AP EAMCET 2022 Mathematics Question Paper with Answer and Solution

(D) Given the limit as $n \rightarrow \infty$:
Since $2-\sqrt{2} \approx 0.586 < 1$,we have $\lim _{n \rightarrow \infty} (2-\sqrt{2})^n = 0$.
Dividing the numerator and denominator by $(2+\sqrt{2})^n$:
$= \lim _{n}$ ${\rightarrow \infty} \sqrt{2} \left[ \frac{1 + \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)^n}{1 - \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)^n} \right] = \sqrt{2} \left[ \frac{1+0}{1-0} \right] = \sqrt{2}$.

(A) Case $1$: When $x < 0$,as $n \rightarrow \infty$,$nx \rightarrow -\infty$. Thus,$e^{nx} \rightarrow 0$. The limit becomes $\lim _{n \rightarrow \infty} \frac{A+0}{x+A(0)} = \frac{A}{x}$.
Case $2$: When $x > 0$,as $n \rightarrow \infty$,$nx \rightarrow \infty$. Thus,$e^{nx} \rightarrow \infty$. Dividing numerator and denominator by $e^{nx}$,we get $\lim _{n \rightarrow \infty} \frac{A/e^{nx} + 1}{x/e^{nx} + A} = \frac{0+1}{0+A} = \frac{1}{A}$ (assuming $A \neq 0$).

(A) For $k = \lim _{x \rightarrow 0^{+}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right)$,let $t = 1/x$. As $x \rightarrow 0^{+}$,$t \rightarrow \infty$.
$k = \lim _{t \rightarrow \infty} \frac{1}{t^2} \left( \frac{e^t - e^{-t}}{e^t + e^{-t}} \right) = \lim _{t \rightarrow \infty} \frac{1}{t^2} \left( \frac{1 - e^{-2t}}{1 + e^{-2t}} \right) = 0 \times 1 = 0$.
For $l = \lim _{x \rightarrow 0^{-}} x^2 \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right)$,let $y = 1/x$. As $x \rightarrow 0^{-}$,$y \rightarrow -\infty$.
$l = \lim _{y \rightarrow -\infty} \frac{1}{y^2} \left( \frac{e^y - e^{-y}}{e^y + e^{-y}} \right) = \lim _{y \rightarrow -\infty} \frac{1}{y^2} \left( \frac{e^{2y} - 1}{e^{2y} + 1} \right) = 0 \times (-1) = 0$.
Therefore,$k = l = 0$.

(D) To find $\lim_{x \rightarrow 0^+} f(x)$,let $x \rightarrow 0^+$,then $\frac{1}{x} \rightarrow \infty$.
Thus,$\lim_{x \rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \rightarrow 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} = \frac{1 - 0}{1 + 0} = 1$.
To find $\lim_{x \rightarrow 0^-} f(x)$,let $x \rightarrow 0^-$,then $\frac{1}{x} \rightarrow -\infty$.
Thus,$\lim_{x \rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$.
Since the left-hand limit and right-hand limit are not equal,$\lim_{x \rightarrow 0} f(x)$ does not exist.
For $\lim_{x \rightarrow \infty} f(x)$,$\frac{1}{x} \rightarrow 0$,so $\lim_{x \rightarrow \infty} f(x) = \frac{e^0 - 1}{e^0 + 1} = \frac{1 - 1}{1 + 1} = 0$.
Therefore,option $D$ is correct.

(C) Let $L = \lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n$.
Taking the natural logarithm on both sides:
$\ln L = \lim _{n \rightarrow \infty} n \ln \left(1 + \frac{b^{1 / n} - 1}{a}\right)$.
Using the limit formula $\lim _{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1$,let $x = \frac{b^{1 / n} - 1}{a}$. As $n \rightarrow \infty$,$x \rightarrow 0$.
$\ln L = \lim _{n \rightarrow \infty} n \cdot \left(\frac{b^{1 / n} - 1}{a}\right) \cdot \frac{\ln(1 + x)}{x} = \lim _{n \rightarrow \infty} \frac{n}{a} (b^{1 / n} - 1)$.
Let $t = 1/n$. As $n \rightarrow \infty$,$t \rightarrow 0$.
$\ln L = \frac{1}{a} \lim _{t \rightarrow 0} \frac{b^t - 1}{t}$.
Since $\lim _{t \rightarrow 0} \frac{b^t - 1}{t} = \ln b$,we have:
$\ln L = \frac{1}{a} \ln b = \ln(b^{1 / a})$.
Therefore,$L = b^{1 / a}$.

(B) Let $L = \lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$.
Since $\lim _{x \rightarrow 0} \frac{e^x-1}{x} = 1$ and the exponent $\frac{x}{x+1-e^x} \to \infty$ as $x \to 0$,this is an indeterminate form of type $1^\infty$.
We use the formula $\lim _{x \to a} f(x)^{g(x)} = e^{\lim _{x \to a} g(x)(f(x)-1)}$.
$L = e^{\lim _{x \to 0} \left( \frac{x}{x+1-e^x} \right) \left( \frac{e^x-1}{x} - 1 \right)}$.
$L = e^{\lim _{x \to 0} \left( \frac{x}{x+1-e^x} \right) \left( \frac{e^x-1-x}{x} \right)}$.
$L = e^{\lim _{x \to 0} \frac{e^x-1-x}{x+1-e^x}}$.
$L = e^{\lim _{x \to 0} \frac{-(x+1-e^x)}{x+1-e^x}} = e^{-1}$.

(D) Given,$f(x) = \lim_{y \rightarrow \infty} y(x^{1/y} - 1)$.
First,we simplify $f(x)$:
$f(x) = \lim_{y \rightarrow \infty} \frac{x^{1/y} - 1}{1/y}$.
Let $t = 1/y$. As $y \rightarrow \infty$,$t \rightarrow 0^+$.
$f(x) = \lim_{t \rightarrow 0} \frac{x^t - 1}{t}$.
Using the standard limit $\lim_{t \rightarrow 0} \frac{a^t - 1}{t} = \ln(a)$,we get $f(x) = \ln(x)$.
Now,substitute $f(x) = \ln(x)$ into the given equation $2022 f(\frac{1}{x}) + P f(x) = f(x^2)$:
$2022 \ln(\frac{1}{x}) + P \ln(x) = \ln(x^2)$.
Since $\ln(\frac{1}{x}) = -\ln(x)$ and $\ln(x^2) = 2 \ln(x)$:
$2022(-\ln(x)) + P \ln(x) = 2 \ln(x)$.
$-2022 \ln(x) + P \ln(x) = 2 \ln(x)$.
Dividing by $\ln(x)$ (for $x \neq 1$):
$-2022 + P = 2$.
$P = 2024$.

(A) Let $a_r = \frac{r+2}{r(r+1)(r+3)}$. Using partial fractions,we write $\frac{r+2}{r(r+1)(r+3)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+3}$.
Solving for coefficients: $r+2 = A(r+1)(r+3) + Br(r+3) + Cr(r+1)$.
For $r=0$: $2 = 3A \implies A = \frac{2}{3}$.
For $r=-1$: $1 = B(-1)(2) \implies B = -\frac{1}{2}$.
For $r=-3$: $-1 = C(-3)(-2) \implies C = -\frac{1}{6}$.
Thus,$a_r = \frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)}$.
The sum $S_n = \sum_{r=1}^n (\frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)})$.
As $n \rightarrow \infty$,the sum converges to $\sum_{r=1}^{\infty} (\frac{2}{3r} - \frac{1}{2(r+1)} - \frac{1}{6(r+3)})$.
Rearranging terms: $\frac{2}{3} \sum \frac{1}{r} - \frac{1}{2} \sum \frac{1}{r+1} - \frac{1}{6} \sum \frac{1}{r+3} = \frac{2}{3}(1 + \frac{1}{2} + \frac{1}{3} + \dots) - \frac{1}{2}(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots) - \frac{1}{6}(\frac{1}{4} + \frac{1}{5} + \dots)$.
$= \frac{2}{3}(1) + (\frac{2}{3} - \frac{1}{2})(\frac{1}{2}) + (\frac{2}{3} - \frac{1}{2} - \frac{1}{6})(\frac{1}{3}) = \frac{2}{3} + \frac{1}{6}(\frac{1}{2}) + 0 = \frac{2}{3} + \frac{1}{12} = \frac{8+1}{12} = \frac{9}{12} = \frac{3}{4}$.
Wait,re-evaluating the partial fraction sum: $\sum_{r=1}^{\infty} [(\frac{1}{3r} - \frac{1}{2(r+1)}) + (\frac{1}{3r} - \frac{1}{6(r+3)})]$.
$= (\frac{1}{3} - \frac{1}{4} + \frac{1}{6} - \frac{1}{6} + \dots) + (\frac{1}{3} - \frac{1}{12} + \frac{1}{6} - \frac{1}{18} + \dots) = \frac{29}{36}$.

(A) The range of a data set is calculated as the difference between the highest value and the lowest value in the set.
Given data: $35, 12, 21, 24, 15, 7, 16, 12, 30, 32, 13, 17$.
Highest value = $35$.
Lowest value = $7$.
Range = $\text{Highest value} - \text{Lowest value} = 35 - 7 = 28$.

(B) Step $1$: Calculate the mean $(\bar{x})$:
$\bar{x} = \frac{5+6+7+8+6+9+13+12+15}{9} = \frac{81}{9} = 9$
Step $2$: Calculate the mean deviation about the mean using the formula $\frac{\sum |x_i - \bar{x}|}{N}$:
$|5-9| + |6-9| + |7-9| + |8-9| + |6-9| + |9-9| + |13-9| + |12-9| + |15-9|$
$= 4 + 3 + 2 + 1 + 3 + 0 + 4 + 3 + 6 = 26$
Step $3$: Divide by the total number of observations $(N=9)$:
$\text{Mean Deviation} = \frac{26}{9} \approx 2.88$

(B) The mean of the data $p, 6, 6, 7, 8, 11, 15, 16$ is given by $\bar{x} = \frac{p + 6 + 6 + 7 + 8 + 11 + 15 + 16}{8} = \frac{p + 69}{8}$.
Given that the mean is $3p$,we have $\frac{p + 69}{8} = 3p$.
$p + 69 = 24p$ $\Rightarrow 23p = 69$ $\Rightarrow p = 3$.
Thus,the data set is $3, 6, 6, 7, 8, 11, 15, 16$ and the mean is $\bar{x} = 3 \times 3 = 9$.
The mean deviation about the mean is given by $M.D. = \frac{1}{n} \sum_{i=1}^n |x_i - \bar{x}|$.
$M.D. = \frac{|3-9| + |6-9| + |6-9| + |7-9| + |8-9| + |11-9| + |15-9| + |16-9|}{8}$.
$M.D. = \frac{6 + 3 + 3 + 2 + 1 + 2 + 6 + 7}{8} = \frac{30}{8} = 3.75$.

(D) First,find the mid-values $(x_i)$ for each class interval:
$0-20: 10$
$20-40: 30$
$40-60: 50$
$60-80: 70$
$80-100: 90$
Calculate the mean $(\bar{x})$:
$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{(10 \times 10) + (8 \times 30) + (12 \times 50) + (9 \times 70) + (11 \times 90)}{10+8+12+9+11} = \frac{100+240+600+630+990}{50} = \frac{2560}{50} = 51.2$
Now,calculate the mean deviation about the mean using the formula: $\text{M.D.}(\bar{x}) = \frac{\Sigma f_i |x_i - \bar{x}|}{\Sigma f_i}$
$\Sigma f_i |x_i - 51.2| = 10|10-51.2| + 8|30-51.2| + 12|50-51.2| + 9|70-51.2| + 11|90-51.2|$
$= 10(41.2) + 8(21.2) + 12(1.2) + 9(18.8) + 11(38.8)$
$= 412 + 169.6 + 14.4 + 169.2 + 426.8 = 1192$
$\text{M.D.}(\bar{x}) = \frac{1192}{50} = 23.84$

(C) The first $10$ multiples of $4$ are $4, 8, 12, 16, 20, 24, 28, 32, 36, 40$.
Let $X = 4i$ where $i = 1, 2, \dots, 10$.
The standard deviation of the first $n$ natural numbers is $\sigma_n = \sqrt{\frac{n^2 - 1}{12}}$.
Since each term is multiplied by $4$,the standard deviation of the multiples is $4 \times \sigma_{10}$.
$\sigma_{10} = \sqrt{\frac{10^2 - 1}{12}} = \sqrt{\frac{99}{12}} = \sqrt{8.25}$.
Standard deviation $= 4 \times \sqrt{8.25} = \sqrt{16 \times 8.25} = \sqrt{132} \approx 11.489 \approx 11.5$.

(B) The given data is $6, 3, 4, 9, 2, 7, 11$.
First,arrange the data in ascending order: $2, 3, 4, 6, 7, 9, 11$.
There are $n = 7$ observations,which is odd.
The median $M$ is the $\left(\frac{n+1}{2}\right)^{th}$ observation,which is the $4^{th}$ observation.
Thus,$M = 6$.
The mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - M|$.
Mean deviation $= \frac{1}{7} [|2-6| + |3-6| + |4-6| + |6-6| + |7-6| + |9-6| + |11-6|]$.
Mean deviation $= \frac{1}{7} [|-4| + |-3| + |-2| + 0 + |1| + |3| + |5|]$.
Mean deviation $= \frac{1}{7} [4 + 3 + 2 + 0 + 1 + 3 + 5] = \frac{18}{7} \approx 2.57$.

(A) First,we calculate the cumulative frequency $(cf)$ and the median $(M)$:

$x_i$	$f_i$	$cf$	$|x_i - M|$	$f_i|x_i - M|$
$10$	$6$	$6$	$|10-12|=2$	$12$
$11$	$12$	$18$	$|11-12|=1$	$12$
$12$	$18$	$36$	$|12-12|=0$	$0$
$13$	$12$	$48$	$|13-12|=1$	$12$

Total frequency $N = \sum f_i = 6 + 12 + 18 + 12 = 48$.
The median is the value corresponding to the $\frac{N}{2} = \frac{48}{2} = 24$th observation. Looking at the $cf$ column,the $24$th observation falls in the class where $x = 12$. Thus,$M = 12$.
The mean deviation about the median is given by:
$MD = \frac{\sum f_i|x_i - M|}{N} = \frac{12 + 12 + 0 + 12}{48} = \frac{36}{48} = 0.75$.

(A) The given data is an arithmetic progression with $n = 101$ terms,where the first term $a = 1$ and the common difference is $d$.
The mean $\bar{x} = \frac{1}{101} \sum_{i=0}^{100} (1 + id) = 1 + \frac{d}{101} \times \frac{100 \times 101}{2} = 1 + 50d$.
The mean deviation about the mean is given by $MD = \frac{1}{n} \sum_{i=0}^{100} |x_i - \bar{x}|$.
$MD = \frac{1}{101} \sum_{i=0}^{100} |(1 + id) - (1 + 50d)| = \frac{1}{101} \sum_{i=0}^{100} |(i - 50)d| = \frac{d}{101} \sum_{i=0}^{100} |i - 50|$.
The sum $\sum_{i=0}^{100} |i - 50| = |0-50| + |1-50| + \ldots + |50-50| + \ldots + |100-50| = 50 + 49 + \ldots + 0 + \ldots + 50 = 2 \times \frac{50 \times 51}{2} = 2550$.
Given $MD = 255$,we have $255 = \frac{d}{101} \times 2550$.
$255 = d \times \frac{2550}{101} \Rightarrow d = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1$.

(C) Given,number of observations $n = 20$,$\sum x_i = 1000$,and $\sum x_i^2 = 84000$.
The mean $\overline{x} = \frac{\sum x_i}{n} = \frac{1000}{20} = 50$.
The variance $\sigma^2$ is given by the formula:
$\sigma^2 = \frac{1}{n} \sum x_i^2 - (\overline{x})^2$.
Substituting the values:
$\sigma^2 = \frac{84000}{20} - (50)^2$.
$\sigma^2 = 4200 - 2500$.
$\sigma^2 = 1700$.

(A) Let the original numbers be $w, x, y, z$ with variance $\sigma^2 = 9$.
The variance of a set of numbers is defined as $\text{Var}(X) = E[X^2] - (E[X])^2$.
If each number is multiplied by a constant $k$,the new variance is given by the property $\text{Var}(kX) = k^2 \text{Var}(X)$.
Here,$k = 5$ and $\text{Var}(X) = 9$.
Therefore,the new variance is $5^2 \times 9 = 25 \times 9 = 225$.

(C) We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Also,$\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} = \frac{s(s-a)}{\Delta}$.
Substituting these into the expression:
$r_1 \cot \frac{A}{2} + r_2 \cot \frac{B}{2} + r_3 \cot \frac{C}{2} = \left( \frac{\Delta}{s-a} \cdot \frac{s(s-a)}{\Delta} \right) + \left( \frac{\Delta}{s-b} \cdot \frac{s(s-b)}{\Delta} \right) + \left( \frac{\Delta}{s-c} \cdot \frac{s(s-c)}{\Delta} \right)$
$= s + s + s = 3s$.

(B) We know that $r = (s-a) \tan(A/2) = (s-b) \tan(B/2) = (s-c) \tan(C/2)$ and $r_1 = s \tan(A/2)$,$r_2 = s \tan(B/2)$,$r_3 = s \tan(C/2)$.
Then,$r_1 - r = s \tan(A/2) - (s-a) \tan(A/2) = a \tan(A/2)$.
Thus,$\frac{r_1-r}{a} = \tan(A/2)$.
Similarly,$\frac{r_2-r}{b} = \tan(B/2)$ and $\frac{r_3-r}{c} = \tan(C/2)$.
Substituting these into the expression,we get $s[\tan(A/2) + \tan(B/2) + \tan(C/2)]$.
Since $r_1 = s \tan(A/2)$,$r_2 = s \tan(B/2)$,and $r_3 = s \tan(C/2)$,the expression becomes $r_1 + r_2 + r_3$.

(A) Using the Law of Sines: $\frac{\sin A}{a} = \frac{\sin B}{b}$.
Given $a=7, b=6, A=120^{\circ}$.
$\sin B = \frac{b \sin A}{a} = \frac{6 \sin 120^{\circ}}{7}$.
Since $\sin 120^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866$.
$\sin B = \frac{6 \times 0.866}{7} = \frac{5.196}{7} \approx 0.7423$.
$B = \arcsin(0.7423) \approx 47.9^{\circ}$.

(C) Using the sine rule,we have $a = k \sin A$ and $b = k \sin B$.
Substituting these into the expression:
$\frac{a-b}{a+b} = \frac{k \sin A - k \sin B}{k \sin A + k \sin B} = \frac{\sin A - \sin B}{\sin A + \sin B}$
Using the sum-to-product formulas:
$= \frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$
$= \cot \left(\frac{A+B}{2}\right) \tan \left(\frac{A-B}{2}\right)$
Since $A+B+C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$.
Therefore,$\cot \left(\frac{A+B}{2}\right) = \cot \left(90^{\circ} - \frac{C}{2}\right) = \tan \frac{C}{2}$.
Thus,$\frac{a-b}{a+b} = \tan \frac{C}{2} \tan \left(\frac{A-B}{2}\right)$.

(D) Given equations are:
$3 \sin A + 4 \cos B = 6$ ... $(i)$
$4 \sin B + 3 \cos A = 1$ ... (ii)
Squaring and adding both equations:
$(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$
$(9 \sin^2 A + 16 \cos^2 B + 24 \sin A \cos B) + (16 \sin^2 B + 9 \cos^2 A + 24 \sin B \cos A) = 37$
$9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$
$9(1) + 16(1) + 24 \sin(A + B) = 37$
$25 + 24 \sin(A + B) = 37$
$24 \sin(A + B) = 12$
$\sin(A + B) = \frac{1}{2}$
Since $A+B+C = \pi$,we have $\sin(A+B) = \sin(\pi - C) = \sin C$.
Therefore,$\sin C = \frac{1}{2}$.
Since $A, B, C$ are angles of a triangle,$C$ can be $\frac{\pi}{6}$ or $\frac{5\pi}{6}$.
However,if $C = \frac{5\pi}{6}$,then $A+B = \frac{\pi}{6}$,which makes $\sin A$ and $\sin B$ very small,contradicting the initial equations.
Thus,$C = \frac{\pi}{6}$.

(A) Given: $a=2, b=3, \sin A=\frac{2}{3}$.
Using the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$.
Substituting the values: $\frac{2}{2/3} = \frac{3}{\sin B}$.
$\Rightarrow 3 = \frac{3}{\sin B}$.
$\Rightarrow \sin B = 1$.
Therefore,$B = \frac{\pi}{2}$.

(A) Given sides are $a=4, b=5, c=7$.
First,calculate the semi-perimeter $s$:
$s = \frac{a+b+c}{2} = \frac{4+5+7}{2} = \frac{16}{2} = 8$.
The formula for $\sin \left(\frac{A}{2}\right)$ is:
$\sin \left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{bc}}$.
Substituting the values:
$\sin \left(\frac{A}{2}\right) = \sqrt{\frac{(8-5)(8-7)}{5 \times 7}} = \sqrt{\frac{3 \times 1}{35}} = \sqrt{\frac{3}{35}}$.

(B) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Using the identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we have $\sin \frac{A}{2} = \frac{a}{2R \cdot 2 \cos \frac{A}{2}} = \frac{a}{4R \cos \frac{A}{2}}$.
Alternatively,using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s$ is the semi-perimeter.
By the $AM$-$GM$ inequality,$\sqrt{(s-b)(s-c)} \leq \frac{(s-b)+(s-c)}{2} = \frac{2s-b-c}{2} = \frac{a}{2}$.
Therefore,$\sin \frac{A}{2} = \frac{\sqrt{(s-b)(s-c)}}{\sqrt{bc}} \leq \frac{a}{2 \sqrt{bc}}$.
Thus,$\sin \frac{A}{2} \leq \frac{a}{2 \sqrt{bc}}$.

(A) Given sides are $a=13, b=14, c=15$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$.
Using the formula $\sin(\frac{A}{2}) = \sqrt{\frac{(s-b)(s-c)}{bc}}$:
$s-b = 21-14 = 7$
$s-c = 21-15 = 6$
$\sin(\frac{A}{2}) = \sqrt{\frac{7 \times 6}{14 \times 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.

(A) Given,in triangle $ABC$,we need to evaluate $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$.
Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$.
Substituting these values:
$= \frac{2R \sin A \cos A - 2R \sin B \cos B}{2R \sin A \cos B - 2R \sin B \cos A} + \cos C$
$= \frac{\sin 2A - \sin 2B}{2(\sin A \cos B - \sin B \cos A)} + \cos C$
Using the formula $\sin 2A - \sin 2B = 2 \sin(A - B) \cos(A + B)$ and $\sin(A - B) = \sin A \cos B - \sin B \cos A$:
$= \frac{2 \sin(A - B) \cos(A + B)}{2 \sin(A - B)} + \cos C$
$= \cos(A + B) + \cos C$
Since $A + B = \pi - C$,we have $\cos(A + B) = \cos(\pi - C) = -\cos C$.
$= -\cos C + \cos C = 0$.

(D) Given,in a triangle $ABC$,$a=2$,$b=3$,and $c=4$.
We know that the semi-perimeter $s = \frac{a+b+c}{2} = \frac{2+3+4}{2} = 4.5$.
The formula for $\tan \left(\frac{A}{2}\right)$ is $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.
Substituting the values:
$\tan \left(\frac{A}{2}\right) = \sqrt{\frac{(4.5-3)(4.5-4)}{4.5(4.5-2)}}$
$= \sqrt{\frac{1.5 \times 0.5}{4.5 \times 2.5}}$
$= \sqrt{\frac{0.75}{11.25}}$
$= \sqrt{\frac{75}{1125}} = \sqrt{\frac{1}{15}}$.

(C) Given that the angles of a triangle are in the ratio $1: 2: 3$.
From the angle sum property,the sum of all angles of a triangle is $180^{\circ}$.
Let the angles be $x, 2x, 3x$.
Then $x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
So,the angles are $A = 30^{\circ}, B = 60^{\circ}, C = 90^{\circ}$.
According to the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Substituting the values: $\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$.
$\Rightarrow \frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$.
Multiplying by $1/2$,we get $a : b : c = 1 : \sqrt{3} : 2$.

(D) Given the expression $2(bc \cos A + ca \cos B + ab \cos C) = 2bc \cos A + 2ca \cos B + 2ab \cos C$.
Using the Law of Cosines,we know that $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{c^2 + a^2 - b^2}{2ca}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting these values into the expression:
$= 2bc \left(\frac{b^2 + c^2 - a^2}{2bc}\right) + 2ca \left(\frac{c^2 + a^2 - b^2}{2ca}\right) + 2ab \left(\frac{a^2 + b^2 - c^2}{2ab}\right)$
$= (b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)$
$= a^2 + b^2 + c^2$.
Thus,the value is $a^2 + b^2 + c^2$.

(C) Given: $\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$.
Since $A+B+C=180^{\circ}$,we have $A+B=120^{\circ}$ $(1)$.
By the sine rule,$\frac{\sin A}{\sin B} = \frac{a}{b} = 2+\sqrt{3}$.
Applying componendo and dividendo:
$\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2+\sqrt{3}+1}{2+\sqrt{3}-1} = \frac{3+\sqrt{3}}{\sqrt{3}+1} = \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = \sqrt{3}$.
Using the sum-to-product formulas:
$\frac{2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})}{2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})} = \sqrt{3}$.
$\tan(\frac{A+B}{2}) \cot(\frac{A-B}{2}) = \sqrt{3}$.
Since $\frac{A+B}{2} = 60^{\circ}$,$\tan(60^{\circ}) \cot(\frac{A-B}{2}) = \sqrt{3}$.
$\sqrt{3} \cot(\frac{A-B}{2}) = \sqrt{3} \Rightarrow \cot(\frac{A-B}{2}) = 1$.
Thus,$\frac{A-B}{2} = 45^{\circ} \Rightarrow A-B = 90^{\circ}$ $(2)$.
Adding $(1)$ and $(2)$: $2A = 210^{\circ} \Rightarrow A = 105^{\circ}$.

(B) Given: $a=2, b=3, c=4$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos C = \frac{2^2 + 3^2 - 4^2}{2 \times 2 \times 3}$.
$\cos C = \frac{4 + 9 - 16}{12}$.
$\cos C = \frac{13 - 16}{12} = \frac{-3}{12}$.
Therefore,$\cos C = -\frac{1}{4}$.

(B) Given expression: $(b+c) \cos A + (c+a) \cos B + (a+b) \cos C$
Expand the terms: $(b \cos A + c \cos A) + (c \cos B + a \cos B) + (a \cos C + b \cos C)$
Rearrange the terms: $(b \cos A + a \cos B) + (c \cos A + a \cos C) + (c \cos B + b \cos C)$
Using the projection formula for a triangle:
$c = a \cos B + b \cos A$
$b = a \cos C + c \cos A$
$a = b \cos C + c \cos B$
Substituting these into the rearranged expression:
$= c + b + a = a + b + c$

(B) In any $\triangle ABC$,by the Law of Cosines,we have:
$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting these into the expression:
$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$
$= \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{2abc}$
$= \frac{a^2+b^2+c^2}{2abc}$.

(A) Given in $\triangle ABC$: $a=6$,$b=5$,$c=4$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\cos A = \frac{5^2 + 4^2 - 6^2}{2 \times 5 \times 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8}$.
Now,using the formula $\cos 2A = 2\cos^2 A - 1$:
$\cos 2A = 2 \left(\frac{1}{8}\right)^2 - 1 = 2 \left(\frac{1}{64}\right) - 1 = \frac{1}{32} - 1 = \frac{1 - 32}{32} = -\frac{31}{32}$.

(C) Given expression: $a(b \cos C - c \cos B)$.
Using the projection formula or cosine rule:
$\cos C = \frac{a^2+b^2-c^2}{2ab}$ and $\cos B = \frac{a^2+c^2-b^2}{2ac}$.
Substituting these values into the expression:
$a \left( b \left( \frac{a^2+b^2-c^2}{2ab} \right) - c \left( \frac{a^2+c^2-b^2}{2ac} \right) \right)$
$= a \left( \frac{a^2+b^2-c^2}{2a} - \frac{a^2+c^2-b^2}{2a} \right)$
$= \frac{1}{2} (a^2+b^2-c^2 - a^2 - c^2 + b^2)$
$= \frac{1}{2} (2b^2 - 2c^2)$
$= b^2 - c^2$.

(C) Apply the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Given $\angle A = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2}$.
$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$ $\Rightarrow bc = b^2+c^2-a^2$ $\Rightarrow b^2+c^2 = bc+a^2 \dots (i)$.
Now,consider the expression $\frac{b}{c+a} + \frac{c}{a+b} = \frac{b(a+b) + c(c+a)}{(c+a)(a+b)}$.
$= \frac{ab+b^2+c^2+ac}{ac+a^2+bc+ab}$.
Substitute $b^2+c^2 = bc+a^2$ from $(i)$:
$= \frac{ab+(bc+a^2)+ac}{ac+a^2+bc+ab} = \frac{ab+bc+a^2+ac}{ab+bc+a^2+ac} = 1$.

(D) Given $a \cos A = b \cos B$.
Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$:
$a \left(\frac{b^2 + c^2 - a^2}{2bc}\right) = b \left(\frac{a^2 + c^2 - b^2}{2ac}\right)$
$\Rightarrow \frac{a}{b} (b^2 + c^2 - a^2) = \frac{b}{a} (a^2 + c^2 - b^2)$
$\Rightarrow a^2 (b^2 + c^2 - a^2) = b^2 (a^2 + c^2 - b^2)$
$\Rightarrow a^2b^2 + a^2c^2 - a^4 = a^2b^2 + b^2c^2 - b^4$
$\Rightarrow a^4 - b^4 + b^2c^2 - a^2c^2 = 0$
$\Rightarrow (a^2 - b^2)(a^2 + b^2) - c^2(a^2 - b^2) = 0$
$\Rightarrow (a^2 - b^2)(a^2 + b^2 - c^2) = 0$
Since $a \neq b$,we have $a^2 - b^2 \neq 0$.
Therefore,$a^2 + b^2 - c^2 = 0$,which implies $a^2 + b^2 = c^2$.
Thus,$\triangle ABC$ is a right-angled triangle.

(D) Given sides are $a=3, b=5, c=3$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\cos A = \frac{5^2 + 3^2 - 3^2}{2 \times 5 \times 3}$.
$\cos A = \frac{25 + 9 - 9}{30} = \frac{25}{30}$.
$\cos A = \frac{5}{6}$.

(B) We know that $\cos 2A = 1 - 2\sin^2 A$ and $\cos 2B = 1 - 2\sin^2 B$.
Substituting these into the expression:
$\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \frac{1}{a^2} - \frac{2\sin^2 A}{a^2} - \left(\frac{1}{b^2} - \frac{2\sin^2 B}{b^2}\right)$
$= \frac{1}{a^2} - \frac{1}{b^2} - 2 \left( \frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2} \right)$
By the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = k$,so $\frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = k^2$.
Thus,$\frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2} = 0$.
Therefore,the expression simplifies to $\frac{1}{a^2} - \frac{1}{b^2}$.

(A) Given $\angle A = 60^{\circ}$. By the Cosine Rule,$a^2 = b^2 + c^2 - 2bc \cos A$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $a^2 = b^2 + c^2 - 2bc \left(\frac{1}{2}\right) = b^2 + c^2 - bc$.
Now,consider the expression $(a+b+c)(b+c-a)$.
This can be rewritten as $((b+c)+a)((b+c)-a) = (b+c)^2 - a^2$.
Expanding this,we get $b^2 + c^2 + 2bc - a^2$.
Substituting $a^2 = b^2 + c^2 - bc$,we get:
$b^2 + c^2 + 2bc - (b^2 + c^2 - bc) = b^2 + c^2 + 2bc - b^2 - c^2 + bc = 3bc$.

(B) Given sides are $a = 4 \text{ cm}$,$b = 5 \text{ cm}$,and $c = 7 \text{ cm}$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{4 + 5 + 7}{2} = \frac{16}{2} = 8 \text{ cm}$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{8(8 - 4)(8 - 5)(8 - 7)}$
$= \sqrt{8 \times 4 \times 3 \times 1}$
$= \sqrt{96}$
$= \sqrt{16 \times 6}$
$= 4 \sqrt{6} \text{ cm}^2$.

(A) Given $\tan \frac{A}{2} = \frac{1}{\sqrt{2}}$.
Using the formula $\cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}}$:
$\cos A = \frac{1 - (\frac{1}{\sqrt{2}})^2}{1 + (\frac{1}{\sqrt{2}})^2} = \frac{1 - 1/2}{1 + 1/2} = \frac{1/2}{3/2} = \frac{1}{3}$.
Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
$\frac{1}{3} = \frac{5^2 + 6^2 - a^2}{2(5)(6)}$.
$\frac{1}{3} = \frac{25 + 36 - a^2}{60}$.
$20 = 61 - a^2$.
$a^2 = 41$.
$a = \sqrt{41}$.

(C) We know that the area $R = \frac{1}{2}ab \sin C$,so $2R = ab \sin C$,which implies $4R^2 = a^2b^2 \sin^2 C$.
Substituting this into the first term: $\sqrt{a^2b^2 - 4R^2} = \sqrt{a^2b^2 - a^2b^2 \sin^2 C} = \sqrt{a^2b^2(1 - \sin^2 C)} = \sqrt{a^2b^2 \cos^2 C} = ab \cos C$.
Similarly,$\sqrt{b^2c^2 - 4R^2} = bc \cos A$ and $\sqrt{c^2a^2 - 4R^2} = ca \cos B$.
Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$,so $ab \cos C = \frac{a^2+b^2-c^2}{2}$.
Thus,the expression becomes: $\frac{a^2+b^2-c^2}{2} + \frac{b^2+c^2-a^2}{2} + \frac{c^2+a^2-b^2}{2} = \frac{a^2+b^2+c^2}{2}$.

(C) We know that $\tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$.
Substituting $y = \log x$,we get:
$\tanh(\log x) = \frac{e^{\log x} - e^{-\log x}}{e^{\log x} + e^{-\log x}}$
Since $e^{\log x} = x$ and $e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}$,the expression becomes:
$\tanh(\log x) = \frac{x - \frac{1}{x}}{x + \frac{1}{x}}$
Multiplying the numerator and denominator by $x$,we get:
$\tanh(\log x) = \frac{x^2 - 1}{x^2 + 1}$

(D) Given,$r_1=36, r_2=18, r_3=12$.
We know that in a $\triangle ABC$,the relation between the inradius $r$ and exradii $r_1, r_2, r_3$ is $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.
Substituting the values: $\frac{1}{r} = \frac{1}{36} + \frac{1}{18} + \frac{1}{12} = \frac{1+2+3}{36} = \frac{6}{36} = \frac{1}{6}$.
Thus,$r=6$.
We also know that $\Delta^2 = r \cdot r_1 \cdot r_2 \cdot r_3$.
$\Delta^2 = 6 \times 36 \times 18 \times 12 = 6 \times 6^2 \times (6 \times 3) \times (6 \times 2) = 6^4 \times 36 = 6^4 \times 6^2 = 6^6$.
Therefore,$\Delta = 6^3 = 216$.
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r} = \frac{216}{6} = 36$.

(B) Given vertices are $A(0,0)$,$B(0,2)$,and $C(2,0)$.
Since $\overline{AC}$ lies on the $x$-axis and $\overline{AB}$ lies on the $y$-axis,the triangle is a right-angled triangle with the right angle at vertex $A(0,0)$.
In a right-angled triangle,the orthocentre is the vertex where the right angle is formed,so the orthocentre is $H(0,0)$.
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $\overline{BC}$.
Circumcentre $O = \left(\frac{0+2}{2}, \frac{2+0}{2}\right) = (1,1)$.
The distance between the orthocentre $(0,0)$ and the circumcentre $(1,1)$ is $\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ units}$.

(D) Given,$\frac{a}{4} = \frac{b}{5} = \frac{c}{6} = k$ (let).
So,$a = 4k, b = 5k, c = 6k$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{15k}{2}$.
The area of the triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \frac{15\sqrt{7}k^2}{4}$.
The inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}k^2 / 4}{15k / 2} = \frac{\sqrt{7}k}{2}$.
The circumradius $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \cdot (15\sqrt{7}k^2 / 4)} = \frac{120k^3}{15\sqrt{7}k^2} = \frac{8k}{\sqrt{7}}$.
Therefore,the ratio $\frac{R}{r} = \frac{8k}{\sqrt{7}} \div \frac{\sqrt{7}k}{2} = \frac{8k}{\sqrt{7}} \cdot \frac{2}{\sqrt{7}k} = \frac{16}{7}$.

(B) In a right-angled triangle $\triangle ABC$ with $\angle C = \frac{\pi}{2}$,the sides are $a$ (opposite to $A$),$b$ (opposite to $B$),and $c$ (hypotenuse opposite to $C$).
From the properties of the incircle,the distance from vertex $C$ to the points of tangency on sides $AC$ and $BC$ is $r$.
Thus,the sides can be expressed as $b = x+r$ and $a = y+r$,where $x$ and $y$ are the lengths of the tangents from vertices $A$ and $B$ to the incircle respectively.
The hypotenuse $c = x+y$.
Since $c$ is the diameter of the circumcircle,$c = 2R$,so $x+y = 2R$.
Now,adding the expressions for $a$ and $b$:
$a+b = (y+r) + (x+r) = (x+y) + 2r = 2R + 2r$.
Dividing by $2$,we get:
$R+r = \frac{a+b}{2}$.