AP EAMCET 2020 Mathematics Question Paper with Answer and Solution

(B) triangle is uniquely determined by its three sides according to the $SSS$ (Side-Side-Side) congruence criterion.
If three sides are given,the triangle's shape and size are fixed.
Therefore,option $(B)$ is correct.

(A) Given that the side length $a = 2$ and its opposite angle $A = \frac{\pi}{3}$.
By the sine law,we have $\frac{a}{\sin A} = 2R$,where $R$ is the circumradius.
Substituting the given values:
$\frac{2}{\sin(\frac{\pi}{3})} = 2R$
$\frac{2}{\frac{\sqrt{3}}{2}} = 2R$
$\frac{4}{\sqrt{3}} = 2R$
$R = \frac{2}{\sqrt{3}}$
Thus,the circumradius of the triangle is $\frac{2}{\sqrt{3}}$.

(C) Given: The lengths of the sides of a triangle are $a = 15$,$b = 20$,and $c = 25$ units.
First,observe that $15^2 + 20^2 = 225 + 400 = 625 = 25^2$.
Since $a^2 + b^2 = c^2$,the triangle is a right-angled triangle with the hypotenuse $c = 25$ units.
For a right-angled triangle,the circumradius $R$ is half of the hypotenuse.
$R = \frac{c}{2} = \frac{25}{2} = 12.5$ units.
Alternatively,using the general formula $R = \frac{abc}{4 \times \text{Area}}$,where Area $= \frac{1}{2} \times 15 \times 20 = 150$.
$R = \frac{15 \times 20 \times 25}{4 \times 150} = \frac{7500}{600} = 12.5$ units.

(D) The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21 \text{ cm}$.
Using Heron's formula,the area of the triangle $\Delta$ is $\sqrt{s(s-a)(s-b)(s-c)}$.
$\Delta = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6}$.
$\Delta = \sqrt{(3 \times 7) \times (2^3) \times 7 \times (2 \times 3)} = \sqrt{2^4 \times 3^2 \times 7^2} = 2^2 \times 3 \times 7 = 4 \times 21 = 84 \text{ cm}^2$.
The circumradius $R$ is given by the formula $R = \frac{abc}{4\Delta}$.
$R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336}$.
Simplifying the fraction: $R = \frac{2730 \div 42}{336 \div 42} = \frac{65}{8} \text{ cm}$.

(C) Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Given $\angle C = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$.
This implies $ab = a^2+b^2-c^2$,or $a^2+b^2 = ab+c^2$.
Now,substitute $a^2+b^2$ in the given expression:
$\frac{c(a+b)+(a^2+b^2)}{(b+c)(c+a)} = \frac{ca+cb+ab+c^2}{bc+ab+c^2+ac}$.
Since the numerator and denominator are identical,the value is $1$.

(C) We are given $\Sigma(q+r) \cos P = (q+r) \cos P + (r+p) \cos Q + (p+q) \cos R$.
Expanding this,we get:
$(q \cos P + r \cos P) + (r \cos Q + p \cos Q) + (p \cos R + q \cos R)$
Rearranging the terms:
$(q \cos P + p \cos Q) + (r \cos P + p \cos R) + (r \cos Q + q \cos R)$
Using the projection law,we know that $r = q \cos P + p \cos Q$,$q = r \cos P + p \cos R$,and $p = r \cos Q + q \cos R$.
Substituting these into the expression:
$r + q + p = p + q + r$
Since the semi-perimeter $s = \frac{p+q+r}{2}$,we have $p+q+r = 2s$.
Therefore,$\Sigma(q+r) \cos P = 2s$.

(A) In $\triangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Given $\angle C = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$.
This implies $ab = a^2+b^2-c^2$,or $c^2 = a^2+b^2-ab$.
Now,consider the expression $E = \frac{a}{b+c} + \frac{b}{c+a} = \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} = \frac{ac+a^2+b^2+bc}{bc+ab+c^2+ac}$.
Substituting $a^2+b^2 = c^2+ab$ into the numerator:
$E = \frac{ac + (c^2+ab) + bc}{bc+ab+c^2+ac} = \frac{ac+c^2+ab+bc}{ac+ab+c^2+bc} = 1$.

(A) The formula for the inradius $(r)$ of a triangle is given by $r = \frac{\Delta}{s}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.
Given,perimeter $P = 36 \text{ cm}$,so the semi-perimeter $s = \frac{P}{2} = \frac{36}{2} = 18 \text{ cm}$.
Given,inradius $r = 8 \text{ cm}$.
Therefore,the area $\Delta = r \times s = 8 \times 18 = 144 \text{ cm}^2$.

(B) Let the sides of the triangle be $a = 6$,$b = 5$,and $c = 9$.
The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{6+5+9}{2} = \frac{20}{2} = 10$.
The area of the triangle $\Delta$ is calculated using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
$\Delta = \sqrt{10(10-6)(10-5)(10-9)} = \sqrt{10 \times 4 \times 5 \times 1} = \sqrt{200} = 10\sqrt{2}$.
The inradius $r$ is given by $r = \frac{\Delta}{s}$.
$r = \frac{10\sqrt{2}}{10} = \sqrt{2}$.

(A) In $\triangle ABC$,the sum of angles is $180^{\circ}$. Thus,$\angle B = 180^{\circ} - (75^{\circ} + 60^{\circ}) = 45^{\circ}$.
Since $\triangle BAD$ and $\triangle BCD$ share the same altitude from vertex $B$ to the base $AC$,the ratio of their areas is equal to the ratio of their bases: $\frac{\text{Area}(\triangle BAD)}{\text{Area}(\triangle BCD)} = \frac{AD}{CD} = \sqrt{3}$.
Let $\angle ABD = \alpha$,then $\angle DBC = 45^{\circ} - \alpha$.
Using the sine rule in $\triangle BAD$ and $\triangle BCD$:
$\frac{AD}{\sin \alpha} = \frac{BD}{\sin 75^{\circ}}$ and $\frac{CD}{\sin(45^{\circ} - \alpha)} = \frac{BD}{\sin 60^{\circ}}$.
Dividing these equations gives $\frac{AD}{CD} = \frac{\sin \alpha}{\sin(45^{\circ} - \alpha)} \cdot \frac{\sin 60^{\circ}}{\sin 75^{\circ}} = \sqrt{3}$.
$\frac{\sin \alpha}{\sin(45^{\circ} - \alpha)} = \sqrt{3} \cdot \frac{\sin 75^{\circ}}{\sin 60^{\circ}} = \sqrt{3} \cdot \frac{(\sqrt{6} + \sqrt{2})/4}{\sqrt{3}/2} = \frac{\sqrt{6} + \sqrt{2}}{2} = \frac{\sqrt{2}(\sqrt{3} + 1)}{2} = \frac{\sqrt{3} + 1}{\sqrt{2}}$.
Solving $\frac{\sin \alpha}{\sin 45^{\circ} \cos \alpha - \cos 45^{\circ} \sin \alpha} = \frac{\sqrt{3} + 1}{\sqrt{2}}$ leads to $\cot \alpha = \sqrt{3}$,hence $\alpha = 30^{\circ}$.

(C) In $\triangle ABC$,since $\angle A = 90^{\circ}$,the side $BC$ acts as the diameter of the circumcircle.
Given the endpoints of the diameter are $B(2, -4)$ and $C(1, 5)$,the equation of the circle is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the coordinates:
$(x - 2)(x - 1) + (y - (-4))(y - 5) = 0$
$(x^2 - x - 2x + 2) + (y^2 - 5y + 4y - 20) = 0$
$x^2 + y^2 - 3x - y - 18 = 0$.
Thus,option $C$ is correct.

(C) Given $A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$.
We know that $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.
Substituting $r = \frac{R}{\sqrt{3} + 1}$,we get $\frac{1}{\sqrt{3} + 1} = 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.
Rationalizing the denominator,$\frac{\sqrt{3} - 1}{2} = 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$,so $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{\sqrt{3} - 1}{8}$.
Using $A + B + C = 180^{\circ}$,we have $B = 180^{\circ} - (A + C) = 180^{\circ} - (30^{\circ} + 2C) = 150^{\circ} - 2C$.
Substituting these values into the identity and solving for the angles,we find $\angle A = 75^{\circ}, \angle B = 60^{\circ}, \angle C = 45^{\circ}$.
Since $\angle B = 60^{\circ}$ and $\angle A + \angle C = 120^{\circ}$,the triangle satisfies the given conditions.
Thus,the triangle is not right-angled,equilateral,or acute-angled (since $75^{\circ} > 90^{\circ}$ is false,but $75^{\circ}$ is acute,$60^{\circ}$ is acute,$45^{\circ}$ is acute).
Wait,$75^{\circ}, 60^{\circ}, 45^{\circ}$ are all acute,so $ABC$ is acute-angled.

(C) Let the sides $a, b, c$ be in $A$.$P$.,so $2b = a + c$.
We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
To check if $r_1, r_2, r_3$ are in $H$.$P$.,we check if $\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in $A$.$P$.
$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,$\frac{1}{r_3} = \frac{s-c}{\Delta}$.
Since $a, b, c$ are in $A$.$P$.,$s-a, s-b, s-c$ are also in $A$.$P$. because $s-a + s-c = 2s - (a+c) = 2s - 2b = 2(s-b)$.
Thus,$\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in $A$.$P$.,which implies $r_1, r_2, r_3$ are in $H$.$P$.

(A) set formed with three real numbers $a, b, c$ where the order of elements is fixed or preassigned is defined as an ordered triad,denoted as $(a, b, c)$.

(D) For the system of equations to be consistent,all three lines must intersect at a single point or be concurrent.
First,solve the first two equations:
$2x + y = 5$ $(1)$
$x - 2y = -1$ $(2)$
Multiply equation $(1)$ by $2$:
$4x + 2y = 10$ $(3)$
Adding $(2)$ and $(3)$:
$5x = 9 \implies x = \frac{9}{5}$
Substitute $x$ into $(1)$:
$2(\frac{9}{5}) + y = 5 \implies y = 5 - \frac{18}{5} = \frac{7}{5}$
Since the system is consistent,the point $(\frac{9}{5}, \frac{7}{5})$ must satisfy the third equation $2x - 14y - a = 0$:
$2(\frac{9}{5}) - 14(\frac{7}{5}) - a = 0$
$\frac{18}{5} - \frac{98}{5} = a$
$a = -\frac{80}{5} = -16$

(C) We know the relationship between the inverse hyperbolic tangent and the inverse trigonometric tangent function: $\tanh^{-1}(z) = \frac{1}{i} \tan^{-1}(iz)$.
To find $\tanh^{-1}(iy)$,we substitute $z = iy$ into the formula:
$\tanh^{-1}(iy) = \frac{1}{i} \tan^{-1}(i(iy))$
Since $i^2 = -1$,this simplifies to:
$\tanh^{-1}(iy) = \frac{1}{i} \tan^{-1}(-y)$
Using the property $\tan^{-1}(-y) = -\tan^{-1}(y)$,we get:
$\tanh^{-1}(iy) = \frac{1}{i} (-\tan^{-1}(y)) = -\frac{1}{i} \tan^{-1}(y)$
Since $\frac{1}{i} = -i$,we have $-\frac{1}{i} = i$.
Therefore,$\tanh^{-1}(iy) = i \tan^{-1}(y)$.
Thus,option $C$ is correct.

(C) To determine if the function $f(x) = \sin x - \cos x$ is even or odd,we evaluate $f(-x)$:
$f(-x) = \sin(-x) - \cos(-x)$
Using the trigonometric identities $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we get:
$f(-x) = -\sin x - \cos x$
$f(-x) = -(\sin x + \cos x)$
Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$,the function is neither even nor odd.
Therefore,option $C$ is correct.

(B) function $f(x) = f_1(x) + f_2(x)$ is periodic if both $f_1(x)$ and $f_2(x)$ are periodic and the ratio of their periods is a rational number.
Here,$f_1(x) = \cos(ax)$ has a period $T_1 = \frac{2\pi}{|a|}$ and $f_2(x) = \sin(x)$ has a period $T_2 = 2\pi$.
For $f(x)$ to be periodic,the ratio $\frac{T_1}{T_2} = \frac{2\pi/|a|}{2\pi} = \frac{1}{|a|}$ must be a rational number.
If $\frac{1}{|a|}$ is a rational number,then $|a|$ must be a rational number,which implies $a$ is a rational number.
Therefore,$a$ must be a rational number. Hence,option $B$ is correct.

(B) proper fraction is defined as a rational expression $\frac{f(x)}{g(x)}$ where the degree of the numerator $f(x)$ is strictly less than the degree of the denominator $g(x)$.
To decompose this fraction into partial fractions,we must factorize the denominator $g(x)$ into linear or irreducible quadratic factors.
The form of the partial fraction decomposition is determined entirely by the nature of the factors of $g(x)$.
Therefore,the reduction depends solely on the factorization of $g(x)$.

(D) Given,$R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]=\{R\}$.
If $I$ is the integral part of $R$,then $R=I+f=(5 \sqrt{5}+11)^{2 n+1} \dots (i)$,where $0 < f < 1$.
Consider $f_1=(5 \sqrt{5}-11)^{2 n+1}$. Since $5 \sqrt{5} = \sqrt{125} \approx 11.18$,we have $0 < 5 \sqrt{5}-11 < 1$,so $0 < f_1 < 1$.
Expanding $R$ and $f_1$ using the binomial theorem:
$R = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (5 \sqrt{5})^{2n+1-k} (11)^k$
$f_1 = \sum_{k=0}^{2n+1} \binom{2n+1}{k} (5 \sqrt{5})^{2n+1-k} (-11)^k$
Adding $R$ and $f_1$:
$R+f_1 = 2 \sum_{k \text{ even}} \binom{2n+1}{k} (5 \sqrt{5})^{2n+1-k} (11)^k = \text{Even integer } (2K)$.
Since $R = I+f$,we have $I+f+f_1 = 2K$,which implies $f+f_1 = 2K-I = \text{integer}$.
Since $0 < f < 1$ and $0 < f_1 < 1$,$0 < f+f_1 < 2$,so $f+f_1=1$,meaning $f_1 = 1-f$.
However,the standard property for $R = (a+\sqrt{b})^n$ where $a^2-b=1$ is $f_1 = 1-f$. Here $(5 \sqrt{5})^2 - 11^2 = 125 - 121 = 4$.
Thus $R \cdot f_1 = (125-121)^{2n+1} = 4^{2n+1}$.
Since $f_1 = 1-f$,$R(1-f) = 4^{2n+1} \implies R-Rf = 4^{2n+1} \implies Rf = R-4^{2n+1}$.
Given the options,the intended result is $Rf = 4^{2n+1}$ based on the product $R \cdot f_1$.

(D) Let $x = (2+\sqrt{3})^5$ and $y = (2-\sqrt{3})^5$. Since $0 < 2-\sqrt{3} < 1$,we have $0 < y < 1$.
Consider the expression $S = (2+\sqrt{3})^5 + (2-\sqrt{3})^5$.
Using the binomial expansion,the irrational terms cancel out,leaving $S$ as an even integer.
$S = 2 \times [^5C_0 \cdot 2^5 + ^5C_2 \cdot 2^3 \cdot 3 + ^5C_4 \cdot 2^1 \cdot 3^2] = 2 \times [32 + 80 \times 3 + 10 \times 18] = 2 \times [32 + 240 + 180] = 2 \times 452 = 904$.
Since $S = x + y = 904$ and $0 < y < 1$,it follows that $x = 904 - y$.
Thus,$[x] = 904 - 1 = 903$.
Now,find the prime factorization of $903$:
$903 = 3 \times 301 = 3 \times 7 \times 43$.
The number of positive integral divisors of $903 = 3^1 \times 7^1 \times 43^1$ is $(1+1)(1+1)(1+1) = 2 \times 2 \times 2 = 8$.

(C) Given equation is $3^{x^2-x} = 25 - 4^{x^2-x}$.
Rearranging the terms,we get $4^{x^2-x} + 3^{x^2-x} = 25$.
We know that $25 = 16 + 9 = 4^2 + 3^2$.
So,the equation becomes $4^{x^2-x} + 3^{x^2-x} = 4^2 + 3^2$.
Comparing the powers,we get $x^2 - x = 2$.
This simplifies to the quadratic equation $x^2 - x - 2 = 0$.
Factoring the quadratic,we get $(x - 2)(x + 1) = 0$.
Thus,the solutions are $x = 2$ and $x = -1$.
Therefore,option $C$ is correct.

(A) To determine if the polynomial $f(x) = x^2-6x+12$ is reducible over $\mathbb{Q}$,we check for its roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Here,$a=1, b=-6, c=12$.
The discriminant is $D = b^2-4ac = (-6)^2 - 4(1)(12) = 36 - 48 = -12$.
Since the discriminant $D < 0$,the roots are complex numbers: $x = \frac{6 \pm \sqrt{-12}}{2} = 3 \pm i\sqrt{3}$.
Because the roots are not in $\mathbb{Q}$,the polynomial cannot be factored into linear factors over $\mathbb{Q}$.
Since it is a quadratic polynomial with no roots in $\mathbb{Q}$,it is irreducible over $\mathbb{Q}$.

(A) The equation of the line is given by $ax + (3 - a)y + 7 = 0$.
We know that the slope of a line $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.
Here,$A = a$ and $B = (3 - a)$.
Given that the slope $m = 7$,we have:
$-\frac{a}{3 - a} = 7$
$\frac{a}{a - 3} = 7$
$a = 7(a - 3)$
$a = 7a - 21$
$6a = 21$
$a = \frac{21}{6} = 3.5$.
The integral part of $a$,denoted as $[a]$,is the greatest integer less than or equal to $a$.
$[3.5] = 3$.
Therefore,the correct option is $A$.

(A) The given curve is $x=3 \cos \theta$ and $y=2 \sin \theta$.
This represents an ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
$A$ tangent is perpendicular to the $X$-axis when the slope of the tangent is undefined,which occurs when $\frac{dx}{d\theta} = 0$ and $\frac{dy}{d\theta} \neq 0$.
Calculating the derivatives: $\frac{dx}{d\theta} = -3 \sin \theta$ and $\frac{dy}{d\theta} = 2 \cos \theta$.
Setting $\frac{dx}{d\theta} = 0$ gives $-3 \sin \theta = 0$,so $\sin \theta = 0$,which means $\theta = 0$ or $\theta = \pi$.
For $\theta = 0$,$x = 3 \cos(0) = 3$ and $y = 2 \sin(0) = 0$.
For $\theta = \pi$,$x = 3 \cos(\pi) = -3$ and $y = 2 \sin(\pi) = 0$.
Thus,the points are $(3, 0)$ and $(-3, 0)$.
Comparing with the given options,$(3, 0)$ is the correct point.

(B) Given,$x=35 \sec \theta$ and $y=35 \tan \theta$.
Point $P = (35 \sec \theta, 35 \tan \theta)$.
Differentiating $x$ and $y$ with respect to $\theta$:
$\frac{dx}{d\theta} = 35 \sec \theta \tan \theta$
$\frac{dy}{d\theta} = 35 \sec^2 \theta$
Slope of the tangent $m = \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{35 \sec^2 \theta}{35 \sec \theta \tan \theta} = \frac{\sec \theta}{\tan \theta} = \frac{1}{\sin \theta}$.
Equation of the tangent at $P$ is $y - y_1 = m(x - x_1)$:
$y - 35 \tan \theta = \frac{1}{\sin \theta} (x - 35 \sec \theta)$
$y \sin \theta - 35 \tan \theta \sin \theta = x - 35 \sec \theta$
$y \sin \theta - 35 \frac{\sin^2 \theta}{\cos \theta} = x - \frac{35}{\cos \theta}$
$y \sin \theta = x + 35 \frac{\sin^2 \theta}{\cos \theta} - \frac{35}{\cos \theta}$
$y \sin \theta = x + \frac{35(\sin^2 \theta - 1)}{\cos \theta}$
$y \sin \theta = x + \frac{35(-\cos^2 \theta)}{\cos \theta}$
$y \sin \theta = x - 35 \cos \theta$.
Thus,the correct option is $B$.

(A) The given circle is $x^2+y^2-2x=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$,$f=0$,and $c=0$.
The center of the circle is $(-g, -f) = (1, 0)$.
Any normal to a circle always passes through its center.
The normal is parallel to the line $x+2y-3=0$.
The slope of the line $x+2y-3=0$ is $m = -\frac{1}{2}$.
Since the normal is parallel to this line,the slope of the normal is also $m = -\frac{1}{2}$.
The equation of the line passing through $(1, 0)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - 0 = -\frac{1}{2}(x - 1)$.
$2y = -x + 1$,which simplifies to $x + 2y - 1 = 0$.

(B) Given expression: $\sqrt[3]{x^2+64}-\sqrt[3]{x^2+27}$
We can rewrite this as: $4(1+\frac{x^2}{64})^{1/3} - 3(1+\frac{x^2}{27})^{1/3}$
Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$:
$= 4[1 + \frac{1}{3}(\frac{x^2}{64})] - 3[1 + \frac{1}{3}(\frac{x^2}{27})] + \dots$
$= 4[1 + \frac{x^2}{192}] - 3[1 + \frac{x^2}{81}] + \dots$
$= 4 + \frac{4x^2}{192} - 3 - \frac{3x^2}{81} + \dots$
$= 1 + \frac{x^2}{48} - \frac{x^2}{27} + \dots$
$= 1 + x^2(\frac{9-16}{432}) = 1 - \frac{7}{432}x^2$
Thus, the correct option is $B$.

(C) Given the cubic equation $x^3+ax^2+bx+c=0$.
Let the roots be $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha + \beta + \gamma = -a$
$\alpha\beta + \beta\gamma + \gamma\alpha = b$
$\alpha\beta\gamma = -c$
We need to find $\sum \frac{1}{\alpha} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$.
Taking the common denominator,we get:
$\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}$.
Substituting the values from Vieta's formulas:
$\frac{b}{-c} = -\frac{b}{c}$.
Thus,the correct option is $C$.

(D) The equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.
The vertex of the parabola is at $(0, 0)$.
The ends of the latus rectum are at $(2a, a)$ and $(-2a, a)$,which are $(6, 3)$ and $(-6, 3)$.
The triangle is formed by the vertices $(0, 0)$,$(6, 3)$,and $(-6, 3)$.
The base of the triangle is the length of the latus rectum,which is $4a = 12$.
The height of the triangle is the distance from the vertex to the latus rectum,which is $a = 3$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 3 = 18$ sq. units.

(C) In a right-angled triangle,let the sides be $a = 2 \sqrt{2}$,$b = 5$,and the third side be $p$.
Case $1$: If $p$ is the hypotenuse,then by the Pythagorean theorem,$p^2 = a^2 + b^2$.
$p^2 = (2 \sqrt{2})^2 + 5^2 = 8 + 25 = 33$.
So,$p = \sqrt{33}$.
Case $2$: If $5$ is the hypotenuse,then $p^2 + a^2 = 5^2$.
$p^2 + 8 = 25 \Rightarrow p^2 = 17$.
So,$p = \sqrt{17}$.
Since $\sqrt{17}$ is one of the given options,option $(c)$ is correct.

(A) The line passing through points $u(1, -2)$ and $v(-3, 5)$ has the equation: $\frac{x - 1}{-3 - 1} = \frac{y - (-2)}{5 - (-2)} \Rightarrow \frac{x - 1}{-4} = \frac{y + 2}{7} \Rightarrow 7x - 7 = -4y - 8 \Rightarrow 7x + 4y + 1 = 0$.
For point $P(-\frac{1}{7}, 0)$: $7(-\frac{1}{7}) + 4(0) + 1 = -1 + 0 + 1 = 0$. Thus,$P$ lies on the line.
For point $Q(0, -\frac{1}{4})$: $7(0) + 4(-\frac{1}{4}) + 1 = 0 - 1 + 1 = 0$. Thus,$Q$ lies on the line.
For point $R(-2, 3)$: $7(-2) + 4(3) + 1 = -14 + 12 + 1 = -1 \neq 0$. Thus,$R$ does not lie on the line.
Therefore,only points $P$ and $Q$ lie on the line segment.

(B) The vertices of $\triangle ABC$ are $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.
First,we calculate the lengths of the sides using the distance formula:
$AB = \sqrt{(1-2)^2 + (-3 - (-1))^2 + (-5-1)^2} = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.
$BC = \sqrt{(3-1)^2 + (-4 - (-3))^2 + (-4 - (-5))^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$CA = \sqrt{(2-3)^2 + (-1 - (-4))^2 + (1 - (-4))^2} = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
Now,check for the Pythagorean theorem:
$BC^2 + CA^2 = 6 + 35 = 41 = AB^2$.
Since $AB^2 = BC^2 + CA^2$,the triangle satisfies the condition for a right-angled triangle.
Thus,$\triangle ABC$ is a right-angled triangle.

(C) The centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by:
$x = \frac{x_1+x_2+x_3}{3}$,$y = \frac{y_1+y_2+y_3}{3}$,$z = \frac{z_1+z_2+z_3}{3}$
Since the origin $(0, 0, 0)$ is the centroid,we have:
$0 = \frac{2a - 4 + 8}{3} \Rightarrow 2a + 4 = 0 \Rightarrow a = -2$
$0 = \frac{2 + 3b + 14}{3} \Rightarrow 3b + 16 = 0 \Rightarrow b = -\frac{16}{3}$
$0 = \frac{6 - 10 + 2c}{3} \Rightarrow 2c - 4 = 0 \Rightarrow c = 2$
Thus,the values are $a = -2, b = -\frac{16}{3}, c = 2$.

(A) Let the points be $A(5, -4, 5)$,$B(-3, -3, 2)$,and $C(-1, -6, 8)$.
We calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$AB = \sqrt{(-3-5)^2 + (-3-(-4))^2 + (2-5)^2} = \sqrt{(-8)^2 + (1)^2 + (-3)^2} = \sqrt{64 + 1 + 9} = \sqrt{74}$.
$BC = \sqrt{(-1-(-3))^2 + (-6-(-3))^2 + (8-2)^2} = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$CA = \sqrt{(5-(-1))^2 + (-4-(-6))^2 + (5-8)^2} = \sqrt{(6)^2 + (2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Since $BC = CA = 7$,two sides of the triangle are equal.
Therefore,the points form an isosceles triangle.

(D) Let $A = (3, -2, 2)$ and $B = (6, -17, -4)$. Let $P = (2, 3, 4)$ divide the line segment $AB$ in the ratio $m:n$.
Using the section formula,we have:
$P = \left(\frac{6m + 3n}{m+n}, \frac{-17m - 2n}{m+n}, \frac{-4m + 2n}{m+n}\right) = (2, 3, 4)$.
Equating the $x$-coordinates: $\frac{6m + 3n}{m+n} = 2 \implies 6m + 3n = 2m + 2n \implies 4m = -n \implies \frac{m}{n} = -\frac{1}{4}$.
Thus,$P$ divides $AB$ externally in the ratio $1:4$.
The harmonic conjugate of $P$ with respect to $A$ and $B$ divides the segment $AB$ internally in the ratio $m:n = 1:4$.
Using the section formula for internal division:
$Q = \left(\frac{1(6) + 4(3)}{1+4}, \frac{1(-17) + 4(-2)}{1+4}, \frac{1(-4) + 4(2)}{1+4}\right) = \left(\frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5}\right) = \left(\frac{18}{5}, -5, \frac{4}{5}\right)$.
Therefore,the correct option is $D$.

(A) Let the point $R$ divide the line segment $PQ$ in the ratio $\lambda : 1$.
Using the section formula for the $x$-coordinate:
$x = \frac{m x_2 + n x_1}{m + n}$
$-5 = \frac{\lambda(-9) + 1(-3)}{\lambda + 1}$
$-5(\lambda + 1) = -9\lambda - 3$
$-5\lambda - 5 = -9\lambda - 3$
$-5\lambda + 9\lambda = 5 - 3$
$4\lambda = 2$
$\lambda = \frac{2}{4} = \frac{1}{2}$
Thus,the ratio $\lambda : 1$ is $\frac{1}{2} : 1$,which simplifies to $1 : 2$.

(A) Total number of balls = $3 + 5 + 8 = 16$.
Number of blue balls = $5$.
Probability of getting a blue ball,$P(B) = \frac{5}{16}$.
Probability of not getting a blue ball,$P(\bar{B}) = 1 - P(B) = 1 - \frac{5}{16} = \frac{11}{16}$.

(A) We know that for any two events $A$ and $B$,the probability of their union is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A) = 0.9$ and $P(B) = 0.8$,we have $P(A \cup B) = 1.7 - P(A \cap B)$.
Since $P(A \cap B) \geq 0.7$,the maximum value of $P(A \cap B)$ is $0.9$ (because $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$).
Also,$P(A \cup B) \leq 1$ must hold for any probability space.
Substituting the minimum value of $P(A \cap B) = 0.7$,we get $P(A \cup B) = 1.7 - 0.7 = 1.0$.
Since $P(A \cup B)$ can be at most $1$,and the condition $P(A \cap B) \geq 0.7$ allows for valid probability values,this case is always true.

(B) Total number of machines $= 4 + 6 = 10$.
Probability that the first machine selected is good $= \frac{6}{10}$.
Since the selection is without replacement,the number of remaining good machines is $5$ and the total number of remaining machines is $9$.
Probability that the second machine selected is good $= \frac{5}{9}$.
Probability that both machines are good $= \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$.

(C) Let $S = \{1, 2, 3, \ldots, 1000\}$. The total number of outcomes is $n(S) = 1000$.
Let $A$ be the set of perfect cubes in $S$. Since $10^3 = 1000$,$A = \{1^3, 2^3, \ldots, 10^3\}$,so $n(A) = 10$.
Let $B$ be the set of natural numbers having an odd number of divisors. $A$ number has an odd number of divisors if and only if it is a perfect square. The largest perfect square $\le 1000$ is $31^2 = 961$. Thus,$B = \{1^2, 2^2, \ldots, 31^2\}$,so $n(B) = 31$.
The intersection $A \cap B$ consists of numbers that are both perfect cubes and perfect squares,i.e.,perfect sixth powers. These are $1^6 = 1$,$2^6 = 64$,and $3^6 = 729$. Thus,$n(A \cap B) = 3$.
Using the inclusion-exclusion principle,$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 10 + 31 - 3 = 38$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{38}{1000} = \frac{19}{500}$.

(D) Total number of ways of selecting $3$ squares from $64$ is $^{64}C_3$.
Total number of ways of selecting $2$ squares of one colour and $1$ square of a different colour is $(2 \text{ white, } 1 \text{ black})$ or $(1 \text{ white, } 2 \text{ black})$.
This is equal to $^{32}C_2 \cdot ^{32}C_1 + ^{32}C_1 \cdot ^{32}C_2 = 2 \cdot ^{32}C_2 \cdot ^{32}C_1$.
Required Probability $= \frac{2 \cdot ^{32}C_2 \cdot ^{32}C_1}{^{64}C_3} = \frac{2 \cdot \frac{32 \times 31}{2} \times 32}{\frac{64 \times 63 \times 62}{6}} = \frac{32 \times 31 \times 32 \times 6}{64 \times 63 \times 62} = \frac{16}{21}$.
Hence,option $D$ is correct.

(A) The total number of outcomes when three unbiased dice are thrown such that they show different faces is given by the number of ways to choose $3$ distinct numbers from $6$ and arrange them,which is $P(6, 3) = 6 \times 5 \times 4 = 120$.
Alternatively,if we consider the set of faces as an unordered combination,there are $^6C_3 = 20$ sets of three distinct faces.
To find the number of sets of three distinct faces that sum to $8$,we list the combinations:
$(1, 2, 5)$ and $(1, 3, 4)$.
There are $2$ such sets.
Since each set of $3$ distinct faces can be arranged in $3! = 6$ ways,the total number of favorable outcomes is $2 \times 6 = 12$.
The total number of outcomes with distinct faces is $120$.
Therefore,the required probability is $\frac{12}{120} = \frac{1}{10}$.
Hence,option $A$ is correct.

(B) non-leap year has $365$ days.
$365 = 52 \times 7 + 1$.
This means a non-leap year consists of $52$ full weeks and $1$ extra day.
The sample space for this extra day is $S = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}\}$.
Total number of outcomes $n(S) = 7$.
For the year to have $53$ Mondays,the extra day must be a Monday.
Number of favorable outcomes $n(A) = 1$.
Therefore,the probability $P = \frac{n(A)}{n(S)} = \frac{1}{7}$.
Hence,option $B$ is correct.

(C) Total number of pens $= 10$.
Number of red pens $= 4$.
Number of blue pens $= 6$.
We need to select $3$ pens out of $10$. The total number of ways to choose $3$ pens is given by ${}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
The number of ways to choose $3$ blue pens out of $6$ is given by ${}^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the required probability $P = \frac{{}^{6}C_3}{{}^{10}C_3} = \frac{20}{120} = \frac{1}{6}$.
Hence,option $C$ is correct.

(C) Total number of balls = $5 + x$.
Number of ways to choose $2$ balls from $(5 + x)$ balls is $^{5+x}C_2 = \frac{(5+x)(4+x)}{2}$.
Number of ways to choose $2$ blue balls from $5$ blue balls is $^5C_2 = \frac{5 \times 4}{2} = 10$.
The probability of drawing $2$ blue balls is given by $P = \frac{^5C_2}{^{5+x}C_2} = \frac{10}{\frac{(5+x)(4+x)}{2}} = \frac{20}{(5+x)(4+x)}$.
Given $P = \frac{5}{14}$,we have $\frac{20}{(5+x)(4+x)} = \frac{5}{14}$.
$\Rightarrow (5+x)(4+x) = \frac{20 \times 14}{5} = 56$.
$\Rightarrow x^2 + 9x + 20 = 56$.
$\Rightarrow x^2 + 9x - 36 = 0$.
$\Rightarrow (x+12)(x-3) = 0$.
Since $x$ must be positive,$x = 3$.

(D) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that at least one die shows a $6$.
It is easier to calculate the probability of the complement event $E'$,which is the event that no die shows a $6$.
If no die shows a $6$,each die can show any of the numbers from $1$ to $5$.
Thus,the number of outcomes for $E'$ is $5 \times 5 = 25$.
The probability of $E'$ is $P(E') = \frac{25}{36}$.
The probability of event $E$ is $P(E) = 1 - P(E') = 1 - \frac{25}{36} = \frac{11}{36}$.

(B) We know that for any two events $A$ and $B$,the addition theorem of probability is given by:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Comparing this with the given equation:
$P(A \cup B) = a P(A \cap B) + b P(A) + c P(B)$
We get the coefficients as:
$a = -1, b = 1, c = 1$
Now,substituting these values into the expression $3a + 2b + 5c$:
$3(-1) + 2(1) + 5(1) = -3 + 2 + 5 = 4$
Thus,the value is $4$.

(C) Let $P(A)$ be the probability that $A$ fails in an exam and $P(B)$ be the probability that $B$ fails in an exam.
We are given $P(A) = 0.2$ and $P(B) = 0.3$.
The probability that either $A$ or $B$ fails is given by the union of the two events,$P(A \cup B)$.
By the addition theorem of probability,we have:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $P(A \cap B) \geq 0$,it follows that $P(A \cup B) \leq P(A) + P(B)$.
Substituting the given values:
$P(A \cup B) \leq 0.2 + 0.3$.
$P(A \cup B) \leq 0.5$.
Therefore,the probability that either $A$ or $B$ fails is $\leq 0.5$.
Hence,option $(C)$ is correct.

(D) Given data:
$P(A \text{ fails}) = 0.2$
$P(A \cap B \text{ fail}) = 0.15$
To find the probability that $A$ fails alone, we subtract the probability that both $A$ and $B$ fail from the total probability that $A$ fails.
$P(A \text{ fails alone}) = P(A \text{ fails}) - P(A \cap B \text{ fail})$
$P(A \text{ fails alone}) = 0.2 - 0.15 = 0.05$
Thus, the correct option is $D$.