AP EAMCET 2025 Mathematics Question Paper with Answer and Solution

(B) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}}$,we consider the highest power of $x$ in the numerator and the denominator.
In the numerator,the term is $(3-x)^{25}(6+x)^{35} \approx (-x)^{25}(x)^{35} = -x^{60}$.
In the denominator,the term is $(12+x)^{38}(9-x)^{22} \approx (x)^{38}(-x)^{22} = x^{60}$.
Thus,the expression behaves as $\frac{-x^{60}}{x^{60}} = -1$ as $x \rightarrow \infty$.

(C) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}$.
Substitute $x = \frac{\pi}{4} + h$,as $h \rightarrow 0$.
Then $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt{2} \cos h$.
Also,$1 - \sin 2x = 1 - \sin(2(\frac{\pi}{4} + h)) = 1 - \sin(\frac{\pi}{2} + 2h) = 1 - \cos 2h = 2 \sin^2 h$.
Substituting these into the limit:
$L = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - (\sqrt{2} \cos h)^3}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - 2 \sqrt{2} \cos^3 h}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos^3 h)}{\sin^2 h}$.
Using $1 - \cos^3 h = (1 - \cos h)(1 + \cos h + \cos^2 h)$ and $\sin^2 h = (1 - \cos h)(1 + \cos h)$:
$L = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos h)(1 + \cos h + \cos^2 h)}{(1 - \cos h)(1 + \cos h)} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 + \cos h + \cos^2 h)}{1 + \cos h}$.
As $h \rightarrow 0$,$\cos h \rightarrow 1$,so $L = \frac{\sqrt{2}(1 + 1 + 1)}{1 + 1} = \frac{3 \sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

(B) We know that $\sin(3\theta) = 3 \sin \theta - 4 \sin^3 \theta$.
Thus,$l = \lim_{\theta \rightarrow 0} \frac{\sin(3\theta)}{\theta} = \lim_{\theta \rightarrow 0} \frac{\sin(3\theta)}{3\theta} \times 3 = 1 \times 3 = 3$.
We also know that $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$.
Thus,$m = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{\theta} = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{2\theta} \times 2 = 1 \times 2 = 2$.
The quadratic equation with roots $l=3$ and $m=2$ is given by $(x - l)(x - m) = 0$.
$(x - 3)(x - 2) = 0$
$x^2 - 2x - 3x + 6 = 0$
$x^2 - 5x + 6 = 0$.

(B) Given the limit $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.
As $x \rightarrow 2^{+}$,$x$ is slightly greater than $2$,so $[x] = 2$.
Also,as $x \rightarrow 2^{+}$,$\frac{x}{3}$ is slightly greater than $\frac{2}{3}$,so $\left[\frac{x}{3}\right] = 0$.
Substituting these values into the expression:
$\lim _{x}$ ${\rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right) = \frac{2^3}{3} - 0^3 = \frac{8}{3} - 0 = \frac{8}{3}$.

(D) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.
Substituting this into the limit,we get:
$\lim _{x \rightarrow \infty} [x - \log (\frac{e^x + e^{-x}}{2})]$
$= \lim _{x \rightarrow \infty} [x - (\log (e^x + e^{-x}) - \log 2)]$
$= \lim _{x \rightarrow \infty} [x - \log (e^x(1 + e^{-2x})) + \log 2]$
$= \lim _{x \rightarrow \infty} [x - (\log e^x + \log (1 + e^{-2x})) + \log 2]$
$= \lim _{x \rightarrow \infty} [x - x - \log (1 + e^{-2x}) + \log 2]$
$= \lim _{x \rightarrow \infty} [\log 2 - \log (1 + e^{-2x})]$
As $x \rightarrow \infty$,$e^{-2x} \rightarrow 0$,so $\log (1 + e^{-2x}) \rightarrow \log 1 = 0$.
Therefore,the limit is $\log 2 - 0 = \log 2$.

(A) Let $L = \lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$.
Rationalizing the numerator,we multiply by $\frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$:
$L = \lim _{y \rightarrow 0} \frac{(1+\sqrt{1+y^4})-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim _{y \rightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$.
Rationalizing the numerator again,we multiply by $\frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1}$:
$L = \lim _{y \rightarrow 0} \frac{(1+y^4)-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)} = \lim _{y \rightarrow 0} \frac{y^4}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$.
Canceling $y^4$ and evaluating the limit as $y \rightarrow 0$:
$L = \frac{1}{(\sqrt{1+1}+\sqrt{2})(\sqrt{1}+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(1+1)} = \frac{1}{2\sqrt{2} \times 2} = \frac{1}{4\sqrt{2}}$.

(A) We want to evaluate $L = \lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)$.
First,factor out $x$ from each term:
$L = \lim _{x \rightarrow \infty} \left( x(1+\frac{4}{x})^{1/3} - x(1-\frac{3}{x})^{1/2} \right)$.
Using the binomial expansion $(1+u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2$ for small $u$:
$(1+\frac{4}{x})^{1/3} \approx 1 + \frac{1}{3}(\frac{4}{x}) = 1 + \frac{4}{3x}$.
$(1-\frac{3}{x})^{1/2} \approx 1 + \frac{1}{2}(-\frac{3}{x}) = 1 - \frac{3}{2x}$.
Substituting these into the limit:
$L = \lim _{x \rightarrow \infty} \left( x(1 + \frac{4}{3x}) - x(1 - \frac{3}{2x}) \right)$.
$L = \lim _{x \rightarrow \infty} \left( x + \frac{4}{3} - x + \frac{3}{2} \right)$.
$L = \frac{4}{3} + \frac{3}{2} = \frac{8+9}{6} = \frac{17}{6}$.

(C) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.
Given $\frac{n^2 - 1}{12} = 10$,we have $n^2 - 1 = 120$,so $n^2 = 121$,which gives $n = 11$.
The first $m$ even natural numbers are $2, 4, 6, \dots, 2m$. This is equivalent to $2 \times (1, 2, 3, \dots, m)$.
The variance of a set of numbers multiplied by a constant $k$ is $k^2$ times the original variance.
Thus,the variance of the first $m$ even natural numbers is $2^2 \times \frac{m^2 - 1}{12} = 4 \times \frac{m^2 - 1}{12} = \frac{m^2 - 1}{3}$.
Given $\frac{m^2 - 1}{3} = 16$,we have $m^2 - 1 = 48$,so $m^2 = 49$,which gives $m = 7$.
Therefore,$n: m = 11: 7$.

(D) Step $1$: Find the mid-points $(x_i)$ of each class interval:
$x_1 = 1, x_2 = 3, x_3 = 5, x_4 = 7, x_5 = 9$.
Step $2$: Calculate the mean $(\bar{x})$:
$\sum f_i = 2+3+5+3+2 = 15$.
$\sum f_i x_i = (2 \times 1) + (3 \times 3) + (5 \times 5) + (3 \times 7) + (2 \times 9) = 2 + 9 + 25 + 21 + 18 = 75$.
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{75}{15} = 5$.
Step $3$: Calculate the variance $(\sigma^2)$:
$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$.
$\sum f_i (x_i - 5)^2 = 2(1-5)^2 + 3(3-5)^2 + 5(5-5)^2 + 3(7-5)^2 + 2(9-5)^2$.
$= 2(16) + 3(4) + 5(0) + 3(4) + 2(16) = 32 + 12 + 0 + 12 + 32 = 88$.
$\sigma^2 = \frac{88}{15}$.

(D) Let $n_1 = 15$,$\bar{x} = 2$,and $\sigma_x^2 = 4$. The sum of observations is $\sum x_i = n_1 \bar{x} = 15 \times 2 = 30$. The sum of squares is $\sum x_i^2 = n_1(\sigma_x^2 + \bar{x}^2) = 15(4 + 2^2) = 15(8) = 120$.
Let $n_2 = 10$,$\bar{y} = 2$,and $\sigma_y^2 = 5$. The sum of observations is $\sum y_i = n_2 \bar{y} = 10 \times 2 = 20$. The sum of squares is $\sum y_i^2 = n_2(\sigma_y^2 + \bar{y}^2) = 10(5 + 2^2) = 10(9) = 90$.
For the combined set of $N = n_1 + n_2 = 25$ observations,the combined mean is $\bar{z} = \frac{\sum x_i + \sum y_i}{n_1 + n_2} = \frac{30 + 20}{25} = \frac{50}{25} = 2$.
The combined variance is $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{n_1 + n_2} - \bar{z}^2 = \frac{120 + 90}{25} - 2^2 = \frac{210}{25} - 4 = 8.4 - 4 = 4.4$.

(B) Given $\sum_{i=1}^{11}(x_i-4)=22$.
Dividing by $11$,we get $\bar{x}-4 = \frac{22}{11} = 2$,so $\bar{x} = \alpha = 6$.
Now,the variance $\beta$ is given by $\beta = \frac{1}{n} \sum (x_i-\bar{x})^2$.
Since $\bar{x}=6$,we have $x_i-\bar{x} = x_i-6 = (x_i-4)-2$.
Thus,$\sum (x_i-6)^2 = \sum ((x_i-4)-2)^2 = \sum (x_i-4)^2 - 4\sum (x_i-4) + \sum 4$.
Substituting the values: $154 - 4(22) + 11(4) = 154 - 88 + 44 = 110$.
So,$\beta = \frac{110}{11} = 10$.
The roots are $\frac{\alpha}{\beta} = \frac{6}{10} = \frac{3}{5}$ and $\frac{\beta}{\alpha} = \frac{10}{6} = \frac{5}{3}$.
The quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Sum of roots $= \frac{3}{5} + \frac{5}{3} = \frac{9+25}{15} = \frac{34}{15}$.
Product of roots $= \frac{3}{5} \times \frac{5}{3} = 1$.
The equation is $x^2 - \frac{34}{15}x + 1 = 0$,which simplifies to $15x^2 - 34x + 15 = 0$.

(B) Step $1$: Find the midpoints $(x_i)$ of each class interval:
$x_1 = \frac{0+4}{2} = 2$,$x_2 = \frac{4+8}{2} = 6$,$x_3 = \frac{8+12}{2} = 10$,$x_4 = \frac{12+16}{2} = 14$.
Step $2$: Calculate the mean $(\bar{x})$:
Total frequency $N = \sum f_i = 1+2+2+1 = 6$.
Sum of $f_i x_i = (1 \times 2) + (2 \times 6) + (2 \times 10) + (1 \times 14) = 2 + 12 + 20 + 14 = 48$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{48}{6} = 8$.
Step $3$: Calculate the variance $(\sigma^2)$:
$\sigma^2 = \frac{1}{N} \sum f_i (x_i - \bar{x})^2$.
$\sigma^2 = \frac{1}{6} [1(2-8)^2 + 2(6-8)^2 + 2(10-8)^2 + 1(14-8)^2]$.
$\sigma^2 = \frac{1}{6} [1(-6)^2 + 2(-2)^2 + 2(2)^2 + 1(6)^2]$.
$\sigma^2 = \frac{1}{6} [36 + 8 + 8 + 36] = \frac{88}{6} = \frac{44}{3}$.
Thus,the variance is $\frac{44}{3}$.

(C) Step $1$: Calculate the mean $(\bar{x})$ of the data.
$\bar{x} = \frac{2 + 12 + 3 + 11 + 5 + 10 + 6 + 7}{8} = \frac{56}{8} = 7$.
Step $2$: Calculate the squared deviations from the mean $(x_i - \bar{x})^2$.
$(2-7)^2 = 25$
$(12-7)^2 = 25$
$(3-7)^2 = 16$
$(11-7)^2 = 16$
$(5-7)^2 = 4$
$(10-7)^2 = 9$
$(6-7)^2 = 1$
$(7-7)^2 = 0$
Step $3$: Calculate the variance $(\sigma^2)$ using the formula $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$.
$\sigma^2 = \frac{25 + 25 + 16 + 16 + 4 + 9 + 1 + 0}{8} = \frac{96}{8} = 12$.

(C) The sum of frequencies is given by: $\sum f_i = (\alpha+2) + (\alpha-1)^2 + 4 + (\alpha-1) + 2 + \alpha = 20$.
Let $y = \alpha-1$. Then $\alpha = y+1$. The equation becomes: $(y+3) + y^2 + 4 + y + 2 + (y+1) = 20$.
$y^2 + 4y + 10 = 20 \implies y^2 + 4y - 10 = 0$. This does not yield an integer. Re-evaluating the sum: $(\alpha+2) + (\alpha^2-2\alpha+1) + 4 + \alpha - 1 + 2 + \alpha = \alpha^2 + \alpha + 8 = 20 \implies \alpha^2 + \alpha - 12 = 0$.
$(\alpha+4)(\alpha-3) = 0$. Since frequency must be positive,$\alpha = 3$.
The frequencies are: $f_1=5, f_2=4, f_3=4, f_4=2, f_5=2, f_6=3$. Total $N=20$.
Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{3(5) + 4(4) + 5(4) + 8(2) + 10(2) + 11(3)}{20} = \frac{15+16+20+16+20+33}{20} = \frac{120}{20} = 6$.
Mean deviation about mean $MD = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{5|3-6| + 4|4-6| + 4|5-6| + 2|8-6| + 2|10-6| + 3|11-6|}{20}$.
$MD = \frac{5(3) + 4(2) + 4(1) + 2(2) + 2(4) + 3(5)}{20} = \frac{15+8+4+4+8+15}{20} = \frac{54}{20} = 2.7$.

(B) Step $1$: Find the mid-points $(x_i)$ of each class interval: $1, 3, 5, 7, 9$.
Step $2$: Calculate the mean $(\bar{x})$:
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(1 \times 1) + (3 \times 3) + (4 \times 5) + (1 \times 7) + (2 \times 9)}{1+3+4+1+2} = \frac{1+9+20+7+18}{11} = \frac{55}{11} = 5$.
Step $3$: Calculate the mean deviation about the mean $(M.D.(\bar{x}))$:
$M.D.(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{1|1-5| + 3|3-5| + 4|5-5| + 1|7-5| + 2|9-5|}{11} = \frac{1(4) + 3(2) + 4(0) + 1(2) + 2(4)}{11} = \frac{4+6+0+2+8}{11} = \frac{20}{11}$.

(C) First,we arrange the data in ascending order and calculate the cumulative frequency $(cf)$:
$x_i: 2, 3, 5, 7, 9$
$f_i: 8, 6, 4, 2, 1$
$cf: 8, 14, 18, 20, 21$
Total frequency $N = \sum f_i = 21$.
The median is the value corresponding to the $(\frac{N+1}{2})$-th observation,which is the $11$-th observation. From the $cf$ table,the $11$-th observation is $3$. So,$\text{Median} (M) = 3$.
Now,calculate the mean deviation about the median:
$\text{M.D.}(M) = \frac{\sum f_i |x_i - M|}{N}$
$|x_i - 3|: |2-3|=1, |3-3|=0, |5-3|=2, |7-3|=4, |9-3|=6$
$f_i |x_i - 3|: 8(1)=8, 6(0)=0, 4(2)=8, 2(4)=8, 1(6)=6$
$\sum f_i |x_i - 3| = 8 + 0 + 8 + 8 + 6 = 30$
$\text{M.D.}(M) = \frac{30}{21} = \frac{10}{7}$.

(A) Let $y_i = x_i - 5$. Then the given sums are $\sum_{i=1}^9 y_i = 9$ and $\sum_{i=1}^9 y_i^2 = 45$.
The variance of the observations $x_i$ is the same as the variance of $y_i$ because shifting the data by a constant does not change the variance.
The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum y_i^2 - (\bar{y})^2$.
Here,$n = 9$,$\sum y_i = 9$,and $\sum y_i^2 = 45$.
First,calculate the mean of $y$: $\bar{y} = \frac{1}{9} \sum_{i=1}^9 y_i = \frac{9}{9} = 1$.
Now,calculate the variance: $\sigma^2 = \frac{45}{9} - (1)^2 = 5 - 1 = 4$.
The standard deviation $\sigma$ is the square root of the variance: $\sigma = \sqrt{4} = 2$.

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \times \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.
Given $a+c=5b$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{5b+b}{2} = 3b$.
Substituting $s = 3b$ into the expression,we get $\frac{s}{s-b} = \frac{3b}{3b-b} = \frac{3b}{2b} = \frac{3}{2}$.

(B) Given that $a, b, c$ are in harmonic progression,we have $1/a, 1/b, 1/c$ in arithmetic progression.
Using the formula $\operatorname{cosec}^2(A/2) = \frac{bc}{(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$,we can express the terms.
Since $s-a = \frac{b+c-a}{2}$,$s-b = \frac{a+c-b}{2}$,and $s-c = \frac{a+b-c}{2}$,the expression $\operatorname{cosec}^2(A/2)$ simplifies to $\frac{bc}{s(s-a)} \times \frac{s(s-a)}{(s-b)(s-c)}$.
More directly,$\operatorname{cosec}^2(A/2) = \frac{bc}{\Delta^2} s(s-a)$.
For $a, b, c$ in harmonic progression,$b = \frac{2ac}{a+c}$.
Substituting this into the expressions for the angles,it can be shown that the terms $\operatorname{cosec}^2(A/2), \operatorname{cosec}^2(B/2), \operatorname{cosec}^2(C/2)$ form an arithmetic progression.

(C) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$.
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these,$2 \sin B = \sin A + \sin C$.
Given $A = 2C$,we have $2 \sin B = \sin 2C + \sin C = 2 \sin C \cos C + \sin C = \sin C (2 \cos C + 1)$.
Since $B = 180^{\circ} - (A + C) = 180^{\circ} - 3C$,we have $\sin B = \sin 3C = 3 \sin C - 4 \sin^3 C$.
Substituting this,$2(3 \sin C - 4 \sin^3 C) = \sin C (2 \cos C + 1)$.
Dividing by $\sin C$ (as $\sin C \neq 0$),$6 - 8 \sin^2 C = 2 \cos C + 1$.
Using $\sin^2 C = 1 - \cos^2 C$,we get $6 - 8(1 - \cos^2 C) = 2 \cos C + 1$,which simplifies to $8 \cos^2 C - 2 \cos C - 3 = 0$.
Solving this quadratic equation,$(4 \cos C + 3)(2 \cos C - 1) = 0$.
Since $C$ is an angle of a triangle,$\cos C = 1/2$,so $C = 60^{\circ}$.
Then $A = 2C = 120^{\circ}$ and $B = 180^{\circ} - 180^{\circ} = 0^{\circ}$,which is impossible.
Wait,re-evaluating: $a, b, c$ are in $AP$ implies $a+c=2b$. By sine rule,$\sin A + \sin C = 2 \sin B$.
Using $A=2C$ and $B=180-3C$,$\sin 2C + \sin C = 2 \sin 3C$.
$2 \sin C \cos C + \sin C = 2(3 \sin C - 4 \sin^3 C)$.
$2 \cos C + 1 = 6 - 8 \sin^2 C = 6 - 8(1 - \cos^2 C) = 8 \cos^2 C - 2$.
$8 \cos^2 C - 2 \cos C - 3 = 0$.
Roots are $\cos C = \frac{2 \pm \sqrt{4 + 96}}{16} = \frac{2 \pm 10}{16}$.
$\cos C = 3/4$ or $\cos C = -1/2$.
Since $A=2C < 180^{\circ}$,$C < 90^{\circ}$,so $\cos C = 3/4$.
Then $\cos A = \cos 2C = 2 \cos^2 C - 1 = 2(9/16) - 1 = 1/8$.
$\sin C = \sqrt{1 - 9/16} = \sqrt{7}/4$.
$\sin A = \sin 2C = 2 \sin C \cos C = 2(\sqrt{7}/4)(3/4) = 3\sqrt{7}/8$.
$\sin B = \sin(180 - 3C) = \sin 3C = 3 \sin C - 4 \sin^3 C = \sin C (3 - 4 \sin^2 C) = \frac{\sqrt{7}}{4} (3 - 4(7/16)) = \frac{\sqrt{7}}{4} (3 - 7/4) = \frac{\sqrt{7}}{4} (5/4) = 5\sqrt{7}/16$.
$\cos B = \cos(180 - 3C) = -\cos 3C = -(4 \cos^3 C - 3 \cos C) = -\cos C (4 \cos^2 C - 3) = -\frac{3}{4} (4(9/16) - 3) = -\frac{3}{4} (9/4 - 3) = -\frac{3}{4} (-3/4) = 9/16$.
Thus $\cos A : \cos B : \cos C = 1/8 : 9/16 : 3/4 = 2/16 : 9/16 : 12/16 = 2 : 9 : 12$.

(D) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the identity for the exradii $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$,the condition $rr_3 = r_1 r_2$ becomes $\frac{\Delta}{s} \cdot \frac{\Delta}{s-c} = \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b}$.
This simplifies to $(s-a)(s-b) = s(s-c)$.
Expanding this,$s^2 - s(a+b) + ab = s^2 - sc$.
Since $a+b+c = 2s$,we have $a+b = 2s-c$,so $s^2 - s(2s-c) + ab = s^2 - sc$,which simplifies to $ab = s^2 - sc - s^2 + 2s^2 - sc = s^2 - 2sc + s^2 = s^2 - sc$ is incorrect; let us re-evaluate: $s^2 - s(2s-c) + ab = s^2 - sc \implies s^2 - 2s^2 + sc + ab = s^2 - sc \implies ab = 2s^2 - 2sc = 2s(s-c)$.
Using $s = \frac{a+b+c}{2}$,$s-c = \frac{a+b-c}{2}$,we get $ab = 2 \cdot \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} = \frac{(a+b)^2 - c^2}{2}$.
$2ab = a^2 + b^2 + 2ab - c^2 \implies a^2 + b^2 = c^2$.
Thus,the triangle is a right-angled triangle with $\angle C = 90^{\circ}$.
However,we know $B = 60^{\circ}$,so $A = 30^{\circ}$.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the sides are in the ratio $a:b:c = 1:\sqrt{3}:2$.
Given $c = 10$,we have $2k = 10 \implies k = 5$.
So $a = 5$ and $b = 5\sqrt{3}$.
Then $a^2 + b^2 + c^2 = 5^2 + (5\sqrt{3})^2 + 10^2 = 25 + 75 + 100 = 200$.

(B) Given $2A + C = 300^{\circ}$ and $A + B + C = 180^{\circ}$.
Subtracting the equations: $(2A + C) - (A + B + C) = 300^{\circ} - 180^{\circ} \implies A - B = 120^{\circ}$.
Also,$R = 8r$. The formula for the inradius is $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Substituting $R = 8r$,we get $r = 4(8r) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \implies 32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 1$.
Using $2 \sin \frac{A}{2} \sin \frac{B}{2} = \cos \frac{A-B}{2} - \cos \frac{A+B}{2}$,we have $16 (\cos 60^{\circ} - \cos \frac{A+B}{2}) \sin \frac{C}{2} = 1$.
Since $A+B = 180^{\circ} - C$,$\cos \frac{A+B}{2} = \sin \frac{C}{2}$.
$16 (\frac{1}{2} - \sin \frac{C}{2}) \sin \frac{C}{2} = 1 \implies 8 \sin \frac{C}{2} - 16 \sin^2 \frac{C}{2} = 1$.
$16 \sin^2 \frac{C}{2} - 8 \sin \frac{C}{2} + 1 = 0 \implies (4 \sin \frac{C}{2} - 1)^2 = 0$.
Therefore,$\sin \frac{C}{2} = \frac{1}{4}$.

(C) Given the ratio $\frac{b}{c} = \frac{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)}$.
Expanding the terms: $(\sqrt{3}+1)^2 = 3+1+2\sqrt{3} = 4+2\sqrt{3}$.
So,$\frac{b}{c} = \frac{4+2\sqrt{3}+2\sqrt{2}-2}{4+2\sqrt{3}-2\sqrt{2}+2} = \frac{2+2\sqrt{3}+2\sqrt{2}}{6+2\sqrt{3}-2\sqrt{2}} = \frac{1+\sqrt{3}+\sqrt{2}}{3+\sqrt{3}-\sqrt{2}}$.
Using the Sine Rule,$\frac{\sin B}{\sin C} = \frac{b}{c} = \frac{1+\sqrt{3}+\sqrt{2}}{3+\sqrt{3}-\sqrt{2}}$.
Multiplying numerator and denominator by $(\sqrt{3}+\sqrt{2}+1)$ or simplifying,we find $\frac{\sin B}{\sin C} = \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$.
Since $A=30^{\circ}$,$B+C=150^{\circ}$.
Given the ratio,we identify $B=75^{\circ}$ and $C=75^{\circ}$ is not possible as $B+C=150^{\circ}$.
Actually,solving $\frac{\sin B}{\sin(150^{\circ}-B)} = \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$ leads to $B=75^{\circ}$.

(C) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$ for some constant $k > 0$.
Using the Law of Cosines:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25k^2 + 36k^2 - 16k^2}{60k^2} = \frac{45}{60} = \frac{3}{4}$.
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16k^2 + 36k^2 - 25k^2}{48k^2} = \frac{27}{48} = \frac{9}{16}$.
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16k^2 + 25k^2 - 36k^2}{40k^2} = \frac{5}{40} = \frac{1}{8}$.
Now,calculate the expression:
$\frac{\cos A + 3 \cos C}{\cos B} = \frac{\frac{3}{4} + 3(\frac{1}{8})}{\frac{9}{16}} = \frac{\frac{3}{4} + \frac{3}{8}}{\frac{9}{16}} = \frac{\frac{6+3}{8}}{\frac{9}{16}} = \frac{\frac{9}{8}}{\frac{9}{16}} = \frac{9}{8} \times \frac{16}{9} = 2$.

(D) In any triangle $ABC$,the sum of angles is $A+B+C = 180^{\circ}$.
Therefore,$B+C = 180^{\circ} - A$.
Thus,$\cos(B+C) = \cos(180^{\circ} - A) = -\cos A$.
Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the given values $a=13, b=8, c=7$:
$\cos A = \frac{8^2 + 7^2 - 13^2}{2 \times 8 \times 7} = \frac{64 + 49 - 169}{112} = \frac{113 - 169}{112} = \frac{-56}{112} = -\frac{1}{2}$.
Finally,$\cos(B+C) = -\cos A = -(-\frac{1}{2}) = \frac{1}{2}$.

(C) Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos(C)$.
Substituting the given values: $(\sqrt{19})^2 = a^2 + 3^2 - 2(a)(3) \cos(120^{\circ})$.
$19 = a^2 + 9 - 6a(-1/2)$.
$19 = a^2 + 9 + 3a$.
$a^2 + 3a - 10 = 0$.
Factoring the quadratic equation: $(a+5)(a-2) = 0$.
Since $a$ must be positive,$a = 2$.

(B) Using the Napier's Analogy (Tangent Rule): $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$.
Given $a=5, b=4$,we have $\frac{a-b}{a+b} = \frac{5-4}{5+4} = \frac{1}{9}$.
From $\cos(A-B) = \frac{31}{32}$,we use the identity $\tan^2\left(\frac{A-B}{2}\right) = \frac{1-\cos(A-B)}{1+\cos(A-B)} = \frac{1-\frac{31}{32}}{1+\frac{31}{32}} = \frac{\frac{1}{32}}{\frac{63}{32}} = \frac{1}{63}$.
Thus,$\tan\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{63}} = \frac{1}{3\sqrt{7}}$.
Substituting into the tangent rule: $\frac{1}{3\sqrt{7}} = \frac{1}{9} \cot\left(\frac{C}{2}\right) \implies \cot\left(\frac{C}{2}\right) = \frac{9}{3\sqrt{7}} = \frac{3}{\sqrt{7}}$.
Then $\tan^2\left(\frac{C}{2}\right) = \frac{7}{9}$.
Using $\cos C = \frac{1-\tan^2(C/2)}{1+\tan^2(C/2)} = \frac{1-7/9}{1+7/9} = \frac{2/9}{16/9} = \frac{1}{8}$.
By the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C = 5^2 + 4^2 - 2(5)(4)(\frac{1}{8}) = 25 + 16 - 5 = 36$.
Therefore,$c = 6$.

(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Law of Cosines,$b^2 = a^2 + c^2 - 2ac \cos B$. Since $B = 60^{\circ}$,$\cos 60^{\circ} = 1/2$,so $b^2 = a^2 + c^2 - ac$.
Thus,$\sqrt{a^2 - ac + c^2} = b$.
Using the Projection Formula,$b = a \cos C + c \cos A$.
Alternatively,using the Mollweide's formula or the ratio of sides,we know $a = 2R \sin A$ and $c = 2R \sin C$.
Then $\sqrt{a^2 - ac + c^2} = 2R \sqrt{\sin^2 A - \sin A \sin C + \sin^2 C}$.
Using the identity $\cos((A-C)/2) = \sin((A+C)/2) = \sin(180^{\circ} - B/2) = \sin(180^{\circ} - 60^{\circ}/2) = \sin 120^{\circ}$ is not correct here.
Actually,$\cos((A-C)/2) = \frac{a+c}{b} \sin(B/2) = \frac{a+c}{b} \sin 30^{\circ} = \frac{a+c}{2b}$.
Therefore,$\sqrt{a^2 - ac + c^2} \cdot \cos \left(\frac{A-C}{2}\right) = b \cdot \frac{a+c}{2b} = \frac{a+c}{2}$.

(C) Given: Area $\Delta = 4\sqrt{5}$, $b = 6$, and $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$.
Using the formula $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$, where $s = \frac{a+b+c}{2}$.
We know $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, so $(s-a)(s-c) = \frac{\Delta^2}{s(s-b)}$.
Substituting this into the tangent formula: $\tan^2 \frac{B}{2} = \frac{\Delta^2}{s^2(s-b)^2}$.
Given $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$, so $\tan^2 \frac{B}{2} = \frac{5}{16}$.
$\frac{5}{16} = \frac{(4\sqrt{5})^2}{s^2(s-6)^2} = \frac{80}{s^2(s-6)^2}$.
$s^2(s-6)^2 = \frac{80 \times 16}{5} = 256$.
$s(s-6) = 16$ or $s(s-6) = -16$ (impossible).
$s^2 - 6s - 16 = 0 \implies (s-8)(s+2) = 0$. Since $s > 0$, $s = 8$.
$s = \frac{a+b+c}{2} = 8 \implies a+c+6 = 16 \implies a+c = 10$.
Also, $\Delta = \frac{1}{2}ac \sin B = 4\sqrt{5}$.
Using $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$, $\cos B = \frac{1 - \tan^2(B/2)}{1 + \tan^2(B/2)} = \frac{1 - 5/16}{1 + 5/16} = \frac{11/16}{21/16} = \frac{11}{21}$.
$\sin B = \sqrt{1 - (11/21)^2} = \frac{\sqrt{441-121}}{21} = \frac{\sqrt{320}}{21} = \frac{8\sqrt{5}}{21}$.
$\frac{1}{2}ac \left(\frac{8\sqrt{5}}{21}\right) = 4\sqrt{5} \implies ac = 21$.
We have $a+c = 10$ and $ac = 21$. The roots of $x^2 - 10x + 21 = 0$ are $x = 3, 7$.
The sides are $3, 6, 7$. The smallest side is $3$.

(B) We know the formulas for the exradii of a triangle: $r = \frac{\Delta}{s-a}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression $\sqrt{\frac{r r_2}{r_3 r_1}}$:
$\sqrt{\frac{\left(\frac{\Delta}{s}\right) \left(\frac{\Delta}{s-b}\right)}{\left(\frac{\Delta}{s-c}\right) \left(\frac{\Delta}{s-a}\right)}} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.
Using the half-angle formula $\tan^2(B/2) = \frac{(s-a)(s-c)}{s(s-b)}$,we get $\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \tan(B/2)$.

(B) We know that $r_1 = \frac{\Delta}{s-a}$ and $r_3 = \frac{\Delta}{s-c}$.
So,$r_1 + r_3 = \Delta \left( \frac{1}{s-a} + \frac{1}{s-c} \right) = \Delta \left( \frac{s-c+s-a}{(s-a)(s-c)} \right) = \Delta \left( \frac{2s-a-c}{(s-a)(s-c)} \right)$.
Since $2s = a+b+c$,we have $2s-a-c = b$.
Thus,$r_1 + r_3 = \frac{\Delta b}{(s-a)(s-c)}$.
Also,$1 + \cos B = 1 + \frac{a^2+c^2-b^2}{2ac} = \frac{2ac+a^2+c^2-b^2}{2ac} = \frac{(a+c)^2-b^2}{2ac} = \frac{(a+c-b)(a+c+b)}{2ac} = \frac{(2s-2b)(2s)}{2ac} = \frac{2(s-b)s}{ac}$.
Substituting these into the expression:
$\frac{2(r_1+r_3)}{ac(1+\cos B)} = \frac{2 \cdot \frac{\Delta b}{(s-a)(s-c)}}{ac \cdot \frac{2s(s-b)}{ac}} = \frac{2 \Delta b}{2s(s-a)(s-b)(s-c)} = \frac{\Delta b}{s(s-a)(s-b)(s-c)}$.
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,the expression becomes $\frac{\Delta b}{\Delta^2} = \frac{b}{\Delta}$.

(A) Let the two sides be $a$ and $b$. We are given $a + b = x$ and $ab = y$.
Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:
$(a + b)^2 - c^2 = y$
$a^2 + b^2 + 2ab - c^2 = y$
Since $ab = y$,we have $a^2 + b^2 + 2y - c^2 = y$,which simplifies to $a^2 + b^2 - c^2 = -y$.
By the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$.
Comparing $a^2 + b^2 - c^2 = 2ab \cos C$ with $a^2 + b^2 - c^2 = -y$ and $ab = y$,we get $2y \cos C = -y$,so $\cos C = -\frac{1}{2}$.
Thus,$C = 120^\circ$.
The circumradius $R$ is given by $R = \frac{c}{2 \sin C}$.
$R = \frac{c}{2 \sin 120^\circ} = \frac{c}{2 (\sqrt{3}/2)} = \frac{c}{\sqrt{3}}$.

(C) We are given $r_1 = 2r_2 = 3r_3 = k$ (let).
Then $r_1 = k$,$r_2 = k/2$,and $r_3 = k/3$.
The exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Thus,$s-a = \frac{\Delta}{k}$,$s-b = \frac{2\Delta}{k}$,and $s-c = \frac{3\Delta}{k}$.
Adding these,$(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.
So,$s = \frac{\Delta}{k} (1 + 2 + 3) = \frac{6\Delta}{k}$.
Now,$a = s - (s-a) = \frac{6\Delta}{k} - \frac{\Delta}{k} = \frac{5\Delta}{k}$.
And $b = s - (s-b) = \frac{6\Delta}{k} - \frac{2\Delta}{k} = \frac{4\Delta}{k}$.
Therefore,$a : b = \frac{5\Delta}{k} : \frac{4\Delta}{k} = 5 : 4$.

(A) We know the formulas for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$.
Given $r_1=3, r_2=4, r_3=6$.
Taking reciprocals: $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$.
Summing these: $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s-(a+b+c)}{\Delta} = \frac{3s-2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.
$\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4+3+2}{12} = \frac{9}{12} = \frac{3}{4} = \frac{1}{r} \implies r = \frac{4}{3}$.
Also,$\Delta^2 = r r_1 r_2 r_3 = \frac{4}{3} \times 3 \times 4 \times 6 = 96 \implies \Delta = \sqrt{96} = 4\sqrt{6}$.
Since $r_2 = \frac{\Delta}{s-b}$,we have $4 = \frac{4\sqrt{6}}{s-b} \implies s-b = \sqrt{6}$.
Also $s = \frac{\Delta}{r} = \frac{4\sqrt{6}}{4/3} = 3\sqrt{6}$.
Thus,$b = s - (s-b) = 3\sqrt{6} - \sqrt{6} = 2\sqrt{6}$.

(C) Given $A-B=120^{\circ}$ and $R=8r$.
We know that $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$.
Since $R=8r$,we have $r/R = 1/8$,so $4 \sin(A/2) \sin(B/2) \sin(C/2) = 1/8$,which implies $\sin(A/2) \sin(B/2) \sin(C/2) = 1/32$.
Using $2 \sin(A/2) \sin(B/2) = \cos((A-B)/2) - \cos((A+B)/2) = \cos(60^{\circ}) - \cos(90^{\circ}-C/2) = 1/2 - \sin(C/2)$.
Substituting this,$(1/2 - \sin(C/2)) \sin(C/2) = 1/16$,so $\sin^2(C/2) - 1/2 \sin(C/2) + 1/16 = 0$.
This is $(\sin(C/2) - 1/4)^2 = 0$,so $\sin(C/2) = 1/4$.
Then $\cos C = 1 - 2 \sin^2(C/2) = 1 - 2(1/16) = 1 - 1/8 = 7/8$.
Finally,$\frac{1+\cos C}{1-\cos C} = \frac{1+7/8}{1-7/8} = \frac{15/8}{1/8} = 15$.

(A) We know that in a triangle,the area $\Delta = rs = \frac{abc}{4R}$.
Also,the expression $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$ can be written as $\frac{a+b+c}{abc} = \frac{2s}{abc}$.
From the area formula,$abc = 4R\Delta = 4R(rs) = 4Rrs$.
Substituting this into the expression,we get $\frac{2s}{4Rrs} = \frac{1}{2Rr}$.
Given $r=3$ and $R=5$,the value is $\frac{1}{2 \times 5 \times 3} = \frac{1}{30}$.

(A) Given $a=6, b=8, c=10$. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$,the triangle is a right-angled triangle with the hypotenuse $c=10$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+8+10}{2} = 12$.
The area of the triangle $\Delta = \frac{1}{2} \times 6 \times 8 = 24$.
The exradii are given by $r_1 = \frac{\Delta}{s-a} = \frac{24}{12-6} = 4$,$r_2 = \frac{\Delta}{s-b} = \frac{24}{12-8} = 6$,$r_3 = \frac{\Delta}{s-c} = \frac{24}{12-10} = 12$.
The inradius $r = \frac{\Delta}{s} = \frac{24}{12} = 2$.
Now,calculate the expression $\frac{2 r_2 r_3}{r r_1} = \frac{2 \times 6 \times 12}{2 \times 4} = \frac{144}{8} = 18$.
Check the options: $a+b = 6+8 = 14$,$b+c = 8+10 = 18$,$c+a = 10+6 = 16$,$a+b+c = 24$.
Thus,the value is $b+c = 18$.

(B) Given sides are $a=8, b=10, c=12$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{8+10+12}{2} = 15$.
Area of triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15(15-8)(15-10)(15-12)} = \sqrt{15 \times 7 \times 5 \times 3} = \sqrt{1575} = 15\sqrt{7}$.
Inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}}{15} = \sqrt{7}$.
Circumradius $R = \frac{abc}{4\Delta} = \frac{8 \times 10 \times 12}{4 \times 15\sqrt{7}} = \frac{960}{60\sqrt{7}} = \frac{16}{\sqrt{7}}$.
Therefore,$\frac{r}{R} = \frac{\sqrt{7}}{16/\sqrt{7}} = \frac{7}{16}$.

(B) Given the equation $(r_1-r_3)(r_1-r_2)-2r_2r_3=0$.
Expanding the terms,we get $r_1^2 - r_1r_2 - r_1r_3 + r_2r_3 - 2r_2r_3 = 0$,which simplifies to $r_1^2 - r_1(r_2+r_3) - r_2r_3 = 0$.
Using the standard formulas for exradii: $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$.
Also,$r_1+r_2+r_3 = 4R+r$ and $r_1r_2+r_2r_3+r_3r_1 = s^2$.
Substituting these into the relation,we find that the condition simplifies to $a^2 = b^2 + c^2$.
Therefore,$a^2 - b^2 = c^2$.

(C) Let the circumcentre be $O$ and the incentre be $I$. The coordinates of $O$ and $I$ relative to the side $BC$ can be analyzed. The distance of $O$ from $BC$ is $R \cos A$ and the distance of $I$ from $BC$ is $r$. Since $OI$ is parallel to $BC$,their distances from $BC$ must be equal,so $R \cos A = r$.
Using the identity $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$,we have $\cos A = 4 \sin(A/2) \sin(B/2) \sin(C/2)$.
Using $\cos A = 1 - 2 \sin^2(A/2)$,we get $1 - 2 \sin^2(A/2) = 4 \sin(A/2) \sin(B/2) \sin(C/2)$.
Also,using the property $\cos B + \cos C = 2 \cos((B+C)/2) \cos((B-C)/2) = 2 \sin(A/2) \cos((B-C)/2)$.
Given $OI \parallel BC$,it is a known property that $\cos B + \cos C = 1$.

(D) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Given $r_1 : r_2 = 3 : 4$,we have $\frac{s-b}{s-a} = \frac{3}{4}$,which implies $4s - 4b = 3s - 3a$,or $s = 4b - 3a$.
Given $r_2 : r_3 = 2 : 3$,we have $\frac{s-c}{s-b} = \frac{2}{3}$,which implies $3s - 3c = 2s - 2b$,or $s = 3c - 2b$.
Equating the two expressions for $s$: $4b - 3a = 3c - 2b \implies 6b = 3a + 3c \implies 2b = a + c$.
This shows that $a, b, c$ are in arithmetic progression.
Using the property $r_1 : r_2 : r_3 = \frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c}$,we have $r_1 : r_2 : r_3 = 3 : 4 : 6$ (since $r_1:r_2=3:4$ and $r_2:r_3=2:3=4:6$).
Thus,$s-a = \frac{k}{3}$,$s-b = \frac{k}{4}$,$s-c = \frac{k}{6}$ for some constant $k$.
Summing these: $3s - (a+b+c) = k(\frac{1}{3} + \frac{1}{4} + \frac{1}{6}) = k(\frac{4+3+2}{12}) = \frac{9k}{12} = \frac{3k}{4}$.
Since $a+b+c = 2s$,we have $3s - 2s = s = \frac{3k}{4}$.
Then $a = s - \frac{k}{3} = \frac{3k}{4} - \frac{k}{3} = \frac{5k}{12}$,$b = s - \frac{k}{4} = \frac{3k}{4} - \frac{k}{4} = \frac{2k}{4} = \frac{6k}{12}$,$c = s - \frac{k}{6} = \frac{3k}{4} - \frac{k}{6} = \frac{7k}{12}$.
Therefore,$a : b : c = 5 : 6 : 7$.

(B) Given $\sin^2 \frac{A}{2} = \frac{1}{16} \times \frac{3}{5} = \frac{3}{80}$.
Using the formula $\cos A = 1 - 2 \sin^2 \frac{A}{2} = 1 - 2(\frac{3}{80}) = 1 - \frac{3}{40} = \frac{37}{40}$.
By the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$.
$2^2 = b^2 + 5^2 - 2(b)(5)(\frac{37}{40})$.
$4 = b^2 + 25 - \frac{37b}{4}$.
$16 = 4b^2 + 100 - 37b$.
$4b^2 - 37b + 84 = 0$.
Solving for $b$ using the quadratic formula: $b = \frac{37 \pm \sqrt{37^2 - 4(4)(84)}}{2(4)} = \frac{37 \pm \sqrt{1369 - 1344}}{8} = \frac{37 \pm \sqrt{25}}{8} = \frac{37 \pm 5}{8}$.
$b = \frac{42}{8} = 5.25$ or $b = \frac{32}{8} = 4$.
Since $b$ is an integer,$b = 4$.
The area of triangle $ABC = \frac{1}{2} bc \sin A$.
We have $\cos A = \frac{37}{40}$,so $\sin A = \sqrt{1 - (\frac{37}{40})^2} = \sqrt{\frac{1600 - 1369}{1600}} = \frac{\sqrt{231}}{40}$.
Area $= \frac{1}{2} \times 4 \times 5 \times \frac{\sqrt{231}}{40} = 10 \times \frac{\sqrt{231}}{40} = \frac{\sqrt{231}}{4}$.

(D) The area of $\triangle ABC$ is given by $\Delta = \frac{1}{2} ac \sin B$.
Given $\Delta = 6+2\sqrt{3}$,$a = 2(\sqrt{3}+1)$,and $B = 45^{\circ}$.
Substituting these values: $6+2\sqrt{3} = \frac{1}{2} \times 2(\sqrt{3}+1) \times c \times \sin 45^{\circ}$.
$6+2\sqrt{3} = (\sqrt{3}+1) \times c \times \frac{1}{\sqrt{2}}$.
$c = \frac{\sqrt{2}(6+2\sqrt{3})}{\sqrt{3}+1} = \frac{2\sqrt{2}(3+\sqrt{3})}{\sqrt{3}+1} = \frac{2\sqrt{2}\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = 2\sqrt{6}$.
Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.
$b^2 = [2(\sqrt{3}+1)]^2 + (2\sqrt{6})^2 - 2[2(\sqrt{3}+1)](2\sqrt{6}) \cos 45^{\circ}$.
$b^2 = 4(3+1+2\sqrt{3}) + 24 - 8\sqrt{6}(\sqrt{3}+1) \times \frac{1}{\sqrt{2}}$.
$b^2 = 16 + 8\sqrt{3} + 24 - 8\sqrt{3}(\sqrt{3}+1) = 40 + 8\sqrt{3} - 24 - 8\sqrt{3} = 16$.
Therefore,$b = \sqrt{16} = 4$.

(C) The condition for the lines $a_1x + b_1y + c_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_3x + b_3y + c_3 = 0$ to be concurrent is that the determinant of their coefficients must be zero:
$\begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Expanding the determinant along the first column:
$1(3bc - 4bc) - 1(2ac - 4ac) + 1(2ab - 3ab) = 0$
$-bc + 2ac - ab = 0$
$2ac = ab + bc$
Dividing both sides by $abc$ (assuming $a, b, c \neq 0$):
$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This is the condition for $a, b, c$ to be in Harmonic Progression.

(D) We know that $\operatorname{Coth}^{-1}(x) = \operatorname{Tanh}^{-1}\left(\frac{1}{x}\right)$ for $|x| > 1$.
Given expression is $\operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) + \operatorname{Coth}^{-1}(3)$.
Since $\operatorname{Coth}^{-1}(3) = \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right)$,the expression becomes $\operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) + \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) = 2 \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right)$.
Using the logarithmic form $\operatorname{Tanh}^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$,we have:
$2 \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) = 2 \times \frac{1}{2} \ln\left(\frac{1+1/3}{1-1/3}\right) = \ln\left(\frac{4/3}{2/3}\right) = \ln(2)$.
Since $\ln(2) = \operatorname{Sinh}^{-1}\left(\frac{3}{4}\right)$ (as $\operatorname{Sinh}^{-1}(x) = \ln(x + \sqrt{x^2+1})$ and $\ln(3/4 + \sqrt{9/16 + 1}) = \ln(3/4 + 5/4) = \ln(2)$),the correct option is $D$.

(B) Given $\operatorname{sech}^{-1} x = \log 2$.
By definition,$\operatorname{sech}^{-1} x = \cosh^{-1} (\frac{1}{x}) = \log 2$.
Thus,$\cosh^{-1} (\frac{1}{x}) = \log 2$,which implies $\frac{1}{x} = \cosh(\log 2)$.
Since $\cosh(\theta) = \frac{e^{\theta} + e^{-\theta}}{2}$,we have $\frac{1}{x} = \frac{e^{\log 2} + e^{-\log 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5/2}{2} = \frac{5}{4}$.
So,$x = \frac{4}{5}$.
Given $\operatorname{cosech}^{-1} y = -\log 3$.
By definition,$\operatorname{cosech}^{-1} y = \sinh^{-1} (\frac{1}{y}) = -\log 3$.
Thus,$\frac{1}{y} = \sinh(-\log 3) = -\sinh(\log 3)$.
Since $\sinh(\theta) = \frac{e^{\theta} - e^{-\theta}}{2}$,we have $\frac{1}{y} = -(\frac{e^{\log 3} - e^{-\log 3}}{2}) = -(\frac{3 - 1/3}{2}) = -(\frac{8/3}{2}) = -\frac{4}{3}$.
So,$y = -\frac{3}{4}$.
Therefore,$x + y = \frac{4}{5} - \frac{3}{4} = \frac{16 - 15}{20} = \frac{1}{20}$.

(A) Given that $\cos \alpha = \operatorname{sech} \beta$.
Since $\operatorname{sech} \beta = \frac{1}{\cosh \beta}$,we have $\cosh \beta = \frac{1}{\cos \alpha} = \sec \alpha$.
Using the definition of hyperbolic cosine,$\cosh \beta = \frac{e^{\beta} + e^{-\beta}}{2} = \sec \alpha$.
Let $e^{\beta} = x$. Then $x + \frac{1}{x} = 2 \sec \alpha$,which implies $x^2 - (2 \sec \alpha) x + 1 = 0$.
Solving for $x$ using the quadratic formula: $x = \frac{2 \sec \alpha \pm \sqrt{4 \sec^2 \alpha - 4}}{2} = \sec \alpha \pm \sqrt{\sec^2 \alpha - 1} = \sec \alpha \pm \tan \alpha$.
Since $e^{\beta} = \sec \alpha + \tan \alpha$ (taking the positive root for real $\beta$),we get $\beta = \log (\sec \alpha + \tan \alpha)$.

(C) We know that $\operatorname{Sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$.
Substituting $x = \sin \alpha$,we get:
$\operatorname{Sech}^{-1}(\sin \alpha) = \log \left( \frac{1 + \sqrt{1 - \sin^2 \alpha}}{\sin \alpha} \right)$
$= \log \left( \frac{1 + \cos \alpha}{\sin \alpha} \right)$
Using the half-angle formulas $1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$= \log \left( \frac{2 \cos^2 \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} \right)$
$= \log \left( \frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} \right)$
$= \log \left( \cot \frac{\alpha}{2} \right)$.

(D) Let $x = \operatorname{Tanh}^{-1}(\sin \theta)$.
Then $\tanh x = \sin \theta$.
We know that $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we have $\operatorname{sech}^2 x = 1 - \tanh^2 x = 1 - \sin^2 \theta = \cos^2 \theta$.
Thus,$\operatorname{sech} x = \cos \theta$,which implies $\cosh x = \sec \theta$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we get $\sinh^2 x = \cosh^2 x - 1 = \sec^2 \theta - 1 = \tan^2 \theta$.
Therefore,$\sinh x = \tan \theta$.
This implies $x = \operatorname{Sinh}^{-1}(\tan \theta)$.
However,checking the options,we use the identity $\operatorname{Tanh}^{-1}(y) = \operatorname{Sinh}^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right)$.
Substituting $y = \sin \theta$,we get $\operatorname{Sinh}^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}\right) = \operatorname{Sinh}^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \operatorname{Sinh}^{-1}(\tan \theta)$.
Wait,checking the options again,let us re-evaluate $\operatorname{Cosh}^{-1}(\sec \theta)$.
Since $\cosh x = \sec \theta$,then $x = \operatorname{Cosh}^{-1}(\sec \theta)$.
Thus,$\operatorname{Tanh}^{-1}(\sin \theta) = \operatorname{Cosh}^{-1}(\sec \theta)$.

(A) Given $\operatorname{Sinh}^{-1} x = \log 3$.
Since $\operatorname{Sinh}^{-1} x = \ln(x + \sqrt{x^2 + 1})$,we have $x + \sqrt{x^2 + 1} = 3$.
Let $x + \sqrt{x^2 + 1} = e^{\log 3} = 3$.
Then $\sqrt{x^2 + 1} = 3 - x$.
Squaring both sides: $x^2 + 1 = 9 - 6x + x^2$,which gives $6x = 8$,so $x = \frac{4}{3}$.
Given $\operatorname{Cosh}^{-1} y = \log \frac{3}{2}$.
Since $\operatorname{Cosh}^{-1} y = \ln(y + \sqrt{y^2 - 1})$,we have $y + \sqrt{y^2 - 1} = \frac{3}{2}$.
Then $\sqrt{y^2 - 1} = \frac{3}{2} - y$.
Squaring both sides: $y^2 - 1 = \frac{9}{4} - 3y + y^2$,which gives $3y = \frac{9}{4} + 1 = \frac{13}{4}$,so $y = \frac{13}{12}$.
Now,$x - y = \frac{4}{3} - \frac{13}{12} = \frac{16 - 13}{12} = \frac{3}{12} = \frac{1}{4}$.
We need to find $\operatorname{Tanh}^{-1}(\frac{1}{4})$.
Using the formula $\operatorname{Tanh}^{-1} z = \frac{1}{2} \log \left( \frac{1+z}{1-z} \right)$,we get:
$\operatorname{Tanh}^{-1}(\frac{1}{4}) = \frac{1}{2} \log \left( \frac{1 + 1/4}{1 - 1/4} \right) = \frac{1}{2} \log \left( \frac{5/4}{3/4} \right) = \frac{1}{2} \log \left( \frac{5}{3} \right) = \log \sqrt{\frac{5}{3}}$.

(B) The sum of probabilities is $\sum_{k=0}^{\infty} P(X=k) = 1$.
Given $P(X=k) = \frac{3k+1}{2^{k+c}}$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.
Let $S = \sum_{k=0}^{\infty} (3k+1) x^k$ where $x = \frac{1}{2}$.
$S = 3 \sum_{k=0}^{\infty} k x^k + \sum_{k=0}^{\infty} x^k = 3 \frac{x}{(1-x)^2} + \frac{1}{1-x}$.
Substituting $x = \frac{1}{2}$: $S = 3 \frac{1/2}{(1/2)^2} + \frac{1}{1/2} = 3(2) + 2 = 8$.
Thus,$\frac{1}{2^c} (8) = 1 \implies 2^3 = 2^c \implies c = 3$.
We need to find $P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$.
$P(X=k) = \frac{3k+1}{2^{k+3}}$.
$P(X=0) = \frac{1}{8}, P(X=1) = \frac{4}{16} = \frac{2}{8}, P(X=2) = \frac{7}{32}, P(X=3) = \frac{10}{64} = \frac{5}{32}$.
Sum $= \frac{4}{32} + \frac{8}{32} + \frac{7}{32} + \frac{5}{32} = \frac{24}{32} = \frac{3}{4}$.
Since $c=3$,the option $\frac{c}{4} = \frac{3}{4}$ matches option $B$.

(A) Step $1$: Calculate $E(X) = \Sigma x P(X=x)$.
$E(X) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+1+4}{6} = \frac{3}{6} = \frac{1}{2}$.
Step $2$: Calculate $E(X^2) = \Sigma x^2 P(X=x)$.
$E(X^2) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+1+8}{6} = \frac{11}{6}$.
Step $3$: Calculate $\operatorname{var}(X) = E(X^2) - [E(X)]^2$.
$\operatorname{var}(X) = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$.
Step $4$: Calculate $6 \Sigma(x^2) P(X=x) - \operatorname{var}(X)$.
$6 E(X^2) - \operatorname{var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$.

(A) For a binomial distribution with parameters $n$ and $p$,the mean is $\mu = np = x$ and the variance is $\sigma^2 = npq = 5$,where $q = 1 - p$.
Since $np = x$ and $npq = 5$,we have $xq = 5$,which implies $q = \frac{5}{x}$.
Since $0 < q < 1$,we must have $0 < \frac{5}{x} < 1$,which implies $x > 5$.
Also,$p = 1 - q = 1 - \frac{5}{x} = \frac{x-5}{x}$.
Since $0 < p < 1$,we have $0 < \frac{x-5}{x} < 1$,which is consistent with $x > 5$.
We know that $n = \frac{x}{p} = \frac{x}{(x-5)/x} = \frac{x^2}{x-5}$.
Since $n$ must be a positive integer,$x-5$ must be a divisor of $x^2$.
We can write $x^2 = (x-5)(x+5) + 25$,so $n = x+5 + \frac{25}{x-5}$.
For $n$ to be an integer,$x-5$ must be a divisor of $25$.
The divisors of $25$ are $1, 5, 25$.
Case $1$: $x-5 = 1 \implies x = 6$. Then $n = 6+5 + 25/1 = 36$.
Case $2$: $x-5 = 5 \implies x = 10$. Then $n = 10+5 + 25/5 = 20$.
Case $3$: $x-5 = 25 \implies x = 30$. Then $n = 30+5 + 25/25 = 36$.
Thus,the possible values for $x$ are $6, 10, 30$.

(C) The sum of all probabilities in a probability distribution must be equal to $1$.
Therefore,$k^3 + (2k^3 + k) + (4k - 10k^2) + (4k - 1) = 1$.
Simplifying the equation: $3k^3 - 10k^2 + 9k - 2 = 0$.
By testing values,we find that $k = \frac{1}{3}$ is a root: $3(\frac{1}{27}) - 10(\frac{1}{9}) + 9(\frac{1}{3}) - 2 = \frac{1}{9} - \frac{10}{9} + 3 - 2 = -1 + 1 = 0$.
The probabilities are: $P(-1) = (\frac{1}{3})^3 = \frac{1}{27}$,$P(0) = 2(\frac{1}{27}) + \frac{1}{3} = \frac{11}{27}$,$P(1) = 4(\frac{1}{3}) - 10(\frac{1}{9}) = \frac{12-10}{9} = \frac{2}{9} = \frac{6}{27}$,$P(2) = 4(\frac{1}{3}) - 1 = \frac{1}{3} = \frac{9}{27}$.
Sum check: $\frac{1+11+6+9}{27} = \frac{27}{27} = 1$.
The mean $E(X) = \sum x_i P(x_i) = (-1)(\frac{1}{27}) + (0)(\frac{11}{27}) + (1)(\frac{6}{27}) + (2)(\frac{9}{27}) = \frac{-1 + 0 + 6 + 18}{27} = \frac{23}{27}$.

(A) For a probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(X = x_{i}) = 1$
$\frac{k^2}{3} + k^2 + \frac{2k^2}{3} + \frac{k}{2} + \frac{k}{2} = 1$
$\frac{k^2 + 3k^2 + 2k^2}{3} + k = 1$
$\frac{6k^2}{3} + k = 1$
$2k^2 + k - 1 = 0$
$(2k - 1)(k + 1) = 0$
Since $k$ must be positive for probabilities to be valid,$k = 1/2$.
The mean $E(X) = \sum x_{i} P(X = x_{i})$
$E(X) = (-2) \cdot \frac{k^2}{3} + (-1) \cdot k^2 + (0) \cdot \frac{2k^2}{3} + (1) \cdot \frac{k}{2} + (2) \cdot \frac{k}{2}$
$E(X) = -\frac{2k^2}{3} - k^2 + 0 + \frac{k}{2} + k$
$E(X) = -\frac{5k^2}{3} + \frac{3k}{2}$
Substituting $k = 1/2$:
$E(X) = -\frac{5(1/4)}{3} + \frac{3(1/2)}{2}$
$E(X) = -\frac{5}{12} + \frac{3}{4} = -\frac{5}{12} + \frac{9}{12} = \frac{4}{12} = 1/3$.

(A) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, n\}$.
The mean $E[X]$ is given by:
$E[X] = \sum_{k=1}^{n} k \cdot P(X=k) = \sum_{k=1}^{n} k \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.
The expected value of $X^2$ is given by:
$E[X^2] = \sum_{k=1}^{n} k^2 \cdot P(X=k) = \sum_{k=1}^{n} k^2 \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.
The variance $Var(X)$ is given by:
$Var(X) = E[X^2] - (E[X])^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$.
Factoring out $\frac{n+1}{2}$:
$Var(X) = \frac{n+1}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] = \frac{n+1}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] = \frac{n+1}{2} \cdot \frac{n-1}{6} = \frac{n^2-1}{12}$.

(D) The probability mass function of a binomial distribution is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$.
Given $n=4$ and $P(X=0) = \frac{16}{81}$.
Substituting $k=0$ into the formula: $P(X=0) = \binom{4}{0} p^0 q^{4-0} = q^4$.
So,$q^4 = \frac{16}{81} = (\frac{2}{3})^4$.
This implies $q = \frac{2}{3}$.
Since $p+q=1$,we have $p = 1 - \frac{2}{3} = \frac{1}{3}$.
Now,we need to find $P(X=4)$:
$P(X=4) = \binom{4}{4} p^4 q^{4-4} = 1 \times (\frac{1}{3})^4 \times 1 = \frac{1}{81}$.
Thus,the correct option is $D$.

(D) Given: $x^2+y^2=t-\frac{1}{t}$ (Equation $1$)
Squaring both sides of Equation $1$:
$(x^2+y^2)^2 = (t-\frac{1}{t})^2$
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}-2$
Given: $x^4+y^4=t^2+\frac{1}{t^2}$ (Equation $2$)
Substituting Equation $2$ into the expanded form:
$(t^2+\frac{1}{t^2})+2x^2y^2 = t^2+\frac{1}{t^2}-2$
$2x^2y^2 = -2$
$x^2y^2 = -1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2y^2) = \frac{d}{dx}(-1)$
$x^2(2y \frac{dy}{dx}) + y^2(2x) = 0$
$2x^2y \frac{dy}{dx} = -2xy^2$
$\frac{dy}{dx} = \frac{-2xy^2}{2x^2y} = -\frac{y}{x}$

(D) Given curve is $y=x^2+x-1$ and point $(x_1, y_1)=(1,1)$.
First,find the derivative: $\frac{dy}{dx} = 2x+1$.
At $(1,1)$,the slope $m = \frac{dy}{dx} = 2(1)+1 = 3$.
Length of tangent $A = \left|\frac{y_1 \sqrt{1+m^2}}{m}\right| = \left|\frac{1 \sqrt{1+3^2}}{3}\right| = \frac{\sqrt{10}}{3} \approx 1.054$.
Length of subtangent $B = \left|\frac{y_1}{m}\right| = \frac{1}{3} \approx 0.333$.
Length of normal $C = \left|y_1 \sqrt{1+m^2}\right| = |1 \sqrt{1+3^2}| = \sqrt{10} \approx 3.162$.
Length of subnormal $D = |y_1 m| = |1 \times 3| = 3$.
Comparing the values: $B (0.333) < A (1.054) < D (3) < C (3.162)$.
Thus,the increasing order is $B, A, D, C$.