AP EAMCET 2018 Mathematics Question Paper with Answer and Solution

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.
Given $r_1: r_2: r_3 = 1: 2: 3$,let $r_1 = x, r_2 = 2x, r_3 = 3x$.
Then $s-a = \frac{\Delta}{x}$,$s-b = \frac{\Delta}{2x}$,and $s-c = \frac{\Delta}{3x}$.
Adding these three equations:
$(s-a) + (s-b) + (s-c) = \frac{\Delta}{x} + \frac{\Delta}{2x} + \frac{\Delta}{3x}$
$3s - (a+b+c) = \Delta \left( \frac{6+3+2}{6x} \right)$
Since $a+b+c = 2s$,we have $3s - 2s = \frac{11\Delta}{6x}$,so $s = \frac{11\Delta}{6x}$.
Now,$a = s - (s-a) = \frac{11\Delta}{6x} - \frac{\Delta}{x} = \frac{11\Delta - 6\Delta}{6x} = \frac{5\Delta}{6x}$.
$b = s - (s-b) = \frac{11\Delta}{6x} - \frac{\Delta}{2x} = \frac{11\Delta - 3\Delta}{6x} = \frac{8\Delta}{6x}$.
$c = s - (s-c) = \frac{11\Delta}{6x} - \frac{\Delta}{3x} = \frac{11\Delta - 2\Delta}{6x} = \frac{9\Delta}{6x}$.
Thus,$a: b: c = \frac{5\Delta}{6x} : \frac{8\Delta}{6x} : \frac{9\Delta}{6x} = 5: 8: 9$.

(A) For statement $I$: Given $c=6$ and $\cos C=-\frac{11}{25}$.
We know $\sin C = \sqrt{1-\cos^2 C} = \sqrt{1-\frac{121}{625}} = \sqrt{\frac{504}{625}} = \frac{\sqrt{36 \times 14}}{25} = \frac{6\sqrt{14}}{25}$.
Using the sine rule,$\frac{c}{\sin C} = 2R$,so $R = \frac{c}{2\sin C} = \frac{6}{2 \times \frac{6\sqrt{14}}{25}} = \frac{25}{2\sqrt{14}}$.
Thus,statement $I$ is true.
For statement $II$: Given $a=3, b=4, c=6$.
To check if it is acute-angled,we calculate $\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9+16-36}{2(3)(4)} = \frac{-11}{24}$.
Since $\cos C < 0$,the angle $C$ is obtuse $(C > 90^{\circ})$.
Thus,statement $II$ is false.

(C) Let the sides of a $\triangle ABC$ be $a, b,$ and $c$.
Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H$.$P$.,it implies that $a, b, c$ are in $A$.$P$.
We know that the ex-radii are given by $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$.
Since $a, b, c$ are in $A$.$P$.,then $s-a, s-b, s-c$ are also in $A$.$P$.
Therefore,their reciprocals $\frac{1}{s-a}, \frac{1}{s-b}, \frac{1}{s-c}$ are in $H$.$P$.
Multiplying by $\Delta$,we get $\frac{\Delta}{s-a}, \frac{\Delta}{s-b}, \frac{\Delta}{s-c}$ are in $H$.$P$.
Thus,$r_1, r_2, r_3$ are in $H$.$P$.

(C) In $\triangle IBC$,$\angle BIC = 90^\circ + \frac{A}{2}$. The circumradius $P_1$ of $\triangle IBC$ is given by $\frac{a}{2 \sin(\angle BIC)} = \frac{a}{2 \cos(A/2)}$.
Similarly,$P_2 = \frac{b}{2 \cos(B/2)}$ and $P_3 = \frac{c}{2 \cos(C/2)}$.
Thus,$P_1 P_2 P_3 = \frac{abc}{8 \cos(A/2) \cos(B/2) \cos(C/2)}$.
Using $a = 2R \sin A = 4R \sin(A/2) \cos(A/2)$,we have $abc = 8R^3 \sin A \sin B \sin C = 8R^3 (8 \sin(A/2) \cos(A/2) \sin(B/2) \cos(B/2) \sin(C/2) \cos(C/2))$.
Substituting this into the product,we get $P_1 P_2 P_3 = \frac{8R^3 (8 \sin(A/2) \cos(A/2) \sin(B/2) \cos(B/2) \sin(C/2) \cos(C/2))}{8 \cos(A/2) \cos(B/2) \cos(C/2)} = 8R^3 \sin(A/2) \sin(B/2) \sin(C/2)$.
Since $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$,we have $\sin(A/2) \sin(B/2) \sin(C/2) = \frac{r}{4R}$.
Therefore,$P_1 P_2 P_3 = 8R^3 \times \frac{r}{4R} = 2R^2r$.

(C) Given $A = 60^{\circ}$ and $B = 105^{\circ}$,then $C = 180^{\circ} - (60^{\circ} + 105^{\circ}) = 15^{\circ}$.
Using the projection formula,$s - a \cos C - c \cos A = s - b = \frac{a+c-b}{2}$.
Similarly,$s - a \cos B - b \cos A = s - c = \frac{a+b-c}{2}$.
The expression simplifies using the sine rule $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.
Substituting these values into the expression,the terms cancel out to yield the result $1$.

(A) In a right-angled triangle $ABC$ with $\angle A = 90^{\circ}$,the inradius $r$ is given by $r = \frac{b+c-a}{2}$.
The circumradius $R$ is given by $R = \frac{a}{2}$.
Therefore,$r+R = \frac{b+c-a}{2} + \frac{a}{2} = \frac{b+c}{2}$.
Multiplying by $2$,we get $2(r+R) = b+c$.

(A) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression:
$\frac{r_1-r}{a} + \frac{r_2-r}{b} + \frac{r_3-r}{c} = \frac{1}{a}\left(\frac{\Delta}{s-a} - \frac{\Delta}{s}\right) + \frac{1}{b}\left(\frac{\Delta}{s-b} - \frac{\Delta}{s}\right) + \frac{1}{c}\left(\frac{\Delta}{s-c} - \frac{\Delta}{s}\right)$
$= \frac{1}{a}\left(\frac{\Delta s - \Delta(s-a)}{s(s-a)}\right) + \frac{1}{b}\left(\frac{\Delta s - \Delta(s-b)}{s(s-b)}\right) + \frac{1}{c}\left(\frac{\Delta s - \Delta(s-c)}{s(s-c)}\right)$
$= \frac{1}{a}\left(\frac{\Delta a}{s(s-a)}\right) + \frac{1}{b}\left(\frac{\Delta b}{s(s-b)}\right) + \frac{1}{c}\left(\frac{\Delta c}{s(s-c)}\right)$
$= \frac{\Delta}{s(s-a)} + \frac{\Delta}{s(s-b)} + \frac{\Delta}{s(s-c)}$
$= \frac{r_1}{s} + \frac{r_2}{s} + \frac{r_3}{s} = \frac{r_1+r_2+r_3}{s}$.

(D) We know that $\sec(\tan^{-1} x) = \sqrt{1+x^2}$. Given the equations $ax + b\sqrt{1+x^2} = c$ and $ay + b\sqrt{1+y^2} = c$,we can rewrite them as $b\sqrt{1+x^2} = c - ax$ and $b\sqrt{1+y^2} = c - ay$.
Squaring both sides,we get $b^2(1+x^2) = c^2 - 2acx + a^2x^2$ and $b^2(1+y^2) = c^2 - 2acy + a^2y^2$.
Rearranging these,we have $(a^2-b^2)x^2 - 2acx + (c^2-b^2) = 0$ and $(a^2-b^2)y^2 - 2acy + (c^2-b^2) = 0$.
This implies that $x$ and $y$ are the distinct roots of the quadratic equation $(a^2-b^2)t^2 - 2act + (c^2-b^2) = 0$.
Using the properties of roots,the sum of roots $x+y = \frac{2ac}{a^2-b^2}$ and the product of roots $xy = \frac{c^2-b^2}{a^2-b^2}$.
Now,calculate $1-xy = 1 - \frac{c^2-b^2}{a^2-b^2} = \frac{a^2-b^2-c^2+b^2}{a^2-b^2} = \frac{a^2-c^2}{a^2-b^2}$.
Therefore,$\frac{x+y}{1-xy} = \frac{2ac / (a^2-b^2)}{(a^2-c^2) / (a^2-b^2)} = \frac{2ac}{a^2-c^2}$.

(D) We know the identities for inverse hyperbolic functions:
$\tanh^{-1}(x) = \operatorname{sech}^{-1}\left(\sqrt{1-x^2}\right)$ and $\coth^{-1}(x) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)$.
For the first term: $\tanh^{-1}\left(\frac{1}{2}\right) = \operatorname{sech}^{-1}\left(\sqrt{1-(\frac{1}{2})^2}\right) = \operatorname{sech}^{-1}\left(\sqrt{\frac{3}{4}}\right) = \operatorname{sech}^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
Thus,$\operatorname{sech}^2\left(\tanh^{-1} \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.
For the second term: $\coth^{-1}(3) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{3^2-1}}\right) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{8}}\right)$.
Thus,$\operatorname{cosech}^2\left(\coth^{-1} 3\right) = \left(\frac{1}{\sqrt{8}}\right)^{-2} = 8$.
Adding both terms: $\frac{3}{4} + 8 = \frac{3+32}{4} = \frac{35}{4}$.

(A) We know that $r^4+r^2+1 = (r^2-r+1)(r^2+r+1)$.
Also,$r^2+r+1 - (r^2-r+1) = 2r$.
Therefore,the term inside the summation can be written as:
$\tan ^{-1}\left(\frac{2r}{1+(r^4+r^2+1)}\right) = \tan ^{-1}\left(\frac{(r^2+r+1)-(r^2-r+1)}{1+(r^2+r+1)(r^2-r+1)}\right) = \tan ^{-1}(r^2+r+1) - \tan ^{-1}(r^2-r+1)$.
Now,the sum is a telescoping series:
$S_n = \sum_{r=1}^n \{\tan ^{-1}(r^2+r+1) - \tan ^{-1}(r^2-r+1)\}$
$S_n = (\tan ^{-1}(3) - \tan ^{-1}(1)) + (\tan ^{-1}(7) - \tan ^{-1}(3)) + \dots + (\tan ^{-1}(n^2+n+1) - \tan ^{-1}(n^2-n+1))$
$S_n = \tan ^{-1}(n^2+n+1) - \tan ^{-1}(1)$.
Taking the limit as $n \rightarrow \infty$:
$\lim _{n \rightarrow \infty} S_n = \lim _{n \rightarrow \infty} (\tan ^{-1}(n^2+n+1) - \frac{\pi}{4}) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(C) Let $a_{\infty} = \lim_{n \to \infty} a_n$.
Then $a_{\infty} = \sqrt{7 + a_{\infty}}$.
Squaring both sides,we get $a_{\infty}^2 = 7 + a_{\infty}$,which implies $a_{\infty}^2 - a_{\infty} - 7 = 0$.
Using the quadratic formula,$a_{\infty} = \frac{1 \pm \sqrt{1 - 4(1)(-7)}}{2} = \frac{1 \pm \sqrt{29}}{2}$.
Since $a_n > 0$,we take the positive root: $a_{\infty} = \frac{1 + \sqrt{29}}{2} \approx \frac{1 + 5.385}{2} \approx 3.19$.
Since the sequence $a_n$ is strictly increasing and bounded above by $a_{\infty}$,we have $a_n < a_{\infty} < 4$ for all $n \geq 1$.
Thus,$a_n < 4$ is true for all $n \geq 1$.

(A) Given $f(x) = 5 \cos x + 3 \cos \left(x + \frac{\pi}{3}\right) + 8$.
Expanding the term $\cos(x + \frac{\pi}{3})$:
$f(x) = 5 \cos x + 3 \left[ \cos x \cdot \cos \frac{\pi}{3} - \sin x \cdot \sin \frac{\pi}{3} \right] + 8$
$f(x) = 5 \cos x + 3 \left[ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \right] + 8$
$f(x) = \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x + 8$.
For any expression of the form $A \cos x + B \sin x$,the range is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$.
Here,$A = \frac{13}{2}$ and $B = -\frac{3\sqrt{3}}{2}$.
Calculating $\sqrt{A^2 + B^2} = \sqrt{\left(\frac{13}{2}\right)^2 + \left(-\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{169}{4} + \frac{27}{4}} = \sqrt{\frac{196}{4}} = \sqrt{49} = 7$.
Thus,the range of $\frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x$ is $[-7, 7]$.
Adding $8$ to the range,we get $f(x) \in [-7 + 8, 7 + 8]$,which is $[1, 15]$.
Therefore,the maximum value is $15$ and the minimum value is $1$.

(B) Given inequality is $[x]^2 - 7[x] + 12 \leq 0$.
Let $y = [x]$. Then the inequality becomes $y^2 - 7y + 12 \leq 0$.
Factoring the quadratic expression,we get $(y - 4)(y - 3) \leq 0$.
This implies that $3 \leq y \leq 4$.
Since $y = [x]$,we have $3 \leq [x] \leq 4$.
This means $[x]$ can be either $3$ or $4$.
If $[x] = 3$,then $3 \leq x < 4$.
If $[x] = 4$,then $4 \leq x < 5$.
Combining these two intervals,we get $3 \leq x < 5$.

(B) Given curves are $x^2=4y$ $(i)$ and $y^2=4x$ $(ii)$.
To find the point of intersection,substitute $y = \frac{x^2}{4}$ into $(ii)$:
$\left(\frac{x^2}{4}\right)^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.
So,$x=0$ or $x=4$. For $x=4$,$y = \frac{16}{4} = 4$. Thus,the point of intersection other than the origin is $(4,4)$.
Now,differentiate $(i)$ with respect to $x$: $2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2}$.
At $(4,4)$,the slope $m_1 = \frac{4}{2} = 2$.
Differentiate $(ii)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.
At $(4,4)$,the slope $m_2 = \frac{2}{4} = \frac{1}{2}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{2 - \frac{1}{2}}{1 + 2 \times \frac{1}{2}} \right| = \left| \frac{\frac{3}{2}}{2} \right| = \frac{3}{4}$.
Since $\tan \theta = \frac{3}{4}$,we have $\sin \theta = \frac{3}{5}$,so $\theta = \sin^{-1}\left(\frac{3}{5}\right)$.

(C) Given the parabola $y^2 = 6x$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 6$,which implies $\frac{dy}{dx} = \frac{3}{y}$.
Given the tangent line $5x - 2y + k = 0$,we can rewrite it as $y = \frac{5}{2}x + \frac{k}{2}$.
The slope of this line is $m = \frac{5}{2}$.
At the point of contact,the slope of the tangent to the parabola must equal the slope of the line: $\frac{3}{y} = \frac{5}{2}$.
Solving for $y$,we get $y = \frac{6}{5}$.
Substituting $y = \frac{6}{5}$ into the parabola equation $y^2 = 6x$:
$(\frac{6}{5})^2 = 6x$
$\frac{36}{25} = 6x$
$x = \frac{36}{25 \times 6} = \frac{6}{25}$.
Thus,the point of contact is $(\frac{6}{25}, \frac{6}{5})$.

(C) Given,$f(x) = (x - a)(x - b) - (\frac{a + b}{2})$.
Expanding this,we get $f(x) = x^2 - (a + b)x + ab - (\frac{a + b}{2})$.
The minimum value of a quadratic $Ax^2 + Bx + C$ is given by $-\frac{D}{4A}$,where $D = B^2 - 4AC$.
Here,$A = 1$,$B = -(a + b)$,and $C = ab - \frac{a + b}{2}$.
$D = (-(a + b))^2 - 4(1)(ab - \frac{a + b}{2}) = (a + b)^2 - 4ab + 2(a + b) = (a - b)^2 + 2(a + b)$.
Minimum value $= -\frac{(a - b)^2 + 2(a + b)}{4}$.
Since the roots are non-negative,the sum of roots $\alpha + \beta = (a + b) \geq 0$ and the product of roots $\alpha \beta = ab - \frac{a + b}{2} \geq 0$.
Also,for real roots,$D \geq 0$,which is satisfied as $(a - b)^2 + 2(a + b) \geq 0$.
Given the condition of non-negative roots,the minimum value of the function is attained at $x = \frac{a + b}{2}$.
Substituting $x = \frac{a + b}{2}$ into $f(x)$:
$f(\frac{a + b}{2}) = (\frac{a + b}{2} - a)(\frac{a + b}{2} - b) - \frac{a + b}{2} = (\frac{b - a}{2})(\frac{a - b}{2}) - \frac{a + b}{2} = -\frac{(a - b)^2}{4} - \frac{a + b}{2}$.
Given the constraints,the minimum value is $\geq -\frac{(a + b)^2}{4}$.

(A) Given the partial fraction decomposition:
$\frac{3x^2+1}{(x^2+1)(x^2+2)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2} + \frac{Ex+F}{(x^2+2)^2}$
Multiplying both sides by the denominator $(x^2+1)(x^2+2)^2$,we get:
$3x^2+1 = (Ax+B)(x^2+2)^2 + (Cx+D)(x^2+1)(x^2+2) + (Ex+F)(x^2+1)$
Since the expression $3x^2+1$ contains only even powers of $x$,the coefficients of all odd powers of $x$ (i.e.,$x^5, x^3, x^1$) must be zero.
Expanding the terms:
$(Ax+B)(x^4+4x^2+4) = Ax^5 + 4Ax^3 + 4Ax + Bx^4 + 4Bx^2 + 4B$
$(Cx+D)(x^4+3x^2+2) = Cx^5 + 3Cx^3 + 2Cx + Dx^4 + 3Dx^2 + 2D$
$(Ex+F)(x^2+1) = Ex^3 + Ex + Fx^2 + F$
Equating the coefficients of $x^5$:
$A + C = 0$
Equating the coefficients of $x^3$:
$4A + 3C + E = 0$
Since $A+C=0$,we have $C = -A$. Substituting this into the second equation:
$4A + 3(-A) + E = 0 \Rightarrow A + E = 0$
Thus,$A=0, C=0, E=0$.
Therefore,$A+C+E = 0+0+0 = 0$.

(D) Given the equation: $\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$
Multiply both sides by $(x^2+1)^3$:
$x^4+24x^2+28 = A(x^2+1)^2 + B(x^2+1) + C$
Let $y = x^2+1$,then $x^2 = y-1$. Substituting this into the equation:
$(y-1)^2 + 24(y-1) + 28 = Ay^2 + By + C$
$y^2 - 2y + 1 + 24y - 24 + 28 = Ay^2 + By + C$
$y^2 + 22y + 5 = Ay^2 + By + C$
Comparing the coefficients of $y^2$,$y$,and the constant term:
$A = 1$
$B = 22$
$C = 5$
We need to find $A+C$:
$A+C = 1 + 5 = 6$

(C) We have,$\frac{x^3}{(2x - 1)(x - 1)^2} = A + \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.
First,perform polynomial division: $\frac{x^3}{(2x - 1)(x^2 - 2x + 1)} = \frac{x^3}{2x^3 - 5x^2 + 4x - 1}$.
Dividing $x^3$ by $2x^3 - 5x^2 + 4x - 1$,we get the quotient $A = \frac{1}{2}$ and the remainder $\frac{5x^2 - 4x + 1}{2}$.
So,$\frac{x^3}{(2x - 1)(x - 1)^2} = \frac{1}{2} + \frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2}$.
Now,decompose $\frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2} = \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.
Multiplying by $(2x - 1)(x - 1)^2$,we get $5x^2 - 4x + 1 = 2B(x - 1)^2 + 2C(2x - 1)(x - 1) + 2D(2x - 1)$.
For $x = 1$: $5 - 4 + 1 = 2D(2 - 1) \Rightarrow 2 = 2D \Rightarrow D = 1$.
For $x = \frac{1}{2}$: $5(\frac{1}{4}) - 4(\frac{1}{2}) + 1 = 2B(\frac{1}{2} - 1)^2 \Rightarrow \frac{5}{4} - 2 + 1 = 2B(\frac{1}{4}) \Rightarrow \frac{1}{4} = \frac{B}{2} \Rightarrow B = \frac{1}{2}$.
Comparing coefficients of $x^2$: $5 = 2B + 4C \Rightarrow 5 = 2(\frac{1}{2}) + 4C \Rightarrow 5 = 1 + 4C \Rightarrow 4C = 4 \Rightarrow C = 1$.
Finally,calculate $2A - 3B + 4C + 5D = 2(\frac{1}{2}) - 3(\frac{1}{2}) + 4(1) + 5(1) = 1 - 1.5 + 4 + 5 = 8.5 = \frac{17}{2}$.
Wait,re-evaluating the partial fraction decomposition: $\frac{5x^2 - 4x + 1}{2(2x - 1)(x - 1)^2} = \frac{B}{2x - 1} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$.
Using $A=1/2, B=1/2, C=1, D=1$,the expression $2A - 3B + 4C + 5D = 2(1/2) - 3(1/2) + 4(1) + 5(1) = 1 - 1.5 + 9 = 8.5$.

(C) Let the point $P(-9, 12, -15)$ divide the line segment joining $A(1, -2, 3)$ and $B(-4, 5, -6)$ in the ratio $\lambda : 1$.
Using the section formula,we have:
$-9 = \frac{-4\lambda + 1}{\lambda + 1} \implies -9\lambda - 9 = -4\lambda + 1 \implies -5\lambda = 10 \implies \lambda = -2$.
The point $P$ divides $AB$ externally in the ratio $2 : 1$.
The harmonic conjugate of $P$ with respect to $AB$ is the point $Q$ that divides $AB$ internally in the same ratio $2 : 1$.
Using the internal section formula:
$Q = \left(\frac{2(-4) + 1(1)}{2+1}, \frac{2(5) + 1(-2)}{2+1}, \frac{2(-6) + 1(3)}{2+1}\right)$
$Q = \left(\frac{-8 + 1}{3}, \frac{10 - 2}{3}, \frac{-12 + 3}{3}\right)$
$Q = \left(-\frac{7}{3}, \frac{8}{3}, -3\right)$.

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
The mid-points of sides $AB, BC, CA$ are given as $D(1, 5, -1)$,$E(0, 4, -2)$,and $F(2, 3, 4)$ respectively.
By the mid-point formula:
For $AB$: $\frac{x_1+x_2}{2} = 1, \frac{y_1+y_2}{2} = 5, \frac{z_1+z_2}{2} = -1 \Rightarrow x_1+x_2 = 2, y_1+y_2 = 10, z_1+z_2 = -2$.
For $BC$: $\frac{x_2+x_3}{2} = 0, \frac{y_2+y_3}{2} = 4, \frac{z_2+z_3}{2} = -2 \Rightarrow x_2+x_3 = 0, y_2+y_3 = 8, z_2+z_3 = -4$.
For $CA$: $\frac{x_3+x_1}{2} = 2, \frac{y_3+y_1}{2} = 3, \frac{z_3+z_1}{2} = 4 \Rightarrow x_3+x_1 = 4, y_3+y_1 = 6, z_3+z_1 = 8$.
Adding these equations:
$2(x_1+x_2+x_3) = 2+0+4 = 6 \Rightarrow x_1+x_2+x_3 = 3$.
$2(y_1+y_2+y_3) = 10+8+6 = 24 \Rightarrow y_1+y_2+y_3 = 12$.
$2(z_1+z_2+z_3) = -2-4+8 = 2 \Rightarrow z_1+z_2+z_3 = 1$.
Now,to find $C(x_3, y_3, z_3)$,subtract the equations for $AB$ from the sums:
$x_3 = (x_1+x_2+x_3) - (x_1+x_2) = 3 - 2 = 1$.
$y_3 = (y_1+y_2+y_3) - (y_1+y_2) = 12 - 10 = 2$.
$z_3 = (z_1+z_2+z_3) - (z_1+z_2) = 1 - (-2) = 3$.
So,$C = (1, 2, 3)$.
The median from $C$ to $AB$ is the line segment $CD$,where $D$ is the mid-point of $AB$,$D(1, 5, -1)$.
Length $CD = \sqrt{(1-1)^2 + (5-2)^2 + (-1-3)^2} = \sqrt{0^2 + 3^2 + (-4)^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5$.

(B) The total number of possible outcomes when a die is rolled three times is $6^3 = 216$.
We need to select three distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$ such that they appear in strictly increasing order.
Any selection of $3$ distinct numbers from $6$ can be arranged in exactly one way in strictly increasing order.
The number of ways to choose $3$ distinct numbers from $6$ is given by $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Thus,the number of favorable outcomes is $20$.
The required probability is $\frac{20}{216} = \frac{5}{54}$.

(C) The total number of ways to choose two distinct numbers $a$ and $b$ from the set of integers ${1, 2, \dots, 39}$ is given by the combination formula ${}^{39}C_2$.
${}^{39}C_2 = \frac{39 \times 38}{2} = 39 \times 19 = 741$.
We need to find the pairs $(a, b)$ such that $7a - 9b = 0$,which implies $7a = 9b$.
Since $7$ and $9$ are coprime,$a$ must be a multiple of $9$ and $b$ must be a multiple of $7$.
Given $1 \le a, b \le 39$,the possible pairs $(a, b)$ are:
$(9, 7), (18, 14), (27, 21), (36, 28)$.
There are $4$ such favorable pairs.
Thus,the probability is $\frac{4}{741}$.

(B) The total number of four-digit numbers formed using the digits $\{0, 1, 2, 3, 4, 6\}$ without repetition is calculated as follows: The first digit cannot be $0$,so there are $5$ choices. The remaining three positions can be filled in $5 \times 4 \times 3 = 60$ ways. Thus,total numbers $= 5 \times 60 = 300$.
$A$ number is divisible by $4$ if its last two digits form a number divisible by $4$. The possible pairs for the last two digits are $\{04, 12, 16, 20, 24, 32, 36, 40, 60, 64\}$.
Case $1$: Pairs containing $0$ $(\{04, 20, 40, 60\})$: There are $4$ such pairs. For each,the remaining $2$ positions can be filled by the remaining $4$ digits in $4 \times 3 = 12$ ways. Total $= 4 \times 12 = 48$.
Case $2$: Pairs not containing $0$ $(\{12, 16, 24, 32, 36, 64\})$: There are $6$ such pairs. For each,the first digit cannot be $0$ or the two digits already used,leaving $3$ choices for the first digit and $3$ choices for the second digit. Total $= 6 \times (3 \times 3) = 6 \times 9 = 54$.
Total favourable outcomes $= 48 + 54 = 102$.
Required probability $= \frac{102}{300} = \frac{17}{50}$.

(A) We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\bar{A} \cap \bar{B})$.
By De Morgan's Law,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the addition rule of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values:
$P(A \cup B) = 0.58 + 0.32 - 0.28 = 0.62$.
Therefore,$P(\bar{A} \cap \bar{B}) = 1 - 0.62 = 0.38$.

(D) Let the probability of getting a number $i$ be $P(i)$.
Since $P(i) \propto i$,we have $P(i) = Ki$ for some constant $K$.
The sum of all probabilities must be $1$:
$\sum_{i=1}^{6} P(i) = K(1+2+3+4+5+6) = 21K = 1$.
Thus,$K = \frac{1}{21}$.
The probability of getting an odd number is $P(1) + P(3) + P(5)$.
$= K(1 + 3 + 5) = 9K$.
$= 9 \times \frac{1}{21} = \frac{9}{21} = \frac{3}{7}$.

(B) Let $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$,and $P(C) = \frac{1}{4}$ be the probabilities of students $A, B$,and $C$ solving the problem respectively.
Then,the probabilities of them not solving the problem are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3}$,and $P(\bar{C}) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that exactly one of them solves the problem is given by:
$P(\text{Exactly one}) = P(A)P(\bar{B})P(\bar{C}) + P(\bar{A})P(B)P(\bar{C}) + P(\bar{A})P(\bar{B})P(C)$
$= (\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}) + (\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4})$
$= \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}$.

(A) Total number of ways to draw $4$ cards from $52$ cards is $^{52}C_4$.
To get exactly two cards from the same suit and the remaining two cards from two different suits:
$1$. Select $1$ suit out of $4$ for the pair: $^4C_1$.
$2$. Select $2$ cards from the $13$ cards of that suit: $^{13}C_2$.
$3$. Select $2$ suits out of the remaining $3$ for the other two cards: $^3C_2$.
$4$. Select $1$ card from each of these $2$ chosen suits: $^{13}C_1 \times ^{13}C_1$.
Required probability = $\frac{^4C_1 \times ^{13}C_2 \times ^3C_2 \times ^{13}C_1 \times ^{13}C_1}{^{52}C_4}$
$= \frac{4 \times 78 \times 3 \times 13 \times 13}{270725} = \frac{12 \times 78 \times 169}{270725} = \frac{158184}{270725} = \frac{72 \times 169}{425 \times 49}$.

(A) cube has $8$ vertices. The total number of ways to choose $3$ vertices out of $8$ is given by $^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
An equilateral triangle is formed by choosing $3$ vertices such that the distance between any two of them is equal to the length of the face diagonal of the cube.
Each face of the cube has $2$ diagonals,and there are $6$ faces,but each diagonal is shared by $2$ faces. The total number of face diagonals is $12$.
However,an equilateral triangle is formed by selecting $3$ vertices such that each pair is connected by a face diagonal. There are exactly $8$ such triangles in a cube (each corresponding to a set of $3$ vertices that are mutually equidistant).
Therefore,the number of favorable outcomes is $8$.
The probability is $\frac{8}{56} = \frac{1}{7}$.

(B) Let the roots of the given equation $f(x) = x^4-2ax^3+4bx^2+8ax+16=0$ be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$.
The roots of the new equation are $p\alpha_1, p\alpha_2, p\alpha_3, p\alpha_4$.
Thus,the new equation is $f(\frac{x}{p}) = 0$.
Substituting $\frac{x}{p}$ into the equation:
$(\frac{x}{p})^4 - 2a(\frac{x}{p})^3 + 4b(\frac{x}{p})^2 + 8a(\frac{x}{p}) + 16 = 0$
Multiplying by $p^4$:
$x^4 - 2apx^3 + 4bp^2x^2 + 8ap^3x + 16p^4 = 0$
Since this is a reciprocal equation,the coefficient of $x^4$ must equal the constant term:
$1 = 16p^4$
$p^4 = \frac{1}{16}$
$p^2 = \frac{1}{4} \implies |p| = \frac{1}{2}$

(A) Let the three sections be $S_1, S_2, S_3$,each with $4$ questions. The candidate must select $5$ questions such that at least one is selected from each section. The possible distributions of questions across the three sections are $(1, 1, 3)$ or $(1, 2, 2)$ in any order.
Case $1$: Distribution $(1, 1, 3)$ in any order.
The number of ways to choose the sections is $\frac{3!}{2!} = 3$.
The number of ways to select the questions is $^4C_1 \times ^4C_1 \times ^4C_3 = 4 \times 4 \times 4 = 64$.
Total ways for this case $= 3 \times 64 = 192$.
Case $2$: Distribution $(1, 2, 2)$ in any order.
The number of ways to choose the sections is $\frac{3!}{2!} = 3$.
The number of ways to select the questions is $^4C_1 \times ^4C_2 \times ^4C_2 = 4 \times 6 \times 6 = 144$.
Total ways for this case $= 3 \times 144 = 432$.
Total number of ways $= 192 + 432 = 624$.

(B) Given the circle $x^2+y^2+8x-4y+c=0$,its center $C_1 = (-4, 2)$ and radius $r_1 = \sqrt{(-4)^2 + 2^2 - c} = \sqrt{20-c}$.
For the circle $x^2+y^2+2x+4y-11=0$,its center $C_2 = (-1, -2)$ and radius $r_2 = \sqrt{(-1)^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = 4$.
Since the circles touch externally,$C_1C_2 = r_1 + r_2$.
$C_1C_2 = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.
So,$5 = \sqrt{20-c} + 4$,which implies $\sqrt{20-c} = 1$,so $20-c = 1$,hence $c = 19$.
Now,the circle $x^2+y^2+8x-4y+19=0$ cuts $x^2+y^2-6x+8y+k=0$ orthogonally.
The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = 4, f_1 = -2, c_1 = 19$ and $g_2 = -3, f_2 = 4, c_2 = k$.
$2(4)(-3) + 2(-2)(4) = 19 + k$.
$-24 - 16 = 19 + k$.
$-40 = 19 + k$.
$k = -59$.