TS EAMCET 2020 Mathematics Question Paper with Answer and Solution

(B) The general equation of a parabola is defined as $PF^2 = e^2 PM^2$,where $e=1$ for a parabola.
Given the equation $4[(x-a)^2+(y-b)^2]=(\alpha x+\beta y+7)^2$,we can rewrite it as:
$(x-a)^2+(y-b)^2 = \left(\frac{\alpha x+\beta y+7}{2}\right)^2$
To match the form $PF^2 = PM^2$,we normalize the line equation:
$(x-a)^2+(y-b)^2 = \left(\frac{\alpha x+\beta y+7}{\sqrt{\alpha^2+\beta^2}}\right)^2 \cdot \frac{\alpha^2+\beta^2}{4}$
For this to represent a parabola,the eccentricity $e$ must be $1$,which implies the coefficient $\frac{\alpha^2+\beta^2}{4} = 1^2$.
Therefore,$\alpha^2+\beta^2=4$.

(C) The length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$.
The distance between the foci is $2ae = 4\sqrt{2}$,so $ae = 2\sqrt{2}$.
Squaring both sides,$a^2e^2 = 8$.
Using the relation $a^2e^2 = a^2 - b^2$,we have $a^2 - b^2 = 8$.
Substituting $b^2 = 2a$,we get $a^2 - 2a - 8 = 0$.
Solving the quadratic equation $(a - 4)(a + 2) = 0$,we get $a = 4$ (since $a > 0$).
Then $b^2 = 2(4) = 8$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.
Multiplying by $16$,we get $x^2 + 2y^2 = 16$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through $(3 \sqrt{2}, \sqrt{10})$,we have $\frac{(3 \sqrt{2})^2}{a^2} + \frac{(\sqrt{10})^2}{b^2} = 1$,which simplifies to $\frac{18}{a^2} + \frac{10}{b^2} = 1$ $(i)$.
The foci are at $(\pm 4, 0)$,so $ae = 4$,which means $a^2 e^2 = 16$.
Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2 e^2$,we get $b^2 = a^2 - 16$.
Substituting $b^2$ into equation $(i)$: $\frac{18}{a^2} + \frac{10}{a^2 - 16} = 1$.
Multiplying by $a^2(a^2 - 16)$,we get $18(a^2 - 16) + 10a^2 = a^2(a^2 - 16)$.
$18a^2 - 288 + 10a^2 = a^4 - 16a^2$.
$a^4 - 44a^2 + 288 = 0$.
Factoring the quadratic in $a^2$: $(a^2 - 36)(a^2 - 8) = 0$.
Since $b^2 = a^2 - 16 > 0$,we must have $a^2 > 16$,so $a^2 = 36$.
Finally,$e^2 = \frac{16}{a^2} = \frac{16}{36} = \frac{4}{9}$.
Therefore,$e = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

(A) Since the foci of the given ellipse lie on the $Y$-axis,it is a vertical ellipse.
Let the required equation be $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,where $a^2 > b^2$.
The foci are $(0, \pm c) = (0, \pm 1)$,which implies $c = 1$.
The length of the major axis is $2a = \sqrt{5}$,so $a = \frac{\sqrt{5}}{2}$ and $a^2 = \frac{5}{4}$.
Using the relation $c^2 = a^2 - b^2$,we have $1 = \frac{5}{4} - b^2$.
Thus,$b^2 = \frac{5}{4} - 1 = \frac{1}{4}$.
Substituting $a^2$ and $b^2$ into the equation,we get $\frac{x^2}{1/4} + \frac{y^2}{5/4} = 1$.
This simplifies to $4x^2 + \frac{4y^2}{5} = 1$,or $20x^2 + 4y^2 = 5$.

(B) By the definition of an ellipse,the distance from a point $(x, y)$ to the focus is $e$ times the distance from the point to the directrix: $\sqrt{(x-1)^2+(y-2)^2} = \frac{1}{2} \frac{|3x+4y-5|}{\sqrt{3^2+4^2}}$.
Squaring both sides,we get: $(x-1)^2+(y-2)^2 = \frac{1}{4} \frac{(3x+4y-5)^2}{25}$.
$100(x^2-2x+1+y^2-4y+4) = (3x+4y-5)^2$.
$100(x^2+y^2-2x-4y+5) = 9x^2+16y^2+25+24xy-30x-40y$.
$100x^2+100y^2-200x-400y+500 = 9x^2+16y^2+24xy-30x-40y+25$.
Rearranging terms: $91x^2+84y^2-24xy-170x-360y+475=0$.

(D) Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad (a>b)$.
According to the given information,the distance from each focus to the centre is $ae$,so the sum of distances is $ae + ae = 2ae = 8\sqrt{6} \Rightarrow ae = 4\sqrt{6}$.
Squaring both sides,$a^2e^2 = 16 \times 6 = 96$.
Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 96$,which simplifies to $a^2 - b^2 = 96 \quad (i)$.
The smallest rectangle in which the ellipse is inscribed has dimensions $2a \times 2b$. Thus,$4ab = 80$ $\Rightarrow ab = 20$ $\Rightarrow a^2b^2 = 400 \quad (ii)$.
Using the identity $(a^2+b^2)^2 = (a^2-b^2)^2 + 4a^2b^2$,we get $(a^2+b^2)^2 = (96)^2 + 4(400) = 9216 + 1600 = 10816$.
Taking the square root,$a^2+b^2 = \sqrt{10816} = 104 \quad (iii)$.
Adding $(i)$ and $(iii)$,$2a^2 = 200 \Rightarrow a^2 = 100$.
Subtracting $(i)$ from $(iii)$,$2b^2 = 8 \Rightarrow b^2 = 4$.
Therefore,the equation of the ellipse is $\frac{x^2}{100}+\frac{y^2}{4}=1$.

(A) Given the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,we have $a^2=16$ and $b^2=12$.
The eccentricity $e$ is given by $b^2=a^2(1-e^2)$,so $12=16(1-e^2)$,which gives $1-e^2=\frac{3}{4}$,so $e^2=\frac{1}{4}$ and $e=\frac{1}{2}$.
For a focal chord with eccentric angles $\alpha$ and $\beta$,the condition is $\tan(\frac{\alpha}{2}) \tan(\frac{\beta}{2}) = \frac{e-1}{e+1}$.
Here $\alpha = \frac{\pi}{3}$,so $\frac{\alpha}{2} = \frac{\pi}{6}$.
$\tan(\frac{\pi}{6}) \tan(\frac{\theta}{2}) = \frac{1/2 - 1}{1/2 + 1} = \frac{-1/2}{3/2} = -\frac{1}{3}$.
Since $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} \tan(\frac{\theta}{2}) = -\frac{1}{3}$,so $\tan(\frac{\theta}{2}) = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}$.
Thus,$\frac{\theta}{2} = -\frac{\pi}{6}$,which means $\theta = -\frac{\pi}{3}$.
Therefore,$\tan \theta = \tan(-\frac{\pi}{3}) = -\sqrt{3}$.

(A) The given equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$. Here $a^2=2$ and $b^2=1$,so $a=\sqrt{2}$ and $b=1$.
The equation of the tangent to the ellipse at point $P(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
Substituting the values,we get $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$.
For the $x$-intercept $(A)$,set $y=0$: $\frac{x \cos \theta}{\sqrt{2}} = 1 \Rightarrow x = \sqrt{2} \sec \theta$. So,$A = (\sqrt{2} \sec \theta, 0)$.
For the $y$-intercept $(B)$,set $x=0$: $y \sin \theta = 1 \Rightarrow y = \operatorname{cosec} \theta$. So,$B = (0, \operatorname{cosec} \theta)$.
Let $M(h, k)$ be the midpoint of $AB$. Then:
$h = \frac{\sqrt{2} \sec \theta + 0}{2} = \frac{\sec \theta}{\sqrt{2}} \Rightarrow \sec \theta = \sqrt{2}h$
$k = \frac{0 + \operatorname{cosec} \theta}{2} = \frac{\operatorname{cosec} \theta}{2} \Rightarrow \operatorname{cosec} \theta = 2k$
We know that $\cos^2 \theta + \sin^2 \theta = 1$,which implies $\frac{1}{\sec^2 \theta} + \frac{1}{\operatorname{cosec}^2 \theta} = 1$.
Substituting the values of $\sec \theta$ and $\operatorname{cosec} \theta$:
$\frac{1}{(\sqrt{2}h)^2} + \frac{1}{(2k)^2} = 1$
$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(C) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Let $P(5 \cos \theta, 4 \sin \theta)$ be a point on the ellipse.
The equation of the tangent at $P$ is $\frac{x(5 \cos \theta)}{25} + \frac{y(4 \sin \theta)}{16} = 1$,which simplifies to $\frac{x \cos \theta}{5} + \frac{y \sin \theta}{4} = 1$.
This tangent intersects the $X$-axis at point $A$ by setting $y=0$,giving $A = (\frac{5}{\cos \theta}, 0) = (5 \sec \theta, 0)$.
The image of $A(5 \sec \theta, 0)$ with respect to the line $y=x$ is $A^{\prime} = (0, 5 \sec \theta)$.
The equation of the circle with diameter $AA^{\prime}$ is $(x - 5 \sec \theta)(x - 0) + (y - 0)(y - 5 \sec \theta) = 0$.
This simplifies to $x^2 + y^2 - 5 \sec \theta (x + y) = 0$.
For this circle to pass through a fixed point independent of $\theta$,we observe that for any $\theta$,the point $(0, 0)$ satisfies the equation $0^2 + 0^2 - 5 \sec \theta (0 + 0) = 0$.

(D) We know that the product of the lengths of the perpendiculars drawn from the foci to any tangent of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is equal to $b^2$.
Given that the product is $9$,we have $b^2 = 9$.
The equation of the tangent is $y = \frac{-3}{4}x + 3\sqrt{2}$.
Comparing this with the standard tangent equation $y = mx \pm \sqrt{a^2m^2 + b^2}$,where $m = \frac{-3}{4}$,we have $3\sqrt{2} = \sqrt{a^2(\frac{-3}{4})^2 + b^2}$.
Squaring both sides,we get $18 = a^2(\frac{9}{16}) + 9$.
Subtracting $9$ from both sides,$9 = \frac{9a^2}{16}$,which gives $a^2 = 16$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.

(A) Given the ellipse equation: $2x^2+y^2=2$,which can be written as $x^2+\frac{y^2}{2}=1$.
Here,$a^2=1$ and $b^2=2$.
The equation of the tangent is $x+2y+k=0$,which implies $y=-\frac{1}{2}x-\frac{k}{2}$.
Comparing this with $y=mx+c$,we get $m=-\frac{1}{2}$ and $c=-\frac{k}{2}$.
The condition for a line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$.
Substituting the values: $(-\frac{k}{2})^2 = (1)(-\frac{1}{2})^2 + 2$.
$\frac{k^2}{4} = \frac{1}{4} + 2 = \frac{9}{4}$.
$k^2=9$,so $k=\pm 3$. Since $k>0$,we have $k=3$.
The point of contact is $\left(\frac{1}{\sqrt{2}}, \frac{3}{3}\right) = \left(\frac{1}{\sqrt{2}}, 1\right)$.
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 $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$.
Substituting $a^2=1, b^2=2, x_1=\frac{1}{\sqrt{2}}, y_1=1$:
$\frac{1 \cdot x}{1/\sqrt{2}} - \frac{2 \cdot y}{1} = 1-2$.
$\sqrt{2}x - 2y = -1$,which simplifies to $\sqrt{2}x-2y+1=0$.

(D) The equation of a line $y = mx + c$ is a common tangent to the circle $x^2 + y^2 = 16$ and the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
For the circle $x^2 + y^2 = 16$,the condition for tangency is $c^2 = r^2(1 + m^2)$,so $c^2 = 16(1 + m^2)$.
For the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$,the condition for tangency is $c^2 = a^2m^2 + b^2$,so $c^2 = 25m^2 + 9$.
Equating the two expressions for $c^2$:
$16(1 + m^2) = 25m^2 + 9$
$16 + 16m^2 = 25m^2 + 9$
$9m^2 = 7$
$m^2 = \frac{7}{9}$.
Since $m = \tan \theta$,we have $\tan^2 \theta = \frac{7}{9}$.
Using the formula $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$:
$\cos 2\theta = \frac{1 - \frac{7}{9}}{1 + \frac{7}{9}} = \frac{\frac{2}{9}}{\frac{16}{9}} = \frac{2}{16} = \frac{1}{8}$.

(B) Let the equation of the common tangent be $y=mx+c$.
For the circle $x^2+y^2=4$,the condition for tangency is $c^2 = r^2(1+m^2)$,where $r^2=4$. So,$c^2 = 4(1+m^2)$.
For the ellipse $\frac{x^2}{25} + \frac{y^2}{2} = 1$,the condition for tangency is $c^2 = a^2m^2 + b^2$,where $a^2=25$ and $b^2=2$. So,$c^2 = 25m^2 + 2$.
Equating the two expressions for $c^2$:
$4(1+m^2) = 25m^2 + 2$
$4 + 4m^2 = 25m^2 + 2$
$21m^2 = 2$ $\Rightarrow m^2 = \frac{2}{21}$ $\Rightarrow m = \pm \sqrt{\frac{2}{21}}$.
Substituting $m^2$ back into $c^2 = 4(1+m^2)$:
$c^2 = 4(1 + \frac{2}{21}) = 4(\frac{23}{21}) = \frac{92}{21}$.
Thus,$c = \pm \sqrt{\frac{92}{21}} = \pm \frac{2\sqrt{23}}{\sqrt{21}}$.
The equation of the tangent is $y = \pm \sqrt{\frac{2}{21}}x \pm \frac{2\sqrt{23}}{\sqrt{21}}$.
Multiplying by $\sqrt{21}$: $\sqrt{21}y = \pm \sqrt{2}x \pm 2\sqrt{23}$.
Rearranging gives $\sqrt{2}x - \sqrt{21}y + 2\sqrt{23} = 0$.

(D) Given ellipse is $\frac{x^2}{16}+\frac{y^2}{12}=1$. Here $a^2=16$ and $b^2=12$.
Eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.
The end points of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 4 \times \frac{1}{2}, \pm \frac{12}{4}) = (\pm 2, \pm 3)$.
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 $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$.
For the point $(2, 3)$,the tangent is $\frac{2x}{16}+\frac{3y}{12}=1$,which simplifies to $\frac{x}{8}+\frac{y}{4}=1$.
This line intersects the $x$-axis at $(8, 0)$ and the $y$-axis at $(0, 4)$.
Due to symmetry,the quadrilateral formed by the four tangents at $(\pm 2, \pm 3)$ is a rhombus with vertices at $(\pm 8, 0)$ and $(0, \pm 4)$.
The area of this rhombus is $4 \times (\text{Area of triangle with vertices } (0,0), (8,0), (0,4)) = 4 \times (\frac{1}{2} \times 8 \times 4) = 64$ sq. units.

(C) The equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$.
Any point $P$ on the ellipse is $(\sqrt{2}\cos\theta, \sin\theta)$.
The equation of the tangent at $P$ is $\frac{x(\sqrt{2}\cos\theta)}{2} + \frac{y(\sin\theta)}{1} = 1$,which simplifies to $\frac{x\cos\theta}{\sqrt{2}} + y\sin\theta = 1$.
The intercepts made by the tangent on the coordinate axes are $A\left(\frac{\sqrt{2}}{\cos\theta}, 0\right)$ and $B\left(0, \frac{1}{\sin\theta}\right)$.
Let $(h, k)$ be the mid-point of $AB$. Then $h = \frac{\sqrt{2}}{2\cos\theta}$ and $k = \frac{1}{2\sin\theta}$.
This implies $\cos\theta = \frac{1}{\sqrt{2}h}$ and $\sin\theta = \frac{1}{2k}$.
Using the identity $\cos^2\theta + \sin^2\theta = 1$,we get $\left(\frac{1}{\sqrt{2}h}\right)^2 + \left(\frac{1}{2k}\right)^2 = 1$.
Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(C) Given the ellipse equation: $9x^2+4y^2-18x+16y-11=0$.
Since $P(1, k)$ lies on the ellipse,substitute $x=1$:
$9(1)^2+4k^2-18(1)+16k-11=0$
$9+4k^2-18+16k-11=0$
$4k^2+16k-20=0$
$k^2+4k-5=0$
$(k+5)(k-1)=0$.
Since $k>0$,we have $k=1$. Thus,$P$ is $(1, 1)$.
Rewrite the ellipse equation in standard form:
$9(x-1)^2+4(y+2)^2 = 11+9+16 = 36$
$\frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$.
Here $a^2=4$ and $b^2=9$,so $b>a$.
Eccentricity $e = \sqrt{1-\frac{a^2}{b^2}} = \sqrt{1-\frac{4}{9}} = \frac{\sqrt{5}}{3}$.
The directrices are $y+2 = \pm \frac{b}{e} = \pm \frac{3}{\sqrt{5}/3} = \pm \frac{9}{\sqrt{5}}$.
So,$y = -2 \pm \frac{9}{\sqrt{5}}$.
The distances from $P(1, 1)$ to the directrices $y = -2 + \frac{9}{\sqrt{5}}$ and $y = -2 - \frac{9}{\sqrt{5}}$ are:
$d_1 = |1 - (-2 + \frac{9}{\sqrt{5}})| = |3 - \frac{9}{\sqrt{5}}| = \frac{9}{\sqrt{5}} - 3$ (since $\frac{9}{\sqrt{5}} \approx 4.02 > 3$).
$d_2 = |1 - (-2 - \frac{9}{\sqrt{5}})| = |3 + \frac{9}{\sqrt{5}}| = 3 + \frac{9}{\sqrt{5}}$.
The shortest distance is $\frac{9}{\sqrt{5}} - 3$.

(C) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.
Here,$a_1^2 = \frac{144}{25}$ and $b_1^2 = \frac{81}{25}$.
The eccentricity $e_1 = \sqrt{1 + \frac{b_1^2}{a_1^2}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.
The foci of the hyperbola are $(\pm a_1 e_1, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm a_2 e_2, 0) = (\pm 4 e_2, 0)$.
Since the foci coincide,$4 e_2 = 3$,so $e_2 = \frac{3}{4}$.
For an ellipse,$e_2^2 = 1 - \frac{b^2}{a_2^2} = 1 - \frac{b^2}{16}$.
Substituting $e_2 = \frac{3}{4}$,we get $\frac{9}{16} = 1 - \frac{b^2}{16}$.
Thus,$\frac{b^2}{16} = 1 - \frac{9}{16} = \frac{7}{16}$,which implies $b^2 = 7$.

(D) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given that the eccentricity $e = \sqrt{2}$ and the distance between the foci is $2ae = 16$.
Substituting $e = \sqrt{2}$ into the equation $2ae = 16$,we get $2a(\sqrt{2}) = 16$.
Thus,$a = \frac{16}{2\sqrt{2}} = 4\sqrt{2}$,so $a^2 = (4\sqrt{2})^2 = 32$.
For a hyperbola,$b^2 = a^2(e^2 - 1)$.
Substituting the values,$b^2 = 32((\sqrt{2})^2 - 1) = 32(2 - 1) = 32$.
Therefore,the equation of the hyperbola is $\frac{x^2}{32} - \frac{y^2}{32} = 1$,which simplifies to $x^2 - y^2 = 32$.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given that the length of the transverse axis is $2a = 12$,so $a = 6$.
Since the point $(8, 2)$ lies on the hyperbola,it must satisfy the equation:
$\frac{8^2}{a^2} - \frac{2^2}{b^2} = 1$
Substituting $a = 6$:
$\frac{64}{36} - \frac{4}{b^2} = 1$
$\frac{16}{9} - 1 = \frac{4}{b^2}$
$\frac{7}{9} = \frac{4}{b^2} \implies b^2 = \frac{36}{7}$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
$e = \sqrt{1 + \frac{36/7}{36}} = \sqrt{1 + \frac{1}{7}} = \sqrt{\frac{8}{7}} = \frac{2 \sqrt{2}}{\sqrt{7}}$.

(B) The given equation of the hyperbola is $x^2-y^2-4x+2y+c=0$.
Completing the square,we get $(x-2)^2 - (y-1)^2 = 3-c$.
Let $a^2 = 3-c$. The equation becomes $\frac{(x-2)^2}{a^2} - \frac{(y-1)^2}{a^2} = 1$.
For this rectangular hyperbola,the eccentricity $e = \sqrt{2}$.
The focus is given by $x = 2 \pm ae = 2 \pm \sqrt{a^2} \cdot \sqrt{2} = 2 \pm \sqrt{2a^2}$.
Given the focus $S(2+2\sqrt{2}, k)$,we have $\sqrt{2a^2} = 2\sqrt{2} \implies 2a^2 = 8 \implies a^2 = 4$.
Since $a^2 = 3-c$,we have $4 = 3-c$,which gives $c = -1$.
The directrix is $x = 2 \pm \frac{a}{e} = 2 \pm \frac{2}{\sqrt{2}} = 2 \pm \sqrt{2}$.
The directrix adjacent to the focus $x = 2+2\sqrt{2}$ is $x = 2+\sqrt{2}$,which matches the given condition.
Thus,$c = -1$.

(A) Given that $p$ and $q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively.
We know that $p^2 = 1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2}$ and $q^2 = 1 + \frac{a^2}{b^2} = \frac{a^2+b^2}{b^2}$.
The equation of the ellipse is $\frac{x^2}{p^2} + \frac{y^2}{q^2} = 1$.
Substituting the values of $p^2$ and $q^2$,we get $\frac{x^2 a^2}{a^2+b^2} + \frac{y^2 b^2}{a^2+b^2} = 1$,which simplifies to $a^2 x^2 + b^2 y^2 = a^2 + b^2$.
Given the pair of lines $x^2 - y^2 = 0$,which implies $y^2 = x^2$.
Substituting $y^2 = x^2$ into the ellipse equation: $a^2 x^2 + b^2 x^2 = a^2 + b^2$.
$(a^2 + b^2) x^2 = a^2 + b^2 \implies x^2 = 1 \implies x = \pm 1$.
Since $y^2 = x^2$,we have $y = \pm 1$.
The points of intersection are $(1, 1), (1, -1), (-1, 1), (-1, -1)$.
These points form a square with side length $s = \sqrt{(1 - (-1))^2 + (1 - 1)^2} = 2$.
Area of the square = $s^2 = 2^2 = 4$ sq. units.

(D) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The equation of the normal at a point $P(a \sec \theta, b \tan \theta)$ is $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2+b^2$,which simplifies to $ax \cos \theta + by \cot \theta = a^2+b^2$.
Similarly,the equation of the normal at $Q(a \sec \phi, b \tan \phi)$ is $ax \cos \phi + by \cot \phi = a^2+b^2$.
Given $\theta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - \theta$. Thus,$\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
The two equations are:
$(1) \quad ax \cos \theta + by \cot \theta = a^2+b^2$
$(2) \quad ax \sin \theta + by \tan \theta = a^2+b^2$
To find $k$ (the $y$-coordinate),we eliminate $x$ by multiplying $(1)$ by $\sin \theta$ and $(2)$ by $\cos \theta$:
$ax \cos \theta \sin \theta + by \cot \theta \sin \theta = (a^2+b^2) \sin \theta$
$ax \sin \theta \cos \theta + by \tan \theta \cos \theta = (a^2+b^2) \cos \theta$
Subtracting the two equations:
$by(\cot \theta \sin \theta - \tan \theta \cos \theta) = (a^2+b^2)(\sin \theta - \cos \theta)$
$by(\cos \theta - \sin \theta) = (a^2+b^2)(\sin \theta - \cos \theta)$
$by = -(a^2+b^2)$
$y = -\frac{a^2+b^2}{b}$
Therefore,$k = -\frac{a^2+b^2}{b}$.

(D) The asymptotes of a rectangular hyperbola are parallel to the coordinate axes,so its equation is of the form $(x - h)(y - k) = c$.
Since the point of intersection of the two perpendicular tangents is the center of the hyperbola,we have $(h, k) = (1, 1)$.
Thus,the equation is $(x - 1)(y - 1) = c$.
Since the hyperbola passes through $(3, 2)$,we substitute these coordinates: $(3 - 1)(2 - 1) = c$ $\Rightarrow 2 \times 1 = c$ $\Rightarrow c = 2$.
Substituting $c = 2$ into the equation,we get $(x - 1)(y - 1) = 2$.
Expanding this,we get $xy - x - y + 1 = 2$,which simplifies to $xy = x + y + 1$.

(C) The equation of the polar of a point $(\alpha, \beta)$ with respect to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\frac{\alpha x}{a^2} - \frac{\beta y}{b^2} = 1$.
Given hyperbola is $9x^2 - 16y^2 = 144$,which can be written as $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
So,the equation of the polar of $(\alpha, \beta)$ is $\frac{\alpha x}{16} - \frac{\beta y}{9} = 1$,or $9\alpha x - 16\beta y - 144 = 0$.
Comparing this with the given line $3x - 16y + 48 = 0$,we have:
$\frac{9\alpha}{3} = \frac{-16\beta}{-16} = \frac{-144}{48}$.
$3\alpha = \beta = -3$.
Thus,$\alpha = -1$ and $\beta = -3$.
Therefore,$\alpha - \beta = -1 - (-3) = -1 + 3 = 2$.

(C) We have the equations:
$x^2+y^2=a^2$ $\dots(i)$
$xy=b^2$ $\dots(ii)$
From equation $(ii)$,we have $x = \frac{b^2}{y}$.
Substituting this into equation $(i)$:
$\left(\frac{b^2}{y}\right)^2 + y^2 = a^2$
$\frac{b^4}{y^2} + y^2 = a^2$
Multiplying by $y^2$:
$b^4 + y^4 = a^2 y^2$
$y^4 - a^2 y^2 + b^4 = 0$
This is a biquadratic equation in $y$. Let its roots be $y_1, y_2, y_3, y_4$.
According to the properties of the roots of a polynomial equation,the product of the roots $y_1 y_2 y_3 y_4$ is equal to the constant term divided by the leading coefficient.
Thus,$y_1 y_2 y_3 y_4 = \frac{b^4}{1} = b^4$.

(C) We have,$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi x+2\pi\right)}{x^2}$
Since $\cos(2\pi + \theta) = \cos \theta$,the expression becomes $\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi x\right)}{x^2}$
Using the identity $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$,we get:
$\lim _{x \rightarrow 0} \frac{2 \sin^2 \left(\frac{x^2+\pi x}{2}\right)}{x^2}$
$= 2 \lim _{x}$ ${\rightarrow 0} \left[ \frac{\sin \left(\frac{x^2+\pi x}{2}\right)}{\frac{x^2+\pi x}{2}} \right]^2 \times \left( \frac{x^2+\pi x}{2} \right)^2 \times \frac{1}{x^2}$
$= 2 \times 1^2 \times \lim _{x \rightarrow 0} \frac{x^2(x+\pi)^2}{4x^2}$
$= 2 \times \frac{1}{4} \times \pi^2 = \frac{\pi^2}{2}$

(C) We know that for small $x$,$\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \tan^{-1} x$.
Substituting this into the limit,we get:
$L = \lim_{x \rightarrow 0} \frac{x^4+x^3+x^2}{(\tan^{-1} x)(\tan^{-1} x)} = \lim_{x \rightarrow 0} \frac{x^2(x^2+x+1)}{(\tan^{-1} x)^2}$.
Since $\lim_{x \rightarrow 0} \frac{\tan^{-1} x}{x} = 1$,we can write $(\tan^{-1} x)^2 \approx x^2$ as $x \rightarrow 0$.
$L = \lim_{x \rightarrow 0} \frac{x^2(x^2+x+1)}{x^2} = \lim_{x \rightarrow 0} (x^2+x+1) = 0^2+0+1 = 1$.

(B) We use the Taylor series expansion for $\tan \theta = \theta + \frac{\theta^3}{3} + \frac{2\theta^5}{15} + \dots$ and $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2}) \approx 2(\frac{\theta}{2})^2 = \frac{\theta^2}{2}$.
The denominator is $(1 - \cos 4x)^2 \approx (\frac{(4x)^2}{2})^2 = (8x^2)^2 = 64x^4$.
The numerator is $x(4x + \frac{(4x)^3}{3} + \dots) - 2x(2x + \frac{(2x)^3}{3} + \dots)$.
$= (4x^2 + \frac{64x^4}{3}) - (4x^2 + \frac{16x^4}{3}) = \frac{48x^4}{3} = 16x^4$.
Thus,the limit is $\lim _{x \rightarrow 0} \frac{16x^4}{64x^4} = \frac{16}{64} = \frac{1}{4}$.

(D) Given that,$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$
$= \lim _{x}$ ${\rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \left(\frac{1-\cos x}{x^2}\right)^2 \times \left(\frac{x}{\sin x}\right)^4$
As $x \rightarrow 0$,$(1-\cos x) \rightarrow 0$.
Using the standard limit $\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2} = \frac{1}{2}$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$= \frac{1}{2} \times \left(\frac{1}{2}\right)^2 \times (1)^4$
$= \frac{1}{2} \times \frac{1}{4} \times 1 = \frac{1}{8}$

(C) Let $f(x) = \sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right) + \sin k$.
As $x \rightarrow k$,the expression $[x] - [x/k]$ takes integer values.
Specifically,for $x$ in a small neighborhood of $k$,$[x] = k$ and $[x/k] = 1$ (since $k \geq 2$).
Thus,the term $2 \pi ([x] - [x/k])$ is an even multiple of $\pi$,say $2 \pi m$.
So,$\sin(2 \pi m - x) = \sin(-x) = -\sin x$.
The limit becomes $\lim_{x \rightarrow k} \frac{-\sin x + \sin k}{x - k}$.
Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim_{x \rightarrow k} \frac{-\cos x}{1} = -\cos k$.

(C) Given the limit expression: $\lim _{x \rightarrow 0} \left[ \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x} \right] = k$
Using the property $\log(a^b) = b \log a$,we get:
$\lim _{x \rightarrow 0} \left[ \frac{(1+x) \log (1+x) - x}{x^2} \right] = k$
Since this is a $\frac{0}{0}$ form,we apply $L$' Hospital's Rule:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx} [(1+x) \log (1+x) - x]}{\frac{d}{dx} [x^2]} = k$
$\lim _{x \rightarrow 0} \frac{[1 \cdot \log (1+x) + (1+x) \cdot \frac{1}{1+x}] - 1}{2x} = k$
$\lim _{x \rightarrow 0} \frac{\log (1+x) + 1 - 1}{2x} = k$
$\lim _{x \rightarrow 0} \frac{\log (1+x)}{2x} = k$
$\frac{1}{2} \lim _{x \rightarrow 0} \frac{\log (1+x)}{x} = k$
Since $\lim _{x \rightarrow 0} \frac{\log (1+x)}{x} = 1$,we have $k = \frac{1}{2} \times 1 = \frac{1}{2}$.
Therefore,$12k = 12 \times \frac{1}{2} = 6$.

(A) Given that $f$ is an increasing function and $f(x) > 0$ for all $x \in R^{+}$.
We are given that $\lim _{x \rightarrow \infty} \frac{f(9 x)}{f(3 x)}=1$.
Since $f$ is an increasing function,for any $x > 0$,we have the inequality:
$3x < 6x < 9x$
$\Rightarrow f(3x) < f(6x) < f(9x)$
Dividing the entire inequality by $f(3x)$ (which is positive),we get:
$\frac{f(3x)}{f(3x)} < \frac{f(6x)}{f(3x)} < \frac{f(9x)}{f(3x)}$
$1 < \frac{f(6x)}{f(3x)} < \frac{f(9x)}{f(3x)}$
Now,taking the limit as $x \rightarrow \infty$ on all sides:
$\lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} \leq \lim _{x \rightarrow \infty} \frac{f(9x)}{f(3x)}$
$1 \leq \lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} \leq 1$
By the Sandwich Theorem (or Squeeze Theorem),we conclude that:
$\lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} = 1$.

(A) The coefficient of variation $(CV)$ is defined as $CV = \frac{\sigma}{\bar{x}} \times 100$.
For group $A$ to be more consistent than group $B$,the coefficient of variation of $A$ must be less than the coefficient of variation of $B$.
Thus,$CV_A < CV_B$.
Substituting the values,we get $\frac{\sigma_A}{\bar{x}} < \frac{\sigma_B}{\bar{y}}$.
Given $\sigma_A = 2$ and $\sigma_B = 3$,we have $\frac{2}{\bar{x}} < \frac{3}{\bar{y}}$.
Rearranging the inequality to solve for $\frac{\bar{y}}{\bar{x}}$,we get $\frac{\bar{y}}{\bar{x}} < \frac{3}{2}$.

(C) First,we find the midpoints $(x_i)$ of each class interval and calculate the mean $(\bar{x})$:

Class Interval	Frequency $(f_i)$	Midpoint $(x_i)$	$f_i x_i$	$|x_i - \bar{x}|$	$f_i |x_i - \bar{x}|$
$0-4$	$4$	$2$	$8$	$|2-6|=4$	$16$
$4-8$	$3$	$6$	$18$	$|6-6|=0$	$0$
$8-12$	$2$	$10$	$20$	$|10-6|=4$	$8$
$12-16$	$1$	$14$	$14$	$|14-6|=8$	$8$
Total	$N=10$	-	$\sum f_i x_i = 60$	-	$\sum f_i |x_i - \bar{x}| = 32$

Mean $(\bar{x})$ = $\frac{\sum f_i x_i}{N} = \frac{60}{10} = 6$
Mean Deviation about the mean = $\frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{32}{10} = 3.2$

(B) We have,$\text{Mean} = 6$.
$\frac{8+9+6+5+x+4+6+5}{8} = 6$
$\Rightarrow 43+x = 48$ $\Rightarrow x = 5$.
So,the data set is $8, 9, 6, 5, 5, 4, 6, 5$.
The sum of observations $\Sigma x_i = 48$ and the sum of squares $\Sigma x_i^2 = 8^2 + 9^2 + 6^2 + 5^2 + 5^2 + 4^2 + 6^2 + 5^2 = 64 + 81 + 36 + 25 + 25 + 16 + 36 + 25 = 308$.
The variance is given by $\sigma^2 = \frac{1}{N} \Sigma x_i^2 - (\bar{x})^2 = \frac{308}{8} - (6)^2 = 38.5 - 36 = 2.5$.
The standard deviation is $\sigma = \sqrt{2.5} \approx 1.58$.

(C) To find the variance,we first calculate the midpoint $x_i$ for each class interval and then compute the necessary sums:

Class Interval	$f_i$	$x_i$	$f_i x_i$	$f_i x_i^2$
$0$-$5$	$4$	$2.5$	$10$	$25$
$5$-$10$	$1$	$7.5$	$7.5$	$56.25$
$10$-$15$	$10$	$12.5$	$125$	$1562.5$
$15$-$20$	$3$	$17.5$	$52.5$	$918.75$
$20$-$25$	$2$	$22.5$	$45$	$1012.5$
Total	$N=20$		$\Sigma f_i x_i = 240$	$\Sigma f_i x_i^2 = 3575$

The variance $\sigma^2$ is given by the formula:
$\sigma^2 = \frac{1}{N} \Sigma f_i x_i^2 - \left(\frac{1}{N} \Sigma f_i x_i\right)^2$
Substituting the values:
$\sigma^2 = \frac{3575}{20} - \left(\frac{240}{20}\right)^2$
$\sigma^2 = 178.75 - (12)^2$
$\sigma^2 = 178.75 - 144 = 34.75$

(C) The first five prime numbers are $2, 3, 5, 7, 11$.
Mean $(\bar{x}) = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$.
Mean deviation about mean $(\alpha) = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6|}{5} = \frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac{13.6}{5} = 2.72$.
Variance $(\beta) = \frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{2^2 + 3^2 + 5^2 + 7^2 + 11^2}{5} - (5.6)^2 = \frac{4 + 9 + 25 + 49 + 121}{5} - 31.36 = \frac{208}{5} - 31.36 = 41.6 - 31.36 = 10.24$.
Thus,the ordered pair $(\alpha, \beta)$ is $(2.72, 10.24)$.

(C) Let the variance of $x_1, x_2, \ldots, x_n$ be $\sigma^2$.
By the property of variance,$\text{Var}(ax + b) = a^2 \text{Var}(x)$.
For the assertion,$a = 4$ and $b = 0$,so $\text{Var}(4x) = 4^2 \text{Var}(x) = 16 \sigma^2$. Thus,$(A)$ is true.
For the reason,the property states $\text{Var}(ax + b) = a^2 \text{Var}(x)$. The given statement says $\text{Var}(y) = a \text{Var}(x) + b$,which is incorrect. Thus,$(R)$ is false.

(D) First,we calculate the mean $(\bar{x})$ of the given data:
$\bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{(1 \times 3) + (3 \times 3) + (5 \times 4) + (7 \times 14) + (9 \times 7) + (11 \times 4) + (13 \times 3) + (15 \times 4)}{3 + 3 + 4 + 14 + 7 + 4 + 3 + 4}$
$\bar{x} = \frac{3 + 9 + 20 + 98 + 63 + 44 + 39 + 60}{42} = \frac{336}{42} = 8$
Now,we calculate the mean deviation about the mean using the formula: $\text{M.D.}(\bar{x}) = \frac{\Sigma f|x - \bar{x}|}{N}$
Calculating $|x - \bar{x}|$ for each $x$:
$|1-8|=7, |3-8|=5, |5-8|=3, |7-8|=1, |9-8|=1, |11-8|=3, |13-8|=5, |15-8|=7$
Calculating $\Sigma f|x - \bar{x}|$:
$(3 \times 7) + (3 \times 5) + (4 \times 3) + (14 \times 1) + (7 \times 1) + (4 \times 3) + (3 \times 5) + (4 \times 7)$
$= 21 + 15 + 12 + 14 + 7 + 12 + 15 + 28 = 124$
$\therefore \text{M.D.}(\bar{x}) = \frac{124}{42} \approx 2.95$

(A) The first $5$ prime numbers are $2, 3, 5, 7, 11$.
The mean $\bar{x} = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$.
The sum of squares $\Sigma x_i^2 = 2^2 + 3^2 + 5^2 + 7^2 + 11^2 = 4 + 9 + 25 + 49 + 121 = 208$.
The standard deviation $\sigma = \sqrt{\frac{\Sigma x_i^2}{n} - (\bar{x})^2} = \sqrt{\frac{208}{5} - (\frac{28}{5})^2} = \sqrt{\frac{1040 - 784}{25}} = \sqrt{\frac{256}{25}} = \frac{16}{5} = 3.2$.
The coefficient of variation $C.V. = \frac{\sigma}{\bar{x}} \times 100 = \frac{16/5}{28/5} \times 100 = \frac{16}{28} \times 100 = \frac{4}{7} \times 100 = \frac{400}{7}$.

(B) Given,$n = 100$,$\bar{x} = 40$,and $\sigma = 5.1$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substituting the values: $(5.1)^2 = \frac{\sum x_i^2}{100} - (40)^2$.
$26.01 = \frac{\sum x_i^2}{100} - 1600$.
$\sum x_i^2 = (26.01 + 1600) \times 100 = 162601$.
This sum includes the incorrect observation $50$. To find the correct sum of squares,we subtract the square of the incorrect value and add the square of the correct value:
Correct $\sum x_i^2 = 162601 - (50)^2 + (40)^2$.
Correct $\sum x_i^2 = 162601 - 2500 + 1600 = 161701$.

(B) To find the variance,we first calculate the mid-values $(x_i)$ for each class and then compute $\Sigma f_i x_i$ and $\Sigma f_i x_i^2$.

Class	Mid value $(x_i)$	$f_i$	$f_i x_i$	$f_i x_i^2$
$0-10$	$5$	$11$	$55$	$275$
$10-20$	$15$	$29$	$435$	$6525$
$20-30$	$25$	$18$	$450$	$11250$
$30-40$	$35$	$4$	$140$	$4900$
$40-50$	$45$	$5$	$225$	$10125$
$50-60$	$55$	$3$	$165$	$9075$
Total	-	$70$	$1470$	$42150$

Here,$N = \Sigma f_i = 70$ and $\Sigma f_i x_i = 1470$.
The mean $\bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{1470}{70} = 21$.
The variance $\sigma^2 = \frac{1}{N} \Sigma f_i x_i^2 - (\bar{x})^2$.
$\sigma^2 = \frac{42150}{70} - (21)^2$.
$\sigma^2 = 602.14 - 441 = 161.14$.
Rounding to one decimal place,the variance is $161.1$.

(A) To find the variance,we first calculate the midpoint $(x_i)$ for each class interval and then compute the required sums:

Class Interval	$f_i$	$x_i$	$f_i x_i$	$f_i x_i^2$
$0-6$	$10$	$3$	$30$	$90$
$6-12$	$8$	$9$	$72$	$648$
$12-18$	$6$	$15$	$90$	$1350$
$18-24$	$4$	$21$	$84$	$1764$
$24-30$	$2$	$27$	$54$	$1458$
Total	$N = 30$	-	$\Sigma f_i x_i = 330$	$\Sigma f_i x_i^2 = 5310$

The formula for variance is $\sigma^2 = \frac{1}{N} \Sigma f_i x_i^2 - \left(\frac{\Sigma f_i x_i}{N}\right)^2$.
Substituting the values:
$\sigma^2 = \frac{5310}{30} - \left(\frac{330}{30}\right)^2$
$\sigma^2 = 177 - (11)^2$
$\sigma^2 = 177 - 121 = 56$.

(B) Given observations are $2, 7, 5, 6, 4, 3, 11, 17, 8$.
First,calculate the arithmetic mean $(\bar{x})$:
$\bar{x} = \frac{2 + 7 + 5 + 6 + 4 + 3 + 11 + 17 + 8}{9} = \frac{63}{9} = 7$.
Now,calculate the absolute deviations $d_i = |x_i - \bar{x}|$ for each observation:
$|2 - 7| = 5$
$|7 - 7| = 0$
$|5 - 7| = 2$
$|6 - 7| = 1$
$|4 - 7| = 3$
$|3 - 7| = 4$
$|11 - 7| = 4$
$|17 - 7| = 10$
$|8 - 7| = 1$
Sum of deviations $\Sigma d_i = 5 + 0 + 2 + 1 + 3 + 4 + 4 + 10 + 1 = 30$.
Mean deviation ($M$.$D$.) = $\frac{\Sigma d_i}{N} = \frac{30}{9} = \frac{10}{3}$.

(C) Given standard deviations $\sigma_x = 5$ and $\sigma_y = 6$ for $n = 100$ observations.
$\sum_{i=1}^{100}(x_i-\bar{x})^2 = n \sigma_x^2 = 100 \times 25 = 2500$.
$\sum_{i=1}^{100}(y_i-\bar{y})^2 = n \sigma_y^2 = 100 \times 36 = 3600$.
Given $\sum_{i=1}^{100}(x_i-\bar{x})(y_i-\bar{y}) = 600$.
Let $z_i = x_i - y_i$. Then $\bar{z} = \bar{x} - \bar{y}$.
The variance of $Z$ is $\sigma_z^2 = \frac{1}{n} \sum_{i=1}^{100}(z_i - \bar{z})^2$.
$\sigma_z^2 = \frac{1}{100} \sum_{i=1}^{100}((x_i - y_i) - (\bar{x} - \bar{y}))^2 = \frac{1}{100} \sum_{i=1}^{100}((x_i - \bar{x}) - (y_i - \bar{y}))^2$.
Expanding the square: $\sigma_z^2 = \frac{1}{100} [\sum(x_i - \bar{x})^2 + \sum(y_i - \bar{y})^2 - 2 \sum(x_i - \bar{x})(y_i - \bar{y})]$.
$\sigma_z^2 = \frac{1}{100} [2500 + 3600 - 2(600)] = \frac{1}{100} [6100 - 1200] = \frac{4900}{100} = 49$.
Therefore,$\sigma_z = \sqrt{49} = 7$.

(A) The variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2 - 1}{12}$.
For the first $2k$ natural numbers,$S_1 = \frac{(2k)^2 - 1}{12} = \frac{4k^2 - 1}{12}$.
For the first $k$ natural numbers,$S_2 = \frac{k^2 - 1}{12}$.
Therefore,the ratio is $\frac{S_1}{S_2} = \frac{4k^2 - 1}{k^2 - 1} = \frac{4(k^2 - 1) + 3}{k^2 - 1} = 4 + \frac{3}{k^2 - 1}$.
Since $k > 1$,$k^2 - 1 > 0$. As $k \to 1^+$,$\frac{3}{k^2 - 1} \to \infty$,and as $k \to \infty$,$\frac{3}{k^2 - 1} \to 0$.
Thus,$4 + \frac{3}{k^2 - 1} \in (4, \infty)$.

(B) First,we calculate the cumulative frequency $(cf)$ and the total frequency $(N)$:

$x_i$	$f_i$	$cf$
$6$	$4$	$4$
$12$	$7$	$11$
$18$	$9$	$20$
$24$	$18$	$38$
$30$	$15$	$53$
$36$	$10$	$63$
$42$	$5$	$68$

Here,$N = 68$,which is even. The median is the average of the $(\frac{N}{2})^{th}$ and $(\frac{N}{2} + 1)^{th}$ observations,i.e.,the $34^{th}$ and $35^{th}$ observations.
Looking at the cumulative frequency,both the $34^{th}$ and $35^{th}$ observations fall in the class where $x_i = 24$.
Therefore,$\text{Median} = 24$.
Now,we calculate the mean deviation about the median using the formula $\text{MD}(\text{Median}) = \frac{\sum f_i |x_i - \text{Median}|}{N}$:
$\sum f_i |x_i - 24| = 4|6-24| + 7|12-24| + 9|18-24| + 18|24-24| + 15|30-24| + 10|36-24| + 5|42-24|$
$= 4(18) + 7(12) + 9(6) + 18(0) + 15(6) + 10(12) + 5(18)$
$= 72 + 84 + 54 + 0 + 90 + 120 + 90 = 510$
$\text{MD}(\text{Median}) = \frac{510}{68} = 7.5$.

(C) Let the total number of observations be $n$.
Given that $\frac{n}{4}$ observations are $a$,$\frac{n}{4}$ are $-a$,$\frac{n}{4}$ are $b$,and $\frac{n}{4}$ are $-b$.
The mean $\bar{x} = \frac{\frac{n}{4}(a - a + b - b)}{n} = 0$.
The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substituting the values: $ab = \frac{\frac{n}{4}(a^2 + (-a)^2 + b^2 + (-b)^2)}{n} - 0$.
$ab = \frac{a^2 + a^2 + b^2 + b^2}{4} = \frac{2a^2 + 2b^2}{4} = \frac{a^2 + b^2}{2}$.
$2ab = a^2 + b^2 \Rightarrow a^2 + b^2 - 2ab = 0$.
$(a - b)^2 = 0 \Rightarrow a = b$.

(A) Given that in $\triangle ABC$,$\angle C = 90^{\circ}$.
Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle A + \angle B = 90^{\circ}$.
Using the sine rule,$a = k \sin A$ and $b = k \sin B$.
Substituting these into the expression:
$\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = \frac{k^2 \sin^2 A + k^2 \sin^2 B}{k^2 \sin^2 A - k^2 \sin^2 B} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin^2 A - \sin^2 B} \sin(A-B)$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:
$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B) \sin(A-B)} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin(A+B)}$.
Since $A+B = 90^{\circ}$,$\sin(A+B) = 1$ and $B = 90^{\circ}-A$,so $\sin B = \cos A$.
$= \frac{\sin^2 A + \cos^2 A}{1} = 1$.

(A) Given,$\cos A \cos B + \sin A \sin B \sin C = 1$.
We can rewrite this as $\cos A \cos B + \sin A \sin B - \sin A \sin B + \sin A \sin B \sin C = 1$.
This simplifies to $\cos(A - B) - \sin A \sin B(1 - \sin C) = 1$.
Rearranging gives $1 - \cos(A - B) + \sin A \sin B(1 - \sin C) = 0$.
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$,we get $2 \sin^2(\frac{A - B}{2}) + \sin A \sin B(1 - \sin C) = 0$.
Since the sum of two non-negative terms is zero,each term must be zero: $\sin(\frac{A - B}{2}) = 0$ and $\sin A \sin B(1 - \sin C) = 0$.
This implies $A = B$ and $\sin C = 1$ (as $\sin A, \sin B \neq 0$ in a triangle).
Thus,$C = 90^{\circ}$ and $A = B = 45^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 90^{\circ}}$.
$a : b : c = \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{2}} : 1 = 1 : 1 : \sqrt{2}$.

(C) The equation of line $L_1$ passing through $A_1 = \hat{i}-2 \hat{j}-\hat{k}$ and $B_1 = 4 \hat{i}-3 \hat{k}$ is given by $r = A_1 + \lambda(B_1 - A_1)$.
$B_1 - A_1 = (4-1)\hat{i} + (0-(-2))\hat{j} + (-3-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.
So,$L_1: r = (\hat{i}-2 \hat{j}-\hat{k}) + \lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})$.
The equation of line $L_2$ passing through $A_2 = \hat{i}+2 \hat{j}-\hat{k}$ and $B_2 = 2 \hat{i}-4 \hat{j}-5 \hat{k}$ is given by $r = A_2 + \mu(B_2 - A_2)$.
$B_2 - A_2 = (2-1)\hat{i} + (-4-2)\hat{j} + (-5-(-1))\hat{k} = \hat{i} - 6\hat{j} - 4\hat{k}$.
So,$L_2: r = (\hat{i}+2 \hat{j}-\hat{k}) + \mu(\hat{i}-6 \hat{j}-4 \hat{k})$.
The shortest distance $D$ between two skew lines $r = a_1 + \lambda b_1$ and $r = a_2 + \mu b_2$ is $D = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|}$.
$a_2 - a_1 = (\hat{i}+2 \hat{j}-\hat{k}) - (\hat{i}-2 \hat{j}-\hat{k}) = 4\hat{j}$.
$b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 1 & -6 & -4 \end{vmatrix} = \hat{i}(-8 - 12) - \hat{j}(-12 - (-2)) + \hat{k}(-18 - 2) = -20\hat{i} + 10\hat{j} - 20\hat{k}$.
$|b_1 \times b_2| = \sqrt{(-20)^2 + 10^2 + (-20)^2} = \sqrt{400 + 100 + 400} = \sqrt{900} = 30$.
$(a_2 - a_1) \cdot (b_1 \times b_2) = (4\hat{j}) \cdot (-20\hat{i} + 10\hat{j} - 20\hat{k}) = 4 \times 10 = 40$.
$D = \frac{|40|}{30} = \frac{4}{3}$.

(B) The shortest distance $d$ between two skew lines $r = a_1 + t b_1$ and $r = a_2 + s b_2$ is given by $d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|}$.
Here,$a_1 = -\hat{i} + 3\hat{k}$,$b_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$,$a_2 = 3\hat{i} + \hat{j} - \hat{k}$,and $b_2 = 2\hat{i} - \hat{j} + 2\hat{k}$.
First,calculate $a_2 - a_1 = (3 - (-1))\hat{i} + (1 - 0)\hat{j} + (-1 - 3)\hat{k} = 4\hat{i} + \hat{j} - 4\hat{k}$.
Next,calculate the cross product $b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & -1 & 2 \end{vmatrix} = \hat{i}(6 - (-6)) - \hat{j}(4 - 12) + \hat{k}(-2 - 6) = 12\hat{i} + 8\hat{j} - 8\hat{k}$.
The magnitude $|b_1 \times b_2| = \sqrt{12^2 + 8^2 + (-8)^2} = \sqrt{144 + 64 + 64} = \sqrt{272} = \sqrt{16 \times 17} = 4\sqrt{17}$.
The dot product $(a_2 - a_1) \cdot (b_1 \times b_2) = (4\hat{i} + \hat{j} - 4\hat{k}) \cdot (12\hat{i} + 8\hat{j} - 8\hat{k}) = (4 \times 12) + (1 \times 8) + (-4 \times -8) = 48 + 8 + 32 = 88$.
Therefore,$d = \frac{|88|}{4\sqrt{17}} = \frac{22}{\sqrt{17}}$.

(A) The given equation is $S = 2 x^2 - 6 y^2 - 12 z^2 + 18 y z + 2 z x + x y = 0$.
Factorizing the quadratic form,we get $S = (x + 2 y - 2 z)(2 x - 3 y + 6 z) = 0$.
Thus,the two planes are $P_1: x + 2 y - 2 z = 0$ and $P_2: 2 x - 3 y + 6 z = 0$.
The normal vectors to these planes are $\vec{n_1} = (1, 2, -2)$ and $\vec{n_2} = (2, -3, 6)$.
The angle $\theta$ between the planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (1)(2) + (2)(-3) + (-2)(6) = 2 - 6 - 12 = -16$.
Calculating the magnitudes: $|\vec{n_1}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$ and $|\vec{n_2}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7$.
Therefore,$\cos \theta = \frac{|-16|}{3 \times 7} = \frac{16}{21}$.
Hence,the acute angle is $\theta = \cos ^{-1}\left(\frac{16}{21}\right)$.

(A) The equation of the plane passing through $A(0,0,2)$,$B(1,0,1)$,and $C(3,1,1)$ is given by the determinant equation:
$\begin{vmatrix} x-0 & y-0 & z-2 \\ 1-0 & 0-0 & 1-2 \\ 3-0 & 1-0 & 1-2 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} x & y & z-2 \\ 1 & 0 & -1 \\ 3 & 1 & -1 \end{vmatrix} = 0$
Expanding along the first row: $x(0 - (-1)) - y(-1 - (-3)) + (z-2)(1 - 0) = 0$
$x(1) - y(2) + (z-2)(1) = 0$
$x - 2y + z - 2 = 0$.
The normal vector to the plane is $\vec{n} = \langle 1, -2, 1 \rangle$.
The $XY$-plane has normal $\vec{n}_1 = \langle 0, 0, 1 \rangle$. The angle $\alpha$ between the planes is given by $\cos \alpha = \frac{|\vec{n} \cdot \vec{n}_1|}{|\vec{n}| |\vec{n}_1|} = \frac{|1|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{1^2}} = \frac{1}{\sqrt{6}}$.
Thus,$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{6} = \frac{5}{6}$.
The $XZ$-plane has normal $\vec{n}_2 = \langle 0, 1, 0 \rangle$. The angle $\beta$ between the planes is given by $\cos \beta = \frac{|\vec{n} \cdot \vec{n}_2|}{|\vec{n}| |\vec{n}_2|} = \frac{|-2|}{\sqrt{6} \sqrt{1^2}} = \frac{2}{\sqrt{6}}$.
Thus,$\sin^2 \beta = 1 - \cos^2 \beta = 1 - \frac{4}{6} = \frac{2}{6}$.
Therefore,$\sin^2 \alpha + \sin^2 \beta = \frac{5}{6} + \frac{2}{6} = \frac{7}{6}$.

(C) The two given lines pass through the point with position vector $a=\hat{i}+\hat{j}$ and are parallel to the vectors $b_1=\hat{i}+2 \hat{j}-\hat{k}$ and $b_2=-\hat{i}+\hat{j}-2 \hat{k}$ respectively. The plane containing these lines passes through the point $a=\hat{i}+\hat{j}$ and is normal to the vector $n = b_1 \times b_2$.
Calculating the normal vector $n$:
$n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix} = \hat{i}(-4+1) - \hat{j}(-2-1) + \hat{k}(1+2) = -3\hat{i} + 3\hat{j} + 3\hat{k}$.
The vector equation of the plane is $r \cdot n = a \cdot n$.
$r \cdot (-3\hat{i} + 3\hat{j} + 3\hat{k}) = (\hat{i} + \hat{j}) \cdot (-3\hat{i} + 3\hat{j} + 3\hat{k}) = -3 + 3 + 0 = 0$.
Dividing by $-3$,we get $r \cdot (\hat{i} - \hat{j} - \hat{k}) = 0$.

(B) The equation of the plane $OAB$ is given by the determinant form:
$\begin{vmatrix} x-0 & y-0 & z-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = 0$
Expanding the determinant: $x(6-1) - y(3-2) + z(1-4) = 0 \Rightarrow 5x - y - 3z = 0$ ... $(i)$
The equation of the plane $ABC$ is given by:
$\begin{vmatrix} x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = 0$
Expanding the determinant: $(x-1)(-1+2) - (y-2)(1+4) + (z-1)(-1-2) = 0$
$(x-1)(1) - (y-2)(5) + (z-1)(-3) = 0 \Rightarrow x - 1 - 5y + 10 - 3z + 3 = 0 \Rightarrow x - 5y - 3z + 12 = 0$ ... $(ii)$
The angle $\theta$ between the planes $5x - y - 3z = 0$ and $x - 5y - 3z + 12 = 0$ is given by:
$\cos \theta = \frac{|(5)(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2} \sqrt{1^2 + (-5)^2 + (-3)^2}}$
$\cos \theta = \frac{|5 + 5 + 9|}{\sqrt{25 + 1 + 9} \sqrt{1 + 25 + 9}} = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$

(A) The normal vector $\vec{n_1}$ to the plane $\pi_1$ is the vector joining the point $(1, 1, 1)$ and the foot of the perpendicular $(1, 3, 5)$.
$\vec{n_1} = (1-1, 3-1, 5-1) = (0, 2, 4)$.
We can simplify this to $\vec{n_1} = (0, 1, 2)$.
The equation of plane $\pi_1$ passing through $(1, 3, 5)$ is $0(x-1) + 1(y-3) + 2(z-5) = 0$,which simplifies to $y + 2z - 13 = 0$.
For plane $\pi_2$,it passes through points $A(2, 2, -1), B(3, 4, 2), C(3, 3, 0)$.
The vectors $\vec{AB} = (1, 2, 3)$ and $\vec{AC} = (1, 1, 1)$ lie on the plane $\pi_2$.
The normal vector $\vec{n_2}$ is $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(1-3) + \hat{k}(1-2) = -\hat{i} + 2\hat{j} - \hat{k}$.
So,$\vec{n_2} = (-1, 2, -1)$.
The angle $\theta$ between the planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\vec{n_1} \cdot \vec{n_2} = (0)(-1) + (1)(2) + (2)(-1) = 0 + 2 - 2 = 0$.
Since the dot product is $0$,the angle $\theta = \frac{\pi}{2}$.

(C) The equation of any plane passing through the line of intersection of planes $\pi_1 = 0$ and $\pi_2 = 0$ is given by $\pi_1 + \lambda \pi_2 = 0$.
Substituting the given planes:
$(2x + 6y + 4z - 7) + \lambda(x - y - 2z - 2) = 0$
$(2 + \lambda)x + (6 - \lambda)y + (4 - 2\lambda)z - (7 + 2\lambda) = 0 \quad \dots(i)$
Since this plane is perpendicular to the plane $x + y + 2z - 5 = 0$,the dot product of their normal vectors must be zero.
The normal vectors are $\vec{n_1} = (2 + \lambda, 6 - \lambda, 4 - 2\lambda)$ and $\vec{n_2} = (1, 1, 2)$.
$(2 + \lambda)(1) + (6 - \lambda)(1) + (4 - 2\lambda)(2) = 0$
$2 + \lambda + 6 - \lambda + 8 - 4\lambda = 0$
$16 - 4\lambda = 0 \implies \lambda = 4$.
Substituting $\lambda = 4$ into equation $(i)$:
$(2 + 4)x + (6 - 4)y + (4 - 8)z - (7 + 8) = 0$
$6x + 2y - 4z - 15 = 0$.

(C) The plane passes through point $A$ with position vector $\vec{a}$ and is parallel to vectors $\vec{b}$ and $\vec{c}$.
Therefore,the normal vector to the plane is $\vec{n} = \vec{b} \times \vec{c}$.
The unit normal vector is $\hat{n} = \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$.
The equation of the plane in normal form is given by $\vec{r} \cdot \hat{n} = \vec{a} \cdot \hat{n}$.
Substituting $\hat{n}$,we get $\vec{r} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} = \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$.
Since the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$,the equation becomes $\vec{r} \cdot \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} = \frac{[\vec{a} \vec{b} \vec{c}]}{|\vec{b} \times \vec{c}|}$.

(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ and perpendicular to two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ is given by the determinant form:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Substituting the given values:
$\begin{vmatrix} x+1 & y-6 & z-2 \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix} = 0$
Expanding the determinant:
$(x+1)(4-6) - (y-6)(2-6) + (z-2)(3-6) = 0$
$-2(x+1) + 4(y-6) - 3(z-2) = 0$
$-2x - 2 + 4y - 24 - 3z + 6 = 0$
$-2x + 4y - 3z - 20 = 0$ or $2x - 4y + 3z + 20 = 0$
The perpendicular distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $Ax+By+Cz+D=0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
For the point $(1, -1, 1)$ and the plane $2x - 4y + 3z + 20 = 0$:
$d = \frac{|2(1) - 4(-1) + 3(1) + 20|}{\sqrt{2^2 + (-4)^2 + 3^2}}$
$d = \frac{|2 + 4 + 3 + 20|}{\sqrt{4 + 16 + 9}} = \frac{29}{\sqrt{29}} = \sqrt{29}$

(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the $x, y, z$ intercepts respectively.
The coordinates of the vertices are $A(a, 0, 0)$,$B(0, b, 0)$,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 $(2, 3, 5)$,we have:
$\frac{a}{3} = 2 \implies a = 6$
$\frac{b}{3} = 3 \implies b = 9$
$\frac{c}{3} = 5 \implies c = 15$
Substituting these values into the intercept form of the plane equation:
$\frac{x}{6} + \frac{y}{9} + \frac{z}{15} = 1$
To simplify,multiply the entire equation by the least common multiple of $6, 9, 15$,which is $90$:
$15x + 10y + 6z = 90$.

(A) Let the origin be $O(0, 0, 0)$. The vertices of the rectangular parallelopiped are $O(0, 0, 0)$,$A(a, 0, 0)$,$E(0, b, 0)$,$D(0, 0, c)$,and others.
$d_1$ is a diagonal of the face in the $XY$-plane not passing through the origin. The vertices of this face are $(0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0)$. The diagonal not passing through the origin is the line segment connecting $(a, 0, 0)$ and $(0, b, 0)$,i.e.,$AE$.
The vector along $d_1$ is $\vec{v_1} = (0-a)\hat{i} + (b-0)\hat{j} + (0-0)\hat{k} = -a\hat{i} + b\hat{j}$.
$d_2$ is a diagonal of the plane $\Pi_2$ (which is parallel to $ZX$ plane at distance $b$) coterminous with $d_1$. The plane $\Pi_2$ contains points $(0, b, 0), (a, b, 0), (0, b, c), (a, b, c)$. The diagonal coterminous with $d_1$ (which starts at $A(a, 0, 0)$) is $AD$,where $D$ is $(0, 0, c)$.
Wait,looking at the provided image,$d_1$ is $AE$ and $d_2$ is $AD$.
The vector $\vec{AE} = (0-a)\hat{i} + (b-0)\hat{j} + (0-0)\hat{k} = -a\hat{i} + b\hat{j}$.
The vector $\vec{AD} = (0-a)\hat{i} + (0-0)\hat{j} + (c-0)\hat{k} = -a\hat{i} + c\hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{\vec{AE} \cdot \vec{AD}}{|\vec{AE}| |\vec{AD}|}$.
$\vec{AE} \cdot \vec{AD} = (-a)(-a) + (b)(0) + (0)(c) = a^2$.
$|\vec{AE}| = \sqrt{(-a)^2 + b^2} = \sqrt{a^2+b^2}$.
$|\vec{AD}| = \sqrt{(-a)^2 + c^2} = \sqrt{a^2+c^2}$.
Thus,$\cos \theta = \frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\right)$.

(C) The line is given by $r = a + \lambda b$,where $a = 2\hat{i} - 2\hat{j} + 3\hat{k}$ and $b = \hat{i} - \hat{j} + 4\hat{k}$.
The plane is given by $r \cdot n = d$,where $n = \hat{i} + 5\hat{j} + \hat{k}$ and $d = 5$.
First,check if the line is parallel to the plane by calculating $b \cdot n = (1)(1) + (-1)(5) + (4)(1) = 1 - 5 + 4 = 0$.
Since $b \cdot n = 0$,the line is parallel to the plane.
The shortest distance $D$ between a parallel line and a plane is given by the formula $D = \frac{|a \cdot n - d|}{|n|}$.
Calculate $a \cdot n = (2)(1) + (-2)(5) + (3)(1) = 2 - 10 + 3 = -5$.
Calculate $|n| = \sqrt{1^2 + 5^2 + 1^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}$.
Substitute the values into the formula: $D = \frac{|-5 - 5|}{|3\sqrt{3}|} = \frac{|-10|}{3\sqrt{3}} = \frac{10}{3\sqrt{3}}$.

(A) Given,the position vector of point $P$ is $\vec{p} = 2 \hat{i}+\hat{j}+3 \hat{k}$.
The plane $\pi$ is determined by vectors $\vec{a} = -\hat{i}-2 \hat{k}$ and $\vec{b} = \hat{i}+\hat{j}+2 \hat{k}$.
The normal vector to the plane is $\vec{n} = \vec{a} \times \vec{b}$.
The line lies on the plane and is normal to $\vec{b}$,so its direction vector $\vec{v}$ must be perpendicular to both $\vec{n}$ and $\vec{b}$.
Thus,$\vec{v} = \vec{n} \times \vec{b} = (\vec{a} \times \vec{b}) \times \vec{b}$.
Using the vector triple product formula $(\vec{a} \times \vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}$.
Calculate $\vec{a} \cdot \vec{b} = (-1)(1) + (0)(1) + (-2)(2) = -1 - 4 = -5$.
Calculate $\vec{b} \cdot \vec{b} = (1)^2 + (1)^2 + (2)^2 = 1 + 1 + 4 = 6$.
So,$\vec{v} = -5(\hat{i}+\hat{j}+2 \hat{k}) - 6(-\hat{i}-2 \hat{k}) = -5\hat{i}-5\hat{j}-10\hat{k} + 6\hat{i} + 12\hat{k} = \hat{i}-5\hat{j}+2\hat{k}$.
Note: The direction vector can be scaled by $-1$,giving $-\hat{i}+5\hat{j}-2\hat{k}$.
The equation of the line is $\vec{r} = \vec{p} + \lambda \vec{v} = 2 \hat{i}+\hat{j}+3 \hat{k} + \lambda(-\hat{i}+5 \hat{j}-2 \hat{k})$.

(C) The equation of the plane is given by $r = b + c + x(a - b) + y(c + a) = (x + y)a + (1 - x)b + (1 + y)c$ $\ldots(i)$.
The equation of the line is given by $r = a + t(b - c) = a + tb - tc$ $\ldots(ii)$.
Since the point of intersection satisfies both equations,we equate the coefficients of $a, b,$ and $c$ because $a, b, c$ are non-coplanar:
$x + y = 1$ $\ldots(iii)$
$1 - x = t$ $\ldots(iv)$
$1 + y = -t$ $\ldots(v)$
Adding equations $(iv)$ and $(v)$,we get $2 - x + y = 0$,so $x - y = 2$ $\ldots(vi)$.
Adding $(iii)$ and $(vi)$,we get $2x = 3$,so $x = \frac{3}{2}$.
Substituting $x = \frac{3}{2}$ into $(iii)$,we get $y = 1 - \frac{3}{2} = -\frac{1}{2}$.
Substituting $x = \frac{3}{2}$ into $(iv)$,we get $t = 1 - \frac{3}{2} = -\frac{1}{2}$.
Now,the point of intersection $r$ is $a + t(b - c) = a - \frac{1}{2}(b - c) = a - \frac{1}{2}b + \frac{1}{2}c$.
Comparing this with $la + mb + nc$,we get $l = 1, m = -\frac{1}{2}, n = \frac{1}{2}$.
Finally,$3l + 4m + 2n = 3(1) + 4(-\frac{1}{2}) + 2(\frac{1}{2}) = 3 - 2 + 1 = 2$.

(B) The equation of the plane $\Pi$ passing through points $A(0,-5,-1), B(1,-2,5), C(-3,5,0)$ is given by the determinant equation:
$\begin{vmatrix} x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1 \end{vmatrix} = 0$
$\begin{vmatrix} x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$x(3-60) - (y+5)(1+18) + (z+1)(10+9) = 0$
$-57x - 19(y+5) + 19(z+1) = 0$
Dividing by $-19$:
$3x + y + 5 - z - 1 = 0 \Rightarrow 3x + y - z + 4 = 0$
The normal vector to the plane is $\vec{n} = 3\hat{i} + \hat{j} - \hat{k}$.
The unit normal vector is $\hat{n} = \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{3^2 + 1^2 + (-1)^2}} = \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{11}}$.
The line $L$ is parallel to the vector $\vec{v} = \hat{i} + 5\hat{j} - 6\hat{k}$.
The unit vector along the line $L$ is $\hat{u} = \frac{\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{1^2 + 5^2 + (-6)^2}} = \frac{\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{62}}$.
The length of the projection of $\hat{n}$ on line $L$ is $|\hat{n} \cdot \hat{u}|$:
$|\hat{n} \cdot \hat{u}| = \left| \left( \frac{3\hat{i} + \hat{j} - \hat{k}}{\sqrt{11}} \right) \cdot \left( \frac{\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{62}} \right) \right|$
$= \left| \frac{3(1) + 1(5) + (-1)(-6)}{\sqrt{11} \cdot \sqrt{62}} \right| = \left| \frac{3 + 5 + 6}{\sqrt{682}} \right| = \frac{14}{\sqrt{682}}$.

(B) Let $l, m, n$ be $\hat{i}, \hat{j}, \hat{k}$.
Points are $A(p, 7, -6), B(2, 5, -4), C(1, 4, -3)$.
Since $A, B, C$ are collinear,the vector $\vec{AB}$ must be parallel to $\vec{BC}$.
$\vec{AB} = (2-p)\hat{i} - 2\hat{j} + 2\hat{k}$.
$\vec{BC} = (1-2)\hat{i} + (4-5)\hat{j} + (-3+4)\hat{k} = -\hat{i} - \hat{j} + \hat{k}$.
Since $\vec{AB} = k\vec{BC}$,we have $\frac{2-p}{-1} = \frac{-2}{-1} = \frac{2}{1} = 2$.
Thus,$2-p = -2 \implies p = 4$.
The point is $(-p, p, p+1) = (-4, 4, 5)$.
The line $L$ passes through $B(2, 5, -4)$ and has direction vector $\vec{v} = (-1, -1, 1)$.
The plane contains the line $L$ and the point $P(-4, 4, 5)$.
The normal to the plane $\vec{n}$ is $\vec{PB} \times \vec{v}$.
$\vec{PB} = (2 - (-4))\hat{i} + (5 - 4)\hat{j} + (-4 - 5)\hat{k} = 6\hat{i} + \hat{j} - 9\hat{k}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & -9 \\ -1 & -1 & 1 \end{vmatrix} = \hat{i}(1-9) - \hat{j}(6-9) + \hat{k}(-6+1) = -8\hat{i} + 3\hat{j} - 5\hat{k}$.
The equation of the plane is $-8(x-2) + 3(y-5) - 5(z+4) = 0$.
$-8x + 16 + 3y - 15 - 5z - 20 = 0 \implies -8x + 3y - 5z = 19$.
Dividing by $19$,we get $-\frac{8}{19}x + \frac{3}{19}y - \frac{5}{19}z = 1$.
Here $a = -\frac{8}{19}, b = \frac{3}{19}, c = -\frac{5}{19}$.
$p(a+b+c) = 4 \times (\frac{-8+3-5}{19}) = 4 \times (\frac{-10}{19}) = -\frac{40}{19}$.

(A) The given line is $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$.
This line passes through the point $P(4, 2, 7)$ and has direction ratios $(1, 1, 2)$.
Since the line lies in the plane $ax+by+z=7$,the point $P(4, 2, 7)$ must satisfy the plane equation:
$a(4) + b(2) + 7 = 7$
$4a + 2b = 0$
$2a + b = 0 \quad \dots(i)$
Also,the direction vector of the line $\vec{v} = (1, 1, 2)$ must be perpendicular to the normal vector of the plane $\vec{n} = (a, b, 1)$.
Thus,their dot product is zero:
$(1)(a) + (1)(b) + (2)(1) = 0$
$a + b + 2 = 0$
$a + b = -2$.

(A) Given the position vectors of $A$ and $B$ are $A(1, 2, 0)$ and $B(2, 1, 1)$.
The equation of the plane is $x+y+z=3$.
The normal vector to the plane is $\vec{n} = \hat{i}+\hat{j}+\hat{k}$.
The line passing through $A$ and perpendicular to the plane is given by $\vec{r} = (\hat{i}+2\hat{j}) + \lambda(\hat{i}+\hat{j}+\hat{k})$.
Any point on this line is $(1+\lambda, 2+\lambda, \lambda)$. Since this point lies on the plane $x+y+z=3$,we have $(1+\lambda) + (2+\lambda) + \lambda = 3$,which gives $3\lambda + 3 = 3$,so $\lambda = 0$. Thus,$P = (1, 2, 0)$.
The line passing through $B$ and perpendicular to the plane is given by $\vec{r} = (2\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}+\hat{j}+\hat{k})$.
Any point on this line is $(2+\mu, 1+\mu, 1+\mu)$. Since this point lies on the plane $x+y+z=3$,we have $(2+\mu) + (1+\mu) + (1+\mu) = 3$,which gives $3\mu + 4 = 3$,so $\mu = -\frac{1}{3}$.
Thus,$Q = (2-\frac{1}{3}, 1-\frac{1}{3}, 1-\frac{1}{3}) = (\frac{5}{3}, \frac{2}{3}, \frac{2}{3})$.
The distance $PQ$ is given by $\sqrt{(\frac{5}{3}-1)^2 + (\frac{2}{3}-2)^2 + (\frac{2}{3}-0)^2} = \sqrt{(\frac{2}{3})^2 + (-\frac{4}{3})^2 + (\frac{2}{3})^2} = \sqrt{\frac{4}{9} + \frac{16}{9} + \frac{4}{9}} = \sqrt{\frac{24}{9}} = \frac{\sqrt{24}}{3} = \frac{2\sqrt{6}}{3} = \frac{2\sqrt{2}}{\sqrt{3}}$.

(A) The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
The outcomes where the sum is at most $7$ are:
Sum $= 2: (1,1)$ ($1$ outcome)
Sum $= 3: (1,2), (2,1)$ ($2$ outcomes)
Sum $= 4: (1,3), (2,2), (3,1)$ ($3$ outcomes)
Sum $= 5: (1,4), (2,3), (3,2), (4,1)$ ($4$ outcomes)
Sum $= 6: (1,5), (2,4), (3,3), (4,2), (5,1)$ ($5$ outcomes)
Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ ($6$ outcomes)
Total favorable outcomes $= 1+2+3+4+5+6 = 21$.
Thus,$x = \frac{21}{36} = \frac{7}{12}$.
For $y$,the probability of getting a sum of $7$ in one roll is $p = \frac{6}{36} = \frac{1}{6}$. The probability of not getting a sum of $7$ is $q = 1 - \frac{1}{6} = \frac{5}{6}$.
When rolled $n$ times,the probability of getting a sum of $7$ at least once is $y = 1 - q^n = 1 - (\frac{5}{6})^n$.
We want $y > x$,so $1 - (\frac{5}{6})^n > \frac{7}{12} \Rightarrow (\frac{5}{6})^n < 1 - \frac{7}{12} = \frac{5}{12}$.
For $n=1: \frac{5}{6} \approx 0.833 > 0.416$
For $n=2: \frac{25}{36} \approx 0.694 > 0.416$
For $n=3: \frac{125}{216} \approx 0.578 > 0.416$
For $n=4: \frac{625}{1296} \approx 0.482 > 0.416$
For $n=5: \frac{3125}{7776} \approx 0.401 < 0.416$.
Thus,the minimum $n$ is $5$.

(C) Let the probability of a bomb hitting the target be $p = \frac{3}{4}$.
Thus,the probability of a bomb missing the target is $q = 1 - p = \frac{1}{4}$.
Let $n$ be the number of bombs dropped. The target is destroyed if there are at least $2$ hits.
Let $X$ be the number of hits. We want $P(X \geq 2) \geq 0.99$.
This is equivalent to $1 - [P(X = 0) + P(X = 1)] \geq 0.99$.
$P(X = 0) + P(X = 1) \leq 0.01 = \frac{1}{100}$.
Using the binomial distribution: $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.
${}^{n}C_{0} (\frac{3}{4})^{0} (\frac{1}{4})^{n} + {}^{n}C_{1} (\frac{3}{4})^{1} (\frac{1}{4})^{n-1} \leq \frac{1}{100}$.
$\frac{1}{4^{n}} + n \cdot \frac{3}{4} \cdot \frac{4}{4^{n}} \leq \frac{1}{100}$.
$\frac{1 + 3n}{4^{n}} \leq \frac{1}{100} \Rightarrow 4^{n} \geq 100(3n + 1) = 300n + 100$.
For $n = 5$: $4^{5} = 1024$ and $300(5) + 100 = 1600$. ($1024 < 1600$,false).
For $n = 6$: $4^{6} = 4096$ and $300(6) + 100 = 1900$. ($4096 \geq 1900$,true).
Thus,the minimum number of bombs is $6$.

(B) Let $S = \{1, 2, \ldots, 100\}$. The total number of elements is $100$.
Let $A$ be the event that the number is divisible by $2$. The elements of $A$ are $\{2, 4, 6, \ldots, 100\}$. The number of elements in $A$ is $n(A) = 50$.
Let $B$ be the event that the number is divisible by $3$ or $5$. We want to find the conditional probability $P(B|A) = \frac{n(A \cap B)}{n(A)}$.
$A \cap B$ is the set of numbers in $\{1, 2, \ldots, 100\}$ that are divisible by $2$ $AND$ (divisible by $3$ $OR$ divisible by $5$).
This is equivalent to numbers divisible by $6$ or $10$.
Numbers divisible by $6$ up to $100$: $\lfloor \frac{100}{6} \rfloor = 16$.
Numbers divisible by $10$ up to $100$: $\lfloor \frac{100}{10} \rfloor = 10$.
Numbers divisible by both $6$ and $10$ (i.e.,divisible by $30$): $\lfloor \frac{100}{30} \rfloor = 3$.
By the Principle of Inclusion-Exclusion,$n(A \cap B) = 16 + 10 - 3 = 23$.
The required probability is $\frac{n(A \cap B)}{n(A)} = \frac{23}{50}$.

(C) The probability that exactly $5$ cards are drawn before the first ace appears means the first $5$ cards are non-aces and the $6$th card is an ace.
$P = \frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times \frac{45}{49} \times \frac{44}{48} \times \frac{4}{47}$
Simplifying the expression:
$P = \frac{4}{49} \times \left( \frac{48 \times 47 \times 46 \times 45 \times 44}{52 \times 51 \times 50 \times 48 \times 47} \right) = \frac{4}{49} \times \left( \frac{46 \times 45 \times 44}{52 \times 51 \times 50} \right)$
$P = \frac{4}{49} \times \left( \frac{23 \times 2 \times 9 \times 5 \times 11 \times 4}{13 \times 4 \times 17 \times 3 \times 25 \times 2} \right) = \frac{4}{49} \times \left( \frac{23 \times 3 \times 11}{13 \times 17 \times 5} \right)$
Here,the prime factors are $p_1=23, p_2=11, p_3=3$ and $p_4=13, p_5=17, p_6=5$.
Thus,$\max \{p_i\} = 23$ and $\min \{p_i\} = 3$.
Therefore,$\max \{p_i\} - \min \{p_i\} = 23 - 3 = 20$.

(B) Given that a fair coin is tossed $K$ times,the probability of getting $X$ heads follows a binomial distribution with $p = \frac{1}{2}$ and $q = \frac{1}{2}$.
Given $P(X=4) = P(X=6)$.
Using the binomial probability formula $P(X=r) = {}^K C_r p^r q^{K-r}$,we have:
${}^K C_4 (\frac{1}{2})^K = {}^K C_6 (\frac{1}{2})^K$
${}^K C_4 = {}^K C_6$
Since ${}^n C_x = {}^n C_y$ implies $n = x + y$ (for $x \neq y$),we get $K = 4 + 6 = 10$.
For a binomial distribution with $p = q = \frac{1}{2}$,the probability $P(X=r)$ is maximum at the mean value.
For $n = 10$,the maximum probability occurs at $r = \frac{n}{2} = \frac{10}{2} = 5$.

(B) Given $n = 2000$ and $p = 0.001$.
Using the Poisson distribution,the parameter $\lambda = np = 2000 \times 0.001 = 2$.
The probability of $X$ individuals suffering a reaction is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need to find $P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
Calculating the individual probabilities:
$P(X=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$
$P(X=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$
$P(X=2) = \frac{e^{-2} 2^2}{2!} = 2e^{-2}$
Summing these: $P(X \le 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}$.
Therefore,$P(X > 2) = 1 - 5e^{-2} = 1 - \frac{5}{e^2}$.

(A) Let $E_1, E_2, E_3$ be three independent events with probabilities $P_1, P_2, P_3$.
The probability that none of the events occur is $P(\bar{E}_1 \cap \bar{E}_2 \cap \bar{E}_3) = (1 - P_1)(1 - P_2)(1 - P_3)$.
The probability that at least one event occurs is $1 - P(\text{none occur}) = 1 - [(1 - P_1)(1 - P_2)(1 - P_3)]$. Thus,$(A)$ is true.
For independent events $A, B, C$,the probability of their union is given by the inclusion-exclusion principle: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(A \cap C) + P(B \cap C)] + P(A \cap B \cap C)$.
Since the events are independent,$P(A \cap B) = P(A)P(B)$,etc. Thus,$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)$. Thus,$(R)$ is true.
Since the formula in $(R)$ is the standard expansion used to derive the result in $(A)$,$(R)$ is the correct explanation for $(A)$.

(A) Let $E$ be the event of getting a number greater than $3$ in a single throw of a die. The outcomes are $\{4, 5, 6\}$.
$P(E) = \frac{3}{6} = \frac{1}{2}$.
Let $F$ be the event of getting a $5$ in a single throw. $P(F) = \frac{1}{6}$.
Let $S$ be the event of getting a number $\leq 3$ in a single throw. $P(S) = \frac{3}{6} = \frac{1}{2}$.
The man stops when he gets a number $> 3$. The possible sequences for the last throw being $5$ are:
$1$. First throw is $5$: Probability $= \frac{1}{6}$.
$2$. First throw is $\leq 3$,second throw is $5$: Probability $= \frac{1}{2} \times \frac{1}{6}$.
$3$. First two throws are $\leq 3$,third throw is $5$: Probability $= (\frac{1}{2})^2 \times \frac{1}{6}$.
This is an infinite geometric series: $\frac{1}{6} + \frac{1}{6}(\frac{1}{2}) + \frac{1}{6}(\frac{1}{2})^2 + \dots$
The sum is $S = \frac{a}{1-r} = \frac{1/6}{1-1/2} = \frac{1/6}{1/2} = \frac{1}{3}$.

(C) The set is $S = \{1, 2, 3, \ldots, 13\}$. There are $7$ odd numbers $\{1, 3, 5, 7, 9, 11, 13\}$ and $6$ even numbers $\{2, 4, 6, 8, 10, 12\}$.
Let $E$ be the event that the sum of the two selected numbers is even. This happens if both numbers are odd or both numbers are even.
Number of ways to select two odd numbers = $^7C_2 = \frac{7 \times 6}{2} = 21$.
Number of ways to select two even numbers = $^6C_2 = \frac{6 \times 5}{2} = 15$.
Total ways for the sum to be even = $21 + 15 = 36$.
Let $A$ be the event that both numbers are odd. We want to find $P(A|E) = \frac{n(A \cap E)}{n(E)}$.
Since $A \cap E$ is the event that both numbers are odd,$n(A \cap E) = 21$.
Therefore,$P(A|E) = \frac{21}{36} = \frac{7}{12}$.

(C) We have,$P(E_1 \cap E_2) = P(E_1) \cdot P(E_2 | E_1)$.
$\therefore P(E_1 \cap E_2) = \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$.
Now,$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$.
$\Rightarrow \frac{1}{2} = \frac{1/12}{P(E_2)}$.
$\Rightarrow P(E_2) = \frac{1}{12} \times 2 = \frac{1}{6}$.
$\therefore P(\bar{E}_2) = 1 - P(E_2) = 1 - \frac{1}{6} = \frac{5}{6}$.
Now,$P(E_1 | \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{P(\bar{E}_2)}$.
$= \frac{\frac{1}{4} - \frac{1}{12}}{\frac{5}{6}} = \frac{\frac{3-1}{12}}{\frac{5}{6}} = \frac{\frac{2}{12}}{\frac{5}{6}} = \frac{1}{6} \times \frac{6}{5} = \frac{1}{5}$.

(B) Given that $P(A) = \frac{4}{9}$ and $P(A \cap \bar{B}) = \frac{3}{7}$.
We know that $P(A) = P(A \cap B) + P(A \cap \bar{B})$.
Therefore,$P(A \cap B) = P(A) - P(A \cap \bar{B})$.
Substituting the values,we get $P(A \cap B) = \frac{4}{9} - \frac{3}{7} = \frac{28 - 27}{63} = \frac{1}{63}$.
Now,the conditional probability $P\left(\frac{B}{A}\right)$ is defined as $P\left(\frac{B}{A}\right) = \frac{P(A \cap B)}{P(A)}$.
Substituting the values,$P\left(\frac{B}{A}\right) = \frac{\frac{1}{63}}{\frac{4}{9}} = \frac{1}{63} \times \frac{9}{4} = \frac{1}{7 \times 4} = \frac{1}{28}$.

(C) Given: $P(X)=\frac{1}{3}$,$P(X|Y)=\frac{1}{2}$,and $P(Y|X)=\frac{2}{5}$.
Using the definition of conditional probability,$P(Y|X) = \frac{P(X \cap Y)}{P(X)}$.
Therefore,$P(X \cap Y) = P(Y|X) \times P(X) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$.
Now,using $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$,we have $\frac{1}{2} = \frac{2/15}{P(Y)}$.
This implies $P(Y) = 2 \times \frac{2}{15} = \frac{4}{15}$.
Thus,option $C$ is correct.

(C) Let $E_1$ be the event that the person suffers from the disease and $E_2$ be the event that the person does not suffer from the disease. Let $A$ be the event that the diagnostic test is positive.
Given:
$P(E_1) = 0.5 \% = 0.005$
$P(E_2) = 99.5 \% = 0.995$
$P(A|E_1) = 0.95$
$P(A|E_2) = 0.10$
We need to find $P(E_1|A)$. Using Bayes' Theorem:
$P(E_1|A) = \frac{P(E_1) \times P(A|E_1)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)}$
$P(E_1|A) = \frac{0.005 \times 0.95}{(0.005 \times 0.95) + (0.995 \times 0.10)}$
$P(E_1|A) = \frac{0.00475}{0.00475 + 0.0995}$
$P(E_1|A) = \frac{0.00475}{0.10425} = \frac{475}{10425} \approx 0.0455$

(A) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.
Since $E_1$ and $E_2$ are independent events,$P(E_1 \cap E_2) = P(E_1) \times P(E_2)$.
Let $P(E_2) = x$. Then $P(E_1 \cap E_2) = \frac{1}{2}x$.
Using the formula $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$:
$\frac{2}{3} = \frac{1}{2} + x - \frac{x}{2}$
$\frac{2}{3} = \frac{1}{2} + \frac{x}{2}$
$\frac{x}{2} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$
$x = \frac{1}{3}$. Thus,$P(E_2) = \frac{1}{3}$. $(A \rightarrow III)$
Now,$P(E_1 \cap E_2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.
$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1/6}{1/3} = \frac{1}{2}$. $(B \rightarrow IV)$
$P(\bar{E}_2 | E_1) = 1 - P(E_2 | E_1) = 1 - P(E_2) = 1 - \frac{1}{3} = \frac{2}{3}$. $(C \rightarrow I)$
$P(\bar{E}_1 \cup \bar{E}_2) = P(\overline{E_1 \cap E_2}) = 1 - P(E_1 \cap E_2) = 1 - \frac{1}{6} = \frac{5}{6}$. $(D \rightarrow II)$
Therefore,the correct match is $A-III, B-IV, C-I, D-II$.

(C) Given that $E_1, E_2, \ldots, E_n$ are independent events with $P(E_r) = \frac{1}{1+r}$.
First,we find the probability of the complement event $\bar{E}_r$ for each $r$:
$P(\bar{E}_r) = 1 - P(E_r) = 1 - \frac{1}{1+r} = \frac{r}{1+r}$.
The probability that at least one of the events occurs is given by $1 - P(\text{none of the events occur})$.
Since the events are independent,the probability that none of the events occur is the product of the probabilities of their complements:
$P(\text{none}) = P(\bar{E}_1) \times P(\bar{E}_2) \times \cdots \times P(\bar{E}_n)$.
Substituting the values:
$P(\text{none}) = \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{3}{4}\right) \times \cdots \times \left(\frac{n}{n+1}\right)$.
This is a telescoping product where the numerator of each term cancels with the denominator of the previous term:
$P(\text{none}) = \frac{1}{n+1}$.
Therefore,the probability that at least one event happens is:
$1 - P(\text{none}) = 1 - \frac{1}{n+1} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

(D) Let $E_1, E_2, E_3, E_4$ be the events of selecting boxes $A, B, C, D$ respectively. Let $F$ be the event that the selected fuse is defective.
$P(E_1) = \frac{5000}{11000} = \frac{5}{11}, P(E_2) = \frac{3000}{11000} = \frac{3}{11}, P(E_3) = \frac{2000}{11000} = \frac{2}{11}, P(E_4) = \frac{1000}{11000} = \frac{1}{11}$.
The conditional probabilities of selecting a defective fuse are:
$P(F|E_1) = \frac{3}{100}, P(F|E_2) = \frac{2}{100}, P(F|E_3) = \frac{1}{100}, P(F|E_4) = \frac{0.5}{100} = \frac{1}{200}$.
Using Bayes' Theorem,the probability that the defective fuse came from box $D$ is $P(E_4|F) = \frac{P(E_4)P(F|E_4)}{\sum_{i=1}^{4} P(E_i)P(F|E_i)}$.
$P(E_4|F) = \frac{\frac{1}{11} \times \frac{1}{200}}{\frac{5}{11} \times \frac{3}{100} + \frac{3}{11} \times \frac{2}{100} + \frac{2}{11} \times \frac{1}{100} + \frac{1}{11} \times \frac{1}{200}}$.
$P(E_4|F) = \frac{\frac{1}{2200}}{\frac{15}{1100} + \frac{6}{1100} + \frac{2}{1100} + \frac{1}{2200}} = \frac{\frac{1}{2200}}{\frac{30+12+4+1}{2200}} = \frac{1}{47}$.

(C) Let $E_1$ be the event that the student answers without guessing,and $E_2$ be the event that the student answers by guessing. Given $P(E_2) = 3/7$,so $P(E_1) = 1 - 3/7 = 4/7$.
Let $A$ be the event that at least $3$ questions are answered correctly.
For $E_1$ (without guessing),the probability of success is $p = 2/3$. Using binomial distribution $B(4, 2/3)$:
$P(A|E_1) = \binom{4}{3} (2/3)^3 (1/3)^1 + \binom{4}{4} (2/3)^4 = 4 \cdot (8/27) \cdot (1/3) + 16/81 = 32/81 + 16/81 = 48/81 = 16/27$.
For $E_2$ (guessing),the probability of success is $p = 1/2$. Using binomial distribution $B(4, 1/2)$:
$P(A|E_2) = \binom{4}{3} (1/2)^4 + \binom{4}{4} (1/2)^4 = 5/16$.
Using Bayes' Theorem,we need $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$.
$P(E_1|A) = \frac{(4/7) \cdot (16/27)}{(4/7) \cdot (16/27) + (3/7) \cdot (5/16)} = \frac{64/189}{64/189 + 15/112} = \frac{64/189}{7168/21168 + 2835/21168} = \frac{64/189}{10003/21168} = \frac{64 \cdot 112}{10003} = \frac{7168}{10003} = \frac{1024}{1429}$.

(A) The average arrival rate,$\lambda$,is $240 \text{ vehicles/hour} = \frac{240}{3600} \text{ vehicles/second} = \frac{1}{15} \text{ vehicles/second}$.
For a time interval $t = 30 \text{ seconds}$,the expected number of arrivals is $\mu = \lambda t = \frac{1}{15} \times 30 = 2$.
According to the Poisson distribution,the probability of $n$ arrivals is $P(n) = \frac{\mu^n e^{-\mu}}{n!}$.
We need to find the probability of more than two vehicles arriving,i.e.,$P(n > 2) = 1 - [P(0) + P(1) + P(2)]$.
Calculating individual probabilities:
$P(0) = \frac{2^0 e^{-2}}{0!} = e^{-2}$
$P(1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2}$
$P(2) = \frac{2^2 e^{-2}}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$
Summing these probabilities: $P(n \leq 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}$.
Therefore,$P(n > 2) = 1 - 5e^{-2} = 1 - \frac{5}{e^2} = \frac{e^2 - 5}{e^2}$.

(A) Given the probability mass function $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \{0, 1, 2, \dots\}$.
Since the sum of all probabilities must be $1$,we have $\sum_{n=0}^{\infty} P(X=n) = 1$.
$k \sum_{n=0}^{\infty} \frac{n+1}{3^n} = k \left( \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots \right) = 1$.
This is an arithmetico-geometric series with $a=1$,$d=1$,and $r=\frac{1}{3}$.
The sum of the series is $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{1}{1-1/3} + \frac{1 \cdot (1/3)}{(1-1/3)^2} = \frac{3}{2} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$.
Thus,$k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}$.
We need to find $P(X < 2) = P(X=0) + P(X=1)$.
$P(X=0) = k \cdot \frac{0+1}{3^0} = k \cdot 1 = \frac{4}{9}$.
$P(X=1) = k \cdot \frac{1+1}{3^1} = k \cdot \frac{2}{3} = \frac{4}{9} \cdot \frac{2}{3} = \frac{8}{27}$.
$P(X < 2) = \frac{4}{9} + \frac{8}{27} = \frac{12+8}{27} = \frac{20}{27}$.

(A) Let $A_i$ (for $i=1, 2, 3, 4$) be the event that the urn contains $i+1$ white balls. Let $B$ be the event that two white balls are drawn.
We need to find $P(A_4 | B)$.
Since the four events $A_1, A_2, A_3, A_4$ are equally likely,$P(A_1) = P(A_2) = P(A_3) = P(A_4) = \frac{1}{4}$.
$P(B | A_i)$ is the probability that two white balls are drawn given the urn contains $i+1$ white balls.
$P(B | A_1) = \frac{^2C_2}{^5C_2} = \frac{1}{10}$.
$P(B | A_2) = \frac{^3C_2}{^5C_2} = \frac{3}{10}$.
$P(B | A_3) = \frac{^4C_2}{^5C_2} = \frac{6}{10} = \frac{3}{5}$.
$P(B | A_4) = \frac{^5C_2}{^5C_2} = 1$.
By Bayes' Theorem:
$P(A_4 | B) = \frac{P(A_4) P(B | A_4)}{\sum_{i=1}^4 P(A_i) P(B | A_i)} = \frac{\frac{1}{4} \cdot 1}{\frac{1}{4} \left( \frac{1}{10} + \frac{3}{10} + \frac{6}{10} + \frac{10}{10} \right)} = \frac{1}{\frac{20}{10}} = \frac{1}{2}$.

(B) Let $E_1, E_2, E_3$ be the events of choosing bags $B_1, B_2, B_3$ respectively.
The probabilities of choosing the bags are:
$P(E_1) = \frac{3}{6} = \frac{1}{2}$ (for outcomes $2, 3, 5$)
$P(E_2) = \frac{2}{6} = \frac{1}{3}$ (for outcomes $4, 6$)
$P(E_3) = \frac{1}{6}$ (for outcome $1$)
Let $G$ be the event of drawing a green ball. The conditional probabilities are:
$P(G|E_1) = \frac{3}{2+3+2} = \frac{3}{7}$
$P(G|E_2) = \frac{2}{3+2+2} = \frac{2}{7}$
$P(G|E_3) = \frac{2}{2+2+3} = \frac{2}{7}$
Using the Law of Total Probability:
$P(G) = P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)$
$P(G) = \left(\frac{3}{6} \times \frac{3}{7}\right) + \left(\frac{2}{6} \times \frac{2}{7}\right) + \left(\frac{1}{6} \times \frac{2}{7}\right)$
$P(G) = \frac{9}{42} + \frac{4}{42} + \frac{2}{42} = \frac{15}{42} = \frac{5}{14}$

(B) When $2n$ unbiased coins are tossed,the total number of outcomes is $2^{2n}$.
The probability of getting $r$ heads in $2n$ tosses is given by the binomial distribution: $P(r) = \frac{1}{2^{2n}} \binom{2n}{r}$.
The number of heads is equal to the number of tails when the number of heads is exactly $n$.
The probability of getting exactly $n$ heads is $P(n) = \frac{1}{2^{2n}} \binom{2n}{n} = \frac{1}{2^{2n}} \cdot \frac{(2n)!}{(n!)^2}$.
The probability that the number of heads is not equal to the number of tails is $1 - P(n)$.
Therefore,the required probability is $1 - \frac{(2n)!}{(n!)^2} \cdot \frac{1}{2^{2n}}$.

(C) Let $X$ be the number of defective bolts in a sample of $n=4$. The probability of a bolt being defective is $p = 20\% = 0.2 = \frac{1}{5}$.
Thus,the probability of a bolt being non-defective is $q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$.
Since the selection is random,$X$ follows a binomial distribution $B(n, p)$ with $n=4$ and $p=\frac{1}{5}$.
The probability mass function is given by $P(X=k) = {}^nC_k p^k q^{n-k}$.
We need to find the probability that less than $2$ bolts are defective,i.e.,$P(X < 2) = P(X=0) + P(X=1)$.
$P(X=0) = {}^4C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^4 = 1 \times 1 \times \frac{256}{625} = \frac{256}{625}$.
$P(X=1) = {}^4C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^3 = 4 \times \frac{1}{5} \times \frac{64}{125} = \frac{256}{625}$.
Therefore,$P(X < 2) = \frac{256}{625} + \frac{256}{625} = \frac{512}{625} = 0.8192$.

(C) This is a binomial distribution problem where the number of trials $n = 3$.
Success is defined as getting $1$ or $6$ on a die.
The probability of success in a single trial is $p = \frac{2}{6} = \frac{1}{3}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
For a binomial distribution,the variance is given by the formula $Var(X) = npq$.
Substituting the values,we get:
$Var(X) = 3 \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{3}$.

(D) Let the probability of getting a head be $p$.
Then the probability of not getting a head is $1-p$.
The person wins in an odd number of tosses if the first head appears on the $1^{st}, 3^{rd}, 5^{th}, \dots$ toss.
This forms a geometric series: $p + (1-p)^2 p + (1-p)^4 p + \dots = 3/4$.
Using the sum formula for an infinite geometric series $S = \frac{a}{1-r}$,where $a = p$ and $r = (1-p)^2$:
$\frac{p}{1-(1-p)^2} = \frac{3}{4}$
$\frac{p}{1-(1-2p+p^2)} = \frac{3}{4}$
$\frac{p}{2p-p^2} = \frac{3}{4}$
$\frac{1}{2-p} = \frac{3}{4}$
$4 = 6 - 3p \implies 3p = 2 \implies p = 2/3$.
If $5$ such coins are tossed at a time,the probability that a head appears on all $5$ coins is $p^5 = (2/3)^5 = 32/243$.

(C) The sum of all probabilities in a distribution must be $1$:
$\Sigma P(X = x) = 0.15 + 0.23 + k + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1$
$0.88 + k = 1$
$k = 0.12$
The event $E$ consists of prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$,so $E = \{2, 3, 5, 7\}$.
The event $F$ consists of values less than $4$,so $F = \{1, 2, 3\}$.
The union $E \cup F = \{1, 2, 3, 5, 7\}$.
The probability $P(E \cup F) = P(X=1) + P(X=2) + P(X=3) + P(X=5) + P(X=7)$
$P(E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77$.

(B) The probability mass function of a Poisson distribution is given by $p(x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$.
To find the value of $x$ for which $p(x)$ is maximum,we examine the ratio $\frac{p(x)}{p(x-1)}$.
$\frac{p(x)}{p(x-1)} = \frac{e^{-\lambda} \lambda^x / x!}{e^{-\lambda} \lambda^{x-1} / (x-1)!} = \frac{\lambda}{x}$.
For $p(x)$ to be maximum,we need $\frac{p(x)}{p(x-1)} \geq 1$ and $\frac{p(x+1)}{p(x)} \leq 1$.
This implies $\frac{\lambda}{x} \geq 1 \implies x \leq \lambda$ and $\frac{\lambda}{x+1} \leq 1 \implies x+1 \geq \lambda$.
Thus,the mode $x$ satisfies $\lambda - 1 \leq x \leq \lambda$.
Given $\lambda = 3.725$,we have $3.725 - 1 \leq x \leq 3.725$,which means $2.725 \leq x \leq 3.725$.
Since $x$ must be an integer,the value of $x$ that maximizes $p(x)$ is $3$.

(B) We know that the sum of all probabilities in a probability distribution is $1$,so $\sum P(X=x_i) = 1$.
$0.1 + 0.15 + 0.3 + 0.25 + k + k = 1$
$0.8 + 2k = 1 \implies 2k = 0.2 \implies k = 0.1$.
Now,we calculate the mean $E(X) = \sum x_i P(x_i)$ and $E(X^2) = \sum x_i^2 P(x_i)$:

$x_i$	$P(x_i)$	$x_i P(x_i)$	$x_i^2 P(x_i)$
$1$	$0.1$	$0.1$	$0.1$
$2$	$0.15$	$0.3$	$0.6$
$3$	$0.3$	$0.9$	$2.7$
$4$	$0.25$	$1.0$	$4.0$
$5$	$0.1$	$0.5$	$2.5$
$6$	$0.1$	$0.6$	$3.6$
Total	$1.0$	$3.4$	$13.5$

The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 13.5 - (3.4)^2 = 13.5 - 11.56 = 1.94$.

(C) For a Poisson distribution,the variance is equal to the parameter $\lambda$. Given that the variance is $3$,we have $\lambda = 3$.
The probability mass function is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!}$.
For a Poisson distribution,the probability $P(X=r)$ is maximum when $r = \lfloor \lambda \rfloor$ if $\lambda$ is not an integer,and it takes two maximum values at $r = \lambda$ and $r = \lambda - 1$ if $\lambda$ is an integer.
Here,$\lambda = 3$,which is an integer.
Therefore,$P(X=r)$ is maximum at $r = 3$ and $r = 3 - 1 = 2$.
Comparing this with the given options,$r=2$ and $r=3$ are both valid,but since $r=2$ is provided in the options,it is a correct value.

(A) Given that the experiment is conducted $n = 5$ times. Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure. The mean of a binomial distribution is $np$ and the variance is $npq$.
Given $np - npq = \frac{5}{9}$.
Substituting $n = 5$: $5p - 5pq = \frac{5}{9} \implies p - pq = \frac{1}{9}$.
Since $1 - q = p$,we have $p(1 - q) = p^2 = \frac{1}{9}$,which gives $p = \frac{1}{3}$.
Thus,$q = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability of getting at most two successes is $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.
Using the formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 0) = {}^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5 = 1 \times 1 \times \frac{32}{243} = \frac{32}{243}$.
$P(X = 1) = {}^5C_1 (\frac{1}{3})^1 (\frac{2}{3})^4 = 5 \times \frac{1}{3} \times \frac{16}{81} = \frac{80}{243}$.
$P(X = 2) = {}^5C_2 (\frac{1}{3})^2 (\frac{2}{3})^3 = 10 \times \frac{1}{9} \times \frac{8}{27} = \frac{80}{243}$.
Summing these probabilities: $P(X \leq 2) = \frac{32}{243} + \frac{80}{243} + \frac{80}{243} = \frac{192}{243} = \frac{64}{81}$.

(B) The number of errors per page follows a Poisson distribution with parameter $\lambda = \frac{60}{600} = 0.1$.
The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!} = \frac{e^{-0.1} (0.1)^x}{x!}$.
We need to find the probability that a page contains at most two errors,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \frac{e^{-0.1} (0.1)^0}{0!} = e^{-0.1}$.
$P(X=1) = \frac{e^{-0.1} (0.1)^1}{1!} = 0.1 e^{-0.1}$.
$P(X=2) = \frac{e^{-0.1} (0.1)^2}{2!} = \frac{0.01}{2} e^{-0.1} = 0.005 e^{-0.1}$.
Adding these probabilities: $P(X \le 2) = e^{-0.1} (1 + 0.1 + 0.005) = e^{-0.1} (1.105) = e^{-0.1} \left(\frac{1105}{1000}\right) = e^{-0.1} \left(\frac{221}{200}\right) = \frac{1}{e^{0.1}} \left(\frac{221}{200}\right)$.