AP EAMCET 2021 Mathematics Question Paper with Answer and Solution

(D) rational fraction $\frac{p(x)}{q(x)}$ is called an improper rational fraction if the degree of the numerator $p(x)$ is greater than or equal to the degree of the denominator $q(x)$.
$(a)$ For $\frac{x^2+1}{(x^2+2)(x^2+x+1)}$,degree of $p(x) = 2$ and degree of $q(x) = 4$. Since $2 < 4$,it is a proper fraction.
$(b)$ For $\frac{x^2+1}{(x+3)(x^2-x+1)}$,degree of $p(x) = 2$ and degree of $q(x) = 3$. Since $2 < 3$,it is a proper fraction.
$(c)$ For $\frac{x}{x^2+3x+1}$,degree of $p(x) = 1$ and degree of $q(x) = 2$. Since $1 < 2$,it is a proper fraction.
$(d)$ For $\frac{x^2+1}{x^2-1}$,degree of $p(x) = 2$ and degree of $q(x) = 2$. Since the degree of the numerator is equal to the degree of the denominator,it is an improper rational fraction.

(A) Let the vertices of the parallelogram be $A(2,4,-1)$,$B(3,6,-1)$,and $C(4,5,1)$.
Let the fourth vertex be $D(x, y, z)$.
In a parallelogram,the diagonals bisect each other,which means the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
Midpoint of $AC = \left( \frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2} \right) = \left( 3, \frac{9}{2}, 0 \right)$.
Midpoint of $BD = \left( \frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2} \right)$.
Equating the midpoints:
$\frac{3+x}{2} = 3 \Rightarrow 3+x = 6 \Rightarrow x = 3$.
$\frac{6+y}{2} = \frac{9}{2} \Rightarrow 6+y = 9 \Rightarrow y = 3$.
$\frac{-1+z}{2} = 0 \Rightarrow -1+z = 0 \Rightarrow z = 1$.
Thus,the fourth vertex $D$ is $(3,3,1)$.

(A) Let the points be $A(1, 2)$,$B(2, -1)$,$C(-1, 0)$,and $D(2, 0)$.
The equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(y - y_1) = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
For line $AB$ passing through $(1, 2)$ and $(2, -1)$:
$(y - 2) = \frac{-1 - 2}{2 - 1}(x - 1) \Rightarrow (y - 2) = -3(x - 1) \Rightarrow y - 2 = -3x + 3 \Rightarrow 3x + y = 5$.
For line $CD$ passing through $(-1, 0)$ and $(2, 0)$:
Since the $y$-coordinates are both $0$,the line is the $x$-axis,which is $y = 0$.
To find the intersection,substitute $y = 0$ into the equation $3x + y = 5$:
$3x + 0 = 5 \Rightarrow x = \frac{5}{3}$.
Thus,the point of intersection is $(\frac{5}{3}, 0)$,which in vector form is $\frac{5}{3} \hat{i}$.

(A) Let the vertices of the triangle be $A(x, y, z)$,$B(-2, 3, 4)$,and $C(3, -1, 5)$.
The centroid $G$ 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 $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given that the centroid is the origin $(0, 0, 0)$,we have:
$\frac{x-2+3}{3} = 0 \Rightarrow x+1 = 0 \Rightarrow x = -1$
$\frac{y+3-1}{3} = 0 \Rightarrow y+2 = 0 \Rightarrow y = -2$
$\frac{z+4+5}{3} = 0 \Rightarrow z+9 = 0 \Rightarrow z = -9$
Therefore,the third vertex is $(-1, -2, -9)$.

(B) Total number of balls = $5 + 7 = 12$.
We need to draw $9$ balls from $12$ balls.
The total number of ways to select $9$ balls is $n(S) = {}^{12}C_9 = {}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
We need to select $3$ white balls from $5$ white balls and $6$ black balls from $7$ black balls.
The number of ways to select $3$ white balls is ${}^5C_3 = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
The number of ways to select $6$ black balls is ${}^7C_6 = {}^7C_1 = 7$.
The number of favorable outcomes is $n(P) = 10 \times 7 = 70$.
The probability is $P = \frac{n(P)}{n(S)} = \frac{70}{220} = \frac{7}{22}$.

(C) Let $A$ be the event of getting a road contract and $B$ be the event of getting a building contract.
Given: $P(A) = \frac{2}{9}$,$P(B) = \frac{5}{9}$,and $P(A \cap B) = \frac{1}{6}$.
We need to find the probability of getting neither contract,which is $P(A' \cap B') = P((A \cup B)')$.
First,calculate the probability of getting at least one contract:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = \frac{2}{9} + \frac{5}{9} - \frac{1}{6} = \frac{7}{9} - \frac{1}{6}$
Finding a common denominator $(18)$:
$P(A \cup B) = \frac{14}{18} - \frac{3}{18} = \frac{11}{18}$.
Now,the probability of getting neither contract is:
$P((A \cup B)') = 1 - P(A \cup B) = 1 - \frac{11}{18} = \frac{7}{18}$.
Thus,the probability is $\frac{7}{18}$.

(C) Given,$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$.
First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{2} = \frac{4+3-6}{12} = \frac{1}{12}$.
Check for independence: $P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$. Since $P(A \cap B) = P(A) \cdot P(B)$,$A$ and $B$ are independent. (Option $A$ is correct).
Calculate $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{1/4} = \frac{1}{3}$. (Option $B$ is correct).
Calculate $P(A^C \cap B) = P(B) - P(A \cap B) = \frac{1}{4} - \frac{1}{12} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$. (Option $C$ is incorrect,as it states $1/3$).
Calculate $P(A \cap B^C) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{12} = \frac{4-1}{12} = \frac{3}{12} = \frac{1}{4}$. (Option $D$ is correct).
Thus,the incorrect statement is $P(A^C \cap B) = \frac{1}{3}$.

(B) The set $A$ is given by $A = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. The total number of elements in $A$ is $n(A) = 10$.
The total number of ways to choose $3$ numbers from $10$ is $n(E) = {}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
For the minimum to be $3$ and the maximum to be $7$,the chosen set must contain $3$ and $7$. The third number must be chosen from the set $\{4, 5, 6\}$.
Thus,the favorable outcomes are $\{3, 4, 7\}, \{3, 5, 7\}, \{3, 6, 7\}$.
The number of favorable outcomes is $n(F) = 3$.
The required probability is $P = \frac{n(F)}{n(E)} = \frac{3}{120} = \frac{1}{40}$.

(C) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Rearranging the terms,we get $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.
Substituting the given values,$P(A) + P(B) = 0.65 + 0.15 = 0.8$.
We need to find $P(A^C) + P(B^C)$.
Using the complement rule,$P(A^C) = 1 - P(A)$ and $P(B^C) = 1 - P(B)$.
Therefore,$P(A^C) + P(B^C) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the sum $P(A) + P(B) = 0.8$,we get $2 - 0.8 = 1.2$.

(C) The set $S$ consists of equations $x^2+bx+c=0$ where $b, c \in \{1, 2, 3, 4, 5, 6\}$.
The total number of such equations is $n(S) = 6 \times 6 = 36$.
$A$ quadratic equation $x^2+bx+c=0$ has real roots if the discriminant $D = b^2-4ac \geq 0$.
Since $a=1$,the condition becomes $b^2 \geq 4c$.
We test values of $b$ from $1$ to $6$:
If $b=1$,$1 \geq 4c$ (No values for $c$)
If $b=2$,$4 \geq 4c \implies c \leq 1$ ($c=1$,$1$ case)
If $b=3$,$9 \geq 4c \implies c \leq 2.25$ ($c=1, 2$,$2$ cases)
If $b=4$,$16 \geq 4c \implies c \leq 4$ ($c=1, 2, 3, 4$,$4$ cases)
If $b=5$,$25 \geq 4c \implies c \leq 6.25$ ($c=1, 2, 3, 4, 5, 6$,$6$ cases)
If $b=6$,$36 \geq 4c \implies c \leq 9$ ($c=1, 2, 3, 4, 5, 6$,$6$ cases)
Total favorable outcomes $n(E) = 0 + 1 + 2 + 4 + 6 + 6 = 19$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{19}{36}$.

(C) Total number of items = $2 \text{ (bolts)} + 2 \text{ (nuts)} + 3 \text{ (needles)} = 7 \text{ items}$.
Total number of ways to choose $2$ parts from $7$ is given by $n(S) = {}^{7}C_{2} = \frac{7 \times 6}{2 \times 1} = 21$.
Let $E$ be the event that one part is a bolt and one part is a needle.
The number of ways to choose $1$ bolt from $2$ and $1$ needle from $3$ is $n(E) = {}^{2}C_{1} \times {}^{3}C_{1} = 2 \times 3 = 6$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{21}$.

(B) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $P(A) = 1 - P(A^C) = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting the values into the addition theorem of probability:
$\frac{5}{6} = \frac{1}{2} + P(B) - \frac{1}{3}$.
$\frac{5}{6} = \frac{3-2}{6} + P(B) = \frac{1}{6} + P(B)$.
$P(B) = \frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.
Therefore,$P(B^C) = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3}$.

(A) Since $A$ and $B$ are mutually exclusive events,$A \cap B = \phi$,which implies $P(A \cap B) = 0$.
Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{3}{7}$.
We need to find $P(A / A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$.
Since $A \subset (A \cup B)$,we have $A \cap (A \cup B) = A$,so $P(A \cap (A \cup B)) = P(A) = \frac{1}{4}$.
Also,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{3}{7} - 0 = \frac{7 + 12}{28} = \frac{19}{28}$.
Therefore,$P(A / A \cup B) = \frac{1/4}{19/28} = \frac{1}{4} \times \frac{28}{19} = \frac{7}{19}$.

(A) Since $A, B, C$ are mutually exclusive,we have $P(A \cap B) = P(B \cap C) = P(C \cap A) = P(A \cap B \cap C) = 0$.
For any event $E$,$0 \leq P(E) \leq 1$.
$1$. For $P(A) = \frac{3x+1}{3}$: $0 \leq \frac{3x+1}{3} \leq 1 \implies 0 \leq 3x+1 \leq 3 \implies -1 \leq 3x \leq 2 \implies -\frac{1}{3} \leq x \leq \frac{2}{3}$.
$2$. For $P(B) = \frac{1-x}{4}$: $0 \leq \frac{1-x}{4} \leq 1 \implies 0 \leq 1-x \leq 4 \implies -1 \leq -x \leq 3 \implies -3 \leq x \leq 1$.
$3$. For $P(C) = \frac{1-2x}{2}$: $0 \leq \frac{1-2x}{2} \leq 1 \implies 0 \leq 1-2x \leq 2 \implies -1 \leq -2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2}$.
Intersection of these intervals: $[-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] = [-\frac{1}{3}, \frac{1}{2}]$.
Also,for mutually exclusive events,$P(A \cup B \cup C) = P(A) + P(B) + P(C) \leq 1$.
$\frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \leq 1$
$\frac{4(3x+1) + 3(1-x) + 6(1-2x)}{12} \leq 1$
$\frac{12x+4+3-3x+6-12x}{12} \leq 1 \implies \frac{13-3x}{12} \leq 1 \implies 13-3x \leq 12 \implies -3x \leq -1 \implies x \geq \frac{1}{3}$.
Combining $x \geq \frac{1}{3}$ with the intersection $[-\frac{1}{3}, \frac{1}{2}]$,we get $x \in [\frac{1}{3}, \frac{1}{2}]$.

(B) Let the winning probability of $C$ be $P(C) = p$.
Since $A$ and $B$ are twice as likely to win as $C$,we have $P(A) = 2p$ and $P(B) = 2p$.
Since the sum of probabilities of all possible outcomes is $1$,we have $P(A) + P(B) + P(C) = 1$.
Substituting the values,we get $2p + 2p + p = 1$,which implies $5p = 1$,so $p = \frac{1}{5}$.
The probability that $B$ or $C$ wins is $P(B \cup C) = P(B) + P(C)$ (since the events are mutually exclusive).
$P(B \cup C) = 2p + p = 3p = 3 \times \frac{1}{5} = \frac{3}{5}$.

(B) Total number of cards $n(S) = 27$.
Let $E$ be the event that the number is even. The even numbers between $1$ and $27$ are $\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26\}$.
Thus,$n(E) = 13$.
Let $F$ be the event that the number is divisible by $5$. The numbers divisible by $5$ are $\{5, 10, 15, 20, 25\}$.
Thus,$n(F) = 5$.
The intersection $E \cap F$ contains numbers that are both even and divisible by $5$,which are multiples of $10$. These are $\{10, 20\}$.
Thus,$n(E \cap F) = 2$.
Using the addition rule for probability,$n(E \cup F) = n(E) + n(F) - n(E \cap F) = 13 + 5 - 2 = 16$.
The probability is $P(E \cup F) = \frac{n(E \cup F)}{n(S)} = \frac{16}{27}$.

(D) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.
Given,$P(A) = \frac{4}{5}$ and $P(B) = \frac{3}{4}$.
The probability that $A$ does not speak the truth is $P(\overline{A}) = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5}$.
The probability that $B$ does not speak the truth is $P(\overline{B}) = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$.
They contradict each other if one speaks the truth and the other lies.
Thus,the probability of contradiction is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = P(A) \cdot P(\overline{B}) + P(\overline{A}) \cdot P(B)$.
Substituting the values: $(\frac{4}{5} \times \frac{1}{4}) + (\frac{1}{5} \times \frac{3}{4}) = \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

(D) The set is $S = \{5, 6, \ldots, 35\}$. The number of elements is $35 - 5 + 1 = 31$.
The total number of ways to choose $2$ integers from $31$ is $^{31}C_2 = \frac{31 \times 30}{2} = 465$.
The difference between two integers is odd if and only if one integer is even and the other is odd.
In the set $\{5, 6, \ldots, 35\}$,the number of odd integers is $16$ (i.e.,$5, 7, \ldots, 35$) and the number of even integers is $15$ (i.e.,$6, 8, \ldots, 34$).
The number of ways to choose one odd and one even integer is $^{16}C_1 \times ^{15}C_1 = 16 \times 15 = 240$.
The probability that the difference is odd is $P = \frac{240}{465}$.
Dividing both numerator and denominator by $15$,we get $P = \frac{16}{31}$.

(B) $A, B, C$ are pairwise independent events.
$\Rightarrow P(A \cap B) = P(A) \cdot P(B) = P^2$
$P(B \cap C) = P(B) \cdot P(C) = P^2$
$P(C \cap A) = P(C) \cdot P(A) = P^2$
Given that all of them cannot occur simultaneously,$P(A \cap B \cap C) = 0$.
Using the inclusion-exclusion principle:
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$
$P(A \cup B \cup C) = P + P + P - [P^2 + P^2 + P^2] + 0$
$P(A \cup B \cup C) = 3P - 3P^2 = 3P(1-P)$.

(B) The prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ are $2, 3, 5, 7$.
$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.
The event $F = \{X < 4\}$ corresponds to $X \in \{1, 2, 3\}$.
$P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.12 = 0.50$.
The intersection $E \cap F$ consists of values that are both prime and less than $4$,which are $\{2, 3\}$.
$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.12 = 0.35$.
Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

(B) Since the axes are rotated through an angle $\theta = 45^{\circ}$,we replace $(x, y)$ with $(x \cos 45^{\circ} - y \sin 45^{\circ}, x \sin 45^{\circ} + y \cos 45^{\circ})$,which is $\left(\frac{x-y}{\sqrt{2}}, \frac{x+y}{\sqrt{2}}\right)$.
Substituting these into the equation $3x^2 + 3y^2 + 2xy = 2$:
$3\left(\frac{x-y}{\sqrt{2}}\right)^2 + 3\left(\frac{x+y}{\sqrt{2}}\right)^2 + 2\left(\frac{x-y}{\sqrt{2}}\right)\left(\frac{x+y}{\sqrt{2}}\right) = 2$
$\frac{3}{2}(x^2 + y^2 - 2xy) + \frac{3}{2}(x^2 + y^2 + 2xy) + \frac{2}{2}(x^2 - y^2) = 2$
$\frac{3}{2}(2x^2 + 2y^2) + (x^2 - y^2) = 2$
$3x^2 + 3y^2 + x^2 - y^2 = 2$
$4x^2 + 2y^2 = 2$
Dividing by $2$,we get $2x^2 + y^2 = 1$.

(C) We use the standard limit formula: $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$.
Given the expression: $\lim _{x \rightarrow \infty} (\frac{x+6}{x+1})^{x+4}$.
Rewrite the base: $\frac{x+6}{x+1} = \frac{x+1+5}{x+1} = 1 + \frac{5}{x+1}$.
So,the limit becomes $\lim _{x \rightarrow \infty} (1 + \frac{5}{x+1})^{x+4}$.
Using the property $\lim _{x \rightarrow \infty} (1 + \frac{k}{f(x)})^{g(x)} = e^{\lim _{x \rightarrow \infty} k \cdot \frac{g(x)}{f(x)}}$:
$= e^{\lim _{x \rightarrow \infty} 5 \cdot \frac{x+4}{x+1}}$.
$= e^{5 \cdot \lim _{x \rightarrow \infty} \frac{1 + 4/x}{1 + 1/x}} = e^{5 \cdot 1} = e^5$.