AP EAMCET 2024 Mathematics Question Paper with Answer and Solution

(D) The given curves are $x^2+y^2-x-y=0$ $(i)$ and $x^2+y^2-2y=0$ $(ii)$.
To find the intersection points,subtract $(i)$ from $(ii)$: $(x^2+y^2-2y) - (x^2+y^2-x-y) = 0 \Rightarrow x-y=0 \Rightarrow x=y$.
Substitute $x=y$ into $(ii)$: $x^2+x^2=2x \Rightarrow 2x^2-2x=0 \Rightarrow 2x(x-1)=0$. Thus,the intersection points are $(0,0)$ and $(1,1)$.
For curve $(i)$,differentiating w.r.t. $x$: $2x+2y\frac{dy}{dx}=1+\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1-2x}{2y-1} = m_1$.
For curve $(ii)$,differentiating w.r.t. $x$: $2x+2y\frac{dy}{dx}=2\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{1-y} = m_2$.
At $(1,1)$,$m_1 = \frac{1-2}{2-1} = -1$ and $m_2 = \frac{1}{1-1}$ (undefined,vertical tangent).
Since one tangent is vertical,the angle $\theta$ is given by $|\tan \theta| = |\frac{1}{m_1}| = |\frac{1}{-1}| = 1$.
Therefore,$\theta = \frac{\pi}{4}$.

(D) Given curves are $2x^2 + ky^2 = 30$ ...$(i)$ and $3y^2 = 28x$ ...(ii).
Differentiating $(i)$ with respect to $x$: $4x + 2ky \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-2x}{ky} = m_1$.
Differentiating (ii) with respect to $x$: $6y \frac{dy}{dx} = 28 \Rightarrow \frac{dy}{dx} = \frac{14}{3y} = m_2$.
Since the curves cut orthogonally,$m_1 m_2 = -1$.
$\left( \frac{-2x}{ky} \right) \left( \frac{14}{3y} \right) = -1 \Rightarrow \frac{28x}{3ky^2} = 1$.
From (ii),$3y^2 = 28x$,so substitute this into the equation:
$\frac{28x}{k(28x)} = 1 \Rightarrow \frac{1}{k} = 1 \Rightarrow k = 1$.

(C) Given $f(x) = \sin x + \frac{1-\tan^2 x}{1+\tan^2 x}$.
Using the identity $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we have $f(x) = \sin x + \cos 2x$.
Since $\cos 2x = 1 - 2\sin^2 x$,we can write $f(x) = \sin x + 1 - 2\sin^2 x$.
Let $t = \sin x$,where $t \in [-1, 1]$. Then $g(t) = -2t^2 + t + 1$.
To find the maximum,we find the derivative $g'(t) = -4t + 1$.
Setting $g'(t) = 0$,we get $t = \frac{1}{4}$.
Since $g''(t) = -4 < 0$,$t = \frac{1}{4}$ is a point of local maxima.
We need to find the number of values of $x \in [0, 2\pi]$ such that $\sin x = \frac{1}{4}$.
Since $\frac{1}{4} > 0$,$\sin x = \frac{1}{4}$ has two solutions in the interval $[0, 2\pi]$ (one in the first quadrant and one in the second quadrant).
Thus,there are $2$ such values of $x$.

(A) Given that $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$.
Multiplying both sides by $(x^4+1)$,we get $(A x+B)(x^2-\sqrt{2} x+1)+(C x+D)(x^2+\sqrt{2} x+1)=1$.
Comparing coefficients:
For $x^3$: $A+C=0 \Rightarrow C=-A$.
For $x^0$ (constant term): $B+D=1$.
For $x^2$: $B-\sqrt{2} A+D+\sqrt{2} C=0 \Rightarrow (B+D)-\sqrt{2}(A-C)=0$.
Since $B+D=1$ and $C=-A$,we have $1-\sqrt{2}(2A)=0 \Rightarrow A=\frac{1}{2\sqrt{2}}$ and $C=-\frac{1}{2\sqrt{2}}$.
For $x^1$: $A-\sqrt{2} B+C+\sqrt{2} D=0 \Rightarrow (A+C)+\sqrt{2}(D-B)=0$.
Since $A+C=0$,we have $\sqrt{2}(D-B)=0 \Rightarrow D=B$.
Since $B+D=1$,we get $2B=1 \Rightarrow B=\frac{1}{2}$ and $D=\frac{1}{2}$.
Now,$B D-A C = (\frac{1}{2})(\frac{1}{2}) - (\frac{1}{2\sqrt{2}})(-\frac{1}{2\sqrt{2}}) = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$.

(C) Given the equation: $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$
Multiplying both sides by $(x-a)(x^2+b^2)$,we get:
$A(x^2+b^2)+(B x+C)(x-a)=1$
Expanding the terms:
$(A+B)x^2+(C-a B)x+(A b^2-a C)=1$
Comparing the coefficients of $x^2$,$x$,and the constant term on both sides:
$A+B=0$ ....$(i)$
$C-a B=0$ ....$(ii)$
$A b^2-a C=1$ ....$(iii)$
From $(i)$,$B=-A$. Substituting this into $(ii)$:
$C-a(-A)=0 \Rightarrow C+a A=0 \Rightarrow A=\frac{-C}{a}$
Substituting $A=\frac{-C}{a}$ into $(iii)$:
$(\frac{-C}{a})b^2-a C=1$
$-C(\frac{b^2+a^2}{a})=1$
$C=\frac{-a}{a^2+b^2}$

(D) We decompose the expression using partial fractions:
$\frac{x+2}{(x^2+3)(x^2)(x^2+1)(x^2+2)} = \frac{A'}{x} + \frac{B'}{x^2} + \frac{C'x+D'}{x^2+1} + \frac{E'x+F'}{x^2+2} + \frac{G'x+H'}{x^2+3}$.
By equating coefficients and solving the system of equations,we obtain:
$A' = \frac{1}{6}, B' = \frac{1}{3}, C' = -\frac{1}{2}, D' = -1, E' = \frac{1}{2}, F' = 1, G' = -\frac{1}{6}, H' = -\frac{1}{3}$.
Substituting these back,we group the terms to match the form given in the question:
$\frac{x+2}{(x^2+3)(x^4+x^2)(x^2+2)} = \frac{-\frac{1}{6}x - \frac{1}{3}}{x^2+3} + \frac{\frac{1}{2}x + 1}{x^2+2} + \frac{-\frac{1}{3}x^3 - \frac{2}{3}x^2 + \frac{1}{6}x + \frac{1}{3}}{x^4+x^2}$.
Comparing this with the given expression,we identify:
$A = -\frac{1}{6}, B = -\frac{1}{3}, C = \frac{1}{2}, D = 1, E = -\frac{1}{3}, F = -\frac{2}{3}, G = \frac{1}{6}, H = \frac{1}{3}$.
Finally,calculating the required value:
$(E+F)(C+D)(A) = (-\frac{1}{3} - \frac{2}{3})(\frac{1}{2} + 1)(-\frac{1}{6}) = (-1)(\frac{3}{2})(-\frac{1}{6}) = \frac{3}{12} = \frac{1}{4}$.

(A) Given the partial fraction decomposition: $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$
Combining the terms on the right side: $\frac{13x+43}{2x^2+17x+30} = \frac{A(x+6) + B(2x+5)}{(2x+5)(x+6)}$
Since the denominators are equal,we equate the numerators: $13x + 43 = A(x+6) + B(2x+5)$
Expanding the right side: $13x + 43 = (A + 2B)x + (6A + 5B)$
Comparing the coefficients of $x$ and the constant terms,we get the system of equations:
$A + 2B = 13$ ... $(i)$
$6A + 5B = 43$ ... $(ii)$
From $(i)$,$A = 13 - 2B$. Substituting this into $(ii)$:
$6(13 - 2B) + 5B = 43$
$78 - 12B + 5B = 43$
$-7B = 43 - 78$
$-7B = -35 \Rightarrow B = 5$
Substituting $B = 5$ back into $(i)$:
$A + 2(5) = 13 \Rightarrow A + 10 = 13 \Rightarrow A = 3$
Therefore,$A + B = 3 + 5 = 8$.

(C) Let $x-2 = t$,so $x = t+2$.
Substituting this into the expression: $4(t+2)^2 + 5 = 4(t^2 + 4t + 4) + 5 = 4t^2 + 16t + 21$.
Now,$\frac{4t^2 + 16t + 21}{t^4} = \frac{4}{t^2} + \frac{16}{t^3} + \frac{21}{t^4}$.
Comparing this with $\frac{A}{t} + \frac{B}{t^2} + \frac{C}{t^3} + \frac{D}{t^4}$,we get:
$A = 0$,$B = 4$,$C = 16$,and $D = 21$.
Now,calculate $\sqrt{\frac{A+B+D}{C}} = \sqrt{\frac{0+4+21}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(B) The given equations represent pairs of straight lines.
For the first curve $3x^2-y^2-2xy+4x+1=0$,we can rewrite it as $3x^2+(4-2y)x+(1-y^2)=0$. Solving for $x$ using the quadratic formula:
$x = \frac{-(4-2y) \pm \sqrt{(4-2y)^2 - 4(3)(1-y^2)}}{6} = \frac{(2y-4) \pm \sqrt{16-16y+4y^2-12+12y^2}}{6} = \frac{(2y-4) \pm \sqrt{16y^2-16y+4}}{6} = \frac{(2y-4) \pm 2(2y-1)}{6} = \frac{(y-2) \pm (2y-1)}{3}$.
This gives two lines: $L_1: x-y+1=0$ and $L_2: 3x+y+1=0$.
For the second curve $3x^2-y^2-2xy+6x+2y=0$,we rewrite it as $3x^2+(6-2y)x+(2y-y^2)=0$. Solving for $x$:
$x = \frac{-(6-2y) \pm \sqrt{(6-2y)^2 - 4(3)(2y-y^2)}}{6} = \frac{(2y-6) \pm \sqrt{36-24y+4y^2-24y+12y^2}}{6} = \frac{(2y-6) \pm \sqrt{16y^2-48y+36}}{6} = \frac{(2y-6) \pm 2(2y-3)}{6} = \frac{(y-3) \pm (2y-3)}{3}$.
This gives two lines: $L_3: x-y+2=0$ and $L_4: 3x+y=0$.
The region is a parallelogram formed by the intersection of these four lines. The vertices are:
$A = L_3 \cap L_4 = (-1/2, 3/2)$
$B = L_1 \cap L_4 = (-1/4, 3/4)$
$C = L_1 \cap L_2 = (-1/2, 1/2)$
$D = L_2 \cap L_3 = (-3/4, 5/4)$
The area of a parallelogram formed by lines $a_1x+b_1y+c_1=0, a_1x+b_1y+c_2=0, a_2x+b_2y+d_1=0, a_2x+b_2y+d_2=0$ is given by $\frac{|(c_1-c_2)(d_1-d_2)|}{|a_1b_2-a_2b_1|}$.
Here,$L_1: x-y+1=0, L_3: x-y+2=0 \implies |c_1-c_2| = |1-2| = 1$.
$L_2: 3x+y+1=0, L_4: 3x+y=0 \implies |d_1-d_2| = |1-0| = 1$.
The denominator is $|(1)(1) - (3)(-1)| = |1+3| = 4$.
Area $= \frac{1 \times 1}{4} = \frac{1}{4}$.

(A) Given points are $P(\alpha, 4, 7)$ and $Q(3, \beta, 8)$.
Since the $YZ$-plane divides the line segment $PQ$ in the ratio $2:3$,the $x$-coordinate of the point of division is zero.
Using the section formula: $\frac{2(3) + 3(\alpha)}{2+3} = 0 \Rightarrow 6 + 3\alpha = 0 \Rightarrow \alpha = -2$.
Since the $ZX$-plane divides the line segment $PQ$ in the ratio $4:5$,the $y$-coordinate of the point of division is zero.
Using the section formula: $\frac{4(\beta) + 5(4)}{4+5} = 0 \Rightarrow 4\beta + 20 = 0 \Rightarrow \beta = -5$.
Thus,the points are $P(-2, 4, 7)$ and $Q(3, -5, 8)$.
The length of the line segment $PQ$ is given by the distance formula:
$PQ = \sqrt{(3 - (-2))^2 + (-5 - 4)^2 + (8 - 7)^2}$
$PQ = \sqrt{(5)^2 + (-9)^2 + (1)^2}$
$PQ = \sqrt{25 + 81 + 1} = \sqrt{107}$.

(A) Let the point $C$ divide the line segment $AB$ externally in the ratio $k: 1$.
The coordinates of point $C$ are given by the section formula for external division:
$C = \left( \frac{k x_2 - x_1}{k - 1}, \frac{k y_2 - y_1}{k - 1}, \frac{k z_2 - z_1}{k - 1} \right)$
Substituting the coordinates $A(1, 2, 3)$ and $B(3, 4, 7)$:
$C = \left( \frac{3k - 1}{k - 1}, \frac{4k - 2}{k - 1}, \frac{7k - 3}{k - 1} \right)$
Given $C = (-3, -2, -5)$,we equate the $x$-coordinates:
$\frac{3k - 1}{k - 1} = -3$
$3k - 1 = -3(k - 1)$
$3k - 1 = -3k + 3$
$6k = 4$
$k = \frac{4}{6} = \frac{2}{3}$
Thus,the ratio $k: 1$ is $\frac{2}{3}: 1$,which is $2: 3$.

(D) The equation of a family of lines passing through the intersection of $x-2y+3=0$ and $2x-y-1=0$ is given by $(x-2y+3) + K(2x-y-1) = 0$.
Rearranging terms,we get $(1+2K)x - (2+K)y + (3-K) = 0$.
This can be written as $\frac{(1+2K)}{K-3}x + \frac{-(2+K)}{K-3}y = 1$.
The intercepts on the $X$ and $Y$ axes are $A\left(\frac{K-3}{1+2K}, 0\right)$ and $B\left(0, \frac{K-3}{-(2+K)}\right)$.
Let the point dividing $AB$ in the ratio $-2:3$ be $(x, y)$. Using the section formula:
$x = \frac{-2(0) + 3(\frac{K-3}{1+2K})}{3-2} = \frac{3(K-3)}{1+2K}$ and $y = \frac{-2(\frac{K-3}{-(2+K)}) + 3(0)}{3-2} = \frac{2(K-3)}{2+K}$.
From $x = \frac{3K-9}{2K+1}$,we get $x(2K+1) = 3K-9 \Rightarrow K(2x-3) = -9-x \Rightarrow K = \frac{x+9}{3-2x}$.
From $y = \frac{2K-6}{K+2}$,we get $y(K+2) = 2K-6 \Rightarrow K(y-2) = -6-2y \Rightarrow K = \frac{6+2y}{2-y}$.
Equating the two expressions for $K$: $\frac{x+9}{3-2x} = \frac{6+2y}{2-y}$.
$(x+9)(2-y) = (6+2y)(3-2x) \Rightarrow 2x - xy + 18 - 9y = 18 - 12x + 6y - 4xy$.
Simplifying,we get $14x + 3xy - 15y = 0$.

(D) Let the points of intersection be $A(x_1, y_1)$ and $B(x_2, y_2)$.
Since the point $(1, 2)$ bisects the line segment $AB$,we have:
$\frac{x_1 + x_2}{2} = 1 \Rightarrow x_2 = 2 - x_1$
$\frac{y_1 + y_2}{2} = 2 \Rightarrow y_2 = 4 - y_1$
Since $A$ lies on $3x - 2y - 1 = 0$,we have:
$3x_1 - 2y_1 - 1 = 0$ --- $(i)$
Since $B$ lies on $x + 2y + 1 = 0$,we substitute $x_2$ and $y_2$:
$(2 - x_1) + 2(4 - y_1) + 1 = 0$
$2 - x_1 + 8 - 2y_1 + 1 = 0$
$x_1 + 2y_1 - 11 = 0$ --- (ii)
Solving equations $(i)$ and (ii) by adding them:
$(3x_1 - 2y_1 - 1) + (x_1 + 2y_1 - 11) = 0$
$4x_1 - 12 = 0 \Rightarrow x_1 = 3$
Substituting $x_1 = 3$ in $(i)$:
$3(3) - 2y_1 - 1 = 0 \Rightarrow 8 = 2y_1 \Rightarrow y_1 = 4$
Thus,$A = (3, 4)$.
Then $x_2 = 2 - 3 = -1$ and $y_2 = 4 - 4 = 0$,so $B = (-1, 0)$.
The equation of line $L$ passing through $(3, 4)$ and $(-1, 0)$ is:
$y - 0 = \frac{4 - 0}{3 - (-1)}(x - (-1))$
$y = \frac{4}{4}(x + 1) \Rightarrow y = x + 1 \Rightarrow x - y = -1$
Dividing by $-1$,we get $\frac{x}{-1} + \frac{y}{1} = 1$.
Comparing with $\frac{x}{a} + \frac{y}{b} = 1$,we get $a = -1$ and $b = 1$.
Therefore,$a + 2b + 1 = -1 + 2(1) + 1 = -1 + 2 + 1 = 2$.

(C) The slope of the given line $2x-y+3=0$ is $m_1 = 2$.
Let the slope of line $L$ be $m$.
The angle between the lines is $60^{\circ}$,so $\tan 60^{\circ} = |\frac{m-m_1}{1+m m_1}|$.
$\sqrt{3} = |\frac{m-2}{1+2m}|$.
This gives two cases:
Case $1$: $\frac{m-2}{1+2m} = \sqrt{3} \Rightarrow m-2 = \sqrt{3} + 2\sqrt{3}m \Rightarrow m(1-2\sqrt{3}) = 2+\sqrt{3} \Rightarrow m = \frac{2+\sqrt{3}}{1-2\sqrt{3}} = \frac{(2+\sqrt{3})(1+2\sqrt{3})}{1-12} = \frac{2+4\sqrt{3}+\sqrt{3}+6}{-11} = \frac{8+5\sqrt{3}}{-11}$.
Case $2$: $\frac{m-2}{1+2m} = -\sqrt{3} \Rightarrow m-2 = -\sqrt{3} - 2\sqrt{3}m \Rightarrow m(1+2\sqrt{3}) = 2-\sqrt{3} \Rightarrow m = \frac{2-\sqrt{3}}{1+2\sqrt{3}} = \frac{(2-\sqrt{3})(1-2\sqrt{3})}{1-12} = \frac{2-4\sqrt{3}-\sqrt{3}+6}{-11} = \frac{8-5\sqrt{3}}{-11}$.
Since $L$ makes an acute angle with the positive $X$-axis,$m > 0$.
Comparing the two slopes,$\frac{8-5\sqrt{3}}{-11} = \frac{5\sqrt{3}-8}{11} \approx \frac{5(1.732)-8}{11} = \frac{8.66-8}{11} > 0$.
Thus,$m = \frac{5\sqrt{3}-8}{11}$.
The equation of line $L$ is $y - 0 = m(x - 2) \Rightarrow y = \frac{5\sqrt{3}-8}{11}(x - 2)$.
The $Y$-intercept is found by setting $x = 0$: $y = \frac{5\sqrt{3}-8}{11}(-2) = \frac{16-10\sqrt{3}}{11}$.

(D) The total number of ways to place $5$ letters in $5$ addressed envelopes is $5! = 120$.
The number of ways in which all letters are placed in the correct envelopes is $1$.
Thus,the probability that all letters are placed correctly is $\frac{1}{120}$.
The probability that at least one letter is placed in the wrongly addressed envelope is $1 - P(\text{all correct}) = 1 - \frac{1}{120} = \frac{119}{120}$.

(C) The set of the first $5$ natural numbers is $S = \{1, 2, 3, 4, 5\}$.
The total number of ways to choose two different numbers $x$ and $y$ from $S$ is $^5C_2 = \frac{5 \times 4}{2} = 10$.
According to Fermat's Little Theorem,for any integer $a$ not divisible by $5$,$a^4 \equiv 1 \pmod{5}$.
If $a$ is divisible by $5$,then $a^4 \equiv 0 \pmod{5}$.
Let $x, y \in \{1, 2, 3, 4, 5\}$. We want $x^4 - y^4$ to be divisible by $5$,i.e.,$x^4 \equiv y^4 \pmod{5}$.
Case $1$: Both $x$ and $y$ are not divisible by $5$. Then $x^4 \equiv 1$ and $y^4 \equiv 1$,so $x^4 - y^4 \equiv 0 \pmod{5}$.
The numbers not divisible by $5$ are $\{1, 2, 3, 4\}$. The number of ways to choose $2$ distinct numbers from these $4$ is $^4C_2 = 6$.
Case $2$: One of the numbers is $5$. If $x=5$,then $x^4 \equiv 0$. For $x^4 - y^4$ to be divisible by $5$,$y^4$ must also be $0$,which is impossible as there is only one multiple of $5$ in the set.
Thus,the favorable outcomes are $6$.
The probability is $\frac{6}{10} = \frac{3}{5}$.

(C) Step $1$: Define the variables.
Total articles $= 15$.
Defective articles $= 3$.
Non-defective articles $= 15 - 3 = 12$.
Sample size $= 5$.
We need the probability of choosing exactly $2$ defective articles in this sample.
Step $2$: Use combinations to calculate possible outcomes.
The probability is calculated using the hypergeometric distribution formula:
$P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$
$1$. Number of ways to choose $2$ defective articles out of $3$:
$\binom{3}{2} = \frac{3!}{2!1!} = 3$.
$2$. Number of ways to choose $3$ non-defective articles out of $12$:
$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
$3$. Total number of ways to choose $5$ articles out of $15$:
$\binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3003$.
Step $3$: Calculate the probability.
$P = \frac{3 \times 220}{3003} = \frac{660}{3003} = \frac{220}{1001}$.
Thus,the correct answer is $\frac{220}{1001}$.

(D) The total number of ways to choose $3$ integers from $20$ is ${}^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
We categorize the numbers from $1$ to $20$ based on their remainder when divided by $3$:
$R_0 = \{3, 6, 9, 12, 15, 18\}$ (count $6$)
$R_1 = \{1, 4, 7, 10, 13, 16, 19\}$ (count $7$)
$R_2 = \{2, 5, 8, 11, 14, 17, 20\}$ (count $7$)
The sum is divisible by $3$ in the following cases:
$(I)$ All three numbers from $R_0$: ${}^{6}C_3 = 20$.
$(II)$ All three numbers from $R_1$: ${}^{7}C_3 = 35$.
$(III)$ All three numbers from $R_2$: ${}^{7}C_3 = 35$.
$(IV)$ One number from each set $R_0, R_1, R_2$: ${}^{6}C_1 \times {}^{7}C_1 \times {}^{7}C_1 = 6 \times 7 \times 7 = 294$.
Total favorable outcomes $= 20 + 35 + 35 + 294 = 384$.
Probability $= \frac{384}{1140} = \frac{32}{85}$.

(C) The total number of outcomes when throwing three dice is $6^3 = 216$.
First,we find the number of ways to get a sum $S$ with three dice,denoted by $n(S)$.
The number of ways to get a sum $S = 13$ is $n(13) = 21$.
The number of ways to get a sum $S > 13$ is $n(14) + n(15) + n(16) + n(17) + n(18)$.
Using the formula for the number of ways to get a sum $S$ with $n$ dice,each having $f$ faces:
$n(14) = 15, n(15) = 10, n(16) = 6, n(17) = 3, n(18) = 1$.
Summing these: $15 + 10 + 6 + 3 + 1 = 35$.
Thus,the probability that $B$ gets a sum higher than $13$ is $\frac{35}{216}$.

(D) The equation is $ax^2 + bx + c = 0$.
For the equation to have equal roots,the discriminant must be zero,i.e.,$D = b^2 - 4ac = 0$,which implies $b^2 = 4ac$.
Since $a, b, c$ are determined by throwing a die,each can take values from the set $\{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6 \times 6 \times 6 = 216$.
We need to find the number of triplets $(a, b, c)$ such that $b^2 = 4ac$.

$b$	$(a, c)$	Count
$b=1$	$1 = 4ac$ (No solution)	$0$
$b=2$	$4 = 4ac \implies ac = 1 \implies (1, 1)$	$1$
$b=3$	$9 = 4ac$ (No solution)	$0$
$b=4$	$16 = 4ac \implies ac = 4 \implies (1, 4), (4, 1), (2, 2)$	$3$
$b=5$	$25 = 4ac$ (No solution)	$0$
$b=6$	$36 = 4ac \implies ac = 9 \implies (3, 3)$	$1$

Total favorable outcomes $= 1 + 3 + 1 = 5$.
Required probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{216}$.

(B) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The outcomes where the sum is $10$ are $(4, 6), (5, 5), (6, 4)$.
The outcomes where the sum is $11$ are $(5, 6), (6, 5)$.
The total number of favorable outcomes is $3 + 2 = 5$.
Therefore,the required probability is $\frac{5}{36}$.

(A) The total number of five-digit numbers formed using the digits $0, 1, 2, 3, 4$ exactly once is $4 \times 4! = 4 \times 24 = 96$.
For a number to be divisible by $4$,the last two digits must form a number divisible by $4$.
The possible pairs for the last two digits are:
$04, 20, 40$ (where $0$ is used): Each gives $3! = 6$ numbers. Total $= 3 \times 6 = 18$.
$12, 24, 32$ (where $0$ is not used): For each,the first digit cannot be $0$,so there are $3$ choices for the first digit and $2!$ for the remaining two. Total $= 3 \times (3 \times 2!) = 3 \times 6 = 18$.
Wait,let us re-evaluate:
If last two digits are $04, 20, 40$: The remaining $3$ digits can be arranged in $3! = 6$ ways. $3 \times 6 = 18$.
If last two digits are $12, 24, 32$: The first digit cannot be $0$ and cannot be one of the two digits used. So $4 - 2 = 2$ choices for the first digit,then $2!$ for the remaining. $3 \times (2 \times 2!) = 3 \times 4 = 12$.
Total favorable numbers $= 18 + 12 = 30$.
Probability $= \frac{30}{96} = \frac{5}{16}$.

(A) Let $P, Q, R$ be the events that $P, Q, R$ hit the target respectively.
Given probabilities are $P(P) = \frac{2}{3}, P(Q) = \frac{3}{5}, P(R) = \frac{5}{7}$.
The probabilities of not hitting the target are $P(P') = 1 - \frac{2}{3} = \frac{1}{3}$,$P(Q') = 1 - \frac{3}{5} = \frac{2}{5}$,and $P(R') = 1 - \frac{5}{7} = \frac{2}{7}$.
We need the probability that the target is hit by $P$ or $Q$ but not by $R$. This can happen in three mutually exclusive ways:
$1$. $P$ hits,$Q$ misses,$R$ misses: $P(P) \times P(Q') \times P(R') = \frac{2}{3} \times \frac{2}{5} \times \frac{2}{7} = \frac{8}{105}$.
$2$. $P$ misses,$Q$ hits,$R$ misses: $P(P') \times P(Q) \times P(R') = \frac{1}{3} \times \frac{3}{5} \times \frac{2}{7} = \frac{6}{105}$.
$3$. $P$ hits,$Q$ hits,$R$ misses: $P(P) \times P(Q) \times P(R') = \frac{2}{3} \times \frac{3}{5} \times \frac{2}{7} = \frac{12}{105}$.
Summing these probabilities: $\frac{8}{105} + \frac{6}{105} + \frac{12}{105} = \frac{26}{105}$.

(C) Total balls in box $P = 1 + 3 + 2 = 6$.
Total balls in box $Q = 2 + 3 + 4 = 9$.
Let $W_P, R_P, B_P$ be the events of drawing a white,red,and black ball from box $P$ respectively,and $W_Q, R_Q, B_Q$ be the corresponding events for box $Q$.
The probabilities are:
$P(W_P) = \frac{1}{6}, P(R_P) = \frac{3}{6}, P(B_P) = \frac{2}{6}$
$P(W_Q) = \frac{2}{9}, P(R_Q) = \frac{3}{9}, P(B_Q) = \frac{4}{9}$
The probability that the balls are of the same colour is:
$P(\text{same}) = P(W_P)P(W_Q) + P(R_P)P(R_Q) + P(B_P)P(B_Q)$
$P(\text{same}) = (\frac{1}{6} \times \frac{2}{9}) + (\frac{3}{6} \times \frac{3}{9}) + (\frac{2}{6} \times \frac{4}{9}) = \frac{2 + 9 + 8}{54} = \frac{19}{54}$
The probability that the balls are of different colours is:
$P(\text{different}) = 1 - P(\text{same}) = 1 - \frac{19}{54} = \frac{35}{54}$

(B) The word '$SENSELESSNESS$' contains $13$ letters in total: $S$ ($6$ times),$E$ ($4$ times),$N$ ($2$ times),and $L$ ($1$ time).
Total number of arrangements $= \frac{13!}{6!4!2!} = 180180$.
To find the number of arrangements where all $E$'s come together,we treat the $4$ $E$'s as a single unit.
Now,we have $10$ units to arrange: $(EEEE)$,$S$ ($6$ times),$N$ ($2$ times),and $L$ ($1$ time).
Number of arrangements with all $E$'s together $= \frac{10!}{6!2!1!} = \frac{3628800}{720 \times 2} = 2520$.
Therefore,the required probability $= \frac{2520}{180180} = \frac{252}{18018} = \frac{2}{143}$.

(C) The total number of ways to select two numbers $x$ and $y$ from the set $\{1, 2, 3, \ldots, 10\}$ with replacement is $10 \times 10 = 100$.
We want to find the number of pairs $(x, y)$ such that $|x^2 - y^2|$ is divisible by $6$.
This condition is equivalent to $|(x - y)(x + y)|$ being divisible by $6$.
We analyze the pairs $(x, y)$ by iterating through $x$ from $1$ to $10$ and finding $y$ such that $x^2 \equiv y^2 \pmod{6}$.
The squares modulo $6$ are: $1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 3, 4^2 \equiv 4, 5^2 \equiv 1, 6^2 \equiv 0, 7^2 \equiv 1, 8^2 \equiv 4, 9^2 \equiv 3, 10^2 \equiv 4$.
Grouping by values: $0: \{6\}$,$1: \{1, 5, 7\}$,$3: \{3, 9\}$,$4: \{2, 4, 8, 10\}$.
Number of pairs $(x, y)$ such that $x^2 \equiv y^2 \pmod{6}$ is:
For $x^2 \equiv 0$: $1^2 = 1$ pair $(6, 6)$.
For $x^2 \equiv 1$: $3^2 = 9$ pairs (from $\{1, 5, 7\} \times \{1, 5, 7\}$).
For $x^2 \equiv 3$: $2^2 = 4$ pairs (from $\{3, 9\} \times \{3, 9\}$).
For $x^2 \equiv 4$: $4^2 = 16$ pairs (from $\{2, 4, 8, 10\} \times \{2, 4, 8, 10\}$).
Total favorable outcomes $= 1 + 9 + 4 + 16 = 30$.
Probability $= \frac{30}{100} = \frac{3}{10}$.

(D) Total number of ways to arrange $8$ teachers and $4$ students around a circular table is $(8+4-1)! = 11!$.
To ensure no two students sit together,we first arrange the $8$ teachers in a circle,which can be done in $(8-1)! = 7!$ ways.
This creates $8$ gaps between the teachers. We need to place $4$ students in these $8$ gaps,which can be done in $^8C_4$ ways.
The students can be arranged among themselves in $4!$ ways.
Thus,the number of favorable arrangements is $^8C_4 \times 4! \times 7!$.
The required probability is $\frac{^8C_4 \times 4! \times 7!}{11!} = \frac{\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times 4! \times 7!}{11 \times 10 \times 9 \times 8 \times 7!} = \frac{70 \times 24 \times 7!}{11 \times 10 \times 9 \times 8 \times 7!} = \frac{7}{33}$.

(B) Total number of apples $= 12$. Number of rotten apples $= 3$. Number of good apples $= 9$.
Total ways to draw $3$ apples from $12$ is ${}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
We need the probability of getting at most one rotten apple,which means $0$ or $1$ rotten apple.
Case $1$: No rotten apple is drawn (all $3$ are good).
Number of ways $= {}^{9}C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Case $2$: Exactly $1$ rotten apple is drawn (and $2$ are good).
Number of ways $= {}^{3}C_1 \times {}^{9}C_2 = 3 \times \frac{9 \times 8}{2 \times 1} = 3 \times 36 = 108$.
Total favorable ways $= 84 + 108 = 192$.
Required probability $= \frac{192}{220} = \frac{48}{55}$.

(A) The total number of outcomes when throwing two dice is $36$. The outcomes for a sum of $4$ are $(1,3), (3,1), (2,2)$.
Therefore,the probability of getting a sum of $4$ in a single throw is $p = \frac{3}{36} = \frac{1}{12}$.
The probability of not getting a sum of $4$ is $q = 1 - p = 1 - \frac{1}{12} = \frac{11}{12}$.
Since $A$ starts the game,$B$ wins if $A$ fails on the $1^{st}$ turn,$B$ succeeds on the $2^{nd}$ turn,or $A$ fails on the $1^{st}, 3^{rd}$ turns and $B$ fails on the $2^{nd}$ turn and succeeds on the $4^{th}$ turn,and so on.
The probability that $B$ wins is given by the infinite geometric series:
$P(B \text{ wins}) = qp + q^3p + q^5p + \dots$
This is a geometric series with first term $a = qp = \frac{11}{12} \times \frac{1}{12} = \frac{11}{144}$ and common ratio $r = q^2 = (\frac{11}{12})^2 = \frac{121}{144}$.
The sum of an infinite geometric series is $S = \frac{a}{1-r}$.
$P(B \text{ wins}) = \frac{\frac{11}{144}}{1 - \frac{121}{144}} = \frac{\frac{11}{144}}{\frac{144-121}{144}} = \frac{11}{23}$.

(B) Let $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}$.
Using the sum formula for sines in arithmetic progression,$\sum_{k=1}^{n} \sin(k\theta) = \frac{\sin(n\theta/2) \sin((n+1)\theta/2)}{\sin(\theta/2)}$.
Here $n = 89$ and $\theta = 1^{\circ}$,so $S = \frac{\sin(89^{\circ}/2) \sin(90^{\circ}/2)}{\sin(0.5^{\circ})} = \frac{\sin(44.5^{\circ}) \sin(45^{\circ})}{\sin(0.5^{\circ})}$.
Now consider the denominator $D = 2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1$.
Using the sum formula $\sum_{k=1}^{n} \cos(k\theta) = \frac{\sin(n\theta/2) \cos((n+1)\theta/2)}{\sin(\theta/2)}$,we have $\sum_{k=1}^{44} \cos(k^{\circ}) = \frac{\sin(44^{\circ}/2) \cos(45^{\circ}/2)}{\sin(0.5^{\circ})} = \frac{\sin(22^{\circ}) \cos(22.5^{\circ})}{\sin(0.5^{\circ})}$.
Alternatively,note that $\sin 1^{\circ} + \ldots + \sin 89^{\circ} = (\sin 1^{\circ} + \sin 89^{\circ}) + \ldots + \sin 45^{\circ} = 2 \sin 45^{\circ} \cos 44^{\circ} + 2 \sin 45^{\circ} \cos 43^{\circ} + \ldots + \sin 45^{\circ}$.
This simplifies to $\sqrt{2}(\cos 44^{\circ} + \cos 43^{\circ} + \ldots + \cos 1^{\circ}) + \frac{1}{\sqrt{2}}$.
Multiplying by $\sqrt{2}$,we get $\sqrt{2} \times S = 2(\cos 1^{\circ} + \ldots + \cos 44^{\circ}) + 1$.
Thus,$\frac{S}{2(\cos 1^{\circ} + \ldots + \cos 44^{\circ}) + 1} = \frac{1}{\sqrt{2}}$.

(D) The given equation of the pair of lines is $2x^2 + hxy + 6y^2 = 0$.
Let the slopes of the two lines be $m$ and $3m$.
The equation of the pair of lines can be written as $(y - mx)(y - 3mx) = 0$.
Expanding this,we get $y^2 - 4mxy + 3m^2x^2 = 0$,or $3m^2x^2 - 4mxy + y^2 = 0$.
Dividing the original equation $2x^2 + hxy + 6y^2 = 0$ by $6$,we get $\frac{1}{3}x^2 + \frac{h}{6}xy + y^2 = 0$.
Comparing the coefficients of $x^2$ and $xy$ with $3m^2x^2 - 4mxy + y^2 = 0$,we have:
$3m^2 = \frac{1}{3}$ $\Rightarrow m^2 = \frac{1}{9}$ $\Rightarrow m = \pm \frac{1}{3}$.
Also,$-4m = \frac{h}{6} \Rightarrow h = -24m$.
Substituting $m = \pm \frac{1}{3}$ into the expression for $h$:
$h = -24(\pm \frac{1}{3}) = \mp 8$.
Thus,$h = \pm 8$.