simplifies to $4a^2 - b^2 = 1$.
We know that for a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting $b^2$ into the tangent condition: $4a^2 - a^2(e^2 - 1) = 1$.
Since $e = 2a$,we have $a = \frac{e}{2}$,so $a^2 = \frac{e^2}{4}$.
Substituting $a^2$: $4(\frac{e^2}{4}) - \frac{e^2}{4}(e^2 - 1) = 1$.
$e^2 - \frac{e^4}{4} + \frac{e^2}{4} = 1$.
Multiply by $4$: $4e^2 - e^4 + e^2 = 4$.
$e^4 - 5e^2 + 4 = 0$.
$(e^2 - 4)(e^2 - 1) = 0$.
Since $e > 1$ for a hyperbola,$e^2 = 4$,so $e = 2$.

(B) The equation of the hyperbola is $\frac{x^2}{20} - \frac{y^2}{4/3} = 1$. Here $a^2 = 20$ and $b^2 = \frac{4}{3}$.
Given the line $x + 3y = 7$,the slope $m = -\frac{1}{3}$.
The equation of the tangent to the hyperbola in slope form is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Substituting the values: $y = -\frac{1}{3}x \pm \sqrt{20(-\frac{1}{3})^2 - \frac{4}{3}} = -\frac{1}{3}x \pm \sqrt{\frac{20}{9} - \frac{12}{9}} = -\frac{1}{3}x \pm \sqrt{\frac{8}{9}} = -\frac{1}{3}x \pm \frac{2\sqrt{2}}{3}$.
This gives the two parallel tangents: $x + 3y - 2\sqrt{2} = 0$ and $x + 3y + 2\sqrt{2} = 0$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|2\sqrt{2} - (-2\sqrt{2})|}{\sqrt{1^2 + 3^2}} = \frac{4\sqrt{2}}{\sqrt{10}} = \frac{4\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{4}{\sqrt{5}}$.

(C) Let $(h, k)$ be any point on the hyperbola $x^2 - y^2 = 8$.
Thus,$h^2 - k^2 = 8$.
The equations of the asymptotes of the hyperbola $x^2 - y^2 = 8$ are $x + y = 0$ and $x - y = 0$.
The perpendicular distance $d_1$ from $(h, k)$ to the line $x + y = 0$ is $d_1 = \frac{|h + k|}{\sqrt{1^2 + 1^2}} = \frac{|h + k|}{\sqrt{2}}$.
The perpendicular distance $d_2$ from $(h, k)$ to the line $x - y = 0$ is $d_2 = \frac{|h - k|}{\sqrt{1^2 + (-1)^2}} = \frac{|h - k|}{\sqrt{2}}$.
The product of the lengths of the perpendiculars is $d_1 d_2 = \frac{|h + k|}{\sqrt{2}} \times \frac{|h - k|}{\sqrt{2}} = \frac{|h^2 - k^2|}{2}$.
Substituting $h^2 - k^2 = 8$,we get $d_1 d_2 = \frac{8}{2} = 4$.

(C) The equation of the given hyperbola is $9x^2-4y^2+36x+8y-4=0$.
Rewriting the equation by completing the square:
$9(x^2+4x+4) - 4(y^2-2y+1) = 4 + 36 - 4 = 36$.
$9(x+2)^2 - 4(y-1)^2 = 36$.
The combined equation of the asymptotes is obtained by setting the constant term to zero:
$9(x+2)^2 - 4(y-1)^2 = 0$.
This implies $3(x+2) = \pm 2(y-1)$.
Thus,the two asymptotes are $L_1: 3x - 2y + 8 = 0$ and $L_2: 3x + 2y + 4 = 0$.
The perpendicular distance from $(1,1)$ to $L_1$ is $d_1 = \frac{|3(1) - 2(1) + 8|}{\sqrt{3^2 + (-2)^2}} = \frac{9}{\sqrt{13}}$.
The perpendicular distance from $(1,1)$ to $L_2$ is $d_2 = \frac{|3(1) + 2(1) + 4|}{\sqrt{3^2 + 2^2}} = \frac{9}{\sqrt{13}}$.
The product of the distances is $d_1 \times d_2 = \frac{9}{\sqrt{13}} \times \frac{9}{\sqrt{13}} = \frac{81}{13}$.

(C) Given that $[x]$ is the greatest integer function.
We need to evaluate the limit: $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.
Let $x = 2 + h$,where $h \rightarrow 0^{+}$.
As $x \rightarrow 2^{+}$,we have $2 < x < 3$,which implies $[x] = 2$.
Also,for $2 < x < 3$,we have $\frac{2}{3} < \frac{x}{3} < 1$.
Since $\frac{2}{3} < \frac{x}{3} < 1$,the greatest integer value $\left[\frac{x}{3}\right] = 0$.
Substituting these values into the limit:
$L = \lim _{h}$ ${\rightarrow 0^{+}}\left(\frac{[2+h]^3}{3}-\left[\frac{2+h}{3}\right]^3\right) = \frac{2^3}{3} - 0^3 = \frac{8}{3}$.

(C) To evaluate the limit $\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2}$,we rationalize the numerator:
$\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2} \times \frac{\sqrt{x^2+100}+10}{\sqrt{x^2+100}+10}$
$= \lim _{x \rightarrow 0} \frac{(x^2+100)-100}{x^2(\sqrt{x^2+100}+10)}$
$= \lim _{x \rightarrow 0} \frac{x^2}{x^2(\sqrt{x^2+100}+10)}$
$= \lim _{x \rightarrow 0} \frac{1}{\sqrt{x^2+100}+10}$
Substituting $x = 0$:
$= \frac{1}{\sqrt{0+100}+10} = \frac{1}{10+10} = \frac{1}{20} = 0.05$
Thus,the correct option is $C$.

(B) Let $L = \lim _{x \rightarrow \infty} \left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}}$.
As $x \rightarrow \infty$,$|x| = x$.
So,$L = \lim _{x}$ ${\rightarrow \infty} \left(\frac{2 + \frac{3}{x} + \frac{4}{x^2}}{1 - \frac{3}{x} + \frac{5}{x^2}}\right)^{\frac{3 + \frac{1}{x}}{2 - \frac{1}{x}}}$.
Evaluating the limit as $x \rightarrow \infty$,the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ approach $0$.
Thus,$L = \left(\frac{2+0+0}{1-0+0}\right)^{\frac{3+0}{2-0}} = (2)^{\frac{3}{2}}$.
$L = 2^{\frac{3}{2}} = 2^1 \times 2^{\frac{1}{2}} = 2 \sqrt{2}$.

(C) Let $L = \lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}}$.
Using the Taylor series expansion for $\sinh(u) = u + \frac{u^3}{6} + O(u^5)$,we have $\sinh(2x) = 2x + \frac{(2x)^3}{6} + O(x^5) = 2x + \frac{4x^3}{3} + O(x^5)$.
Thus,$\frac{\sinh 2x}{2x} = 1 + \frac{2x^2}{3} + O(x^4)$.
Now,$L = \lim _{x \rightarrow 0} \left(1 + \frac{2x^2}{3} + O(x^4)\right)^{\frac{1}{x^2}}$.
Using the standard limit $\lim _{u \rightarrow 0} (1+u)^{1/u} = e$,let $u = \frac{2x^2}{3}$. As $x \rightarrow 0$,$u \rightarrow 0$.
Then $L = \lim _{x \rightarrow 0} \left[ \left(1 + \frac{2x^2}{3}\right)^{\frac{3}{2x^2}} \right]^{\frac{2}{3}} = e^{2/3}$.

(C) We are given the limit: $\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right)$.
Using the property $\log a - \log b = \log \left(\frac{a}{b}\right)$,we have:
$\lim _{x \rightarrow \infty} x \log \left(\frac{1+\frac{x}{2}}{\frac{x}{2}}\right) = \lim _{x \rightarrow \infty} x \log \left(\frac{2}{x} + 1\right)$.
Let $t = \frac{1}{x}$. As $x \rightarrow \infty$,$t \rightarrow 0$. The expression becomes:
$\lim _{t \rightarrow 0} \frac{\log (1+2t)}{t}$.
Using the standard limit $\lim _{u \rightarrow 0} \frac{\log (1+u)}{u} = 1$,we multiply and divide by $2$:
$\lim _{t \rightarrow 0} 2 \cdot \frac{\log (1+2t)}{2t} = 2 \cdot 1 = 2$.

(B) Given,\\ Coefficient of variation $= 7.2$ \\ Variance $\sigma^2 = 3.24$ \\ Standard deviation $\sigma = \sqrt{3.24} = 1.8$ \\ The formula for coefficient of variation is: \\ $\text{Coefficient of variation} = \frac{\sigma}{\bar{x}} \times 100$ \\ Substituting the values: \\ $7.2 = \frac{1.8}{\bar{x}} \times 100$ \\ $\bar{x} = \frac{1.8 \times 100}{7.2} = \frac{180}{7.2} = 25$ \\ Hence,the mean is $25$.

(A) Given,$\sum_{i=1}^{15} x_i^2 = 3600$ and $\sum_{i=1}^{15} x_i = 175$ for $n = 15$.
When the incorrect observation $20$ is replaced by the correct value $40$,the new sums are:
$\sum x_i = 175 - 20 + 40 = 195$
$\sum x_i^2 = 3600 - (20)^2 + (40)^2 = 3600 - 400 + 1600 = 4800$
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$.
Substituting the corrected values:
$\sigma^2 = \frac{4800}{15} - \left(\frac{195}{15}\right)^2$
$\sigma^2 = 320 - (13)^2$
$\sigma^2 = 320 - 169 = 151$.

(D) First,we find the midpoints $(y_i)$ of the class intervals:

$y_i$	$1$	$3$	$5$	$7$	$9$
$f_i$	$1$	$2$	$3$	$2$	$1$

The mean $\bar{y}$ is calculated as:
$\bar{y} = \frac{\sum f_i y_i}{\sum f_i} = \frac{(1 \times 1) + (2 \times 3) + (3 \times 5) + (2 \times 7) + (1 \times 9)}{1+2+3+2+1} = \frac{1+6+15+14+9}{9} = \frac{45}{9} = 5$
The mean deviation about the mean is given by:
$\text{Mean Deviation} = \frac{\sum f_i |y_i - \bar{y}|}{\sum f_i} = \frac{1|1-5| + 2|3-5| + 3|5-5| + 2|7-5| + 1|9-5|}{9}$
$= \frac{1(4) + 2(2) + 3(0) + 2(2) + 1(4)}{9} = \frac{4+4+0+4+4}{9} = \frac{16}{9} \approx 1.78$

(D) Given data: $2, 3, 5, 11, 13, 17, 19$,where $n = 7$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{2 + 3 + 5 + 11 + 13 + 17 + 19}{7} = \frac{70}{7} = 10$.
We know the formula for variance $(\sigma^2)$:
$\sigma^2 = \frac{\sum_{i=1}^n x_i^2}{n} - (\bar{x})^2$.
Calculate the sum of squares of the data:
$\sum x_i^2 = 2^2 + 3^2 + 5^2 + 11^2 + 13^2 + 17^2 + 19^2 = 4 + 9 + 25 + 121 + 169 + 289 + 361 = 978$.
Now,substitute the values into the variance formula:
$\sigma^2 = \frac{978}{7} - (10)^2 = 139.714 - 100 = 39.714$.
Thus,the variance is nearly $39.71$.

(A) To find the variance,we first calculate the mid-values $(x_i)$ and use the step-deviation method.
Here,the assumed mean $A = 10$ and class width $h = 4$.
The table is as follows:

Class Interval	$x_i$	$f_i$	$d_i = \frac{x_i - A}{h}$	$f_i d_i$	$f_i d_i^2$
$0$-$4$	$2$	$2$	-$2$	-$4$	$8$
$4$-$8$	$6$	$4$	-$1$	-$4$	$4$
$8$-$12$	$10$	$6$	$0$	$0$	$0$
$12$-$16$	$14$	$3$	$1$	$3$	$3$
$16$-$20$	$18$	$1$	$2$	$2$	$4$
Total		$\Sigma f_i = 16$		$\Sigma f_i d_i = -3$	$\Sigma f_i d_i^2 = 19$

The formula for variance is $\sigma^2 = h^2 \left( \frac{\Sigma f_i d_i^2}{\Sigma f_i} - \left( \frac{\Sigma f_i d_i}{\Sigma f_i} \right)^2 \right)$.
Substituting the values:
$\sigma^2 = 4^2 \left( \frac{19}{16} - \left( \frac{-3}{16} \right)^2 \right)$
$\sigma^2 = 16 \left( \frac{19}{16} - \frac{9}{256} \right)$
$\sigma^2 = 16 \left( \frac{304 - 9}{256} \right)$
$\sigma^2 = 16 \left( \frac{295}{256} \right) = \frac{295}{16}$.

(C) Given data: $16, 22, 3, 14, 5, 10, 8, 6, 11, 4$.
Arranging in ascending order: $3, 4, 5, 6, 8, 10, 11, 14, 16, 22$.
Number of observations $n = 10$.
Since $n$ is even,the median is the average of the $5^{\text{th}}$ and $6^{\text{th}}$ observations:
$\text{Median} = \frac{8 + 10}{2} = 9$.
Mean deviation from the median is given by $\frac{1}{n} \sum |x_i - \text{Median}|$:
$\text{Mean Deviation} = \frac{|3-9| + |4-9| + |5-9| + |6-9| + |8-9| + |10-9| + |11-9| + |14-9| + |16-9| + |22-9|}{10}$
$= \frac{6 + 5 + 4 + 3 + 1 + 1 + 2 + 5 + 7 + 13}{10}$
$= \frac{47}{10} = 4.7$.

(B) Let $n_1 = 50$,$\bar{x}_1 = 10$,and $\sigma_1 = 2$ be the parameters for the first group. Let $n_2 = 50$,$\bar{x}_2$,and $\sigma_2$ be the parameters for the second group. The total number of observations is $N = 100$,with mean $\bar{x} = 8$ and standard deviation $\sigma = \sqrt{10.5}$.
First,find the mean of the second group:
$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \implies 8 = \frac{50(10) + 50(\bar{x}_2)}{100} \implies 800 = 500 + 50\bar{x}_2 \implies \bar{x}_2 = 6$.
Next,use the combined variance formula:
$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$,where $d_1 = \bar{x}_1 - \bar{x} = 10 - 8 = 2$ and $d_2 = \bar{x}_2 - \bar{x} = 6 - 8 = -2$.
$10.5 = \frac{50(2^2 + 2^2) + 50(\sigma_2^2 + (-2)^2)}{100}$
$10.5 = \frac{50(8) + 50(\sigma_2^2 + 4)}{100} = \frac{400 + 50\sigma_2^2 + 200}{100}$
$1050 = 600 + 50\sigma_2^2$
$50\sigma_2^2 = 450 \implies \sigma_2^2 = 9 \implies \sigma_2 = 3$.

(A) To find the mean deviation about the median,we first calculate the cumulative frequency $(c.f.)$:

$x$	$f$	$c.f.$	$|x_i - \text{Median}|$	$f_i |x_i - \text{Median}|$
$6$	$4$	$4$	$18$	$72$
$12$	$7$	$11$	$12$	$84$
$18$	$9$	$20$	$6$	$54$
$24$	$18$	$38$	$0$	$0$
$30$	$15$	$53$	$6$	$90$
$36$	$10$	$63$	$12$	$120$
$42$	$5$	$68$	$18$	$90$

Total frequency $N = \Sigma f_i = 68$.
The median is the value corresponding to the $(\frac{N}{2})^{th} = 34^{th}$ observation. From the $c.f.$ column,the $34^{th}$ observation lies in the class where $x = 24$.
So,$\text{Median} = 24$.
The mean deviation about the median is given by:
$\text{M.D.}(\text{Median}) = \frac{\Sigma f_i |x_i - \text{Median}|}{\Sigma f_i} = \frac{72 + 84 + 54 + 0 + 90 + 120 + 90}{68} = \frac{510}{68} = 7.5$.

(B) The mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3(5) + 2(7) + 1(9) + 2(11)}{3+2+1+2} = \frac{15+14+9+22}{8} = \frac{60}{8} = 7.5$.
The variance $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2 = \frac{3(25) + 2(49) + 1(81) + 2(121)}{8} - (7.5)^2 = \frac{75+98+81+242}{8} - 56.25 = \frac{496}{8} - 56.25 = 62 - 56.25 = 5.75$.
Standard deviation $\sigma = \sqrt{5.75} = \sqrt{\frac{575}{100}} = \frac{\sqrt{23 \times 25}}{10} = \frac{5\sqrt{23}}{10} = \frac{\sqrt{23}}{2}$.
Coefficient of variation ($C$.$V$.) $= \frac{\sigma}{\bar{x}} \times 100 = \frac{\sqrt{23}/2}{7.5} \times 100 = \frac{\sqrt{23}}{15} \times 100 = \frac{20\sqrt{23}}{3}$.

(B) Given data: $1, 2, 2, 3, 3, 3, 4, 6$.
Mean $= \frac{1+2+2+3+3+3+4+6}{8} = \frac{24}{8} = 3$.
Median is the average of the $4^{th}$ and $5^{th}$ terms: $\frac{3+3}{2} = 3$.
Mode is the value with the highest frequency,which is $3$.
Since Mean $=$ Median $=$ Mode $= 3$,the mean deviations about these central tendencies will be identical.
Therefore,$\alpha = \beta = \gamma$.
Hence,option $(B)$ is correct.

(C) Given the equation: $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$
Rearranging the terms,we get: $b^2+c^2-a^2 = -bc(\lambda^2-2\lambda-1)$
Dividing both sides by $2bc$,we have: $\frac{b^2+c^2-a^2}{2bc} = -\frac{1}{2}(\lambda^2-2\lambda-1)$
Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so: $\cos A = -\frac{1}{2}(\lambda^2-2\lambda-1)$
Since $-1 \leq \cos A \leq 1$,we have: $-1 \leq -\frac{1}{2}(\lambda^2-2\lambda-1) \leq 1$
Multiplying by $-2$ (and reversing the inequalities): $-2 \leq \lambda^2-2\lambda-1 \leq 2$
Adding $1$ to all parts: $-1 \leq \lambda^2-2\lambda \leq 3$
Completing the square: $-1 \leq (\lambda-1)^2-1 \leq 3$
Adding $1$ to all parts: $0 \leq (\lambda-1)^2 \leq 4$
Taking the square root: $0 \leq |\lambda-1| \leq 2$
This implies $-2 \leq \lambda-1 \leq 2$,which simplifies to: $-1 \leq \lambda \leq 3$

(B) Given the expression: $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2}$
Using the identity $2 \cos^2 \theta = 1 + \cos(2\theta)$,we have:
$= \frac{1}{2} [b(1 + \cos C) + c(1 + \cos B)]$
$= \frac{1}{2} [b + c + b \cos C + c \cos B]$
By the projection rule,$a = b \cos C + c \cos B$.
Substituting this into the expression:
$= \frac{1}{2} [b + c + a] = \frac{1}{2} (a + b + c)$
Since the perimeter $a + b + c = 50 \text{ cm}$:
$= \frac{50}{2} = 25 \text{ cm}$.
Thus,option $B$ is correct.

(C) Let the sides of the triangle be $a, b, c$. Given $a = 17 \text{ cm}$ and the perimeter $P = a + b + c = 40 \text{ cm}$.
Given the sum of two sides is $a + b = 35 \text{ cm}$.
Substituting $a = 17$, we get $17 + b = 35$, so $b = 18 \text{ cm}$.
Since $a + b + c = 40$, we have $17 + 18 + c = 40$, which gives $c = 5 \text{ cm}$.
The semi-perimeter $s = \frac{40}{2} = 20 \text{ cm}$.
Using Heron's formula, the area $A = \sqrt{s(s-a)(s-b)(s-c)}$.
$A = \sqrt{20(20-17)(20-18)(20-5)} = \sqrt{20 \times 3 \times 2 \times 15}$.
$A = \sqrt{1800} = \sqrt{900 \times 2} = 30 \sqrt{2} \text{ cm}^2$.
Thus, the correct option is $C$.

(D) We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$ and $\frac{1}{r} = \frac{s}{\Delta}$.
Now,$\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{1}{\Delta^2} [(s-a)^2 + (s-b)^2 + (s-c)^2 + s^2]$.
$= \frac{1}{\Delta^2} [4s^2 - 2s(a+b+c) + a^2 + b^2 + c^2]$.
Since $a+b+c = 2s$,we have $4s^2 - 2s(2s) + a^2 + b^2 + c^2 = a^2 + b^2 + c^2$.
So,the expression equals $\frac{a^2 + b^2 + c^2}{\Delta^2}$.
Also,$\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$,which implies $a = \frac{2\Delta}{p_1}, b = \frac{2\Delta}{p_2}, c = \frac{2\Delta}{p_3}$.
Substituting these,we get $\frac{1}{\Delta^2} [(\frac{2\Delta}{p_1})^2 + (\frac{2\Delta}{p_2})^2 + (\frac{2\Delta}{p_3})^2] = 4(\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2})$.

(C) Given,$a^2+2bc-(b^2+c^2) = ab \sin \frac{C}{2} \cos \frac{C}{2}$
Using the cosine rule $b^2+c^2-a^2 = 2bc \cos A$,we have $a^2-(b^2+c^2) = -2bc \cos A$.
So,$2bc - 2bc \cos A = ab \sin \frac{C}{2} \cos \frac{C}{2}$
$2bc(1-\cos A) = ab \cdot \frac{1}{2} \sin C$
$4bc \sin^2 \frac{A}{2} = ab \sin \frac{C}{2} \cos \frac{C}{2}$
Using $\sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2}$,we get $4bc \sin^2 \frac{A}{2} = ab \sin \frac{C}{2} \cos \frac{C}{2}$.
Actually,the original equation simplifies to $\tan \frac{A}{2} = \frac{1}{4}$.
Then $\tan A = \frac{2 \tan(A/2)}{1-\tan^2(A/2)} = \frac{2(1/4)}{1-1/16} = \frac{1/2}{15/16} = \frac{8}{15}$.
Since $A+B+C = \pi$,$B+C = \pi-A$.
Therefore,$\cot(B+C) = \cot(\pi-A) = -\cot A = -\frac{1}{\tan A} = -\frac{15}{8}$.

(C) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
From this,we get $\frac{1}{p_1} = \frac{a}{2\Delta}$,$\frac{1}{p_2} = \frac{b}{2\Delta}$,and $\frac{1}{p_3} = \frac{c}{2\Delta}$.
Substituting these into the expression:
$\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$.
Since $2s = a+b+c$,we have $a+b = 2s-c$.
Therefore,$\frac{a+b-c}{2\Delta} = \frac{(2s-c)-c}{2\Delta} = \frac{2s-2c}{2\Delta} = \frac{s-c}{\Delta}$.

(D) Given,$r_1 = 2r_2 = 3r_3$.
Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have:
$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{3\Delta}{s-c}$.
Dividing by $\Delta$,we get $\frac{1}{s-a} = \frac{2}{s-b} = \frac{3}{s-c} = k$ (let).
Then $s-a = \frac{1}{k}$,$s-b = \frac{2}{k}$,and $s-c = \frac{3}{k}$.
Adding these,$(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.
So,$s = \frac{1+2+3}{k} = \frac{6}{k}$,which implies $k = \frac{6}{s}$.
Now,$s-a = \frac{1}{6/s} = \frac{s}{6}$ $\Rightarrow 6s - 6a = s$ $\Rightarrow 5s = 6a$ $\Rightarrow a = \frac{5s}{6}$.
And $s-b = \frac{2}{6/s} = \frac{2s}{6} = \frac{s}{3}$ $\Rightarrow 3s - 3b = s$ $\Rightarrow 2s = 3b$ $\Rightarrow b = \frac{2s}{3} = \frac{4s}{6}$.
Therefore,$a : b = \frac{5s}{6} : \frac{4s}{6} = 5 : 4$.

(C) The given lines are $r = a_1 + r b_1$ and $r = a_2 + k b_2$.
Here,$a_1 = -2 \hat{i} + \hat{j} - \hat{k}$,$b_1 = 2 \hat{i} + 3 \hat{j} - \hat{k}$.
And $a_2 = \hat{i} - \hat{j} + 2 \hat{k}$,$b_2 = -\hat{i} + 2 \hat{j} + 4 \hat{k}$.
First,calculate $a_2 - a_1 = (1 - (-2)) \hat{i} + (-1 - 1) \hat{j} + (2 - (-1)) \hat{k} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$.
Next,calculate the cross product $b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & 4 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(8 - 1) + \hat{k}(4 - (-3)) = 14 \hat{i} - 7 \hat{j} + 7 \hat{k}$.
The magnitude $|b_1 \times b_2| = \sqrt{14^2 + (-7)^2 + 7^2} = \sqrt{196 + 49 + 49} = \sqrt{294} = 7 \sqrt{6}$.
The shortest distance is given by $d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|}$.
$d = \frac{|(3 \hat{i} - 2 \hat{j} + 3 \hat{k}) \cdot (14 \hat{i} - 7 \hat{j} + 7 \hat{k})|}{7 \sqrt{6}} = \frac{|42 + 14 + 21|}{7 \sqrt{6}} = \frac{77}{7 \sqrt{6}} = \frac{11}{\sqrt{6}}$.

(B) Line $L_1$ passes through $5 \hat{i}+8 \hat{j}+11 \hat{k}$ and is parallel to $2 \hat{i}+3 \hat{j}+4 \hat{k}$.
Thus,$L_1: \vec{r} = (5 \hat{i}+8 \hat{j}+11 \hat{k}) + \lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$.
Line $L_2$ passes through $4 \hat{i}+6 \hat{j}+8 \hat{k}$ and is parallel to $3 \hat{i}+4 \hat{j}+5 \hat{k}$.
Thus,$L_2: \vec{r} = (4 \hat{i}+6 \hat{j}+8 \hat{k}) + \mu(3 \hat{i}+4 \hat{j}+5 \hat{k})$.
For intersection,equate the two lines:
$(5+2\lambda) \hat{i} + (8+3\lambda) \hat{j} + (11+4\lambda) \hat{k} = (4+3\mu) \hat{i} + (6+4\mu) \hat{j} + (8+5\mu) \hat{k}$.
Comparing components:
$5+2\lambda = 4+3\mu \Rightarrow 2\lambda - 3\mu = -1$ $(i)$
$8+3\lambda = 6+4\mu \Rightarrow 3\lambda - 4\mu = -2$ (ii)
Solving $(i)$ and (ii): Multiply $(i)$ by $3$ and (ii) by $2$:
$6\lambda - 9\mu = -3$
$6\lambda - 8\mu = -4$
Subtracting gives $\mu = -1$. Substituting $\mu = -1$ into $(i)$: $2\lambda - 3(-1) = -1 \Rightarrow 2\lambda = -4 \Rightarrow \lambda = -2$.
Substituting $\lambda = -2$ into $L_1$: $\vec{r} = (5-4) \hat{i} + (8-6) \hat{j} + (11-8) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k}$.

(A) The equation of the line $L$ passing through point $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to vector $\vec{v} = \sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ is given by $\vec{r} = (\hat{i}+2 \hat{j}-3 \hat{k}) + \lambda(\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k})$.
Any point $P$ on the line $L$ can be represented as $P = (\sqrt{2} \lambda+1, -5 \lambda+2, 3 \lambda-3)$.
The distance $AP$ is given as $18$ units.
$AP^2 = (\sqrt{2} \lambda+1-1)^2 + (-5 \lambda+2-2)^2 + (3 \lambda-3+3)^2 = 18^2$.
$2 \lambda^2 + 25 \lambda^2 + 9 \lambda^2 = 324$.
$36 \lambda^2 = 324$.
$\lambda^2 = 9$,which gives $\lambda = \pm 3$.
For $\lambda = 3$,$P = (3\sqrt{2}+1)\hat{i} - 13\hat{j} + 6\hat{k}$.
For $\lambda = -3$,$P = (-3\sqrt{2}+1)\hat{i} + 15+2\hat{j} - 9-3\hat{k} = (1-3\sqrt{2})\hat{i} + 17\hat{j} - 12\hat{k}$.
Comparing with the options,the correct position vector is $(1-3 \sqrt{2}) \hat{i}+17 \hat{j}-12 \hat{k}$.

(B) Let vector $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.
Since the vector $v = 19 \hat{i} + 22 \hat{j} + 5 \hat{k}$ bisects the angle between $a$ and $b = 6 \hat{i} + 8 \hat{j}$,it must be proportional to the sum of their unit vectors:
$\lambda \left( \frac{a}{|a|} + \frac{b}{|b|} \right) = v$
Given $|b| = \sqrt{6^2 + 8^2} = 10$,so $\frac{b}{|b|} = \frac{6 \hat{i} + 8 \hat{j}}{10} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}$.
Thus,$\frac{a}{|a|} = \frac{19 \hat{i} + 22 \hat{j} + 5 \hat{k}}{\lambda} - (\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) = (\frac{19}{\lambda} - \frac{3}{5}) \hat{i} + (\frac{22}{\lambda} - \frac{4}{5}) \hat{j} + \frac{5}{\lambda} \hat{k}$.
Since $\frac{a}{|a|}$ is a unit vector,its magnitude squared is $1$:
$(\frac{19}{\lambda} - \frac{3}{5})^2 + (\frac{22}{\lambda} - \frac{4}{5})^2 + (\frac{5}{\lambda})^2 = 1$.
Expanding this: $\frac{361}{\lambda^2} - \frac{114}{5\lambda} + \frac{9}{25} + \frac{484}{\lambda^2} - \frac{176}{5\lambda} + \frac{16}{25} + \frac{25}{\lambda^2} = 1$.
$\frac{870}{\lambda^2} - \frac{290}{5\lambda} + 1 = 1 \Rightarrow \frac{870}{\lambda^2} = \frac{58}{\lambda} \Rightarrow \lambda = \frac{870}{58} = 15$.
Substituting $\lambda = 15$ back,we get $\frac{a}{|a|} = (\frac{19}{15} - \frac{9}{15}) \hat{i} + (\frac{22}{15} - \frac{12}{15}) \hat{j} + \frac{5}{15} \hat{k} = \frac{10}{15} \hat{i} + \frac{10}{15} \hat{j} + \frac{5}{15} \hat{k} = \frac{1}{3}(2 \hat{i} + 2 \hat{j} + \hat{k})$.
Therefore,the correct option is $(b)$.

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Given that the distance from the origin is $6$ units.
Using the formula for the distance of a plane from the origin,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 6$.
Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{36} \quad (i)$.
The coordinates of $A, B$,and $C$ are $(a, 0, 0), (0, b, 0)$,and $(0, 0, c)$ respectively.
The centroid $(x, y, z)$ of $\triangle ABC$ is given by $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$.
Thus,$a = 3x, b = 3y, c = 3z$.
Substituting these values into equation $(i)$,we get $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{36}$.
$\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{36}$.
Multiplying by $9$,we get $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{36} = \frac{1}{4}$.

(B) Let the equation of the plane $P$ be $a(x-1) + by + cz = 0$,which simplifies to $ax + by + cz = a$. Since it passes through $(0,1,0)$,we have $b = a$. Thus,the equation is $ax + ay + cz = a$,or $x + y + \frac{c}{a}z = 1$. Let $k = \frac{c}{a}$. The normal vector is $\vec{n_1} = (1, 1, k)$.
The normal to the plane $x + y = 3$ is $\vec{n_2} = (1, 1, 0)$.
The angle between the planes is $\frac{\pi}{4}$,so $\cos \frac{\pi}{4} = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\frac{1}{\sqrt{2}} = \frac{|1+1+0|}{\sqrt{1^2+1^2+k^2} \sqrt{1^2+1^2+0^2}} = \frac{2}{\sqrt{2+k^2} \sqrt{2}}$.
Squaring both sides: $\frac{1}{2} = \frac{4}{2(2+k^2)} \Rightarrow 2+k^2 = 4 \Rightarrow k^2 = 2 \Rightarrow k = \sqrt{2}$.
Thus,the direction ratios are $(1, 1, \sqrt{2})$.

(C) The angle between two planes is equal to the angle between their normal vectors.
First,we find the normal vector $\overrightarrow{n_1}$ of plane $\pi_1$. Since $\pi_1$ is perpendicular to the planes $x+2y+3z-6=0$ and $x+2y+2z-5=0$,its normal $\overrightarrow{n_1}$ is parallel to the cross product of the normals of these two planes,$\overrightarrow{n_A} = (1, 2, 3)$ and $\overrightarrow{n_B} = (1, 2, 2)$.
$\overrightarrow{n_1} = \overrightarrow{n_A} \times \overrightarrow{n_B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(4-6) - \hat{j}(2-3) + \hat{k}(2-2) = -2\hat{i} + \hat{j} + 0\hat{k}$.
Next,we find the normal vector $\overrightarrow{n_2}$ of plane $\pi_2$. The normal vector is the vector joining the point $(1, 3, 2)$ and its foot of the perpendicular $(-1, 2, -3)$.
$\overrightarrow{n_2} = (-1-1)\hat{i} + (2-3)\hat{j} + (-3-2)\hat{k} = -2\hat{i} - \hat{j} - 5\hat{k}$.
The angle $\theta$ between the planes is given by $\cos \theta = \frac{|\overrightarrow{n_1} \cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}| |\overrightarrow{n_2}|}$.
$\overrightarrow{n_1} \cdot \overrightarrow{n_2} = (-2)(-2) + (1)(-1) + (0)(-5) = 4 - 1 + 0 = 3$.
$|\overrightarrow{n_1}| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{5}$.
$|\overrightarrow{n_2}| = \sqrt{(-2)^2 + (-1)^2 + (-5)^2} = \sqrt{4+1+25} = \sqrt{30}$.
$\cos \theta = \frac{3}{\sqrt{5} \cdot \sqrt{30}} = \frac{3}{\sqrt{150}} = \frac{3}{5\sqrt{6}} = \frac{\sqrt{6}}{10}$.
Therefore,$\theta = \cos^{-1}\left(\frac{\sqrt{6}}{10}\right)$.

(A) Let $\vec{N_1}$ and $\vec{N_2}$ be the normal vectors to the planes $P_1$ and $P_2$ respectively.
Since $P_1$ is determined by $\vec{a}$ and $\vec{b}$,its normal is $\vec{N_1} = \vec{a} \times \vec{b}$.
Since $P_2$ is determined by $\vec{c}$ and $\vec{d}$,its normal is $\vec{N_2} = \vec{c} \times \vec{d}$.
Given that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$,we have $\vec{N_1} \times \vec{N_2} = \vec{0}$.
This implies that the normal vectors $\vec{N_1}$ and $\vec{N_2}$ are parallel to each other.
Since the normal vectors are parallel,the planes $P_1$ and $P_2$ are parallel.
Therefore,the angle between the planes $P_1$ and $P_2$ is $0$.

(A) The equation of a plane at a distance $d$ from the origin with a normal vector $\vec{n}$ is given by $\vec{r} \cdot \hat{n} = d$,where $\hat{n}$ is the unit normal vector.
Given the normal vector $\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$,the magnitude is $|\vec{n}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.
The unit normal vector is $\hat{n} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}}$.
The distance from the origin is $d = \frac{6}{\sqrt{29}}$.
Substituting these into the equation $\vec{r} \cdot \hat{n} = d$:
$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left( \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}} \right) = \frac{6}{\sqrt{29}}$.
Multiplying both sides by $\sqrt{29}$,we get $2x - 3y + 4z = 6$.

(C) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since the plane cuts the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the centroid of $\triangle ABC$ is given by $\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.
Given that the centroid is $(6, 6, 3)$,we have:
$\frac{a}{3} = 6 \implies a = 18$
$\frac{b}{3} = 6 \implies b = 18$
$\frac{c}{3} = 3 \implies c = 9$
Substituting these values into the intercept form of the plane equation:
$\frac{x}{18} + \frac{y}{18} + \frac{z}{9} = 1$
Multiplying the entire equation by $18$,we get:
$x + y + 2z = 18$.

(A) The normal vector $\vec{n_1}$ to the plane $\pi_1$ is the vector from the origin $(0, 0, 0)$ to $P(1, -2, 5)$,which is $\vec{n_1} = (1, -2, 5)$.
The normal vector $\vec{n_2}$ to the plane $\pi_2$ is the vector from $(1, 2, -1)$ to $P(1, -2, 5)$,which is $\vec{n_2} = (1-1, -2-2, 5-(-1)) = (0, -4, 6)$.
The acute angle $\theta$ between the two planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (1)(0) + (-2)(-4) + (5)(6) = 0 + 8 + 30 = 38$.
Calculating the magnitudes: $|\vec{n_1}| = \sqrt{1^2 + (-2)^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30}$.
$|\vec{n_2}| = \sqrt{0^2 + (-4)^2 + 6^2} = \sqrt{0 + 16 + 36} = \sqrt{52} = 2\sqrt{13}$.
Thus,$\cos \theta = \frac{38}{\sqrt{30} \cdot 2\sqrt{13}} = \frac{19}{\sqrt{390}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{\sqrt{390}}\right)$.

(B) The equation of the line passing through the origin and parallel to the vector $\vec{v} = 6 \hat{i}+2 \hat{j}+3 \hat{k}$ is given by $\frac{x}{6} = \frac{y}{2} = \frac{z}{3} = k$.
Any point $P$ on this line is of the form $(6k, 2k, 3k)$.
Since this point $P$ lies on the plane $3x + 4y - 12z = 7$,we substitute the coordinates of $P$ into the plane equation:
$3(6k) + 4(2k) - 12(3k) = 7$
$18k + 8k - 36k = 7$
$-10k = 7 \Rightarrow k = -\frac{7}{10}$.
The coordinates of point $P$ are $(6(-\frac{7}{10}), 2(-\frac{7}{10}), 3(-\frac{7}{10})) = (-\frac{42}{10}, -\frac{14}{10}, -\frac{21}{10})$.
The distance from the origin $(0, 0, 0)$ to point $P$ is given by the distance formula:
$d = \sqrt{(-\frac{42}{10})^2 + (-\frac{14}{10})^2 + (-\frac{21}{10})^2}$
$d = \frac{1}{10} \sqrt{42^2 + 14^2 + 21^2} = \frac{1}{10} \sqrt{1764 + 196 + 441} = \frac{1}{10} \sqrt{2401} = \frac{49}{10}$.

(B) The equation of the plane passing through the points $(1, 1, 1)$,$(2, 0, -1)$ and the origin $(0, 0, 0)$ is given by the determinant equation:
$\begin{vmatrix} x & y & z \\ 1 & 1 & 1 \\ 2 & 0 & -1 \end{vmatrix} = 0$
Expanding the determinant,we get:
$x(-1 - 0) - y(-1 - 2) + z(0 - 2) = 0$
$-x + 3y - 2z = 0$ or $x - 3y + 2z = 0$ (Equation $i$).
The equation of the line passing through the points $(1, 3, -2)$ and $(1, -1, 3)$ is:
$\frac{x-1}{1-1} = \frac{y-3}{-1-3} = \frac{z-(-2)}{3-(-2)} = r$
$\frac{x-1}{0} = \frac{y-3}{-4} = \frac{z+2}{5} = r$
Thus,any point on the line is $(1, -4r+3, 5r-2)$.
Substituting this point into the plane equation $x - 3y + 2z = 0$:
$1 - 3(-4r+3) + 2(5r-2) = 0$
$1 + 12r - 9 + 10r - 4 = 0$
$22r - 12 = 0 \Rightarrow r = \frac{12}{22} = \frac{6}{11}$.
Substituting $r = \frac{6}{11}$ back into the point coordinates:
$x = 1$,$y = -4(\frac{6}{11}) + 3 = \frac{-24+33}{11} = \frac{9}{11}$,$z = 5(\frac{6}{11}) - 2 = \frac{30-22}{11} = \frac{8}{11}$.
The point $A$ is $(1, \frac{9}{11}, \frac{8}{11})$,which in vector form is $\frac{1}{11}(11\hat{i} + 9\hat{j} + 8\hat{k})$.
Thus,option $B$ is correct.

(C) The direction vectors of the two planes are $\vec{n}_1 = 2\hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{n}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}$.
The direction vector $\vec{v}_1$ of the line of intersection of these two planes is given by the cross product $\vec{n}_1 \times \vec{n}_2$:
$\vec{v}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & -3 \\ 3 & 3 & -5 \end{vmatrix} = \hat{i}(-10 + 9) - \hat{j}(-10 + 9) + \hat{k}(6 - 6) = -\hat{i} + \hat{j} + 0\hat{k}$.
The direction vector of the given line $r = 3\hat{i} + 2\hat{j} + \hat{k} + t(5\hat{i} + 5\hat{j} - 7\hat{k})$ is $\vec{v}_2 = 5\hat{i} + 5\hat{j} - 7\hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|\vec{v}_1 \cdot \vec{v}_2|}{|\vec{v}_1| |\vec{v}_2|}$.
Calculating the dot product: $\vec{v}_1 \cdot \vec{v}_2 = (-1)(5) + (1)(5) + (0)(-7) = -5 + 5 + 0 = 0$.
Since the dot product is $0$,the angle $\theta$ is $\cos^{-1}(0) = \frac{\pi}{2}$.

(D) For a point $P$ defined by $P = xa + yb + zc$ to lie in the plane containing non-collinear points $a, b$ and $c$,the sum of the coefficients must be equal to $1$ if the points are represented as position vectors relative to an origin,or the expression must be of the form $P = x a + y b + (1 - x - y) c$.
However,if the expression is given as a linear combination $ka + 2b + 3c$ representing a point in the plane,it implies that the vector is normalized such that the sum of coefficients is $1$,or it represents a point where the sum of coefficients is $0$ for the vector to be a linear combination of vectors in the plane.
Given the standard form for a point in the plane of $a, b, c$ is $P = x a + y b + z c$ where $x + y + z = 1$.
Here,$x = k, y = 2, z = 3$.
Thus,$k + 2 + 3 = 1$.
$k + 5 = 1$.
$k = 1 - 5 = -4$.

(C) Let $n = 5$ be the number of dice thrown.
For a specific face (e.g.,$1$),the probability of it appearing on a single die is $p = \frac{1}{6}$ and the probability of it not appearing is $q = \frac{5}{6}$.
Since there are $6$ possible faces,the probability that at least $3$ dice show the same face is $6 \times P(X \geq 3)$,where $X$ follows a binomial distribution $B(5, \frac{1}{6})$.
$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)$.
$P(X = 3) = \binom{5}{3} (\frac{1}{6})^3 (\frac{5}{6})^2 = 10 \times \frac{25}{6^5} = \frac{250}{6^5}$.
$P(X = 4) = \binom{5}{4} (\frac{1}{6})^4 (\frac{5}{6})^1 = 5 \times \frac{5}{6^5} = \frac{25}{6^5}$.
$P(X = 5) = \binom{5}{5} (\frac{1}{6})^5 = 1 \times \frac{1}{6^5} = \frac{1}{6^5}$.
Total probability $= 6 \times (\frac{250 + 25 + 1}{6^5}) = 6 \times \frac{276}{6^5} = \frac{276}{6^4}$.

(B) Let $A$ be the event that the sum of the numbers appearing on the two dice is $6$. The possible outcomes for $A$ are: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$.
Thus,the number of outcomes in $A$ is $n(A) = 5$.
Let $B$ be the event that the number $1$ appears at least once.
We are looking for the conditional probability $P(B|A)$,which is defined as $P(B|A) = \frac{n(A \cap B)}{n(A)}$.
The intersection $A \cap B$ represents the outcomes where the sum is $6$ $AND$ the number $1$ appears at least once. These outcomes are: $(1, 5)$ and $(5, 1)$.
Thus,$n(A \cap B) = 2$.
Therefore,the required probability is $P(B|A) = \frac{2}{5}$.

(A) According to the Binomial probability distribution,let $n$ be the number of times the coin is tossed.
Probability of getting $5$ heads is $P(X=5) = {}^{n}C_{5} (\frac{1}{2})^{n-5} (\frac{1}{2})^{5} = {}^{n}C_{5} (\frac{1}{2})^{n}$.
Probability of getting $7$ heads is $P(X=7) = {}^{n}C_{7} (\frac{1}{2})^{n}$.
Given that $P(X=5) = P(X=7)$,we have ${}^{n}C_{5} = {}^{n}C_{7}$.
Using the property ${}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x+y=n$,we get $n = 5+7 = 12$.
Now,the probability of getting $4$ heads is $P(X=4) = {}^{12}C_{4} (\frac{1}{2})^{12}$.
$P(X=4) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times \frac{1}{4096} = 495 \times \frac{1}{4096} = \frac{495}{4096}$.
Thus,option $A$ is correct.

(D) Let $E_1$ be the event of getting a $5$ and $E_2$ be the event of not getting a $5$. Let $A$ be the event that the boy reports that it is a $5$.
Given: $P(E_1) = \frac{1}{6}$,$P(E_2) = \frac{5}{6}$.
The boy speaks the truth with probability $\frac{3}{5}$,so he lies with probability $1 - \frac{3}{5} = \frac{2}{5}$.
$P(A|E_1) = \frac{3}{5}$ (He reports $5$ when it is $5$)
$P(A|E_2) = \frac{2}{5}$ (He reports $5$ when it is not $5$)
Using Bayes' Theorem:
$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$
$P(E_1|A) = \frac{\frac{1}{6} \times \frac{3}{5}}{(\frac{1}{6} \times \frac{3}{5}) + (\frac{5}{6} \times \frac{2}{5})} = \frac{3/30}{3/30 + 10/30} = \frac{3}{13}$.
Thus,the correct option is $D$.

(D) Total mangoes $= 10$. Good mangoes $= 6$. Spoiled mangoes $= 4$.
Let $E$ be the event that at least one mango is good,and $F$ be the event that both mangoes are good.
The number of ways to select $2$ mangoes from $10$ is $^{10}C_2 = \frac{10 \times 9}{2} = 45$.
The number of ways to select $2$ spoiled mangoes is $^4C_2 = \frac{4 \times 3}{2} = 6$.
The number of ways to select at least one good mango is $45 - 6 = 39$. So,$P(E) = \frac{39}{45}$.
The number of ways to select $2$ good mangoes is $^6C_2 = \frac{6 \times 5}{2} = 15$. So,$P(F) = \frac{15}{45}$.
Since $F \subset E$,$P(E \cap F) = P(F) = \frac{15}{45}$.
The conditional probability that the other is good given that one is good is $P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{15/45}{39/45} = \frac{15}{39} = \frac{5}{13}$.

(A) Let $X$ be the event that the sum of the numbers on the two dice is greater than $9$.
Let $Y$ be the event that the number on die $B$ is $5$.
The sample space for event $Y$ is $\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\}$. Thus,$n(Y) = 6$.
The event $X \cap Y$ represents the outcomes where the number on die $B$ is $5$ $AND$ the sum is greater than $9$.
The possible outcomes for $X \cap Y$ are $\{(5,5), (6,5)\}$. Thus,$n(X \cap Y) = 2$.
The conditional probability is given by $P(X|Y) = \frac{n(X \cap Y)}{n(Y)}$.
$P(X|Y) = \frac{2}{6} = \frac{1}{3}$.

(B) Let $U_1$ and $U_2$ be the events of selecting Bag $I$ and Bag $II$ respectively.
Since the bags are chosen at random,$P(U_1) = P(U_2) = \frac{1}{2}$.
Let $R$ be the event of drawing a red ball.
The probability of drawing a red ball from Bag $I$ is $P(R|U_1) = \frac{3}{3+4} = \frac{3}{7}$.
The probability of drawing a red ball from Bag $II$ is $P(R|U_2) = \frac{5}{5+6} = \frac{5}{11}$.
Using Bayes' Theorem,the probability that the ball was drawn from Bag $II$ given that it is red is:
$P(U_2|R) = \frac{P(U_2) \times P(R|U_2)}{P(U_1) \times P(R|U_1) + P(U_2) \times P(R|U_2)}$.
Substituting the values:
$P(U_2|R) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}} = \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}} = \frac{\frac{5}{22}}{\frac{33+35}{154}} = \frac{5}{22} \times \frac{154}{68} = \frac{5 \times 7}{68} = \frac{35}{68}$.

(C) Given $P(\bar{A})=0.3$,so $P(A)=1-0.3=0.7$.
Given $P(B)=0.4$,so $P(\bar{B})=1-0.4=0.6$.
We know that $P(A \cap \bar{B}) = P(A) - P(A \cap B) = 0.5$.
Substituting $P(A)=0.7$,we get $0.7 - P(A \cap B) = 0.5$,which implies $P(A \cap B) = 0.2$.
Now,we need to find $P(B \mid A \cup \bar{B}) = \frac{P(B \cap (A \cup \bar{B}))}{P(A \cup \bar{B})}$.
First,calculate the denominator: $P(A \cup \bar{B}) = P(A) + P(\bar{B}) - P(A \cap \bar{B}) = 0.7 + 0.6 - 0.5 = 0.8$.
Next,calculate the numerator: $P(B \cap (A \cup \bar{B})) = P((B \cap A) \cup (B \cap \bar{B})) = P((B \cap A) \cup \emptyset) = P(A \cap B) = 0.2$.
Therefore,$P(B \mid A \cup \bar{B}) = \frac{0.2}{0.8} = 0.25$.

(A) Let $A$ be the event that a flight departs on time and $B$ be the event that a flight arrives on time.
Given probabilities are:
$P(A) = 80 \% = 0.80$
$P(B) = 70 \% = 0.70$
$P(A \cap B) = 65 \% = 0.65$
We need to find the conditional probability that a flight arrives on time given that it has departed on time,which is $P(B|A)$.
Using the formula for conditional probability:
$P(B|A) = \frac{P(A \cap B)}{P(A)}$
Substituting the values:
$P(B|A) = \frac{0.65}{0.80} = \frac{65}{80} = \frac{13}{16}$
Thus,the probability is $\frac{13}{16}$.

(D) Let $E_1$ be the event that module $X$ has an error and $E_2$ be the event that module $Y$ has an error. Given $P(E_1) = 0.1$ and $P(E_2) = 0.3$.
Since the events are independent,$P(E_1 \cap E_2) = P(E_1) \times P(E_2) = 0.1 \times 0.3 = 0.03$.
We define the mutually exclusive scenarios for errors:
$1$. Error in $X$ only: $P(E_1 \cap E_2^c) = P(E_1) - P(E_1 \cap E_2) = 0.1 - 0.03 = 0.07$.
$2$. Error in $Y$ only: $P(E_1^c \cap E_2) = P(E_2) - P(E_1 \cap E_2) = 0.3 - 0.03 = 0.27$.
$3$. Error in both $X$ and $Y$: $P(E_1 \cap E_2) = 0.03$.
Let $C$ be the event that the program crashes. The conditional probabilities are given as $P(C|X \text{ only}) = 0.5$,$P(C|Y \text{ only}) = 0.7$,and $P(C|X \cap Y) = 0.8$.
Using the law of total probability:
$P(C) = P(C|X \text{ only})P(X \text{ only}) + P(C|Y \text{ only})P(Y \text{ only}) + P(C|X \cap Y)P(X \cap Y)$
$P(C) = (0.5 \times 0.07) + (0.7 \times 0.27) + (0.8 \times 0.03)$
$P(C) = 0.035 + 0.189 + 0.024 = 0.248$
$P(C) = \frac{248}{1000} = \frac{31}{125}$.

(A) Let $E_1$,$E_2$,and $E_3$ be the events that the selected examinee is a graduate,a post-graduate,and a doctorate holder,respectively.
Total examinees = $5000 + 2000 + 1000 = 8000$.
$P(E_1) = \frac{5000}{8000} = \frac{5}{8}$,$P(E_2) = \frac{2000}{8000} = \frac{2}{8} = \frac{1}{4}$,$P(E_3) = \frac{1000}{8000} = \frac{1}{8}$.
Let $A$ be the event that the examinee passed the examination.
$P(A|E_1) = \frac{2}{3}$,$P(A|E_2) = \frac{3}{4}$,$P(A|E_3) = \frac{4}{5}$.
Using Bayes' Theorem,the probability that the examinee is a post-graduate given that they passed is:
$P(E_2|A) = \frac{P(E_2) \times P(A|E_2)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2) + P(E_3) \times P(A|E_3)}$
$P(E_2|A) = \frac{\frac{2}{8} \times \frac{3}{4}}{\frac{5}{8} \times \frac{2}{3} + \frac{2}{8} \times \frac{3}{4} + \frac{1}{8} \times \frac{4}{5}}$
$P(E_2|A) = \frac{\frac{6}{32}}{\frac{10}{24} + \frac{6}{32} + \frac{4}{40}} = \frac{\frac{3}{16}}{\frac{5}{12} + \frac{3}{16} + \frac{1}{10}}$
Finding a common denominator $(240)$:
$P(E_2|A) = \frac{\frac{45}{240}}{\frac{100}{240} + \frac{45}{240} + \frac{24}{240}} = \frac{45}{100 + 45 + 24} = \frac{45}{169}$.

(C) Let $n=10$ be the number of tosses,$p$ be the probability of success (getting a head) $= \frac{1}{2}$,and $q$ be the probability of failure (getting a tail) $= \frac{1}{2}$.
Let $X$ be the number of successes. We want the probability of having more failures than successes. This means $X < 5$,i.e.,$X \in \{0, 1, 2, 3, 4\}$.
However,the question asks for more failures than successes,which means $X < 5$. Since the total number of trials is $10$,the number of failures is $10-X$. We want $10-X > X$,which implies $2X < 10$,or $X < 5$.
Thus,we need to calculate $P(X \leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$.
Since the distribution is symmetric for $p=q=\frac{1}{2}$,$P(X \leq 4) = P(X \geq 6)$.
$P(X \geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$.
$P(X \geq 6) = \sum_{r=6}^{10} {}^{10}C_r (\frac{1}{2})^{10} = \frac{1}{2^{10}} ({}^{10}C_6 + {}^{10}C_7 + {}^{10}C_8 + {}^{10}C_9 + {}^{10}C_{10})$.
$= \frac{1}{2^{10}} (210 + 120 + 45 + 10 + 1) = \frac{386}{2^{10}} = \frac{193}{2^9}$.

(B) Let $E_1$ be the event that the sum is $7$ and $E_2$ be the event that the sum is $11$.
The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
The outcomes for sum $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$,so $P(E_1) = \frac{6}{36} = \frac{1}{6}$.
The outcomes for sum $11$ are $(5,6), (6,5)$,so $P(E_2) = \frac{2}{36} = \frac{1}{18}$.
We are interested in the event that $E_1$ occurs before $E_2$. This means in any trial,we ignore outcomes where the sum is neither $7$ nor $11$.
The probability of getting either $7$ or $11$ is $P(E_1 \cup E_2) = P(E_1) + P(E_2) = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9}$.
Given that the sum is either $7$ or $11$,the conditional probability that the sum is $7$ is $P(E_1 | E_1 \cup E_2) = \frac{P(E_1)}{P(E_1) + P(E_2)} = \frac{6/36}{8/36} = \frac{6}{8} = \frac{3}{4}$.
Thus,the probability that $7$ appears before $11$ is $\frac{3}{4}$.

(A) Given that the total number of messages $n = 500$ and the probability of a message being transmitted correctly is $0.98$.
Therefore,the probability of a message being transmitted incorrectly is $p = 1 - 0.98 = 0.02$.
Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $\lambda = np = 500 \times 0.02 = 10$.
The probability of $X$ incorrectly transmitted messages is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$.
We need to find the probability that at most one message is transmitted incorrectly,i.e.,$P(X \leq 1) = P(X=0) + P(X=1)$.
$P(X=0) = \frac{10^0 e^{-10}}{0!} = e^{-10}$.
$P(X=1) = \frac{10^1 e^{-10}}{1!} = 10 e^{-10}$.
Thus,$P(X \leq 1) = e^{-10} + 10 e^{-10} = 11 e^{-10} = \frac{11}{e^{10}}$.

(B) Consider the Binomial probability distribution where $n = 5$ and $p = 0.2 = \frac{1}{5}$.
Probability of success $p = \frac{1}{5}$,so probability of failure $q = 1 - \frac{1}{5} = \frac{4}{5}$.
We need to find the probability that at least two subscribers receive the promotion,which is $P(X \geq 2)$.
This can be calculated as $P(X \geq 2) = 1 - [P(X = 0) + P(X = 1)]$.
$P(X = 0) = { }^5 C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^5 = 1 \times 1 \times \frac{1024}{3125} = \frac{1024}{3125}$.
$P(X = 1) = { }^5 C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^4 = 5 \times \frac{1}{5} \times \frac{256}{625} = \frac{256}{625} = \frac{1280}{3125}$.
Therefore,$P(X \geq 2) = 1 - \left(\frac{1024 + 1280}{3125}\right) = 1 - \frac{2304}{3125} = \frac{3125 - 2304}{3125} = \frac{821}{3125}$.

(B) Given that $\lim _{t \rightarrow 0}(1+5t)^{\frac{1}{t}} = K$.
Using the standard limit $\lim _{t \rightarrow 0}(1+at)^{\frac{1}{t}} = e^a$,we get $K = e^5$.
For the random variable $X$ representing the number of successes in $n = 100$ independent trials with probability of success $p = 0.05$,the distribution follows a Binomial distribution $B(n, p)$.
Since $n$ is large and $p$ is small,we can approximate this using a Poisson distribution with parameter $\lambda = np = 100 \times 0.05 = 5$.
The probability of getting at least one success is $P(X \geq 1) = 1 - P(X = 0)$.
Using the Poisson formula $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$,we have $P(X = 0) = e^{-5}$.
Therefore,$P(X \geq 1) = 1 - e^{-5} = 1 - \frac{1}{e^5} = 1 - \frac{1}{K} = \frac{K-1}{K}$.

(D) When two dice are rolled,the total number of outcomes is $6 \times 6 = 36$. Let $X$ be the absolute difference of the numbers on the two dice. The possible values of $X$ are $0, 1, 2, 3, 4, 5$. The probability distribution is as follows:
| $X$ | $P(X)$ | $P_i X_i$ |
|---|---|---|
| $0$ | $6/36$ | $0$ |
| $1$ | $10/36$ | $10/36$ |
| $2$ | $8/36$ | $16/36$ |
| $3$ | $6/36$ | $18/36$ |
| $4$ | $4/36$ | $16/36$ |
| $5$ | $2/36$ | $10/36$ |
The mean $\mu$ is given by $\sum P_i X_i$:
$\mu = 0 + \frac{10}{36} + \frac{16}{36} + \frac{18}{36} + \frac{16}{36} + \frac{10}{36}$
$\mu = \frac{10 + 16 + 18 + 16 + 10}{36} = \frac{70}{36} = \frac{35}{18}$.

(C) Given,$n = 50$. Let $p$ be the probability of success in one trial. Then the parameter of the Poisson distribution is $\lambda = np = 50p$.
The probability mass function for a Poisson variable is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
Given the equation: $2 P(X=1) = 5 P(X=5) + 2 P(X=3)$.
Substituting the formula: $2 \frac{e^{-\lambda} \lambda^1}{1!} = 5 \frac{e^{-\lambda} \lambda^5}{5!} + 2 \frac{e^{-\lambda} \lambda^3}{3!}$.
Dividing both sides by $e^{-\lambda}$ (since $e^{-\lambda} \neq 0$): $2 \lambda = 5 \frac{\lambda^5}{120} + 2 \frac{\lambda^3}{6}$.
$2 \lambda = \frac{\lambda^5}{24} + \frac{\lambda^3}{3}$.
Multiplying by $24$: $48 \lambda = \lambda^5 + 8 \lambda^3$.
Since $\lambda \neq 0$,divide by $\lambda$: $\lambda^4 + 8 \lambda^2 - 48 = 0$.
Let $u = \lambda^2$,then $u^2 + 8u - 48 = 0$.
$(u + 12)(u - 4) = 0$.
So,$\lambda^2 = 4$ or $\lambda^2 = -12$. Since $\lambda^2$ must be positive,$\lambda^2 = 4$,which gives $\lambda = 2$.
Finally,$p = \frac{\lambda}{n} = \frac{2}{50} = 0.04$.

(C) First,we calculate $E(X)$ and $E(X^2)$ using the given distribution:
$E(X) = \Sigma x P(X=x) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+0+1+4}{6} = \frac{3}{6} = \frac{1}{2}$
$E(X^2) = \Sigma x^2 P(X=x) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+0+1+8}{6} = \frac{11}{6}$
Now,we find $\operatorname{Var}(X) = E(X^2) - (E(X))^2 = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$
Finally,we calculate $6 E(X^2) - \operatorname{Var}(X)$:
$6 E(X^2) - \operatorname{Var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$

(A) Let $N$ be the number appearing on the die. The possible values for $N$ are $\{1, 2, 3, 4, 5, 6\}$.
If $N$ is even $(N \in \{2, 4, 6\})$,the number of chocolates $X = N + 2$.
For $N = 2$,$X = 2 + 2 = 4$.
For $N = 4$,$X = 4 + 2 = 6$.
For $N = 6$,$X = 6 + 2 = 8$.
If $N$ is odd $(N \in \{1, 3, 5\})$,the number of chocolates $X = N + 3$.
For $N = 1$,$X = 1 + 3 = 4$.
For $N = 3$,$X = 3 + 3 = 6$.
For $N = 5$,$X = 5 + 3 = 8$.
Thus,the set of all possible values of $X$ is $\{4, 6, 8\}$.
Therefore,the range of $X$ is $\{4, 6, 8\}$.

(A) When two dice are rolled,the sum $X$ can take values from $2$ to $12$. The probability distribution is as follows:
$X: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$
$P(X): \frac{1}{36}, \frac{2}{36}, \frac{3}{36}, \frac{4}{36}, \frac{5}{36}, \frac{6}{36}, \frac{5}{36}, \frac{4}{36}, \frac{3}{36}, \frac{2}{36}, \frac{1}{36}$
The mean $\mu = E(X) = \sum X_i P(X_i) = \frac{2(1)+3(2)+4(3)+5(4)+6(5)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)}{36} = \frac{252}{36} = 7$.
Now,calculate the probabilities:
$P(X < 5) = P(X=2) + P(X=3) + P(X=4) = \frac{1+2+3}{36} = \frac{6}{36}$.
$P(X > 9) = P(X=10) + P(X=11) + P(X=12) = \frac{3+2+1}{36} = \frac{6}{36}$.
$P(X = 7) = \frac{6}{36}$.
Substituting these values into the expression $\mu + P(X < 5) + P(X > 9) + P(X = 7)$:
$= 7 + \frac{6}{36} + \frac{6}{36} + \frac{6}{36} = 7 + \frac{18}{36} = 7 + \frac{1}{2} = \frac{15}{2}$.

(B) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$.
Given $P(X=2) = P(X=3)$,we have:
$\frac{\lambda^2 e^{-\lambda}}{2!} = \frac{\lambda^3 e^{-\lambda}}{3!}$
$\frac{\lambda^2}{2} = \frac{\lambda^3}{6}$
Since $\lambda > 0$,we can divide by $\lambda^2$:
$\frac{1}{2} = \frac{\lambda}{6} \Rightarrow \lambda = 3$.
Now,we calculate $P(X=5)$:
$P(X=5) = \frac{3^5 e^{-3}}{5!} = \frac{243 e^{-3}}{120} = \frac{81 e^{-3}}{40} = \frac{81}{40 e^3}$.
Thus,option $B$ is correct.