AP EAMCET 2020 Mathematics Question Paper with Answer and Solution

(B) The equation of the given ellipse is $x^2+4y^2=64$,which can be written as $\frac{x^2}{64}+\frac{y^2}{16}=1$.
Let the vertex $P$ of the rectangle in the first quadrant be $(8\cos\theta, 4\sin\theta)$.
The coordinates of the four vertices of the rectangle are $(8\cos\theta, 4\sin\theta)$,$(-8\cos\theta, 4\sin\theta)$,$(-8\cos\theta, -4\sin\theta)$,and $(8\cos\theta, -4\sin\theta)$.
The length of the sides of the rectangle are $L = 2(8\cos\theta) = 16\cos\theta$ and $W = 2(4\sin\theta) = 8\sin\theta$.
The area $A$ of the rectangle is $A = L \times W = (16\cos\theta)(8\sin\theta) = 128\sin\theta\cos\theta = 64\sin(2\theta)$.
For the area to be maximum,$\sin(2\theta)$ must be $1$,which implies $2\theta = \frac{\pi}{2}$ or $\theta = \frac{\pi}{4}$.
Substituting $\theta = \frac{\pi}{4}$ into the expressions for the sides:
$L = 16\cos(\frac{\pi}{4}) = 16(\frac{1}{\sqrt{2}}) = 8\sqrt{2}$.
$W = 8\sin(\frac{\pi}{4}) = 8(\frac{1}{\sqrt{2}}) = 4\sqrt{2}$.
Thus,the lengths of the sides are $8\sqrt{2}$ and $4\sqrt{2}$.

(B) The equation of the ellipse is $9x^2 + y^2 = 9$.
Substituting $x = m^2$ into the equation of the ellipse,we get $9(m^2)^2 + y^2 = 9$.
This simplifies to $9m^4 + y^2 = 9$,which gives $y^2 = 9 - 9m^4 = 9(1 - m^4)$.
For the points to be real and distinct,we must have $y^2 > 0$.
Therefore,$9(1 - m^4) > 0$,which implies $1 - m^4 > 0$ or $m^4 < 1$.
Taking the fourth root on both sides,we get $|m| < 1$.
Thus,the correct option is $B$.

(A) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{32} = 1$,where $a^2 = 18$ and $b^2 = 32$. Since $b^2 > a^2$,the major axis is along the $y$-axis.
The equation of a tangent with slope $m = -\frac{4}{3}$ is given by $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Substituting the values: $y = -\frac{4}{3}x \pm \sqrt{18(-\frac{4}{3})^2 + 32} = -\frac{4}{3}x \pm \sqrt{18(\frac{16}{9}) + 32} = -\frac{4}{3}x \pm \sqrt{32 + 32} = -\frac{4}{3}x \pm 8$.
This simplifies to $4x + 3y = \pm 24$.
For the tangent $4x + 3y = 24$:
The $x$-intercept (where $y=0$) is $Q(6,0)$.
The $y$-intercept (where $x=0$) is $P(0,8)$.
Thus,the points are $P(0,8)$ and $Q(6,0)$.

(C) The condition for the line $y=mx+c$ to touch the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$.
Here,$a^2=4$,$b^2=1$,and $m=4$.
Substituting these values,we get $c^2 = 4(4)^2 + 1 = 4(16) + 1 = 64 + 1 = 65$.
Thus,$c = \pm \sqrt{65}$.
There are $2$ possible values for '$c$'.
Therefore,the correct option is $C$.

(A) The given ellipse is $x^2+16y^2=16$,which can be written as $\frac{x^2}{16}+\frac{y^2}{1}=1$.
Here,$a^2=16$ and $b^2=1$.
The slope of the tangent is $m = \tan 60^{\circ} = \sqrt{3}$.
The equation of the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with slope $m$ is given by $y = mx \pm \sqrt{a^2m^2+b^2}$.
Substituting the values,we get $y = \sqrt{3}x \pm \sqrt{16(\sqrt{3})^2+1}$.
$y = \sqrt{3}x \pm \sqrt{16(3)+1} = \sqrt{3}x \pm \sqrt{48+1} = \sqrt{3}x \pm \sqrt{49}$.
$y = \sqrt{3}x \pm 7$.
Thus,the equations are $\sqrt{3}x-y+7=0$ or $\sqrt{3}x-y-7=0$.
Comparing with the given options,$\sqrt{3}x-y+7=0$ is the correct choice.

(B) The given equation of the curve is $x^2+4y^2=4$. Dividing by $4$,we get the standard form of the ellipse: $\frac{x^2}{4}+\frac{y^2}{1}=1$.
Here,$a^2=4$ and $b^2=1$.
The condition for the line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$.
Comparing $y=2x+c$ with $y=mx+c$,we have $m=2$.
Substituting the values $a^2=4$,$b^2=1$,and $m=2$ into the condition:
$c^2 = (4)(2)^2 + 1$
$c^2 = 4(4) + 1$
$c^2 = 16 + 1 = 17$.

(C) Let a point $P(2 \cos \theta, 6 \sin \theta)$ be on the ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$. The equation of the normal to the ellipse at point $P$ is given by $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Here $a^2 = 4$ and $b^2 = 36$. Substituting the coordinates $(2 \cos \theta, 6 \sin \theta)$,we get: $\frac{4x}{2 \cos \theta} - \frac{36y}{6 \sin \theta} = 4 - 36$. Simplifying,we have $2x \sec \theta - 6y \operatorname{cosec} \theta = -32$,or $2x \sec \theta - 6y \operatorname{cosec} \theta + 32 = 0$. Comparing this with $ax + by + c = 0$,we have $\frac{a}{2 \sec \theta} = \frac{b}{-6 \operatorname{cosec} \theta} = \frac{c}{32}$. This gives $\cos \theta = \frac{2c}{32a} = \frac{c}{16a}$ and $\sin \theta = -\frac{6c}{32b} = -\frac{3c}{16b}$. Using $\cos^2 \theta + \sin^2 \theta = 1$,we get $\frac{c^2}{256a^2} + \frac{9c^2}{256b^2} = 1$,which simplifies to $\frac{1}{a^2} + \frac{9}{b^2} = \frac{256}{c^2}$.

(B) Let $P(h, k)$ be any point on the locus.
Let $A$ be on the $X$-axis and $B$ be on the $Y$-axis.
Let $\theta$ be the angle that the line segment $AB$ makes with the $X$-axis.
From the geometry of the figure,in $\triangle PMA$,we have $\sin \theta = \frac{k}{b}$,which implies $k = b \sin \theta$.
In $\triangle BNP$,we have $\cos \theta = \frac{h}{a}$,which implies $h = a \cos \theta$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we substitute the values:
$(\frac{k}{b})^2 + (\frac{h}{a})^2 = 1$.
Replacing $(h, k)$ with $(x, y)$,the equation of the locus is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$. The latus rectum through $F_1$ has endpoints $A(-ae, b^2/a)$ and $A'(-ae, -b^2/a)$.
The angle subtended by the latus rectum at the other focus $F_2$ is $60^{\circ}$.
Considering the right-angled triangle formed by the focus $F_2$,the midpoint of the latus rectum $B(-ae, 0)$,and one endpoint $A(-ae, b^2/a)$,the angle at $F_2$ is $30^{\circ}$.
In $\triangle ABF_2$,we have $AB = \frac{b^2}{a}$ and $BF_2 = ae - (-ae) = 2ae$.
Thus,$\tan 30^{\circ} = \frac{AB}{BF_2} = \frac{b^2/a}{2ae} = \frac{b^2}{2a^2e}$.
Since $\frac{1}{\sqrt{3}} = \frac{b^2}{2a^2e}$ and $b^2 = a^2(e^2 - 1)$,we substitute $b^2$:
$\frac{1}{\sqrt{3}} = \frac{a^2(e^2 - 1)}{2a^2e} = \frac{e^2 - 1}{2e}$.
This gives $2e = \sqrt{3}(e^2 - 1)$,or $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.
Solving the quadratic equation: $e = \frac{2 \pm \sqrt{4 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} = \frac{2 \pm \sqrt{4 + 12}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}}$.
Since $e > 1$,we take $e = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

(B) The given equation is $(x-3)^2+(y+1)^2=(4x+3y)^2$.
We can rewrite this as $\sqrt{(x-3)^2+(y+1)^2} = 5 \left| \frac{4x+3y}{5} \right|$.
This is of the form $SP = ePM$,where $S(3, -1)$ is the focus,$e = 5$ is the eccentricity,and $4x+3y=0$ is the directrix.
The transverse axis is a line passing through the focus $(3, -1)$ and is perpendicular to the directrix $4x+3y=0$.
The slope of the directrix is $m_1 = -4/3$.
Since the transverse axis is perpendicular to the directrix,its slope is $m_2 = 3/4$.
The equation of the transverse axis is $y - (-1) = \frac{3}{4}(x - 3)$.
$4(y+1) = 3(x-3) \implies 4y+4 = 3x-9 \implies 3x-4y = 13$.
Thus,option $B$ is correct.

(D) The given equation of the hyperbola is $7x^2 - 49y^2 = 343$.
Dividing both sides by $343$,we get:
$\frac{7x^2}{343} - \frac{49y^2}{343} = \frac{343}{343}$
$\frac{x^2}{49} - \frac{y^2}{7} = 1$.
Comparing this with the standard equation of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 49$,which implies $a = 7$.
The vertices of the hyperbola are given by $(\pm a, 0)$.
Therefore,the vertices are $(\pm 7, 0)$.
Thus,option $D$ is correct.

(A) Given,Focus $(S) = (1, 2)$,Eccentricity $(e) = \sqrt{3}$,and Directrix $2x + y - 1 = 0$.
By the definition of a conic,$SP = e \cdot PM$,where $P(x, y)$ is a point on the hyperbola.
$SP^2 = e^2 \cdot PM^2$
$(x - 1)^2 + (y - 2)^2 = 3 \cdot \frac{(2x + y - 1)^2}{2^2 + 1^2}$
$x^2 - 2x + 1 + y^2 - 4y + 4 = \frac{3}{5}(4x^2 + y^2 + 1 + 4xy - 4x - 2y)$
$5(x^2 + y^2 - 2x - 4y + 5) = 3(4x^2 + y^2 + 4xy - 4x - 2y + 1)$
$5x^2 + 5y^2 - 10x - 20y + 25 = 12x^2 + 3y^2 + 12xy - 12x - 6y + 3$
Rearranging the terms:
$12x^2 - 5x^2 + 3y^2 - 5y^2 + 12xy - 12x + 10x - 6y + 20y + 3 - 25 = 0$
$7x^2 - 2y^2 + 12xy - 2x + 14y - 22 = 0$
Multiplying by $-1$:
$-7x^2 + 2y^2 - 12xy + 2x - 14y + 22 = 0$
Thus,the equation is $2y^2 - 12xy - 7x^2 + 2x - 14y + 22 = 0$.

(A) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The coordinates of the vertex $A^{\prime}$ are $(-a, 0)$.
The focus $S$ is $(ae, 0)$ and the endpoints of the latus rectum $L$ and $L^{\prime}$ are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.
The length of the side $LL^{\prime} = \frac{2b^2}{a}$.
Since $\triangle A^{\prime}LL^{\prime}$ is equilateral,the height from $A^{\prime}$ to $LL^{\prime}$ is $\frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times \frac{2b^2}{a} = \frac{\sqrt{3}b^2}{a}$.
The distance from $A^{\prime}(-a, 0)$ to the line $x = ae$ is $ae - (-a) = a(e+1)$.
Thus,$a(e+1) = \frac{\sqrt{3}b^2}{a}$.
Using $b^2 = a^2(e^2-1)$,we get $a(e+1) = \frac{\sqrt{3}a^2(e^2-1)}{a} = \sqrt{3}a(e-1)(e+1)$.
Dividing by $a(e+1)$,we get $1 = \sqrt{3}(e-1)$,which implies $e-1 = \frac{1}{\sqrt{3}}$,so $e = 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}}$.

(B) The given curve is $x=a \cosh(t)$ and $y=b \sinh(t)$,which represents the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The slope of the tangent at point $t$ is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cosh(t)}{a \sinh(t)}$.
The slope of the normal is $-\frac{dx}{dy} = -\frac{a \sinh(t)}{b \cosh(t)}$.
The equation of the normal at point $(a \cosh(t), b \sinh(t))$ is given by:
$y - b \sinh(t) = -\frac{a \sinh(t)}{b \cosh(t)} (x - a \cosh(t))$.
Multiplying by $b \cosh(t)$:
$by \cosh(t) - b^2 \sinh(t) \cosh(t) = -ax \sinh(t) + a^2 \sinh(t) \cosh(t)$.
Rearranging the terms:
$ax \sinh(t) + by \cosh(t) = (a^2+b^2) \sinh(t) \cosh(t)$.
Dividing both sides by $\sinh(t) \cosh(t)$:
$\frac{ax}{\cosh(t)} + \frac{by}{\sinh(t)} = a^2+b^2$.
This can be written as $ax \operatorname{sech}(t) + by \operatorname{cosech}(t) = a^2+b^2$.

(A) The locus of the point of intersection of perpendicular tangents to a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is known as the director circle,given by the equation $x^2 + y^2 = a^2 - b^2$.
For the given hyperbola $\frac{x^2}{4} - \frac{y^2}{2} = 1$,we have $a^2 = 4$ and $b^2 = 2$.
Substituting these values into the director circle equation,we get $x^2 + y^2 = 4 - 2 = 2$.
Thus,the intersection point lies on the circle $x^2 + y^2 = 2$.
Hence,option $A$ is correct.

(B) The equation of the asymptotes of the hyperbola $2 x^2+5 x y+2 y^2-11 x-7 y-4=0$ is of the form $2 x^2+5 x y+2 y^2-11 x-7 y+\lambda=0$.
Since this represents a pair of straight lines,the determinant of the general second-degree equation must be zero:
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Here,$a=2, b=2, c=\lambda, h=\frac{5}{2}, g=-\frac{11}{2}, f=-\frac{7}{2}$.
Substituting these values:
$2(2)(\lambda) + 2(-\frac{7}{2})(-\frac{11}{2})(\frac{5}{2}) - 2(-\frac{7}{2})^2 - 2(-\frac{11}{2})^2 - \lambda(\frac{5}{2})^2 = 0$.
$4\lambda + \frac{385}{4} - \frac{49}{2} - \frac{121}{2} - \frac{25\lambda}{4} = 0$.
Multiplying by $4$:
$16\lambda + 385 - 98 - 242 - 25\lambda = 0$.
$-9\lambda + 45 = 0 \Rightarrow \lambda = 5$.
Thus,the equation is $2 x^2+5 x y+2 y^2-11 x-7 y+5=0$.

(D) The limit of a continuous function $f(x) = x^n$ as $x$ approaches $a$ is simply the value of the function at $x = a$.
Since $f(x) = x^n$ is continuous for $x > 0$,we have:
$\lim_{x \rightarrow a} x^n = a^n$

(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !-n !}$
First,simplify the denominator by factoring out $n!$:
$(n+1)! - n! = n!(n+1) - n!(1) = n!(n+1-1) = n!(n)$
Now,substitute this back into the expression:
$\lim _{n \rightarrow \infty} \frac{n!}{n!(n)} = \lim _{n \rightarrow \infty} \frac{1}{n}$
As $n \rightarrow \infty$,the value of $\frac{1}{n}$ approaches $0$.
Therefore,the limit is $0$.

(D) Given,$\lim _{x \rightarrow 3} \frac{x^n - 3^n}{x - 3} = 108$.
Using the standard limit formula $\lim _{x \rightarrow a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1}$:
Here,$a = 3$,so $n \cdot 3^{n-1} = 108$.
$n \cdot 3^{n-1} = 108$.
Multiply both sides by $3$:
$n \cdot 3^n = 108 \times 3 = 324$.
We can write $324$ as $4 \times 81 = 4 \times 3^4$.
Comparing $n \cdot 3^n = 4 \times 3^4$,we get $n = 4$.
Thus,the value of $n$ is $4$.

(C) To evaluate the limit $\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x}$,we observe that it is in the indeterminate form $\frac{0}{0}$.
We can apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}\left(\left(1+\frac{x}{2}\right)^{5 / 7}-1\right)}{\frac{d}{dx}(x)}$
$= \lim _{x}$ ${\rightarrow 0} \frac{\frac{5}{7}\left(1+\frac{x}{2}\right)^{5 / 7 - 1} \cdot \frac{d}{dx}\left(1+\frac{x}{2}\right)}{1}$
$= \lim _{x \rightarrow 0} \frac{5}{7}\left(1+\frac{x}{2}\right)^{-2 / 7} \cdot \frac{1}{2}$
$= \frac{5}{7} \cdot (1+0)^{-2 / 7} \cdot \frac{1}{2}$
$= \frac{5}{14}$
Thus,the correct option is $C$.

(A) For the limit $\lim_{x \rightarrow 2} f(x)$ to exist,the Left Hand Limit $(LHL)$ must be equal to the Right Hand Limit $(RHL)$ at $x = 2$.
$LHL = \lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2} (4x - 5) = 4(2) - 5 = 3$.
$RHL = \lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2} (x - k) = 2 - k$.
Equating $LHL$ and $RHL$:
$3 = 2 - k$
$k = 2 - 3$
$k = -1$.
Thus,the correct option is $A$.

(A) We are given the limit $\lim _{x \rightarrow 0} \frac{a^x-1}{\sin x}$.
Dividing both the numerator and the denominator by $x$,we get:
$\lim _{x \rightarrow 0} \frac{\frac{a^x-1}{x}}{\frac{\sin x}{x}}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log _e a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$\frac{\log _e a}{1} = \log _e a$.
Thus,the correct option is $A$.

(D) Let $L = \lim _{x \rightarrow 1}(1-x) \tan \left(\frac{\pi x}{2}\right)$.
This is an indeterminate form of type $0 \times \infty$.
We can rewrite the limit as:
$L = \lim _{x \rightarrow 1} \frac{1-x}{\cot \left(\frac{\pi x}{2}\right)}$.
Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(1-x)}{\frac{d}{dx}(\cot \left(\frac{\pi x}{2}\right))}$
$L = \lim _{x \rightarrow 1} \frac{-1}{-\csc^2 \left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}}$
$L = \frac{1}{\frac{\pi}{2} \csc^2 \left(\frac{\pi}{2}\right)}$
Since $\csc \left(\frac{\pi}{2}\right) = 1$,we get:
$L = \frac{1}{\frac{\pi}{2} \cdot 1^2} = \frac{2}{\pi}$.
Thus,the correct option is $D$.

(B) To evaluate the limit $\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}$,we multiply and divide by the conjugate expression:
$= \lim _{n \rightarrow \infty} \frac{(n-\sqrt{n^2-4 n})(n+\sqrt{n^2-4 n})}{n+\sqrt{n^2-4 n}}$
$= \lim _{n \rightarrow \infty} \frac{n^2-(n^2-4 n)}{n+\sqrt{n^2-4 n}}$
$= \lim _{n \rightarrow \infty} \frac{4 n}{n+\sqrt{n^2-4 n}}$
Divide the numerator and denominator by $n$:
$= \lim _{n \rightarrow \infty} \frac{4}{1+\sqrt{1-\frac{4}{n}}}$
As $n \rightarrow \infty$,$\frac{4}{n} \rightarrow 0$,so the expression becomes:
$= \frac{4}{1+\sqrt{1-0}} = \frac{4}{1+1} = \frac{4}{2} = 2$.

(C) Let $x = \pi + h$. As $x \rightarrow \pi$,$h \rightarrow 0$.
The expression becomes $\lim _{h \rightarrow 0} \frac{1-\sin (\frac{\pi+h}{2})}{\cos (\frac{\pi+h}{2}) (\cos (\frac{\pi+h}{4})-\sin (\frac{\pi+h}{4}))}$.
Using trigonometric identities,$\sin (\frac{\pi}{2} + \frac{h}{2}) = \cos (\frac{h}{2})$ and $\cos (\frac{\pi}{2} + \frac{h}{2}) = -\sin (\frac{h}{2})$.
The expression simplifies to $\lim _{h \rightarrow 0} \frac{1-\cos (h/2)}{-\sin (h/2) (\cos (\frac{\pi}{4} + \frac{h}{4}) - \sin (\frac{\pi}{4} + \frac{h}{4}))}$.
Since $\cos (\frac{\pi}{4} + \frac{h}{4}) - \sin (\frac{\pi}{4} + \frac{h}{4}) = \sqrt{2} \sin (\frac{\pi}{4} - (\frac{\pi}{4} + \frac{h}{4})) = \sqrt{2} \sin (-h/4) = -\sqrt{2} \sin (h/4)$.
Substituting this back: $\lim _{h \rightarrow 0} \frac{2 \sin^2 (h/4)}{-\sin (h/2) \cdot (-\sqrt{2} \sin (h/4))} = \lim _{h \rightarrow 0} \frac{2 \sin (h/4)}{\sqrt{2} \sin (h/2)}$.
Using $\sin (h/2) = 2 \sin (h/4) \cos (h/4)$,we get $\lim _{h \rightarrow 0} \frac{2 \sin (h/4)}{\sqrt{2} \cdot 2 \sin (h/4) \cos (h/4)} = \lim _{h \rightarrow 0} \frac{1}{\sqrt{2} \cos (h/4)} = \frac{1}{\sqrt{2} \cdot 1} = \frac{1}{\sqrt{2}}$.

(C) We are given the limit $L = \lim _{x \rightarrow 0} \frac{\sin (x^m)}{(\sin x)^n}$ where $n < m$.
By multiplying and dividing by $x^m$ in the numerator and $x^n$ in the denominator,we get:
$L = \lim _{x \rightarrow 0} \left( \frac{\sin (x^m)}{x^m} \cdot x^m \right) / \left( \frac{\sin x}{x} \cdot x \right)^n$
$L = \lim _{x \rightarrow 0} \left( \frac{\sin (x^m)}{x^m} \right) \cdot \left( \frac{x}{\sin x} \right)^n \cdot x^{m-n}$
Since $\lim _{x \rightarrow 0} \frac{\sin (x^m)}{x^m} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,the expression becomes:
$L = 1 \cdot (1)^n \cdot \lim _{x \rightarrow 0} x^{m-n}$
Given $n < m$,we have $m - n > 0$.
Therefore,$L = 0^{m-n} = 0$.
Hence,option $C$ is correct.

(C) We know that the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$.
Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$,the expression becomes:
$\lim _{x \rightarrow 0} \frac{2 \sin^2(\frac{x}{2})}{x^2}$
$= 2 \lim _{x \rightarrow 0} \left( \frac{\sin(\frac{x}{2})}{x} \right)^2$
$= 2 \lim _{x \rightarrow 0} \left( \frac{\sin(\frac{x}{2})}{2 \cdot \frac{x}{2}} \right)^2$
$= 2 \cdot \frac{1}{4} \lim _{x \rightarrow 0} \left( \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \right)^2$
Since $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get:
$= 2 \cdot \frac{1}{4} \cdot (1)^2 = \frac{1}{2}$.

(A) Given,$\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x}=e^9$.
Using the standard limit formula $\lim _{x \rightarrow \infty}(1+\frac{a}{x})^x = e^a$,we have:
$\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x} = \left[ \lim _{x \rightarrow \infty} \left(1+\frac{p}{x}\right)^{x} \right]^{q} = (e^p)^q = e^{pq}$.
Comparing this with the given expression $e^9$,we get $pq = 9$.
Since $p, q \in \mathbb{N}$ (natural numbers),the possible pairs $(p, q)$ are $(1, 9), (3, 3), (9, 1)$.
In these cases,$p+q$ can be $1+9=10$ or $3+3=6$.
Given the options,the correct value is $6$.

(B) We know that the standard limit formula is $\lim _{x \rightarrow 0}(1+ax)^{\frac{1}{ax}} = e$ for $a \neq 0$.
Given expression is $\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}}$.
We can rewrite the exponent as $\frac{2}{x} = 6 \times \frac{1}{3x}$.
Substituting this into the limit,we get $\lim _{x \rightarrow 0}(1+3x)^{\frac{2}{x}} = \lim _{x \rightarrow 0} \left((1+3x)^{\frac{1}{3x}}\right)^6$.
Using the standard limit property,this expression evaluates to $e^6$.

(A) We use the standard limit formula: $\lim _{x \rightarrow \infty} (1 + \frac{a}{x})^{bx} = e^{ab}$.
Given expression: $\lim _{x \rightarrow \infty} (1 + \frac{2}{x})^{3x}$.
Here,$a = 2$ and $b = 3$.
Applying the formula: $e^{2 \times 3} = e^6$.

(B) Let the expression be $L = \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\}$.
Factoring the expression inside the curly brackets:
$1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right) = \left(1-\cos \left(\frac{x^2}{2}\right)\right) - \cos \left(\frac{x^2}{4}\right) \left(1-\cos \left(\frac{x^2}{2}\right)\right) = \left(1-\cos \left(\frac{x^2}{2}\right)\right) \left(1-\cos \left(\frac{x^2}{4}\right)\right)$.
Using the identity $1-\cos \theta = 2 \sin ^2 \left(\frac{\theta}{2}\right)$:
$L = \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left(2 \sin ^2 \left(\frac{x^2}{4}\right)\right) \left(2 \sin ^2 \left(\frac{x^2}{8}\right)\right) = \lim _{x \rightarrow 0} \frac{32 \sin ^2 \left(\frac{x^2}{4}\right) \sin ^2 \left(\frac{x^2}{8}\right)}{\sin ^8 x}$.
Using the limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$L = 32 \lim _{x \rightarrow 0} \left[ \frac{\sin ^2 \left(\frac{x^2}{4}\right)}{\left(\frac{x^2}{4}\right)^2} \cdot \left(\frac{x^2}{4}\right)^2 \cdot \frac{\sin ^2 \left(\frac{x^2}{8}\right)}{\left(\frac{x^2}{8}\right)^2} \cdot \left(\frac{x^2}{8}\right)^2 \cdot \frac{1}{\left(\frac{\sin x}{x}\right)^8 \cdot x^8} \right]$.
$L = 32 \cdot 1 \cdot \frac{x^4}{16} \cdot 1 \cdot \frac{x^4}{64} \cdot \frac{1}{x^8} = 32 \cdot \frac{1}{16 \cdot 64} = \frac{32}{1024} = \frac{1}{32}$.

(C) The total frequency is given as $120$. So,$17 + f_1 + 32 + f_2 + 19 = 120$.
$f_1 + f_2 + 68 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).
The class marks $(x_i)$ are $10, 30, 50, 70, 90$.
The mean is $\frac{\sum f_i x_i}{\sum f_i} = 50$.
$\frac{17(10) + f_1(30) + 32(50) + f_2(70) + 19(90)}{120} = 50$.
$170 + 30f_1 + 1600 + 70f_2 + 1710 = 6000$.
$30f_1 + 70f_2 + 3480 = 6000$.
$30f_1 + 70f_2 = 2520 \implies 3f_1 + 7f_2 = 252$ (Equation $2$).
From Equation $1$,$f_1 = 52 - f_2$. Substituting into Equation $2$:
$3(52 - f_2) + 7f_2 = 252$.
$156 - 3f_2 + 7f_2 = 252$.
$4f_2 = 96 \implies f_2 = 24$.
Then $f_1 = 52 - 24 = 28$.

(B) The formula for the mean of $n$ observations is given by $\text{Mean} = \frac{\sum x_i}{n}$.
Given the observations are $8, 6, 7, 5, x, 4$ and the mean is $7$.
Number of observations $n = 6$.
$\frac{8 + 6 + 7 + 5 + x + 4}{6} = 7$
$\frac{30 + x}{6} = 7$
$30 + x = 42$
$x = 42 - 30$
$x = 12$
Hence,option $B$ is correct.

(A) Let the number of boys and girls in the class be $m$ and $n$ respectively. According to the given information:
Total marks of boys $= 40m$
Total marks of girls $= 45n$
Total marks of boys and girls combined $= 42(m + n)$
Equating the total marks:
$40m + 45n = 42(m + n)$
$40m + 45n = 42m + 42n$
$3n = 2m$
$\frac{m}{n} = \frac{3}{2}$
The percentage of boys in the class is given by:
$\frac{m}{m + n} \times 100 = \frac{3}{3 + 2} \times 100$
$= \frac{3}{5} \times 100 = 60 \%$
Therefore,the percentage of boys is $60 \%$.
Hence,option $A$ is correct.

(C) The mean and the median of a data set are measures of central tendency.
They are not necessarily equal for all data sets.
However,they can be equal for certain symmetric distributions or specific data sets.
Therefore,they are sometimes equal.
Hence,option $C$ is correct.

(A, B) Given the numbers are $2, 3, 2x, 11$. The mean $\bar{x} = \frac{2+3+2x+11}{4} = \frac{16+2x}{4} = 4 + \frac{x}{2}$.
Given standard deviation $\sigma = 3.5 = \frac{7}{2}$,so $\sigma^2 = \frac{49}{4}$.
The variance formula is $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$\frac{49}{4} = \frac{2^2 + 3^2 + (2x)^2 + 11^2}{4} - (4 + \frac{x}{2})^2$.
$\frac{49}{4} = \frac{4 + 9 + 4x^2 + 121}{4} - (16 + 4x + \frac{x^2}{4})$.
$\frac{49}{4} = \frac{134 + 4x^2}{4} - \frac{64 + 16x + x^2}{4}$.
$49 = 134 + 4x^2 - 64 - 16x - x^2$.
$49 = 3x^2 - 16x + 70$.
$3x^2 - 16x + 21 = 0$.
$(3x - 7)(x - 3) = 0$.
Thus,$x = \frac{7}{3}$ or $x = 3$.

(D) The mean $\bar{x}$ of the first $n$ natural numbers is given by $\bar{x} = \frac{1+2+\ldots+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2}$.
Given that $n$ is an even number,the mean deviation about the mean is calculated as:
$M.D.(\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = \frac{1}{n} \sum_{i=1}^{n} |i - \frac{n+1}{2}|$.
Since $n$ is even,the sum splits into two equal parts of absolute differences:
$M.D.(\bar{x}) = \frac{1}{n} \left[ \sum_{i=1}^{n/2} (\frac{n+1}{2} - i) + \sum_{i=n/2+1}^{n} (i - \frac{n+1}{2}) \right]$.
Evaluating this sum,we get $M.D.(\bar{x}) = \frac{n^2-1}{4n}$.

(C) Given: $\bar{x} = 50$,$n = 100$,and $\sigma = 5$.
We know that $\bar{x} = \frac{\Sigma x_i}{n}$,so $\Sigma x_i = n \cdot \bar{x} = 100 \cdot 50 = 5000$.
The formula for standard deviation is $\sigma = \sqrt{\frac{\Sigma x_i^2}{n} - (\bar{x})^2}$.
Squaring both sides,we get $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$.
Substituting the given values: $5^2 = \frac{\Sigma x_i^2}{100} - (50)^2$.
$25 = \frac{\Sigma x_i^2}{100} - 2500$.
$\frac{\Sigma x_i^2}{100} = 2525$.
$\Sigma x_i^2 = 252500$.

(C) The range of a data set is defined as the difference between the maximum value and the minimum value.
Given data: $15, 14, k, 25, 30, 35$.
Range $= 23$.
Case $1$: If $35$ is the maximum value,then the minimum value must be $35 - 23 = 12$.
If $k = 12$,the data set becomes $12, 14, 15, 25, 30, 35$. The range is $35 - 12 = 23$. This is a valid case.
Case $2$: If $k$ is the maximum value,then $k - 14 = 23$,which gives $k = 37$.
Since we are looking for the least positive value of $k$,we compare $12$ and $37$.
The least positive value is $12$.

(B) The given observations are $20, 28, 40, 12, 30, 15, 50$.
To find the range,we identify the maximum and minimum values in the data set.
Maximum value $= 50$.
Minimum value $= 12$.
Range $= \text{Maximum value} - \text{Minimum value}$.
Range $= 50 - 12 = 38$.

(D) The coefficient of variation $(CV)$ is a statistical measure of the dispersion of data points in a data series around the mean.
It is defined as the ratio of the standard deviation $\sigma$ to the mean $\bar{x}$,expressed as a percentage.
The formula is given by:
$CV = \frac{\sigma}{\bar{x}} \times 100$
where $\bar{x} \neq 0$.
Thus,option $D$ is correct.

(D) Let the observations be $x_1, x_2, x_3, x_4, x_5$. The mean $\bar{x} = \frac{\sum_{i=1}^{5} x_i}{5} = 15$,so $\sum x_i = 75$.
Given variance $\sigma^2 = \frac{\sum x_i^2}{5} - (\bar{x})^2 = 9$.
$\Rightarrow \frac{\sum x_i^2}{5} - 225 = 9$ $\Rightarrow \sum x_i^2 = 5(234) = 1170$.
Now,two new observations $-5$ and $13$ are added. The new sum of observations is $75 - 5 + 13 = 83$.
The new sum of squares is $1170 + (-5)^2 + (13)^2 = 1170 + 25 + 169 = 1364$.
The new mean is $\bar{x}' = \frac{83}{7}$.
The new variance is $\sigma'^2 = \frac{\sum x_i^2 + 25 + 169}{7} - (\bar{x}')^2 = \frac{1364}{7} - (\frac{83}{7})^2$.
$\sigma'^2 = \frac{1364 \times 7 - 6889}{49} = \frac{9548 - 6889}{49} = \frac{2659}{49}$.
Thus,the correct option is $D$.

(B) The given series is an arithmetic progression with $2n+1$ terms: $a, a+d, a+2d, \ldots, a+2nd$.
The mean $m$ of the series is the middle term: $m = a + nd$.
The mean deviation from the mean is given by $\frac{1}{2n+1} \sum_{i=0}^{2n} |x_i - m|$.
Substituting the terms:
$\text{Mean Deviation} = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a+kd) - (a+nd)| = \frac{1}{2n+1} \sum_{k=0}^{2n} |(k-n)d|$.
This sum is $\frac{d}{2n+1} [| -n | + | -(n-1) | + \ldots + | 0 | + \ldots + | n |]$.
The sum inside the bracket is $2 \times (1 + 2 + \ldots + n) = 2 \times \frac{n(n+1)}{2} = n(n+1)$.
Thus,the mean deviation is $\frac{n(n+1)d}{2n+1}$.
Therefore,option $B$ is correct.

(B) Let the five observations be $x_1, x_2, x_3, x_4, x_5$.
Given that the mean is $9$,we have $\frac{x_1+x_2+x_3+x_4+x_5}{5} = 9$,which implies $x_1+x_2+x_3+x_4+x_5 = 45$.
Since the standard deviation is $0$,all observations must be equal to the mean. Thus,$x_1 = x_2 = x_3 = x_4 = x_5 = 9$.
Now,one observation $x_5$ is changed to $y$ such that the new mean is $10$.
So,$\frac{x_1+x_2+x_3+x_4+y}{5} = 10$,which implies $x_1+x_2+x_3+x_4+y = 50$.
Substituting $x_1+x_2+x_3+x_4 = 36$ (since $x_1=x_2=x_3=x_4=9$),we get $36+y = 50$,so $y = 14$.
The new set of observations is ${9, 9, 9, 9, 14}$.
The new mean is $10$.
The new standard deviation is $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{(9-10)^2 + (9-10)^2 + (9-10)^2 + (9-10)^2 + (14-10)^2}{5}}$.
$= \sqrt{\frac{(-1)^2 + (-1)^2 + (-1)^2 + (-1)^2 + (4)^2}{5}} = \sqrt{\frac{1+1+1+1+16}{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$.

(A) Given that the number of observations $n = 60$,$\Sigma x_i = 960$,and $\Sigma x_i^2 = 18000$.
The formula for variance $\sigma^2$ is given by $\sigma^2 = \frac{\Sigma x_i^2}{n} - \left(\frac{\Sigma x_i}{n}\right)^2$.
Substituting the given values:
$\sigma^2 = \frac{18000}{60} - \left(\frac{960}{60}\right)^2$.
$\sigma^2 = 300 - (16)^2$.
$\sigma^2 = 300 - 256$.
$\sigma^2 = 44$.

(D) Given: $n_1 = 20, \bar{x}_1 = 25, \sigma_1^2 = 9$ and $n_2 = 30, \bar{x}_2 = 10, \sigma_2^2 = 16$.
Combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{20 \times 25 + 30 \times 10}{20 + 30} = \frac{500 + 300}{50} = \frac{800}{50} = 16$.
Let $d_1 = \bar{x}_1 - \bar{x} = 25 - 16 = 9$ and $d_2 = \bar{x}_2 - \bar{x} = 10 - 16 = -6$.
Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$.
$\sigma^2 = \frac{20(9 + 9^2) + 30(16 + (-6)^2)}{20 + 30} = \frac{20(9 + 81) + 30(16 + 36)}{50} = \frac{20(90) + 30(52)}{50} = \frac{1800 + 1560}{50} = \frac{3360}{50} = 67.2$.

(B) In a $\triangle ABC$,we are given that $\frac{\tan A}{2} = \frac{\tan B}{3} = \frac{\tan C}{4} = k$.
Thus,$\tan A = 2k$,$\tan B = 3k$,and $\tan C = 4k$.
In any $\triangle ABC$,the sum of angles is $180^\circ$,which implies $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Substituting the values,we get $2k + 3k + 4k = (2k)(3k)(4k)$,which simplifies to $9k = 24k^3$.
Since $k \neq 0$,we have $k^2 = \frac{9}{24} = \frac{3}{8}$.
We need to find $\sec^2 A + \sec^2 B + \sec^2 C = (1 + \tan^2 A) + (1 + \tan^2 B) + (1 + \tan^2 C) = 3 + \tan^2 A + \tan^2 B + \tan^2 C$.
Substituting the values,we get $3 + (2k)^2 + (3k)^2 + (4k)^2 = 3 + k^2(4 + 9 + 16) = 3 + 29k^2$.
Substituting $k^2 = \frac{3}{8}$,we get $3 + 29 \times \frac{3}{8} = 3 + \frac{87}{8} = \frac{24 + 87}{8} = \frac{111}{8}$.

(D) Given that in a $\triangle ABC$,angles $A, B, C$ are in $AP$,so $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,we get $\frac{b}{c} = \frac{\sin B}{\sin C}$.
Substituting the given values,$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}/2}{\sin C}$.
This simplifies to $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,so $C = 45^{\circ}$.
Finally,$\angle A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 75^{\circ}$.

(C) Given $A+B+C=0$,we have $A+B = -C$.
Taking the sine of both sides,$\sin(A+B) = \sin(-C) = -\sin(C)$.
Now,consider the expression $\sin(2A) + \sin(2B) + \sin(2C)$:
$\sin(2A) + \sin(2B) + \sin(2C) = 2 \sin(A+B) \cos(A-B) + 2 \sin(C) \cos(C)$
Since $A+B = -C$,then $\cos(A+B) = \cos(-C) = \cos(C)$.
Substituting $\sin(A+B) = -\sin(C)$ and $\cos(C) = \cos(A+B)$:
$= 2(-\sin(C)) \cos(A-B) + 2 \sin(C) \cos(A+B)$
$= -2 \sin(C) [\cos(A-B) - \cos(A+B)]$
Using the identity $\cos(x-y) - \cos(x+y) = 2 \sin(x) \sin(y)$:
$= -2 \sin(C) [2 \sin(A) \sin(B)]$
$= -4 \sin(A) \sin(B) \sin(C)$.

(C) In a $\triangle ABC$,it is given that $a=1$,$b=\sqrt{3}$ and $\angle C=\frac{\pi}{6}$.
Using the Law of Cosines:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos \frac{\pi}{6} = \frac{1^2 + (\sqrt{3})^2 - c^2}{2(1)(\sqrt{3})}$
$\frac{\sqrt{3}}{2} = \frac{1 + 3 - c^2}{2\sqrt{3}}$
$\frac{\sqrt{3}}{2} = \frac{4 - c^2}{2\sqrt{3}}$
Multiply both sides by $2\sqrt{3}$:
$3 = 4 - c^2$
$c^2 = 4 - 3 = 1$
Since $c$ represents the length of a side,$c = 1$.
Therefore,option $C$ is correct.