TS EAMCET 2021 Mathematics Question Paper with Answer and Solution

(D) By the definition of an ellipse,the distance from any point $P(x, y)$ on the ellipse to the focus $S(0, 0)$ is $e$ times the distance to the directrix $x = 4$.
$SP^2 = e^2 \times (\text{distance to directrix})^2$
$x^2 + y^2 = (\frac{1}{2})^2 (x - 4)^2$
$x^2 + y^2 = \frac{1}{4} (x^2 - 8x + 16)$
$4x^2 + 4y^2 = x^2 - 8x + 16$
$3x^2 + 8x + 4y^2 = 16$
Multiply by $3$ to complete the square for $x$:
$9x^2 + 24x + 12y^2 = 48$
$(3x + 4)^2 - 16 + 12y^2 = 48$
$(3x + 4)^2 + 12y^2 = 64$

(A) The given equation is $4(x-2y+1)^2 + 9(2x+y+2)^2 = 25$.
Divide by $25$: $\frac{(x-2y+1)^2}{25/4} + \frac{(2x+y+2)^2}{25/9} = 1$.
Let $X = \frac{x-2y+1}{\sqrt{1^2+(-2)^2}} = \frac{x-2y+1}{\sqrt{5}}$ and $Y = \frac{2x+y+2}{\sqrt{2^2+1^2}} = \frac{2x+y+2}{\sqrt{5}}$.
The equation becomes $\frac{5X^2}{25/4} + \frac{5Y^2}{25/9} = 1$,which simplifies to $\frac{X^2}{5/4} + \frac{Y^2}{5/9} = 1$.
Here $a^2 = 5/4$ and $b^2 = 5/9$,so $a = \sqrt{5}/2$ and $b = \sqrt{5}/3$.
The major axis corresponds to the term with the larger denominator,which is $X=0$,i.e.,$x-2y+1=0$.
The eccentricity $e = \sqrt{1 - b^2/a^2} = \sqrt{1 - (5/9)/(5/4)} = \sqrt{1 - 4/9} = \sqrt{5}/3$.
The length of the major axis is $2a = 2(\sqrt{5}/2) = \sqrt{5}$.
The centre is the intersection of $x-2y+1=0$ and $2x+y+2=0$,which is $(-4/5, -2/5)$.

(B) For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the coordinates of the vertices of an inscribed rectangle are $(\pm a \cos \theta, \pm b \sin \theta)$.
Area of the rectangle $A = (2a \cos \theta)(2b \sin \theta) = 2ab \sin 2\theta$.
For the area to be maximum,$\sin 2\theta = 1$,which implies $\theta = \frac{\pi}{4}$.
Given length $L = 2a \cos \theta = 2a \cos(\frac{\pi}{4}) = 2a(\frac{1}{\sqrt{2}}) = a\sqrt{2} = 8\sqrt{2}$,so $a = 8$.
Given breadth $B = 2b \sin \theta = 2b \sin(\frac{\pi}{4}) = 2b(\frac{1}{\sqrt{2}}) = b\sqrt{2} = 4\sqrt{2}$,so $b = 4$.
Thus,$\frac{b}{a} = \frac{4}{8} = \frac{1}{2}$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$.
The length of the latus rectum is $\frac{2a^2}{b}$ and the length of the minor axis is $2a$.
Given that the length of the latus rectum is $\frac{2}{3}$ times the length of the minor axis:
$\frac{2a^2}{b} = \frac{2}{3}(2a)$
$\frac{a^2}{b} = \frac{2a}{3}$
$\frac{a}{b} = \frac{2}{3}$
$\frac{a^2}{b^2} = \frac{4}{9}$
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}}$.
$e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.

(A) Given equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(i)$ and equation of parabola is $y^2=8ax$ $(ii)$.
Differentiating $(i)$ with respect to $x$,we get $\frac{2x}{a^2}+\frac{2yy'}{b^2}=0 \Rightarrow y'=-\frac{b^2x}{a^2y}$.
Differentiating $(ii)$ with respect to $x$,we get $2yy'=8a \Rightarrow y'=\frac{4a}{y}$.
Since the curves intersect orthogonally,the product of their slopes at the point of intersection is $-1$.
$\left(-\frac{b^2x}{a^2y}\right) \times \left(\frac{4a}{y}\right) = -1$
$\Rightarrow \frac{4b^2x}{a^2y^2} = 1$ $\Rightarrow 4b^2x = a^2y^2$.
Substituting $y^2=8ax$ from $(ii)$ into this equation:
$4b^2x = a^2(8ax)$
$4b^2x = 8a^3x$
Since $x \neq 0$ at the point of intersection,we have $4b^2 = 8a^3$,which implies $b^2 = 2a^3$. Wait,re-evaluating the condition: $\frac{4b^2x}{a^2(8ax)} = 1$ $\Rightarrow \frac{4b^2}{8a^3} = 1$ $\Rightarrow b^2 = 2a^3$. This seems inconsistent with the standard form. Let's re-check: $\frac{4b^2x}{a^2y^2} = 1$ $\Rightarrow 4b^2x = a^2(8ax)$ $\Rightarrow 4b^2 = 8a^3$. Actually,the condition for orthogonality is $\frac{4b^2x}{a^2y^2} = 1$. Substituting $y^2=8ax$ gives $\frac{4b^2x}{a^2(8ax)} = 1$ $\Rightarrow \frac{4b^2}{8a^2} = 1$ $\Rightarrow \frac{b^2}{2a^2} = 1$ $\Rightarrow b^2 = 2a^2$.
For an ellipse with $b>a$,$e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{a^2}{2a^2} = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$e^4 = (e^2)^2 = (\frac{1}{2})^2 = \frac{1}{4}$.

(A) Given,equation of ellipse $16 x^2+25 y^2=400$ $\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1$.
Comparing to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we get $a=5$ and $b=4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Let $P(5 \cos \theta, 4 \sin \theta)$ be any point on the ellipse.
The focal distances are $P F_1 = a - ex = 5 - (\frac{3}{5})(5 \cos \theta) = 5 - 3 \cos \theta$ and $P F_2 = a + ex = 5 + 3 \cos \theta$.
Now,the product $P F_1 \cdot P F_2 = (5 - 3 \cos \theta)(5 + 3 \cos \theta) = 25 - 9 \cos^2 \theta$.
Since $0 \leq \cos^2 \theta \leq 1$,we have:
$25 - 9(1) \leq 25 - 9 \cos^2 \theta \leq 25 - 9(0)$
$16 \leq P F_1 \cdot P F_2 \leq 25$.
Thus,the value lies in the interval $[16, 25]$.

(D) The given equation of the ellipse is $9x^2+4y^2=36$.
Dividing both sides by $36$,we get $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
Comparing this with the standard form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (since $9 > 4$),we have $a^2 = 9$ and $b^2 = 4$,which gives $a = 3$ and $b = 2$.
For any point $P$ on an ellipse,the sum of its focal distances $PF_1 + PF_2$ is equal to the length of the major axis,which is $2a$.
Here,the major axis is along the $y$-axis because $a > b$.
Thus,the sum of the focal distances is $2a = 2 \times 3 = 6$.

(D) For the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$,$a^2=9$ and $b^2=5$. The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(\pm ae, 0) = (\pm 2, 0)$.
For the ellipse $\frac{x^2}{5}+\frac{y^2}{9}=1$,$a^2=9$ and $b^2=5$ (with major axis along the $y$-axis). The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(0, \pm ae) = (0, \pm 2)$.
The vertices of the quadrilateral are $(2, 0), (0, 2), (-2, 0),$ and $(0, -2)$.
This quadrilateral is a rhombus with diagonals of length $d_1 = 4$ and $d_2 = 4$.
The area of the quadrilateral is $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4 \times 4 = 8$ square units.

(D) Given the ellipse $E: 4x^2 + 9y^2 - 36 = 0$ and the circle $C: x^2 + y^2 - 9 = 0$.
For point $A(1, 2)$:
$E(1, 2) = 4(1)^2 + 9(2)^2 - 36 = 4 + 36 - 36 = 4 > 0$,so $A$ lies outside $E$.
$C(1, 2) = (1)^2 + (2)^2 - 9 = 1 + 4 - 9 = -4 < 0$,so $A$ lies inside $C$.
For point $B(2, 1)$:
$E(2, 1) = 4(2)^2 + 9(1)^2 - 36 = 16 + 9 - 36 = -11 < 0$,so $B$ lies inside $E$.
$C(2, 1) = (2)^2 + (1)^2 - 9 = 4 + 1 - 9 = -4 < 0$,so $B$ lies inside $C$.
Thus,$A$ lies inside $C$ but outside $E$ is the correct statement.

(B) The foci are $(-2,0)$ and $(8,0)$.
The distance between the foci is $2ae = 8 - (-2) = 10$.
$\Rightarrow ae = 5$.
Given $e = \frac{1}{\sqrt{2}}$,we have $a \left(\frac{1}{\sqrt{2}}\right) = 5 \Rightarrow a = 5\sqrt{2}$.
Now,$b^2 = a^2(1 - e^2) = (5\sqrt{2})^2 \left(1 - \frac{1}{2}\right) = 50 \left(\frac{1}{2}\right) = 25$.
Thus,$b = 5$.
The centre of the ellipse is the mid-point of the line joining the two foci: $\left(\frac{-2+8}{2}, \frac{0+0}{2}\right) = (3,0)$.
The equation of the ellipse is $\frac{(x-3)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$,which is $\frac{(x-3)^2}{50} + \frac{y^2}{25} = 1$.
The parametric coordinates are $(x, y) = (h + a \cos \theta, k + b \sin \theta)$,where $(h, k)$ is the centre.
Substituting the values,we get $(3 + 5\sqrt{2} \cos \theta, 5 \sin \theta)$.

(A) Given circle is $(x-1)^2+y^2=r^2 \dots (i)$
Here,the centre of the circle is $(1,0)$ and the radius is $r$.
Given ellipse is $x^2+4y^2=16$,which can be written as $\frac{x^2}{16}+\frac{y^2}{4}=1$.
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we get $a=4$ and $b=2$.
Since the circle touches the ellipse internally,the normal to the ellipse at the point of contact $P$ must pass through the centre of the circle $(1,0)$.
Let the point of contact be $P(4\cos\theta, 2\sin\theta)$.
The equation of the normal to the ellipse at $P$ is given by $ax\sec\theta - by\operatorname{cosec}\theta = a^2-b^2$.
Substituting $a=4, b=2$ and $a^2-b^2 = 16-4=12$,we get $4x\sec\theta - 2y\operatorname{cosec}\theta = 12$.
Since this normal passes through $(1,0)$,we have $4(1)\sec\theta - 2(0)\operatorname{cosec}\theta = 12$,which implies $4\sec\theta = 12$,so $\sec\theta = 3$.
Thus,$\cos\theta = \frac{1}{3}$ and $\sin\theta = \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.
The point $P$ is $\left(4\left(\frac{1}{3}\right), 2\left(\frac{2\sqrt{2}}{3}\right)\right) = \left(\frac{4}{3}, \frac{4\sqrt{2}}{3}\right)$.
The radius $r$ is the distance between the centre $(1,0)$ and the point $P\left(\frac{4}{3}, \frac{4\sqrt{2}}{3}\right)$.
$r^2 = \left(\frac{4}{3}-1\right)^2 + \left(\frac{4\sqrt{2}}{3}-0\right)^2 = \left(\frac{1}{3}\right)^2 + \frac{32}{9} = \frac{1}{9} + \frac{32}{9} = \frac{33}{9} = \frac{11}{3}$.
Therefore,$r = \sqrt{\frac{11}{3}}$.

(B) Given the equation of the conic:
$a^2 x^2 + b^2 y^2 = a^2(a^2 + b^2 - y^2)$
$\Rightarrow a^2 x^2 + b^2 y^2 + a^2 y^2 = a^2(a^2 + b^2)$
$\Rightarrow a^2 x^2 + y^2(a^2 + b^2) = a^2(a^2 + b^2)$
Dividing by $a^2(a^2 + b^2)$:
$\frac{x^2}{a^2 + b^2} + \frac{y^2}{a^2} = 1$
This is an ellipse with $A^2 = a^2 + b^2$ and $B^2 = a^2$. Since $A^2 > B^2$,the eccentricity $e$ is given by:
$e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{a^2}{a^2 + b^2}} = \sqrt{\frac{b^2}{a^2 + b^2}} = \frac{b}{\sqrt{a^2 + b^2}}$
By the definition of a conic section,the distance from a point $P$ to the focus $S$ is $SP = e \cdot PM$,where $PM$ is the perpendicular distance to the directrix.
Thus,$PM = \frac{1}{e} SP$.
Comparing this with $PM = K SP$,we get $K = \frac{1}{e}$.
$K = \frac{\sqrt{a^2 + b^2}}{b}$.

(B) The equation of the hyperbola is $S = 4x^2 - 3y^2 - 1 = 0$.
For a point $(\alpha, \beta)$ to be in the interior of the hyperbola,the condition is $S_1 < 0$ if the point is between the branches,but for the standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the interior region containing the foci is defined by $S_1 < 0$.
However,checking the standard definition for $4x^2 - 3y^2 - 1 = 0$ at $(0,0)$,we get $S_1 = -1 < 0$,which is the interior region.
Substituting $(\alpha, -1)$ into $S < 0$:
$4\alpha^2 - 3(-1)^2 - 1 < 0$
$4\alpha^2 - 3 - 1 < 0$
$4\alpha^2 - 4 < 0$
$\alpha^2 < 1$
$-1 < \alpha < 1$
Thus,$\alpha \in (-1, 1)$.

(D) For the equation $\frac{x^2}{\alpha+3} + \frac{y^2}{2-\alpha} = 1$ to represent a hyperbola,the coefficients of $x^2$ and $y^2$ must have opposite signs.
This implies that their product must be negative:
$(\alpha+3)(2-\alpha) < 0$
Multiplying both sides by $-1$ reverses the inequality sign:
$(\alpha+3)(\alpha-2) > 0$
Solving this inequality,we find the roots are $\alpha = -3$ and $\alpha = 2$.
The expression is positive outside the interval between the roots.
Therefore,$\alpha \in (-\infty, -3) \cup (2, \infty)$.

(D) The foci of the hyperbola are $S(ae, 0)$ and $S'(-ae, 0)$.
The endpoints of the latus rectum passing through $S$ are $A(ae, \frac{b^2}{a})$ and $B(ae, -\frac{b^2}{a})$.
The angle subtended by $AB$ at $S'$ is $60^{\circ}$.
Since the triangle $\triangle AS'B$ is isosceles,the line $S'S$ bisects the angle $\angle AS'B$,so $\angle AS'S = 30^{\circ}$.
The slope of $S'A$ is $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$.
Thus,$\frac{b^2/a}{ae - (-ae)} = \frac{b^2}{2a^2e} = \frac{1}{\sqrt{3}}$.
This gives $\frac{b^2}{a^2} = \frac{2e}{\sqrt{3}}$.
Using the relation $b^2 = a^2(e^2 - 1)$,we have $\frac{b^2}{a^2} = e^2 - 1$.
Equating the two expressions: $e^2 - 1 = \frac{2e}{\sqrt{3}}$,which simplifies to $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.
Solving the quadratic equation: $(\sqrt{3}e + 1)(e - \sqrt{3}) = 0$.
Since the eccentricity $e > 1$,we get $e = \sqrt{3}$.

(D) Given equation: $5x^2 - 6y^2 - 10x - 24y - 34 = 0$
Rearranging terms: $5(x^2 - 2x) - 6(y^2 + 4y) = 34$
Completing the square: $5(x^2 - 2x + 1) - 6(y^2 + 4y + 4) = 34 + 5 - 24$
$5(x - 1)^2 - 6(y + 2)^2 = 15$
Dividing by $15$: $\frac{(x - 1)^2}{3} - \frac{(y + 2)^2}{2.5} = 1$
Here,$a^2 = 3$ and $b^2 = 2.5 = \frac{5}{2}$.
Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5/2}{3}} = \sqrt{1 + \frac{5}{6}} = \sqrt{\frac{11}{6}}$.
Foci are given by $(h \pm ae, k)$,where $(h, k) = (1, -2)$.
$ae = \sqrt{3} \times \sqrt{\frac{11}{6}} = \sqrt{\frac{33}{6}} = \sqrt{\frac{11}{2}}$.
Thus,the foci are $\left(1 \pm \sqrt{\frac{11}{2}}, -2\right)$.

(C) The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.
Given $e = \frac{5}{4}$,we have $1 + \frac{b^2}{a^2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$.
Thus,$\frac{b^2}{a^2} = \frac{25}{16} - 1 = \frac{9}{16}$,which implies $b^2 = \frac{9}{16} a^2$.
The length of the latus rectum is $\frac{2b^2}{a} = 9$.
Substituting $b^2 = \frac{9}{16} a^2$ into the latus rectum equation:
$\frac{2}{a} \left(\frac{9}{16} a^2\right) = 9$
$\frac{9}{8} a = 9 \Rightarrow a = 8$.
Now,$b^2 = \frac{9}{16} (8)^2 = \frac{9}{16} \times 64 = 36$,so $b = 6$.
Therefore,$ab = 8 \times 6 = 48$.

(C) The equation of the straight lines joining the origin to the points of intersection of the line $x \cos \alpha + y \sin \alpha = P$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{36} = 1$ is obtained by homogenizing the hyperbola equation using the line equation: $\frac{x^2}{9} - \frac{y^2}{36} = (\frac{x \cos \alpha + y \sin \alpha}{P})^2$.
Expanding this,we get $x^2(\frac{1}{9} - \frac{\cos^2 \alpha}{P^2}) + y^2(-\frac{1}{36} - \frac{\sin^2 \alpha}{P^2}) - \frac{2xy \cos \alpha \sin \alpha}{P^2} = 0$.
Since the lines subtend a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(\frac{1}{9} - \frac{\cos^2 \alpha}{P^2}) + (-\frac{1}{36} - \frac{\sin^2 \alpha}{P^2}) = 0$.
This simplifies to $\frac{1}{12} - \frac{1}{P^2} = 0$,so $P^2 = 12$.
The polar of a point $(h, k)$ with respect to $\frac{x^2}{9} - \frac{y^2}{36} = 1$ is $\frac{xh}{9} - \frac{yk}{36} = 1$.
Comparing this with $x \cos \alpha + y \sin \alpha = P$,we have $\frac{h/9}{\cos \alpha} = \frac{-k/36}{\sin \alpha} = \frac{1}{P}$.
Thus,$\cos \alpha = \frac{Ph}{9}$ and $\sin \alpha = -\frac{Pk}{36}$.
Using $\cos^2 \alpha + \sin^2 \alpha = 1$,we get $\frac{P^2 h^2}{81} + \frac{P^2 k^2}{1296} = 1$.
Substituting $P^2 = 12$,we get $\frac{12h^2}{81} + \frac{12k^2}{1296} = 1$,which simplifies to $\frac{4h^2}{27} + \frac{k^2}{108} = 1$.
Multiplying by $108$,we get $16h^2 + k^2 = 108$.
Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2 + y^2 = 108$.

(B) The given limit is $\lim _{x \rightarrow 1} \frac{\log x}{1-x}$.
Substituting $x = 1$,we get the indeterminate form $\frac{\log 1}{1-1} = \frac{0}{0}$.
Applying $L'\text{Hospital's Rule}$,we differentiate the numerator and the denominator with respect to $x$:
$\lim _{x \rightarrow 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(1-x)} = \lim _{x \rightarrow 1} \frac{\frac{1}{x}}{-1}$.
Evaluating the limit as $x \rightarrow 1$:
$\frac{\frac{1}{1}}{-1} = -1$.

(D) Given $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$.
Since the limit exists and the denominator approaches $0$ as $x \rightarrow 2$,the numerator must also approach $0$ at $x=2$.
$3(2)^2 - a(2) + 5b = 0$ $\Rightarrow 12 - 2a + 5b = 0$ $\Rightarrow 2a - 5b = 12$.
Applying $L$'Hospital's rule:
$\lim _{x \rightarrow 2} \frac{d}{dx}(3x^2 - ax + 5b) / \frac{d}{dx}(x-2) = 17$.
$\lim _{x}$ ${\rightarrow 2} (6x - a) = 17$ $\Rightarrow 6(2) - a = 17$ $\Rightarrow 12 - a = 17$ $\Rightarrow a = -5$.
Substitute $a = -5$ into $2a - 5b = 12$:
$2(-5) - 5b = 12$ $\Rightarrow -10 - 5b = 12$ $\Rightarrow -5b = 22$ $\Rightarrow b = -\frac{22}{5}$.
Therefore,$ab = (-5) \times (-\frac{22}{5}) = 22$.

(C) Given $f(x) = -(\sin^2 x + \cos^5 x)$.
To find: $\lim_{x \rightarrow 0} \frac{f'(x)}{x}$.
Differentiating $f(x)$ with respect to $x$ using the chain rule:
$f'(x) = -[2 \sin x \cos x + 5 \cos^4 x(-\sin x)]$
$f'(x) = -\sin x (2 \cos x - 5 \cos^4 x)$
Now,evaluate the limit:
$\lim_{x \rightarrow 0} \frac{f'(x)}{x} = \lim_{x \rightarrow 0} \frac{-\sin x (2 \cos x - 5 \cos^4 x)}{x}$
$= -(\lim_{x \rightarrow 0} \frac{\sin x}{x}) \times (\lim_{x \rightarrow 0} (2 \cos x - 5 \cos^4 x))$
$= -1 \times (2(1) - 5(1)^4)$
$= -1 \times (2 - 5) = -1 \times (-3) = 3$.
Thus,the limit exists and is equal to $3$.

(D) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{3^x(1 - \cos x) - (1 - \cos x)}$
$= \lim _{x \rightarrow 0} \frac{x^2(2^x - \sin x - 1)}{(3^x - 1)(1 - \cos x)}$
$= \lim _{x \rightarrow 0} \frac{2^x - \sin x - 1}{(3^x - 1) \frac{(1 - \cos x)}{x^2}}$
Since $\lim _{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,the expression becomes:
$= \lim _{x \rightarrow 0} \frac{2(2^x - \sin x - 1)}{3^x - 1}$
Applying $L$'Hospital's Rule:
$= \lim _{x \rightarrow 0} \frac{2(2^x \ln 2 - \cos x)}{3^x \ln 3}$
$= \frac{2(\ln 2 - 1)}{\ln 3}$
$= \frac{2}{\log 3}(\log 2 - 1)$

(D) Given the limit: $\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7}$
Since $x \rightarrow -\infty$,we have $|x| = -x$. Let $x = -t$,where $t \rightarrow \infty$. Then $|x| = t$.
Substituting these into the expression:
$= \lim _{t \rightarrow \infty} \frac{3t^3 - (-t)^2 + 2t - 5}{-5t^3 + 3(-t)^2 - 2t + 7}$
$= \lim _{t \rightarrow \infty} \frac{3t^3 - t^2 + 2t - 5}{-5t^3 + 3t^2 - 2t + 7}$
Divide the numerator and denominator by $t^3$:
$= \lim _{t}$ ${\rightarrow \infty} \frac{3 - \frac{1}{t} + \frac{2}{t^2} - \frac{5}{t^3}}{-5 + \frac{3}{t} - \frac{2}{t^2} + \frac{7}{t^3}}$
As $t \rightarrow \infty$,all terms with $t$ in the denominator approach $0$.
$= \frac{3 - 0 + 0 - 0}{-5 + 0 - 0 + 0} = -\frac{3}{5}$

(B) Given,$\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x \tan 2 x+\frac{2 x}{3} \tan 3 x}$
Using the identity $1-\cos 2x = 2 \sin^2 x$,we get:
$= \lim _{x \rightarrow 0} \frac{2 \sin^2 x}{x \tan 2 x + \frac{2 x}{3} \tan 3 x}$
Divide numerator and denominator by $x^2$:
$= \lim _{x \rightarrow 0} \frac{2 (\frac{\sin x}{x})^2}{\frac{\tan 2 x}{x} + \frac{2}{3} \frac{\tan 3 x}{x}}$
$= \lim _{x \rightarrow 0} \frac{2 (\frac{\sin x}{x})^2}{2 (\frac{\tan 2 x}{2 x}) + 2 (\frac{\tan 3 x}{3 x})}$
Using $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\tan kx}{kx} = 1$:
$= \frac{2(1)^2}{2(1) + 2(1)} = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$

(B) We need to evaluate $\lim_{x \to 0^-} f(x)$.
For $x \in (-1, 0)$,the greatest integer function $[x] = -1$.
Therefore,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin[x]}{[x]}$.
Substituting $[x] = -1$,we get $\frac{\sin(-1)}{-1}$.
Since $\sin(-\theta) = -\sin\theta$,we have $\frac{-\sin 1}{-1} = \sin 1$.

(A) Let $L = \lim _{x \rightarrow 0} \frac{2 \sin x - \sin 2x}{x^3}$ (which is of the form $\frac{0}{0}$).
Applying $L$-Hospital rule:
$L = \lim _{x \rightarrow 0} \frac{2 \cos x - 2 \cos 2x}{3x^2}$ (which is of the form $\frac{0}{0}$).
Applying $L$-Hospital rule again:
$L = \lim _{x \rightarrow 0} \frac{-2 \sin x + 4 \sin 2x}{6x}$ (which is of the form $\frac{0}{0}$).
Applying $L$-Hospital rule again:
$L = \lim _{x \rightarrow 0} \frac{-2 \cos x + 8 \cos 2x}{6}$.
Substituting $x = 0$:
$L = \frac{-2 \cos(0) + 8 \cos(0)}{6} = \frac{-2(1) + 8(1)}{6} = \frac{6}{6} = 1$.

(B) Given the limit $L = \lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$.
Using the identity $1-\cos ^2 x = \sin ^2 x$,we have:
$L = \lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2} \sin ^2 x}}{x} = \lim _{x \rightarrow 0^{-}} \frac{\frac{1}{\sqrt{2}} |\sin x|}{x}$.
Since $x \rightarrow 0^{-}$,$x < 0$,which implies $\sin x < 0$. Therefore,$|\sin x| = -\sin x$.
$L = \frac{1}{\sqrt{2}} \lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x} = -\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we get $L = -\frac{1}{\sqrt{2}} \times 1 = -\frac{1}{\sqrt{2}}$.

(D) Let $f(x) = px^2 + qx + r$. Since $a$ and $b$ are roots,$f(x) = p(x - a)(x - b)$.
We need to evaluate $L = \lim_{x \rightarrow b} \frac{1 - \cos 2(f(x))}{2(px - pb)^2}$.
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$,we have $1 - \cos 2(f(x)) = 2 \sin^2(f(x))$.
So,$L = \lim_{x \rightarrow b} \frac{2 \sin^2(f(x))}{2 p^2(x - b)^2} = \frac{1}{p^2} \lim_{x \rightarrow b} \left( \frac{\sin(f(x))}{x - b} \right)^2$.
Since $f(x) = p(x - a)(x - b)$,we have $\frac{f(x)}{x - b} = p(x - a)$.
As $x \rightarrow b$,$p(x - a) \rightarrow p(b - a)$.
Thus,$L = \frac{1}{p^2} \lim_{x \rightarrow b} \left( \frac{\sin(p(x - a)(x - b))}{p(x - a)(x - b)} \cdot p(x - a) \right)^2$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get $L = \frac{1}{p^2} \cdot (p(b - a))^2 = \frac{p^2(b - a)^2}{p^2} = (b - a)^2$.
Note: The provided options seem to have a typo. $(b - a)^2$ is equivalent to $a^2 - 2ab + b^2$,which is option $D$.

(C) Let $y = \lim _{n \rightarrow \infty} P\left(1+\frac{r}{100 n}\right)^{t n}$.
We know the standard limit formula $\lim _{k \rightarrow \infty} (1+\frac{x}{k})^k = e^x$.
Here,let $k = n$. Then the expression becomes $P \left[ \lim _{n \rightarrow \infty} (1+\frac{r/100}{n})^n \right]^t$.
Using the limit formula,$\lim _{n \rightarrow \infty} (1+\frac{r/100}{n})^n = e^{r/100}$.
Therefore,$y = P \left( e^{r/100} \right)^t = P e^{\frac{rt}{100}}$.

(B) Given the mean $\bar{x} = 10$ for the $10$ observations:
$\frac{2+3+5+18+17+15+13+x+9+7}{10} = 10$
$\Rightarrow 89 + x = 100$
$\Rightarrow x = 11$
The data set is ${2, 3, 5, 18, 17, 15, 13, 11, 9, 7}$.
The mean deviation about the mean is calculated as $\frac{1}{n} \sum |x_i - \bar{x}|$:

$x_i$	$|x_i - 10|$
$2$	$8$
$3$	$7$
$5$	$5$
$18$	$8$
$17$	$7$
$15$	$5$
$13$	$3$
$11$	$1$
$9$	$1$
$7$	$3$

Sum of absolute deviations $= 8+7+5+8+7+5+3+1+1+3 = 48$.
Mean Deviation $= \frac{48}{10} = 4.8$.

(C) The given data set is $50, 70, 60, B, 20, 40$. The range is defined as the difference between the maximum and minimum values,which is given as $65$.
Case $1$: If $B$ is the maximum value,then $B - 20 = 65$,which implies $B = 85$.
Case $2$: If $B$ is the minimum value,then $70 - B = 65$,which implies $B = 5$.
The possible values for $B$ are $85$ and $5$.
The absolute difference between these values is $|85 - 5| = 80$.

(B) The first five odd natural numbers are $1, 3, 5, 7, 9$. Their mean $\bar{x} = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$.
The mean deviation $O = \frac{\sum |x_i - \bar{x}|}{5} = \frac{|1-5| + |3-5| + |5-5| + |7-5| + |9-5|}{5} = \frac{4+2+0+2+4}{5} = \frac{12}{5} = 2.4$.
The first five prime numbers are $2, 3, 5, 7, 11$. Their mean $\bar{y} = \frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6$.
The mean deviation $P = \frac{\sum |y_i - \bar{y}|}{5} = \frac{|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6|}{5} = \frac{3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac{13.6}{5} = 2.72$.
Therefore,$P - O = 2.72 - 2.4 = 0.32$.

(B) The first $15$ even integers are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30$.
Mean $\bar{x} = \frac{2+4+6+\dots+30}{15} = \frac{2(1+2+\dots+15)}{15} = \frac{2 \times \frac{15 \times 16}{2}}{15} = 16$.
Median $\tilde{x} = \left(\frac{15+1}{2}\right)^{\text{th}}$ term $= 8^{\text{th}}$ term $= 16$.
$M_1 = \text{mean deviation about mean} = \frac{\sum_{i=1}^{15} |x_i - \bar{x}|}{15} = \frac{|2-16| + |4-16| + \dots + |30-16|}{15}$.
$M_1 = \frac{14+12+10+8+6+4+2+0+2+4+6+8+10+12+14}{15} = \frac{112}{15}$.
Since the mean and median are equal,$M_2 = \text{mean deviation about median} = M_1 = \frac{112}{15}$.
Therefore,$M_1 + M_2 = \frac{112}{15} + \frac{112}{15} = \frac{224}{15}$.

(A) Given data in ascending order is $4, 4, 7, 12, 14, 15, 15, 23$.
Here,the number of terms $n = 8$,which is even.
So,the median is the average of the $4^{th}$ and $5^{th}$ terms.
Median $= \frac{12 + 14}{2} = \frac{26}{2} = 13$.
Now,we calculate the mean deviation about the median using the formula $\frac{\sum |x_i - \text{Median}|}{n}$.

$x_i$	$|x_i - 13|$
$4$	$9$
$4$	$9$
$7$	$6$
$12$	$1$
$14$	$1$
$15$	$2$
$15$	$2$
$23$	$10$

Sum of absolute deviations $\sum |d_i| = 9 + 9 + 6 + 1 + 1 + 2 + 2 + 10 = 40$.
Mean deviation about median $= \frac{40}{8} = 5$.

(D) Given,$\angle B = \frac{\pi}{4}$,$\angle C = \frac{\pi}{3}$.
In $\triangle ABC$,$\angle A = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} = 75^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$.
The area of the triangle is $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} (2R \sin B)(2R \sin C) \sin A = 2R^2 \sin A \sin B \sin C$.
Given $\Delta = 54 + 18\sqrt{3} = 18(3 + \sqrt{3})$.
$\sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
$\Delta = 2R^2 \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) = 2R^2 \left(\frac{\sqrt{3} + 1}{4\sqrt{2}}\right) \left(\frac{\sqrt{3}}{\sqrt{2}}\right) = 2R^2 \frac{3 + \sqrt{3}}{8} = R^2 \frac{3 + \sqrt{3}}{4}$.
Equating the areas: $R^2 \frac{3 + \sqrt{3}}{4} = 18(3 + \sqrt{3})$ $\Rightarrow R^2 = 72$ $\Rightarrow R = 6\sqrt{2}$.
Finally,$a = 2R \sin A = 2(6\sqrt{2}) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = 12\sqrt{2} \left(\frac{\sqrt{2}(\sqrt{3} + 1)}{4}\right) = 3(2)(\sqrt{3} + 1) = 6(\sqrt{3} + 1)$.

(A) We have to find the value of $\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2}$.
Using the sine rule,$b = 2R \sin B$ and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the expression:
$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(2R(\sin B + \sin C))^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(2R(\sin B - \sin C))^2}$
$= \frac{1}{4R^2} \left[ \frac{\cos^2 \left( \frac{B - C}{2} \right)}{(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2})^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(2 \cos \frac{B+C}{2} \sin \frac{B-C}{2})^2} \right]$
$= \frac{1}{4R^2} \left[ \frac{1}{4 \sin^2 \frac{B+C}{2}} + \frac{1}{4 \cos^2 \frac{B+C}{2}} \right]$
$= \frac{1}{16R^2} \left[ \frac{\cos^2 \frac{B+C}{2} + \sin^2 \frac{B+C}{2}}{\sin^2 \frac{B+C}{2} \cos^2 \frac{B+C}{2}} \right]$
$= \frac{1}{16R^2} \left[ \frac{1}{\sin^2 \frac{B+C}{2} \cos^2 \frac{B+C}{2}} \right] = \frac{1}{4R^2} \left[ \frac{1}{\sin^2 (B+C)} \right]$
Since $A+B+C = \pi$,$\sin(B+C) = \sin(\pi - A) = \sin A$.
Thus,the expression becomes $\frac{1}{4R^2 \sin^2 A} = \frac{1}{(2R \sin A)^2} = \frac{1}{a^2}$.

(D) Given the expression $\frac{1+\cos(A-B) \cdot \cos C}{1+\cos(A-C) \cdot \cos B}$.
Since $A+B+C = 180^{\circ}$,we have $C = 180^{\circ} - (A+B)$ and $B = 180^{\circ} - (A+C)$.
Thus,$\cos C = -\cos(A+B)$ and $\cos B = -\cos(A+C)$.
Substituting these into the expression:
$= \frac{1 - \cos(A-B)\cos(A+B)}{1 - \cos(A-C)\cos(A+C)}$
Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$:
$= \frac{1 - (\cos^2 A - \sin^2 B)}{1 - (\cos^2 A - \sin^2 C)}$
$= \frac{1 - \cos^2 A + \sin^2 B}{1 - \cos^2 A + \sin^2 C}$
$= \frac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C}$
Using the Sine Rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$.
$= \frac{(a/k)^2 + (b/k)^2}{(a/k)^2 + (c/k)^2} = \frac{a^2+b^2}{a^2+c^2}$.

(A) Given that $a, b, c$ are consecutive natural numbers with $a < b < c$,we have $a = b - 1$ and $c = b + 1$.
Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting these into the expression $(\cos A + \cos B + \cos C) 2abc$:
$= (\frac{b^2 + c^2 - a^2}{2bc} + \frac{a^2 + c^2 - b^2}{2ac} + \frac{a^2 + b^2 - c^2}{2ab}) 2abc$
$= a(b^2 + c^2 - a^2) + b(a^2 + c^2 - b^2) + c(a^2 + b^2 - c^2)$
Substitute $a = b - 1$ and $c = b + 1$:
$= (b - 1)(b^2 + (b + 1)^2 - (b - 1)^2) + b((b - 1)^2 + (b + 1)^2 - b^2) + (b + 1)((b - 1)^2 + b^2 - (b + 1)^2)$
$= (b - 1)(b^2 + 4b) + b(b^2 - 2b + 1 + b^2 + 2b + 1 - b^2) + (b + 1)(b^2 - 2b + 1 + b^2 - b^2 - 2b - 1)$
$= (b - 1)(b^2 + 4b) + b(b^2 + 2) + (b + 1)(b^2 - 4b)$
$= (b^3 + 4b^2 - b^2 - 4b) + (b^3 + 2b) + (b^3 - 4b^2 + b^2 - 4b)$
$= 3b^3 - 6b = 3b(b^2 - 2)$.

(D) Given expression: $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right) = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right)$
$= \frac{(b+c)^2-a^2}{bc} = \frac{b^2+c^2-a^2+2bc}{bc}$
$= \frac{b^2+c^2-a^2}{bc} + 2 = 2\left(\frac{b^2+c^2-a^2}{2bc}\right) + 2$
$= 2\cos A + 2$
Since $A+B+C = 180^{\circ}$ and $B+C = 72^{\circ}$,we have $A = 180^{\circ} - 72^{\circ} = 108^{\circ}$.
So,the expression becomes $2\cos 108^{\circ} + 2 = 2\cos(90^{\circ} + 18^{\circ}) + 2$
$= -2\sin 18^{\circ} + 2 = -2\left(\frac{\sqrt{5}-1}{4}\right) + 2$
$= \frac{1-\sqrt{5}}{2} + 2 = \frac{1-\sqrt{5}+4}{2} = \frac{5-\sqrt{5}}{2}$.

(A) Given,$a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have:
$a \left( \frac{1 + \cos C}{2} \right) + c \left( \frac{1 + \cos A}{2} \right) = \frac{3b}{2}$
Multiplying by $2$:
$a(1 + \cos C) + c(1 + \cos A) = 3b$
$a + a \cos C + c + c \cos A = 3b$
$(a + c) + (a \cos C + c \cos A) = 3b$
By the projection rule,$a \cos C + c \cos A = b$.
Substituting this into the equation:
$(a + c) + b = 3b$
$a + c = 2b$

(B) Using the projection formula,we know that $b = c \cos A + a \cos C$ and $c = b \cos A + a \cos B$.
Substituting these into the expression:
$\frac{b - c \cos A}{c - b \cos A} = \frac{(c \cos A + a \cos C) - c \cos A}{(b \cos A + a \cos B) - b \cos A}$
$= \frac{a \cos C}{a \cos B}$
$= \frac{\cos C}{\cos B}$

(B) Let the angles of $\triangle ABC$ be $(A-d), A, (A+d)$.
Since the sum of angles in a triangle is $180^{\circ}$,we have:
$(A-d) + A + (A+d) = 180^{\circ}$
$3A = 180^{\circ} \Rightarrow A = 60^{\circ}$.
Using the cosine rule for angle $A$:
$\cos A = \frac{b^2+c^2-a^2}{2bc}$
$\cos 60^{\circ} = \frac{b^2+c^2-a^2}{2bc}$
$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$
$bc = b^2+c^2-a^2$
$a^2 = b^2+c^2-bc$.
Thus,the correct relation is $a^2 = b^2+c^2-bc$.

(C) Given the equation: $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$
Multiply both sides by $(a+b+c)$:
$\frac{a+b+c}{a+b} + \frac{a+b+c}{c+a} = 3$
$1 + \frac{c}{a+b} + 1 + \frac{b}{c+a} = 3$
$\frac{c}{a+b} + \frac{b}{c+a} = 1$
$c(c+a) + b(a+b) = (a+b)(c+a)$
$c^2 + ac + ab + b^2 = ac + a^2 + bc + ab$
$b^2 + c^2 - a^2 = bc$
Using the Cosine Rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos A = \frac{bc}{2bc} = \frac{1}{2}$
Since $\cos A = \frac{1}{2}$,$A = 60^{\circ}$
Therefore,$\sin A = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$

(A) Let the sides of the triangle $ABC$ be $a, b, c$ units and the corresponding altitudes be $h_1, h_2, h_3$ units respectively.
Then the perimeter is $a + b + c = 2S$ and the area is $P = \frac{1}{2}(ah_1) = \frac{1}{2}(bh_2) = \frac{1}{2}(ch_3)$.
This implies $h_1 = \frac{2P}{a}, h_2 = \frac{2P}{b}, h_3 = \frac{2P}{c}$.
Now,consider the expression $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right]$.
Substituting the values of $h_1, h_2, h_3$:
$= P^2 \left[ \frac{(\frac{4P^2}{ab} + \frac{4P^2}{bc} + \frac{4P^2}{ca})^2}{(\frac{8P^3}{abc})^2} - 2 \right]$
$= P^2 \left[ \frac{16P^4 (\frac{c+a+b}{abc})^2}{\frac{64P^6}{a^2b^2c^2}} - 2 \right]$
$= P^2 \left[ \frac{16P^4 (\frac{2S}{abc})^2}{\frac{64P^6}{a^2b^2c^2}} - 2 \right]$
$= P^2 \left[ \frac{16P^4 \cdot \frac{4S^2}{a^2b^2c^2}}{\frac{64P^6}{a^2b^2c^2}} - 2 \right]$
$= P^2 \left[ \frac{64P^4S^2}{64P^6} - 2 \right]$
$= P^2 \left[ \frac{S^2}{P^2} - 2 \right] = S^2 - 2P^2$.

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$.
Multiplying these,we get $\cot \frac{A}{2} \cot \frac{B}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-b)}{(s-a)(s-c)}} = \sqrt{\frac{s^2}{(s-c)^2}} = \frac{s}{s-c}$.
Given $\frac{s}{s-c} = K$,we can write $K = \frac{s-c+c}{s-c} = 1 + \frac{c}{s-c}$.
Since $s = \frac{a+b+c}{2}$,we have $s-c = \frac{a+b-c}{2}$.
Thus,$K = 1 + \frac{c}{\frac{a+b-c}{2}} = 1 + \frac{2c}{a+b-c}$.
By the triangle inequality,$a+b > c$,so $a+b-c > 0$,which implies $K > 1$.
As $C \to 180^{\circ}$,$a+b-c \to 0^{+}$,so $K \to \infty$.
Therefore,the range of $K$ is $(1, \infty)$.

(D) Given,$4 \Delta = c^2 - (a - b)^2$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get:
$4 \Delta = (c - (a - b))(c + (a - b)) = (c - a + b)(c + a - b)$.
Dividing by $4$,we have:
$\Delta = \left(\frac{b + c - a}{2}\right) \left(\frac{a + c - b}{2}\right) = (s - a)(s - b)$,where $s$ is the semi-perimeter.
From Heron's formula,$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
Thus,$\sqrt{s(s - a)(s - b)(s - c)} = (s - a)(s - b)$.
Squaring both sides: $s(s - a)(s - b)(s - c) = (s - a)^2(s - b)^2$.
$\frac{s(s - c)}{(s - a)(s - b)} = 1$.
We know that $\tan^2\left(\frac{C}{2}\right) = \frac{(s - a)(s - b)}{s(s - c)}$.
Therefore,$\tan^2\left(\frac{C}{2}\right) = 1$,which implies $\tan\left(\frac{C}{2}\right) = 1$.
$\frac{C}{2} = \frac{\pi}{4} \Rightarrow C = \frac{\pi}{2}$.
Hence,$\sin C = \sin\left(\frac{\pi}{2}\right) = 1$.

(B) We know that $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.
Multiplying these,we get $\tan \frac{B}{2} \tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-a)^2}{s^2}} = \frac{s-a}{s}$.
Substituting $s = \frac{a+b+c}{2}$,we have $\frac{s-a}{s} = \frac{\frac{a+b+c}{2} - a}{\frac{a+b+c}{2}} = \frac{b+c-a}{b+c+a}$.
Given $b+c = 4a$,we substitute this into the expression:
$\frac{4a-a}{4a+a} = \frac{3a}{5a} = \frac{3}{5}$.

(A) We have $r r_2 = r_1 r_3$.
Using the formulas $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$:
$\left(\frac{\Delta}{s}\right) \left(\frac{\Delta}{s-b}\right) = \left(\frac{\Delta}{s-a}\right) \left(\frac{\Delta}{s-c}\right)$
$\Rightarrow s(s-b) = (s-a)(s-c)$
$\Rightarrow s^2 - sb = s^2 - s(a+c) + ac$
$\Rightarrow s(a+c-b) = ac$
Since $2s = a+b+c$,we have $a+c = 2s-b$.
Substituting this: $s(2s-b-b) = ac$ $\Rightarrow s(2s-2b) = ac$ $\Rightarrow 2s(s-b) = ac$.
Using the identity $\cos B = \frac{a^2+c^2-b^2}{2ac}$,we know $\sin^2(\frac{B}{2}) = \frac{(s-a)(s-c)}{ac}$ and $\cos^2(\frac{B}{2}) = \frac{s(s-b)}{ac}$.
From $s(s-b) = (s-a)(s-c)$,we get $\frac{s(s-b)}{ac} = \frac{(s-a)(s-c)}{ac} \Rightarrow \cos^2(\frac{B}{2}) = \sin^2(\frac{B}{2})$.
$\Rightarrow \tan^2(\frac{B}{2}) = 1$ $\Rightarrow \frac{B}{2} = \frac{\pi}{4}$ $\Rightarrow B = \frac{\pi}{2}$.
Therefore,$\cos 2B = \cos \pi = -1$.

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a) \cdot s(s-c)}{(s-b)(s-c) \cdot (s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.
Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.
Thus,$\frac{s}{s-b} = \frac{2s}{2s-2b} = \frac{a+b+c}{a+b+c-2b} = \frac{(a+c)+b}{(a+c)-b}$.
Given $a+c=5b$,substituting this into the expression gives $\frac{5b+b}{5b-b} = \frac{6b}{4b} = \frac{3}{2}$.

(B) Given sides of the triangle are $a=12, b=16, c=20$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{12+16+20}{2} = \frac{48}{2} = 24$.
Area of the triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{24(24-12)(24-16)(24-20)} = \sqrt{24 \times 12 \times 8 \times 4} = \sqrt{9216} = 96$.
The exradii are given by $r_A = \frac{\Delta}{s-a}, r_B = \frac{\Delta}{s-b}, r_C = \frac{\Delta}{s-c}$.
$r_C = \frac{96}{24-20} = \frac{96}{4} = 24$.
$r_B = \frac{96}{24-16} = \frac{96}{8} = 12$.
$r_A = \frac{96}{24-12} = \frac{96}{12} = 8$.
The ratio $r_C : r_B : r_A = 24 : 12 : 8 = 6 : 3 : 2$.