AP EAMCET 2024 Mathematics Question Paper with Answer and Solution

(D) Let $L = \lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$.
Divide the numerator and denominator by $x^4$ inside the terms or analyze the growth rates.
As $x \rightarrow \infty$,$\sqrt{2x+1} + \sqrt{2x-1} \approx 2\sqrt{2x}$ and $\sqrt{2x+1} - \sqrt{2x-1} = \frac{2}{\sqrt{2x+1} + \sqrt{2x-1}} \approx \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$.
Thus,$(\sqrt{2x+1} + \sqrt{2x-1})^8 \approx (2\sqrt{2x})^8 = 2^4 \cdot (2x)^4 = 16 \cdot 16x^4 = 256x^4$.
Also,$(x+\sqrt{x^2-2})^8 \approx (2x)^8 = 256x^8$.
The expression becomes $\lim _{x \rightarrow \infty} \frac{256x^4 + (\frac{1}{\sqrt{2x}})^8(Px^4-16)}{256x^8} = 1$.
Since $(\frac{1}{\sqrt{2x}})^8 = \frac{1}{16x^4}$,the expression is $\lim _{x \rightarrow \infty} \frac{256x^4 + \frac{1}{16x^4}(Px^4-16)}{256x^8} = 1$.
This simplifies to $\lim _{x \rightarrow \infty} \frac{256x^4 + \frac{P}{16} - \frac{1}{x^4}}{256x^8} = 1$.
For the limit to be $1$,the leading coefficients must match.
$256x^4 \cdot Px^4$ terms dominate,leading to $P = \frac{1}{16}$.

(A) Let $L = \lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right)$.
We can rewrite the numerator as: $(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)$.
So,$L = \lim _{x \rightarrow 1} \left( \frac{(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)}{x-1} \right)$.
$L = \lim _{x}$ ${\rightarrow 1} \left( \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right)$.
Using the standard limit $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,we get:
$L = 1 + 2(1)^{2-1} + 3(1)^{3-1} + \ldots + n(1)^{n-1}$.
$L = 1 + 2 + 3 + \ldots + n$.
$L = \frac{n(n+1)}{2}$.

(A) Using $L'H\hat{o}pital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\cos 2x - \cos 3x)}{\frac{d}{dx}(\cos 4x - \cos 5x)} = \lim _{x \rightarrow 0} \frac{-2 \sin 2x + 3 \sin 3x}{-4 \sin 4x + 5 \sin 5x}$
Since this is still in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital's$ rule again:
$\lim _{x \rightarrow 0} \frac{-4 \cos 2x + 9 \cos 3x}{-16 \cos 4x + 25 \cos 5x}$
Substituting $x = 0$:
$= \frac{-4 \cos(0) + 9 \cos(0)}{-16 \cos(0) + 25 \cos(0)} = \frac{-4(1) + 9(1)}{-16(1) + 25(1)} = \frac{5}{9}$

(B) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x}$.
Using the identity $\cos 2x = 1 - 2\sin^2 x$,we rewrite the numerator:
$1 - \cos x(1 - 2\sin^2 x) = 1 - \cos x + 2\sin^2 x \cos x$.
So,$L = \lim _{x \rightarrow 0} \frac{1 - \cos x + 2\sin^2 x \cos x}{\sin^2 x}$.
$L = \lim _{x \rightarrow 0} \left( \frac{1 - \cos x}{\sin^2 x} + \frac{2\sin^2 x \cos x}{\sin^2 x} \right)$.
Using $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$:
$L = \lim _{x \rightarrow 0} \left( \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} + 2\cos x \right)$.
$L = \lim _{x \rightarrow 0} \left( \frac{1}{1 + \cos x} + 2\cos x \right)$.
Substituting $x = 0$:
$L = \frac{1}{1 + 1} + 2(1) = \frac{1}{2} + 2 = \frac{5}{2}$.

(C) Given $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$.
Substituting $x^2$ for $x$,we get:
$f(x^2) = \frac{5x^2 \operatorname{cosec}(x) - 1}{(x^2 - 2) \operatorname{cosec}(x)} = \frac{5x^2 \operatorname{cosec}(x)}{(x^2 - 2) \operatorname{cosec}(x)} - \frac{1}{(x^2 - 2) \operatorname{cosec}(x)}$.
$f(x^2) = \frac{5x^2}{x^2 - 2} - \frac{\sin(x)}{x^2 - 2}$.
Now,taking the limit as $x \rightarrow \infty$:
$\lim_{x \rightarrow \infty} f(x^2) = \lim_{x \rightarrow \infty} \left( \frac{5x^2}{x^2 - 2} - \frac{\sin(x)}{x^2 - 2} \right)$.
Since $\lim_{x \rightarrow \infty} \frac{5x^2}{x^2 - 2} = 5$ and $\lim_{x \rightarrow \infty} \frac{\sin(x)}{x^2 - 2} = 0$ (as $-1 \leq \sin(x) \leq 1$ and $x^2 - 2 \rightarrow \infty$),
$\lim_{x \rightarrow \infty} f(x^2) = 5 - 0 = 5$.

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$
Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \sin ^2 x$.
Using the identity $\sin (\pi - \theta) = \sin \theta$,we get $\sin (\pi - \pi \sin ^2 x) = \sin (\pi \sin ^2 x)$.
Now,the limit becomes: $\lim _{x \rightarrow 0} \frac{\sin (\pi \sin ^2 x)}{x^2}$.
Multiplying and dividing by $\pi \sin ^2 x$:
$\lim _{x}$ ${\rightarrow 0} \left( \frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} \right) \times \left( \frac{\pi \sin ^2 x}{x^2} \right)$.
As $x \rightarrow 0$,$\sin ^2 x \rightarrow 0$,so $\frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} \rightarrow 1$.
Also,$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} = 1$.
Therefore,the limit is $1 \times \pi \times 1 = \pi$.

(D) Given $\lim _{x \rightarrow 0} \frac{e^x-a-\log (1+x)}{\sin x}=0$.
For the limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$ because the denominator $\sin x \rightarrow 0$.
Substituting $x=0$ in the numerator: $e^0 - a - \log(1+0) = 0$.
$1 - a - 0 = 0$.
Therefore,$a = 1$.
Checking with $a=1$: $\lim _{x \rightarrow 0} \frac{e^x-1-\log (1+x)}{\sin x}$.
Using $L$'Hopital's rule: $\lim _{x \rightarrow 0} \frac{e^x - \frac{1}{1+x}}{\cos x} = \frac{1-1}{1} = 0$.
Thus,the condition holds for $a=1$.

(D) Given limit: $L = \lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = \lim _{x \rightarrow 0} \left[ \frac{e^x - 1 - x}{x(e^x - 1)} \right]$
This is in the $\frac{0}{0}$ indeterminate form.
Applying $L'\text{Hopital's rule}$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(xe^x - x)} = \lim _{x \rightarrow 0} \frac{e^x - 1}{e^x + xe^x - 1}$
This is still in the $\frac{0}{0}$ form.
Applying $L'\text{Hopital's rule}$ again:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(e^x + xe^x - 1)} = \lim _{x \rightarrow 0} \frac{e^x}{e^x + e^x + xe^x} = \lim _{x \rightarrow 0} \frac{e^x}{2e^x + xe^x}$
Substituting $x = 0$:
$= \frac{e^0}{2e^0 + 0 \cdot e^0} = \frac{1}{2 + 0} = \frac{1}{2}$

(C) The given limit is of the form $1^\infty$.
We use the formula $\lim _{x \rightarrow \infty} (f(x))^{g(x)} = e^{\lim _{x \rightarrow \infty} (f(x)-1)g(x)}$.
Here,$f(x) = \frac{3x^2-2x+3}{3x^2+x-2}$ and $g(x) = 3x-2$.
$(f(x)-1) = \frac{3x^2-2x+3 - (3x^2+x-2)}{3x^2+x-2} = \frac{-3x+5}{3x^2+x-2}$.
Now,$\lim _{x \rightarrow \infty} (f(x)-1)g(x) = \lim _{x \rightarrow \infty} \left(\frac{-3x+5}{3x^2+x-2}\right)(3x-2)$.
$= \lim _{x \rightarrow \infty} \frac{-9x^2+6x+15x-10}{3x^2+x-2} = \lim _{x \rightarrow \infty} \frac{-9x^2+21x-10}{3x^2+x-2}$.
Dividing the numerator and denominator by $x^2$,we get $\frac{-9}{3} = -3$.
Therefore,the limit is $e^{-3}$.

(D) We know that for any real number $y$,$y - 1 < [y] \leq y$.
Substituting $y = 2x - 3$,we get:
$(2x - 3) - 1 < [2x - 3] \leq 2x - 3$
$2x - 4 < [2x - 3] \leq 2x - 3$
Dividing the entire inequality by $x$ (for $x > 0$):
$\frac{2x - 4}{x} < \frac{[2x - 3]}{x} \leq \frac{2x - 3}{x}$
$2 - \frac{4}{x} < \frac{[2x - 3]}{x} \leq 2 - \frac{3}{x}$
Applying the limit as $x \rightarrow \infty$:
$\lim_{x \rightarrow \infty} (2 - \frac{4}{x}) \leq \lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} \leq \lim_{x \rightarrow \infty} (2 - \frac{3}{x})$
$2 - 0 \leq \lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} \leq 2 - 0$
By the Squeeze Theorem,$\lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} = 2$.

(A) The given data is $20, 5, 15, 2, 7, 3, 11$. The number of observations $n = 7$.
First,calculate the mean $\bar{x} = \frac{20+5+15+2+7+3+11}{7} = \frac{63}{7} = 9$.
Mean deviation about mean $m = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|20-9|+|5-9|+|15-9|+|2-9|+|7-9|+|3-9|+|11-9|}{7} = \frac{11+4+6+7+2+4+2}{7} = \frac{36}{7}$.
Wait,let us re-calculate: $11+4+6+7+2+4+2 = 36$. So $m = \frac{36}{7}$.
Now,for median,arrange data in ascending order: $2, 3, 5, 7, 11, 15, 20$.
Median $= \left(\frac{7+1}{2}\right)^{\text{th}}$ term $= 4^{\text{th}}$ term $= 7$.
Mean deviation about median $M = \frac{\sum |x_i - \text{Median}|}{n} = \frac{|2-7|+|3-7|+|5-7|+|7-7|+|11-7|+|15-7|+|20-7|}{7} = \frac{5+4+2+0+4+8+13}{7} = \frac{36}{7}$.
Wait,let us re-calculate: $5+4+2+0+4+8+13 = 36$. So $M = \frac{36}{7}$.
Actually,$m = \frac{36}{7}$ and $M = \frac{36}{7}$.
The mean of $m$ and $M$ is $\bar{x}^{\prime} = \frac{m+M}{2} = \frac{36/7 + 36/7}{2} = \frac{36}{7}$.
The mean deviation about the mean of $m$ and $M$ is $\frac{|m-\bar{x}^{\prime}| + |M-\bar{x}^{\prime}|}{2} = \frac{|36/7 - 36/7| + |36/7 - 36/7|}{2} = 0$.
Given the options,there might be a calculation error in the provided question's data or options. Re-evaluating the sum for $m$: $|20-9|=11, |5-9|=4, |15-9|=6, |2-9|=7, |7-9|=2, |3-9|=6, |11-9|=2$. Sum $= 11+4+6+7+2+6+2 = 38$. So $m = \frac{38}{7}$.
Re-evaluating the sum for $M$: $|20-7|=13, |5-7|=2, |15-7|=8, |2-7|=5, |7-7|=0, |3-7|=4, |11-7|=4$. Sum $= 13+2+8+5+0+4+4 = 36$. So $M = \frac{36}{7}$.
Mean of $m$ and $M$ is $\bar{x}^{\prime} = \frac{38/7 + 36/7}{2} = \frac{74/7}{2} = \frac{37}{7}$.
Mean deviation about mean of $m$ and $M$ is $\frac{|38/7 - 37/7| + |36/7 - 37/7|}{2} = \frac{1/7 + 1/7}{2} = \frac{2/7}{2} = \frac{1}{7}$.

(A) Consistency is measured by the coefficient of variation $(CV)$. $A$ lower $CV$ implies higher consistency.
Batsman $A$ is more consistent than $B$,so $CV_{A} < CV_{B}$.
This implies $\frac{\sigma_{A}}{\bar{x}} < \frac{\sigma_{B}}{\bar{y}}$.
Rearranging this gives $\frac{\sigma_{A}}{\sigma_{B}} < \frac{\bar{x}}{\bar{y}}$.
Since $\sigma_{A}, \sigma_{B}, \bar{x}, \bar{y} > 0$,we have $0 < \frac{\sigma_{A}}{\sigma_{B}} < \frac{\bar{x}}{\bar{y}}$.
For $A$ to be a higher run scorer,we must have $\bar{x} > \bar{y}$,which implies $\frac{\bar{x}}{\bar{y}} > 1$.

(D) Given the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$.
Mean $= \frac{7+8+9+7+8+7+\lambda+8}{8} = 8$
$\Rightarrow \frac{54+\lambda}{8} = 8$
$\Rightarrow 54+\lambda = 64$
$\Rightarrow \lambda = 10$
Now,the data set is $7, 8, 9, 7, 8, 7, 10, 8$.
Variance $(\sigma^2) = \frac{1}{n} \sum (x_i - \bar{x})^2$
Variance $= \frac{(7-8)^2 + (8-8)^2 + (9-8)^2 + (7-8)^2 + (8-8)^2 + (7-8)^2 + (10-8)^2 + (8-8)^2}{8}$
Variance $= \frac{(-1)^2 + 0^2 + 1^2 + (-1)^2 + 0^2 + (-1)^2 + 2^2 + 0^2}{8}$
Variance $= \frac{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0}{8} = \frac{8}{8} = 1$.

(B) Step $1$: Find the class midpoints $(x_i)$ :
The midpoint for a class interval is calculated as: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$
- $x_1 = \frac{0+6}{2} = 3$
- $x_2 = \frac{6+12}{2} = 9$
- $x_3 = \frac{12+18}{2} = 15$
- $x_4 = \frac{18+24}{2} = 21$
- $x_5 = \frac{24+30}{2} = 27$
Thus,$x_i = 3, 9, 15, 21, 27$.
Step $2$: Compute the mean $(\bar{x})$ :
The mean is calculated as: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
- $\sum f_i = 1+2+3+2+1 = 9$
- $\sum f_i x_i = (1 \cdot 3) + (2 \cdot 9) + (3 \cdot 15) + (2 \cdot 21) + (1 \cdot 27) = 3 + 18 + 45 + 42 + 27 = 135$
- $\bar{x} = \frac{135}{9} = 15$
Step $3$: Find $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$ :
- For $x_1 = 3: |3-15| = 12, f_1 |x_i - \bar{x}| = 1 \cdot 12 = 12$
- For $x_2 = 9: |9-15| = 6, f_2 |x_i - \bar{x}| = 2 \cdot 6 = 12$
- For $x_3 = 15: |15-15| = 0, f_3 |x_i - \bar{x}| = 3 \cdot 0 = 0$
- For $x_4 = 21: |21-15| = 6, f_4 |x_i - \bar{x}| = 2 \cdot 6 = 12$
- For $x_5 = 27: |27-15| = 12, f_5 |x_i - \bar{x}| = 1 \cdot 12 = 12$
Step $4$: Calculate mean deviation about the mean:
- $\text{Mean deviation} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{12+12+0+12+12}{9} = \frac{48}{9} = \frac{16}{3}$

(D)

$x_i$	$f_i$	$f_i x_i$	$D = x_i - \mu$	$f_i D^2$
$4$	$1$	$4$	$2$	$4$
$3$	$3$	$9$	$1$	$3$
$1$	$5$	$5$	$-1$	$5$
Total	$9$	$18$	-	$12$

Mean $(\mu) = \frac{\sum f_i x_i}{\sum f_i} = \frac{18}{9} = 2$
Standard deviation $(\sigma) = \sqrt{\frac{\sum f_i (x_i - \mu)^2}{\sum f_i}} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$
Coefficient of variation $= \frac{\sigma}{\mu} \times 100 = \frac{2/\sqrt{3}}{2} \times 100 = \frac{100}{\sqrt{3}}$.

(D) The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.
The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 \times n(n+1)}{2n} = n+1$.
The variance is given by $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (2i)^2 - (n+1)^2$.
$\sigma^2 = \frac{4}{n} \times \frac{n(n+1)(2n+1)}{6} - (n+1)^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2$.
$\sigma^2 = (n+1) \left[ \frac{4n+2-3n-3}{3} \right] = \frac{(n+1)(n-1)}{3} = \frac{n^2-1}{3}$.
Thus,Statement-$I$ is false.
For $n=20$,the mean is $20+1 = 21$.
The variance is $\frac{20^2-1}{3} = \frac{399}{3} = 133$.
The difference is $133 - 21 = 112$.
Thus,Statement-$II$ is true.

(B) We know that the coefficient of variation $(CV)$ is given by the formula:
$CV = \frac{\sigma}{|\bar{x}|} \times 100$
Given $CV = 25$ and mean $\bar{x} = 44$.
Substituting the values:
$25 = \frac{\sigma}{44} \times 100$
$\sigma = \frac{25 \times 44}{100} = \frac{1100}{100} = 11$
Now,the variance is defined as the square of the standard deviation:
$\text{Variance} = \sigma^2 = 11^2 = 121$

(D) First,we calculate the mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4(1) + 24(3) + 28(5) + 16(7) + 8(9)}{4 + 24 + 28 + 16 + 8} = \frac{4 + 72 + 140 + 112 + 72}{80} = \frac{400}{80} = 5$.
Next,the mean deviation about the mean $m = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{4|1-5| + 24|3-5| + 28|5-5| + 16|7-5| + 8|9-5|}{80} = \frac{4(4) + 24(2) + 28(0) + 16(2) + 8(4)}{80} = \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = \frac{8}{5} = 1.6$.
The variance $\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - (\bar{x})^2 = \frac{4(1)^2 + 24(3)^2 + 28(5)^2 + 16(7)^2 + 8(9)^2}{80} - 5^2 = \frac{4 + 216 + 700 + 784 + 648}{80} - 25 = \frac{2352}{80} - 25 = 29.4 - 25 = 4.4 = \frac{22}{5}$.
Finally,$m + \sigma^2 = \frac{8}{5} + \frac{22}{5} = \frac{30}{5} = 6$.

(A) Given that $\sum(x_i+2)^2 = 28n$
$\Rightarrow \sum x_i^2 + 4\sum x_i + 4n = 28n$
$\Rightarrow \sum x_i^2 + 4\sum x_i = 24n$ $... (i)$
Similarly,$\sum(x_i-2)^2 = 12n$
$\Rightarrow \sum x_i^2 - 4\sum x_i + 4n = 12n$
$\Rightarrow \sum x_i^2 - 4\sum x_i = 8n$ $... (ii)$
Adding $(i)$ and $(ii)$:
$2\sum x_i^2 = 32n \Rightarrow \sum x_i^2 = 16n$
Subtracting $(ii)$ from $(i)$:
$8\sum x_i = 16n \Rightarrow \sum x_i = 2n$
The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$
$\sigma^2 = \frac{16n}{n} - \left(\frac{2n}{n}\right)^2 = 16 - 4 = 12$

(D) Given,in $\triangle ABC$,$r_1 = 2r_2 = 3r_3$.
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Let $\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{3\Delta}{s-c} = \frac{1}{K}$.
Then $s-a = K$,$s-b = 2K$,and $s-c = 3K$.
Adding these,$(s-a) + (s-b) + (s-c) = K + 2K + 3K = 6K$.
$3s - (a+b+c) = 6K$ $\Rightarrow 3s - 2s = 6K$ $\Rightarrow s = 6K$.
Thus,$a = s - K = 5K$,$b = s - 2K = 4K$,and $c = s - 3K = 3K$.
Using the sine rule,$\sin A : \sin B : \sin C = a : b : c = 5K : 4K : 3K = 5 : 4 : 3$.

(C) Given,in $\triangle ABC$,$B=90^{\circ}$.
Using the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.
Since $B=90^{\circ}$,$\frac{\sin 90^{\circ}}{b} = \frac{1}{2R} \implies b = 2R$.
The inradius $r$ is given by $r = (s-b) \tan(\frac{B}{2})$.
Substituting $B=90^{\circ}$,$r = (s-b) \tan(45^{\circ}) = s-b$.
Since $s = \frac{a+b+c}{2}$,we have $r = \frac{a+b+c}{2} - b = \frac{a-b+c}{2}$.
Thus,$2r = a-b+c$.
Finally,$2(r+R) = 2r + 2R = (a-b+c) + b = a+c$.

(B) Given $(a+c)^2 = b^2 + 3ca$,we expand the left side: $a^2 + c^2 + 2ac = b^2 + 3ca$.
Rearranging gives $a^2 + c^2 - b^2 = ca$.
Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{ca}{2ac} = \frac{1}{2}$.
Thus,$B = 60^{\circ}$,which implies $\frac{B}{2} = 30^{\circ}$.
Using the Sine Rule,$\frac{a}{2R} = \sin A$ and $\frac{c}{2R} = \sin C$,so $\frac{a+c}{2R} = \sin A + \sin C$.
Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$.
Since $A+B+C = 180^{\circ}$,$\frac{A+C}{2} = 90^{\circ} - \frac{B}{2}$.
Therefore,$\frac{a+c}{2R} = 2 \sin \left(90^{\circ} - \frac{B}{2}\right) \cos \left(\frac{A-C}{2}\right) = 2 \cos \left(\frac{B}{2}\right) \cos \left(\frac{A-C}{2}\right)$.
Substituting $B/2 = 30^{\circ}$,we get $2 \cos 30^{\circ} \cos \left(\frac{A-C}{2}\right) = 2 \left(\frac{\sqrt{3}}{2}\right) \cos \left(\frac{A-C}{2}\right) = \sqrt{3} \cos \left(\frac{A-C}{2}\right)$.

(D) Given $r: R: r_2 = 1: 3: 7$. Let $r = k, R = 3k, r_2 = 7k$.
We know that $r_2 - r = 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \sin \frac{B}{2} \cos \left( \frac{A+C}{2} \right)$.
Substituting the values: $7k - k = 4(3k) \sin \frac{B}{2} \cos \left( \frac{\pi - B}{2} \right)$.
$6k = 12k \sin \frac{B}{2} \sin \frac{B}{2} = 12k \sin^2 \frac{B}{2}$.
$\sin^2 \frac{B}{2} = \frac{6k}{12k} = \frac{1}{2}$.
Since $\sin^2 \frac{B}{2} = \frac{1 - \cos B}{2}$,we have $\frac{1 - \cos B}{2} = \frac{1}{2}$,which implies $\cos B = 0$,so $B = 90^{\circ}$.
Now,$\sin(A+C) + \sin B = \sin(\pi - B) + \sin B = \sin B + \sin B = 2 \sin 90^{\circ} = 2(1) = 2$.

(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = K$.
Thus,$a = K \sin A$ and $b = K \sin B$.
Substituting these into the expression:
$a^2 \sin 2B + b^2 \sin 2A = (K \sin A)^2 (2 \sin B \cos B) + (K \sin B)^2 (2 \sin A \cos A)$
$= 2K^2 \sin A \sin B (\sin A \cos B + \cos A \sin B)$
$= 2(K \sin A)(K \sin B) \sin(A + B)$
$= 2ab \sin(A + B)$
Since $A + B + C = \pi$,we have $\sin(A + B) = \sin(\pi - C) = \sin C$.
Therefore,the expression equals $2ab \sin C$.

(D) We know that in any $\triangle ABC$,$\cos A + \cos B + \cos C = 1 + 4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$.
Using the identity for the inradius $r = 4R \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$,we can substitute the product of sines.
Thus,$4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right) = \frac{r}{R}$.
Therefore,$\cos A + \cos B + \cos C = 1 + \frac{r}{R}$.

(D) Given in $\triangle ABC$:
$a=26, b=30, \cos C=\frac{63}{65}$.
Using the Law of Cosines:
$\cos C = \frac{a^2+b^2-c^2}{2ab}$
$\frac{63}{65} = \frac{26^2+30^2-c^2}{2 \times 26 \times 30}$
$\frac{63}{65} = \frac{676+900-c^2}{1560}$
$c^2 = 1576 - \frac{63 \times 1560}{65}$
$c^2 = 1576 - (63 \times 24)$
$c^2 = 1576 - 1512 = 64$
$c = \sqrt{64} = 8$.

(B) Let the angles of the triangle be $A - d, A, A + d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.
Thus,the angles are $60^{\circ} - d, 60^{\circ}, 60^{\circ} + d$.
The sides opposite to these angles are $a, b, c$. Given two sides are $7$ and $8$. Let the sides be $a, 7, 8$ where $a$ is the smallest side.
Using the Law of Cosines for the angle $60^{\circ}$:
Case $1$: If the angle $60^{\circ}$ is between sides $7$ and $8$,then $a^2 = 7^2 + 8^2 - 2(7)(8) \cos 60^{\circ} = 49 + 64 - 56 = 57$. This gives $a = \sqrt{57} \approx 7.55$,which is not smaller than $7$.
Case $2$: If the angle $60^{\circ}$ is opposite to side $7$,then $8^2 = a^2 + 7^2 - 2(a)(7) \cos 60^{\circ}$ $\Rightarrow 64 = a^2 + 49 - 7a$ $\Rightarrow a^2 - 7a - 15 = 0$. This yields $a = \frac{7 \pm \sqrt{49 + 60}}{2} = \frac{7 \pm \sqrt{109}}{2}$. Only the positive value is valid,$a \approx 8.72 > 7$.
Case $3$: If the angle $60^{\circ}$ is opposite to side $8$,then $7^2 = a^2 + 8^2 - 2(a)(8) \cos 60^{\circ}$ $\Rightarrow 49 = a^2 + 64 - 8a$ $\Rightarrow a^2 - 8a + 15 = 0$.
Solving $a^2 - 8a + 15 = 0$ gives $(a - 3)(a - 5) = 0$,so $a = 3$ or $a = 5$.
Since $a$ is the smallest side,both $3 < 7$ and $5 < 7$ are valid.
Thus,$a_1 = 3$ and $a_2 = 5$.
Finally,$2 a_1 + 3 a_2 = 2(3) + 3(5) = 6 + 15 = 21$.

(C) Given $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$.
Using the formula $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$,we have $\frac{s(s-c)}{ab} = \frac{9}{13}$.
Substituting $a=13, b=14$,we get $\frac{s(s-c)}{182} = \frac{9}{13}$,so $s(s-c) = 126$.
Since $s = \frac{a+b+c}{2} = \frac{27+c}{2}$,then $s-c = \frac{27-c}{2}$.
Thus,$\left(\frac{27+c}{2}\right)\left(\frac{27-c}{2}\right) = 126$ $\Rightarrow 729-c^2 = 504$ $\Rightarrow c^2 = 225$ $\Rightarrow c=15$.
Then $s = \frac{13+14+15}{2} = 21$.
The exradius $r_1$ is given by $r_1 = \frac{\Delta}{s-a}$.
Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.
So,$r_1 = \frac{84}{21-13} = \frac{84}{8} = 10.5$.
Therefore,$2r_1 = 2 \times 10.5 = 21 = s$.

(A) Given $b+c = 7k$,$c+a = 8k$,and $a+b = 9k$.
Adding these equations,we get $2(a+b+c) = 24k$,which implies $a+b+c = 12k$.
Subtracting the given equations from $a+b+c = 12k$:
$a = (a+b+c) - (b+c) = 12k - 7k = 5k$.
$b = (a+b+c) - (c+a) = 12k - 8k = 4k$.
$c = (a+b+c) - (a+b) = 12k - 9k = 3k$.
Since $c < b < a$,the smallest angle is $C$ (opposite to side $c$).
Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
$\cos C = \frac{(5k)^2 + (4k)^2 - (3k)^2}{2(5k)(4k)} = \frac{25k^2 + 16k^2 - 9k^2}{40k^2} = \frac{32k^2}{40k^2} = \frac{4}{5}$.
Therefore,the smallest angle is $C = \cos^{-1}\left(\frac{4}{5}\right)$.

(C) Given $(a-b)(s-c)=(b-c)(s-a)$.
Using the relation $s-a = \frac{\Delta}{r_1}$,$s-b = \frac{\Delta}{r_2}$,and $s-c = \frac{\Delta}{r_3}$,we have $a-b = (s-b)-(s-a) = \Delta(\frac{1}{r_2} - \frac{1}{r_1}) = \Delta \frac{r_1-r_2}{r_1 r_2}$.
Substituting these into the given equation:
$\Delta \frac{r_1-r_2}{r_1 r_2} \cdot \frac{\Delta}{r_3} = \Delta \frac{r_2-r_3}{r_2 r_3} \cdot \frac{\Delta}{r_1}$.
Canceling $\frac{\Delta^2}{r_1 r_2 r_3}$ from both sides,we get:
$r_1-r_2 = r_2-r_3$.
Therefore,$r_1+r_3 = 2r_2$.

(B) Given $r_1=4, r_2=8, r_3=24$.
We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6+3+1}{24} = \frac{10}{24} = \frac{5}{12}$.
Thus,$r = \frac{12}{5}$.
We know that $\Delta^2 = r r_1 r_2 r_3 = \frac{12}{5} \times 4 \times 8 \times 24 = \frac{9216}{5}$.
So,$\Delta = \sqrt{\frac{9216}{5}} = \frac{96}{\sqrt{5}}$.
Using $r = \frac{\Delta}{s}$,we get $s = \frac{\Delta}{r} = \frac{96}{\sqrt{5}} \times \frac{5}{12} = 8\sqrt{5}$.
Since $r_1 = \frac{\Delta}{s-a}$,we have $4 = \frac{96/\sqrt{5}}{8\sqrt{5}-a}$.
$4(8\sqrt{5}-a) = \frac{96}{\sqrt{5}} \Rightarrow 32\sqrt{5} - 4a = \frac{96}{\sqrt{5}}$.
$4a = 32\sqrt{5} - \frac{96}{\sqrt{5}} = \frac{160-96}{\sqrt{5}} = \frac{64}{\sqrt{5}}$.
$a = \frac{16}{\sqrt{5}}$.

(B) $\sum \cot A = \cot A + \cot B + \cot C = \frac{b^2+c^2-a^2}{4\Delta} + \frac{c^2+a^2-b^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta} = \frac{a^2+b^2+c^2}{4\Delta}$. Thus,$(A)$ matches $(ii)$.
$(B)$ $\sum \cot \frac{A}{2} = \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta} = \frac{s}{\Delta}(3s - (a+b+c)) = \frac{s}{\Delta}(3s - 2s) = \frac{s^2}{\Delta} = \frac{(a+b+c)^2}{4\Delta}$. Thus,$(B)$ matches $(i)$.
$(C)$ Given $\tan A : \tan B : \tan C = 1 : 2 : 3$. Let $\tan A = k, \tan B = 2k, \tan C = 3k$. Since $A+B+C = \pi$,$\tan A + \tan B + \tan C = \tan A \tan B \tan C \Rightarrow 6k = 6k^3 \Rightarrow k=1$. So $\tan A = 1, \tan B = 2, \tan C = 3$. Then $\sin A = \frac{1}{\sqrt{2}}, \sin B = \frac{2}{\sqrt{5}}, \sin C = \frac{3}{\sqrt{10}}$. Ratio $\sin A : \sin B : \sin C = \frac{1}{\sqrt{2}} : \frac{2}{\sqrt{5}} : \frac{3}{\sqrt{10}} = \sqrt{5} : 2\sqrt{2} : 3$. Thus,$(C)$ matches $(v)$.
$(D)$ Given $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$. Since $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,we have $(s-a) : (s-b) : (s-c) = 3 : 7 : 9$. Let $s-a=3k, s-b=7k, s-c=9k$. Adding gives $3s - (a+b+c) = 19k \Rightarrow s = 19k$. Then $a = 16k, b = 12k, c = 10k$. Ratio $a : b : c = 16 : 12 : 10 = 8 : 6 : 5$. Thus,$(D)$ matches $(iii)$.

(A) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Consider the expression $rr_1 + r_2 r_3 = \frac{\Delta^2}{s(s-a)} + \frac{\Delta^2}{(s-b)(s-c)}$.
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $\frac{\Delta^2}{s(s-a)} = (s-b)(s-c)$ and $\frac{\Delta^2}{(s-b)(s-c)} = s(s-a)$.
Thus,$rr_1 + r_2 r_3 = (s-b)(s-c) + s(s-a)$.
Expanding this,we get $s^2 - s(b+c) + bc + s^2 - sa = 2s^2 - s(a+b+c) + bc$.
Since $2s = a+b+c$,we have $2s^2 - s(2s) + bc = 2s^2 - 2s^2 + bc = bc$.
Therefore,$rr_1 + r_2 r_3 = bc$,which implies $bc - r_2 r_3 = rr_1$.

(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$ and $\Delta^2 = s(s-a)(s-b)(s-c)$.
Substituting these into the expression:
$\frac{r_2(r_1+r_3)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} = \frac{\frac{\Delta}{s-b}(\frac{\Delta}{s-a} + \frac{\Delta}{s-c})}{\sqrt{\frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}}}$
$= \frac{\frac{\Delta^2}{s-b} \cdot \frac{s-c+s-a}{(s-a)(s-c)}}{\Delta \sqrt{\frac{s-c+s-a+s-b}{(s-a)(s-b)(s-c)}}}$
$= \frac{\Delta \cdot b}{(s-a)(s-b)(s-c)} \cdot \sqrt{\frac{(s-a)(s-b)(s-c)}{3s-(a+b+c)}}$
Since $a+b+c = 2s$,the denominator inside the square root is $3s-2s = s$.
$= \frac{\Delta \cdot b}{(s-a)(s-b)(s-c)} \cdot \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}$
$= \frac{\Delta \cdot b}{\sqrt{s(s-a)(s-b)(s-c)}} = \frac{\Delta \cdot b}{\Delta} = b$.

(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.
Also,$\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$.
Thus,$(r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} = \left(\frac{\Delta}{s-b} + \frac{\Delta}{s-c}\right) \times \frac{bc}{(s-b)(s-c)}$.
$= \Delta \left(\frac{s-c+s-b}{(s-b)(s-c)}\right) \times \frac{bc}{(s-b)(s-c)} = \Delta \left(\frac{a}{(s-b)(s-c)}\right) \times \frac{bc}{(s-b)(s-c)}$.
$= \frac{\Delta abc}{(s-b)^2(s-c)^2} = \frac{4R \Delta^2}{(s-b)^2(s-c)^2}$.
$= 4R \left(\frac{\Delta}{(s-b)(s-c)}\right)^2 = 4R \left(\cot \frac{A}{2}\right)^2 = 4R \cot^2 \frac{A}{2}$.

(B) The line $2x + 5y + \alpha = 0$ intersects the coordinate axes at points $A$ and $B$. Since the triangle is formed with the positive coordinate axes,the intercepts must be positive. Let $\alpha = -k$ where $k > 0$. The equation becomes $2x + 5y = k$,or $\frac{x}{k/2} + \frac{y}{k/5} = 1$.
Thus,the vertices of the right-angled triangle are $O(0, 0)$,$A(\frac{k}{2}, 0)$,and $B(0, \frac{k}{5})$.
The hypotenuse $AB$ is the diameter of the circum-circle. The length of the hypotenuse is $d = \sqrt{(\frac{k}{2})^2 + (\frac{k}{5})^2} = \sqrt{\frac{k^2}{4} + \frac{k^2}{25}} = \sqrt{\frac{29k^2}{100}} = \frac{k\sqrt{29}}{10}$.
The radius $r$ of the circum-circle is $\frac{d}{2} = \frac{k\sqrt{29}}{20}$.
The area of the circum-circle is $\pi r^2 = \pi \left(\frac{k^2 \cdot 29}{400}\right) = \frac{29\pi k^2}{400}$.
Given the area is $\frac{29\pi}{4}$,we have $\frac{29\pi k^2}{400} = \frac{29\pi}{4}$.
This simplifies to $k^2 = 100$,so $k = 10$.
Since $k = |\alpha|$,we have $|\alpha| = 10$.

(D) The area of an equilateral triangle with side $a$ is $\Delta = \frac{\sqrt{3}}{4} a^2$.
The semi-perimeter of the triangle is $s = \frac{3a}{2}$.
The radius $r$ of the incircle is given by $r = \frac{\Delta}{s} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{3a}{2}} = \frac{a}{2 \sqrt{3}}$.
$A$ square inscribed in a circle of radius $r$ has a diagonal equal to the diameter of the circle,which is $2r$.
Diagonal of the square $= 2 \times \frac{a}{2 \sqrt{3}} = \frac{a}{\sqrt{3}}$.
The area of a square with diagonal $d$ is $\frac{d^2}{2}$.
Area of the square $= \frac{(\frac{a}{\sqrt{3}})^2}{2} = \frac{\frac{a^2}{3}}{2} = \frac{a^2}{6}$.

(D) Given that $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$.
Dividing by $r_2 r_3$,we get $(1-\frac{r_1}{r_2})(1-\frac{r_1}{r_3})=2$.
Using $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$,we have $(1-\frac{s-b}{s-a})(1-\frac{s-c}{s-a})=2$.
This simplifies to $(\frac{s-a-s+b}{s-a})(\frac{s-a-s+c}{s-a})=2$,which is $(b-a)(c-a)=2(s-a)^2$.
Since $2(s-a) = b+c-a$,we have $(b-a)(c-a) = \frac{1}{2}(b+c-a)^2$.
Expanding this leads to $b^2+c^2-a^2=0$,so $a^2=b^2+c^2$,meaning $\angle A=90^{\circ}$.
We know $2(r+R) = 2r+2R = (b+c-a) + a = b+c$.
Since $b=2R \sin B$ and $c=2R \sin C$,$b+c = 2R(\sin B + \sin C) = 2R(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2})$.
Since $A=90^{\circ}$,$B+C=90^{\circ}$,so $\sin \frac{B+C}{2} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Thus,$2(r+R) = 4R(\frac{1}{\sqrt{2}}) \cos \frac{B-C}{2} = 2 \sqrt{2} R \cos \frac{B-C}{2}$.

(C) Given that $A, B, C$ are in arithmetic progression,so $A+C = 2B$. Since $A+B+C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Given $r_1 r_2 = r_3 r$,we use the formulas $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.
Substituting these,we get $\frac{\Delta^2}{(s-a)(s-b)} = \frac{\Delta^2}{s(s-c)}$,which simplifies to $(s-a)(s-b) = s(s-c)$.
This is equivalent to $\tan^2 \frac{C}{2} = 1$,so $\frac{C}{2} = 45^{\circ}$,which means $C = 90^{\circ}$.
Since $B = 60^{\circ}$ and $C = 90^{\circ}$,then $A = 30^{\circ}$.
The area $\Delta = \frac{1}{2} ab \sin C = \frac{1}{2} (2R \sin A)(2R \sin B) \sin 90^{\circ} = 2R^2 \sin 30^{\circ} \sin 60^{\circ} = 2R^2 (\frac{1}{2}) (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} R^2$.
Given $\Delta = \frac{\sqrt{3}}{2}$,we have $\frac{\sqrt{3}}{2} R^2 = \frac{\sqrt{3}}{2}$,so $R^2 = 1$,which gives $R = 1$.

(B) Given sides of the triangle are $a=13, b=14, c=15$.
First,calculate the semi-perimeter $s$:
$s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$.
Next,calculate the area of the triangle $\Delta$ using Heron's formula:
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.
The exradius $r_1$ is given by the formula $r_1 = \frac{\Delta}{s-a}$:
$r_1 = \frac{84}{21-13} = \frac{84}{8} = \frac{21}{2}$.

(C) In $\triangle ABC$,$A+B+C=\pi$.
We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$ and $r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$.
Adding these,$r_1+r_2 = 4R \cos \frac{C}{2} [\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}]$.
Using the identity $\sin(x+y) = \sin x \cos y + \cos x \sin y$,we get $r_1+r_2 = 4R \cos \frac{C}{2} \sin(\frac{A+B}{2})$.
Since $A+B = \pi - C$,$\sin(\frac{A+B}{2}) = \sin(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2}$.
Thus,$r_1+r_2 = 4R \cos^2 \frac{C}{2}$.
Now,$(r_1+r_2) \operatorname{cosec}^2 \frac{C}{2} = \frac{4R \cos^2 \frac{C}{2}}{\sin^2 \frac{C}{2}} = 4R \cot^2 \frac{C}{2}$.

(C) Given $n(A) = 8$.
The number of subsets of $A$ containing at least $6$ elements is given by the sum of combinations:
$^8C_6 + ^8C_7 + ^8C_8$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$:
$^8C_6 = \frac{8 \times 7}{2 \times 1} = 28$
$^8C_7 = \frac{8}{1} = 8$
$^8C_8 = 1$
Total subsets $= 28 + 8 + 1 = 37$.

(A) We know that the formula for the inverse hyperbolic cosine function is given by:
$\cosh^{-1}(x) = \log(x + \sqrt{x^2 - 1})$
Substituting $x = 2$ into the formula:
$\cosh^{-1}(2) = \log(2 + \sqrt{2^2 - 1})$
$\cosh^{-1}(2) = \log(2 + \sqrt{4 - 1})$
$\cosh^{-1}(2) = \log(2 + \sqrt{3})$

(C) Given $\cosh x = K$ and $\sinh x = \tan \theta$.
We know the identity for hyperbolic functions: $\cosh^2 x - \sinh^2 x = 1$.
Substituting the given values: $K^2 - \sinh^2 x = 1$.
Therefore,$\sinh^2 x = K^2 - 1$,which implies $\sinh x = \sqrt{K^2 - 1}$.
Since $\sinh x = \tan \theta$,we have $\tan \theta = \sqrt{K^2 - 1} = \frac{\sqrt{K^2 - 1}}{1}$.
In a right-angled triangle,$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{\sqrt{K^2 - 1}}{1}$.
The Hypotenuse $H = \sqrt{(\sqrt{K^2 - 1})^2 + 1^2} = \sqrt{K^2 - 1 + 1} = \sqrt{K^2} = K$.
Thus,$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{\sqrt{K^2 - 1}}{K}$.

(D) $I$. Evaluate $(S1): \sin 55^{\circ} + \sin 53^{\circ} - \sin 19^{\circ} - \sin 17^{\circ}$
Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= (\sin 55^{\circ} - \sin 17^{\circ}) + (\sin 53^{\circ} - \sin 19^{\circ})$
$= 2 \cos 36^{\circ} \sin 19^{\circ} + 2 \cos 36^{\circ} \sin 17^{\circ}$
$= 2 \cos 36^{\circ} (\sin 19^{\circ} + \sin 17^{\circ})$
$= 2 \cos 36^{\circ} (2 \sin 18^{\circ} \cos 1^{\circ})$
Since $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:
$= 2 \left(\frac{\sqrt{5}+1}{4}\right) \cdot 2 \left(\frac{\sqrt{5}-1}{4}\right) \cos 1^{\circ}$
$= 4 \left(\frac{5-1}{16}\right) \cos 1^{\circ} = \cos 1^{\circ}$.
Thus,$(S1)$ is false as $\cos 1^{\circ} \neq \cos 2^{\circ}$.
$II$. Evaluate $(S2): f(x) = \frac{1}{3 - \cos 2x}$
Since $-1 \leq \cos 2x \leq 1$,we have $-1 \leq -\cos 2x \leq 1$.
Adding $3$: $2 \leq 3 - \cos 2x \leq 4$.
Taking reciprocals: $\frac{1}{4} \leq \frac{1}{3 - \cos 2x} \leq \frac{1}{2}$.
Thus,$(S2)$ is true.

(B) For the first expression,we know that $-\sqrt{11^2 + 60^2} \leq 11 \cos 2x + 60 \sin 2x \leq \sqrt{11^2 + 60^2}$.
This simplifies to $-61 \leq 11 \cos 2x + 60 \sin 2x \leq 61$.
To maximize $\frac{1}{11 \cos 2x + 60 \sin 2x + 69}$,we need to minimize the denominator.
The minimum value of the denominator is $69 - 61 = 8$.
Thus,$M_1 = \frac{1}{8}$.
For the second expression,$3 \cos^2 5x + 4 \sin^2 5x = 3(\cos^2 5x + \sin^2 5x) + \sin^2 5x = 3 + \sin^2 5x$.
The maximum value occurs when $\sin^2 5x = 1$,so $M_2 = 3 + 1 = 4$.
Therefore,$\frac{M_1}{M_2} = \frac{1/8}{4} = \frac{1}{32}$.

(C) function $f(x)$ is odd if $f(-x) = -f(x)$.
$I. f(x) = x \left( \frac{e^x-1}{e^x+1} \right)$
$f(-x) = (-x) \left( \frac{e^{-x}-1}{e^{-x}+1} \right) = (-x) \left( \frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+1} \right) = (-x) \left( \frac{1-e^x}{1+e^x} \right) = x \left( \frac{e^x-1}{e^x+1} \right) = f(x)$.
Since $f(-x) = f(x)$,$I$ is an even function.
$II. f(x) = k^x + k^{-x} + \cos x$
$f(-x) = k^{-x} + k^x + \cos(-x) = k^{-x} + k^x + \cos x = f(x)$.
Since $f(-x) = f(x)$,$II$ is an even function.
$III. f(x) = \log \left(x+\sqrt{x^2+1}\right)$
$f(-x) = \log \left(-x+\sqrt{(-x)^2+1}\right) = \log \left(\sqrt{x^2+1}-x\right)$
Multiplying and dividing by $\sqrt{x^2+1}+x$:
$f(-x) = \log \left( \frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x} \right) = \log \left( \frac{x^2+1-x^2}{\sqrt{x^2+1}+x} \right) = \log \left( \frac{1}{\sqrt{x^2+1}+x} \right)$
$f(-x) = \log \left( (x+\sqrt{x^2+1})^{-1} \right) = -\log \left( x+\sqrt{x^2+1} \right) = -f(x)$.
Since $f(-x) = -f(x)$,$III$ is an odd function.
Therefore,only $III$ is an odd function.

(C) The periods of $\tan 5x$,$\cos 3x$,and $\sin 6x$ are $\frac{\pi}{5}$,$\frac{2\pi}{3}$,and $\frac{\pi}{3}$ respectively.
To find the period $\alpha$ of $f(x) = \frac{\tan 5x \cos 3x}{\sin 6x}$,we simplify the expression:
$f(x) = \frac{\tan 5x \cos 3x}{2 \sin 3x \cos 3x} = \frac{\tan 5x}{2 \sin 3x}$.
The period of $\tan 5x$ is $\frac{\pi}{5}$ and the period of $\sin 3x$ is $\frac{2\pi}{3}$.
The period $\alpha$ of $f(x)$ is the least common multiple of $\frac{\pi}{5}$ and $\frac{2\pi}{3}$,which is $\frac{2\pi}{\gcd(1, 1)} = 2\pi$.
Thus,$\alpha = 2\pi$.
We need to evaluate $f\left(\frac{\alpha}{8}\right) = f\left(\frac{2\pi}{8}\right) = f\left(\frac{\pi}{4}\right)$.
$f\left(\frac{\pi}{4}\right) = \frac{\tan(5\pi/4)}{2 \sin(3\pi/4)} = \frac{1}{2 \times (1/\sqrt{2})} = \frac{1}{\sqrt{2}}$.

(A) Given the equation $f(x) = x^4 + 2x^3 - 19x^2 - 8x + 60 = 0$.
To remove the $x^3$ term,we substitute $x$ with $(x+h)$.
The term containing $x^3$ in the expansion of $(x+h)^4 + 2(x+h)^3 - 19(x+h)^2 - 8(x+h) + 60 = 0$ is obtained from the binomial expansion.
$(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$.
$2(x+h)^3 = 2(x^3 + 3x^2h + 3xh^2 + h^3) = 2x^3 + 6x^2h + 6xh^2 + 2h^3$.
The coefficient of $x^3$ in the transformed equation is $4h + 2$.
For the $x^3$ term to be removed,we set the coefficient to zero:
$4h + 2 = 0$.
$4h = -2$.
$h = -\frac{2}{4} = -\frac{1}{2}$.

(A) The $n$-th term of the series is $T_n = (2n-1)(2n+1)(2n+3) = (4n^2-1)(2n+3) = 8n^3 + 12n^2 - 2n - 3$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = 8 \sum k^3 + 12 \sum k^2 - 2 \sum k - 3 \sum 1$.
$S_n = 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n$.
$S_n = 2n^2(n+1)^2 + 2n(n+1)(2n+1) - n(n+1) - 3n$.
$S_n = n(n+1) [2n(n+1) + 2(2n+1) - 1] - 3n$.
$S_n = n(n+1) [2n^2 + 2n + 4n + 2 - 1] - 3n = n(n+1)(2n^2 + 6n + 1) - 3n$.
Comparing with $n(n+1)f(n) - 3n$,we get $f(n) = 2n^2 + 6n + 1$.
Therefore,$f(1) = 2(1)^2 + 6(1) + 1 = 2 + 6 + 1 = 9$.

(C) Since the lines $L_1$ and $L_2$ pass through the origin $(0, 0, 0)$,the plane containing them also passes through the origin. The equation of such a plane is given by $ax + by + cz = 0$.
The normal vector $\vec{n}$ to the plane is the cross product of the direction vectors of the two lines,$\vec{v}_1 = 3\hat{i} + \hat{j} - 5\hat{k}$ and $\vec{v}_2 = 2\hat{i} + 3\hat{j} - \hat{k}$.
$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -5 \\ 2 & 3 & -1 \end{vmatrix}$
$\vec{n} = \hat{i}(-1 - (-15)) - \hat{j}(-3 - (-10)) + \hat{k}(9 - 2)$
$\vec{n} = 14\hat{i} - 7\hat{j} + 7\hat{k}$
Dividing by $7$,we get the normal vector as $2\hat{i} - \hat{j} + \hat{k}$.
Thus,the equation of the plane is $2(x - 0) - 1(y - 0) + 1(z - 0) = 0$,which simplifies to $2x - y + z = 0$.

(C) The direction ratios of the line are $\vec{l} = (1, 2, 2)$.
The normal vector to the plane is $\vec{n} = (2, -1, \sqrt{\lambda})$.
The angle $\theta$ between a line with direction vector $\vec{l}$ and a plane with normal $\vec{n}$ is given by $\sin \theta = \frac{|\vec{l} \cdot \vec{n}|}{|\vec{l}| |\vec{n}|}$.
Calculating the dot product: $\vec{l} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{\lambda}) = 2 - 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda}$.
Calculating the magnitudes: $|\vec{l}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$ and $|\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{5 + \lambda}$.
Substituting into the formula: $\sin \theta = \frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}}$.
Given $\sin \theta = \frac{1}{3}$,we have $\frac{1}{3} = \frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}}$.
Simplifying,$\frac{1}{3} = \frac{2}{3} \sqrt{\frac{\lambda}{5 + \lambda}} \Rightarrow \frac{1}{2} = \sqrt{\frac{\lambda}{5 + \lambda}}$.
Squaring both sides: $\frac{1}{4} = \frac{\lambda}{5 + \lambda}$.
$5 + \lambda = 4\lambda \Rightarrow 3\lambda = 5 \Rightarrow \lambda = \frac{5}{3}$.

(B) The total number of ways to distribute $7$ different balls into $4$ different boxes is $4^7$.
To find the number of ways such that the first box contains exactly $3$ balls,we first choose $3$ balls out of $7$ in ${}^7C_3$ ways.
The remaining $4$ balls can be distributed among the other $3$ boxes in $3^4$ ways.
Thus,the number of favorable ways is ${}^7C_3 \times 3^4$.
The required probability is $\frac{{}^7C_3 \times 3^4}{4^7} = \frac{35 \times 3^4}{4^7} = \frac{35}{4^3} \times \left(\frac{3}{4}\right)^4 = \frac{35}{64} \left(\frac{3}{4}\right)^4$.

(C) Let $A$ be the event of getting two odd numbers and $B$ be the event of getting an even sum.
We need to find the conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$.
There are $5$ odd digits $(1, 3, 5, 7, 9)$ and $4$ even digits $(2, 4, 6, 8)$.
The total number of ways to select $2$ digits from $9$ is ${}^9C_2 = 36$.
The sum is even if both digits are odd or both digits are even.
Number of ways to get an even sum $= {}^5C_2 + {}^4C_2 = 10 + 6 = 16$.
Number of ways to get both odd $= {}^5C_2 = 10$.
Thus,$P(A|B) = \frac{10}{16} = \frac{5}{8}$.

(D) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
For a sum of $6$,the favorable outcomes are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,so there are $5$ outcomes. The probability of getting a sum of $6$ in one trial is $p_1 = \frac{5}{36}$.
The probability of not getting a sum of $6$ in one trial is $q_1 = 1 - \frac{5}{36} = \frac{31}{36}$.
The probability of getting a sum of $6$ at least once in $9$ trials is $P_1 = 1 - (q_1)^9 = 1 - \left(\frac{31}{36}\right)^9$.
For a sum of $8$,the favorable outcomes are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$,so there are $5$ outcomes. The probability of getting a sum of $8$ in one trial is $p_2 = \frac{5}{36}$.
The probability of not getting a sum of $8$ in one trial is $q_2 = 1 - \frac{5}{36} = \frac{31}{36}$.
The probability of getting a sum of $8$ at least once in $9$ trials is $P_2 = 1 - (q_2)^9 = 1 - \left(\frac{31}{36}\right)^9$.
Since $P_1 = P_2$,the ratio $P_1 : P_2 = 1 : 1$.

(C) For a single die,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.
Multiples of $3$ on a die are $\{3, 6\}$.
The probability of getting a multiple of $3$ on one die is $P(M) = \frac{2}{6} = \frac{1}{3}$.
The probability of $NOT$ getting a multiple of $3$ on one die is $P(M') = 1 - \frac{1}{3} = \frac{2}{3}$.
Since $12$ dice are thrown independently,the probability that a multiple of $3$ does not appear on any of the $12$ dice is $\left(\frac{2}{3}\right)^{12}$.

(D) Let $W_A, W_B, W_C$ be the events of drawing a white ball from bags $A, B, C$ respectively,and $R_A, R_B, R_C$ be the events of drawing a red ball from bags $A, B, C$ respectively.
$P(W_A) = \frac{3}{7}, P(R_A) = \frac{4}{7}$
$P(W_B) = \frac{4}{9}, P(R_B) = \frac{5}{9}$
$P(W_C) = \frac{5}{11}, P(R_C) = \frac{6}{11}$
We need one white and two red balls. This can happen in three mutually exclusive cases:
Case $I$: White from $A$,Red from $B$,Red from $C$: $P_1 = \frac{3}{7} \times \frac{5}{9} \times \frac{6}{11} = \frac{90}{693}$
Case $II$: Red from $A$,White from $B$,Red from $C$: $P_2 = \frac{4}{7} \times \frac{4}{9} \times \frac{6}{11} = \frac{96}{693}$
Case $III$: Red from $A$,Red from $B$,White from $C$: $P_3 = \frac{4}{7} \times \frac{5}{9} \times \frac{5}{11} = \frac{100}{693}$
Total probability $= P_1 + P_2 + P_3 = \frac{90+96+100}{693} = \frac{286}{693}$.

(D) The total number of questions is $n = 8$. Since each question is true or false,the probability of answering correctly is $p = \frac{1}{2}$ and incorrectly is $q = \frac{1}{2}$.
Let $X$ be the number of correct answers. $X$ follows a binomial distribution $B(8, \frac{1}{2})$.
The student passes if $X \ge 6$.
$P(\text{Pass}) = P(X=6) + P(X=7) + P(X=8)$
$P(\text{Pass}) = \binom{8}{6}(\frac{1}{2})^8 + \binom{8}{7}(\frac{1}{2})^8 + \binom{8}{8}(\frac{1}{2})^8$
$P(\text{Pass}) = \frac{28 + 8 + 1}{256} = \frac{37}{256}$.
The probability that the student fails is $P(\text{Fail}) = 1 - P(\text{Pass})$.
$P(\text{Fail}) = 1 - \frac{37}{256} = \frac{219}{256}$.

(D) Total number of balls $= 2 + 3 + 5 = 10$.
Let $R_3$ be the event that the third ball drawn is red.
By the symmetry of drawing balls without replacement,the probability that the $k$-th ball drawn is of a certain color is equal to the initial proportion of that color in the bag.
Specifically,for any position $k$ (where $1 \le k \le 10$),the probability that the $k$-th ball is red is given by $P(R_k) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$.
Thus,$P(R_3) = \frac{5}{10} = \frac{1}{2}$.
Alternatively,using the law of total probability:
$P(R_3) = P(R_3|R_1 R_2)P(R_1 R_2) + P(R_3|R_1 R_2^c)P(R_1 R_2^c) + P(R_3|R_1^c R_2)P(R_1^c R_2) + P(R_3|R_1^c R_2^c)P(R_1^c R_2^c) = \frac{1}{2}$.

(D) Using the definition of conditional probability,$P(E/F) = \frac{P(E \cap F)}{P(F)}$.
Given expression: $P(A / A \cap B) + P(B / A \cap B)$
$= \frac{P(A \cap (A \cap B))}{P(A \cap B)} + \frac{P(B \cap (A \cap B))}{P(A \cap B)}$
Since $A \cap (A \cap B) = A \cap B$ and $B \cap (A \cap B) = A \cap B$,we have:
$= \frac{P(A \cap B)}{P(A \cap B)} + \frac{P(A \cap B)}{P(A \cap B)}$
$= 1 + 1 = 2$

(C) Let $E_A$ be the event of getting a sum of $6$ and $E_B$ be the event of getting a sum of $7$.
The probability of getting a sum of $6$ is $P(E_A) = \frac{5}{36}$.
The probability of not getting a sum of $6$ is $P(E_A^c) = 1 - \frac{5}{36} = \frac{31}{36}$.
The probability of getting a sum of $7$ is $P(E_B) = \frac{6}{36} = \frac{1}{6}$.
The probability of not getting a sum of $7$ is $P(E_B^c) = 1 - \frac{1}{6} = \frac{5}{6}$.
$A$ wins if he gets $6$ on his $1^{st}$ turn,or if $A$ fails,$B$ fails,and then $A$ gets $6$ on his $2^{nd}$ turn,and so on.
$P(A \text{ wins}) = P(E_A) + P(E_A^c)P(E_B^c)P(E_A) + P(E_A^c)P(E_B^c)P(E_A^c)P(E_B^c)P(E_A) + \dots$
This is an infinite geometric series with first term $a = \frac{5}{36}$ and common ratio $r = P(E_A^c)P(E_B^c) = \frac{31}{36} \times \frac{5}{6} = \frac{155}{216}$.
$P(A \text{ wins}) = \frac{a}{1-r} = \frac{\frac{5}{36}}{1 - \frac{155}{216}} = \frac{\frac{5}{36}}{\frac{61}{216}} = \frac{5}{36} \times \frac{216}{61} = \frac{30}{61}$.

(A) $I$. $A, B$ are mutually exclusive events $\Rightarrow P(A \cap B) = 0$. Therefore,$P(A \cup B) = P(A) + P(B)$. This matches $(iv)$.
$II$. $A, B$ are independent events $\Rightarrow P(A \cap B) = P(A) \cdot P(B)$. Then $P(\frac{\bar{A}}{B}) = \frac{P(\bar{A} \cap B)}{P(B)} = \frac{P(\bar{A}) \cdot P(B)}{P(B)} = P(\bar{A}) = 1 - P(A)$. This matches $(iii)$.
$III$. $A \cap B = A \Rightarrow A \subset B \Rightarrow P(A) \leq P(B)$. This matches $(ii)$.
$IV$. $A \cup B = S \Rightarrow P(A \cup B) = 1$. Since $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1$,we have $P(A \cap B) = P(A) + P(B) - 1$. Since $P(A) = 1 - P(\bar{A})$,we get $P(A \cap B) = 1 - P(\bar{A}) + P(B) - 1 = P(B) - P(\bar{A})$. This matches $(i)$.
Thus,the correct matching is $(I)$-$(iv)$,$(II)$-$(iii)$,$(III)$-$(ii)$,$(IV)$-$(i)$.

(B) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.
Since $E_1$ and $E_2$ are independent events,$P(E_1 \cap E_2) = P(E_1)P(E_2) = \frac{1}{2}P(E_2)$.
Using the formula $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$:
$\frac{2}{3} = \frac{1}{2} + P(E_2) - \frac{1}{2}P(E_2)$
$\frac{2}{3} - \frac{1}{2} = \frac{1}{2}P(E_2)$
$\frac{1}{6} = \frac{1}{2}P(E_2) \implies P(E_2) = \frac{1}{3}$. (Matches $A-iii$)
Now,$P(E_1 \cap E_2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.
$P(\frac{E_1}{E_2}) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1/6}{1/3} = \frac{1}{2}$. (Matches $B-i$)
$P(\frac{\bar{E}_2}{E_1}) = \frac{P(\bar{E}_2 \cap E_1)}{P(E_1)} = \frac{P(E_1) - P(E_1 \cap E_2)}{P(E_1)} = \frac{1/2 - 1/6}{1/2} = \frac{1/3}{1/2} = \frac{2}{3}$. (Matches $C-v$)
$P(\bar{E}_1 \cup \bar{E}_2) = P(\overline{E_1 \cap E_2}) = 1 - P(E_1 \cap E_2) = 1 - \frac{1}{6} = \frac{5}{6}$. (Matches $D-ii$)
Thus,the correct match is $A-iii, B-i, C-v, D-ii$.

(C) $P(A) = \frac{75}{100} = \frac{3}{4}$,so $P(\bar{A}) = 1 - \frac{3}{4} = \frac{1}{4}$.
$P(B) = \frac{80}{100} = \frac{4}{5}$,so $P(\bar{B}) = 1 - \frac{4}{5} = \frac{1}{5}$.
They contradict each other if one speaks the truth and the other lies.
This can happen in two ways: ($A$ speaks truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks truth).
$P(\text{contradict}) = P(A \cap \bar{B}) + P(\bar{A} \cap B)$
$P(\text{contradict}) = (P(A) \times P(\bar{B})) + (P(\bar{A}) \times P(B))$
$P(\text{contradict}) = (\frac{3}{4} \times \frac{1}{5}) + (\frac{1}{4} \times \frac{4}{5})$
$P(\text{contradict}) = \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$.

(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$.
Given $P(A) = 0.30$ and $P(B) = 0.50$,we find $P(C) = 1 - (0.30 + 0.50) = 0.20$.
The conditional probabilities are $P(E \mid A) = 0.6$,$P(E \mid B) = 0.3$,and $P(E \mid C) = 0.1$.
Using Bayes' Theorem,the probability $P(C \mid E)$ is given by:
$P(C \mid E) = \frac{P(C) P(E \mid C)}{P(A) P(E \mid A) + P(B) P(E \mid B) + P(C) P(E \mid C)}$
Substituting the values:
$P(C \mid E) = \frac{0.20 \times 0.1}{(0.30 \times 0.6) + (0.50 \times 0.3) + (0.20 \times 0.1)}$
$P(C \mid E) = \frac{0.02}{0.18 + 0.15 + 0.02} = \frac{0.02}{0.35} = \frac{2}{35}$.

(C) Total number of ways to choose $2$ numbers from $100$ is $N = {}^{100}C_2 = \frac{100 \times 99}{2} = 4950$.
Event $A$: The product is even. This happens if at least one number is even. The complement is that both numbers are odd. Number of odd numbers is $50$.
$n(A) = {}^{100}C_2 - {}^{50}C_2 = 4950 - 1225 = 3725$.
Event $B$: The product is divisible by $4$. This happens if (both are even and at least one is a multiple of $4$) or (one is odd and one is a multiple of $4$).
Let $E$ be the set of even numbers ${2, 4, \dots, 100}$ ($50$ numbers) and $O$ be the set of odd numbers ${1, 3, \dots, 99}$ ($50$ numbers).
Let $M_4$ be the set of multiples of $4$ ${4, 8, \dots, 100}$ ($25$ numbers).
Let $E_2$ be the set of even numbers not divisible by $4$ ${2, 6, \dots, 98}$ ($25$ numbers).
$n(A \cap B)$ is the number of pairs whose product is divisible by $4$.
Pairs are: (one from $M_4$,any other) or (both from $E_2$).
$n(A \cap B) = {}^{25}C_1 \times {}^{75}C_1 + {}^{25}C_2 = 25 \times 75 + 300 = 1875 + 300 = 2175$.
$n(A \cap \bar{B}) = n(A) - n(A \cap B) = 3725 - 2175 = 1550$.
$P(A \cap \bar{B}) = \frac{1550}{4950} = \frac{155}{495} = \frac{31}{99}$.
Re-evaluating: The product is even but not divisible by $4$ if one number is from $E_2$ and the other is from $O$.
$n(A \cap \bar{B}) = {}^{25}C_1 \times {}^{50}C_1 = 25 \times 50 = 1250$.
$P(A \cap \bar{B}) = \frac{1250}{4950} = \frac{125}{495} = \frac{25}{99}$.

(A) Total number of girls $= 30$.
Given that $40 \%$ of the girls are good at Mathematics.
Number of girls good at Mathematics $= \frac{40}{100} \times 30 = 12$.
Number of girls not good at Mathematics $= 30 - 12 = 18$.
Since the student selected is already known to be a girl,we only consider the sample space of girls.
Required probability $= \frac{\text{Number of girls not good at Mathematics}}{\text{Total number of girls}} = \frac{18}{30} = \frac{3}{5}$.

(C) Let $B_1$ be the first bag and $B_2$ be the second bag.
Bag $B_1$ contains $4$ red and $5$ black balls (Total = $9$).
Bag $B_2$ contains $3$ red and $6$ black balls (Total = $9$).
We draw $1$ ball from $B_1$ and $2$ balls from $B_2$. Total balls drawn = $3$.
We need $2$ black and $1$ red ball. There are two cases:
Case $I$: The ball from $B_1$ is red and the two balls from $B_2$ are black.
$P(Case I) = P(R_1) \times P(B_2, B_2) = \frac{4}{9} \times \frac{\binom{6}{2}}{\binom{9}{2}} = \frac{4}{9} \times \frac{15}{36} = \frac{4}{9} \times \frac{5}{12} = \frac{20}{108} = \frac{5}{27}$.
Case $II$: The ball from $B_1$ is black and the two balls from $B_2$ are one red and one black.
$P(Case II) = P(B_1) \times P(R_2, B_2) = \frac{5}{9} \times \frac{\binom{3}{1} \times \binom{6}{1}}{\binom{9}{2}} = \frac{5}{9} \times \frac{3 \times 6}{36} = \frac{5}{9} \times \frac{18}{36} = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}$.
Total Probability = $P(Case I) + P(Case II) = \frac{5}{27} + \frac{5}{18} = \frac{10 + 15}{54} = \frac{25}{54}$.

(C) Let $C$,$B$,and $T$ be the events that the person travels by car,bus,and train respectively. Let $L$ be the event that the person reaches college late,and $L'$ be the event that the person reaches college in time.
Given probabilities: $P(C) = \frac{1}{5}$,$P(B) = \frac{2}{5}$,$P(T) = \frac{3}{5}$.
Probabilities of being late: $P(L|C) = \frac{2}{7}$,$P(L|B) = \frac{4}{7}$,$P(L|T) = \frac{1}{7}$.
Probabilities of being on time: $P(L'|C) = 1 - \frac{2}{7} = \frac{5}{7}$,$P(L'|B) = 1 - \frac{4}{7} = \frac{3}{7}$,$P(L'|T) = 1 - \frac{1}{7} = \frac{6}{7}$.
Using Bayes' Theorem,the probability that he traveled by car given he reached in time is:
$P(C|L') = \frac{P(L'|C)P(C)}{P(L'|C)P(C) + P(L'|B)P(B) + P(L'|T)P(T)}$
$P(C|L') = \frac{(\frac{5}{7} \times \frac{1}{5})}{(\frac{5}{7} \times \frac{1}{5}) + (\frac{3}{7} \times \frac{2}{5}) + (\frac{6}{7} \times \frac{3}{5})}$
$P(C|L') = \frac{\frac{5}{35}}{\frac{5}{35} + \frac{6}{35} + \frac{18}{35}} = \frac{5}{5 + 6 + 18} = \frac{5}{29}$.

(A) Let $B_1$ be the event of choosing a bag from the first group and $B_2$ be the event of choosing a bag from the second group.
Total bags = $2 + 4 = 6$.
$P(B_1) = \frac{2}{6} = \frac{1}{3}$ and $P(B_2) = \frac{4}{6} = \frac{2}{3}$.
Let $B$ be the event that the drawn ball is black.
For the first group,the probability of drawing a black ball is $P(B|B_1) = \frac{5}{3+5} = \frac{5}{8}$.
For the second group,the probability of drawing a black ball is $P(B|B_2) = \frac{4}{6+4} = \frac{4}{10} = \frac{2}{5}$.
Using Bayes' Theorem,the probability that the ball is from the first set of bags given that it is black is:
$P(B_1|B) = \frac{P(B_1)P(B|B_1)}{P(B_1)P(B|B_1) + P(B_2)P(B|B_2)}$
$P(B_1|B) = \frac{\frac{1}{3} \times \frac{5}{8}}{\frac{1}{3} \times \frac{5}{8} + \frac{2}{3} \times \frac{2}{5}} = \frac{\frac{5}{24}}{\frac{5}{24} + \frac{4}{15}} = \frac{\frac{5}{24}}{\frac{25 + 32}{120}} = \frac{5}{24} \times \frac{120}{57} = \frac{5 \times 5}{57} = \frac{25}{57}$.

(B) Let $K$ be the event that the card drawn is a king,and $K^c$ be the event that the card drawn is not a king. Let $R$ be the event that the person reports that the card is a king.
Given:
$P(K) = \frac{4}{52} = \frac{1}{13}$
$P(K^c) = 1 - \frac{1}{13} = \frac{12}{13}$
Let $T$ be the event that the person speaks the truth. $P(T) = \frac{3}{4}$ and $P(T^c) = \frac{1}{4}$.
The probability that the person reports a king given it is a king is $P(R|K) = P(T) = \frac{3}{4}$.
The probability that the person reports a king given it is not a king is $P(R|K^c) = P(T^c) = \frac{1}{4}$.
Using Bayes' Theorem,the probability that it is actually a king given the report is:
$P(K|R) = \frac{P(R|K)P(K)}{P(R|K)P(K) + P(R|K^c)P(K^c)}$
$P(K|R) = \frac{(\frac{3}{4} \times \frac{1}{13})}{(\frac{3}{4} \times \frac{1}{13}) + (\frac{1}{4} \times \frac{12}{13})}$
$P(K|R) = \frac{\frac{3}{52}}{\frac{3}{52} + \frac{12}{52}} = \frac{3}{15} = \frac{1}{5}$

(C) Let $A$ be the event that we draw three green balls. Let $E_k$ be the event that there are $k$ green balls in the bag,where $k \in \{3, 4, 5, 6\}$. Assuming each number of green balls is equally likely,$P(E_k) = \frac{1}{4}$.
The probability of drawing $3$ green balls given $k$ green balls are present is $P(A|E_k) = \frac{{}^k C_3}{{}^6 C_3}$.
Using Bayes' Theorem,the probability that there are $5$ green balls given that $3$ green balls were drawn is:
$P(E_5|A) = \frac{P(A|E_5)P(E_5)}{\sum_{k=3}^{6} P(A|E_k)P(E_k)}$
Since $P(E_k) = \frac{1}{4}$ for all $k$,this simplifies to:
$P(E_5|A) = \frac{{}^5 C_3}{\sum_{k=3}^{6} {}^k C_3} = \frac{10}{1 + 4 + 10 + 20} = \frac{10}{35} = \frac{2}{7}$.

(A) We know that the conditional probability formula is given by $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.
Substituting the given values,we get $P(A \cap B) = P(B \mid A) \times P(A) = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$.
Now,we use the conditional probability formula for $P(A \mid B)$:
$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Rearranging for $P(B)$,we get $P(B) = \frac{P(A \cap B)}{P(A \mid B)}$.
Substituting the values,$P(B) = \frac{1/6}{1/2} = \frac{1}{6} \times 2 = \frac{1}{3}$.

(C) $E_1$: $A$ ball is drawn from bag $A$.
$E_2$: $A$ ball is drawn from bag $B$.
$F$: $A$ ball is found to be red.
Given:
$P(E_1) = \frac{1}{2}$,$P(E_2) = \frac{1}{2}$.
$P(F|E_1) = \frac{3}{5}$ (probability of drawing a red ball from bag $A$).
$P(F|E_2) = \frac{5}{9}$ (probability of drawing a red ball from bag $B$).
Using Bayes' Theorem:
$P(E_2|F) = \frac{P(F|E_2) \cdot P(E_2)}{P(F|E_1) \cdot P(E_1) + P(F|E_2) \cdot P(E_2)}$
$P(E_2|F) = \frac{\frac{5}{9} \times \frac{1}{2}}{\frac{3}{5} \times \frac{1}{2} + \frac{5}{9} \times \frac{1}{2}}$
$P(E_2|F) = \frac{\frac{5}{9}}{\frac{3}{5} + \frac{5}{9}} = \frac{\frac{5}{9}}{\frac{27 + 25}{45}} = \frac{5}{9} \times \frac{45}{52} = \frac{25}{52}$.

(A) Let $X$ be the number of defective bulbs chosen. This follows a binomial distribution $B(n, p)$ where $n = 5$ and $p = 20\% = \frac{1}{5}$.
Then $q = 1 - p = \frac{4}{5}$.
The probability of getting exactly $k$ defective bulbs is given by $P(X = k) = { }^nC_k \cdot p^k \cdot q^{n-k}$.
For $k = 3$,we have:
$P(X = 3) = { }^5C_3 \cdot (\frac{1}{5})^3 \cdot (\frac{4}{5})^{5-3}$
$P(X = 3) = \frac{5 \times 4}{2 \times 1} \cdot (\frac{1}{125}) \cdot (\frac{16}{25})$
$P(X = 3) = 10 \cdot \frac{16}{3125}$
$P(X = 3) = \frac{160}{3125} = \frac{32}{625}$.

(B) Given that the difference between the mean and variance is $1$:
$np - npq = 1 \Rightarrow np(1-q) = 1 \Rightarrow np^2 = 1$ $\qquad (i)$
Also,the difference between their squares is $11$:
$(np)^2 - (npq)^2 = 11$ $\qquad (ii)$
$n^2p^2 - n^2p^2q^2 = 11 \Rightarrow n^2p^2(1-q^2) = 11$
Dividing $(ii)$ by $(i)$,we get:
$\frac{n^2p^2(1-q^2)}{np^2} = 11 \Rightarrow n(1-q^2) = 11$
Since $n=36$,$36(1-q^2) = 11 \Rightarrow 1-q^2 = \frac{11}{36} \Rightarrow q^2 = 1 - \frac{11}{36} = \frac{25}{36} \Rightarrow q = \frac{5}{6}$
Thus,$p = 1 - q = 1 - \frac{5}{6} = \frac{1}{6}$
Now,$P(X=2) = {}^{36}C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{34} = m\left(\frac{5}{6}\right)^{36}$
$\frac{36 \times 35}{2} \times \frac{1}{36} \times \left(\frac{5}{6}\right)^{34} = m\left(\frac{5}{6}\right)^{36}$
$\frac{35}{2} \times \left(\frac{5}{6}\right)^{34} = m \times \left(\frac{5}{6}\right)^{34} \times \left(\frac{5}{6}\right)^2$
$m = \frac{35}{2} \times \left(\frac{6}{5}\right)^2 = \frac{35}{2} \times \frac{36}{25} = \frac{7 \times 18}{5} = \frac{126}{5} = 25.2$
Wait,re-evaluating $m$: $m = \frac{35}{2} \times \frac{36}{25} = \frac{7 \times 18}{5} = 25.2$.
Given $m : n = 25.2 : 36 = 252 : 360 = 7 : 10$.

(A) Let $p$ be the probability of hitting the target and $q$ be the probability of failing to hit the target.
Given $q = \frac{1}{3}$,so $p = 1 - q = 1 - \frac{1}{3} = \frac{2}{3}$.
The number of trials is $n = 4$.
We need to find the probability of hitting the target at least thrice,which is $P(X \geq 3) = P(X = 3) + P(X = 4)$.
Using the binomial distribution formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 3) = {}^4C_3 \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^1 = 4 \times \frac{8}{27} \times \frac{1}{3} = \frac{32}{81}$.
$P(X = 4) = {}^4C_4 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^0 = 1 \times \frac{16}{81} \times 1 = \frac{16}{81}$.
Therefore,$P(X \geq 3) = \frac{32}{81} + \frac{16}{81} = \frac{48}{81} = \frac{16}{27}$.

(C) For a binomial distribution with $n=7$ and $p=q=\frac{1}{2}$:
$\mu = np = 7 \times \frac{1}{2} = \frac{7}{2}$
$\sigma^2 = npq = 7 \times \frac{1}{2} \times \frac{1}{2} = \frac{7}{4}$
$P(X=3) = {}^{7}C_{3} \times (\frac{1}{2})^{3} \times (\frac{1}{2})^{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times (\frac{1}{2})^{7} = 35 \times \frac{1}{128} = \frac{35}{128}$
$\frac{\mu \sigma^2}{P(X=3)} = \frac{(\frac{7}{2}) \times (\frac{7}{4})}{\frac{35}{128}} = \frac{49}{8} \times \frac{128}{35} = \frac{7}{1} \times \frac{16}{5} = \frac{112}{5}$

(D) Given: $n = 300$,$p = 0.01$.
Let $m$ be the mean number of defective items.
$m = n \times p = 300 \times 0.01 = 3$.
Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $m = 3$.
The probability of $X$ defective items is given by $P(X = k) = \frac{e^{-m} m^k}{k!}$.
We need to find the probability that the number of defective items is at most one,i.e.,$P(X \leq 1)$.
$P(X \leq 1) = P(X = 0) + P(X = 1)$.
$P(X = 0) = \frac{e^{-3} 3^0}{0!} = \frac{e^{-3} \times 1}{1} = \frac{1}{e^3}$.
$P(X = 1) = \frac{e^{-3} 3^1}{1!} = \frac{3}{e^3}$.
Therefore,$P(X \leq 1) = \frac{1}{e^3} + \frac{3}{e^3} = \frac{4}{e^3}$.

(C) Given $X \sim B(5, p)$.
$P(X=3) = P(X=4)$
$\Rightarrow { }^5 C_3 p^3(1-p)^2 = { }^5 C_4 p^4(1-p)$
$\Rightarrow 10(1-p) = 5p$
$\Rightarrow 10 - 10p = 5p \Rightarrow 15p = 10 \Rightarrow p = \frac{2}{3}$.
Now,we need to find $P(|X-3| < 2)$.
$P(|X-3| < 2) = P(-2 < X-3 < 2) = P(1 < X < 5) = P(X=2) + P(X=3) + P(X=4)$.
Since $P(X=3) = P(X=4)$,we have $P(X=2) + 2P(X=3)$.
$P(X=2) = { }^5 C_2 (\frac{2}{3})^2 (\frac{1}{3})^3 = 10 \times \frac{4}{9} \times \frac{1}{27} = \frac{40}{243}$.
$P(X=3) = { }^5 C_3 (\frac{2}{3})^3 (\frac{1}{3})^2 = 10 \times \frac{8}{27} \times \frac{1}{9} = \frac{80}{243}$.
$P(|X-3| < 2) = \frac{40}{243} + 2(\frac{80}{243}) = \frac{40 + 160}{243} = \frac{200}{243}$.

(C) Let the mean be $\mu = np$ and standard deviation be $\sigma = \sqrt{npq}$,where $q = 1-p$.
Given,$\mu - \sigma = 3 \Rightarrow \mu - 3 = \sigma$.
Squaring both sides,$(\mu - 3)^2 = \sigma^2 = npq$.
Also given,$\mu^2 - \sigma^2 = 21$.
Substituting $\sigma^2 = \mu^2 - 21$ into the first equation:
$(\mu - 3)^2 = \mu^2 - 21$
$\mu^2 - 6\mu + 9 = \mu^2 - 21$
$-6\mu = -30 \Rightarrow \mu = 5$.
Since $\mu = np = 5$,then $\sigma^2 = 5^2 - 21 = 25 - 21 = 4$.
We know $\sigma^2 = npq = 5q = 4$,so $q = \frac{4}{5}$ and $p = 1 - \frac{4}{5} = \frac{1}{5}$.
Since $np = 5$,$n(\frac{1}{5}) = 5 \Rightarrow n = 25$.
The probability mass function is $P(X=k) = {}^{n}C_k p^k q^{n-k}$.
$\frac{P(X=1)}{P(X=2)} = \frac{{}^{25}C_1 p^1 q^{24}}{{}^{25}C_2 p^2 q^{23}} = \frac{25 \cdot q}{300 \cdot p} = \frac{1}{12} \cdot \frac{4/5}{1/5} = \frac{1}{12} \cdot 4 = \frac{1}{3}$.

(A) Let $X$ be the number of times the radar detects the plane in $n = 4$ scans. This follows a binomial distribution $B(n, p)$ where $n = 4$ and $p = 0.1$ (probability of detection).
The probability of not detecting the plane is $q = 1 - p = 0.9$.
We want to find the probability that it cannot detect the plane at least two times,which is equivalent to $1 - P(\text{detecting the plane } 2, 3, \text{ or } 4 \text{ times})$.
$P(X \ge 2) = P(X=2) + P(X=3) + P(X=4)$.
$P(X=2) = {}^{4}C_{2} (0.1)^{2} (0.9)^{2} = 6 \times 0.01 \times 0.81 = 0.0486$.
$P(X=3) = {}^{4}C_{3} (0.1)^{3} (0.9)^{1} = 4 \times 0.001 \times 0.9 = 0.0036$.
$P(X=4) = {}^{4}C_{4} (0.1)^{4} (0.9)^{0} = 1 \times 0.0001 \times 1 = 0.0001$.
$P(X \ge 2) = 0.0486 + 0.0036 + 0.0001 = 0.0523$.
The required probability is $1 - P(X \ge 2) = 1 - 0.0523 = 0.9477$.

(C) For a Binomial distribution $B(n, p)$,Mean $\mu = np$ and Variance $\sigma^2 = npq$,where $q = 1-p$.
Given:
Sum: $np + npq = 5 \Rightarrow np(1+q) = 5$ ...$(i)$
Product: $(np)(npq) = n^2p^2q = 6$ ...(ii)
From $(i)$,$np = \frac{5}{1+q}$. Substituting this into (ii):
$\left(\frac{5}{1+q}\right)^2 q = 6 \Rightarrow 25q = 6(1+q)^2 \Rightarrow 25q = 6(1+2q+q^2) \Rightarrow 6q^2 - 13q + 6 = 0$.
Solving the quadratic equation: $6q^2 - 9q - 4q + 6 = 0 \Rightarrow 3q(2q-3) - 2(2q-3) = 0 \Rightarrow (3q-2)(2q-3) = 0$.
Since $q < 1$,we have $q = \frac{2}{3}$. Then $p = 1 - q = \frac{1}{3}$.
Substituting $q = \frac{2}{3}$ into $(i)$: $np(1 + \frac{2}{3}) = 5 \Rightarrow np(\frac{5}{3}) = 5 \Rightarrow np = 3$.
Since $p = \frac{1}{3}$,$n(\frac{1}{3}) = 3 \Rightarrow n = 9$.
Finally,$6(n+p-q) = 6(9 + \frac{1}{3} - \frac{2}{3}) = 6(9 - \frac{1}{3}) = 54 - 2 = 52$.

(D) For a Poisson distribution with mean $\lambda = 5$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
We need to find $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$.
Substituting the values:
$P(X = 0) = \frac{e^{-5} 5^0}{0!} = e^{-5}$
$P(X = 1) = \frac{e^{-5} 5^1}{1!} = 5e^{-5}$
$P(X = 2) = \frac{e^{-5} 5^2}{2!} = \frac{25}{2}e^{-5}$
Summing these probabilities:
$P(X < 3) = e^{-5} \left( 1 + 5 + \frac{25}{2} \right) = e^{-5} \left( 6 + 12.5 \right) = 18.5 e^{-5} = \frac{37}{2} e^{-5}$.

(B) Let $X$ be the random variable representing the number of kings drawn.
Total number of cards = $52$. Number of kings = $4$. Number of non-kings = $48$.
We draw $2$ cards. The possible values for $X$ are $0, 1, 2$.
$P(X=0) = \frac{{}^{48}C_2}{{}^{52}C_2} = \frac{48 \times 47}{52 \times 51} = \frac{1128}{1326} = \frac{188}{221}$.
$P(X=1) = \frac{{}^{4}C_1 \times {}^{48}C_1}{{}^{52}C_2} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221}$.
$P(X=2) = \frac{{}^{4}C_2}{{}^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$.
The mean $E(X) = \sum x_i P(x_i) = 0 \times P(X=0) + 1 \times P(X=1) + 2 \times P(X=2)$.
$E(X) = 0 + \frac{32}{221} + 2 \times \frac{1}{221} = \frac{32+2}{221} = \frac{34}{221} = \frac{2}{13}$.

(B) The sum of all probabilities in a probability distribution must be $1$.
$\sum P(X=x) = \frac{1}{10} + (K + \frac{2}{10}) + (K + \frac{3}{10}) + (K + \frac{3}{10}) + (K + \frac{4}{10}) + (K + \frac{2}{10}) = 1$
$\Rightarrow 5K + \frac{15}{10} = 1$
$\Rightarrow 5K + 1.5 = 1$
$\Rightarrow 5K = -0.5$
$\Rightarrow K = -0.1 = -\frac{1}{10}$
Now,substitute $K = -\frac{1}{10}$ into the table:
For $x = -2, P = \frac{1}{10}$
For $x = -1, P = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10}$
For $x = 0, P = -\frac{1}{10} + \frac{3}{10} = \frac{2}{10}$
For $x = 1, P = -\frac{1}{10} + \frac{3}{10} = \frac{2}{10}$
For $x = 2, P = -\frac{1}{10} + \frac{4}{10} = \frac{3}{10}$
For $x = 3, P = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10}$
The mean $\mu = \sum x P(x) = (-2)(\frac{1}{10}) + (-1)(\frac{1}{10}) + (0)(\frac{2}{10}) + (1)(\frac{2}{10}) + (2)(\frac{3}{10}) + (3)(\frac{1}{10})$
$\mu = \frac{-2 - 1 + 0 + 2 + 6 + 3}{10} = \frac{8}{10} = \frac{4}{5}$

(B) Let $X$ be the number of red balls drawn. The total number of balls is $3 + 5 = 8$. We draw $3$ balls from $8$,so the total number of ways is ${}^8 C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
The random variable $X$ can take values $0, 1, 2, 3$.
$P(X=0) = \frac{{}^5 C_0 \times {}^3 C_3}{56} = \frac{1 \times 1}{56} = \frac{1}{56}$
$P(X=1) = \frac{{}^5 C_1 \times {}^3 C_2}{56} = \frac{5 \times 3}{56} = \frac{15}{56}$
$P(X=2) = \frac{{}^5 C_2 \times {}^3 C_1}{56} = \frac{10 \times 3}{56} = \frac{30}{56}$
$P(X=3) = \frac{{}^5 C_3 \times {}^3 C_0}{56} = \frac{10 \times 1}{56} = \frac{10}{56}$
The mean $E(X) = \sum x_i P(x_i) = 0 \times \frac{1}{56} + 1 \times \frac{15}{56} + 2 \times \frac{30}{56} + 3 \times \frac{10}{56} = \frac{0 + 15 + 60 + 30}{56} = \frac{105}{56} = \frac{15}{8}$.

(D) Given $P(X=k) = k^2 P$ for $k = 1, 2, 3, 4, 5, 6$.
Since the sum of all probabilities must be $1$,we have:
$\sum_{k=1}^6 P(X=k) = 1$
$P(1^2) + P(2^2) + P(3^2) + P(4^2) + P(5^2) + P(6^2) = 1$
$P(1 + 4 + 9 + 16 + 25 + 36) = 1$
$91P = 1 \Rightarrow P = \frac{1}{91}$.
The mean of $X$ is given by $E(X) = \sum_{k=1}^6 k \cdot P(X=k)$.
$E(X) = \sum_{k=1}^6 k \cdot (k^2 P) = P \sum_{k=1}^6 k^3$.
$E(X) = \frac{1}{91} (1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3)$.
$E(X) = \frac{1}{91} (1 + 8 + 27 + 64 + 125 + 216) = \frac{441}{91}$.

(A) For a probability distribution,the sum of all probabilities must be $1$.
$\sum P(X=x) = 1$
$0.05 + 0.1 + 2K + 0 + 0.3 + K + 0.1 = 1$
$0.55 + 3K = 1 \Rightarrow 3K = 0.45 \Rightarrow K = 0.15$.
Now,the distribution is:

$X$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.05$	$0.1$	$0.3$	$0$	$0.3$	$0.15$	$0.1$

Mean $\mu = E(X) = \sum x_i P(x_i) = (-3)(0.05) + (-2)(0.1) + (-1)(0.3) + (0)(0) + (1)(0.3) + (2)(0.15) + (3)(0.1)$
$\mu = -0.15 - 0.2 - 0.3 + 0 + 0.3 + 0.3 + 0.3 = 0.25$.
Variance $\sigma^2 = E(X^2) - \mu^2 = \sum x_i^2 P(x_i) - (0.25)^2$.
$E(X^2) = (-3)^2(0.05) + (-2)^2(0.1) + (-1)^2(0.3) + (0)^2(0) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.1)$
$E(X^2) = 9(0.05) + 4(0.1) + 1(0.3) + 0 + 1(0.3) + 4(0.15) + 9(0.1)$
$E(X^2) = 0.45 + 0.4 + 0.3 + 0 + 0.3 + 0.6 + 0.9 = 2.95$.
Variance $= 2.95 - (0.25)^2 = 2.95 - 0.0625 = 2.8875$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\Sigma P(X=x) = 1$
$2k + 4k + 3k + k = 1$
$10k = 1$
$k = \frac{1}{10}$

(A) Let $r$ be the radius and $A$ be the area of the circle.
The area of the circle is given by $A = \pi r^2$.
Taking the natural logarithm on both sides,we get $\ln A = \ln \pi + 2 \ln r$.
Differentiating both sides with respect to $r$,we get $\frac{dA}{A} = 2 \frac{dr}{r}$.
The percentage error in the radius is given as $\frac{dr}{r} \times 100 = 3\%$.
The percentage error in the area is $\frac{dA}{A} \times 100 = 2 \times (\frac{dr}{r} \times 100)$.
Substituting the given value,we get $\frac{dA}{A} \times 100 = 2 \times 3\% = 6\%$.
Thus,the percentage error in the area is $6\%$.

(A) The position vectors of points $P$ and $Q$ with respect to the origin $O(0,0,0)$ are given by $\vec{OP} = 0\hat{i} + 1\hat{j} + 2\hat{k}$ and $\vec{OQ} = 4\hat{i} - 2\hat{j} + 1\hat{k}$.
To find the angle $\theta = \angle POQ$,we use the dot product formula: $\cos \theta = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}| |\vec{OQ}|}$.
First,calculate the dot product:
$\vec{OP} \cdot \vec{OQ} = (0)(4) + (1)(-2) + (2)(1) = 0 - 2 + 2 = 0$.
Since the dot product is $0$,the vectors $\vec{OP}$ and $\vec{OQ}$ are perpendicular to each other.
Therefore,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.