AP EAMCET 2023 Mathematics Question Paper with Answer and Solution

(A) We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$,$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$,and $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Substituting these into the expression $r_1+r_2+r_3-r$:
$= 4R [\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} + \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} - \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}]$
$= 4R [\cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}) + \sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2})]$
$= 4R [\cos \frac{C}{2} \sin (\frac{A+B}{2}) + \sin \frac{C}{2} \cos (\frac{A+B}{2})]$
Since $A+B+C = \pi$,we have $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.
$= 4R [\cos \frac{C}{2} \sin (\frac{\pi}{2} - \frac{C}{2}) + \sin \frac{C}{2} \cos (\frac{\pi}{2} - \frac{C}{2})]$
$= 4R [\cos \frac{C}{2} \cos \frac{C}{2} + \sin \frac{C}{2} \sin \frac{C}{2}]$
$= 4R [\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}] = 4R(1) = 4R$.

(C) Let $n(M)$ be the number of members proficient in Mathematics and $n(S)$ be the number of members proficient in Statistics.
Given $n(M \cup S) = 25$,$n(M) = 19$,and $n(S) = 16$.
Using the formula $n(M \cup S) = n(M) + n(S) - n(M \cap S)$:
$25 = 19 + 16 - n(M \cap S)$
$25 = 35 - n(M \cap S)$
$n(M \cap S) = 35 - 25 = 10$.
The probability that a person selected at random is proficient in both is $P(M \cap S) = \frac{n(M \cap S)}{n(M \cup S)} = \frac{10}{25} = \frac{2}{5}$.

(C) When only event $A$ occurs,it is represented as $(A \cap B^c)$.
When only event $B$ occurs,it is represented as $(A^c \cap B)$.
Therefore,the event 'exactly one of the events $A$ or $B$ occurs' is the union of these two disjoint sets: $(A \cap B^c) \cup (A^c \cap B)$.

(B) Given the equation $4a = 5b$,we can write $a = \frac{5}{4}b$.
Since $a$ must be an integer,$b$ must be a multiple of $4$.
Given $b \in \{1, 2, 3, \dots, 30\}$,the possible values for $b$ are $4, 8, 12, 16, 20, 24, 28$.
For each $b$,we calculate $a = \frac{5}{4}b$:
If $b = 4, a = 5$.
If $b = 8, a = 10$.
If $b = 12, a = 15$.
If $b = 16, a = 20$.
If $b = 20, a = 25$.
If $b = 24, a = 30$.
If $b = 28, a = 35$ (which is not in the set $\{1, 2, \dots, 30\}$).
Thus,the valid ordered pairs $(a, b)$ are $(5, 4), (10, 8), (15, 12), (20, 16), (25, 20), (30, 24)$.
The total number of such pairs is $6$.

(C) Given $S = \{x \in \mathbb{Z} : x^2 - 7x + 6 \leq 0 \text{ and } x^2 - 3x > 0\}$ ...$(i)$
First,solve the inequality $x^2 - 7x + 6 \leq 0$:
$(x - 6)(x - 1) \leq 0$
This implies $x \in [1, 6]$.
Next,solve the inequality $x^2 - 3x > 0$:
$x(x - 3) > 0$
This implies $x \in (-\infty, 0) \cup (3, \infty)$.
Now,find the intersection of these two intervals:
$S = \{x \in \mathbb{Z} : x \in [1, 6] \cap ((-\infty, 0) \cup (3, \infty))\}$
$S = \{x \in \mathbb{Z} : x \in (3, 6] \cap \mathbb{Z}\}$
$S = \{4, 5, 6\}$.
The number of elements in set $S$ is $3$.

(C) Since $2$ particular students are always to be included in the team,we have already selected $2$ members.
Remaining members to be selected $= 5 - 2 = 3$.
Remaining students available to choose from $= 12 - 2 = 10$.
Therefore,the number of ways to select the remaining $3$ students from $10$ available students is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Number of ways $= {}^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$.

(A) Given the recurrence relation $f(x)=f(x-2)+f(x-3)$ with initial values $f(0)=0, f(1)=1, f(2)=2$.
We calculate the subsequent values step by step:
$f(3)=f(1)+f(0)=1+0=1$
$f(4)=f(2)+f(1)=2+1=3$
$f(5)=f(3)+f(2)=1+2=3$
$f(6)=f(4)+f(3)=3+1=4$
$f(7)=f(5)+f(4)=3+3=6$
$f(8)=f(6)+f(5)=4+3=7$
$f(9)=f(7)+f(6)=6+4=10$
$f(10)=f(8)+f(7)=7+6=13$

(A) Given $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.
We know that $\sin^2 x + \cos^2 x = 1$,so $\cos^2 x = 1 - \sin^2 x$ and $\sin^2 x = 1 - \cos^2 x$.
Substituting these into the numerator and denominator:
Numerator: $\cos^2 x + \sin^4 x = \cos^2 x + \sin^2 x \cdot \sin^2 x = \cos^2 x + \sin^2 x(1 - \cos^2 x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.
Denominator: $\sin^2 x + \cos^4 x = \sin^2 x + \cos^2 x \cdot \cos^2 x = \sin^2 x + \cos^2 x(1 - \sin^2 x) = \sin^2 x + \cos^2 x - \cos^2 x \sin^2 x = 1 - \cos^2 x \sin^2 x$.
Thus,$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \cos^2 x \sin^2 x} = 1$.
Since $f(x) = 1$ for all $x \in R$,it follows that $f(2023) = 1$.

(B) Given $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$ ...$(i)$
Factorize the denominator: $x^4+3 x^2+4 = (x^2+2)^2 - x^2 = (x^2+x+2)(x^2-x+2)$.
Thus,$\frac{x^2-7 x+2}{(x^2+x+2)(x^2-x+2)} = \frac{Px+Q}{x^2+x+2} + \frac{Rx+S}{x^2-x+2}$.
Equating numerators: $x^2-7x+2 = (Px+Q)(x^2-x+2) + (Rx+S)(x^2+x+2)$.
Expanding the right side: $x^2-7x+2 = x^3(P+R) + x^2(-P+Q+R+S) + x(2P-Q+2R+S) + (2Q+2S)$.
Comparing coefficients:
$P+R=0$
$-P+Q+R+S=1$
$2P-Q+2R+S=-7$
$2Q+2S=2 \Rightarrow Q+S=1$
Solving these,we get $P=0, Q=4, R=0, S=-3$.
So,$\frac{x^2-7x+2}{x^4+3x^2+4} = \frac{4}{x^2+x+2} + \frac{-3}{x^2-x+2}$.
Comparing with $(i)$,we have $a=1, b=-1$ (since $a>b$),$A=0, B=4, C=0, D=-3$.
Then $B+D = 4-3 = 1$.
Checking options: $2a+b = 2(1) + (-1) = 1$.
Therefore,$B+D = 2a+b$.

(C) Given the partial fraction decomposition: $\frac{2 x^2+5 x+6}{(x+2)^3}=\frac{a}{x+2}+\frac{b}{(x+2)^2}+\frac{c}{(x+2)^3}$
Multiply both sides by $(x+2)^3$:
$2 x^2+5 x+6 = a(x+2)^2 + b(x+2) + c$
Let $x+2 = t$,then $x = t-2$. Substituting this:
$2(t-2)^2 + 5(t-2) + 6 = a t^2 + b t + c$
$2(t^2 - 4t + 4) + 5t - 10 + 6 = a t^2 + b t + c$
$2t^2 - 8t + 8 + 5t - 4 = a t^2 + b t + c$
$2t^2 - 3t + 4 = a t^2 + b t + c$
Comparing coefficients:
$a = 2, b = -3, c = 4$
Now,calculate $a \cdot b + b \cdot c + c \cdot a$:
$a \cdot b = 2 \cdot (-3) = -6$
$b \cdot c = (-3) \cdot 4 = -12$
$c \cdot a = 4 \cdot 2 = 8$
Sum $= -6 - 12 + 8 = -10$

(D) Given the partial fraction decomposition: $\frac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx-3}{x^2+2}$.
Multiplying both sides by $(x-1)^2(x^2+2)$,we get:
$-x^2+6x+1 = A(x-1)(x^2+2) + B(x^2+2) + (Cx-3)(x-1)^2$.
Setting $x=1$: $-1+6+1 = B(1+2) \Rightarrow 6 = 3B \Rightarrow B=2$.
Expanding the right side: $-x^2+6x+1 = A(x^3-x^2+2x-2) + B(x^2+2) + (Cx-3)(x^2-2x+1)$.
Comparing the coefficients of $x^3$: $0 = A + C \Rightarrow C = -A$.
Comparing the coefficients of $x^2$: $-1 = -A + B + (-2C - 3) = -A + 2 - 2C - 3 = -A - 2C - 1$.
Thus,$A + 2C = 0$. Since $C = -A$,we have $A - 2A = 0 \Rightarrow -A = 0 \Rightarrow A = 0$.
Therefore,$C = -A = 0$.
Finally,$A+B+C = 0 + 2 + 0 = 2$.

(B) The region is bounded by the lines $x=4$,$y=-4$,and $y=x$.
To find the vertices of the triangle,we find the intersection points of these lines:
$1$. Intersection of $x=4$ and $y=x$ is $B(4, 4)$.
$2$. Intersection of $y=-4$ and $y=x$ is $A(-4, -4)$.
$3$. Intersection of $x=4$ and $y=-4$ is $C(4, -4)$.
The region is a right-angled triangle $ABC$ with vertices $A(-4, -4)$,$B(4, 4)$,and $C(4, -4)$.
The length of the base $AC$ is the distance between $(-4, -4)$ and $(4, -4)$,which is $|4 - (-4)| = 8$ units.
The length of the height $BC$ is the distance between $(4, -4)$ and $(4, 4)$,which is $|4 - (-4)| = 8$ units.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Area $= \frac{1}{2} \times 8 \times 8 = 32$ sq. units.

(A) Let the original coordinates be $(x, y, z) = (3, -7, 5)$.
Let the origin be shifted to $(h, k, l) = (-1, -1, -1)$.
The new coordinates $(x', y', z')$ are given by the transformation:
$x' = x - h$
$y' = y - k$
$z' = z - l$
Substituting the values:
$x' = 3 - (-1) = 3 + 1 = 4$
$y' = -7 - (-1) = -7 + 1 = -6$
$z' = 5 - (-1) = 5 + 1 = 6$
Therefore,the new coordinates are $(4, -6, 6)$.

(C) Let the vertices of $\triangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given that the midpoints of $AB, BC, CA$ are $(l, 0, 0), (0, m, 0), (0, 0, n)$ respectively.
For midpoint of $AB$: $\frac{x_1+x_2}{2}=l, \frac{y_1+y_2}{2}=0, \frac{z_1+z_2}{2}=0 \Rightarrow x_1+x_2=2l, y_1+y_2=0, z_1+z_2=0$ ... $(i)$
For midpoint of $BC$: $\frac{x_2+x_3}{2}=0, \frac{y_2+y_3}{2}=m, \frac{z_2+z_3}{2}=0 \Rightarrow x_2+x_3=0, y_2+y_3=2m, z_2+z_3=0$ ... (ii)
For midpoint of $CA$: $\frac{x_3+x_1}{2}=0, \frac{y_3+y_1}{2}=0, \frac{z_3+z_1}{2}=n \Rightarrow x_3+x_1=0, y_3+y_1=0, z_3+z_1=2n$ ... (iii)
Solving these equations:
Adding $(i)$,(ii),and (iii): $2(x_1+x_2+x_3)=2l \Rightarrow x_1+x_2+x_3=l$. Similarly,$y_1+y_2+y_3=m$ and $z_1+z_2+z_3=n$.
Subtracting (ii) from the sum: $x_1=l, y_1=-m, z_1=n$ (Wait,let's re-evaluate).
From $(i)$,$x_2=2l-x_1$. Substituting into (ii): $2l-x_1+x_3=0 \Rightarrow x_1-x_3=2l$. From (iii),$x_1+x_3=0$. Adding gives $2x_1=2l \Rightarrow x_1=l, x_3=-l, x_2=l$.
Similarly,$y_1=m, y_2=-m, y_3=m$ and $z_1=-n, z_2=n, z_3=n$.
Thus,$A(l, m, -n), B(l, -m, n), C(-l, m, n)$.
$AB^2 = (l-l)^2 + (m-(-m))^2 + (-n-n)^2 = 0 + 4m^2 + 4n^2 = 4(m^2+n^2)$.
$BC^2 = (l-(-l))^2 + (-m-m)^2 + (n-n)^2 = 4l^2 + 4m^2 + 0 = 4(l^2+m^2)$.
$CA^2 = (-l-l)^2 + (m-m)^2 + (n-(-n))^2 = 4l^2 + 0 + 4n^2 = 4(l^2+n^2)$.
$AB^2+BC^2+CA^2 = 4(m^2+n^2+l^2+m^2+l^2+n^2) = 8(l^2+m^2+n^2)$.
Therefore,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2} = \frac{8(l^2+m^2+n^2)}{l^2+m^2+n^2} = 8$.

(B) The formula for the coordinates of a point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ externally is given by:
$\left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n}\right)$
Given points are $(2, 3, 4)$ and $(3, -4, 7)$ with ratio $m:n = 2:4$.
Substituting the values:
$x = \frac{2(3) - 4(2)}{2 - 4} = \frac{6 - 8}{-2} = \frac{-2}{-2} = 1$
$y = \frac{2(-4) - 4(3)}{2 - 4} = \frac{-8 - 12}{-2} = \frac{-20}{-2} = 10$
$z = \frac{2(7) - 4(4)}{2 - 4} = \frac{14 - 16}{-2} = \frac{-2}{-2} = 1$
Thus,the coordinates are $(1, 10, 1)$.

(B) Let the side of the square be $a$.
The length of the diagonal $d$ of a square with side $a$ is given by $d = a\sqrt{2}$.
The distance between the two given points $(1, -2, 3)$ and $(2, -3, 5)$ is the length of the diagonal $d$.
$d = \sqrt{(2-1)^2 + (-3 - (-2))^2 + (5-3)^2}$
$d = \sqrt{(1)^2 + (-1)^2 + (2)^2}$
$d = \sqrt{1 + 1 + 4} = \sqrt{6}$
Since $d = a\sqrt{2}$,we have $a\sqrt{2} = \sqrt{6}$.
Dividing both sides by $\sqrt{2}$,we get $a = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$.

(A) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left(\frac{\alpha+7}{2}, \frac{3+5}{2}, \frac{3+\beta}{2}\right) = \left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$.
The direction ratios of the median $AD$ are given by the difference of coordinates of $D$ and $A$:
$\left(\frac{\alpha+7}{2} - 2, 4 - 3, \frac{3+\beta}{2} - 5\right) = \left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction cosines are equal,which implies that its direction ratios must be proportional to $(1, 1, 1)$.
Thus,$\frac{\alpha+3}{2} = 1$ and $\frac{\beta-7}{2} = 1$.
Solving for $\alpha$: $\alpha+3 = 2 \Rightarrow \alpha = -1$.
Solving for $\beta$: $\beta-7 = 2 \Rightarrow \beta = 9$.
Therefore,$\frac{\beta}{\alpha} = \frac{9}{-1} = -9$.

(A) Let $S$ be a set with $n$ elements. Each element $x \in S$ can belong to one of the following four disjoint sets:
$1$. $x \in A$ and $x \in B$
$2$. $x \in A$ and $x \notin B$
$3$. $x \notin A$ and $x \in B$
$4$. $x \notin A$ and $x \notin B$
Since there are $n$ elements,there are $4^n$ total ways to choose subsets $A$ and $B$.
We are given the conditions $A \cap B = \phi$ and $A \cup B = S$.
For each element $x \in S$,it must belong to either $A$ or $B$ (since $A \cup B = S$) but not both (since $A \cap B = \phi$).
Thus,for each element,there are only $2$ choices: either $x \in A$ (which implies $x \notin B$) or $x \in B$ (which implies $x \notin A$).
Therefore,the number of favorable outcomes is $2^n$.
The probability is $\frac{2^n}{4^n} = \frac{2^n}{(2^2)^n} = \frac{2^n}{2^{2n}} = \frac{1}{2^n}$.

(A) The sample space $S$ for tossing a coin three times contains $2^3 = 8$ outcomes: $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Event $A$ (getting three heads) is $\{HHH\}$,so $P(A) = \frac{1}{8}$.
Event $B$ (getting a head on the first toss) is $\{HHH, HHT, HTH, HTT\}$,so $P(B) = \frac{4}{8} = \frac{1}{2}$.
The intersection $A \cap B$ is $\{HHH\}$,so $P(A \cap B) = \frac{1}{8}$.
For events to be independent,we must have $P(A \cap B) = P(A) \times P(B)$.
Calculating $P(A) \times P(B) = \frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$.
Since $P(A \cap B) = \frac{1}{8} \neq \frac{1}{16}$,the events are dependent.

(C) The total number of outcomes when drawing two cards with replacement from $8$ cards is $8 \times 8 = 64$.
Let the two numbers drawn be $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6, 7, 8\}$.
We need the product $xy$ to be a perfect square.
The pairs $(x, y)$ whose product is a perfect square are:
$(1, 1), (1, 4), (2, 2), (2, 8), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6), (7, 7), (8, 2), (8, 8)$.
There are $12$ such favorable outcomes.
Therefore,the probability $P = \frac{12}{64} = \frac{3}{16}$.

(D) Total number of outcomes $= 30$.
Multiples of $3$ in the range $1$ to $30$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$ (Total $10$).
Multiples of $5$ in the range $1$ to $30$ are: $5, 10, 15, 20, 25, 30$ (Total $6$).
Multiples of both $3$ and $5$ (i.e.,multiples of $15$) are: $15, 30$ (Total $2$).
Using the inclusion-exclusion principle,the number of favorable outcomes (multiples of $3$ or $5$) is: $10 + 6 - 2 = 14$.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{14}{30} = \frac{7}{15}$.

(A) The total number of outcomes when tossing a coin $3$ times is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
The boy wins if all tosses give the same outcome,which are the events $\{HHH, TTT\}$.
The number of winning outcomes is $2$.
The number of losing outcomes is $8 - 2 = 6$. These outcomes are $\{HHT, HTH, HTT, THH, THT, TTH\}$.
The probability that the boy loses the game is $P(\text{Lose}) = \frac{\text{Number of losing outcomes}}{\text{Total number of outcomes}} = \frac{6}{8} = \frac{3}{4}$.

(B) Total coins = $12 + 7 + 4 = 23$. Total ways to select $3$ coins = ${}^{23}C_3$. \\ The sum of values is $NOT$ an integral multiple of a rupee if the number of $50$-paise coins selected is odd. \\ Possible cases for selecting $3$ coins such that the number of $50$-paise coins is odd: \\ $(i)$ $1$ fifty-paise coin and $2$ other coins (from $12$ two-rupee and $7$ one-rupee coins): ${}^{4}C_1 \times {}^{19}C_2$. \\ $(ii)$ $3$ fifty-paise coins: ${}^{4}C_3$. \\ Total favorable ways = ${}^{4}C_1 \times {}^{19}C_2 + {}^{4}C_3$. \\ Expanding ${}^{19}C_2 = {}^{12}C_2 + {}^{7}C_2 + {}^{12}C_1 \times {}^{7}C_1$. \\ Thus,the probability is $\frac{4({}^{12}C_1 \times {}^{7}C_1 + {}^{12}C_2 + {}^{7}C_2) + {}^{4}C_3}{{}^{23}C_3}$.

(C) Total number of ways to arrange $20$ persons around a round table is $(20-1)! = 19!$.
Fix person $A$ at one position.
There are $19$ remaining seats for person $B$.
For there to be exactly $6$ persons between $A$ and $B$,$B$ must sit in a specific position relative to $A$.
Counting $6$ seats clockwise from $A$,the $7$th seat is occupied by $B$.
Counting $6$ seats counter-clockwise from $A$,the $7$th seat is also occupied by $B$.
Thus,there are $2$ favorable positions for $B$ out of $19$ possible seats.
Therefore,the probability is $\frac{2}{19}$.

(A) non-leap year contains $365$ days,which is equal to $52$ weeks and $1$ extra day.
Since there are $52$ weeks,there are definitely $52$ Sundays.
For the year to have $53$ Sundays,the extra day must be a Sunday.
The set of possible outcomes for the extra day is $\{ \text{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday} \}$.
Total number of possible outcomes $= 7$.
Number of favorable outcomes (the day being a Sunday) $= 1$.
Therefore,the probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{7}$.

(C) We know that for any two events $A$ and $B$,the addition theorem of probability states:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Substituting the given values:
$0.65 = P(A) + P(B) - 0.15$
$P(A) + P(B) = 0.65 + 0.15 = 0.8$
We are asked to find $P(\overline{A}) + P(\overline{B})$.
Using the complement rule $P(\overline{E}) = 1 - P(E)$:
$P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B))$
$= 2 - (P(A) + P(B))$
$= 2 - 0.8 = 1.2$

(A) Total number of cards in a pack $= 52$.
Number of sample space $n(S) = 52$.
Let $A$ be the event of drawing an ace and $B$ be the event of drawing a spade.
Number of aces $n(A) = 4$.
Number of spades $n(B) = 13$.
The number of cards that are both an ace and a spade is $n(A \cap B) = 1$.
Using the addition rule of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$.
Simplifying the fraction,we get $\frac{16}{52} = \frac{4}{13}$.

(B) We know that the probability of the union of two events is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = 0.5 + 0.4 - 0.3 = 0.6$.
The probability that neither $A$ nor $B$ occurs is given by $P(A^c \cap B^c) = 1 - P(A \cup B)$.
Therefore,$P(A^c \cap B^c) = 1 - 0.6 = 0.4$.

(A) Given,$P(A)=0.4, P(B)=0.3$ and $P(A \cap B)=0.2$.
We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\overline{A} \cap \overline{B})$.
By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
First,calculate $P(A \cup B)$ using the formula:
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.2 = 0.5$.
Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.5 = 0.5$.

(C) $A =$ Event that $A$ speaks the truth.
$B =$ Event that $B$ speaks the truth.
$P(A) = 4/5 \implies P(A^c) = 1/5$.
$P(B) = 3/4 \implies P(B^c) = 1/4$.
$A$ and $B$ contradict each other if one speaks the truth and the other lies.
Required probability $= P(A) \cdot P(B^c) + P(A^c) \cdot P(B)$.
$= (4/5 \times 1/4) + (1/5 \times 3/4)$.
$= 4/20 + 3/20 = 7/20$.

(C) Since $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.
Given $P(A \cap B) = \frac{1}{6}$,let $P(A) = x$ and $P(B) = y$. Thus,$xy = \frac{1}{6}$.
The probability that neither occurs is $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{3}$.
Therefore,$P(A \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $x + y - \frac{1}{6} = \frac{2}{3}$,which implies $x + y = \frac{5}{6}$.
Substituting $y = \frac{5}{6} - x$ into $xy = \frac{1}{6}$,we get $x(\frac{5}{6} - x) = \frac{1}{6}$.
This simplifies to $6x^2 - 5x + 1 = 0$.
Factoring the quadratic equation,we get $(2x - 1)(3x - 1) = 0$.
Thus,$x = \frac{1}{2}$ or $x = \frac{1}{3}$.
Hence,the probability of occurrence of $A$ is $\frac{1}{2}$ or $\frac{1}{3}$.

(B) The probability that none of the $n$ independent events $X_1, X_2, \ldots, X_n$ occur is given by $P(X_1' \cap X_2' \cap \ldots \cap X_n')$.
Since the events are independent,this is equal to $P(X_1') P(X_2') \ldots P(X_n')$.
Given $P(X_r) = \frac{1}{r+1}$,the probability of the complement event is $P(X_r') = 1 - P(X_r) = 1 - \frac{1}{r+1} = \frac{r}{r+1}$.
Thus,the required probability is:
$P(X_1') P(X_2') \ldots P(X_n') = \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \ldots \left(1 - \frac{1}{n+1}\right)$
$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{n}{n+1}$
$= \frac{1}{n+1}$.

(A) Let the probabilities of horses $A$,$B$,and $C$ winning be $a$,$b$,and $c$ respectively.
Since one of them must win,the sum of their probabilities is $a + b + c = 1$ ... $(i)$
Given that $a = 2b$ and $b = 2c$.
Substituting $b = 2c$ into the expression for $a$,we get $a = 2(2c) = 4c$.
Now,substitute $a = 4c$ and $b = 2c$ into equation $(i)$:
$4c + 2c + c = 1$
$7c = 1$
$c = \frac{1}{7}$
Now,find $b$ and $a$:
$b = 2c = 2 \times \frac{1}{7} = \frac{2}{7}$
$a = 2b = 2 \times \frac{2}{7} = \frac{4}{7}$
Therefore,the probabilities of horses $A$,$B$,and $C$ winning are $\frac{4}{7}, \frac{2}{7}, \frac{1}{7}$ respectively.

(B) Total number of ways to select $2$ students out of $100$ is ${ }^{100} C_2$.
Number of ways in which both you and your friend are placed in the first section (strength $40$) is ${ }^{40} C_2$.
Number of ways in which both you and your friend are placed in the second section (strength $60$) is ${ }^{60} C_2$.
Since these are mutually exclusive events,the total number of favorable ways is ${ }^{40} C_2 + { }^{60} C_2$.
Therefore,the required probability is $\frac{{ }^{40} C_2 + { }^{60} C_2}{{ }^{100} C_2}$.

(A) The given equation of the pair of lines is $x^2+2 \sqrt{2} x y+k y^2=0$.
Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=\sqrt{2}, b=k$.
The angle $\theta$ between the lines is given by $\tan \theta = \frac{2\sqrt{h^2-ab}}{a+b}$.
Given $\theta = 45^{\circ}$,so $\tan 45^{\circ} = 1 = \frac{2\sqrt{2-k}}{1+k}$.
Squaring both sides,$(1+k)^2 = 4(2-k)$ $\Rightarrow k^2+2k+1 = 8-4k$ $\Rightarrow k^2+6k-7=0$.
Factoring gives $(k+7)(k-1)=0$. Since $k>0$,we have $k=1$.
The equation of the pair of lines becomes $x^2+2\sqrt{2}xy+y^2=0$.
The equation of the angle bisectors is $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$ $\Rightarrow \frac{x^2-y^2}{1-1} = \frac{xy}{\sqrt{2}}$ $\Rightarrow x^2-y^2=0$.
This represents two lines: $x-y=0$ and $x+y=0$.
The third line is $x+2y+1=0$.
The vertices of the triangle are the intersection points of these three lines:
$1$) $x-y=0$ and $x+y=0 \Rightarrow (0,0)$.
$2$) $x-y=0$ and $x+2y+1=0$ $\Rightarrow -3y=1$ $\Rightarrow y=-1/3, x=-1/3$.
$3$) $x+y=0$ and $x+2y+1=0 \Rightarrow y=-1, x=1$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area $= \frac{1}{2} |0(-1/3 - (-1)) + (-1/3)(-1 - 0) + 1(0 - (-1/3))| = \frac{1}{2} |0 + 1/3 + 1/3| = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.

(D) Homogenize the equation of the pair of lines $x^2+y^2-2x-4y+2=0$ using the line $x+y=k$,which can be written as $\frac{x+y}{k}=1$.
Substituting this into the equation,we get:
$x^2+y^2-2x(\frac{x+y}{k})-4y(\frac{x+y}{k})+2(\frac{x+y}{k})^2=0$.
Since the lines $OA$ and $OB$ are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero.
Expanding the equation:
$x^2+y^2-\frac{2}{k}(x^2+xy)-\frac{4}{k}(xy+y^2)+\frac{2}{k^2}(x^2+2xy+y^2)=0$.
Grouping the terms:
$(1-\frac{2}{k}+\frac{2}{k^2})x^2 + (1-\frac{4}{k}+\frac{2}{k^2})y^2 + (\dots)xy = 0$.
Setting the sum of coefficients of $x^2$ and $y^2$ to zero:
$(1-\frac{2}{k}+\frac{2}{k^2}) + (1-\frac{4}{k}+\frac{2}{k^2}) = 0$.
$2 - \frac{6}{k} + \frac{4}{k^2} = 0$.
Multiplying by $k^2$:
$2k^2 - 6k + 4 = 0 \Rightarrow k^2 - 3k + 2 = 0$.
$(k-1)(k-2) = 0$.
Thus,$k=1$ or $k=2$.
Given $k>1$,the value is $k=2$.