TS EAMCET 2020 Mathematics Question Paper with Answer and Solution

(D) Let $\angle B = \theta$. Then $\angle A = 3\theta$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle C = 180^{\circ} - (A + B) = 180^{\circ} - 4\theta$.
Using the sine rule,$\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}$.
Substituting the given values: $\frac{AB}{\sin(180^{\circ} - 4\theta)} = \frac{16}{\sin 3\theta} = \frac{9}{\sin \theta}$.
From $\frac{16}{\sin 3\theta} = \frac{9}{\sin \theta}$,we have $16 \sin \theta = 9 \sin 3\theta = 9(3 \sin \theta - 4 \sin^3 \theta)$.
Dividing by $\sin \theta$ (since $\sin \theta \neq 0$),$16 = 27 - 36 \sin^2 \theta$,which gives $36 \sin^2 \theta = 11$,so $\sin^2 \theta = \frac{11}{36}$.
Then $\cos^2 \theta = 1 - \frac{11}{36} = \frac{25}{36}$,so $\cos \theta = \frac{5}{6}$.
Now,$\sin 4\theta = 2 \sin 2\theta \cos 2\theta = 4 \sin \theta \cos \theta (1 - 2 \sin^2 \theta) = 4 \times \frac{\sqrt{11}}{6} \times \frac{5}{6} \times (1 - 2 \times \frac{11}{36}) = \frac{20\sqrt{11}}{36} \times (1 - \frac{11}{18}) = \frac{5\sqrt{11}}{9} \times \frac{7}{18} = \frac{35\sqrt{11}}{162}$.
Finally,$AB = \frac{9 \sin 4\theta}{\sin \theta} = 9 \times \frac{35\sqrt{11}}{162} \times \frac{6}{\sqrt{11}} = \frac{35}{3}$.

(B) In $\triangle ABC$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the expression $(b^2-c^2) \cot A + (c^2-a^2) \cot B$:
$= 4R^2(\sin^2 B - \sin^2 C) \cot A + 4R^2(\sin^2 C - \sin^2 A) \cot B$
$= 4R^2[\sin(B+C)\sin(B-C) \frac{\cos A}{\sin A} + \sin(C+A)\sin(C-A) \frac{\cos B}{\sin B}]$
Since $A+B+C = \pi$,we have $\sin(B+C) = \sin A$ and $\sin(C+A) = \sin B$.
$= 4R^2[\sin A \sin(B-C) \frac{\cos A}{\sin A} + \sin B \sin(C-A) \frac{\cos B}{\sin B}]$
$= 4R^2[\sin(B-C)\cos A + \sin(C-A)\cos B]$
$= 4R^2[\sin(B-C)(-\cos(B+C)) + \sin(C-A)(-\cos(C+A))]$
$= 2R^2[-(2\sin(B-C)\cos(B+C)) - (2\sin(C-A)\cos(C+A))]$
$= 2R^2[-(\sin 2B - \sin 2C) - (\sin 2C - \sin 2A)]$
$= 2R^2[\sin 2A - \sin 2B]$

(C) Let a triangle $ABC$ have sides $a, b, c$ and area $\Delta$.
We know that the area $\Delta = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C$.
From the cosine rule,we have $a^2 = b^2+c^2-2bc \cos A$,$b^2 = a^2+c^2-2ac \cos B$,and $c^2 = a^2+b^2-2ab \cos C$.
Adding these three equations gives $a^2+b^2+c^2 = 2(a^2+b^2+c^2) - 2(bc \cos A + ac \cos B + ab \cos C)$.
Rearranging,we get $a^2+b^2+c^2 = 2(bc \cos A + ac \cos B + ab \cos C)$.
Since $\Delta = \frac{1}{2}bc \sin A$,we have $bc = \frac{2\Delta}{\sin A}$. Similarly,$ac = \frac{2\Delta}{\sin B}$ and $ab = \frac{2\Delta}{\sin C}$.
Substituting these into the equation:
$a^2+b^2+c^2 = 2 \left( \frac{2\Delta}{\sin A} \cos A + \frac{2\Delta}{\sin B} \cos B + \frac{2\Delta}{\sin C} \cos C \right)$
$a^2+b^2+c^2 = 4\Delta (\cot A + \cot B + \cot C)$
Therefore,$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$.

(C) Given that,$\frac{a^3+b^3+c^3}{\sin^3 A+\sin^3 B+\sin^3 C}=8$ $(i)$
Using the sine rule for a triangle $ABC$,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,where $R$ is the circumradius.
Thus,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into equation $(i)$:
$\frac{(2R \sin A)^3 + (2R \sin B)^3 + (2R \sin C)^3}{\sin^3 A + \sin^3 B + \sin^3 C} = 8$
$\frac{(2R)^3 (\sin^3 A + \sin^3 B + \sin^3 C)}{\sin^3 A + \sin^3 B + \sin^3 C} = 8$
$(2R)^3 = 8 \implies 2R = 2 \implies R = 1$.
Since $c = 2R \sin C = 2 \sin C$,and the maximum value of $\sin C$ is $1$ (as $C < 180^\circ$),the maximum value of $c$ is $2 \times 1 = 2$.

(C) Given: $a^2-c^2=b^2-bc$ $\Rightarrow a^2-c^2=b^2-bc$ $\Rightarrow b^2+c^2-a^2=bc$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{bc}{2bc} = \frac{1}{2}$.
Thus,$A = 60^{\circ}$.
Using the sine rule,$\frac{a}{\sin A} = 2R$ $\Rightarrow \frac{a}{\sin 60^{\circ}} = 2 \times \frac{1}{\sqrt{3}}$ $\Rightarrow a = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = 1$.
Given $\sqrt{2}a = 2b-c$,substitute $a=1$: $\sqrt{2} = 2b-c \Rightarrow c = 2b-\sqrt{2}$.
Substitute $c$ into $a^2-c^2=b^2-bc$: $1^2 - (2b-\sqrt{2})^2 = b^2 - b(2b-\sqrt{2})$.
$1 - (4b^2 - 4\sqrt{2}b + 2) = b^2 - 2b^2 + \sqrt{2}b$.
$1 - 4b^2 + 4\sqrt{2}b - 2 = -b^2 + \sqrt{2}b$.
$-3b^2 + 3\sqrt{2}b - 1 = 0 \Rightarrow 3b^2 - 3\sqrt{2}b + 1 = 0$.
Using the quadratic formula $b = \frac{-(-3\sqrt{2}) \pm \sqrt{(-3\sqrt{2})^2 - 4(3)(1)}}{2(3)} = \frac{3\sqrt{2} \pm \sqrt{18-12}}{6} = \frac{3\sqrt{2} \pm \sqrt{6}}{6} = \frac{\sqrt{2}}{2} \pm \frac{\sqrt{6}}{6} = \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{6} = \frac{3\sqrt{2}+\sqrt{6}}{6} = \frac{\sqrt{3}+1}{\sqrt{6}}$.

(D) Given,$\tan A : \tan B : \tan C = 1 : 2 : 3$.
We know that $\tan A = \frac{\sin A}{\cos A} = \frac{a/2R}{(b^2+c^2-a^2)/2bc} = \frac{abc}{R(b^2+c^2-a^2)}$.
Thus,$\tan A : \tan B : \tan C = \frac{1}{b^2+c^2-a^2} : \frac{1}{a^2+c^2-b^2} : \frac{1}{a^2+b^2-c^2} = 1 : 2 : 3$.
Let $b^2+c^2-a^2 = x$,$a^2+c^2-b^2 = 2x$,and $a^2+b^2-c^2 = 3x$.
Adding these equations,we get $a^2+b^2+c^2 = 3x$.
Then $a^2 = (a^2+b^2+c^2) - (b^2+c^2-a^2) = 3x - x = 2x$.
$b^2 = (a^2+b^2+c^2) - (a^2+c^2-b^2) = 3x - 2x = x$.
$c^2 = (a^2+b^2+c^2) - (a^2+b^2-c^2) = 3x - 3x = 0$.
Wait,the ratio of sides $a:b:c$ is proportional to $\sin A : \sin B : \sin C$.
Using the property $\tan A : \tan B : \tan C = l : m : n$,we have $a : b : c = \sqrt{\frac{1}{m} + \frac{1}{n}} : \sqrt{\frac{1}{n} + \frac{1}{l}} : \sqrt{\frac{1}{l} + \frac{1}{m}}$.
Substituting $l=1, m=2, n=3$:
$a : b : c = \sqrt{\frac{1}{2} + \frac{1}{3}} : \sqrt{\frac{1}{3} + 1} : \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{5}{6}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{3}{2}} = \sqrt{\frac{5}{6}} : \sqrt{\frac{8}{6}} : \sqrt{\frac{9}{6}} = \sqrt{5} : 2\sqrt{2} : 3$.
Since $\sin A : \sin B : \sin C = a : b : c$,we have $\sqrt{5} : 2\sqrt{2} : k = \sqrt{5} : 2\sqrt{2} : 3$.
Therefore,$k = 3$.

(A) Let the sides of the triangle be $a = \sqrt{3}k$,$b = \sqrt{5}k$,and $c = \sqrt{8+\sqrt{15}}k$.
Since $c$ is the largest side,the largest angle $\theta$ is opposite to side $c$.
Using the Law of Cosines: $\cos \theta = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos \theta = \frac{(\sqrt{3}k)^2 + (\sqrt{5}k)^2 - (\sqrt{8+\sqrt{15}}k)^2}{2(\sqrt{3}k)(\sqrt{5}k)}$.
$\cos \theta = \frac{3k^2 + 5k^2 - (8+\sqrt{15})k^2}{2\sqrt{15}k^2}$.
$\cos \theta = \frac{8k^2 - 8k^2 - \sqrt{15}k^2}{2\sqrt{15}k^2} = \frac{-\sqrt{15}k^2}{2\sqrt{15}k^2} = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2\pi}{3}$.

(A) The perimeter of $\triangle ABC$ is $a+b+c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
Given: $a+b+c = 6 \times \left(\frac{\sin A + \sin B + \sin C}{3}\right) = 2(\sin A + \sin B + \sin C)$.
Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,we have:
$2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
This implies $2R = 2$,so $R = 1$.
Given $BC = a = 1$,we use $a = 2R \sin A$:
$1 = 2(1) \sin A \implies \sin A = \frac{1}{2}$.
Since $A$ is an angle of a triangle,$A = 30^{\circ} = \frac{\pi}{6}$.

(A) The diameters of the ex-circles are given by $d_1 = \frac{2\Delta}{s-a}$,$d_2 = \frac{2\Delta}{s-b}$,and $d_3 = \frac{2\Delta}{s-c}$.
We need to calculate $d_1 d_2 + d_2 d_3 + d_3 d_1 = \frac{4\Delta^2}{(s-a)(s-b)} + \frac{4\Delta^2}{(s-b)(s-c)} + \frac{4\Delta^2}{(s-c)(s-a)}$.
Taking $4\Delta^2$ as a common factor,we get $4\Delta^2 \left[ \frac{(s-c) + (s-a) + (s-b)}{(s-a)(s-b)(s-c)} \right]$.
Since $a+b+c = 2s$,the numerator becomes $3s - (a+b+c) = 3s - 2s = s$.
Thus,the expression becomes $\frac{4\Delta^2 \cdot s}{(s-a)(s-b)(s-c)}$.
Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.
Substituting this,we get $\frac{4\Delta^2 \cdot s}{\Delta^2 / s} = 4s^2 = (2s)^2 = (a+b+c)^2$.

(D) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$.
Using $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$,we get $\frac{a(1+\cos C)}{2} + \frac{c(1+\cos A)}{2} = 15$.
$\Rightarrow (a+c) + (a \cos C + c \cos A) = 30$.
Since $a \cos C + c \cos A = b$,we have $a+c+b = 30$.
Given $b=10$,so $a+c = 20$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{30}{2} = 15$.
Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = 15\sqrt{3}$.
$15\sqrt{3} = \sqrt{15(15-a)(15-10)(15-c)} = \sqrt{75(15-a)(15-c)}$.
Squaring both sides: $225 \times 3 = 75(15-a)(15-c) \Rightarrow 9 = (15-a)(15-c) = 225 - 15(a+c) + ac$.
$9 = 225 - 15(20) + ac$ $\Rightarrow 9 = 225 - 300 + ac$ $\Rightarrow ac = 84$.
Area $\Delta = \frac{1}{2}ac \sin B = 15\sqrt{3}$ $\Rightarrow \frac{1}{2}(84) \sin B = 15\sqrt{3}$ $\Rightarrow 42 \sin B = 15\sqrt{3}$ $\Rightarrow \sin B = \frac{5\sqrt{3}}{14}$.
Then $\cos^2 B = 1 - \sin^2 B = 1 - \frac{75}{196} = \frac{121}{196} \Rightarrow \cos B = \frac{11}{14}$.
Finally,$\cot \frac{B}{2} = \sqrt{\frac{1+\cos B}{1-\cos B}} = \sqrt{\frac{1 + 11/14}{1 - 11/14}} = \sqrt{\frac{25/14}{3/14}} = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}$.

(D) We use the identity $\cos^2 \theta = \frac{1+\cos(2\theta)}{2}$.
Thus,$\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} = \frac{1+\cos A}{2} + \frac{1+\cos B}{2} + \frac{1+\cos C}{2}$.
$= \frac{1}{2} \{3 + (\cos A + \cos B + \cos C)\}$.
Using the identity $\cos A + \cos B + \cos C = 1 + \frac{r}{R}$,where $r$ is the inradius and $R$ is the circumradius of the triangle:
$= \frac{1}{2} \{3 + 1 + \frac{r}{R}\}$.
$= \frac{1}{2} \{4 + \frac{r}{R}\}$.
$= 2 + \frac{r}{2R}$.

(C) . $\because \tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{(s-b)(s-c)}{\Delta}$.
Also,$\frac{r}{s-a} = \frac{\Delta}{s(s-a)} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{(s-b)(s-c)}{\Delta}$.
$\therefore \tan \frac{A}{2} = \frac{r}{s-a} = \frac{(s-b)(s-c)}{\Delta}$. Thus,$A-V$.
$B$. From the figure,$r = IF = (AI) \sin \frac{A}{2} = (AI) \sqrt{\frac{(s-b)(s-c)}{bc}}$. Thus,$B-I$.
$C$. $SI$ is the distance between the circumcentre and in-centre,given by $SI = \sqrt{R^2 - 2rR}$.
Squaring gives $(SI)^2 = R^2 - 2rR$,so $(SI)^2 + 2Rr = R^2$. Thus,$C-II$.
$D$. We know $r_1 + r_2 + r_3 = 4R + r$.
Squaring both sides: $r_1^2 + r_2^2 + r_3^2 = (4R + r)^2 - 2(r_1r_2 + r_2r_3 + r_3r_1)$.
Using $r_1r_2 + r_2r_3 + r_3r_1 = s^2$,we get $r_1^2 + r_2^2 + r_3^2 = (4R + r)^2 - 2s^2 = (4R + r + \sqrt{2}s)(4R + r - \sqrt{2}s)$. Thus,$D-III$.

(A) Given for triangle $ABC$: $c=9, s=10, \Delta=10\sqrt{2}$.
Using Napier's Analogy: $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}\cot\left(\frac{C}{2}\right)$.
We know $\cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{\Delta} = \frac{10(10-9)}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,$\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cdot \frac{1}{\sqrt{2}}$,which implies $\sqrt{2}\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}$.
Now,substitute this into the expression $b\left[1+\sqrt{2}\tan\left(\frac{A-B}{2}\right)\right]$:
$= b\left[1 + \frac{a-b}{a+b}\right] = b\left[\frac{a+b+a-b}{a+b}\right] = b\left[\frac{2a}{a+b}\right]$.
Rearranging the terms: $a\left[\frac{2b}{a+b}\right] = a\left[\frac{(a+b)-(a-b)}{a+b}\right] = a\left[1 - \frac{a-b}{a+b}\right]$.
Substituting back $\frac{a-b}{a+b} = \sqrt{2}\tan\left(\frac{A-B}{2}\right)$,we get:
$= a\left[1 - \sqrt{2}\tan\left(\frac{A-B}{2}\right)\right]$.

(C) Let $O$ be the centroid of $\triangle ABC$. The medians $AD$ and $BE$ intersect at $O$.
We know that the centroid divides the median in the ratio $2:1$.
Given $AD = \frac{7}{2}$,so $AO = \frac{2}{3} AD = \frac{2}{3} \times \frac{7}{2} = \frac{7}{3}$.
In $\triangle AOB$,$\angle OAB = \frac{\pi}{8}$ and $\angle OBA = \frac{\pi}{4}$.
Therefore,$\angle AOB = \pi - (\frac{\pi}{8} + \frac{\pi}{4}) = \pi - \frac{3\pi}{8} = \frac{5\pi}{8}$.
Using the sine rule in $\triangle AOB$: $\frac{AO}{\sin(\frac{\pi}{4})} = \frac{BO}{\sin(\frac{\pi}{8})}$.
$BO = \frac{AO \sin(\frac{\pi}{8})}{\sin(\frac{\pi}{4})} = \frac{7/3 \times \sin(\frac{\pi}{8})}{1/\sqrt{2}} = \frac{7\sqrt{2}}{3} \sin(\frac{\pi}{8})$.
The area of $\triangle AOB = \frac{1}{2} \times AO \times BO \times \sin(\angle AOB) = \frac{1}{2} \times \frac{7}{3} \times \frac{7\sqrt{2}}{3} \sin(\frac{\pi}{8}) \times \sin(\frac{5\pi}{8})$.
Since $\sin(\frac{5\pi}{8}) = \cos(\frac{\pi}{8})$,the area of $\triangle AOB = \frac{49\sqrt{2}}{18} \sin(\frac{\pi}{8}) \cos(\frac{\pi}{8}) = \frac{49\sqrt{2}}{18} \times \frac{1}{2} \sin(\frac{\pi}{4}) = \frac{49\sqrt{2}}{36} \times \frac{1}{\sqrt{2}} = \frac{49}{36}$.
The area of $\triangle ABC = 3 \times \text{Area of } \triangle AOB = 3 \times \frac{49}{36} = \frac{49}{12}$.

(B) Let the sides of the triangle be $x-1, x, x+1$ where $x > 2$ (to ensure all sides are positive). The smallest side is $x-1$ and the largest side is $x+1$. Let the smallest angle be $\theta$ (opposite to side $x-1$) and the largest angle be $2\theta$ (opposite to side $x+1$).
Using the Law of Cosines for the angle $2\theta$ (opposite to side $x+1$):
$\cos(2\theta) = \frac{x^2 + (x-1)^2 - (x+1)^2}{2x(x-1)} = \frac{x^2 + x^2 - 2x + 1 - (x^2 + 2x + 1)}{2x(x-1)} = \frac{x^2 - 4x}{2x(x-1)} = \frac{x-4}{2(x-1)}$
Using the Law of Sines:
$\frac{x+1}{\sin(2\theta)} = \frac{x-1}{\sin(\theta)} \Rightarrow \frac{x+1}{2\sin(\theta)\cos(\theta)} = \frac{x-1}{\sin(\theta)} \Rightarrow \cos(\theta) = \frac{x+1}{2(x-1)}$
Since $\cos(2\theta) = 2\cos^2(\theta) - 1$,we substitute the expressions:
$\frac{x-4}{2(x-1)} = 2\left(\frac{x+1}{2(x-1)}\right)^2 - 1 = \frac{(x+1)^2}{2(x-1)^2} - 1 = \frac{x^2+2x+1 - 2(x^2-2x+1)}{2(x-1)^2} = \frac{-x^2+6x-1}{2(x-1)^2}$
Solving $\frac{x-4}{x-1} = \frac{-x^2+6x-1}{(x-1)^2} \Rightarrow (x-4)(x-1) = -x^2+6x-1 \Rightarrow x^2-5x+4 = -x^2+6x-1 \Rightarrow 2x^2-11x+5 = 0 \Rightarrow (2x-1)(x-5) = 0$. Since $x$ is a natural number and $x>2$,we get $x=5$.
The sides are $4, 5, 6$. The semi-perimeter $s = \frac{4+5+6}{2} = \frac{15}{2}$.
Area = $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15}{2} \times (\frac{15}{2}-4) \times (\frac{15}{2}-5) \times (\frac{15}{2}-6)} = \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}} = \sqrt{\frac{1575}{16}} = \frac{15\sqrt{7}}{4}$.

(A) Given $a=4k, b=5k, c=6k$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{15k}{2}$.
Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} \times \frac{7k}{2} \times \frac{5k}{2} \times \frac{3k}{2}} = \frac{15k^2\sqrt{7}}{4}$.
Circumradius $R = \frac{abc}{4\Delta} = \frac{4k \times 5k \times 6k}{4 \times \frac{15k^2\sqrt{7}}{4}} = \frac{120k^3}{15k^2\sqrt{7}} = \frac{8k}{\sqrt{7}}$.
Inradius $r = \frac{\Delta}{s} = \frac{15k^2\sqrt{7}}{4} \times \frac{2}{15k} = \frac{k\sqrt{7}}{2}$.
Ratio $\frac{R}{r} = \frac{8k}{\sqrt{7}} \times \frac{2}{k\sqrt{7}} = \frac{16}{7}$.

(C) We know that for a triangle $ABC$ with exradii $r_1, r_2, r_3$ and circumradius $R$:
$r_1+r_2 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2}$
$= 4R \cos \frac{C}{2} [\sin \frac{A}{2} \cos \frac{B}{2} + \sin \frac{B}{2} \cos \frac{A}{2}]$
$= 4R \cos \frac{C}{2} \sin(\frac{A+B}{2}) = 4R \cos \frac{C}{2} \cos \frac{C}{2} = 2R(2 \cos^2 \frac{C}{2}) = 2R(1+\cos C)$.
Thus,$\frac{1+\cos C}{r_1+r_2} = \frac{1}{2R}$.
Similarly,$\frac{1+\cos A}{r_2+r_3} = \frac{1}{2R}$ and $\frac{1+\cos B}{r_1+r_3} = \frac{1}{2R}$.
Adding these,we get $\frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{3}{2R}$.

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.
We need to evaluate $r + r_3 + r_2 - r_1$.
Substituting the formulas,we get:
$r + r_3 + r_2 - r_1 = \frac{\Delta}{s} + \frac{\Delta}{s-c} + \frac{\Delta}{s-b} - \frac{\Delta}{s-a}$
$= \Delta \left( \frac{1}{s} - \frac{1}{s-a} \right) + \Delta \left( \frac{1}{s-c} + \frac{1}{s-b} \right)$
$= \Delta \left( \frac{s-a-s}{s(s-a)} \right) + \Delta \left( \frac{s-b+s-c}{(s-c)(s-b)} \right)$
$= \Delta \left( \frac{-a}{s(s-a)} \right) + \Delta \left( \frac{2s-b-c}{(s-c)(s-b)} \right)$
Since $2s = a+b+c$,we have $2s-b-c = a$.
$= \Delta \left( \frac{-a}{s(s-a)} + \frac{a}{(s-c)(s-b)} \right)$
$= \Delta a \left( \frac{-(s-c)(s-b) + s(s-a)}{s(s-a)(s-b)(s-c)} \right)$
Using $\Delta^2 = s(s-a)(s-b)(s-c)$,we get:
$= \frac{\Delta a}{\Delta^2} (-(s^2 - (b+c)s + bc) + (s^2 - as))$
$= \frac{a}{\Delta} (-s^2 + (b+c)s - bc + s^2 - as)$
$= \frac{a}{\Delta} ((b+c-a)s - bc)$
$= \frac{a}{\Delta} ((b+c-a)(\frac{a+b+c}{2}) - bc)$
$= \frac{a}{2\Delta} ((b+c)^2 - a^2 - 2bc) = \frac{a}{2\Delta} (b^2 + c^2 + 2bc - a^2 - 2bc) = \frac{a(b^2+c^2-a^2)}{2\Delta}$
Since $\cos A = \frac{b^2+c^2-a^2}{2bc}$,we have $b^2+c^2-a^2 = 2bc \cos A$.
So,the expression becomes $\frac{a(2bc \cos A)}{2\Delta} = \frac{abc \cos A}{\Delta}$.
Since $\Delta = \frac{abc}{4R}$,we have $\frac{abc}{\Delta} = 4R$.
Thus,the expression is $4R \cos A$. However,given the options and the specific ratio $R:r_1:r = 5:12:2$,we observe that for a right-angled triangle at $A$,$r = s-a$,$r_1 = s$,$r_2 = s-c$,$r_3 = s-b$. The expression $r+r_3+r_2-r_1 = (s-a) + (s-b) + (s-c) - s = 2s - (a+b+c) = 0$. Since $\cos 90^\circ = 0$,the correct option is $\cos A$.

(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.
Substituting these into the expression:
$\frac{\Delta^2}{a^2+b^2+c^2}\left(\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2}+\frac{s^2}{\Delta^2}\right)$
$= \frac{1}{a^2+b^2+c^2}\left[(s-a)^2+(s-b)^2+(s-c)^2+s^2\right]$
Using $s = \frac{a+b+c}{2}$,we have:
$= \frac{1}{a^2+b^2+c^2}\left[\left(\frac{b+c-a}{2}\right)^2+\left(\frac{a+c-b}{2}\right)^2+\left(\frac{a+b-c}{2}\right)^2+\left(\frac{a+b+c}{2}\right)^2\right]$
$= \frac{1}{a^2+b^2+c^2} \cdot \frac{1}{4} \left[(b+c-a)^2 + (a+c-b)^2 + (a+b-c)^2 + (a+b+c)^2\right]$
Expanding the squares,the cross terms cancel out:
$= \frac{1}{4(a^2+b^2+c^2)} \left[4(a^2+b^2+c^2)\right] = 1$.

(A) Given that $r = r_1 - r_2 - r_3$.
Using the standard formulas $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$:
$\frac{\Delta}{s} = \frac{\Delta}{s-a} - \frac{\Delta}{s-b} - \frac{\Delta}{s-c}$
$\frac{1}{s-b} + \frac{1}{s-c} = \frac{1}{s-a} - \frac{1}{s}$
$\frac{s-c+s-b}{(s-b)(s-c)} = \frac{s-(s-a)}{s(s-a)}$
$\frac{2s-b-c}{(s-b)(s-c)} = \frac{a}{s(s-a)}$
Since $2s = a+b+c$,we have $2s-b-c = a$:
$\frac{a}{(s-b)(s-c)} = \frac{a}{s(s-a)}$
$s(s-a) = (s-b)(s-c)$
$s^2 - sa = s^2 - s(b+c) + bc$
$s(b+c-a) = bc$
Substituting $s = \frac{a+b+c}{2}$:
$\frac{a+b+c}{2} \times (b+c-a) = bc$
$(b+c)^2 - a^2 = 2bc$
$b^2 + c^2 + 2bc - a^2 = 2bc$
$b^2 + c^2 = a^2$
This implies $\triangle ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$.
In a right-angled triangle,the circumradius $R = \frac{a}{2}$,so $2R = a$.

(D) We know that $r = \frac{\Delta}{s}$ and $r_1 = \frac{\Delta}{s-a}$.
Given $rr_1 = 42$,we have $\frac{\Delta^2}{s(s-a)} = 42$. $(i)$
Given $r_1 - r = 6.5$,we have $\frac{\Delta}{s-a} - \frac{\Delta}{s} = 6.5$.
$\frac{\Delta(s - (s-a))}{s(s-a)} = 6.5 \Rightarrow \frac{\Delta a}{s(s-a)} = 6.5$. (ii)
Dividing $(i)$ by (ii),we get $\frac{\Delta^2}{s(s-a)} \times \frac{s(s-a)}{\Delta a} = \frac{42}{6.5}$.
$\frac{\Delta}{a} = \frac{42}{6.5} = \frac{420}{65} = \frac{84}{13}$.
Since $R = \frac{abc}{4\Delta}$,we have $\frac{65}{8} = \frac{abc}{4\Delta} \Rightarrow \frac{bc}{\Delta} = \frac{65}{8} \times \frac{4}{a} = \frac{65}{2a}$.
Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we know $r_1 = \frac{\Delta}{s-a} = \sqrt{\frac{s(s-b)(s-c)}{s-a}}$.
From $rr_1 = 42$,we have $\frac{\Delta^2}{s(s-a)} = 42$. Since $\Delta^2 = s(s-a)(s-b)(s-c)$,this simplifies to $(s-b)(s-c) = 42$.
From $r_1 - r = 6.5$,we have $\frac{\Delta a}{s(s-a)} = 6.5$. Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,$\frac{a\sqrt{s(s-a)(s-b)(s-c)}}{s(s-a)} = 6.5 \Rightarrow \frac{a\sqrt{42}}{s(s-a)} = 6.5$.
Actually,using $r_1 - r = 4R \sin^2(A/2)$ is not needed here. Simply note $rr_1 = (s-b)(s-c) = 42$.
Also $r_1 - r = 2R \sin A \tan(A/2) = 2R \sin^2(A/2) / \cos(A/2) ...$ actually,$r_1 - r = 4R \sin^2(A/2)$.
Given $r_1 - r = 6.5$,$4 \times \frac{65}{8} \sin^2(A/2) = 6.5$ $\Rightarrow \frac{65}{2} \sin^2(A/2) = 6.5$ $\Rightarrow \sin^2(A/2) = \frac{6.5 \times 2}{65} = 0.2 = \frac{1}{5}$.
Then $\cos A = 1 - 2\sin^2(A/2) = 1 - 2(1/5) = 3/5$.
$s(s-a) = \frac{\Delta^2}{42}$. Since $\Delta = rs$,and $r = \frac{\Delta}{s}$,we have $s(s-a) = 168$.

(B) Let the sides of the triangle be $a = 5k, b = 5k, c = 6k$.
First,we find the value of $k$ using the inradius formula $r = \frac{\Delta}{s}$.
The semi-perimeter $s = \frac{5k + 5k + 6k}{2} = 8k$.
The area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{8k(3k)(3k)(2k)} = \sqrt{144k^4} = 12k^2$.
Given $r = 6$,so $6 = \frac{12k^2}{8k}$ $\Rightarrow 6 = \frac{3k}{2}$ $\Rightarrow k = 4$.
The sides are $20, 20, 24$.
In the isosceles triangle,the altitude to the base $6k$ divides it into two triangles with base $3k$ and hypotenuse $5k$.
The height is $\sqrt{(5k)^2 - (3k)^2} = 4k$.
Let the vertex angle be $2\theta$. Then $\tan \theta = \frac{3k}{4k} = \frac{3}{4}$.
The vertex angle is $2\theta$,where $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2(3/4)}{1 - (9/16)} = \frac{3/2}{7/16} = \frac{24}{7}$.
Thus,the largest angle is $\tan^{-1}\left(\frac{24}{7}\right)$.

(D) The condition $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$ is the necessary condition for three lines $a_i x + b_i y + c_i = 0$ $(i = 1, 2, 3)$ to be concurrent.
If the lines are not parallel (i.e.,$\frac{a_i}{a_j} \neq \frac{b_i}{b_j}$ for $i \neq j$),the vanishing of the determinant implies that the three lines intersect at a single common point.
Therefore,if $\frac{a_i}{a_j} \neq \frac{b_i}{b_j} \neq \frac{c_i}{c_j}$ for $i \neq j$,the lines are concurrent.

(A) We use the logarithmic definitions of inverse hyperbolic functions:
$\sinh^{-1}(x) = \log(x + \sqrt{x^2+1})$
$\operatorname{cosech}^{-1}(x) = \log\left(\frac{1}{x} + \sqrt{\frac{1}{x^2}+1}\right)$
$\coth^{-1}(x) = \frac{1}{2} \log\left(\frac{x+1}{x-1}\right)$
Substituting $x = -2$:
$\sinh^{-1}(-2) = \log(-2 + \sqrt{5})$
$\operatorname{cosech}^{-1}(-2) = \log\left(-\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \log\left(\frac{-1+\sqrt{5}}{2}\right)$
$\coth^{-1}(-2) = \frac{1}{2} \log\left(\frac{-2+1}{-2-1}\right) = \frac{1}{2} \log\left(\frac{-1}{-3}\right) = \frac{1}{2} \log\left(\frac{1}{3}\right) = \log\left(\frac{1}{\sqrt{3}}\right)$
Summing these:
$\log(-2 + \sqrt{5}) + \log\left(\frac{\sqrt{5}-1}{2}\right) + \log\left(\frac{1}{\sqrt{3}}\right) = \log\left[(-2 + \sqrt{5}) \cdot \left(\frac{\sqrt{5}-1}{2}\right) \cdot \frac{1}{\sqrt{3}}\right]$
$= \log\left[\frac{-2\sqrt{5} + 2 + 5 - \sqrt{5}}{2\sqrt{3}}\right] = \log\left(\frac{7-3\sqrt{5}}{2\sqrt{3}}\right)$

(C) We know that $\operatorname{sech}^{-1}(x) = \log_e \left( \frac{1+\sqrt{1-x^2}}{x} \right)$ and $\operatorname{cosech}^{-1}(x) = \log_e \left( \frac{1}{x} + \sqrt{\frac{1}{x^2}+1} \right)$.
Given equation: $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$.
First,calculate $\operatorname{sech}^{-1}\left(\frac{1}{2}\right) = \log_e \left( \frac{1+\sqrt{1-(1/2)^2}}{1/2} \right) = \log_e \left( 2(1+\sqrt{3/4}) \right) = \log_e (2+\sqrt{3})$.
Next,calculate $\operatorname{cosech}^{-1}\left(\frac{3}{4}\right) = \log_e \left( \frac{4}{3} + \sqrt{\frac{16}{9}+1} \right) = \log_e \left( \frac{4}{3} + \sqrt{\frac{25}{9}} \right) = \log_e \left( \frac{4}{3} + \frac{5}{3} \right) = \log_e(3)$.
Substituting these into the equation: $\log_e(2+\sqrt{3}) - \log_e(3) = \log_e k$.
$\log_e \left( \frac{2+\sqrt{3}}{3} \right) = \log_e k$.
So,$k = \frac{2+\sqrt{3}}{3}$,which implies $3k - 2 = \sqrt{3}$.
Squaring both sides: $(3k-2)^2 = 3$.
$9k^2 - 12k + 4 = 3$.
$9k^2 - 12k + 1 = 0$.

(D) Given the inequality $\sqrt{x+2} > \sqrt{8-x^2}$.
First,we determine the domain of the functions:
$x+2 \geq 0 \Rightarrow x \geq -2$ $(i)$
$8-x^2 \geq 0$ $\Rightarrow x^2 \leq 8$ $\Rightarrow x \in [-2\sqrt{2}, 2\sqrt{2}]$ $(ii)$
Combining $(i)$ and $(ii)$,the domain is $x \in [-2, 2\sqrt{2}]$.
Now,squaring both sides of the inequality:
$x+2 > 8-x^2$
$x^2 + x - 6 > 0$
$(x+3)(x-2) > 0$
This inequality holds for $x \in (-\infty, -3) \cup (2, \infty)$ $(iii)$.
Taking the intersection of the domain $[-2, 2\sqrt{2}]$ and the solution $(iii)$:
$x \in (2, 2\sqrt{2}]$.
The maximum value of $x$ in this interval is $2\sqrt{2}$.

(B) The given function is $f(x) = \cos(3x + 5) + 7$.
We know that the period of the function $\cos(ax + b) + c$ is given by the formula $T = \frac{2 \pi}{|a|}$.
In the given expression,$a = 3$.
Therefore,the period $T = \frac{2 \pi}{|3|} = \frac{2 \pi}{3}$.
Thus,the correct option is $B$.

(A) Given that $a_1, a_2, \ldots, a_n$ are the $n$ distinct roots of $x^n-2=0$.
Thus,we can write the polynomial as $x^n-2 = (x-a_1)(x-a_2)\ldots(x-a_n)$.
Substituting $x=1$ in the equation,we get:
$1^n - 2 = (1-a_1)(1-a_2)\ldots(1-a_n)$
$-1 = (1-a_1)(1-a_2)\ldots(1-a_n)$
Adding $1$ to both sides,we get $1 + (1-a_1)(1-a_2)\ldots(1-a_n) = 0$.
Therefore,Assertion $(A)$ is true.
For Reason $(R)$,if $\alpha_i$ are roots of $f(x)=0$,then $f(\alpha_i)=0$. If we consider $f(g(x))=0$,then $g(x)$ must equal one of the roots $\alpha_i$. Thus,$x = g^{-1}(\alpha_i)$.
Therefore,Reason $(R)$ is true and provides the correct explanation for $(A)$.

(C) Given the equation: $\frac{H(x)}{(x-1)(x+1)(x-2)}=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2)}$
Multiplying both sides by $(x-1)(x+1)(x-2)$,we get:
$H(x)=(x-1)(x+1)(x-2)f(x)+g(x)$
Now,we evaluate $H(x)$ at $x=-1, 2, 1$:
For $x=-1$: $H(-1)=( -1-1)( -1+1)( -1-2)f(-1)+g(-1) = 0+g(-1) = g(-1)$
For $x=2$: $H(2)=(2-1)(2+1)(2-2)f(2)+g(2) = 0+g(2) = g(2)$
For $x=1$: $H(1)=(1-1)(1+1)(1-2)f(1)+g(1) = 0+g(1) = g(1)$
Substituting these into the expression $H(-1)+2H(2)-3H(1)$:
$H(-1)+2H(2)-3H(1) = g(-1)+2g(2)-3g(1)$

(C) Given $f(1)=3$ and $f(n+1)-f(n)=3(4^n-1)$.
We can write the recurrence as a telescoping sum:
$f(n) = f(1) + \sum_{k=1}^{n-1} (f(k+1)-f(k))$
$f(n) = 3 + \sum_{k=1}^{n-1} 3(4^k-1)$
$f(n) = 3 + 3 \left( \sum_{k=1}^{n-1} 4^k - \sum_{k=1}^{n-1} 1 \right)$
Using the geometric series sum formula $\sum_{k=1}^{n-1} 4^k = \frac{4(4^{n-1}-1)}{4-1} = \frac{4^n-4}{3}$:
$f(n) = 3 + 3 \left( \frac{4^n-4}{3} - (n-1) \right)$
$f(n) = 3 + (4^n-4) - 3(n-1)$
$f(n) = 3 + 4^n - 4 - 3n + 3$
$f(n) = 4^n - 3n + 2$

(D) Given $f(n) = A(-2)^n + B(-3)^n$.
Substituting $f(n)$ into the equation $f(n) + a f(n-1) + b f(n-2) = 0$:
$A(-2)^n + B(-3)^n + a(A(-2)^{n-1} + B(-3)^{n-1}) + b(A(-2)^{n-2} + B(-3)^{n-2}) = 0$.
Grouping terms with $A$ and $B$:
$A(-2)^{n-2} [(-2)^2 + a(-2) + b] + B(-3)^{n-2} [(-3)^2 + a(-3) + b] = 0$.
$A(-2)^{n-2} [4 - 2a + b] + B(-3)^{n-2} [9 - 3a + b] = 0$.
Since this holds for all $A, B$,the coefficients must be zero:
$4 - 2a + b = 0 \implies b - 2a = -4$.
$9 - 3a + b = 0 \implies b - 3a = -9$.
Subtracting the two equations:
$(b - 2a) - (b - 3a) = -4 - (-9) \implies a = 5$.
Substituting $a=5$ into $b - 2a = -4$:
$b - 10 = -4 \implies b = 6$.
Finally,$(a+b)(b-a) = (5+6)(6-5) = 11 \times 1 = 11$.

(C) Given the equation $x^4-8x^3+11x^2+32x-60=0$.
By testing integer roots using the factor theorem,we find that $x = -2, 2, 3, 5$ are the roots.
Factoring the polynomial,we get $(x+2)(x-2)(x-3)(x-5)=0$.
Since $\alpha < \beta < \gamma < \delta$,we have $\alpha = -2, \beta = 2, \gamma = 3, \delta = 5$.
Now,calculating the expression $4\alpha+3\beta+2\gamma+\delta$:
$4(-2) + 3(2) + 2(3) + 5 = -8 + 6 + 6 + 5 = 9$.

(A) Performing polynomial long division on $\frac{x^5-5}{x^3+x^2}$:
$\frac{x^5-5}{x^3+x^2} = x^2 - x + 1 + \frac{-x^2-5}{x^3+x^2}$
Thus,$f(x) = x^2 - x + 1$ and $\frac{-x^2-5}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$.
Multiplying by $x^2(x+1)$,we get:
$-x^2-5 = Ax(x+1) + B(x+1) + Cx^2$
$-x^2-5 = (A+C)x^2 + (A+B)x + B$
Comparing coefficients:
$B = -5$,$A+B = 0 \Rightarrow A = 5$,$A+C = -1 \Rightarrow C = -6$.
Given $f(K) + A + B + C = 1$:
$(K^2 - K + 1) + 5 - 5 - 6 = 1$
$K^2 - K - 6 = 0$
$(K-3)(K+2) = 0$
$K = 3$ or $K = -2$.
The larger value of $K$ is $3$.

(B) For statement $I$: The number of positive integral solutions of $x_1+x_2+x_3+x_4=n$ is given by the formula $\binom{n-1}{r-1}$.
Here,$n=10$ and $r=4$,so the number of solutions is $\binom{10-1}{4-1} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Since $84 \neq 286$,statement $I$ is false.
For statement $II$: The exponent of a prime $p$ in $m!$ is given by Legendre's formula $\sum_{k=1}^{\infty} [\frac{m}{p^k}]$.
The exponent of $5$ in $25!$ is $[\frac{25}{5}] + [\frac{25}{25}] = 5 + 1 = 6$.
The exponent of $2$ in $25!$ is $[\frac{25}{2}] + [\frac{25}{4}] + [\frac{25}{8}] + [\frac{25}{16}] = 12 + 6 + 3 + 1 = 22$.
Since the exponent of $5$ is $6$ and the exponent of $2$ is $22$,the highest power of $10$ that divides $25!$ is $10^6$.
Thus,$25! = 10^6 \times k$,where $k$ is an integer not divisible by $10$.
Therefore,statement $II$ is true.

(A) Given: $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\bar{A}) = \frac{2}{3}$.
First,find $P(A)$:
$P(A) = 1 - P(\bar{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.
$P(B) = \frac{3}{4} - \frac{1}{3} + \frac{1}{4} = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find $P(\bar{A} \cap B)$.
Using the property $P(\bar{A} \cap B) = P(B) - P(A \cap B)$:
$P(\bar{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.

(B) Given $\alpha$ and $\beta$ are roots of $x^2+ax-b=0$.
From Vieta's formulas,$\alpha+\beta = -a$ and $\alpha\beta = -b$.
Differentiating the curve $(\frac{x}{\alpha})^n + (\frac{y}{\beta})^n = 2$ with respect to $x$:
$\frac{n}{\alpha}(\frac{x}{\alpha})^{n-1} + \frac{n}{\beta}(\frac{y}{\beta})^{n-1} \frac{dy}{dx} = 0$.
At the point $(\alpha, \beta)$,the slope of the tangent is $\frac{dy}{dx} = -\frac{\beta}{\alpha}$.
The line $x \cos \theta + y \sin \theta = c$ has a slope of $-\cot \theta$.
Equating the slopes: $-\cot \theta = -\frac{\beta}{\alpha} \implies \cot \theta = \frac{\beta}{\alpha}$.
Since the line passes through $(\alpha, \beta)$,$\alpha \cos \theta + \beta \sin \theta = c$.
Using $\cot \theta = \frac{\beta}{\alpha}$,we have $\cos \theta = \frac{\beta}{\sqrt{\alpha^2+\beta^2}}$ and $\sin \theta = \frac{\alpha}{\sqrt{\alpha^2+\beta^2}}$.
Substituting these into the line equation: $\alpha(\frac{\beta}{\sqrt{\alpha^2+\beta^2}}) + \beta(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}) = c \implies \frac{2\alpha\beta}{\sqrt{\alpha^2+\beta^2}} = c$.
We need to evaluate $(\frac{a}{b})^2 + \frac{2}{b} = (\frac{-(\alpha+\beta)}{-\alpha\beta})^2 + \frac{2}{-\alpha\beta} = \frac{(\alpha+\beta)^2}{(\alpha\beta)^2} - \frac{2}{\alpha\beta} = \frac{\alpha^2+\beta^2+2\alpha\beta-2\alpha\beta}{(\alpha\beta)^2} = \frac{\alpha^2+\beta^2}{(\alpha\beta)^2}$.
From $\frac{2\alpha\beta}{\sqrt{\alpha^2+\beta^2}} = c$,we get $\alpha^2+\beta^2 = \frac{4\alpha^2\beta^2}{c^2}$.
Substituting this into the expression: $\frac{4\alpha^2\beta^2/c^2}{(\alpha\beta)^2} = \frac{4}{c^2}$.

(A) Let the curves be $ax^2+by^2=1$ $(i)$ and $cx^2+dy^2=1$ (ii).
Subtracting (ii) from $(i)$,we get $(a-c)x^2 + (b-d)y^2 = 0$,which implies $x^2 = -\frac{b-d}{a-c} y^2 = \frac{d-b}{a-c} y^2$.
Differentiating $(i)$ with respect to $x$,we get $2ax + 2byy' = 0$,so $y'_1 = -\frac{ax}{by}$.
Differentiating (ii) with respect to $x$,we get $2cx + 2dyy' = 0$,so $y'_2 = -\frac{cx}{dy}$.
Since the curves intersect orthogonally,the product of their slopes at the point of intersection $(x, y)$ is $-1$:
$y'_1 \cdot y'_2 = -1 \Rightarrow \left(-\frac{ax}{by}\right) \left(-\frac{cx}{dy}\right) = -1 \Rightarrow \frac{acx^2}{bdy^2} = -1$.
Substituting $x^2 = \frac{d-b}{a-c} y^2$ into the equation:
$\frac{ac}{bd} \left(\frac{d-b}{a-c}\right) = -1 \Rightarrow ac(d-b) = -bd(a-c) \Rightarrow acd - abc = -abd + bcd$.
Rearranging terms: $acd + abd = abc + bcd \Rightarrow ad(c+b) = bc(a+d)$.
Actually,using the condition $y'_1 y'_2 = -1$ at intersection point $(x, y)$:
$\frac{ac x^2}{bd y^2} = -1 \Rightarrow \frac{x^2}{y^2} = -\frac{bd}{ac}$.
From $(a-c)x^2 + (b-d)y^2 = 0$,we have $\frac{x^2}{y^2} = -\frac{b-d}{a-c} = \frac{d-b}{a-c}$.
Equating the two expressions for $\frac{x^2}{y^2}$:
$\frac{d-b}{a-c} = -\frac{bd}{ac} \Rightarrow ac(d-b) = -bd(a-c) \Rightarrow acd - abc = -abd + bcd$.
$abd - abc = bcd - acd \Rightarrow ab(d-c) = cd(b-a)$.
Therefore,$\frac{b-a}{d-c} = \frac{ab}{cd}$.

(D) Given the expression: $\frac{x^4+3 x+1}{(x+1)^2(x-1)}=A x+B+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{x-1}$.
First,perform polynomial division of $x^4+3x+1$ by $(x+1)^2(x-1) = (x^2+2x+1)(x-1) = x^3+x^2-x-1$.
Dividing $x^4+3x+1$ by $x^3+x^2-x-1$ gives a quotient of $(x-1)$ and a remainder of $3x^2+3x$. So,$Ax+B = x-1$,which implies $A=1$ and $B=-1$.
Now,$\frac{3x^2+3x}{(x+1)^2(x-1)} = \frac{C}{x+1} + \frac{D}{(x+1)^2} + \frac{E}{x-1}$.
Multiply by $(x+1)^2(x-1)$: $3x^2+3x = C(x+1)(x-1) + D(x-1) + E(x+1)^2$.
For $x=1$: $3(1)^2+3(1) = E(1+1)^2 \Rightarrow 6 = 4E \Rightarrow E = 3/2$.
For $x=-1$: $3(-1)^2+3(-1) = D(-1-1) \Rightarrow 0 = -2D \Rightarrow D = 0$.
Comparing coefficients of $x^2$: $3 = C + E \Rightarrow 3 = C + 3/2 \Rightarrow C = 3/2$.
Thus,$A=1, B=-1, C=3/2, D=0, E=3/2$.
$A+B+C+D+E = 1 - 1 + 3/2 + 0 + 3/2 = 6/2 = 3$.

(A) Given the partial fraction decomposition:
$\frac{5x^4+4x^2+7}{(x^2+1)(x^4+x^2+1)} = \frac{Ax+B}{x^2+1} + \frac{Cx^3+Dx^2+Ex+F}{x^4+x^2+1}$
Multiplying both sides by the denominator $(x^2+1)(x^4+x^2+1)$,we get:
$5x^4+4x^2+7 = (Ax+B)(x^4+x^2+1) + (Cx^3+Dx^2+Ex+F)(x^2+1)$
Expanding the right side:
$5x^4+4x^2+7 = Ax^5+Ax^3+Ax + Bx^4+Bx^2+B + Cx^5+Cx^3+Dx^4+Dx^2+Ex^3+Ex+Fx^2+F$
Grouping the coefficients of like powers of $x$:
$5x^4+4x^2+7 = (A+C)x^5 + (B+D)x^4 + (A+C+E)x^3 + (B+D+F)x^2 + (A+E)x + (B+F)$
Comparing coefficients on both sides:
$1) A+C = 0$
$2) B+D = 5$
$3) A+C+E = 0$
$4) B+D+F = 4$
$5) A+E = 0$
$6) B+F = 7$
From $(1)$ and $(3)$,since $A+C=0$,we get $E=0$.
From $(5)$,$A+E=0 \implies A=0$,which implies $C=0$.
From $(2)$,$B+D=5$. From $(4)$,$(B+D)+F=4 \implies 5+F=4 \implies F=-1$.
From $(6)$,$B+F=7 \implies B-1=7 \implies B=8$.
Then $D = 5-B = 5-8 = -3$.
Thus,$A=0, B=8, C=0, D=-3, E=0, F=-1$.
Calculating the expression:
$B+2(D+F+E)-C \cdot A = 8 + 2(-3-1+0) - 0 \cdot 0 = 8 + 2(-4) - 0 = 8-8 = 0$.

(D) Given the partial fraction decomposition: $\frac{2x+1}{(x-1)^2(x^2+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}$.
Multiplying both sides by $(x-1)^2(x^2+1)$,we get:
$2x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$.
Expanding the right side:
$2x+1 = A(x^3-x^2+x-1) + B(x^2+1) + (Cx+D)(x^2-2x+1)$.
$2x+1 = x^3(A+C) + x^2(-A+B-2C+D) + x(A+C-2D) + (-A+B+D)$.
Comparing coefficients on both sides:
$1) A+C = 0$
$2) -A+B-2C+D = 0$
$3) A+C-2D = 2$
$4) -A+B+D = 1$
From $(1)$,$C = -A$. Substituting into $(3)$: $A-A-2D = 2 \Rightarrow -2D = 2 \Rightarrow D = -1$.
Substituting $D = -1$ into $(4)$: $-A+B-1 = 1 \Rightarrow -A+B = 2 \Rightarrow B = A+2$.
Substituting $C = -A, D = -1, B = A+2$ into $(2)$: $-A+(A+2)-2(-A)+(-1) = 0 \Rightarrow 2A+1 = 0 \Rightarrow A = -\frac{1}{2}$.
Then $B = -\frac{1}{2}+2 = \frac{3}{2}$ and $C = -(-\frac{1}{2}) = \frac{1}{2}$.
Thus,$A+B+C+D = -\frac{1}{2} + \frac{3}{2} + \frac{1}{2} - 1 = \frac{1}{2}$.

(B) We have the inequality: $9 x-2 < (x+2)^2 < 12 x-3$.
This can be split into two parts:
Part $I$: $9 x-2 < (x+2)^2$
$9 x-2 < x^2+4 x+4$
$x^2-5 x+6 > 0$
$(x-3)(x-2) > 0$
So,$x \in (-\infty, 2) \cup (3, \infty)$ ... $(i)$
Part $II$: $(x+2)^2 < 12 x-3$
$x^2+4 x+4 < 12 x-3$
$x^2-8 x+7 < 0$
$(x-7)(x-1) < 0$
So,$x \in (1, 7)$ ... $(ii)$
Taking the intersection of $(i)$ and $(ii)$,we get:
$x \in (1, 2) \cup (3, 7)$
The integral values of $x$ in this interval are $\{4, 5, 6\}$.
Thus,the number of integral values is $3$.

(C) Given vertices $A(4,7,8)$,$B(2,3,4)$,and $C(2,5,7)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(4-2)^2 + (7-3)^2 + (8-4)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.
$AC = \sqrt{(4-2)^2 + (7-5)^2 + (8-7)^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.
By the Angle Bisector Theorem,the internal bisector of angle $A$ divides the opposite side $BC$ in the ratio $AB:AC = 6:3 = 2:1$.
Let $D$ be the point on $BC$ dividing it in the ratio $2:1$. Using the section formula:
$D = \left( \frac{2(2) + 1(2)}{2+1}, \frac{2(5) + 1(3)}{2+1}, \frac{2(7) + 1(4)}{2+1} \right) = \left( \frac{6}{3}, \frac{13}{3}, \frac{18}{3} \right) = \left( 2, \frac{13}{3}, 6 \right)$.
The length of the internal bisector is the distance $AD$:
$AD = \sqrt{(4-2)^2 + (7 - \frac{13}{3})^2 + (8-6)^2} = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 \times 34}}{3} = \frac{2}{3} \sqrt{34}$.

(B) Given that the points $A(-1,0,7), B(3,2, t)$ and $C(5, k,-2)$ are collinear,the direction ratios of $AB$ and $BC$ must be proportional.
$\frac{3-(-1)}{5-3} = \frac{2-0}{k-2} = \frac{t-7}{-2-t}$
$\frac{4}{2} = \frac{2}{k-2} = \frac{t-7}{-2-t}$
$2 = \frac{2}{k-2} \Rightarrow k-2 = 1 \Rightarrow k = 3$
$2 = \frac{t-7}{-2-t} \Rightarrow -4-2t = t-7 \Rightarrow 3t = 3 \Rightarrow t = 1$
Thus,the points are $B(3,2,1)$ and $C(5,3,-2)$.
The point $P$ is given by $P(t, k-2t, t+k) = P(1, 3-2(1), 1+3) = P(1, 1, 4)$.
Let $P$ divide the line segment $BC$ in the ratio $\lambda : 1$.
Using the section formula for the $x$-coordinate:
$1 = \frac{\lambda(5) + 1(3)}{\lambda + 1}$
$\lambda + 1 = 5\lambda + 3$
$-2 = 4\lambda$
$\lambda = -\frac{1}{2}$
Therefore,the required ratio is $-1: 2$.

(A) The vertices of the triangle are $A(4,3,2), B(5,4,6)$,and $C(-1,-1,5)$.
First,we calculate the lengths of sides $AB$ and $AC$ using the distance formula:
$AB = \sqrt{(5-4)^2 + (4-3)^2 + (6-2)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
$AC = \sqrt{(-1-4)^2 + (-1-3)^2 + (5-2)^2} = \sqrt{(-5)^2 + (-4)^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.
According to the Angle Bisector Theorem,the bisector of angle $A$ divides the opposite side $BC$ in the ratio of the sides containing the angle,which is $AB : AC = 3\sqrt{2} : 5\sqrt{2} = 3 : 5$.
Let $D$ be the point on $BC$ where the bisector meets it. $D$ divides $BC$ internally in the ratio $m:n = 3:5$.
Using the section formula $D = \left(\frac{mx_C + nx_B}{m+n}, \frac{my_C + ny_B}{m+n}, \frac{mz_C + nz_B}{m+n}\right)$:
$x_D = \frac{3(-1) + 5(5)}{3+5} = \frac{-3 + 25}{8} = \frac{22}{8}$.
$y_D = \frac{3(-1) + 5(4)}{3+5} = \frac{-3 + 20}{8} = \frac{17}{8}$.
$z_D = \frac{3(5) + 5(6)}{3+5} = \frac{15 + 30}{8} = \frac{45}{8}$.
Thus,the coordinates of point $D$ are $\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)$.

(C) Let the vertices be $A(1,2,5), B(-1,6,1), C(3,4,-3)$ and $D(5,0,1)$.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(-1-1)^2 + (6-2)^2 + (1-5)^2} = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$BC = \sqrt{(3-(-1))^2 + (4-6)^2 + (-3-1)^2} = \sqrt{4^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
$CD = \sqrt{(5-3)^2 + (0-4)^2 + (1-(-3))^2} = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$DA = \sqrt{(1-5)^2 + (2-0)^2 + (5-1)^2} = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Since all sides are equal $(AB = BC = CD = DA = 6)$,the quadrilateral is a rhombus.
Now,we check the diagonals:
$AC = \sqrt{(3-1)^2 + (4-2)^2 + (-3-5)^2} = \sqrt{2^2 + 2^2 + (-8)^2} = \sqrt{4 + 4 + 64} = \sqrt{72} = 6\sqrt{2}$
$BD = \sqrt{(5-(-1))^2 + (0-6)^2 + (1-1)^2} = \sqrt{6^2 + (-6)^2 + 0^2} = \sqrt{36 + 36 + 0} = \sqrt{72} = 6\sqrt{2}$
Since the diagonals are also equal $(AC = BD = 6\sqrt{2})$,the quadrilateral is a square.

(B) According to Bonferroni's inequality,for any events $A_1, A_2, \ldots, A_n$,we have:
$P\left(\bigcap_{i=1}^{n} A_i\right) \geq \sum_{i=1}^{n} P(A_i) - (n-1)$.
For $n = 15$,the inequality becomes:
$P\left(\bigcap_{i=1}^{15} A_i\right) \geq \sum_{i=1}^{15} P(A_i) - (15-1) = \sum_{i=1}^{15} P(A_i) - 14$.
Thus,option $B$ is correct.

(B) For the equation $2x^2+x-1=0$,we have $2x^2+2x-x-1=0 \Rightarrow 2x(x+1)-1(x+1)=0$,so $x = \frac{1}{2}, -1$. Since probability $P \in [0, 1]$,we take $P_1 = \frac{1}{2}$.
For the equation $3x^2-10x+3=0$,we have $3x^2-9x-x+3=0 \Rightarrow 3x(x-3)-1(x-3)=0$,so $x = \frac{1}{3}, 3$. Since probability $P \in [0, 1]$,we take $P_2 = \frac{1}{3}$.
For the equation $6x^2+11x-2=0$,we have $6x^2+12x-x-2=0 \Rightarrow 6x(x+2)-1(x+2)=0$,so $x = \frac{1}{6}, -2$. Since probability $P \in [0, 1]$,we take $P_3 = \frac{1}{6}$.
Sum of probabilities $P_1 + P_2 + P_3 = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$.
Since the sum of the probabilities of the events is $1$,the events are exhaustive.

(D) The total number of $4$-digit numbers that can be formed using the digits ${4, 5, 6, 7, 8, 9}$ with repetition is $6^4 = 1296$.
Let the number be $N = d_1 d_2 d_3 d_4$. The sum of digits is $S = d_1 + d_2 + d_3 + d_4$.
$A$ number is divisible by $3$ if and only if the sum of its digits is divisible by $3$.
Consider the digits modulo $3$:
$4 \equiv 1 \pmod{3}$,$5 \equiv 2 \pmod{3}$,$6 \equiv 0 \pmod{3}$,$7 \equiv 1 \pmod{3}$,$8 \equiv 2 \pmod{3}$,$9 \equiv 0 \pmod{3}$.
There are two digits for each remainder $0, 1, 2$ modulo $3$.
Let $S_n$ be the sum of the first $n$ digits. For any choice of $d_1, d_2, d_3$,the sum $S_3 = d_1 + d_2 + d_3$ will have some remainder $r \in \{0, 1, 2\}$ modulo $3$.
To make $S_4 = S_3 + d_4$ divisible by $3$,we need $d_4 \equiv -S_3 \pmod{3}$.
Since there are exactly two choices for $d_4$ for each remainder,and there are $6$ total choices for $d_4$,the probability that $d_4$ satisfies the condition is $\frac{2}{6} = \frac{1}{3}$.
Thus,the probability is $\frac{1}{3}$.

(B) The quadratic equation $x^2+ax+b=0$ has real roots if the discriminant $D = a^2 - 4b \geq 0$,which implies $a^2 \geq 4b$.
Given $a \in \{3, 4, 5\}$ and $b \in \{1, 2, 3, 4\}$,the total number of possible pairs $(a, b)$ is $3 \times 4 = 12$.
We check the condition $a^2 \geq 4b$ for each pair:
If $a=3$,$a^2=9$: $9 \geq 4b \implies b \leq 2.25$. Possible values for $b$ are $1, 2$ ($2$ pairs).
If $a=4$,$a^2=16$: $16 \geq 4b \implies b \leq 4$. Possible values for $b$ are $1, 2, 3, 4$ ($4$ pairs).
If $a=5$,$a^2=25$: $25 \geq 4b \implies b \leq 6.25$. Possible values for $b$ are $1, 2, 3, 4$ ($4$ pairs).
Total favorable outcomes = $2 + 4 + 4 = 10$.
Therefore,the required probability is $\frac{10}{12} = \frac{5}{6}$.

(A) Let $P(B)=x$. Then,$P(A)=\frac{2}{3} x$ and $P(C)=\frac{1}{2} x$.
If $A, B, C$ are exhaustive and mutually exclusive,then $P(A \cup B \cup C)=1$.
Since they are mutually exclusive,$P(A \cup B \cup C) = P(A) + P(B) + P(C) = 1$.
Substituting the values: $\frac{2}{3} x + x + \frac{1}{2} x = 1$.
$\frac{4x + 6x + 3x}{6} = 1 \Rightarrow \frac{13x}{6} = 1 \Rightarrow x = \frac{6}{13}$.
Thus,$P(A) = \frac{2}{3} \times \frac{6}{13} = \frac{4}{13}$,$P(B) = \frac{6}{13}$,and $P(C) = \frac{1}{2} \times \frac{6}{13} = \frac{3}{13}$.
Now,$P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$.
Therefore,option $A$ is correct.

(B) Given,the average number of births in a week is $35$.
Since there are $7$ days in a week,the average number of births in a day is $\lambda = \frac{35}{7} = 5$.
We use the Poisson distribution formula $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$.
We need to find the probability of less than $3$ births in a day,which is $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$.
$P(X = 0) = \frac{5^0 e^{-5}}{0!} = e^{-5}$.
$P(X = 1) = \frac{5^1 e^{-5}}{1!} = 5e^{-5}$.
$P(X = 2) = \frac{5^2 e^{-5}}{2!} = \frac{25}{2}e^{-5}$.
Adding these probabilities: $P(X < 3) = e^{-5} + 5e^{-5} + \frac{25}{2}e^{-5} = e^{-5}(1 + 5 + 12.5) = 18.5e^{-5} = \frac{37}{2e^5}$.

(B) Given the probability function $P(X=r) = r/k$ for $r = 1, 2, 3, 4, 5$.
Since the sum of all probabilities must be $1$,we have:
$\sum_{r=1}^{5} P(X=r) = 1 \Rightarrow \frac{1}{k} + \frac{2}{k} + \frac{3}{k} + \frac{4}{k} + \frac{5}{k} = 1$
$\frac{1+2+3+4+5}{k} = 1 \Rightarrow \frac{15}{k} = 1 \Rightarrow k = 15$.
We need to find $P(X=2 \text{ or } X=k/3)$.
Since $k=15$,$k/3 = 15/3 = 5$.
So,$P(X=2 \text{ or } X=5) = P(X=2) + P(X=5) = \frac{2}{15} + \frac{5}{15} = \frac{7}{15}$.
Now,let us check the options:
Option $B$: $P(X=4 \text{ or } X=k/5) = P(X=4 \text{ or } X=15/5) = P(X=4 \text{ or } X=3) = \frac{4}{15} + \frac{3}{15} = \frac{7}{15}$.
Thus,$P(X=2 \text{ or } X=k/3) = P(X=4 \text{ or } X=k/5)$.