TS EAMCET 2023 Mathematics Question Paper with Answer and Solution

(C) Let $P = (x, y)$,$A = (0, 2)$,and $B = (-1, 0)$.
Given $PA = \frac{1}{\sqrt{2}} PB$,which implies $2 PA^2 = PB^2$.
Substituting the coordinates:
$2[(x - 0)^2 + (y - 2)^2] = (x + 1)^2 + (y - 0)^2$
$2(x^2 + y^2 - 4y + 4) = x^2 + 2x + 1 + y^2$
$2x^2 + 2y^2 - 8y + 8 = x^2 + y^2 + 2x + 1$
$x^2 + y^2 - 2x - 8y + 7 = 0$.
This is the equation of a circle in the form $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing,$2g = -2 \implies g = -1$ and $2f = -8 \implies f = -4$.
The centre is $(-g, -f) = (1, 4)$.
The radius is $\sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-4)^2 - 7} = \sqrt{1 + 16 - 7} = \sqrt{10}$ units.

(A) The directrix is $x+2y-1=0$ and the focus is $S(1,0)$.
Let $P(x, y)$ be any point on the parabola.
By definition,the distance from $P$ to the focus equals the perpendicular distance from $P$ to the directrix $(PS = PM)$.
$PS^2 = PM^2$
$(x-1)^2 + (y-0)^2 = \frac{(x+2y-1)^2}{1^2+2^2}$
$5((x-1)^2 + y^2) = (x+2y-1)^2$
$5(x^2 - 2x + 1 + y^2) = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$
$5x^2 - 10x + 5 + 5y^2 = x^2 + 4y^2 + 4xy - 2x - 4y + 1$
$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$

(C) Given parabolas are $P_1: y^2=5x$ and $P_2: x^2=5y$.
Solving these equations: Substituting $y = \frac{x^2}{5}$ into $y^2=5x$ gives $(\frac{x^2}{5})^2 = 5x$,which simplifies to $x^4 = 125x$. Thus,$x(x^3 - 125) = 0$,giving $x=0$ or $x=5$.
The points of intersection are $(0,0)$ and $(5,5)$.
The line $L$ passing through $(0,0)$ and $(5,5)$ is $y=x$.
The directrix of $P_1$ ($y^2=4ax$ with $4a=5 \Rightarrow a=5/4$) is $x = -\frac{5}{4}$.
The latus rectum of $P_2$ ($x^2=4ay$ with $4a=5 \Rightarrow a=5/4$) is $y = \frac{5}{4}$.
The vertices of the triangle are the intersection points of these three lines:
$1$. Intersection of $x = -\frac{5}{4}$ and $y = \frac{5}{4}$ is $C(-\frac{5}{4}, \frac{5}{4})$.
$2$. Intersection of $x = -\frac{5}{4}$ and $y=x$ is $B(-\frac{5}{4}, -\frac{5}{4})$.
$3$. Intersection of $y = \frac{5}{4}$ and $y=x$ is $A(\frac{5}{4}, \frac{5}{4})$.
The area of the triangle with vertices $A(\frac{5}{4}, \frac{5}{4})$,$B(-\frac{5}{4}, -\frac{5}{4})$,and $C(-\frac{5}{4}, \frac{5}{4})$ is given by:
Area $= \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$
Area $= \frac{1}{2} |\frac{5}{4}(-\frac{5}{4} - \frac{5}{4}) + (-\frac{5}{4})(\frac{5}{4} - \frac{5}{4}) + (-\frac{5}{4})(\frac{5}{4} - (-\frac{5}{4}))|$
Area $= \frac{1}{2} |\frac{5}{4}(-\frac{10}{4}) + 0 + (-\frac{5}{4})(\frac{10}{4})| = \frac{1}{2} |-\frac{50}{16} - \frac{50}{16}| = \frac{1}{2} |-\frac{100}{16}| = \frac{50}{16} = \frac{25}{8}$.

(D) Let the side length of the equilateral triangle be $a$.
Since one vertex is at the origin $(0,0)$ and the triangle is symmetric about the $x$-axis, the other two vertices lie on the parabola $y^2=12x$ at angles of $30^{\circ}$ and $-30^{\circ}$ with the $x$-axis.
The coordinates of the vertex $A$ are $(a \cos 30^{\circ}, a \sin 30^{\circ})$.
Since $A$ lies on the parabola $y^2=12x$, we have:
$(a \sin 30^{\circ})^2 = 12(a \cos 30^{\circ})$
$a^2 \sin^2 30^{\circ} = 12a \cos 30^{\circ}$
$a^2 (1/4) = 12a (\sqrt{3}/2)$
$a/4 = 6\sqrt{3}$
$a = 24\sqrt{3}$.
The area of an equilateral triangle with side $a$ is $\frac{\sqrt{3}}{4} a^2$.
Area $= \frac{\sqrt{3}}{4} (24\sqrt{3})^2 = \frac{\sqrt{3}}{4} (576 \times 3) = \frac{\sqrt{3}}{4} (1728) = 432\sqrt{3}$ sq. units.

(B) The equation of the parabola is $y^2=16x$,which is of the form $y^2=4ax$. Thus,$4a=16$,so $a=4$.
Let the parameters of points $A$ and $B$ be $t_1$ and $t_2$ respectively. Since $AB$ is a focal chord,$t_1 \cdot t_2 = -1$.
For point $A(1, -4)$,we have $at_1^2 = 1$ and $2at_1 = -4$.
Substituting $a=4$,we get $4t_1^2 = 1$ $\Rightarrow t_1^2 = 1/4$ $\Rightarrow t_1 = -1/2$ (since $y < 0$ at $A$).
Since $t_1 \cdot t_2 = -1$,we have $(-1/2) \cdot t_2 = -1$,which gives $t_2 = 2$.
Point $B$ is $(at_2^2, 2at_2) = (4(2)^2, 2(4)(2)) = (16, 16)$.
The equation of the normal to the parabola $y^2=4ax$ at point $t$ is $y + tx = 2at + at^3$.
For $t_2 = 2$ and $a=4$,the equation of the normal at $B$ is $y + 2x = 2(4)(2) + 4(2)^3$.
$y + 2x = 16 + 32$.
$y + 2x = 48$.
Therefore,the equation of the normal is $2x + y - 48 = 0$.

(C) Given line: $x-2y+k=0 \Rightarrow y = \frac{1}{2}x + \frac{k}{2}$.
Comparing with $y=mx+c$,we get $m=\frac{1}{2}$ and $c=\frac{k}{2}$.
Given parabola: $y^2-4x-4y+8=0$.
Rewriting in standard form: $(y-2)^2 = 4x-8+4 = 4(x-1)$.
Let $Y = y-2$ and $X = x-1$,then $Y^2 = 4X$.
The condition for the line $Y=mX+c'$ to be a tangent to $Y^2=4aX$ is $c' = \frac{a}{m}$.
Here $a=1$,$m=\frac{1}{2}$,and $Y = \frac{1}{2}(X+1) - 2 = \frac{1}{2}X - \frac{3}{2}$.
So $c' = -\frac{3}{2}$.
Setting $c' = \frac{a}{m} = \frac{1}{1/2} = 2$ is not applicable directly due to the shift.
Alternatively,substitute $x = 2y-k$ into the parabola equation:
$(y-2)^2 = 4(2y-k-1) \Rightarrow y^2-4y+4 = 8y-4k-4$.
$y^2-12y+(8+4k) = 0$.
For tangency,the discriminant $D=0$:
$(-12)^2 - 4(1)(8+4k) = 0 \Rightarrow 144 - 32 - 16k = 0$.
$112 = 16k \Rightarrow k = 7$.

(D) For the parabola $y^2=16x$,we have $a=4$. The equation of any tangent to the parabola is $y=mx+\frac{a}{m}$,which gives $y=mx+\frac{4}{m}$.
For the circle $x^2+y^2=8$,the radius is $r=\sqrt{8}=2\sqrt{2}$. The equation of any tangent to the circle is $y=mx \pm r\sqrt{1+m^2}$,which gives $y=mx \pm 2\sqrt{2}\sqrt{1+m^2}$.
Since these represent the same line,the constant terms must be equal:
$\frac{4}{m} = \pm 2\sqrt{2}\sqrt{1+m^2}$.
Squaring both sides,we get $\frac{16}{m^2} = 8(1+m^2)$.
Dividing by $8$,we get $\frac{2}{m^2} = 1+m^2$,which implies $m^4+m^2-2=0$.
Let $t=m^2$,then $t^2+t-2=0$,so $(t+2)(t-1)=0$. Since $t=m^2 \ge 0$,we have $m^2=1$,so $m=\pm 1$.
Substituting $m=1$ into the tangent equation: $y=x+\frac{4}{1} \Rightarrow y=x+4$.
Substituting $m=-1$ into the tangent equation: $y=-x+\frac{4}{-1} \Rightarrow y=-x-4$.
Thus,the common tangents are $y=x+4$ and $y=-x-4$.

(D) The focal distance of a point $(x_1, y_1)$ on the parabola $y^2=4ax$ is $x_1+a$.
Given parabola is $y^2=kx$,so $4a=k \Rightarrow a=k/4$.
The focal distance is $x_1+a = 2 + k/4 = 3$.
$k/4 = 1 \Rightarrow k=4$.
The parabola is $y^2=4x$.
Since $P(2, y_1)$ lies on $y^2=4x$,$y_1^2 = 4(2) = 8 \Rightarrow y_1 = \pm 2\sqrt{2}$.
So $P$ is $(2, 2\sqrt{2})$ or $(2, -2\sqrt{2})$.
The equation of the tangent to $y^2=4x$ at $(x_1, y_1)$ is $yy_1 = 2(x+x_1)$.
For $P(2, 2\sqrt{2})$: $y(2\sqrt{2}) = 2(x+2)$ $\Rightarrow \sqrt{2}y = x+2$ $\Rightarrow x - \sqrt{2}y + 2 = 0$.
For $P(2, -2\sqrt{2})$: $y(-2\sqrt{2}) = 2(x+2)$ $\Rightarrow -\sqrt{2}y = x+2$ $\Rightarrow x + \sqrt{2}y + 2 = 0$.
Combining these,the equation is $x \pm \sqrt{2}y + 2 = 0$.

(D) The equation of the tangent to the parabola $y^2 = 8\sqrt{5}x$ is $y = ax + \frac{2\sqrt{5}}{a}$.
Comparing this with $y = ax + c$,we get $c = \frac{2\sqrt{5}}{a}$.
The condition for the line $y = ax + c$ to be a tangent to the circle $x^2 + y^2 = 1$ is $c^2 = 1 + a^2$.
Substituting $c = \frac{2\sqrt{5}}{a}$ into the condition,we get $(\frac{2\sqrt{5}}{a})^2 = 1 + a^2$.
$\frac{20}{a^2} = 1 + a^2$ $\Rightarrow 20 = a^2 + a^4$ $\Rightarrow a^4 + a^2 - 20 = 0$.
Let $t = a^2$,then $t^2 + t - 20 = 0$.
$(t + 5)(t - 4) = 0$. Since $a^2 > 0$,we have $t = 4$,so $a^2 = 4$.
Then $c^2 = 1 + a^2 = 1 + 4 = 5$.
Therefore,$a^2c^2 = 4 \times 5 = 20$.

(D) The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.
For the parabola $y^2=4x$,we have $a=1$,so the equation becomes $y=mx-2m-m^3$.
Since the normal passes through $(5,0)$,we substitute $x=5$ and $y=0$:
$0 = m(5) - 2m - m^3$
$0 = 3m - m^3$
$m(3 - m^2) = 0$
Thus,the slopes are $m=0$ and $m=\pm\sqrt{3}$.
For $m=\sqrt{3}$,the equation of the normal is $y=\sqrt{3}(x-5)$,which simplifies to $\sqrt{3}x-y-5\sqrt{3}=0$.
For $m=-\sqrt{3}$,the equation of the normal is $y=-\sqrt{3}(x-5)$,which simplifies to $\sqrt{3}x+y-5\sqrt{3}=0$.
For $m=0$,the equation of the normal is $y=0$.

(B) Given parabola is $y^2=12x$,so $4a=12 \Rightarrow a=3$.
Equation of normal with slope $m$ is $y=mx-2am-am^3$,which becomes $y=mx-6m-3m^3$.
Since the normal passes through $P(8,0)$,we have $0=8m-6m-3m^3$.
$2m-3m^3=0 \Rightarrow m(2-3m^2)=0$.
The slopes are $m_1=0$,$m_2=\sqrt{\frac{2}{3}}$,and $m_3=-\sqrt{\frac{2}{3}}$.
The non-horizontal normals have slopes $m_2=\sqrt{\frac{2}{3}}$ and $m_3=-\sqrt{\frac{2}{3}}$.
The angle $\theta$ between these two normals is given by $\tan \theta = \left| \frac{m_2-m_3}{1+m_2m_3} \right|$.
$\tan \theta = \left| \frac{\sqrt{\frac{2}{3}} - (-\sqrt{\frac{2}{3}})}{1 + (\sqrt{\frac{2}{3}})(-\sqrt{\frac{2}{3}})} \right| = \left| \frac{2\sqrt{\frac{2}{3}}}{1 - \frac{2}{3}} \right| = \frac{2\sqrt{\frac{2}{3}}}{\frac{1}{3}} = 6 \sqrt{\frac{2}{3}} = 6 \frac{\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}$.

(C) The equation of a tangent to the parabola $y^2 = 4ax$ is $y = mx + \frac{a}{m}$.
Given $y^2 = 8x$,we have $a = 2$,so the tangent is $y = mx + \frac{2}{m}$.
Rewriting the given line $2x + 3y + n = 0$ as $y = -\frac{2}{3}x - \frac{n}{3}$.
Comparing the slopes,$m = -\frac{2}{3}$.
Equating the intercepts,$-\frac{n}{3} = \frac{2}{m} = \frac{2}{-2/3} = -3$,which gives $n = 9$.
The point of contact is $(2n, 4\sqrt{n}) = (18, 12)$.
For the parabola $y^2 = 8x$,the normal at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.
Since $(at^2, 2at) = (18, 12)$ and $a = 2$,we have $2t^2 = 18$ $\Rightarrow t^2 = 9$ $\Rightarrow t = 3$ (taking positive root for the point).
The equation of the normal is $y = -3x + 2(2)(3) + 2(3)^3 = -3x + 12 + 54$.
Thus,$3x + y - 66 = 0$.

(A) Let the foci be $S_1 = (3,3)$ and $S_2 = (-4,4)$,and the point on the ellipse be $P = (0,0)$.
By the definition of an ellipse,the sum of the distances from any point on the ellipse to the two foci is constant and equal to the length of the major axis,$2a$.
$PS_1 + PS_2 = 2a$
$\sqrt{(3-0)^2 + (3-0)^2} + \sqrt{(-4-0)^2 + (4-0)^2} = 2a$
$3\sqrt{2} + 4\sqrt{2} = 2a$
$7\sqrt{2} = 2a \Rightarrow a = \frac{7\sqrt{2}}{2}$.
The distance between the foci is $2ae = S_1S_2$.
$S_1S_2 = \sqrt{(-4-3)^2 + (4-3)^2} = \sqrt{(-7)^2 + (1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$.
$2ae = 5\sqrt{2}$
$2 \cdot \frac{7\sqrt{2}}{2} \cdot e = 5\sqrt{2}$
$7\sqrt{2} \cdot e = 5\sqrt{2}$
$e = \frac{5}{7}$.

(C) The equation of the ellipse $E$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.
The foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the end of the minor axis is $B(0, b)$.
Given $\angle S^{\prime} SB = \frac{\pi}{6}$. In $\triangle OBS$,$\angle OSB = \frac{\pi}{6}$.
Thus,$\tan(\frac{\pi}{6}) = \frac{OB}{OS} = \frac{b}{ae} = \frac{1}{\sqrt{3}}$.
This implies $3b^2 = a^2e^2$.
Since $b^2 = a^2(1 - e^2)$,we have $a^2e^2 = a^2 - b^2$.
Substituting this,$3b^2 = a^2 - b^2$,which gives $a^2 = 4b^2$.
The point $(2\sqrt{3}, 1)$ lies on the ellipse,so $\frac{(2\sqrt{3})^2}{a^2} + \frac{1^2}{b^2} = 1$.
Substituting $a^2 = 4b^2$,we get $\frac{12}{4b^2} + \frac{1}{b^2} = 1$,which simplifies to $\frac{3}{b^2} + \frac{1}{b^2} = 1$.
Thus,$\frac{4}{b^2} = 1$,so $b^2 = 4$ and $a^2 = 16$.
The length of the major axis is $2a$ and the length of the minor axis is $2b$.
The sum of the squares of the lengths is $(2a)^2 + (2b)^2 = 4a^2 + 4b^2 = 4(16) + 4(4) = 64 + 16 = 80$.

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The foci are $S_1(ae, 0)$ and $S_2(-ae, 0)$. The ends of the major axis are $A(a, 0)$ and $A'(-a, 0)$. The ends of the minor axis are $B(0, b)$ and $B'(0, -b)$.
Given the distance from focus $S_1$ to the corresponding end of the major axis $A$ is $a - ae = 4 - \sqrt{7}$.
Given the distance from focus $S_1$ to the end of the minor axis $B$ is $\sqrt{(ae)^2 + b^2} = 4$. Since $b^2 = a^2(1 - e^2)$,we have $\sqrt{(ae)^2 + a^2 - a^2e^2} = 4$,which implies $a = 4$.
Substituting $a = 4$ into $a(1 - e) = 4 - \sqrt{7}$,we get $4(1 - e) = 4 - \sqrt{7}$,so $1 - e = 1 - \frac{\sqrt{7}}{4}$,which gives $e = \frac{\sqrt{7}}{4}$.
Then $b^2 = a^2(1 - e^2) = 16(1 - \frac{7}{16}) = 16(\frac{9}{16}) = 9$,so $b = 3$.
The distance $S_1S_2 = 2ae = 2(4)(\frac{\sqrt{7}}{4}) = 2\sqrt{7}$.
In $\triangle BS_1S_2$,$BS_1 = BS_2 = 4$ and $S_1S_2 = 2\sqrt{7}$.
Using the law of cosines,$\cos \theta = \frac{BS_1^2 + BS_2^2 - S_1S_2^2}{2(BS_1)(BS_2)} = \frac{4^2 + 4^2 - (2\sqrt{7})^2}{2(4)(4)} = \frac{16 + 16 - 28}{32} = \frac{4}{32} = \frac{1}{8}$.

(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since it passes through $(2,2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which simplifies to $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ ... $(i)$.
Since it passes through $(3,1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$ ... $(ii)$.
Subtracting $(i)$ from $(ii)$,we get $\frac{8}{a^2} = 1 - \frac{1}{4} = \frac{3}{4}$.
Thus,$a^2 = \frac{32}{3}$,which implies $3a^2 = 32$.
Substituting $a^2$ into $(ii)$,we get $\frac{9}{32/3} + \frac{1}{b^2} = 1$,so $\frac{27}{32} + \frac{1}{b^2} = 1$.
This gives $\frac{1}{b^2} = 1 - \frac{27}{32} = \frac{5}{32}$,so $b^2 = \frac{32}{5}$,which implies $5b^2 = 32$.
Therefore,$3a^2 + 5b^2 = 32 + 32 = 64$.

(B) The given line is $x \cos \alpha + y \sin \alpha = 2 \sqrt{3}$ and the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.
Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 8$.
The condition for the line $x \cos \alpha + y \sin \alpha = p$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$.
Here $p = 2 \sqrt{3}$,so $p^2 = (2 \sqrt{3})^2 = 12$.
Substituting the values: $12 = 16 \cos^2 \alpha + 8 \sin^2 \alpha$.
Using $\cos^2 \alpha = 1 - \sin^2 \alpha$,we get $12 = 16(1 - \sin^2 \alpha) + 8 \sin^2 \alpha$.
$12 = 16 - 16 \sin^2 \alpha + 8 \sin^2 \alpha$.
$8 \sin^2 \alpha = 4$.
$\sin^2 \alpha = \frac{4}{8} = \frac{1}{2}$.
Since $\alpha$ is an acute angle,$\sin \alpha = \frac{1}{\sqrt{2}}$.
Therefore,$\alpha = \frac{\pi}{4}$.

(C) The equation of the line is $y=mx+c$,where $m=4$.
The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,where $a^2=4$ and $b^2=1$.
$A$ line $y=mx+c$ touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if $c^2=a^2m^2+b^2$.
Substituting the values $a^2=4$,$b^2=1$,and $m=4$:
$c^2 = 4(4)^2 + 1$
$c^2 = 4(16) + 1$
$c^2 = 64 + 1 = 65$
Therefore,$c = \pm \sqrt{65}$.

(B) The equation of the ellipse is $\frac{x^2}{100} + \frac{y^2}{25} = 1$.
The equation of the tangent at point $(x_1, y_1) = (-8, 3)$ is given by $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$.
Substituting the values: $\frac{x(-8)}{100} + \frac{y(3)}{25} = 1$.
Simplifying the equation: $-\frac{2x}{25} + \frac{3y}{25} = 1$,which gives $-2x + 3y = 25$.
To find the point where the particle crosses the $Y$-axis,we set $x = 0$.
Substituting $x = 0$ into the tangent equation: $3y = 25$,which gives $y = \frac{25}{3}$.
Thus,the point is $\left(0, \frac{25}{3}\right)$.

(A) For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the ends of the latus rectum are $(\pm ae, \pm \frac{b^2}{a})$.
Given $\frac{x^2}{9}+\frac{y^2}{5}=1$,we have $a^2=9, b^2=5$.
$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{5}{9} = \frac{4}{9} \Rightarrow e = \frac{2}{3}$.
Ends of latus rectum are $(\pm 2, \pm \frac{5}{3})$.
The equation of the tangent at $P(2, \frac{5}{3})$ is $\frac{2x}{9} + \frac{5y/3}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$.
This line intersects the axes at $A(\frac{9}{2}, 0)$ and $C(0, 3)$.
The area of the triangle formed in the first quadrant is $\frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$.
Since there are four such symmetric triangles formed by the tangents at the four ends of the latus rectum,the total area is $4 \times \frac{27}{4} = 27$ sq. units.

(D) The product of the lengths of the perpendiculars drawn from the two foci of an ellipse to any tangent is equal to the square of the semi-minor axis.
For the given ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$,we have $a^2 = 9$ and $b^2 = 25$. Since $b^2 > a^2$,the semi-minor axis is $b = \sqrt{9} = 3$.
Therefore,the product of the lengths of the perpendiculars is $b^2 = 3^2 = 9$.

(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 $x^2+2y^2=2$,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 $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(D) The equation of the line is $4x + 2y + n = 0$,which can be rewritten as $y = -2x - \frac{n}{2}$.
Comparing this with $y = mx + c$,we get $m = -2$ and $c = -\frac{n}{2}$.
The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$,so $a^2 = 36$ and $b^2 = 16$.
The condition for the line $y = mx + c$ to be a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = \frac{(a^2 - b^2)^2 m^2}{a^2 + b^2 m^2}$.
Substituting the values: $(-\frac{n}{2})^2 = \frac{(36 - 16)^2 (-2)^2}{36 + 16(-2)^2}$.
$\frac{n^2}{4} = \frac{(20)^2 \times 4}{36 + 16(4)} = \frac{400 \times 4}{36 + 64} = \frac{1600}{100} = 16$.
$n^2 = 16 \times 4 = 64$.
Therefore,$n = \pm 8$.

(C) The given equations are $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$.
This represents an ellipse centered at $(h, k) = (1, 2)$ with $a = 2$ and $b = 1$.
The equation of the ellipse is $\frac{(x-1)^2}{4} + \frac{(y-2)^2}{1} = 1$.
The parametric point $P$ at $\theta = \pi/4$ is $(1 + 2 \cos(\pi/4), 2 + \sin(\pi/4)) = (1 + \sqrt{2}, 2 + 1/\sqrt{2})$.
The equation of the normal at point $(a \cos \theta, b \sin \theta)$ for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $ax \sec \theta - by \csc \theta = a^2 - b^2$.
Shifting the origin to $(1, 2)$,the normal equation is $2(x-1) \sec(\pi/4) - 1(y-2) \csc(\pi/4) = 2^2 - 1^2$.
$2(x-1) \sqrt{2} - (y-2) \sqrt{2} = 3$.
$2\sqrt{2}(x-1) - \sqrt{2}(y-2) = 3$.
The major axis of this ellipse is the line $y = 2$.
Substituting $y = 2$ into the normal equation:
$2\sqrt{2}(x-1) - \sqrt{2}(2-2) = 3$.
$2\sqrt{2}(x-1) = 3$.
$x-1 = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4}$.
$x = 1 + \frac{3\sqrt{2}}{4} = \frac{4+3\sqrt{2}}{4}$.
Note: Re-evaluating the normal equation $ax \sec \theta - by \csc \theta = a^2 - b^2$ with $a=2, b=1, \theta=\pi/4$:
$2(x-1)\sqrt{2} - 1(y-2)\sqrt{2} = 3$.
At $y=2$,$2\sqrt{2}(x-1) = 3 \implies x = 1 + \frac{3}{2\sqrt{2}} = 1 + \frac{3\sqrt{2}}{4}$.
Given the options provided,there appears to be a discrepancy in the standard form. Based on the calculation,the intersection point is $(\frac{4+3\sqrt{2}}{4}, 2)$.

(D) Given the tangent line $x+\sqrt{3} y=3$ ... $(i)$ and the ellipse $2 x^2+3 y^2=k$.
The equation of the tangent at point $P(x_1, y_1)$ is $2 x x_1+3 y y_1=k$ ... (ii).
Comparing $(i)$ and (ii),we have $\frac{2 x_1}{1} = \frac{3 y_1}{\sqrt{3}} = \frac{k}{3}$.
Thus,$x_1 = \frac{k}{6}$ and $y_1 = \frac{k}{3 \sqrt{3}}$.
Since $P(x_1, y_1)$ lies on the ellipse,$2(\frac{k}{6})^2 + 3(\frac{k}{3 \sqrt{3}})^2 = k$.
$\frac{k^2}{18} + \frac{k^2}{9} = k$ $\Rightarrow \frac{k^2}{6} = k$ $\Rightarrow k = 6$.
So,$x_1 = 1$ and $y_1 = \frac{6}{3 \sqrt{3}} = \frac{2}{\sqrt{3}}$.
The slope of the tangent is $m_t = -\frac{1}{\sqrt{3}}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = \sqrt{3}$.
The equation of the normal at $P(1, \frac{2}{\sqrt{3}})$ is $y - \frac{2}{\sqrt{3}} = \sqrt{3}(x - 1)$.
$\sqrt{3} y - 2 = 3x - 3 \Rightarrow 3x - \sqrt{3} y = 1$.

(B) The given line is $2x + \sqrt{6}y = 2$,which can be written as $2x + \sqrt{6}y - 2 = 0$.
The equation of the hyperbola is $x^2 - 2y^2 = 4$.
Let the point of contact be $(x_1, y_1)$.
The equation of the tangent to the hyperbola $x^2 - 2y^2 = 4$ at $(x_1, y_1)$ is given by $xx_1 - 2yy_1 = 4$,or $xx_1 - 2yy_1 - 4 = 0$.
Since this line is the same as the given line $2x + \sqrt{6}y - 2 = 0$,the coefficients must be proportional:
$\frac{x_1}{2} = \frac{-2y_1}{\sqrt{6}} = \frac{-4}{-2}$
$\frac{x_1}{2} = 2 \Rightarrow x_1 = 4$
$\frac{-2y_1}{\sqrt{6}} = 2$ $\Rightarrow -2y_1 = 2\sqrt{6}$ $\Rightarrow y_1 = -\sqrt{6}$
Thus,the point of contact is $(4, -\sqrt{6})$.

(B) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
Given the tangent $y = mx + 4$,we have $c = 4$.
Comparing $c^2 = a^2m^2 - b^2$,we get $16 = 25m^2 - 9$.
$25m^2 = 25 \Rightarrow m^2 = 1$. Since $m$ is usually positive for such forms,$m = 1$.
The point of contact for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left(\frac{a^2m}{c}, \frac{b^2}{c}\right)$.
Substituting $a^2 = 25, b^2 = 9, m = 1, c = 4$:
Point of contact = $\left(\frac{25 \times 1}{4}, \frac{9}{4}\right) = \left(\frac{25}{4}, \frac{9}{4}\right)$.

(D) The equation of the normal at point $\theta$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by $ax \cos \theta + by \cot \theta = a^2+b^2$.
For point $P(\theta)$,the normal is $ax \cos \theta + by \cot \theta = a^2+b^2$ (Equation $1$).
For point $Q(\phi)$,the normal is $ax \cos \phi + by \cot \phi = a^2+b^2$.
Since $\phi = \frac{\pi}{2} - \theta$,we have $\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
Thus,the normal at $Q$ is $ax \sin \theta + by \tan \theta = a^2+b^2$ (Equation $2$).
Solving for the intersection $(h, k)$ by eliminating $x$ or $y$:
From Equation $1$,$ax \cos \theta = a^2+b^2 - by \cot \theta$.
From Equation $2$,$ax \sin \theta = a^2+b^2 - by \tan \theta$.
Using Cramer's rule or substitution,we find the $y$-coordinate $k$:
$k = \frac{(a^2+b^2)a(\cos \theta - \sin \theta)}{ab(\sin \theta - \cos \theta)} = \frac{-(a^2+b^2)}{b}$.
Therefore,$k = -\left(\frac{a^2+b^2}{b}\right)$.

(B) The given equation of the line is $x+y+n=0$,which can be written as $y=-x-n$.
Comparing this with $y=mx+c$,we get $m=-1$ and $c=-n$.
The condition for the line $y=mx+c$ to be a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $c^2 = \frac{(a^2+b^2)^2 m^2}{a^2-b^2 m^2}$.
Here,$a^2=6$ and $b^2=2$.
Substituting the values,we get $(-n)^2 = \frac{(6+2)^2 (-1)^2}{6-2(-1)^2}$.
$n^2 = \frac{8^2 \times 1}{6-2} = \frac{64}{4} = 16$.
Therefore,$n = \pm 4$.

(A) Given hyperbola $H: 9x^2 - 16y^2 + 72x - 32y - 16 = 0$.
Rewrite the equation by completing the squares:
$9(x^2 + 8x) - 16(y^2 + 2y) = 16$
$9(x+4)^2 - 144 - 16(y+1)^2 + 16 = 16$
$9(x+4)^2 - 16(y+1)^2 = 144$
Dividing by $144$,we get $\frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = 1$.
The equation of the conjugate hyperbola is $\frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = -1$.
Multiplying by $144$: $9(x+4)^2 - 16(y+1)^2 = -144$.
$9(x^2 + 8x + 16) - 16(y^2 + 2y + 1) = -144$
$9x^2 + 72x + 144 - 16y^2 - 32y - 16 = -144$
$9x^2 - 16y^2 + 72x - 32y + 128 = -144$
$9x^2 - 16y^2 + 72x - 32y + 272 = 0$.

(D) The angle between the asymptotes of a hyperbola is given by $2 \sec^{-1}(e)$.
Given that the angle is $30^{\circ}$,we have:
$2 \sec^{-1}(e) = 30^{\circ}$
$\sec^{-1}(e) = 15^{\circ}$
$e = \sec(15^{\circ})$
Since $\cos(15^{\circ}) = \cos(45^{\circ}-30^{\circ}) = \cos(45^{\circ}) \cos(30^{\circ}) + \sin(45^{\circ}) \sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Therefore,$e = \frac{2\sqrt{2}}{\sqrt{3}+1} = \frac{2\sqrt{2}(\sqrt{3}-1)}{3-1} = \sqrt{6}-\sqrt{2}$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}$.
Using the identity $1 - \cos 4x = 2 \sin^2 2x$,we have $L = \lim _{x \rightarrow 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{2 \sin^2 2x}$.
Divide numerator and denominator by $x^2$ and use standard limits $\lim_{x \to 0} \frac{\sin ax}{ax} = 1$:
$L = \lim _{x \rightarrow 0} \left( \frac{3^{2x} - \sqrt{x+1}}{x} \right) \cdot \left( \frac{\sin 5x}{x} \right) \cdot \left( \frac{x^2}{2 \sin^2 2x} \right) = \lim _{x \rightarrow 0} \left( \frac{3^{2x} - 1 - (\sqrt{x+1} - 1)}{x} \right) \cdot 5 \cdot \frac{1}{2(2)^2}$.
$L = \frac{5}{8} \lim _{x \rightarrow 0} \left( \frac{3^{2x} - 1}{x} - \frac{\sqrt{x+1} - 1}{x} \right)$.
Using $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ and $\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{2}$:
$L = \frac{5}{8} \left( 2 \ln 3 - \frac{1}{2} \right) = \frac{5}{16} (4 \ln 3 - 1) = \frac{5}{16} (\ln 81 - \ln e) = \frac{5}{16} \ln \left( \frac{81}{e} \right)$.

(A) For the first limit: $\lim_{x \rightarrow -2^{+}} [x] = -2$.
Thus,$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) = (-2)^2 - (-2) - 2 = 4 + 2 - 2 = 4$.
For the second limit: $\lim_{x \rightarrow -3^{-}} [x] = -4$.
Thus,$\lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) = (-4)^2 - 4(-4) + 3 = 16 + 16 + 3 = 35$.
Adding both results: $4 + 35 = 39$.

(B) Given limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$
Factorize the denominator: $2x^2+x-3 = (2x+3)(x-1)$
Substitute this into the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(2 x+3)(x-1)}$
Multiply the numerator and denominator by $(\sqrt{x}+1)$: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{(2 x+3)(x-1)(\sqrt{x}+1)}$
Simplify using $(\sqrt{x}-1)(\sqrt{x}+1) = (x-1)$: $\lim _{x \rightarrow 1} \frac{(2 x-3)(x-1)}{(2 x+3)(x-1)(\sqrt{x}+1)}$
Cancel $(x-1)$ from numerator and denominator: $\lim _{x \rightarrow 1} \frac{2 x-3}{(2 x+3)(\sqrt{x}+1)}$
Evaluate the limit by substituting $x=1$: $\frac{2(1)-3}{(2(1)+3)(\sqrt{1}+1)} = \frac{-1}{5(2)} = -\frac{1}{10}$

(B) Given the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital$'s rule:
$\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} = \lim _{x \rightarrow 9} \frac{\frac{d}{dx}(\sqrt{f(x)}-3)}{\frac{d}{dx}(\sqrt{x}-3)}$
$= \lim _{x \rightarrow 9} \frac{\frac{f^{\prime}(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$
$= \lim _{x \rightarrow 9} \frac{f^{\prime}(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$
Substituting the values $f(9)=9$ and $f^{\prime}(9)=4$:
$= \frac{f^{\prime}(9) \cdot \sqrt{9}}{\sqrt{f(9)}} = \frac{4 \cdot 3}{\sqrt{9}} = \frac{12}{3} = 4$.

(A) Let $L = \lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}$.
Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,where $a = \sqrt[3]{6+x}$ and $b = \sqrt[3]{10-x}$:
$L = \lim _{x \rightarrow 2} \frac{(6+x) - (10-x)}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$
$L = \lim _{x \rightarrow 2} \frac{2x - 4}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$
$L = \lim _{x \rightarrow 2} \frac{2(x-2)}{(x-2)((6+x)^{2/3} + (6+x)^{1/3}(10-x)^{1/3} + (10-x)^{2/3})}$
$L = \frac{2}{8^{2/3} + (8 \times 8)^{1/3} + 8^{2/3}} = \frac{2}{4 + 4 + 4} = \frac{2}{12} = \frac{1}{6}$.

(B) Let $h = 1-x$. As $x \rightarrow 1$,$h \rightarrow 0$,so $x = 1-h$.
Substituting this into the limit:
$\lim _{h \rightarrow 0} h \tan \left(\frac{\pi}{2}(1-h)\right) = \lim _{h \rightarrow 0} h \tan \left(\frac{\pi}{2} - \frac{\pi h}{2}\right)$
Since $\tan(\frac{\pi}{2} - \theta) = \cot(\theta)$,we have:
$\lim _{h \rightarrow 0} h \cot \left(\frac{\pi h}{2}\right) = \lim _{h \rightarrow 0} h \frac{1}{\tan \left(\frac{\pi h}{2}\right)}$
Multiply and divide by $\frac{\pi}{2}$:
$\lim _{h \rightarrow 0} \frac{h \cdot \frac{\pi}{2}}{\tan \left(\frac{\pi h}{2}\right)} \cdot \frac{2}{\pi} = 1 \cdot \frac{2}{\pi} = \frac{2}{\pi}$.

(C) We have the limit: $\lim _{x \rightarrow 0} \frac{\tan ^4 x-\sin ^4 x}{x^6}$
Factor out $\sin ^4 x$: $\lim _{x \rightarrow 0} \frac{\sin ^4 x(\sec ^4 x-1)}{x^6}$
Using the identity $\sec ^4 x - 1 = (\sec ^2 x - 1)(\sec ^2 x + 1) = \tan ^2 x (\sec ^2 x + 1)$:
$\lim _{x \rightarrow 0} \frac{\sin ^4 x \cdot \tan ^2 x (\sec ^2 x + 1)}{x^6}$
Rewrite as: $\lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^4 \cdot \left( \frac{\tan x}{x} \right)^2 \cdot (\sec ^2 x + 1)$
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$:
$= (1)^4 \cdot (1)^2 \cdot (\sec ^2 0 + 1) = 1 \cdot 1 \cdot (1 + 1) = 2$

(A) First,we evaluate $\ell$:
$\ell = \lim_{\theta \rightarrow 0} \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} = \lim_{\theta \rightarrow 0} \frac{\sin(3\theta)}{\theta} = \lim_{\theta \rightarrow 0} 3 \left( \frac{\sin(3\theta)}{3\theta} \right) = 3(1) = 3$.
Next,we evaluate $m$:
$m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{\theta} = \lim_{\theta \rightarrow 0} 2 \left( \frac{\tan(2\theta)}{2\theta} \right) = 2(1) = 2$.
The quadratic equation with roots $\ell = 3$ and $m = 2$ is given by $x^2 - (\ell + m)x + \ell m = 0$.
$x^2 - (3 + 2)x + (3 \times 2) = 0$
$x^2 - 5x + 6 = 0$.

(B) We need to evaluate $\lim_{x \rightarrow \pi/4} \frac{\cot^3 x - \tan x}{\cos(x + \pi/4)}$.
Let $t = \tan x$. As $x \rightarrow \pi/4$,$t \rightarrow 1$.
The numerator is $\frac{1}{t^3} - t = \frac{1 - t^4}{t^3} = \frac{(1 - t^2)(1 + t^2)}{t^3} = \frac{(1 - t)(1 + t)(1 + t^2)}{t^3}$.
The denominator is $\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = \frac{\cos x}{\sqrt{2}}(1 - \tan x) = \frac{\cos x}{\sqrt{2}}(1 - t)$.
Substituting these into the limit:
$\lim_{t \rightarrow 1} \frac{(1 - t)(1 + t)(1 + t^2)}{t^3} \cdot \frac{\sqrt{2}}{\cos x(1 - t)} = \lim_{t \rightarrow 1} \frac{\sqrt{2}(1 + t)(1 + t^2)}{t^3 \cos x}$.
As $t \rightarrow 1$,$x \rightarrow \pi/4$,so $\cos x \rightarrow 1/\sqrt{2}$.
$= \frac{\sqrt{2}(1 + 1)(1 + 1^2)}{1^3 \cdot (1/\sqrt{2})} = \frac{\sqrt{2} \cdot 2 \cdot 2}{1/\sqrt{2}} = 4 \cdot 2 = 8$.

(C) Let $L = \lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}}$.
Rationalizing the denominator:
$L = \lim _{x \rightarrow 0} \frac{(2 \tan x + \cos x - 1 + x)(\sqrt{4 \sin ^2 x + 2 \tan x + 1} + \sqrt{3 \tan ^2 x + \sin x + 1})}{(4 \sin ^2 x + 2 \tan x + 1) - (3 \tan ^2 x + \sin x + 1)}$.
As $x \rightarrow 0$,the term $(\sqrt{4 \sin ^2 x + 2 \tan x + 1} + \sqrt{3 \tan ^2 x + \sin x + 1}) \rightarrow \sqrt{1} + \sqrt{1} = 2$.
So,$L = 2 \lim _{x \rightarrow 0} \frac{2 \tan x + x + (\cos x - 1)}{4 \sin ^2 x - 3 \tan ^2 x + 2 \tan x - \sin x}$.
Using small angle approximations $\tan x \approx x$,$\sin x \approx x$,$\cos x - 1 \approx -\frac{x^2}{2}$:
$L = 2 \lim _{x \rightarrow 0} \frac{2x + x - \frac{x^2}{2}}{4x^2 - 3x^2 + 2x - x} = 2 \lim _{x \rightarrow 0} \frac{3x - \frac{x^2}{2}}{x^2 + x} = 2 \lim _{x \rightarrow 0} \frac{3 - \frac{x}{2}}{x + 1} = 2 \times \frac{3}{1} = 6$.

(B) Given $\lim _{x \rightarrow \infty} x\left(a^{1 / x}+b^{1 / x}+c^{1 / x}-3 k^{1 / x}\right)=0$.
Let $y = \frac{1}{x}$. As $x \rightarrow \infty$,$y \rightarrow 0$.
The expression becomes $\lim _{y \rightarrow 0} \frac{a^y+b^y+c^y-3 k^y}{y} = 0$.
This can be rewritten as $\lim _{y \rightarrow 0} \left[ \frac{a^y-1}{y} + \frac{b^y-1}{y} + \frac{c^y-1}{y} - 3 \frac{k^y-1}{y} \right] = 0$.
Using the standard limit $\lim _{y \rightarrow 0} \frac{a^y-1}{y} = \ln a$,we get:
$\ln a + \ln b + \ln c - 3 \ln k = 0$.
$\ln (abc) = 3 \ln k$.
$\ln (abc) = \ln (k^3)$.
Therefore,$k^3 = abc$,which implies $k = (abc)^{1/3}$.

(C) We are given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$
Since $x$ is independent of $k$,we can write: $\lim _{n \rightarrow \infty} \frac{x}{n^3} \sum_{k=1}^n k^2$
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
Substituting this into the expression: $\lim _{n \rightarrow \infty} \frac{x}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$
$= \lim _{n \rightarrow \infty} \frac{x}{6} \cdot \frac{n(n+1)(2n+1)}{n^3}$
$= \frac{x}{6} \lim _{n \rightarrow \infty} \frac{n^3(1 + \frac{1}{n})(2 + \frac{1}{n})}{n^3}$
$= \frac{x}{6} \cdot (1 \cdot 2) = \frac{2x}{6} = \frac{x}{3}$

(A) Given observations: $2, 3, 5, 8, 12$
Mean $\bar{x} = \frac{2+3+5+8+12}{5} = \frac{30}{5} = 6$
Variance $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(2-6)^2 + (3-6)^2 + (5-6)^2 + (8-6)^2 + (12-6)^2}{5}$
$\sigma^2 = \frac{(-4)^2 + (-3)^2 + (-1)^2 + 2^2 + 6^2}{5} = \frac{16 + 9 + 1 + 4 + 36}{5} = \frac{66}{5} = 13.2$
Median $m$ of the data $2, 3, 5, 8, 12$ is the middle term,so $m = 5$.
Mean deviation about median $M = \frac{\sum |x_i - m|}{n} = \frac{|2-5| + |3-5| + |5-5| + |8-5| + |12-5|}{5}$
$M = \frac{3 + 2 + 0 + 3 + 7}{5} = \frac{15}{5} = 3$
Therefore,$\sigma^2 - M = 13.2 - 3 = 10.2$.

(A) The sum of the first $n$ odd natural numbers is given by $\sum_{i=1}^{n} (2i-1) = n^2$.
The sum of the squares of the first $n$ odd natural numbers is $\sum_{i=1}^{n} (2i-1)^2 = \sum_{i=1}^{n} (4i^2 - 4i + 1) = 4 \sum i^2 - 4 \sum i + \sum 1$.
Using standard summation formulas: $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum i = \frac{n(n+1)}{2}$.
Sum $= 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 4 \left[ \frac{n(n+1)}{2} \right] + n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n(4n^2-1)}{3}$.
Thus,Reason $(R)$ is true.
Variance is defined as $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$.
$\sigma^2 = \frac{n(4n^2-1)}{3n} - \left( \frac{n^2}{n} \right)^2 = \frac{4n^2-1}{3} - n^2 = \frac{4n^2-1-3n^2}{3} = \frac{n^2-1}{3}$.
Thus,Assertion $(A)$ is true and $(R)$ is the correct explanation of $(A)$.

(D) Let the original observations be $x_i$ with variance $\sigma^2 = 7$.
When each observation is transformed to $y_i = ax + b$,the new variance is given by $\sigma^2(y) = a^2 \sigma^2(x)$.
Here,$a = 6$ and $b = -5$.
The constant $b$ does not affect the variance.
Therefore,the new variance is $\sigma^2_{new} = 6^2 \times 7 = 36 \times 7 = 252$.

(D) Given: $n = 100$,$\bar{x} = 40$,$\sigma = 5.1$.
We know that $\text{Variance} = \sigma^2 = (5.1)^2 = 26.01$.
The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
Substituting the values: $26.01 = \frac{\sum x_i^2}{100} - (40)^2$.
$26.01 = \frac{\sum x_i^2}{100} - 1600$.
$\frac{\sum x_i^2}{100} = 1626.01$.
$\sum x_i^2 = 162601$ (This is the incorrect sum of squares).
To find the correct sum of squares,we subtract the square of the incorrect observation and add the square of the correct observation:
$\text{Correct } \sum x_i^2 = 162601 - (50)^2 + (40)^2$.
$\text{Correct } \sum x_i^2 = 162601 - 2500 + 1600$.
$\text{Correct } \sum x_i^2 = 161701$.

(D) The given data is $1, 3, 5, 7, 11, 13, 17, 19, 23$. The number of observations $n = 9$.
The mean $\bar{x} = \frac{1+3+5+7+11+13+17+19+23}{9} = \frac{99}{9} = 11$.
The mean deviation from the mean $M = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|1-11| + |3-11| + |5-11| + |7-11| + |11-11| + |13-11| + |17-11| + |19-11| + |23-11|}{9} = \frac{10+8+6+4+0+2+6+8+12}{9} = \frac{56}{9}$.
The variance $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(-10)^2 + (-8)^2 + (-6)^2 + (-4)^2 + 0^2 + 2^2 + 6^2 + 8^2 + 12^2}{9} = \frac{100 + 64 + 36 + 16 + 0 + 4 + 36 + 64 + 144}{9} = \frac{464}{9}$.
Now,$3(\sigma^2 - M) = 3 \left( \frac{464}{9} - \frac{56}{9} \right) = 3 \left( \frac{408}{9} \right) = \frac{408}{3} = 136$.

(D) Let the circumradius be $R_c$. Given $R_c = PQ$. In $\triangle PQR$,$PQ = PR$,so $\angle Q = \angle R$.
Using the sine rule,$\frac{PQ}{\sin R} = 2R_c$.
Substituting $R_c = PQ$,we get $\frac{PQ}{\sin R} = 2PQ$.
This implies $\sin R = \frac{1}{2}$,so $R = 30^{\circ}$.
Since $\angle Q = \angle R$,we have $\angle Q = 30^{\circ}$.
In $\triangle PQR$,the sum of angles is $180^{\circ}$,so $\angle P + \angle Q + \angle R = 180^{\circ}$.
$\angle P + 30^{\circ} + 30^{\circ} = 180^{\circ} \Rightarrow \angle P = 120^{\circ}$.

(A) Let $X$ be the random variable representing the number of spade cards drawn. Since the cards are drawn with replacement,this follows a binomial distribution $B(n, p)$ where $n = 2$ and $p = \frac{13}{52} = \frac{1}{4}$.
The probability of success (getting a spade) is $p = \frac{1}{4}$ and the probability of failure is $q = 1 - p = \frac{3}{4}$.
The variance of a binomial distribution is given by $Var(X) = npq$.
Substituting the values: $Var(X) = 2 \times \frac{1}{4} \times \frac{3}{4} = \frac{6}{16} = \frac{3}{8}$.

(C) The probability mass function of a Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$.
Given the condition $3 P(X=2) = P(X=4)$.
Substituting the formula,we get $3 \cdot \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^4}{4!}$.
Canceling $e^{-\lambda}$ from both sides and simplifying the factorials: $\frac{3 \lambda^2}{2} = \frac{\lambda^4}{24}$.
Multiplying both sides by $24$,we get $36 \lambda^2 = \lambda^4$.
Since $\lambda > 0$,we have $\lambda^2 = 36$,which implies $\lambda = 6$.
Now,we need to find $P(X=6) = \frac{e^{-6} \cdot 6^6}{6!}$.
Calculating the value: $P(X=6) = \frac{e^{-6} \cdot 46656}{720} = \frac{46656}{720 e^6} = \frac{324}{5 e^6}$.

(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 $P(E \cup F) = P(X=1) + P(X=2) + P(X=3) + P(X=5) + P(X=7)$
$P(E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77$.

(C) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X = x_{i}) = 0.1 + k + 0.2 + 2k + 3k + k = 1$
$7k + 0.3 = 1 \Rightarrow 7k = 0.7 \Rightarrow k = 0.1$.
Now,we calculate the mean $\mu = E(X) = \sum x_{i} P(x_{i})$:
$\mu = (-2)(0.1) + (-1)(0.1) + (0)(0.2) + (1)(0.2) + (2)(0.3) + (3)(0.1)$
$\mu = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8$.
Next,we calculate $E(X^2) = \sum x_{i}^2 P(x_{i})$:
$E(X^2) = (-2)^2(0.1) + (-1)^2(0.1) + (0)^2(0.2) + (1)^2(0.2) + (2)^2(0.3) + (3)^2(0.1)$
$E(X^2) = 4(0.1) + 1(0.1) + 0 + 1(0.2) + 4(0.3) + 9(0.1)$
$E(X^2) = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 = 2.8$.
The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.
$Var(X) = 2.8 - (0.8)^2 = 2.8 - 0.64 = 2.16$.