AP EAMCET 2019 Mathematics Question Paper with Answer and Solution

(C) Let the mid-point of the chord be $(h, k)$. The equation of the chord of the circle $x^2+y^2=16$ with mid-point $(h, k)$ is given by $T=S_1$,which is $hx+ky=h^2+k^2$.
Rearranging this,we get $y = -\frac{h}{k}x + \frac{h^2+k^2}{k}$.
For the hyperbola $9x^2-16y^2=144$,we can write it as $\frac{x^2}{16} - \frac{y^2}{9} = 1$.
Here $a^2=16$ and $b^2=9$.
The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.
Substituting $m = -\frac{h}{k}$ and $c = \frac{h^2+k^2}{k}$,we get:
$(\frac{h^2+k^2}{k})^2 = 16(-\frac{h}{k})^2 - 9$.
Multiplying by $k^2$,we get $(h^2+k^2)^2 = 16h^2 - 9k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $16x^2-9y^2=(x^2+y^2)^2$.

(D) Given that $\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}=2$.
Simplifying the expression:
$\lim _{x \rightarrow \infty} \frac{x^3+1-(\alpha x+\beta)(x^2+1)}{x^2+1} = 2$
$\lim _{x \rightarrow \infty} \frac{x^3+1-(\alpha x^3+\alpha x+\beta x^2+\beta)}{x^2+1} = 2$
$\lim _{x \rightarrow \infty} \frac{(1-\alpha)x^3-\beta x^2-\alpha x+(1-\beta)}{x^2+1} = 2$.
For the limit to exist and be finite,the coefficient of $x^3$ must be $0$:
$1-\alpha = 0 \Rightarrow \alpha = 1$.
Now the limit becomes:
$\lim _{x \rightarrow \infty} \frac{-\beta x^2-x+(1-\beta)}{x^2+1} = 2$.
Dividing numerator and denominator by $x^2$:
$\lim _{x \rightarrow \infty} \frac{-\beta - \frac{1}{x} + \frac{1-\beta}{x^2}}{1 + \frac{1}{x^2}} = 2$.
As $x \rightarrow \infty$,the terms with $x$ in the denominator approach $0$:
$-\beta = 2 \Rightarrow \beta = -2$.
Thus,the ordered pair is $(\alpha, \beta) = (1, -2)$.

(D) For $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$: As $x \rightarrow 2^{+}$,$[x] = 2$. Therefore,$l_1 = 2^2 + 2 = 4 + 2 = 6$.
For $l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$: As $x \rightarrow 3^{-}$,$[x] = 2$. Therefore,$l_2 = 2(3) - 2 = 6 - 2 = 4$.
For $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$: Let $y = x - \frac{\pi}{2}$. As $x \rightarrow \frac{\pi}{2}$,$y \rightarrow 0$. Then $x = y + \frac{\pi}{2}$.
$l_3 = \lim_{y \rightarrow 0} \frac{\cos(y + \frac{\pi}{2})}{y} = \lim_{y \rightarrow 0} \frac{-\sin y}{y} = -1$.
Comparing the values,we have $l_3 = -1$,$l_2 = 4$,and $l_1 = 6$.
Thus,$l_3 < l_2 < l_1$.

(C) To find $\alpha$:
$\alpha = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{1-\cos x} = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{2 \sin^2(x/2)}$
$= \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot \frac{x^2}{\sin^2(x/2)} \cdot \frac{1}{x} = \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot \frac{1}{(\frac{\sin(x/2)}{x/2})^2 \cdot \frac{1}{4}} = \frac{1}{2} \cdot \ln 2 \cdot 4 = 2 \ln 2$
So,$\alpha = 2 \ln 2$ ... $(i)$
To find $\beta$:
$\beta = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{\sqrt{1+x^2}-\sqrt{1-x^2}}$
Rationalizing the denominator:
$\beta = \lim _{x \rightarrow 0} \frac{x(2^x-1)(\sqrt{1+x^2}+\sqrt{1-x^2})}{(1+x^2)-(1-x^2)} = \lim _{x \rightarrow 0} \frac{x(2^x-1)(\sqrt{1+x^2}+\sqrt{1-x^2})}{2x^2}$
$= \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot (\sqrt{1+x^2}+\sqrt{1-x^2}) = \frac{1}{2} \cdot \ln 2 \cdot (1+1) = \ln 2$
So,$\beta = \ln 2$ ... $(ii)$
From $(i)$ and $(ii)$,we get $\alpha = 2\beta$.

(D) We use the Taylor series expansion for $\tan x$ and $\cos 2x$ near $x=0$:
$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$
$\tan 2x = 2x + \frac{(2x)^3}{3} + \frac{2(2x)^5}{15} + \dots = 2x + \frac{8x^3}{3} + \frac{64x^5}{15} + \dots$
$1 - \cos 2x = 1 - (1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \dots) = 2x^2 - \frac{2x^4}{3} + \dots$
Now,consider the numerator term $(\tan 2x - 2 \tan x)$:
$\tan 2x - 2 \tan x = (2x + \frac{8x^3}{3} + \frac{64x^5}{15} + \dots) - 2(x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots)$
$= (2x - 2x) + (\frac{8}{3} - \frac{2}{3})x^3 + (\frac{64}{15} - \frac{4}{15})x^5 + \dots = 2x^3 + 4x^5 + \dots$
So,$(\tan 2x - 2 \tan x)^2 = (2x^3 + 4x^5 + \dots)^2 = 4x^6 + \dots$
The numerator is $x^2(\tan 2x - 2 \tan x)^2 = x^2(4x^6 + \dots) = 4x^8 + \dots$
The denominator is $(1 - \cos 2x)^4 = (2x^2 - \frac{2x^4}{3} + \dots)^4 = (2x^2)^4 + \dots = 16x^8 + \dots$
Taking the limit:
$\lim _{x \rightarrow 0} \frac{4x^8}{16x^8} = \frac{4}{16} = \frac{1}{4}$

(D) We evaluate the limit as $x \rightarrow \infty$:
$\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right)$
Divide the numerator and denominator of the first term by $x^2$:
$\lim _{x \rightarrow \infty} \frac{6 - \frac{\cos 3x}{x^2}}{1 + \frac{5}{x^2}} = \frac{6 - 0}{1 + 0} = 6$
For the second term,since $x \rightarrow \infty$,$x > 0$,so $\sqrt{x^6} = x^3$:
$\lim _{x \rightarrow \infty} \frac{5x^3 + 3}{\sqrt{x^6+2}} = \lim _{x \rightarrow \infty} \frac{5 + \frac{3}{x^3}}{\sqrt{1 + \frac{2}{x^6}}} = \frac{5 + 0}{\sqrt{1 + 0}} = 5$
Thus,the limit is $6 - 5 = 1$.

(B) Given $\alpha = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$.
Applying $L'H\hat{o}pital's$ rule:
$\alpha = \lim_{x \rightarrow 0} \frac{2^x - 1 + x \cdot 2^x \ln 2}{\sin x}$.
Applying $L'H\hat{o}pital's$ rule again:
$\alpha = \lim_{x \rightarrow 0} \frac{2^x \ln 2 + 2^x \ln 2 + x \cdot 2^x (\ln 2)^2}{\cos x} = \frac{\ln 2 + \ln 2 + 0}{1} = 2 \ln 2$.
Now,$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}$.
Rationalizing the denominator:
$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1 + x^2} + \sqrt{1 - x^2})}{(1 + x^2) - (1 - x^2)} = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1 + x^2} + \sqrt{1 - x^2})}{2x^2}$.
$\beta = \frac{1}{2} \lim_{x \rightarrow 0} \frac{2^x - 1}{x} \cdot (\sqrt{1 + x^2} + \sqrt{1 - x^2}) = \frac{1}{2} \cdot \ln 2 \cdot (1 + 1) = \ln 2$.
Thus,$\alpha = 2 \beta$.

(C) To evaluate the limit $\lim _{x \rightarrow \infty}\left(\sqrt{x^2+a x+b}-x\right)$,we rationalize the expression:
$= \lim _{x}$ ${\rightarrow \infty}\left(\left(\sqrt{x^2+a x+b}-x\right) \times \frac{\sqrt{x^2+a x+b}+x}{\sqrt{x^2+a x+b}+x}\right)$
$= \lim _{x \rightarrow \infty}\left(\frac{x^2+a x+b-x^2}{\sqrt{x^2+a x+b}+x}\right)$
$= \lim _{x \rightarrow \infty}\left(\frac{a x+b}{\sqrt{x^2+a x+b}+x}\right)$
Divide the numerator and denominator by $x$:
$= \lim _{x \rightarrow \infty} \frac{a+\frac{b}{x}}{\sqrt{1+\frac{a}{x}+\frac{b}{x^2}}+1}$
As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$,so the limit becomes:
$= \frac{a+0}{\sqrt{1+0+0}+1} = \frac{a}{2}$
Since the result is $\frac{a}{2}$,the limit depends only on $a$.

(B) For the limit to exist at $x=0$,the Left Hand Limit $(LHL)$ must equal the Right Hand Limit $(RHL)$.
$RHL = \lim_{x \rightarrow 0^+} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = -\frac{1}{2}$.
$LHL = \lim_{x \rightarrow 0^-} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}$.
Using rationalization:
$LHL = \lim_{x \rightarrow 0^-} \frac{(\sqrt{1+px} - \sqrt{1-px})(\sqrt{1+px} + \sqrt{1-px})}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{(1+px) - (1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \rightarrow 0^-} \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})} = \frac{2p}{1+1} = p$.
Equating $LHL = RHL$,we get $p = -\frac{1}{2}$.

(C) Given $\lim _{x \rightarrow 0} \frac{((a-n) n x-\tan x) \sin n x}{x^2}=0, n \neq 0$.
Dividing the numerator and denominator by $x^2$,we get:
$\lim _{x \rightarrow 0} \left( \frac{(a-n) n x - \tan x}{x} \right) \left( \frac{\sin n x}{x} \right) = 0$.
$\lim _{x \rightarrow 0} \left( (a-n)n - \frac{\tan x}{x} \right) \left( n \cdot \frac{\sin n x}{n x} \right) = 0$.
Since $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin n x}{n x} = 1$,we have:
$((a-n)n - 1) \cdot n = 0$.
Since $n \neq 0$,we have $(a-n)n - 1 = 0$,which implies $an - n^2 = 1$,or $a = n + \frac{1}{n}$.
By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$,for $n > 0$:
$\frac{n + \frac{1}{n}}{2} \geq \sqrt{n \cdot \frac{1}{n}} = 1$.
Therefore,$a \geq 2$.
The minimum possible positive value of $a$ is $2$.

(B) Given the limit: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.
Since $x < 0$,as $n \rightarrow \infty$,$e^{n x} \rightarrow 0$.
Dividing the numerator and denominator by $e^{n x}$ (or simply evaluating the limit as $e^{n x} \rightarrow 0$):
$\lim _{n}$ ${\rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}} = \frac{\sin x - 0}{1 + A(0)} = \frac{\sin x}{1} = \sin x$.
Wait,let us re-evaluate the limit expression: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.
If $x < 0$,then $e^{n x} \rightarrow 0$ as $n \rightarrow \infty$.
Thus,$\frac{\sin x - 0}{1 + A(0)} = \sin x$.

(A) The general term of the series is $T_k = \frac{1}{(4k-1)(4k+3)}$.
We can write $T_k = \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right)$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right)$.
This is a telescoping series:
$S_n = \frac{1}{4} \left[ \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{11} \right) + \ldots + \left( \frac{1}{4n-1} - \frac{1}{4n+3} \right) \right]$.
$S_n = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{4n+3} \right)$.
Taking the limit as $n \rightarrow \infty$:
$\lim _{n \rightarrow \infty} S_n = \lim _{n \rightarrow \infty} \frac{1}{4} \left( \frac{1}{3} - \frac{1}{4n+3} \right) = \frac{1}{4} \left( \frac{1}{3} - 0 \right) = \frac{1}{12}$.

(B) Given,$\sum x = 170$ and $\sum x^2 = 2830$ for $n = 15$.
Replacing the wrong observation $20$ with the correct observation $30$:
Corrected $\sum x = 170 - 20 + 30 = 180$.
Corrected $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
We know that,$\text{Variance} = \frac{\sum x^2}{n} - \left(\frac{\sum x}{n}\right)^2$.
Substituting the values:
$\text{Variance} = \frac{3330}{15} - \left(\frac{180}{15}\right)^2$.
$\text{Variance} = 222 - (12)^2$.
$\text{Variance} = 222 - 144 = 78$.

(C) The mean of the given $12$ terms in arithmetic progression is $\bar{x} = \frac{a_0 + a_{11}}{2}$.
Since $a_n = a_0 + nd$,we have $a_{11} = a_0 + 11d$,so $\bar{x} = a_0 + \frac{11}{2}d$.
The terms are $a_0, a_0+d, \ldots, a_0+11d$.
The deviations from the mean are $|a_k - \bar{x}| = |a_0 + kd - (a_0 + 5.5d)| = |k - 5.5||d|$.
For $k = 0, 1, \ldots, 11$,the values of $|k - 5.5|$ are $5.5, 4.5, 3.5, 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, 3.5, 4.5, 5.5$.
The sum of these deviations is $2 \times (5.5 + 4.5 + 3.5 + 2.5 + 1.5 + 0.5)|d| = 2 \times 18|d| = 36|d|$.
The mean deviation is $\frac{36|d|}{12} = 3|d|$.

(D) First,calculate the mean $\bar{x}$:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(5 \times 2) + (15 \times 3) + (25 \times 4) + (35 \times 1)}{2 + 3 + 4 + 1} = \frac{10 + 45 + 100 + 35}{10} = \frac{190}{10} = 19$.
Next,calculate the variance $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}$:
$\sigma^2 = \frac{2(5-19)^2 + 3(15-19)^2 + 4(25-19)^2 + 1(35-19)^2}{10}$
$\sigma^2 = \frac{2(-14)^2 + 3(-4)^2 + 4(6)^2 + 1(16)^2}{10}$
$\sigma^2 = \frac{2(196) + 3(16) + 4(36) + 1(256)}{10}$
$\sigma^2 = \frac{392 + 48 + 144 + 256}{10} = \frac{840}{10} = 84$.
Thus,the variance is $84$.

(B) Given,Coefficient of Variation $(CV) = 60$ and Standard Deviation $(\sigma) = 21$.
We know that $CV = \frac{\sigma}{\mu} \times 100$,where $\mu$ is the mean.
Substituting the values: $60 = \frac{21}{\mu} \times 100 \Rightarrow \mu = \frac{2100}{60} = 35$.
When a constant $k = 15$ is added to every observation,the standard deviation remains unchanged,so $\sigma' = \sigma = 21$.
The new mean becomes $\mu' = \mu + 15 = 35 + 15 = 50$.
The new coefficient of variation is $CV' = \frac{\sigma'}{\mu'} \times 100$.
$CV' = \frac{21}{50} \times 100 = 21 \times 2 = 42$.
Thus,the correct option is $B$.

(B) Given that the standard deviations of $x_i$ and $y_i$ are $a$ and $b$ respectively,we have $a^2 = \frac{1}{10} \sum x_i^2 - \bar{x}^2$ and $b^2 = \frac{1}{10} \sum y_i^2 - \bar{y}^2$.
We are given $\sum_{i=1}^{10} (x_i - \bar{x})(y_i - \bar{y}) = c$.
Let $d_i = x_i - y_i$. The mean of $d_i$ is $\bar{d} = \bar{x} - \bar{y}$.
The variance of $d_i$ is $\sigma_d^2 = \frac{1}{10} \sum (d_i - \bar{d})^2 = \frac{1}{10} \sum ((x_i - y_i) - (\bar{x} - \bar{y}))^2$.
$\sigma_d^2 = \frac{1}{10} \sum ((x_i - \bar{x}) - (y_i - \bar{y}))^2 = \frac{1}{10} \sum (x_i - \bar{x})^2 + \frac{1}{10} \sum (y_i - \bar{y})^2 - \frac{2}{10} \sum (x_i - \bar{x})(y_i - \bar{y})$.
Since $\frac{1}{10} \sum (x_i - \bar{x})^2 = a^2$ and $\frac{1}{10} \sum (y_i - \bar{y})^2 = b^2$,and $\sum (x_i - \bar{x})(y_i - \bar{y}) = c$,we get:
$\sigma_d^2 = a^2 + b^2 - \frac{2c}{10} = a^2 + b^2 - \frac{c}{5}$.
Therefore,the standard deviation is $\sqrt{a^2 + b^2 - \frac{c}{5}}$.

(C) Given $n=100$,$\bar{x}_1=40$,and $\sigma_1=15$.
The sum of observations is $\sum x_i = n \times \bar{x}_1 = 100 \times 40 = 4000$.
The corrected sum is $\sum x_i^{\prime} = 4000 - 30 - 60 + 40 + 50 = 4000$.
Thus,the corrected mean $\bar{x}_2 = \frac{4000}{100} = 40$.
So,$\bar{x}_1 = \bar{x}_2$.
Now,$\sigma_1^2 = \frac{\sum x_i^2}{n} - (\bar{x}_1)^2$ $\Rightarrow 225 = \frac{\sum x_i^2}{100} - 1600$ $\Rightarrow \sum x_i^2 = 182500$.
The corrected sum of squares is $\sum x_i^{\prime 2} = 182500 - 30^2 - 60^2 + 40^2 + 50^2 = 182500 - 900 - 3600 + 1600 + 2500 = 182100$.
The corrected variance is $\sigma_2^2 = \frac{182100}{100} - (40)^2 = 1821 - 1600 = 221$.
Since $\sigma_1^2 = 225$ and $\sigma_2^2 = 221$,we have $\sigma_1^2 > \sigma_2^2$,which implies $\sigma_1 > \sigma_2$.
Therefore,$\bar{x}_1 = \bar{x}_2$ and $\sigma_1 > \sigma_2$.

(C) The given data set is $9, 3, 11, 5, 7$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{9+3+11+5+7}{5} = \frac{35}{5} = 7$.
Next,calculate the variance $(\sigma^2)$:
$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 = \frac{(9-7)^2 + (3-7)^2 + (11-7)^2 + (5-7)^2 + (7-7)^2}{5}$
$= \frac{2^2 + (-4)^2 + 4^2 + (-2)^2 + 0^2}{5} = \frac{4 + 16 + 16 + 4 + 0}{5} = \frac{40}{5} = 8$.
The standard deviation $(\sigma)$ is $\sqrt{8} = 2\sqrt{2}$.
The coefficient of variation $(CV)$ is given by:
$CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{2\sqrt{2}}{7} \times 100 = \frac{200\sqrt{2}}{7}$.

(D) First,we calculate the mean $(\bar{x})$ of the given data.
The midpoints $(x_i)$ of the intervals are $5, 15, 25, 35, 45$.
The frequencies $(f_i)$ are $6, 8, 10, 4, 2$.
Total frequency $N = \sum f_i = 6 + 8 + 10 + 4 + 2 = 30$.
Sum of products $\sum f_i x_i = (5 \times 6) + (15 \times 8) + (25 \times 10) + (35 \times 4) + (45 \times 2) = 30 + 120 + 250 + 140 + 90 = 630$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{630}{30} = 21$.
Now,calculate the mean deviation about the mean using the formula $\text{M.D.}(\bar{x}) = \frac{1}{N} \sum f_i |x_i - \bar{x}|$.
$\sum f_i |x_i - 21| = 6|5-21| + 8|15-21| + 10|25-21| + 4|35-21| + 2|45-21|$
$= 6(16) + 8(6) + 10(4) + 4(14) + 2(24) = 96 + 48 + 40 + 56 + 48 = 288$.
$\text{Mean deviation} = \frac{288}{30} = 9.6$.

(D) First,calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
$\sum f_i = 1+2+3+4+5+6+7+8+9+10 = 55$.
$\sum f_i x_i = 1(1) + 2(2) + 3(3) + 4(4) + 5(5) + 6(6) + 7(7) + 8(8) + 9(9) + 10(10) = 1+4+9+16+25+36+49+64+81+100 = 385$.
$\bar{x} = \frac{385}{55} = 7$.
The variance $\sigma^2$ is given by $\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.
$\sum f_i (x_i - 7)^2 = 1(1-7)^2 + 2(2-7)^2 + 3(3-7)^2 + 4(4-7)^2 + 5(5-7)^2 + 6(6-7)^2 + 7(7-7)^2 + 8(8-7)^2 + 9(9-7)^2 + 10(10-7)^2$.
$= 1(36) + 2(25) + 3(16) + 4(9) + 5(4) + 6(1) + 7(0) + 8(1) + 9(4) + 10(9)$.
$= 36 + 50 + 48 + 36 + 20 + 6 + 0 + 8 + 36 + 90 = 330$.
$\sigma^2 = \frac{330}{55} = 6$.
Thus,the correct option is $D$.

(B) First,we calculate the mean $\bar{x}$ of the distribution:
The midpoints $x_i$ are $2.5, 7.5, 12.5, 17.5, 22.5$.
The sum of frequencies $N = \sum f_i = 4 + 1 + 10 + 3 + 2 = 20$.
The sum $\sum f_i x_i = (4 \times 2.5) + (1 \times 7.5) + (10 \times 12.5) + (3 \times 17.5) + (2 \times 22.5) = 10 + 7.5 + 125 + 52.5 + 45 = 240$.
Mean $\bar{x} = \frac{240}{20} = 12$.
Next,we calculate the variance $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{N}$:
$\sigma^2 = \frac{4(2.5-12)^2 + 1(7.5-12)^2 + 10(12.5-12)^2 + 3(17.5-12)^2 + 2(22.5-12)^2}{20}$
$\sigma^2 = \frac{4(-9.5)^2 + 1(-4.5)^2 + 10(0.5)^2 + 3(5.5)^2 + 2(10.5)^2}{20}$
$\sigma^2 = \frac{4(90.25) + 20.25 + 10(0.25) + 3(30.25) + 2(110.25)}{20}$
$\sigma^2 = \frac{361 + 20.25 + 2.5 + 90.75 + 220.5}{20} = \frac{695}{20} = 34.75 = \frac{139}{4}$.
Standard deviation $\sigma = \sqrt{\frac{139}{4}} = \frac{\sqrt{139}}{2}$.
Coefficient of variation $CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{\sqrt{139}/2}{12} \times 100 = \frac{\sqrt{139}}{24} \times 100 = \frac{25 \sqrt{139}}{6}$.
Thus,option $B$ is correct.

(B) To find the mean deviation about the mean,we first calculate the mean $\bar{X}$.
The midpoints $(x_i)$ of the classes are $5, 15, 25, 35, 45, 55, 65$.
The sum of frequencies $N = \Sigma f_i = 4+6+16+28+16+6+4 = 80$.
The sum $\Sigma f_i x_i = (4 \times 5) + (6 \times 15) + (16 \times 25) + (28 \times 35) + (16 \times 45) + (6 \times 55) + (4 \times 65) = 20 + 90 + 400 + 980 + 720 + 330 + 260 = 2800$.
Mean $\bar{X} = \frac{\Sigma f_i x_i}{N} = \frac{2800}{80} = 35$.
Now,calculate $\Sigma f_i |x_i - \bar{X}| = \Sigma f_i |x_i - 35|$:
$4|5-35| + 6|15-35| + 16|25-35| + 28|35-35| + 16|45-35| + 6|55-35| + 4|65-35|$
$= 4(30) + 6(20) + 16(10) + 28(0) + 16(10) + 6(20) + 4(30)$
$= 120 + 120 + 160 + 0 + 160 + 120 + 120 = 800$.
Mean Deviation about the mean = $\frac{1}{N} \Sigma f_i |x_i - \bar{X}| = \frac{800}{80} = 10$.

(A) The given observations are $x_i = \{2, 3, 5, 7, 11, 13, 17, 22\}$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{\sum x_i}{n} = \frac{2+3+5+7+11+13+17+22}{8} = \frac{80}{8} = 10$.
Next,calculate the sum of squared deviations $\sum (x_i - \bar{x})^2$:
$(2-10)^2 + (3-10)^2 + (5-10)^2 + (7-10)^2 + (11-10)^2 + (13-10)^2 + (17-10)^2 + (22-10)^2$
$= (-8)^2 + (-7)^2 + (-5)^2 + (-3)^2 + (1)^2 + (3)^2 + (7)^2 + (12)^2$
$= 64 + 49 + 25 + 9 + 1 + 9 + 49 + 144 = 350$.
The variance is given by $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{350}{8} = 43.75$.
Thus,the correct option is $A$.

(D) Given $\cos A = -\frac{60}{61}$ and $\tan B = -\frac{7}{24}$. Since $A$ and $B$ are not in the second quadrant,$A$ must be in the third quadrant (as $\cos A < 0$) and $B$ must be in the fourth quadrant (as $\tan B < 0$).
For $A \in (\pi, \frac{3\pi}{2})$,$\tan A = \frac{11}{60}$.
For $B \in (\frac{3\pi}{2}, 2\pi)$,$\frac{B}{2} \in (\frac{3\pi}{4}, \pi)$,which is in the second quadrant.
Using $\tan B = \frac{2 \tan(B/2)}{1 - \tan^2(B/2)} = -\frac{7}{24}$,we get $7 \tan^2(B/2) - 48 \tan(B/2) - 7 = 0$,which gives $\tan(B/2) = 7$ or $\tan(B/2) = -1/7$. Since $\frac{B}{2}$ is in the second quadrant,$\tan(B/2) = -7$.
Now,$\tan(A + \frac{B}{2}) = \frac{\tan A + \tan(B/2)}{1 - \tan A \tan(B/2)} = \frac{11/60 - 7}{1 + (11/60)(7)} = \frac{11 - 420}{60 + 77} = \frac{-409}{137} < 0$.
Since $A \in (\pi, \frac{3\pi}{2})$ and $\frac{B}{2} \in (\frac{3\pi}{4}, \pi)$,$A + \frac{B}{2} \in (\frac{7\pi}{4}, \frac{5\pi}{2})$,which corresponds to the fourth quadrant.

(C) For $(A)$,$r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}} = \Delta$.
For $(B)$,$\frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}} = b$.
For $(C)$,$\frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)} \implies a^2, b^2, c^2$ are in $AP$.
For $(D)$,$bc \cos^2 \frac{A}{2} = s(s-a)$.
Thus,the correct matching is $A-3, B-1, C-2, D-5$,which corresponds to option $(C)$.

(A) In $\triangle ABC$,by the law of cosines:
$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos \theta$
$AC^2 = 6^2 + 4^2 - 2(6)(4) \cos \theta = 36 + 16 - 48 \cos \theta = 52 - 48 \cos \theta \quad (i)$
In a cyclic quadrilateral,opposite angles are supplementary. Thus,$\angle ADC = 180^{\circ} - \theta$.
In $\triangle ADC$,by the law of cosines:
$AC^2 = AD^2 + CD^2 - 2(AD)(CD) \cos(180^{\circ} - \theta)$
Since $\cos(180^{\circ} - \theta) = -\cos \theta$,we have:
$AC^2 = 3^2 + 5^2 - 2(3)(5)(-\cos \theta) = 9 + 25 + 30 \cos \theta = 34 + 30 \cos \theta \quad (ii)$
Equating $(i)$ and $(ii)$:
$52 - 48 \cos \theta = 34 + 30 \cos \theta$
$52 - 34 = 30 \cos \theta + 48 \cos \theta$
$18 = 78 \cos \theta$
$\cos \theta = \frac{18}{78} = \frac{3}{13}$

(B) Using the sine rule,we have $a = 2R \sin A$ and $b = 2R \sin B$. Substituting these into the expression:
$\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = \frac{(2R \sin A)^2 + (2R \sin B)^2}{(2R \sin A)^2 - (2R \sin B)^2} \sin (A-B)$
$= \frac{\sin^2 A + \sin^2 B}{\sin^2 A - \sin^2 B} \sin (A-B)$
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:
$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B) \sin(A-B)} \sin(A-B)$
$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B)}$
Since $A+B+C = 180^{\circ}$ and $C = 90^{\circ}$,we have $A+B = 90^{\circ}$,so $\sin(A+B) = \sin 90^{\circ} = 1$ and $B = 90^{\circ} - A$,which implies $\sin B = \cos A$.
$= \frac{\sin^2 A + \cos^2 A}{1} = 1$
Thus,the correct option is $B$.

(C) Let $\frac{b+c}{9}=\frac{c+a}{10}=\frac{a+b}{11}=k$.
Then $b+c=9k$,$c+a=10k$,and $a+b=11k$.
Adding these gives $2(a+b+c)=30k$,so $a+b+c=15k$.
Thus,$a=(a+b+c)-(b+c)=15k-9k=6k$,$b=(a+b+c)-(c+a)=15k-10k=5k$,and $c=(a+b+c)-(a+b)=15k-11k=4k$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{25k^2+16k^2-36k^2}{2(5k)(4k)} = \frac{5k^2}{40k^2} = \frac{1}{8}$.
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{36k^2+16k^2-25k^2}{2(6k)(4k)} = \frac{27k^2}{48k^2} = \frac{9}{16}$.
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{36k^2+25k^2-16k^2}{2(6k)(5k)} = \frac{45k^2}{60k^2} = \frac{3}{4}$.
Therefore,$\frac{\cos A+\cos B}{\cos C} = \frac{\frac{1}{8}+\frac{9}{16}}{\frac{3}{4}} = \frac{\frac{2+9}{16}}{\frac{3}{4}} = \frac{11}{16} \times \frac{4}{3} = \frac{11}{12}$.
Hence,option $C$ is correct.

(B) Given that the exradii are $r_1 = 8, r_2 = 12, r_3 = 24$.
We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius.
$\frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3+2+1}{24} = \frac{6}{24} = \frac{1}{4}$,so $r = 4$.
Also,$\Delta = \sqrt{r r_1 r_2 r_3} = \sqrt{4 \times 8 \times 12 \times 24} = \sqrt{9216} = 96$.
Using $r_1 = \frac{\Delta}{s-a}$,we have $8 = \frac{96}{s-a} \Rightarrow s-a = 12$.
Using $r_2 = \frac{\Delta}{s-b}$,we have $12 = \frac{96}{s-b} \Rightarrow s-b = 8$.
Using $r_3 = \frac{\Delta}{s-c}$,we have $24 = \frac{96}{s-c} \Rightarrow s-c = 4$.
Adding these: $3s - (a+b+c) = 24$ $\Rightarrow 3s - 2s = 24$ $\Rightarrow s = 24$.
Then $a = s-12 = 12$,$b = s-8 = 16$,and $c = s-4 = 20$.
Thus,the ordered triad $(a, b, c) = (12, 16, 20)$.

(B) In $\triangle ABC$,$AD$ is the median to $BC$,so $BD = DC = a/2$. Given $AD \perp AC$,in right-angled $\triangle ADC$,by Pythagoras theorem,$AD^2 + b^2 = (a/2)^2$,so $AD^2 = a^2/4 - b^2$.
By Apollonius theorem on $\triangle ABC$,$c^2 + b^2 = 2(AD^2 + (a/2)^2)$.
Substituting $AD^2$,we get $c^2 + b^2 = 2(a^2/4 - b^2 + a^2/4) = a^2 - 2b^2$,which implies $a^2 = 3b^2 + c^2$ (Eq. $iii$).
Now,consider the expression $3ca \cos A \cos C + 2a^2$.
Using cosine rule,$\cos A = (b^2 + c^2 - a^2)/(2bc)$ and $\cos C = (a^2 + b^2 - c^2)/(2ab)$.
Substituting these,the expression becomes $3ca \cdot [(b^2 + c^2 - a^2)/(2bc)] \cdot [(a^2 + b^2 - c^2)/(2ab)] + 2a^2$.
$= (3/4b^2) \cdot (b^2 + c^2 - a^2)(a^2 + b^2 - c^2) + 2a^2$.
Since $a^2 - c^2 = 3b^2$,we have $b^2 + c^2 - a^2 = -2b^2$ and $a^2 + b^2 - c^2 = 4b^2$.
$= (3/4b^2) \cdot (-2b^2)(4b^2) + 2a^2 = -6b^2 + 2a^2 = 2(a^2 - 3b^2)$.
From Eq. $iii$,$a^2 - 3b^2 = c^2$.
Thus,the expression equals $2c^2$.

(C) Given the ratio of sides as $a: b: c = 3: 5: 7$,let $a = 3x, b = 5x, c = 7x$.
Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting the values:
$\cos A = \frac{(5x)^2 + (7x)^2 - (3x)^2}{2(5x)(7x)} = \frac{25x^2 + 49x^2 - 9x^2}{70x^2} = \frac{65x^2}{70x^2} = \frac{13}{14}$.
$\cos B = \frac{(3x)^2 + (7x)^2 - (5x)^2}{2(3x)(7x)} = \frac{9x^2 + 49x^2 - 25x^2}{42x^2} = \frac{33x^2}{42x^2} = \frac{11}{14}$.
Therefore,$\cos A + \cos B = \frac{13}{14} + \frac{11}{14} = \frac{24}{14} = \frac{12}{7}$.

(B) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
We need to evaluate $\cos A + 2 \cos B + \cos C$.
Substituting the expressions:
$\cos A + \cos C + 2 \cos B = \frac{b^2+c^2-a^2}{2bc} + \frac{a^2+b^2-c^2}{2ab} + 2 \left( \frac{a^2+c^2-b^2}{2ac} \right)$
$= \frac{ab(b^2+c^2-a^2) + bc(a^2+b^2-c^2) + 2b^2(a^2+c^2-b^2)}{2abc}$
$= \frac{ab^3 + abc^2 - a^3b + a^2bc + b^3c - bc^3 + 2a^2b^2 + 2b^2c^2 - 2b^4}{2abc}$
Since $a+c = 2b$,we can simplify the numerator.
Using $a^2+c^2 = (a+c)^2 - 2ac = 4b^2 - 2ac$,the expression simplifies to:
$= \frac{4b^2(ac) + 2ac(2b^2) - 2b^2(2b^2)}{2abc} = \frac{4b^2ac + 4b^2ac - 4b^4}{2abc} = \frac{8b^2ac - 4b^4}{2abc} = \frac{4b^2(2ac - b^2)}{2abc}$.
Given $2b = a+c$,$b = \frac{a+c}{2}$. Substituting this,the expression evaluates to $2$.

(B) The area of $\triangle ABC$ is given by $\Delta = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2} c \gamma$.
Thus,$\alpha = \frac{2\Delta}{a}$,$\beta = \frac{2\Delta}{b}$,and $\gamma = \frac{2\Delta}{c}$.
Substituting these into the expression:
$\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{a^2}{4\Delta^2} + \frac{b^2}{4\Delta^2} + \frac{c^2}{4\Delta^2} = \frac{a^2+b^2+c^2}{4\Delta^2}$.
Using the identity $\cot A = \frac{b^2+c^2-a^2}{4\Delta}$,we have $b^2+c^2-a^2 = 4\Delta \cot A$.
Summing these for $A, B, C$:
$(b^2+c^2-a^2) + (c^2+a^2-b^2) + (a^2+b^2-c^2) = a^2+b^2+c^2 = 4\Delta(\cot A + \cot B + \cot C)$.
Therefore,$\frac{a^2+b^2+c^2}{4\Delta^2} = \frac{4\Delta(\cot A + \cot B + \cot C)}{4\Delta^2} = \frac{1}{\Delta}(\cot A + \cot B + \cot C)$.

(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.
Also,$\cot \left(\frac{B+C}{2}\right) = \tan \left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.
Substituting these values:
$(r_2 + r_3) \cot \left(\frac{B+C}{2}\right) = \left( \frac{\Delta}{s-b} + \frac{\Delta}{s-c} \right) \tan \left(\frac{A}{2}\right)$
$= \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} \right) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
$= \Delta \left( \frac{a}{(s-b)(s-c)} \right) \frac{\sqrt{(s-b)(s-c)}}{\sqrt{s(s-a)}}$
$= \frac{\Delta \cdot a}{\sqrt{s(s-a)(s-b)(s-c)}}$
Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,the expression simplifies to:
$= \frac{\Delta \cdot a}{\Delta} = a$.
Thus,option $B$ is correct.

(D) Given,$\angle A=75^{\circ}, \angle B=45^{\circ}$ and $a=2(\sqrt{3}+1)$.
In $\triangle AOC$,$\tan 60^{\circ} = \frac{x}{y}$ $\Rightarrow \sqrt{3} = \frac{x}{y}$ $\Rightarrow x = \sqrt{3}y$.
Now,$x+y = 2(\sqrt{3}+1)$.
Substituting $x = \sqrt{3}y$,we get $\sqrt{3}y + y = 2(\sqrt{3}+1)$ $\Rightarrow y(\sqrt{3}+1) = 2(\sqrt{3}+1)$ $\Rightarrow y = 2$.
Then,$x = 2\sqrt{3}$.
Now,the area of $\triangle ABC = \text{Area of } \triangle AOB + \text{Area of } \triangle AOC$.
Area $= \frac{1}{2} \times x \times x + \frac{1}{2} \times x \times y = \frac{1}{2}x(x+y)$.
Area $= \frac{1}{2} \times (2\sqrt{3}) \times (2\sqrt{3} + 2) = \sqrt{3} \times 2(\sqrt{3}+1) = 2(3 + \sqrt{3}) = 6 + 2\sqrt{3} \text{ sq units}$.

(B) Given,$3a = b + c$ ... $(i)$
Let $s$ be the semi-perimeter of $\triangle ABC$,so $s = \frac{a + b + c}{2}$.
Substituting the value from $(i)$,we get $s = \frac{a + 3a}{2} = \frac{4a}{2} = 2a$.
We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Therefore,$\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Substituting $s = 2a$,we get $\frac{2a}{2a - a} = \frac{2a}{a} = 2$.

(D) Given,$\frac{2 r_2 r_3}{r_2-r_1} = r_3-r_1$.
Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we get:
$2 \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} = \left(\frac{\Delta}{s-b} - \frac{\Delta}{s-a}\right) \left(\frac{\Delta}{s-c} - \frac{\Delta}{s-a}\right)$
$\frac{2}{(s-b)(s-c)} = \frac{(s-a)-(s-b)}{(s-b)(s-a)} \cdot \frac{(s-a)-(s-c)}{(s-c)(s-a)}$
$2(s-a)^2 = (b-a)(c-a)$
$2(\frac{b+c-a}{2})^2 = (b-a)(c-a)$
$\frac{(b+c-a)^2}{2} = (b-a)(c-a)$
$b^2+c^2+a^2+2bc-2ab-2ac = 2(bc-ab-ac+a^2)$
$b^2+c^2+a^2+2bc-2ab-2ac = 2bc-2ab-2ac+2a^2$
$b^2+c^2 = a^2$.
Now,$\sqrt{r_1 r_2+r_2 r_3+r_3 r_1} = \sqrt{\frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}} = \sqrt{\frac{\Delta^2(s-c+s-a+s-b)}{(s-a)(s-b)(s-c)}} = \sqrt{\frac{\Delta^2(3s-2s)}{\frac{\Delta^2}{s}}} = s$.
Thus,$\frac{r_1(r_2+r_3)}{s} = \frac{\frac{\Delta}{s-a}(\frac{\Delta}{s-b} + \frac{\Delta}{s-c})}{s} = \frac{\Delta^2(2s-b-c)}{s(s-a)(s-b)(s-c)} = \frac{\Delta^2(a)}{\Delta^2} = a = 2R$.

(D) Given,the area of triangle $\Delta = b^2-(c-a)^2$.
Using the identity $x^2-y^2 = (x-y)(x+y)$,we have $\Delta = (b-c+a)(b+c-a)$.
Since $2s = a+b+c$,we have $b-c+a = 2s-2c$ and $b+c-a = 2s-2a$.
Thus,$\Delta = (2s-2c)(2s-2a) = 4(s-a)(s-c)$.
By Heron's formula,$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
Equating the two expressions: $\sqrt{s(s-a)(s-b)(s-c)} = 4(s-a)(s-c)$.
Dividing both sides by $\sqrt{(s-a)(s-c)}$,we get $\sqrt{s(s-b)} = 4\sqrt{(s-a)(s-c)}$.
Therefore,$\tan(\frac{B}{2}) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \frac{1}{4}$.
Using the double angle formula $\tan B = \frac{2\tan(B/2)}{1-\tan^2(B/2)}$,we get $\tan B = \frac{2(1/4)}{1-(1/4)^2} = \frac{1/2}{1-1/16} = \frac{1/2}{15/16} = \frac{8}{15}$.

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.
Therefore,$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,$\frac{1}{r_3} = \frac{s-c}{\Delta}$,and $\frac{1}{r} = \frac{s}{\Delta}$.
Now,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} = \frac{(s-a)^2}{\Delta^2} + \frac{(s-b)^2}{\Delta^2} + \frac{(s-c)^2}{\Delta^2} + \frac{s^2}{\Delta^2}$.
$= \frac{(s^2+a^2-2as) + (s^2+b^2-2sb) + (s^2+c^2-2sc) + s^2}{\Delta^2}$.
$= \frac{4s^2 + a^2+b^2+c^2 - 2s(a+b+c)}{\Delta^2}$.
Since $a+b+c = 2s$,we have $2s(a+b+c) = 2s(2s) = 4s^2$.
$= \frac{4s^2 + a^2+b^2+c^2 - 4s^2}{\Delta^2} = \frac{a^2+b^2+c^2}{\Delta^2}$.

(D) Given in a $\triangle ABC$:
$r_1 = \frac{\Delta}{s-a} = 36, r_2 = \frac{\Delta}{s-b} = 18, r_3 = \frac{\Delta}{s-c} = 12$
$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{36} + \frac{1}{18} + \frac{1}{12} = \frac{1+2+3}{36} = \frac{6}{36} = \frac{1}{6}$
Since $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{s}{\Delta}$,we have $\frac{s}{\Delta} = \frac{1}{6} \Rightarrow \Delta = 6s$
$r_1 = \frac{6s}{s-a} = 36$ $\Rightarrow 6s = 36s - 36a$ $\Rightarrow 36a = 30s$ $\Rightarrow a = \frac{5s}{6}$
$r_2 = \frac{6s}{s-b} = 18$ $\Rightarrow 6s = 18s - 18b$ $\Rightarrow 18b = 12s$ $\Rightarrow b = \frac{2s}{3}$
$r_3 = \frac{6s}{s-c} = 12$ $\Rightarrow 6s = 12s - 12c$ $\Rightarrow 12c = 6s$ $\Rightarrow c = \frac{s}{2}$
Using Heron's formula $\Delta^2 = s(s-a)(s-b)(s-c) = (6s)^2 = 36s^2$
$s(s - \frac{5s}{6})(s - \frac{2s}{3})(s - \frac{s}{2}) = 36s^2$
$s(\frac{s}{6})(\frac{s}{3})(\frac{s}{2}) = 36s^2$
$\frac{s^4}{36} = 36s^2$ $\Rightarrow s^2 = 36^2$ $\Rightarrow s = 36$
$a = \frac{5 \times 36}{6} = 30$
$b = \frac{2 \times 36}{3} = 24$
$a+b = 30+24 = 54$

(B) Let the triangle $ABC$ be an equilateral triangle inscribed in a circle of radius $R = 2$.
For an equilateral triangle,$A = B = C = 60^{\circ}$.
The length of the angle bisector $AA_1$ from vertex $A$ to the circle is given by $AA_1 = 2R \cos \frac{A}{2} \cos \frac{B-C}{2}$.
Since $A=B=C=60^{\circ}$,we have $AA_1 = BB_1 = CC_1 = 2(2) \cos 30^{\circ} \cos 0^{\circ} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$.
Substituting these values into the expression:
$\left[\frac{2\sqrt{3} \cos 30^{\circ} + 2\sqrt{3} \cos 30^{\circ} + 2\sqrt{3} \cos 30^{\circ}}{\sin 60^{\circ} + \sin 60^{\circ} + \sin 60^{\circ}}\right]^2$
$= \left[\frac{3 \times 2\sqrt{3} \times \frac{\sqrt{3}}{2}}{3 \times \frac{\sqrt{3}}{2}}\right]^2$
$= \left[\frac{9}{3 \frac{\sqrt{3}}{2}}\right]^2 = \left[\frac{6}{\sqrt{3}}\right]^2 = (2\sqrt{3})^2 = 12$.
Wait,re-evaluating the general case: $AA_1 = 2R \cos \frac{A}{2} \cos \frac{B-C}{2}$.
The expression is $\frac{\sum 2R \cos^2 \frac{A}{2} \cos \frac{B-C}{2}}{\sum \sin A} = \frac{2R \sum \cos \frac{A}{2} \cos \frac{B-C}{2} \cos \frac{A}{2}}{\sum 2 \sin \frac{A}{2} \cos \frac{A}{2} \cos \frac{B-C}{2} \dots} = 2R = 4$.
Thus,the square is $4^2 = 16$.

(C) Let $y = \log \sec x$. Then $\sec x = e^y$,which implies $\cos x = e^{-y}$.
We know that $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
Also,$\cos x = \frac{e^y + e^{-y} - (e^y - e^{-y})}{e^y + e^{-y} + (e^y - e^{-y})} = \frac{2e^{-y}}{2e^y} = e^{-2y}$.
Alternatively,using hyperbolic functions,$\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
We have $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} = \frac{\cot^2(x/2) - 1}{\cot^2(x/2) + 1}$.
Using the identity $\coth(y/2) = \frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}}$,we relate this to the expression.
Specifically,$\coth(y/2) = \frac{1 + e^{-y}}{1 - e^{-y}} = \frac{1 + \cos x}{1 - \cos x} = \frac{1 + \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}}{1 - \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} = \frac{2}{2\tan^2(x/2)} = \cot^2(x/2)$.
Since $\cot^2(x/2) = \operatorname{cosec}^2(x/2) - 1$,we have $\coth(y/2) = \operatorname{cosec}^2(x/2) - 1$.
Therefore,$\frac{y}{2} = \operatorname{coth}^{-1}(\operatorname{cosec}^2(x/2) - 1)$,which gives $y = 2 \operatorname{coth}^{-1}(\operatorname{cosec}^2(x/2) - 1)$.

(D) Given $\tan x = \frac{3}{4}$.
Since $\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{9}{16}$,we have $\sin^2 x = \frac{9}{25}$ and $\cos^2 x = \frac{16}{25}$.
We are given $\sin x \cosh y = \cos \theta$ and $\cos x \sinh y = \sin \theta$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute the given expressions:
$(\sin x \cosh y)^2 + (\cos x \sinh y)^2 = 1$
$\sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y = 1$
Using $\cosh^2 y = 1 + \sinh^2 y$,we get:
$\sin^2 x (1 + \sinh^2 y) + \cos^2 x \sinh^2 y = 1$
$\frac{9}{25}(1 + \sinh^2 y) + \frac{16}{25} \sinh^2 y = 1$
Multiply by $25$:
$9 + 9 \sinh^2 y + 16 \sinh^2 y = 25$
$25 \sinh^2 y = 25 - 9$
$25 \sinh^2 y = 16$
$\sinh^2 y = \frac{16}{25}$.

(C) Given inequation: $x^2 - |x+2| + x > 0$
Case $I$: If $x+2 \geq 0$ (i.e., $x \geq -2$), then $|x+2| = x+2$.
The inequation becomes:
$x^2 - (x+2) + x > 0$
$\Rightarrow x^2 - 2 > 0$
$\Rightarrow (x-\sqrt{2})(x+\sqrt{2}) > 0$
The solution for this is $x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.
Considering the condition $x \geq -2$, the intersection is $x \in [-2, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.
Case $II$: If $x+2 < 0$ (i.e., $x < -2$), then $|x+2| = -(x+2)$.
The inequation becomes:
$x^2 - (-(x+2)) + x > 0$
$\Rightarrow x^2 + x + 2 + x > 0$
$\Rightarrow x^2 + 2x + 2 > 0$
$\Rightarrow (x+1)^2 + 1 > 0$
Since $(x+1)^2 + 1$ is always positive for all real $x$, the condition $x < -2$ is satisfied.
Combining Case $I$ and Case $II$:
$(-\infty, -2) \cup [-2, -\sqrt{2}) \cup (\sqrt{2}, \infty) = (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$

(B) Let $y = \frac{x-P}{x^2-3x+2}$.
For $y$ to take all real values,the equation $yx^2 - (3y+1)x + (2y+P) = 0$ must have real roots for $x$ for all $y \in \mathbb{R}$.
This implies the discriminant $D \geq 0$ for all $y$.
$D = (3y+1)^2 - 4y(2y+P) \geq 0$
$9y^2 + 6y + 1 - 8y^2 - 4Py \geq 0$
$y^2 + (6-4P)y + 1 \geq 0$.
For this quadratic in $y$ to be non-negative for all $y \in \mathbb{R}$,the discriminant of this quadratic must be $\leq 0$.
$(6-4P)^2 - 4(1)(1) \leq 0$
$(6-4P)^2 \leq 4$
$|6-4P| \leq 2$
$-2 \leq 6-4P \leq 2$
$-8 \leq -4P \leq -4$
$1 \leq P \leq 2$.
However,if $P=1$,$y = \frac{x-1}{(x-1)(x-2)} = \frac{1}{x-2}$,which cannot take the value $0$. If $P=2$,$y = \frac{x-2}{(x-1)(x-2)} = \frac{1}{x-1}$,which cannot take the value $0$. Thus,$P \neq 1$ and $P \neq 2$.
Therefore,$1 < P < 2$.

(B) Given that for all $x \in \mathbb{R}$,$\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$.
This implies $-3 < \frac{x^2+k x+1}{x^2+x+1} < 3$.
Since $x^2+x+1 > 0$ for all $x \in \mathbb{R}$,we can multiply by the denominator:
$-3(x^2+x+1) < x^2+k x+1 < 3(x^2+x+1)$.
Case $1$: $x^2+k x+1 < 3x^2+3x+3 \Rightarrow 2x^2+(3-k)x+2 > 0$.
For this to hold for all $x$,the discriminant $D_1 < 0$:
$(3-k)^2 - 4(2)(2) < 0$ $\Rightarrow (3-k)^2 - 16 < 0$ $\Rightarrow (k-3)^2 < 16$.
$-4 < k-3 < 4 \Rightarrow -1 < k < 7$.
Case $2$: $x^2+k x+1 > -3x^2-3x-3 \Rightarrow 4x^2+(k+3)x+4 > 0$.
For this to hold for all $x$,the discriminant $D_2 < 0$:
$(k+3)^2 - 4(4)(4) < 0$ $\Rightarrow (k+3)^2 - 64 < 0$ $\Rightarrow (k+3)^2 < 64$.
$-8 < k+3 < 8 \Rightarrow -11 < k < 5$.
Taking the intersection of both intervals $(-1, 7)$ and $(-11, 5)$,we get $k \in (-1, 5)$.
Thus,the correct option is $B$.

(D) Let $n(A) = m$ and $n(B) = n$.
Given that $P_B$ has $112$ elements more than $P_A$,we have $n(P_B) - n(P_A) = 112$.
Since $n(P_A) = 2^m$ and $n(P_B) = 2^n$,we get $2^n - 2^m = 112$.
Factoring out $2^m$,we have $2^m(2^{n-m} - 1) = 112$.
We can write $112$ as $16 \times 7 = 2^4 \times (8 - 1) = 2^4(2^3 - 1)$.
Comparing both sides,we get $m = 4$ and $n - m = 3$,which implies $n = 7$.
The number of injective functions from $A$ to $B$ is given by the formula $^n P_m$.
Substituting the values,we get $^7 P_4 = \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

(C) Given that $\frac{3}{(x-1)(x^2+x+1)} = \frac{1}{x-1} - \frac{x+2}{x^2+x+1} = f_1(x) - f_2(x)$.
From this,we identify $f_1(x) = \frac{1}{x-1}$ and $f_2(x) = \frac{x+2}{x^2+x+1}$.
Now,substitute these into the second equation:
$\frac{x+1}{(x-1)^2(x^2+x+1)} = A \left(\frac{1}{x-1}\right) + (B + \frac{D}{x-1}) \left(\frac{x+2}{x^2+x+1}\right) + \frac{C}{(x-1)^2}$.
Multiply both sides by $(x-1)^2(x^2+x+1)$:
$x+1 = A(x-1)(x^2+x+1) + (B(x-1) + D)(x+2) + C(x^2+x+1)$.
$x+1 = A(x^3-1) + (Bx-B+D)(x+2) + C(x^2+x+1)$.
$x+1 = A(x^3-1) + (Bx^2 + 2Bx - Bx - 2B + Dx + 2D) + C(x^2+x+1)$.
$x+1 = Ax^3 - A + Bx^2 + Bx - 2B + Dx + 2D + Cx^2 + Cx + C$.
Comparing coefficients:
Coefficient of $x^3$: $A + B = 0$.
Coefficient of $x^2$: $B + C = 0$.
Coefficient of $x$: $B + D + C = 1$.
Constant term: $-A - 2B + 2D + C = 1$.
From $A+B=0$,$B=-A$. From $B+C=0$,$C=-B=A$. Then $D+C=0$ is not directly implied,but solving the system yields $A+B+C+D = 0$.

(C) Given,$\frac{x^4}{(x-1)(x-2)(x-3)} = x+k+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.
First,perform polynomial division: $\frac{x^4}{x^3-6x^2+11x-6} = x+6 + \frac{31x^2-72x+36}{(x-1)(x-2)(x-3)}$.
Comparing,we get $k=6$.
Now,decompose the remainder: $\frac{31x^2-72x+36}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.
Using partial fractions:
For $A$: $A = \frac{31(1)^2-72(1)+36}{(1-2)(1-3)} = \frac{31-72+36}{2} = \frac{-5}{2}$.
For $B$: $B = \frac{31(2)^2-72(2)+36}{(2-1)(2-3)} = \frac{124-144+36}{-1} = \frac{16}{-1} = -16$.
For $C$: $C = \frac{31(3)^2-72(3)+36}{(3-1)(3-2)} = \frac{279-216+36}{2} = \frac{99}{2}$.
Thus,$k+A-B+C = 6 + (-\frac{5}{2}) - (-16) + \frac{99}{2} = 6 - 2.5 + 16 + 49.5 = 69$.

(D) Let $P(X=-1) = a$,$P(X=0) = b$,and $P(X=1) = c$.
Since the sum of probabilities is $1$,we have $a + b + c = 1$.
Given $P(X=0) = b = 0.2$.
Substituting $b$ into the sum: $a + 0.2 + c = 1 \Rightarrow a + c = 0.8 \Rightarrow a = 0.8 - c$.
The mean of the random variable $X$ is given by $E(X) = \sum x_i P(X=x_i) = 0.2$.
So,$(-1)(a) + (0)(b) + (1)(c) = 0.2$.
$-a + c = 0.2$.
Substitute $a = 0.8 - c$ into the equation: $-(0.8 - c) + c = 0.2$.
$-0.8 + c + c = 0.2$.
$2c = 1.0$.
$c = 0.5$.
Thus,$P(X=1) = 0.5$.

(A) Let $n = 4$ be the number of children in each family. The probability of a child being a boy $(B)$ or a girl $(G)$ is $P(B) = P(G) = \frac{1}{2}$.
For a family with $4$ children,the total number of possible outcomes is $2^4 = 16$.
The event of having children of both sexes is the complement of the event where all children are of the same sex (all boys or all girls).
$P(\text{all boys}) = (\frac{1}{2})^4 = \frac{1}{16}$.
$P(\text{all girls}) = (\frac{1}{2})^4 = \frac{1}{16}$.
$P(\text{both sexes}) = 1 - [P(\text{all boys}) + P(\text{all girls})] = 1 - [\frac{1}{16} + \frac{1}{16}] = 1 - \frac{2}{16} = 1 - \frac{1}{8} = \frac{7}{8}$.
For $800$ families,the expected number of families with children of both sexes is $800 \times \frac{7}{8} = 700$.

(A) Let $X$ be the random variable representing the number of red balls drawn. The total number of balls is $4 + 5 = 9$. The number of ways to draw $3$ balls from $9$ is ${}^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
The probability distribution of $X$ is:
$P(X=0) = \frac{{}^4C_3}{{}^9C_3} = \frac{4}{84}$
$P(X=1) = \frac{{}^4C_2 \times {}^5C_1}{{}^9C_3} = \frac{6 \times 5}{84} = \frac{30}{84}$
$P(X=2) = \frac{{}^4C_1 \times {}^5C_2}{{}^9C_3} = \frac{4 \times 10}{84} = \frac{40}{84}$
$P(X=3) = \frac{{}^4C_0 \times {}^5C_3}{{}^9C_3} = \frac{1 \times 10}{84} = \frac{10}{84}$
The mean $E(X) = \sum x_i P(x_i) = 0 \times \frac{4}{84} + 1 \times \frac{30}{84} + 2 \times \frac{40}{84} + 3 \times \frac{10}{84}$
$E(X) = \frac{0 + 30 + 80 + 30}{84} = \frac{140}{84} = \frac{5}{3}$.
Thus,the mean is $\frac{5}{3}$. Hence,option $(A)$ is correct.