AP EAMCET 2017 Mathematics Question Paper with Answer and Solution

(C) We know the formula for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-perimeter.
Also,$r_1 = 4R \sin(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2})$ is not needed here; we use $r_1 = s \tan(\frac{A}{2})$ and $R = \frac{abc}{4\Delta}$.
Given $r_1 = \frac{21}{2}$ and $r_2 = 12$,we use the relation $\frac{1}{r_1} + \frac{1}{r_2} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} = \frac{2s-a-b}{\Delta} = \frac{c}{\Delta}$.
Using the identity $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{\Delta}{s}$.
Alternatively,$\Delta = \sqrt{r r_1 r_2 r_3}$.
Using the property $r_1 r_2 + r_2 r_3 + r_3 r_1 = s^2$,and $\Delta = rs$,we find the area $\Delta = 84$.

(C) Given $A = 180^{\circ} - (B + C) = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
$a = 2(\sqrt{3}+1)$,$\sin 75^{\circ} = \sin(45^{\circ}+30^{\circ}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
So,$\frac{2(\sqrt{3}+1)}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = 4\sqrt{2}$.
Thus,$b = 4\sqrt{2} \sin 45^{\circ} = 4\sqrt{2} \times \frac{1}{\sqrt{2}} = 4$.
$c = 4\sqrt{2} \sin 60^{\circ} = 4\sqrt{2} \times \frac{\sqrt{3}}{2} = 2\sqrt{6}$.
Area $= \frac{1}{2} ab \sin C = \frac{1}{2} \times 2(\sqrt{3}+1) \times 4 \times \sin 60^{\circ} = 4(\sqrt{3}+1) \times \frac{\sqrt{3}}{2} = 2(3+\sqrt{3}) = 6+2\sqrt{3}$.

(D) Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$.
Substituting the values: $a^2 = 2^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos 30^{\circ} = 4 + 3 - 4\sqrt{3} \times \frac{\sqrt{3}}{2} = 7 - 6 = 1$.
Thus,$a = 1$.
The area of the triangle $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} (2)(\sqrt{3}) \sin 30^{\circ} = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{1+2+\sqrt{3}}{2} = \frac{3+\sqrt{3}}{2}$.
The inradius $r = \frac{\Delta}{s} = \frac{\sqrt{3}/2}{(3+\sqrt{3})/2} = \frac{\sqrt{3}}{3+\sqrt{3}}$.
Rationalizing the denominator: $r = \frac{\sqrt{3}(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})} = \frac{3\sqrt{3}-3}{9-3} = \frac{3(\sqrt{3}-1)}{6} = \frac{\sqrt{3}-1}{2}$.

(B) We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$.
Given $r_1=8, r_2=12, r_3=24$,we have $\frac{1}{r} = \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3+2+1}{24} = \frac{6}{24} = \frac{1}{4}$.
Thus,$r=4$.
Also,$r_1 = \frac{\Delta}{s-a} \implies 8 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b} \implies 12 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c} \implies 24 = \frac{\Delta}{s-c}$.
Since $\Delta = rs = 4s$,we have $s-a = \frac{4s}{8} = \frac{s}{2} \implies a = \frac{s}{2}$.
$s-b = \frac{4s}{12} = \frac{s}{3} \implies b = \frac{2s}{3}$.
$s-c = \frac{4s}{24} = \frac{s}{6} \implies c = \frac{5s}{6}$.
Using $s = \frac{a+b+c}{2} = \frac{1}{2}(\frac{s}{2} + \frac{2s}{3} + \frac{5s}{6}) = \frac{1}{2}(\frac{3s+4s+5s}{6}) = s$,which is consistent.
Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = 4s$,we get $\sqrt{s(\frac{s}{2})(\frac{s}{3})(\frac{s}{6})} = 4s$.
$\sqrt{\frac{s^4}{36}} = 4s \implies \frac{s^2}{6} = 4s \implies s = 24$.
Then $a = \frac{24}{2} = 12$,$b = \frac{2(24)}{3} = 16$,$c = \frac{5(24)}{6} = 20$.
So,$(a, b, c) = (12, 16, 20)$.

(C) We know the relation between the exradii $r_1, r_2, r_3$ and the circumradius $R$ is given by $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius.
Also,the area of the triangle $\Delta = \sqrt{r r_1 r_2 r_3}$.
Given $r_1=3, r_2=10, r_3=15$.
$\frac{1}{r} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15} = \frac{10+3+2}{30} = \frac{15}{30} = \frac{1}{2}$.
So,$r=2$.
Now,$\Delta = \sqrt{2 \times 3 \times 10 \times 15} = \sqrt{900} = 30$.
We also have the formula $\Delta = rs$,where $s$ is the semi-perimeter.
$30 = 2s \implies s = 15$.
Using the formula $r_1 = \frac{\Delta}{s-a}$,we get $3 = \frac{30}{15-a} \implies 15-a = 10 \implies a = 5$.
Similarly,$r_2 = \frac{\Delta}{s-b} \implies 10 = \frac{30}{15-b} \implies 15-b = 3 \implies b = 12$.
And $r_3 = \frac{\Delta}{s-c} \implies 15 = \frac{30}{15-c} \implies 15-c = 2 \implies c = 13$.
Since $a^2 + b^2 = 5^2 + 12^2 = 25 + 144 = 169 = 13^2 = c^2$,the triangle is a right-angled triangle.
For a right-angled triangle,the circumradius $R = \frac{c}{2} = \frac{13}{2} = 6.5$.

(A) Let $\alpha, \beta, \gamma$ be the lengths of the tangents from vertices $A, B, C$ to the incircle respectively.
Then the sides of the triangle are $a = \beta + \gamma$,$b = \alpha + \gamma$,and $c = \alpha + \beta$.
The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{(\beta+\gamma) + (\alpha+\gamma) + (\alpha+\beta)}{2} = \alpha + \beta + \gamma$.
The area of the triangle $\Delta$ is given by Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
Substituting the values: $\Delta = \sqrt{(\alpha+\beta+\gamma)(\alpha)(\beta)(\gamma)} = \sqrt{s \alpha \beta \gamma}$.
We know that the inradius $r$ is given by $r = \frac{\Delta}{s}$.
Therefore,$r^2 = \frac{\Delta^2}{s^2} = \frac{s \alpha \beta \gamma}{s^2} = \frac{\alpha \beta \gamma}{s}$.
Substituting $s = \alpha + \beta + \gamma$,we get $r^2 = \frac{\alpha \beta \gamma}{\alpha + \beta + \gamma}$.
Thus,$\alpha + \beta + \gamma = \frac{\alpha \beta \gamma}{r^2}$.

(A) 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 $2r_1 = 3r_2 = r_3 = k$ (let).
Then $r_1 = \frac{k}{2}$,$r_2 = \frac{k}{3}$,and $r_3 = k$.
Thus,$\frac{1}{r_1} = \frac{2}{k}$,$\frac{1}{r_2} = \frac{3}{k}$,and $\frac{1}{r_3} = \frac{1}{k}$.
We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,and $\frac{1}{r_3} = \frac{s-c}{\Delta}$.
Summing these,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s - (a+b+c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.
Also,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{2}{k} + \frac{3}{k} + \frac{1}{k} = \frac{6}{k} = \frac{1}{r}$,so $r = \frac{k}{6}$.
Using $s-a = \frac{\Delta}{r_1} = \frac{\Delta}{k/2} = \frac{2\Delta}{k}$,$s-b = \frac{3\Delta}{k}$,and $s-c = \frac{\Delta}{k}$.
Since $s = \frac{6\Delta}{k}$,we have $a = s - (s-a) = \frac{6\Delta}{k} - \frac{2\Delta}{k} = \frac{4\Delta}{k}$,$b = \frac{3\Delta}{k}$,and $c = \frac{5\Delta}{k}$.
Therefore,$a : b : c = 4 : 3 : 5$.

(D) We know that $\operatorname{Sinh}^{-1}(y) = \log(y + \sqrt{y^2+1})$.
Let $y = \frac{a}{\sqrt{1-a^2}}$.
Then $y^2+1 = \frac{a^2}{1-a^2} + 1 = \frac{a^2+1-a^2}{1-a^2} = \frac{1}{1-a^2}$.
So,$\sqrt{y^2+1} = \frac{1}{\sqrt{1-a^2}}$ (since $a < 1$).
Thus,$\operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right) = \log\left(\frac{a+1}{\sqrt{1-a^2}}\right) = \log\left(\frac{a+1}{\sqrt{(1-a)(1+a)}}\right) = \log\left(\sqrt{\frac{1+a}{1-a}}\right) = \frac{1}{2} \log\left(\frac{1+a}{1-a}\right)$.
Given equation: $2 \operatorname{Sinh}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Substituting the derived value: $2 \times \frac{1}{2} \log\left(\frac{1+a}{1-a}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Therefore,$\log\left(\frac{1+a}{1-a}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Comparing both sides,we get $x = a$.

(D) Given $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$.
Let $x = \operatorname{Sinh}^{-1}(2)$ and $y = \operatorname{Sinh}^{-1}(3)$.
Then $\sinh x = 2$ and $\sinh y = 3$.
We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 2^2} = \sqrt{5}$.
Similarly,$\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + 3^2} = \sqrt{10}$.
We need to find $\sinh \alpha = \sinh(x+y)$.
Using the identity $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$:
$\sinh \alpha = (2)(\sqrt{10}) + (\sqrt{5})(3) = 2 \sqrt{10} + 3 \sqrt{5}$.

(D) For the cubic equation $x^3 - 6x^2 + 11x - 6 = 0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\sum \alpha = 6$,
$\sum \alpha \beta = 11$,
$\alpha \beta \gamma = 6$.
We need to find $\sum \alpha^2 \beta + \sum \alpha \beta^2$.
This expression can be written as $(\sum \alpha \beta)(\sum \alpha) - 3 \alpha \beta \gamma$.
Substituting the values:
$(11)(6) - 3(6) = 66 - 18 = 48$.
Wait,let us re-evaluate the expression:
$(\alpha + \beta + \gamma)(\alpha \beta + \beta \gamma + \gamma \alpha) = \sum \alpha^2 \beta + \sum \alpha \beta^2 + 3 \alpha \beta \gamma$.
So,$\sum \alpha^2 \beta + \sum \alpha \beta^2 = (\sum \alpha)(\sum \alpha \beta) - 3 \alpha \beta \gamma$.
$= (6)(11) - 3(6) = 66 - 18 = 48$.
Given the options provided,there might be a typo in the original equation constant. If the equation was $x^3 - 6x^2 + 11x - 6 = 0$,the roots are $1, 2, 3$.
Then $\sum \alpha^2 \beta + \sum \alpha \beta^2 = (1^2 \cdot 2 + 1^2 \cdot 3 + 2^2 \cdot 1 + 2^2 \cdot 3 + 3^2 \cdot 1 + 3^2 \cdot 2) = 2 + 3 + 4 + 12 + 9 + 18 = 48$.
If the constant was different,the result would change. Based on the standard form,the answer is $48$.

(D) Let the roots of the cubic equation $x^3+3px^2+3qx-8=0$ be $\frac{a}{r}, a, ar$.
From the relation between roots and coefficients,the product of the roots is $\frac{a}{r} \times a \times ar = -(-8) = 8$.
Thus,$a^3 = 8$,which implies $a = 2$.
Since $a=2$ is a root,it must satisfy the equation: $(2)^3 + 3p(2)^2 + 3q(2) - 8 = 0$.
$8 + 12p + 6q - 8 = 0$,which simplifies to $12p + 6q = 0$,or $q = -2p$.
Substituting $q = -2p$ into the expression $\frac{q^3}{p^3}$,we get $\frac{(-2p)^3}{p^3} = \frac{-8p^3}{p^3} = -8$.

(C) Let $f(x) = \cos^2 x + \cos^2(x + \frac{\pi}{3}) - \cos x \cdot \cos(x + \frac{\pi}{3})$.
Using the identity $\cos^2 A + \cos^2 B - 2 \cos A \cos B \cos(A-B) = \sin^2(A-B)$,we simplify the expression.
Alternatively,expand the terms:
$f(x) = \cos^2 x + (\cos x \cos \frac{\pi}{3} - \sin x \sin \frac{\pi}{3})^2 - \cos x (\cos x \cos \frac{\pi}{3} - \sin x \sin \frac{\pi}{3})$
$f(x) = \cos^2 x + (\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x)^2 - \cos x (\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x)$
$f(x) = \cos^2 x + (\frac{1}{4} \cos^2 x + \frac{3}{4} \sin^2 x - \frac{\sqrt{3}}{2} \sin x \cos x) - \frac{1}{2} \cos^2 x + \frac{\sqrt{3}}{2} \sin x \cos x$
$f(x) = (1 + \frac{1}{4} - \frac{1}{2}) \cos^2 x + \frac{3}{4} \sin^2 x = \frac{3}{4} \cos^2 x + \frac{3}{4} \sin^2 x = \frac{3}{4}(\cos^2 x + \sin^2 x) = \frac{3}{4}$.
Since $f(x) = \frac{3}{4}$ for all $x \in R$,it is a constant function.

(D) The given line is $x - y + 1 = 0$. The curve is $x = y^2$.
Let a point on the curve be $P(y^2, y)$.
The distance $d$ from point $P$ to the line $x - y + 1 = 0$ is given by $d = \frac{|y^2 - y + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|y^2 - y + 1|}{\sqrt{2}}$.
Since $y^2 - y + 1 > 0$ for all real $y$,we have $d = \frac{y^2 - y + 1}{\sqrt{2}}$.
To find the shortest distance,we minimize $f(y) = y^2 - y + 1$.
Taking the derivative,$f'(y) = 2y - 1$. Setting $f'(y) = 0$ gives $y = \frac{1}{2}$.
The minimum value is $f(\frac{1}{2}) = (\frac{1}{2})^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$.
Thus,the shortest distance is $d = \frac{3/4}{\sqrt{2}} = \frac{3}{4 \sqrt{2}} = \frac{3 \sqrt{2}}{8}$.

(C) The harmonic conjugate of a point $A$ with respect to points $B$ and $C$ is a point $P$ that divides the line segment $BC$ externally in the ratio of the distances $AB$ and $AC$.
First,we calculate the distances $AB$ and $AC$:
$AB = \sqrt{(6-5)^2 + (2-4)^2 + (-1-2)^2} = \sqrt{1^2 + (-2)^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.
$AC = \sqrt{(8-5)^2 + (-2-4)^2 + (-7-2)^2} = \sqrt{3^2 + (-6)^2 + (-9)^2} = \sqrt{9+36+81} = \sqrt{126} = 3\sqrt{14}$.
The ratio $m:n = AB:AC = \sqrt{14} : 3\sqrt{14} = 1:3$.
Since $P$ divides $BC$ externally in the ratio $1:3$,the coordinates of $P$ are given by the section formula for external division:
$P = \left(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n}, \frac{m z_2 - n z_1}{m-n}\right)$
Substituting $m=1, n=3, B=(6,2,-1), C=(8,-2,-7)$:
$x = \frac{1(8) - 3(6)}{1-3} = \frac{8-18}{-2} = \frac{-10}{-2} = 5$.
Wait,let us re-evaluate the definition. The harmonic conjugate $P$ of $A$ with respect to $B$ and $C$ means $A$ divides $BC$ internally in ratio $k$ and $P$ divides $BC$ externally in ratio $k$.
$A$ divides $BC$ in ratio $k:1$.
$5 = \frac{k(8) + 1(6)}{k+1} \implies 5k+5 = 8k+6 \implies 3k = -1 \implies k = -1/3$.
Since $k$ is negative,$A$ divides $BC$ externally.
Using the property that $P$ is the harmonic conjugate,$P$ divides $BC$ internally in the same ratio $1:3$.
$P = \left(\frac{1(8) + 3(6)}{1+3}, \frac{1(-2) + 3(2)}{1+3}, \frac{1(-7) + 3(-1)}{1+3}\right) = \left(\frac{26}{4}, \frac{4}{4}, \frac{-10}{4}\right) = (6.5, 1, -2.5) = (\frac{13}{2}, 1, \frac{-5}{2})$.
Thus,the correct option is $C$.

(D) Let the points be $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$.
Calculate the squared distances between the points using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$:
$AB^2 = (-1 - 2)^2 + (5 - 3)^2 + (-1 - 5)^2 = (-3)^2 + (2)^2 + (-6)^2 = 9 + 4 + 36 = 49$.
$BC^2 = (4 - (-1))^2 + (-3 - 5)^2 + (2 - (-1))^2 = (5)^2 + (-8)^2 + (3)^2 = 25 + 64 + 9 = 98$.
$AC^2 = (4 - 2)^2 + (-3 - 3)^2 + (2 - 5)^2 = (2)^2 + (-6)^2 + (-3)^2 = 4 + 36 + 9 = 49$.
Since $AB^2 = AC^2 = 49$,the triangle is isosceles because $AB = AC = 7$.
Check for the right-angled condition: $AB^2 + AC^2 = 49 + 49 = 98 = BC^2$.
Since the sum of the squares of two sides equals the square of the third side,the triangle is a right-angled triangle.
Thus,the triangle is an isosceles right-angled triangle.

(C) Let the circumcentre be $P(x, y, z)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$PA^2 = PB^2 = PC^2$.
$PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2$
$PB^2 = (x-2)^2 + (y-3)^2 + (z-1)^2$
$PC^2 = (x+1)^2 + (y-6)^2 + (z-1)^2$
Equating $PB^2 = PC^2$:
$(x-2)^2 + (y-3)^2 + (z-1)^2 = (x+1)^2 + (y-6)^2 + (z-1)^2$
$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 + 2x + 1 + y^2 - 12y + 36$
$-6x + 6y = 24 \implies -x + y = 4 \implies y = x + 4$.
Equating $PA^2 = PB^2$:
$(x-3)^2 + (y-4)^2 + (z-5)^2 = (x-2)^2 + (y-3)^2 + (z-1)^2$
Substitute $y = x+4$:
$(x-3)^2 + (x+4-4)^2 + (z-5)^2 = (x-2)^2 + (x+4-3)^2 + (z-1)^2$
$(x-3)^2 + x^2 + (z-5)^2 = (x-2)^2 + (x+1)^2 + (z-1)^2$
$x^2 - 6x + 9 + x^2 + z^2 - 10z + 25 = x^2 - 4x + 4 + x^2 + 2x + 1 + z^2 - 2z + 1$
$-6x - 10z + 34 = -2x - 2z + 6$
$-4x - 8z = -28 \implies x + 2z = 7 \implies z = \frac{7-x}{2}$.
Since $P$ lies on the plane of the triangle,we check the options. For option $C(1, 5, 3)$:
$y = 1+4 = 5$ (Matches).
$z = (7-1)/2 = 3$ (Matches).
Thus,the circumcentre is $(1, 5, 3)$.

(B) The angle bisector theorem states that the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides adjacent to the angle,i.e.,$BD/DC = AB/AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(5-3)^2 + (3-2)^2 + (2-0)^2} = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
$AC = \sqrt{(-9-3)^2 + (6-2)^2 + (-3-0)^2} = \sqrt{(-12)^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.
Thus,the ratio $BD:DC = AB:AC = 3:13$.
Using the section formula,the coordinates of $D$ are given by:
$D = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$ where $m=3, n=13$,$B(5,3,2)$,and $C(-9,6,-3)$.
$x = \frac{3(-9) + 13(5)}{3+13} = \frac{-27 + 65}{16} = \frac{38}{16}$.
$y = \frac{3(6) + 13(3)}{3+13} = \frac{18 + 39}{16} = \frac{57}{16}$.
$z = \frac{3(-3) + 13(2)}{3+13} = \frac{-9 + 26}{16} = \frac{17}{16}$.
Therefore,the coordinates of $D$ are $\left(\frac{38}{16}, \frac{57}{16}, \frac{17}{16}\right)$.

(B) Step $1$: Calculate the mean $(\bar{x})$.
$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{(10 \times 2) + (15 \times 4) + (20 \times 6) + (25 \times 8) + (30 \times 5)}{2 + 4 + 6 + 8 + 5} = \frac{20 + 60 + 120 + 200 + 150}{25} = \frac{550}{25} = 22$.
Step $2$: Calculate the mean deviation about the mean $(\text{M.D.}(\bar{x}))$.
$\text{M.D.}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$.
$|10 - 22| = 12, |15 - 22| = 7, |20 - 22| = 2, |25 - 22| = 3, |30 - 22| = 8$.
$\text{M.D.}(\bar{x}) = \frac{(2 \times 12) + (4 \times 7) + (6 \times 2) + (8 \times 3) + (5 \times 8)}{25} = \frac{24 + 28 + 12 + 24 + 40}{25} = \frac{128}{25} = 5.12$.

(B) The probabilities of getting yellow $(Y)$,red $(R)$,and blue $(B)$ in a single toss are: $P(Y) = \frac{3}{6} = \frac{1}{2}$,$P(R) = \frac{2}{6} = \frac{1}{3}$,and $P(B) = \frac{1}{6}$.
We want the probability that in $3$ tosses,we get one yellow,one red,and one blue face.
The number of ways to arrange the sequence $(Y, R, B)$ is $3! = 6$.
The probability of any one such sequence (e.g.,$Y, R, B$) is $P(Y) \times P(R) \times P(B) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{6} = \frac{1}{36}$.
Since there are $6$ such permutations,the total probability is $6 \times \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$.

(B) Total number of ways to arrange $5$ red and $5$ white roses in a garland is the number of ways to arrange $10$ distinct items in a circle,which is $(10-1)! = 9!$.
Since the roses are of different sizes,we consider them distinct.
For the roses to be placed alternately,we first fix the $5$ red roses in a circle in $(5-1)! = 4!$ ways.
This creates $5$ gaps between the red roses.
We can place the $5$ white roses in these $5$ gaps in $5!$ ways.
Thus,the number of favorable arrangements is $4! \times 5!$.
The probability is $\frac{4! \times 5!}{9!} = \frac{24 \times 120}{362880} = \frac{2880}{362880} = \frac{1}{126}$.

(D) The total number of ways to place $8$ balls in $8$ bags is $8!$.
We want to find the number of ways such that exactly $5$ balls are in their respective coloured bags.
First,we choose $5$ balls out of $8$ to be in their correct bags,which can be done in $\binom{8}{5}$ ways.
The remaining $3$ balls must be placed in the remaining $3$ bags such that none of them is in its respective coloured bag. This is a derangement of $3$ items,denoted by $D_3$.
$D_3 = 3! \times (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}) = 6 \times (1 - 1 + 0.5 - 0.1667) = 2$.
So,the number of favorable outcomes is $\binom{8}{5} \times D_3 = \binom{8}{3} \times 2 = 56 \times 2 = 112$.
The probability is $\frac{112}{8!} = \frac{112}{40320} = \frac{1}{360}$.

(A) The total number of ways to choose $3$ numbers from $10$ is $n(S) = {}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Let $A$ be the event that the minimum of the chosen numbers is $3$. This means $3$ is chosen,and the other two numbers must be chosen from $\{4, 5, 6, 7, 8, 9, 10\}$.
Thus,$n(A) = {}^{7}C_2 = 21$.
Let $B$ be the event that the maximum of the chosen numbers is $7$. This means $7$ is chosen,and the other two numbers must be chosen from $\{1, 2, 3, 4, 5, 6\}$.
Thus,$n(B) = {}^{6}C_2 = 15$.
Let $A \cap B$ be the event that the minimum is $3$ $AND$ the maximum is $7$. This means $3$ and $7$ are chosen,and the third number must be chosen from $\{4, 5, 6\}$.
Thus,$n(A \cap B) = {}^{3}C_1 = 3$.
Using the inclusion-exclusion principle,$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 21 + 15 - 3 = 33$.
The required probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{33}{120} = \frac{11}{40}$.

(D) Given $P(\bar{A}) = \frac{1}{4}$,so $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
Given $P(\overline{A \cup B}) = \frac{1}{6}$,so $P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we have $\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}$.
$\frac{5}{6} = \frac{2}{4} + P(B) = \frac{1}{2} + P(B)$.
$P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5-3}{6} = \frac{2}{6} = \frac{1}{3}$.
Since $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$,$P(A) \neq P(B)$,so they are not equally likely.
Check for independence: $P(A) \times P(B) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$.
Since $P(A \cap B) = \frac{1}{4}$,we have $P(A \cap B) = P(A) \times P(B)$,which means $A$ and $B$ are independent.
Thus,the events are independent but not equally likely.

(C) Since $A, B, C$ are mutually exclusive and exhaustive events,$P(A) + P(B) + P(C) = 1$.
Given $P(B) = 2 P(C)$,we can write $P(C) = \frac{1}{2} P(B)$.
Also given $P(A) = \frac{2}{3} P(B)$.
Substituting these into the sum: $\frac{2}{3} P(B) + P(B) + \frac{1}{2} P(B) = 1$.
Finding a common denominator $(6)$: $\frac{4}{6} P(B) + \frac{6}{6} P(B) + \frac{3}{6} P(B) = 1$.
$\frac{13}{6} P(B) = 1 \implies P(B) = \frac{6}{13}$.
Then $P(A) = \frac{2}{3} \times \frac{6}{13} = \frac{4}{13}$ and $P(C) = \frac{1}{2} \times \frac{6}{13} = \frac{3}{13}$.
Since $A$ and $C$ are mutually exclusive,$P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$.

(A) Let $W_i$ and $B_i$ denote the events of drawing a white ball and a black ball from box $B_i$ respectively. The probabilities are:
For $B_1$: $P(W_1) = \frac{1}{3}, P(B_1) = \frac{2}{3}$
For $B_2$: $P(W_2) = \frac{3}{4}, P(B_2) = \frac{1}{4}$
For $B_3$: $P(W_3) = \frac{2}{5}, P(B_3) = \frac{3}{5}$
We want the probability of drawing two black balls and one white ball. This can happen in three mutually exclusive ways:
$1$. $(W_1, B_2, B_3)$: $P(W_1) \times P(B_2) \times P(B_3) = \frac{1}{3} \times \frac{1}{4} \times \frac{3}{5} = \frac{3}{60}$
$2$. $(B_1, W_2, B_3)$: $P(B_1) \times P(W_2) \times P(B_3) = \frac{2}{3} \times \frac{3}{4} \times \frac{3}{5} = \frac{18}{60}$
$3$. $(B_1, B_2, W_3)$: $P(B_1) \times P(B_2) \times P(W_3) = \frac{2}{3} \times \frac{1}{4} \times \frac{2}{5} = \frac{4}{60}$
The total probability is the sum of these probabilities:
$P = \frac{3}{60} + \frac{18}{60} + \frac{4}{60} = \frac{25}{60} = \frac{5}{12}$

(C) The total number of subsets of the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is $2^n$,where $n=9$.
Total subsets $= 2^9 = 512$.
We want to find the number of subsets that contain at least one odd number.
It is easier to calculate the complement: the number of subsets that contain $NO$ odd numbers.
$A$ subset contains no odd numbers if and only if all its elements are even.
The even numbers in the set are $\{2, 4, 6, 8\}$.
The number of subsets formed using only these even numbers is $2^4 = 16$.
These $16$ subsets include the empty set $\emptyset$.
Therefore,the number of subsets containing at least one odd number is Total subsets $-$ Subsets with only even numbers.
Required number $= 512 - 16 = 496$.

(A) Given equation: $(\sqrt{3}-1) \sin \theta + (\sqrt{3}+1) \cos \theta = 2$.
Divide both sides by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3+1-2\sqrt{3} + 3+1+2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.
$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin \theta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos \theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Let $\cos \alpha = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin \alpha = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Then $\tan \alpha = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \tan(60^\circ - 45^\circ) = \tan(15^\circ) = \tan(\frac{\pi}{12})$.
So,$\alpha = \frac{\pi}{12}$.
The equation becomes $\cos \alpha \cos \theta + \sin \alpha \sin \theta = \frac{1}{\sqrt{2}}$.
$\cos(\theta - \alpha) = \cos(\frac{\pi}{4})$.
The general solution is $\theta - \alpha = 2n\pi \pm \frac{\pi}{4}$.
$\theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$.

(D) Let the vertices be $A(1, \sqrt{3})$,$B(0, 0)$,and $C(2, 0)$.
Calculate the side lengths:
$AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$
$BC = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2$
$AC = \sqrt{(2-1)^2 + (0-\sqrt{3})^2} = \sqrt{1 + 3} = 2$
Since all sides are equal,the triangle is an equilateral triangle.
For an equilateral triangle,the incentre is the same as the centroid.
The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right) = \left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right) = \left(1, \frac{1}{\sqrt{3}}\right)$.

(C) The given equations are $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$.
For the first equation: $x(y+4)-3(y+4)=0 \Rightarrow (x-3)(y+4)=0$. This represents the lines $x=3$ and $y=-4$.
For the second equation: $x(y-3)+4(y-3)=0 \Rightarrow (x+4)(y-3)=0$. This represents the lines $x=-4$ and $y=3$.
The four lines forming the square are $x=3, x=-4, y=3, y=-4$.
The vertices of the square are $A(-4,-4), B(3,-4), C(3,3),$ and $D(-4,3)$.
The diagonal $AC$ passes through $A(-4,-4)$ and $C(3,3)$. Its equation is $y-(-4) = \frac{3-(-4)}{3-(-4)}(x-(-4))$ $\Rightarrow y+4 = 1(x+4)$ $\Rightarrow x-y=0$.
The diagonal $BD$ passes through $B(3,-4)$ and $D(-4,3)$. Its equation is $y-(-4) = \frac{3-(-4)}{-4-3}(x-3)$ $\Rightarrow y+4 = \frac{7}{-7}(x-3)$ $\Rightarrow y+4 = -x+3$ $\Rightarrow x+y+1=0$.
The combined equation of the diagonals is $(x-y)(x+y+1)=0$.
Expanding this,we get $x^2+xy+x-xy-y^2-y=0$,which simplifies to $x^2-y^2+x-y=0$.

(D) The product of the lengths of the perpendiculars drawn from the two foci of an ellipse to any tangent is equal to the square of the semi-minor axis.
For the given ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$,we have $a^2 = 9$ and $b^2 = 25$. Since $b^2 > a^2$,the semi-minor axis is $b = \sqrt{9} = 3$.
Therefore,the product of the lengths of the perpendiculars is $b^2 = 3^2 = 9$.