TS EAMCET 2022 Mathematics Question Paper with Answer and Solution

(B) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$.
Given the line $2x - 3y + 4 = 0$,we can write $y = \frac{2x + 4}{3}$.
Substituting this into the ellipse equation: $25x^2 + 9(\frac{2x + 4}{3})^2 = 225$.
$25x^2 + (2x + 4)^2 = 225$ $\Rightarrow 25x^2 + 4x^2 + 16x + 16 = 225$ $\Rightarrow 29x^2 + 16x - 209 = 0$.
The midpoint $\alpha$ of the $x$-coordinates is $\frac{x_1 + x_2}{2} = \frac{-16/29}{2} = -\frac{8}{29}$.
Similarly,for $y$,$x = \frac{3y - 4}{2}$. Substituting into the ellipse equation: $25(\frac{3y - 4}{2})^2 + 9y^2 = 225$.
$\frac{25}{4}(9y^2 - 24y + 16) + 9y^2 = 225$ $\Rightarrow 225y^2 - 600y + 400 + 36y^2 = 900$ $\Rightarrow 261y^2 - 600y - 500 = 0$.
The midpoint $\beta$ of the $y$-coordinates is $\frac{y_1 + y_2}{2} = \frac{600/261}{2} = \frac{300}{261} = \frac{100}{87}$.
Now,$3\beta - 2\alpha = 3(\frac{100}{87}) - 2(-\frac{8}{29}) = \frac{100}{29} + \frac{16}{29} = \frac{116}{29} = 4$.

(D) The coordinates of points $P$ and $Q$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $P = (a \cos \theta, b \sin \theta)$ and $Q = (a \cos(\frac{\pi}{2}+\theta), b \sin(\frac{\pi}{2}+\theta)) = (-a \sin \theta, b \cos \theta)$.
Let the midpoint of $PQ$ be $(x, y)$. Then,
$x = \frac{a(\cos \theta - \sin \theta)}{2} \Rightarrow \frac{2x}{a} = \cos \theta - \sin \theta$
$y = \frac{b(\sin \theta + \cos \theta)}{2} \Rightarrow \frac{2y}{b} = \sin \theta + \cos \theta$
Squaring and adding these equations:
$(\frac{2x}{a})^2 + (\frac{2y}{b})^2 = (\cos \theta - \sin \theta)^2 + (\sin \theta + \cos \theta)^2$
$\frac{4x^2}{a^2} + \frac{4y^2}{b^2} = (\cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta) + (\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta) = 2$
$\frac{x^2}{a^2/2} + \frac{y^2}{b^2/2} = 1$
Comparing this with $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$,we get $\alpha = \frac{a}{\sqrt{2}}$ and $\beta = \frac{b}{\sqrt{2}}$.
Therefore,$\frac{a+b}{\alpha+\beta} = \frac{a+b}{\frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}}} = \frac{a+b}{\frac{1}{\sqrt{2}}(a+b)} = \sqrt{2}$.

(C) The given equation is $\frac{x^2}{k-\frac{5}{2}} + \frac{y^2}{\frac{7}{3}-k} = 1$.
For this to represent a hyperbola,the denominators must have opposite signs,i.e.,their product must be negative:
$(k - \frac{5}{2})(\frac{7}{3} - k) < 0$.
Multiplying by $-1$,we get $(k - \frac{5}{2})(k - \frac{7}{3}) > 0$.
Since $\frac{7}{3} \approx 2.33$ and $\frac{5}{2} = 2.5$,we have $\frac{7}{3} < \frac{5}{2}$.
The inequality $(k - \frac{7}{3})(k - \frac{5}{2}) > 0$ holds when $k < \frac{7}{3}$ or $k > \frac{5}{2}$.
Thus,the set of all values of $k$ is $\left(-\infty, \frac{7}{3}\right) \cup \left(\frac{5}{2}, \infty\right)$.

(C) The given hyperbola is $\frac{x^2}{25} - \frac{y^2}{9} = 1$.
Any point on this hyperbola is given by $(5 \sec \theta, 3 \tan \theta)$.
Given $P(\theta) = (x_1, \frac{3 \sqrt{5}}{2})$,we equate the $y$-coordinates: $3 \tan \theta = \frac{3 \sqrt{5}}{2} \implies \tan \theta = \frac{\sqrt{5}}{2}$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have $\sec^2 \theta = 1 + \frac{5}{4} = \frac{9}{4}$,so $\sec \theta = \frac{3}{2}$ (as $0 < \theta < \frac{\pi}{2}$).
Thus,$x_1 = 5 \sec \theta = 5 \times \frac{3}{2} = \frac{15}{2}$.
Also,$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{\sec^2 \theta} = 1 - \frac{4}{9} = \frac{5}{9}$.
Finally,$2 x_1 + 9 \sin^2 \theta = 2(\frac{15}{2}) + 9(\frac{5}{9}) = 15 + 5 = 20$.

(C) Given the hyperbola equation $\frac{x^2}{16}-\frac{y^2}{9}=1$,we have $a^2=16$ and $b^2=9$.
Since point $P(5, y_1)$ lies on the hyperbola,we substitute $x=5$:
$\frac{25}{16}-\frac{y_1^2}{9}=1$
$\Rightarrow \frac{y_1^2}{9}=\frac{25}{16}-1 = \frac{9}{16}$
$\Rightarrow y_1^2 = \frac{81}{16}$ $\Rightarrow y_1 = \pm \frac{9}{4}$.
Thus,$P = (5, \pm \frac{9}{4})$.
The eccentricity $e$ is given by $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{16}} = \frac{5}{4}$.
The focus $S$ on the positive $X$-axis is $(ae, 0) = (4 \times \frac{5}{4}, 0) = (5, 0)$.
Now,the distance $SP$ is:
$SP = \sqrt{(5-5)^2 + (0 - (\pm \frac{9}{4}))^2} = \sqrt{0 + \frac{81}{16}} = \frac{9}{4}$.

(C) The given equations are in parametric form $x = a \sec \theta, y = b \tan \theta$,which represents the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$.
For the first hyperbola,$a = 1$ and $b = \sqrt{2}$,so $e_1^2 = 1 + \frac{(\sqrt{2})^2}{1^2} = 1 + 2 = 3$.
For the second hyperbola,$a = \sqrt{2}$ and $b = 1$,so $e_2^2 = 1 + \frac{1^2}{(\sqrt{2})^2} = 1 + \frac{1}{2} = \frac{3}{2}$.
Therefore,$\frac{e_2^2}{e_1^2} = \frac{3/2}{3} = \frac{1}{2}$.

(C) The equation of the hyperbola with centre at the origin and transverse axis along the $X$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
Given the length of the transverse axis $2a = 8$,we have $a = 4$,so $a^2 = 16$.
The hyperbola passes through $(5, 2)$,so $\frac{25}{16} - \frac{4}{b^2} = 1$.
$\frac{4}{b^2} = \frac{25}{16} - 1 = \frac{9}{16}$,which gives $b^2 = \frac{64}{9}$.
The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.
The eccentricity $e'$ of the conjugate hyperbola is given by $e' = \sqrt{1 + \frac{a^2}{b^2}}$.
$e' = \sqrt{1 + \frac{16}{64/9}} = \sqrt{1 + \frac{16 \times 9}{64}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$.

(B) For the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$,we have $a^2=9$ and $b^2=4$.
Thus,$a=3$ and $b=2$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
The focus $S$ on the positive $X$-axis is $(ae, 0) = (3 \times \frac{\sqrt{5}}{3}, 0) = (\sqrt{5}, 0)$.
$A$ point $P$ on the ellipse is $(3 \cos \theta, 2 \sin \theta)$.
The distance $SP$ is given by the focal distance formula $SP = a - ex$ for a focus on the positive $X$-axis,where $x = 3 \cos \theta$.
So,$SP = 3 - (\frac{\sqrt{5}}{3})(3 \cos \theta) = 3 - \sqrt{5} \cos \theta$.
Given $SP = 1$,we have $1 = 3 - \sqrt{5} \cos \theta$.
$\sqrt{5} \cos \theta = 2$.
$\cos \theta = \frac{2}{\sqrt{5}}$.

(B) For the first hyperbola,the distance between foci is $2ae_1$ and the distance between directrices is $\frac{2a}{e_1}$.
Given $2ae_1 = 2 \times \frac{2a}{e_1}$,which simplifies to $e_1^2 = 2$. Since $e_1 > 1$,we have $e_1 = \sqrt{2}$.
For the second hyperbola,the length of the transverse axis is $2a_2$ and the length of the conjugate axis is $2b_2$.
Given $2a_2 = 2(2b_2)$,so $a_2 = 2b_2$ or $b_2 = \frac{a_2}{2}$.
The eccentricity $e_2$ is given by $e_2 = \sqrt{1 + \frac{b_2^2}{a_2^2}} = \sqrt{1 + \frac{(a_2/2)^2}{a_2^2}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.
Therefore,$e_1 e_2 = \sqrt{2} \times \frac{\sqrt{5}}{2} = \frac{\sqrt{10}}{2}$.
Thus,option $B$ is correct.

(C) For a point $P$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focal distances are $SP = |ex - a|$ and $S^{\prime}P = |ex + a|$.
Given the condition $SP + S^{\prime}P - 2|SP - S^{\prime}P| = 0$,we know that $|SP - S^{\prime}P| = 2a$.
Substituting this into the equation: $SP + S^{\prime}P = 2(2a) = 4a$.
For a hyperbola,the sum of focal distances is $SP + S^{\prime}P = 2ex$ (assuming $x > 0$).
Thus,$2ex = 4a$,which implies $ex = 2a$,or $x = \frac{2a}{e}$.
The point $P$ is given as $P(\frac{\pi}{6})$,which in parametric form is $(a \sec \theta, b \tan \theta)$.
Here,$x = a \sec(\frac{\pi}{6}) = a \cdot \frac{2}{\sqrt{3}}$.
Equating the two expressions for $x$: $\frac{2a}{e} = \frac{2a}{\sqrt{3}}$.
Therefore,$e = \sqrt{3}$.

(C) The equation of the hyperbola is $x^2 - 2y^2 = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 1$ and $b^2 = \frac{1}{2}$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1/2}{1}} = \sqrt{\frac{3}{2}}$.
The focus $S$ is $(ae, 0) = (\sqrt{\frac{3}{2}}, 0)$.
The line passing through $P(-1, 1)$ and $S(\sqrt{\frac{3}{2}}, 0)$ has slope $m = \frac{0 - 1}{\sqrt{\frac{3}{2}} - (-1)} = \frac{-1}{\sqrt{\frac{3}{2}} + 1} = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}$.
The equation of line $PS$ is $y - 0 = m(x - \sqrt{\frac{3}{2}})$,which is $y = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}(x - \sqrt{\frac{3}{2}})$.
To find the $Y$-intercept $(B)$,set $x = 0$: $y = \frac{-\sqrt{2}}{\sqrt{3} + \sqrt{2}}(-\sqrt{\frac{3}{2}}) = \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}}$.
The area of the triangle formed by the line with coordinate axes is $\frac{1}{2} \times |OS| \times |OB| = \frac{1}{2} \times \sqrt{\frac{3}{2}} \times \frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}} = \frac{1}{2} \times \frac{3}{\sqrt{2}(\sqrt{3} + \sqrt{2})} = \frac{3}{2(\sqrt{6} + 2)}$.
Thus,the correct option is $C$.

(C) The given equation $9x^2 - 16y^2 = 9$ can be written as $\frac{x^2}{1^2} - \frac{y^2}{(3/4)^2} = 1$. Here $a=1, b=3/4$.
The parametric points on a hyperbola are $(a \sec \theta, b \tan \theta)$.
For $\theta = \frac{\pi}{4}$,$P_1 = (1 \cdot \sqrt{2}, \frac{3}{4} \cdot 1) = (\sqrt{2}, \frac{3}{4})$.
For $\theta = \frac{\pi}{3}$,$P_2 = (1 \cdot 2, \frac{3}{4} \cdot \sqrt{3}) = (2, \frac{3\sqrt{3}}{4})$.
The distance $D = \sqrt{(2 - \sqrt{2})^2 + (\frac{3\sqrt{3}}{4} - \frac{3}{4})^2} = \sqrt{(4 + 2 - 4\sqrt{2}) + \frac{9}{16}(3 + 1 - 2\sqrt{3})} = \sqrt{6 - 4\sqrt{2} + \frac{9}{4} - \frac{9\sqrt{3}}{8}} = \sqrt{\frac{48 - 32\sqrt{2} + 18 - 9\sqrt{3}}{8}} = \sqrt{\frac{66 - 32\sqrt{2} - 9\sqrt{3}}{8}} = \frac{1}{2\sqrt{2}} \sqrt{66 - 32\sqrt{2} - 9\sqrt{3}}$.
The Assertion is true.
The Reason states $x = a \cosh t, y = b \sinh t$ are parametric equations for $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is incorrect; the correct parametric equations are $x = a \sec \theta, y = b \tan \theta$.
Thus,$(A)$ is true but $(R)$ is false.

(D) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The centre is $(0, 0)$.
The endpoints of the latus rectum are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.
The angle subtended by the latus rectum at the centre is $120^{\circ}$.
Thus,the angle made by the line joining the centre to one endpoint $(ae, \frac{b^2}{a})$ with the $x$-axis is $60^{\circ}$.
Therefore,$\tan(60^{\circ}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.
Since $\sqrt{3} = \frac{b^2}{a^2e}$ and $b^2 = a^2(e^2 - 1)$,we have $\sqrt{3} = \frac{a^2(e^2 - 1)}{a^2e} = \frac{e^2 - 1}{e}$.
This gives $e^2 - \sqrt{3}e - 1 = 0$.
Using the quadratic formula,$e = \frac{\sqrt{3} \pm \sqrt{3 - 4(1)(-1)}}{2} = \frac{\sqrt{3} \pm \sqrt{7}}{2}$.
Since $e > 1$,we take $e = \frac{\sqrt{3} + \sqrt{7}}{2}$.

(B) The equation of the chord joining points $A(\theta_1)$ and $B(\theta_2)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by:
$\frac{x}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right)-\frac{y}{b} \sin \left(\frac{\theta_1+\theta_2}{2}\right)=\cos \left(\frac{\theta_1+\theta_2}{2}\right)$
Since the chord passes through the focus $S(ae, 0)$,we substitute $x=ae$ and $y=0$ into the equation:
$\frac{ae}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right) = \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
$e \cos \left(\frac{\theta_1-\theta_2}{2}\right) = \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
Using the relation $e = \frac{\sqrt{a^2+b^2}}{a}$,we get:
$\frac{\sqrt{a^2+b^2}}{a} \cos \left(\frac{\theta_1-\theta_2}{2}\right) = \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
$\sqrt{a^2+b^2} \cos \left(\frac{\theta_1-\theta_2}{2}\right) = a \cos \left(\frac{\theta_1+\theta_2}{2}\right)$
Comparing this with the given equation $a \cos \left(\frac{\theta_1+\theta_2}{2}\right) = k \cos \left(\frac{\theta_1-\theta_2}{2}\right)$,we find:
$k = \sqrt{a^2+b^2}$

(D) The equation of the hyperbola is $x^2 - 4y^2 = 4$, which can be written as $\frac{x^2}{4} - \frac{y^2}{1} = 1$.
Here, $a = 2$ and $b = 1$.
The parametric coordinates are given by $(a \sec \theta, b \tan \theta) = (2 \sec \theta, \tan \theta)$.
Calculating the points:
$P(\frac{\pi}{4}) = (2 \sec \frac{\pi}{4}, \tan \frac{\pi}{4}) = (2 \sqrt{2}, 1)$
$Q(\frac{5 \pi}{4}) = (2 \sec \frac{5 \pi}{4}, \tan \frac{5 \pi}{4}) = (-2 \sqrt{2}, 1)$
$R(\frac{3 \pi}{4}) = (2 \sec \frac{3 \pi}{4}, \tan \frac{3 \pi}{4}) = (-2 \sqrt{2}, -1)$
$T(\frac{7 \pi}{4}) = (2 \sec \frac{7 \pi}{4}, \tan \frac{7 \pi}{4}) = (2 \sqrt{2}, -1)$
These points form a rectangle with vertices $(2 \sqrt{2}, 1), (-2 \sqrt{2}, 1), (-2 \sqrt{2}, -1),$ and $(2 \sqrt{2}, -1)$.
The length of the sides are:
Width $= |2 \sqrt{2} - (-2 \sqrt{2})| = 4 \sqrt{2}$
Height $= |1 - (-1)| = 2$
Area $= \text{width} \times \text{height} = 4 \sqrt{2} \times 2 = 8 \sqrt{2}$ square units.
Thus, option $D$ is correct.

(D) Let $L = \lim _{x \rightarrow 0} \frac{\tan 2x - 2\tan x}{(1 - \cos x)(2^x - 1)}$.
Using the expansion $\tan x = x + \frac{x^3}{3} + O(x^5)$:
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.
$2\tan x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.
Numerator: $\tan 2x - 2\tan x = (2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3}) = \frac{6x^3}{3} = 2x^3$.
Denominator: $(1 - \cos x)(2^x - 1) \approx (\frac{x^2}{2})(x \ln 2) = \frac{x^3 \ln 2}{2}$.
$L = \lim _{x \rightarrow 0} \frac{2x^3}{\frac{x^3 \ln 2}{2}} = \frac{2 \times 2}{\ln 2} = \frac{4}{\ln 2}$.

(D) We evaluate the limit by splitting it into two parts:
$\lim _{x \rightarrow 2} \left[ (x-2)^2 \cos \left(\frac{2}{x-2}\right) \right] + \lim _{x \rightarrow 2} \left[ \frac{x^2-4}{x^3-2 x-4} \right]$
For the first part,$\lim _{x \rightarrow 2} (x-2)^2 \cos \left(\frac{2}{x-2}\right)$. Since $(x-2)^2 \rightarrow 0$ and $\cos \left(\frac{2}{x-2}\right)$ is bounded in $[-1, 1]$,by the Squeeze Theorem,the limit is $0 \times [-1, 1] = 0$.
For the second part,$\lim _{x \rightarrow 2} \frac{(x-2)(x+2)}{(x-2)(x^2+2x+2)} = \lim _{x \rightarrow 2} \frac{x+2}{x^2+2x+2}$.
Substituting $x=2$,we get $\frac{2+2}{2^2+2(2)+2} = \frac{4}{4+4+2} = \frac{4}{10} = \frac{2}{5}$.
Thus,the total limit is $0 + \frac{2}{5} = \frac{2}{5}$.

(A) Given $f(x) = 2x - [2x]$.
To find $l_1 = \lim_{x \rightarrow 2^{-}} f(x)$:
Let $x = 2 - h$,where $h \rightarrow 0$ and $h > 0$.
$l_1 = \lim_{h \rightarrow 0} (2(2 - h) - [2(2 - h)]) = \lim_{h \rightarrow 0} (4 - 2h - [4 - 2h])$.
Since $h$ is a very small positive number,$4 - 2h$ is slightly less than $4$,so $[4 - 2h] = 3$.
Thus,$l_1 = 4 - 3 = 1$.
To find $l_2 = \lim_{x \rightarrow 2^{+}} f(x)$:
Let $x = 2 + h$,where $h \rightarrow 0$ and $h > 0$.
$l_2 = \lim_{h \rightarrow 0} (2(2 + h) - [2(2 + h)]) = \lim_{h \rightarrow 0} (4 + 2h - [4 + 2h])$.
Since $h$ is a very small positive number,$4 + 2h$ is slightly greater than $4$,so $[4 + 2h] = 4$.
Thus,$l_2 = 4 - 4 = 0$.
Therefore,$l_1 + l_2 = 1 + 0 = 1$.

(B) Given,$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}$
$= \lim _{x \rightarrow 0} \frac{(2^x)^2 - 2 \cdot 2^x + 1 + 1 - \cos 2x}{x^2}$
$= \lim _{x \rightarrow 0} \frac{(2^x - 1)^2 + (1 - \cos 2x)}{x^2}$
$= \lim _{x \rightarrow 0} \left( \frac{2^x - 1}{x} \right)^2 + \lim _{x \rightarrow 0} \frac{1 - \cos 2x}{x^2}$
Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x - 1}{x} = \log _e a$ and $\lim _{x \rightarrow 0} \frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$:
$= (\log _e 2)^2 + \frac{2^2}{2}$
$= (\log _e 2)^2 + 2$

(C) Given,$L = \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3}$.
Checking the form at $x = 3$,we get $\frac{3^3 - 3(3^2) - 4(3) + 12}{2(3^3) - 7(3^2) + 2(3) + 3} = \frac{27 - 27 - 12 + 12}{54 - 63 + 6 + 3} = \frac{0}{0}$ form.
Applying $L'H\hat{o}pital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$L = \lim _{x \rightarrow 3^{-}} \frac{\frac{d}{dx}(x^3-3 x^2-4 x+12)}{\frac{d}{dx}(2 x^3-7 x^2+2 x+3)} = \lim _{x \rightarrow 3^{-}} \frac{3 x^2-6 x-4}{6 x^2-14 x+2}$.
Now,substitute $x = 3$:
$L = \frac{3(3^2) - 6(3) - 4}{6(3^2) - 14(3) + 2} = \frac{27 - 18 - 4}{54 - 42 + 2} = \frac{5}{14}$.

(C) Let $L = \lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right]$.
Rationalizing the expression:
$L = \lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right] \times \frac{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}$
$= \lim _{x \rightarrow \infty} x^3 \frac{x^2 + \sqrt{x^4 + 1} - 2x^2}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x} = \lim _{x \rightarrow \infty} x^3 \frac{\sqrt{x^4 + 1} - x^2}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}$.
Rationalizing again:
$= \lim _{x \rightarrow \infty} x^3 \frac{(\sqrt{x^4 + 1} - x^2)(\sqrt{x^4 + 1} + x^2)}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)} = \lim _{x \rightarrow \infty} x^3 \frac{x^4 + 1 - x^4}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)}$
$= \lim _{x \rightarrow \infty} \frac{x^3}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)}$.
Dividing numerator and denominator by $x^3$ (specifically $x$ in the first bracket and $x^2$ in the second):
$= \lim _{x \rightarrow \infty} \frac{1}{(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2})(\sqrt{1 + \frac{1}{x^4}} + 1)} = \frac{1}{(\sqrt{1 + 1} + \sqrt{2})(1 + 1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{\tan ^2(\pi \sec ^4 x)}{\pi^2 x^4}$.
Since $\sec(0) = 1$,the expression takes the form $\frac{\tan^2(\pi)}{0} = \frac{0}{0}$.
We use the property $\tan(\pi \sec^4 x) = \tan(\pi \sec^4 x - \pi) = \tan(\pi(\sec^4 x - 1))$.
Recall that $\sec^4 x - 1 = (\sec^2 x - 1)(\sec^2 x + 1) = \tan^2 x (\sec^2 x + 1)$.
As $x \rightarrow 0$,$\tan x \approx x$ and $\sec^2 x \approx 1$,so $\sec^4 x - 1 \approx x^2(1+1) = 2x^2$.
Thus,$\tan(\pi \sec^4 x) \approx \tan(2\pi x^2) \approx 2\pi x^2$.
Substituting this into the limit:
$L = \lim _{x \rightarrow 0} \frac{(2\pi x^2)^2}{\pi^2 x^4} = \lim _{x \rightarrow 0} \frac{4\pi^2 x^4}{\pi^2 x^4} = 4$.

(C) Given that $f(x)$ is a differentiable function with $f(0)=0$ and $f^{\prime}(0)=20$.
We need to find $\lim _{x \rightarrow 0} A(x) = \lim _{x \rightarrow 0} [2 f(x) \operatorname{cosec} 4 x + 4 f(x)(\cos ^2 x + 1) - 4 \cos ^2 x]$.
Rewrite the expression as:
$\lim _{x \rightarrow 0} A(x) = \lim _{x \rightarrow 0} \left[ \frac{2 f(x)}{\sin 4x} + 4 f(x)(\cos ^2 x + 1) - 4 \cos ^2 x \right]$.
Since $f(0)=0$,the first term is an indeterminate form of type $\frac{0}{0}$.
Using the limit $\lim _{x \rightarrow 0} \frac{f(x)}{\sin 4x} = \lim _{x \rightarrow 0} \frac{f(x)}{x} \cdot \frac{4x}{\sin 4x} \cdot \frac{1}{4} = f^{\prime}(0) \cdot 1 \cdot \frac{1}{4} = \frac{20}{4} = 5$.
Now,evaluate the limit:
$\lim _{x \rightarrow 0} A(x) = 2 \left( \lim _{x \rightarrow 0} \frac{f(x)}{\sin 4x} \right) + 4 \lim _{x \rightarrow 0} f(x)(\cos ^2 x + 1) - 4 \lim _{x \rightarrow 0} \cos ^2 x$.
$= 2(5) + 4(0)(1+1) - 4(1)^2$.
$= 10 + 0 - 4 = 6$.
Therefore,the correct option is $C$.

(D) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)}$.
First,consider the term $(1+\sin x)^{\frac{2}{\sin x}}$. As $x \rightarrow 0$,$\sin x \rightarrow 0$,so this is of the form $(1+u)^{2/u}$ where $u = \sin x$. We know that $\lim _{u \rightarrow 0} (1+u)^{1/u} = e$,so $\lim _{x \rightarrow 0} (1+\sin x)^{\frac{2}{\sin x}} = e^2$.
Now,rewrite the limit as:
$L = \lim _{x \rightarrow 0} \left( \frac{2^x-1}{\log(1+2x)} \right) \times \lim _{x \rightarrow 0} (1+\sin x)^{\frac{2}{\sin x}}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log a$ and $\lim _{x \rightarrow 0} \frac{\log(1+x)}{x} = 1$:
$\lim _{x \rightarrow 0} \frac{2^x-1}{\log(1+2x)} = \lim _{x \rightarrow 0} \left( \frac{2^x-1}{x} \cdot \frac{2x}{\log(1+2x)} \cdot \frac{1}{2} \right) = \log 2 \cdot 1 \cdot \frac{1}{2} = \frac{\log 2}{2} = \log \sqrt{2}$.
Thus,$L = \log \sqrt{2} \cdot e^2 = e^2 \log \sqrt{2}$.
Therefore,option $(D)$ is correct.

(D) Given the limit: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$
Substituting $x = 2$,we get $\frac{2^3 - 2^2 - 2 - 2}{2(2^3) - 3(2^2) - 3(2) + 2} = \frac{8-4-2-2}{16-12-6+2} = \frac{0}{0}$ form.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(x^3-x^2-x-2) = 3x^2-2x-1$
Denominator derivative: $\frac{d}{dx}(2x^3-3x^2-3x+2) = 6x^2-6x-3$
Now,evaluate the limit: $\lim _{x \rightarrow 2} \frac{3x^2-2x-1}{6x^2-6x-3}$
$= \frac{3(2)^2 - 2(2) - 1}{6(2)^2 - 6(2) - 3} = \frac{12-4-1}{24-12-3} = \frac{7}{9}$

(B) Using the trigonometric identities $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$ and $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:
Numerator: $4[\sin (2022 x) - \sin (2020 x)] = 4 \times 2 \cos(2021 x) \sin(x) = 8 \cos(2021 x) \sin(x)$.
Denominator: $x[\cos (2022 x) + \cos (2020 x) + 2 \cos (2021 x)] = x[2 \cos(2021 x) \cos(x) + 2 \cos(2021 x)] = 2x \cos(2021 x) [\cos(x) + 1]$.
Substituting these into the limit:
$\lim _{x \rightarrow 0} \frac{8 \cos(2021 x) \sin(x)}{2x \cos(2021 x) [\cos(x) + 1]} = \lim _{x \rightarrow 0} \frac{4 \sin(x)}{x [\cos(x) + 1]}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\cos(0) = 1$:
$= \frac{4(1)}{1 + 1} = \frac{4}{2} = 2$.

(C) Let the given expression be $L = \lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$.
We can factor the expression inside the limit as:
$L = \lim _{x \rightarrow 0} \frac{4!}{x^8} (1 - \cos \frac{x^2}{3})(1 - \cos \frac{x^2}{4})$.
Using the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$,we get:
$1 - \cos \frac{x^2}{3} = 2 \sin^2(\frac{x^2}{6})$ and $1 - \cos \frac{x^2}{4} = 2 \sin^2(\frac{x^2}{8})$.
Substituting these into the limit:
$L = 24 \times \lim _{x \rightarrow 0} \frac{1}{x^8} \times 2 \sin^2(\frac{x^2}{6}) \times 2 \sin^2(\frac{x^2}{8})$.
$L = 24 \times 4 \times \lim _{x}$ ${\rightarrow 0} \left(\frac{\sin(\frac{x^2}{6})}{\frac{x^2}{6}}\right)^2 \times (\frac{x^2}{6})^2 \times \left(\frac{\sin(\frac{x^2}{8})}{\frac{x^2}{8}}\right)^2 \times (\frac{x^2}{8})^2 \times \frac{1}{x^8}$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have:
$L = 96 \times 1^2 \times \frac{x^4}{36} \times 1^2 \times \frac{x^4}{64} \times \frac{1}{x^8}$.
$L = 96 \times \frac{1}{36 \times 64} = \frac{96}{2304} = \frac{1}{24}$.

(B) Given $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$.
We know that $(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x)$.
Since $e^x = \cot \left( \frac{\pi}{4} + \theta \right) = \frac{1 - \tan \theta}{1 + \tan \theta}$,we have $e^{2x} = \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right)^2$ and $e^{-2x} = \left( \frac{1 + \tan \theta}{1 - \tan \theta} \right)^2$.
Then $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2} = \frac{1}{2} \left[ \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right)^2 - \left( \frac{1 + \tan \theta}{1 - \tan \theta} \right)^2 \right]$.
Simplifying the expression inside the bracket: $\frac{(1 - \tan \theta)^4 - (1 + \tan \theta)^4}{(1 - \tan^2 \theta)^2} = \frac{-8 \tan \theta - 8 \tan^3 \theta}{(1 - \tan^2 \theta)^2} = \frac{-8 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2}$.
Thus,$(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x) = \frac{-4 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2}$.
Now,the limit is $\lim_{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)} = \lim_{\theta \rightarrow 0} \frac{\theta (1 - \tan^2 \theta)^2}{-4 \tan \theta (1 + \tan^2 \theta)}$.
Since $\lim_{\theta \rightarrow 0} \frac{\theta}{\tan \theta} = 1$,we have $\lim_{\theta \rightarrow 0} \frac{1 \cdot (1 - 0)^2}{-4(1 + 0)} = -\frac{1}{4}$.
Wait,re-checking the identity: $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}$. With $e^x = \frac{1-\tan \theta}{1+\tan \theta}$,$e^{2x} = \frac{1-\sin 2\theta}{1+\sin 2\theta}$.
Actually,$\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\cot(\pi/4+\theta) - \tan(\pi/4+\theta)}{2} = -\tan(2\theta)/2$ and $\cosh x = \frac{\cot(\pi/4+\theta) + \tan(\pi/4+\theta)}{2} = \frac{1}{\sin(2\theta)}$.
So $(\sinh x)(\cosh x) = -\frac{\tan(2\theta)}{2 \sin(2\theta)} = -\frac{1}{2 \cos(2\theta)}$.
Then $\lim_{\theta \rightarrow 0} \frac{\theta}{-1/(2 \cos(2\theta))} = \lim_{\theta \rightarrow 0} -2\theta \cos(2\theta) = 0$.
Re-evaluating: $x = \log_e(\cot(\pi/4+\theta))$. $\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\cot(\pi/4+\theta) - \tan(\pi/4+\theta)}{2} = \frac{\cos^2(\pi/4+\theta) - \sin^2(\pi/4+\theta)}{2 \sin(\pi/4+\theta)\cos(\pi/4+\theta)} = \frac{\cos(\pi/2+2\theta)}{\sin(\pi/2+2\theta)} = -\tan(2\theta)$.
$\cosh x = \frac{\cot(\pi/4+\theta) + \tan(\pi/4+\theta)}{2} = \frac{1}{\sin(\pi/2+2\theta)} = \sec(2\theta)$.
$(\sinh x)(\cosh x) = -\tan(2\theta) \sec(2\theta) = -\frac{\sin(2\theta)}{\cos^2(2\theta)}$.
Limit $\lim_{\theta \rightarrow 0} \frac{\theta}{-\sin(2\theta)/\cos^2(2\theta)} = \lim_{\theta \rightarrow 0} -\frac{\theta}{\sin(2\theta)} \cdot \cos^2(2\theta) = -\frac{1}{2} \cdot 1 = -\frac{1}{2}$.

(C) The mean of the observations is given by the ratio of the sum of observations to the number of observations.
Sum $= 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 78 = 198$.
Mean $(\bar{x}) = \frac{198}{9} = 22$.
Now,the mean deviation from the mean is calculated as $M.D. = \frac{\sum |x_i - \bar{x}|}{n}$.
$M.D. = \frac{|1-22| + |3-22| + |4-22| + |7-22| + |11-22| + |18-22| + |29-22| + |47-22| + |78-22|}{9}$.
$M.D. = \frac{21 + 19 + 18 + 15 + 11 + 4 + 7 + 25 + 56}{9}$.
$M.D. = \frac{176}{9}$.

(C) Given that $x_1, x_2, \ldots, x_n$ are negative numbers in ascending order. Since they are negative,we can write them as $-|x_1|, -|x_2|, \ldots, -|x_n|$.
Because the order is ascending,we have $-|x_1| < -|x_2| < \ldots < -|x_n|$,which implies $|x_1| > |x_2| > \ldots > |x_n|$.
After changing the signs of the first term $(x_1)$ and the last term $(x_n)$,the new set of observations becomes $|x_1|, -|x_2|, -|x_3|, \ldots, -|x_{n-1}|, |x_n|$.
In this new set,the largest value is $|x_1|$ and the smallest value is $-|x_2|$ (since $|x_1| > |x_2| > \ldots > |x_n|$).
The range is defined as $\text{Maximum} - \text{Minimum}$.
Range $= |x_1| - (-|x_2|) = |x_1| + |x_2|$.
Wait,re-evaluating the sequence: The original sequence is $x_1 < x_2 < \ldots < x_n < 0$. After changing signs,the new sequence is $-x_1, x_2, x_3, \ldots, x_{n-1}, -x_n$. Since $x_1$ is the most negative,$-x_1$ is the largest positive. The smallest value is $x_2$. Thus,the range is $-x_1 - x_2 = |x_1| - x_2$.

(C) Let the mean of the observations $x_1, x_2, \ldots, x_n$ be $\bar{x}$.
Given that the mean deviation ($M$.$D$.) of $x_i$ is $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = 10$.
Let the new observations be $y_i = \frac{2x_i + 5}{3}$.
The mean of the new observations is $\bar{y} = \frac{2\bar{x} + 5}{3}$.
The mean deviation of the new observations is given by:
$\text{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{y}|$
$= \frac{1}{n} \sum_{i=1}^{n} |\frac{2x_i + 5}{3} - \frac{2\bar{x} + 5}{3}|$
$= \frac{1}{n} \sum_{i=1}^{n} |\frac{2(x_i - \bar{x})}{3}|$
$= \frac{2}{3} \times (\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|)$
$= \frac{2}{3} \times 10 = \frac{20}{3}$.
Thus,the correct option is $C$.

(A) Statement $(I)$: The range is defined as the difference between the maximum and minimum observations. Since intermediate observations do not affect the maximum or minimum values,the range remains unchanged. Thus,Statement $(I)$ is true.
Statement $(II)$: $A$ well-known property of mean deviation is that it is minimized when calculated about the median. Thus,Statement $(II)$ is true.
Statement $(III)$: For grouped data,the range is defined as the difference between the upper limit of the largest class and the lower limit of the smallest class. The statement provided swaps these,making it false.
Therefore,Statements $I$ and $II$ are true but statement $III$ is false.

(D) The mean of the absolute deviations of observations $x_1, x_2, \ldots, x_n$ from their mean $\bar{x}$ is defined as the mean deviation about the mean.
Mathematically,it is given by:
$\text{Mean Deviation} = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$
where $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$.

(D) The given observations are $1, 3, 5, 7, 11, 13, 17, 19, 23$.
First,we calculate the mean $(\bar{x})$:
$\bar{x} = \frac{1+3+5+7+11+13+17+19+23}{9} = \frac{99}{9} = 11$.
The mean deviation about the mean is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$.
Mean deviation $= \frac{1}{9} [|1-11| + |3-11| + |5-11| + |7-11| + |11-11| + |13-11| + |17-11| + |19-11| + |23-11|]$.
Mean deviation $= \frac{1}{9} [10 + 8 + 6 + 4 + 0 + 2 + 6 + 8 + 12]$.
Mean deviation $= \frac{56}{9} = 6 \frac{2}{9}$.

(D) Given: $A=\frac{\pi}{6}, b=7, c=4\sqrt{3}$.
By the cosine rule,$a^2 = b^2 + c^2 - 2bc \cos A$.
$a^2 = 7^2 + (4\sqrt{3})^2 - 2(7)(4\sqrt{3}) \cos(\frac{\pi}{6})$.
$a^2 = 49 + 48 - 56\sqrt{3} \times \frac{\sqrt{3}}{2} = 97 - 84 = 13$.
So,$a = \sqrt{13}$.
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,$\sin B = \frac{b \sin A}{a} = \frac{7 \sin(\pi/6)}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.
And $\sin C = \frac{c \sin A}{a} = \frac{4\sqrt{3} \sin(\pi/6)}{\sqrt{13}} = \frac{2\sqrt{3}}{\sqrt{13}}$.
Therefore,$a \sin B \sin C = \sqrt{13} \times \frac{7}{2\sqrt{13}} \times \frac{2\sqrt{3}}{\sqrt{13}} = \frac{7\sqrt{3}}{\sqrt{13}}$.

(B) $L.H.S. = (b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right)$
$= (b^2 + c^2 + 2bc) \sin^2\left(\frac{A}{2}\right) + (b^2 + c^2 - 2bc) \cos^2\left(\frac{A}{2}\right)$
$= (b^2 + c^2) \left[\sin^2\left(\frac{A}{2}\right) + \cos^2\left(\frac{A}{2}\right)\right] - 2bc \left[\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)\right]$
$= b^2 + c^2 - 2bc \cos A$
$= a^2$ (Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$)
Since $a = 2R \sin A$,we have $a^2 = 4R^2 \sin^2 A$
$= 4R^2 \left(\frac{1 - \cos 2A}{2}\right)$
$= 2R^2 (1 - \cos 2A)$
Comparing with $K(1 - \cos 2A)$,we get $K = 2R^2$.

(C) Let the sides of the triangle be $a-d, a, a+d$. The perimeter is $(a-d) + a + (a+d) = 42$,which gives $3a = 42$,so $a = 14$.
Since $B < A < C$,the sides follow $b < a < c$. Thus,the sides are $14-d, 14, 14+d$.
The circum-radius $R$ is given by $R = \frac{abc}{4\Delta}$.
Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{42}{2} = 21$.
$\Delta = \sqrt{21(21-14)(21-(14-d))(21-14)(21-(14+d))} = \sqrt{21 \times 7 \times (7+d) \times 7 \times (7-d)} = \sqrt{21 \times 49 \times (49-d^2)} = 7 \sqrt{21(49-d^2)}$.
Given $R = \frac{65}{8}$,we have $\frac{(14-d)(14)(14+d)}{4 \times 7 \sqrt{21(49-d^2)}} = \frac{65}{8}$.
$\frac{14(196-d^2)}{28 \sqrt{21(49-d^2)}} = \frac{65}{8} \implies \frac{196-d^2}{2 \sqrt{21(49-d^2)}} = \frac{65}{8}$.
Let $x = 49-d^2$. Then $196-d^2 = 147+x$. The equation becomes $\frac{147+x}{2 \sqrt{21x}} = \frac{65}{8} \implies \frac{147+x}{\sqrt{x}} = \frac{65\sqrt{21}}{4}$.
Squaring both sides leads to $x = 12$ or $x = 1764/21 = 84$. Testing $x=12$,$49-d^2=12 \implies d^2=37$. Testing $x=84$,$49-d^2=84$ (impossible).
Using the sine rule,$\sin A = \frac{a}{2R} = \frac{14}{2 \times (65/8)} = \frac{14 \times 8}{130} = \frac{112}{130} = \frac{56}{65}$.

(A) Given $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$. In $\triangle ABC$,$\angle C = \pi - (A+B) = \pi - (\frac{\pi}{3} + \frac{\pi}{4}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin(\frac{\pi}{3}) = k \frac{\sqrt{3}}{2}$,$b = k \sin(\frac{\pi}{4}) = k \frac{1}{\sqrt{2}}$,and $c = k \sin(\frac{5\pi}{12}) = k \sin(\frac{\pi}{4} + \frac{\pi}{6}) = k (\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}) = k \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Now,$\frac{a^2-b^2}{c^2} = \frac{k^2(\frac{3}{4} - \frac{1}{2})}{k^2(\frac{(\sqrt{3}+1)^2}{8})} = \frac{1/4}{(\frac{3+1+2\sqrt{3}}{8})} = \frac{1/4}{(\frac{4+2\sqrt{3}}{8})} = \frac{1/4}{(\frac{2+\sqrt{3}}{4})} = \frac{1}{2+\sqrt{3}}$.
Rationalizing the denominator: $\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = 2-\sqrt{3}$.
Thus,the correct option is $A$.

(B) Given $\frac{a}{\tan A} = \frac{b}{\tan B} = \frac{c}{\tan C}$.
Since $\tan A = \frac{\sin A}{\cos A}$,we have $\frac{a \cos A}{\sin A} = \frac{b \cos B}{\sin B} = \frac{c \cos C}{\sin C}$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Substituting this into the given equation,we get $2R \cos A = 2R \cos B = 2R \cos C$.
This implies $\cos A = \cos B = \cos C$.
Since $A, B, C$ are angles of a triangle,$A = B = C = 60^{\circ}$.
Therefore,$\cos^2 60^{\circ} + \cos^2 60^{\circ} + \cos^2 60^{\circ} = (\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.

(B) Given,in $\triangle ABC$,$A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$.
First,calculate $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3\sqrt{3}}{14}\right)^2} = \sqrt{1 - \frac{27}{196}} = \sqrt{\frac{169}{196}} = \frac{13}{14}$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\frac{13}{14} = \frac{5^2 + c^2 - 3^2}{2 \times 5 \times c} = \frac{25 + c^2 - 9}{10c} = \frac{16 + c^2}{10c}$.
Cross-multiplying: $130c = 14(16 + c^2) \Rightarrow 130c = 224 + 14c^2$.
Rearranging: $14c^2 - 130c + 224 = 0 \Rightarrow 7c^2 - 65c + 112 = 0$.
Factoring: $7c^2 - 49c - 16c + 112 = 0$ $\Rightarrow 7c(c - 7) - 16(c - 7) = 0$ $\Rightarrow (7c - 16)(c - 7) = 0$.
So,$c = \frac{16}{7}$ or $c = 7$.
Since $C$ is obtuse,the side $c$ must be the longest side.
Checking the conditions: If $c = \frac{16}{7} \approx 2.28$,then $a=3, b=5, c=2.28$. Here $b$ is the longest side,so $B$ would be obtuse.
If $c = 7$,then $a=3, b=5, c=7$. Here $c$ is the longest side,so $C$ is obtuse.
Therefore,$c = 7$.

(B) We know that $r = (s-a) \tan \frac{A}{2} = (s-b) \tan \frac{B}{2} = (s-c) \tan \frac{C}{2}$ and $r_1 = s \tan \frac{A}{2}$.
Given $\frac{r}{r_1} = \frac{1}{2}$,we have $\frac{(s-a) \tan \frac{A}{2}}{s \tan \frac{A}{2}} = \frac{1}{2}$,which implies $\frac{s-a}{s} = \frac{1}{2}$,so $2s - 2a = s$,or $s = 2a$.
Since $2s = a+b+c$,we have $a+b+c = 4a$,which means $b+c = 3a$.
We use the identity $\tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} + \tan \frac{A}{2} \tan \frac{B}{2} = 1$.
This can be written as $\tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 1 - \tan \frac{B}{2} \tan \frac{C}{2}$.
Using the formula $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$,we get $\tan \frac{B}{2} \tan \frac{C}{2} = \frac{s-a}{s}$.
Since $s = 2a$,$\frac{s-a}{s} = \frac{2a-a}{2a} = \frac{1}{2}$.
Thus,$\tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$4 \tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 4 \times \frac{1}{2} = 2$.

(D) Given that $a=7, b=8$ and $\cos C=\frac{2}{7}$.
Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting the values: $\frac{2}{7} = \frac{7^2+8^2-c^2}{2 \times 7 \times 8}$.
$\frac{2}{7} = \frac{49+64-c^2}{112}$.
Multiply both sides by $112$: $\frac{2 \times 112}{7} = 113 - c^2$.
$2 \times 16 = 113 - c^2$.
$32 = 113 - c^2$.
$c^2 = 113 - 32 = 81$.
Therefore,$c = \sqrt{81} = 9$.

(B) Since $BC$ is the hypotenuse,the triangle is right-angled at $A$,therefore $\angle A = \frac{\pi}{2}$.
We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.
Thus,$r_2 + r_3 = \Delta \left( \frac{1}{s-b} + \frac{1}{s-c} \right) = \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} \right) = \Delta \left( \frac{2s-b-c}{(s-b)(s-c)} \right)$.
Since $2s = a+b+c$,we have $2s-b-c = a$.
Also,$\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-b)(s-c) = \frac{\Delta^2}{s(s-a)}$.
Substituting these,$r_2 + r_3 = \Delta \left( \frac{a}{\Delta^2 / s(s-a)} \right) = \frac{as(s-a)}{\Delta}$.
Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we have $\frac{s(s-a)}{\Delta} = \cot \frac{A}{2}$.
Therefore,$r_2 + r_3 = a \cot \frac{A}{2} = a \cot \frac{\pi}{4} = a(1) = a$.

(D) Given that $\Delta$ is the area of triangle $ABC$.
Using the projection rule,we have $b \cos C + c \cos B = a$.
Using the sine rule,we have $\sin C = \frac{c}{2R}$ and $\sin B = \frac{b}{2R}$,where $R$ is the circumradius.
Thus,$b \sin C + c \sin B = b(\frac{c}{2R}) + c(\frac{b}{2R}) = \frac{bc}{2R} + \frac{cb}{2R} = \frac{2bc}{2R} = \frac{bc}{R}$.
Now,the expression becomes:
$(b \sin C + c \sin B)(b \cos C + c \cos B) = (\frac{bc}{R})(a) = \frac{abc}{R}$.
Since the area of a triangle is given by $\Delta = \frac{abc}{4R}$,we have $\frac{abc}{R} = 4 \Delta$.
Therefore,the correct option is $4 \Delta$.

(C) Given $\sinh x = -\frac{1}{2}$.
We know the identity $\cosh^2 x - \sinh^2 x = 1$.
Substituting the value: $\cosh^2 x = 1 + (-\frac{1}{2})^2 = 1 + \frac{1}{4} = \frac{5}{4}$.
Since $\cosh x > 0$,we have $\cosh x = \frac{\sqrt{5}}{2}$.
Now,$\tanh x = \frac{\sinh x}{\cosh x} = \frac{-1/2}{\sqrt{5}/2} = -\frac{1}{\sqrt{5}}$.
Using the double angle formula $\tanh 2x = \frac{2 \tanh x}{1 + \tanh^2 x}$:
$\tanh 2x = \frac{2(-1/\sqrt{5})}{1 + (-1/\sqrt{5})^2} = \frac{-2/\sqrt{5}}{1 + 1/5} = \frac{-2/\sqrt{5}}{6/5} = -\frac{2}{\sqrt{5}} \times \frac{5}{6} = -\frac{\sqrt{5}}{3}$.

(D) Let $G$ be the centroid of $\triangle ABC$. Since $AD$ and $BE$ are medians,$G$ divides $AD$ and $BE$ in the ratio $2:1$.
Thus,$AG = \frac{2}{3} AD = \frac{2}{3} \times 4 = \frac{8}{3}$ and $BG = \frac{2}{3} BE$.
In $\triangle ABG$,by the Sine Rule: $\frac{AG}{\sin(\angle ABG)} = \frac{BG}{\sin(\angle BAG)}$.
Given $\angle BAG = \frac{\pi}{6}$ and $\angle ABG = \frac{\pi}{3}$,we have $\frac{8/3}{\sin(\pi/3)} = \frac{BG}{\sin(\pi/6)}$.
$BG = \frac{8}{3} \times \frac{\sin(\pi/6)}{\sin(\pi/3)} = \frac{8}{3} \times \frac{1/2}{\sqrt{3}/2} = \frac{8}{3 \sqrt{3}}$.
Area of $\triangle ABG = \frac{1}{2} \times AG \times BG \times \sin(\angle AGB)$.
Since $\angle AGB = \pi - (\pi/6 + \pi/3) = \pi/2$,$\sin(\angle AGB) = 1$.
Area of $\triangle ABG = \frac{1}{2} \times \frac{8}{3} \times \frac{8}{3 \sqrt{3}} = \frac{32}{9 \sqrt{3}}$.
Since the centroid divides the triangle into three triangles of equal area,Area of $\triangle ABC = 3 \times \text{Area of } \triangle ABG = 3 \times \frac{32}{9 \sqrt{3}} = \frac{32}{3 \sqrt{3}}$.

(D) Given $a=3, b=7, c=8$.
Using the Law of Tangents,$\tan \frac{C-A}{2} = \frac{c-a}{c+a} \cot \frac{B}{2}$.
Substituting this into the expression:
$\sin \frac{B}{2} \tan \frac{C-A}{2} = \sin \frac{B}{2} \left( \frac{c-a}{c+a} \right) \cot \frac{B}{2} = \sin \frac{B}{2} \left( \frac{c-a}{c+a} \right) \frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} = \left( \frac{c-a}{c+a} \right) \cos \frac{B}{2}$.
Now,calculate $\cos B$ using the Law of Cosines:
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{3^2+8^2-7^2}{2 \times 3 \times 8} = \frac{9+64-49}{48} = \frac{24}{48} = \frac{1}{2}$.
Using the half-angle formula $\cos \frac{B}{2} = \sqrt{\frac{1+\cos B}{2}}$:
$\cos \frac{B}{2} = \sqrt{\frac{1+1/2}{2}} = \sqrt{\frac{3/2}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Finally,substitute the values:
$\left( \frac{8-3}{8+3} \right) \cos \frac{B}{2} = \frac{5}{11} \times \frac{\sqrt{3}}{2} = \frac{5 \sqrt{3}}{22}$.
Thus,option $(d)$ is correct.

(D) The area of a circle is given by $\pi r^2$. Given the areas of the ex-circles $A_1, A_2, A_3$ as $4, 9, 16$ respectively,we have:
$\pi r_1^2 = 4 \Rightarrow r_1 = \frac{2}{\sqrt{\pi}}$
$\pi r_2^2 = 9 \Rightarrow r_2 = \frac{3}{\sqrt{\pi}}$
$\pi r_3^2 = 16 \Rightarrow r_3 = \frac{4}{\sqrt{\pi}}$
The relationship between the inradius $r$ and the exradii $r_1, r_2, r_3$ is given by $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.
Substituting the values:
$\frac{1}{r} = \frac{\sqrt{\pi}}{2} + \frac{\sqrt{\pi}}{3} + \frac{\sqrt{\pi}}{4} = \sqrt{\pi} \left( \frac{6+4+3}{12} \right) = \frac{13\sqrt{\pi}}{12}$.
Thus,$r = \frac{12}{13\sqrt{\pi}}$.
The area of the in-circle $A$ is $\pi r^2 = \pi \left( \frac{12}{13\sqrt{\pi}} \right)^2 = \pi \left( \frac{144}{169\pi} \right) = \frac{144}{169}$.

(A) We know that in a triangle $ABC$,the cotangent half-angle formulas are $\cot \frac{A}{2} = \frac{s-a}{r}$,$\cot \frac{B}{2} = \frac{s-b}{r}$,and $\cot \frac{C}{2} = \frac{s-c}{r}$.
Substituting these into the expression $r^2 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$:
$= r^2 \left( \frac{s-a}{r} \right) \left( \frac{s-b}{r} \right) \left( \frac{s-c}{r} \right)$
$= r^2 \cdot \frac{(s-a)(s-b)(s-c)}{r^3}$
$= \frac{(s-a)(s-b)(s-c)}{r}$
Since the area of the triangle $\Delta = rs$,we have $r = \frac{\Delta}{s}$.
Also,by Heron's formula,$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,so $\Delta^2 = s(s-a)(s-b)(s-c)$,which implies $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.
Substituting these values:
$= \frac{\Delta^2 / s}{\Delta / s} = \frac{\Delta^2}{s} \cdot \frac{s}{\Delta} = \Delta$.
Thus,the correct option is $A$.

(C) Given $a=7, c=11, \cos A=\frac{17}{22}, \cos C=\frac{1}{14}$.
Using the law of tangents:
$\tan \left(\frac{C-A}{2}\right) = \left(\frac{c-a}{c+a}\right) \cot \left(\frac{B}{2}\right)$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$:
$\frac{17}{22} = \frac{b^2+121-49}{2b(11)} \implies \frac{17}{22} = \frac{b^2+72}{22b} \implies 17b = b^2+72 \implies b^2-17b+72=0$.
Solving the quadratic equation: $(b-8)(b-9)=0$,so $b=8$ or $b=9$.
For $b=9$,the expression becomes:
$b \tan \frac{B}{2} \tan \frac{C-A}{2} = b \tan \frac{B}{2} \left(\frac{c-a}{c+a}\right) \cot \frac{B}{2} = b \left(\frac{c-a}{c+a}\right) = 9 \left(\frac{11-7}{11+7}\right) = 9 \left(\frac{4}{18}\right) = 2$.
Thus,the correct option is $C$.