AP EAMCET 2025 Mathematics Question Paper with Answer and Solution

(C) Let the curves be $C_1: y^2 = 16x$ and $C_2: 9x^2 + \alpha y^2 = 25$.
First,find the point of intersection $(x_1, y_1)$.
From $C_1$,$y^2 = 16x$. Substituting this into $C_2$: $9x^2 + \alpha(16x) = 25 \implies 9x^2 + 16\alpha x - 25 = 0$.
For the curves to intersect at right angles,the product of their slopes at the point of intersection must be $-1$.
Differentiating $C_1$: $2y \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{8}{y}$.
Differentiating $C_2$: $18x + 2\alpha y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{9x}{\alpha y}$.
At the point of intersection,the product of slopes is $(\frac{8}{y})(-\frac{9x}{\alpha y}) = -1 \implies \frac{72x}{\alpha y^2} = 1$.
Since $y^2 = 16x$,we have $\frac{72x}{\alpha(16x)} = 1 \implies \frac{72}{16\alpha} = 1 \implies 16\alpha = 72 \implies \alpha = \frac{72}{16} = \frac{9}{2}$.

(C) Let the height of the tower be $H$ and the height of the building be $h$. The horizontal distance $d = 10 \sqrt{3}$.
From the top of the tower,the angle of depression to the foot of the building is $60^{\circ}$. Thus,$\tan(60^{\circ}) = \frac{H}{d}$.
$H = d \times \tan(60^{\circ}) = 10 \sqrt{3} \times \sqrt{3} = 10 \times 3 = 30$ units.
From the foot of the tower,the angle of elevation to the top of the building is $30^{\circ}$. Thus,$\tan(30^{\circ}) = \frac{h}{d}$.
$h = d \times \tan(30^{\circ}) = 10 \sqrt{3} \times \frac{1}{\sqrt{3}} = 10$ units.
The sum of the heights is $H + h = 30 + 10 = 40$ units.

(B) Let the height of the plane be $h = 5 \text{ km}$.
Let the positions of the plane be $A$ and $B$ at angles $15^{\circ}$ and $30^{\circ}$ respectively.
Let $O$ be the observer on the ground.
In $\triangle ODA$ (where $D$ is the point directly below $A$),$\tan(15^{\circ}) = \frac{h}{OD} \implies OD = \frac{5}{\tan(15^{\circ})} = 5 \cot(15^{\circ})$.
Using $\cot(15^{\circ}) = 2 + \sqrt{3}$,$OD = 5(2 + \sqrt{3}) = 10 + 5\sqrt{3} \text{ km}$.
In $\triangle OEB$ (where $E$ is the point directly below $B$),$\tan(30^{\circ}) = \frac{h}{OE} \implies OE = \frac{5}{\tan(30^{\circ})} = 5\sqrt{3} \text{ km}$.
The distance traveled by the plane is $d = OD - OE = 10 + 5\sqrt{3} - 5\sqrt{3} = 10 \text{ km}$.
The time taken is $t = 50 \text{ seconds} = \frac{50}{3600} \text{ hours} = \frac{1}{72} \text{ hours}$.
Speed $v = \frac{d}{t} = \frac{10}{1/72} = 720 \text{ kmph}$.

(D) Let the coordinates of $C$ be $(x, y, z)$.
Since $G(1,0,1)$ is the centroid of $\triangle ABC$,we have:
$\frac{1+3+x}{3} = 1 \implies 4+x = 3 \implies x = -1$
$\frac{-4+1+y}{3} = 0 \implies -3+y = 0 \implies y = 3$
$\frac{2+0+z}{3} = 1 \implies 2+z = 3 \implies z = 1$
So,$C = (-1, 3, 1)$.
Now,calculate $AG^2$:
$AG^2 = (1-1)^2 + (0-(-4))^2 + (1-2)^2 = 0^2 + 4^2 + (-1)^2 = 16 + 1 = 17$.
Calculate $CG^2$:
$CG^2 = (-1-1)^2 + (3-0)^2 + (1-1)^2 = (-2)^2 + 3^2 + 0^2 = 4 + 9 = 13$.
Thus,$AG^2 + CG^2 = 17 + 13 = 30$.
Calculate $BG^2$:
$BG^2 = (3-1)^2 + (1-0)^2 + (0-1)^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6$.
Comparing the values,$AG^2 + CG^2 = 30 = 5 \times 6 = 5 BG^2$.

(A) Let the points be $A(2, -5, 3)$ and $B(-1, -8, 0)$. The point $P(0, -7, 1)$ divides the line segment $AB$ in the ratio $k:1$.
Using the section formula:
$0 = \frac{k(-1) + 1(2)}{k+1} \implies -k + 2 = 0 \implies k = 2$.
So,$P$ divides $AB$ internally in the ratio $2:1$.
Since $Q$ is the harmonic conjugate of $P$ with respect to $AB$,$Q$ divides $AB$ externally in the ratio $2:1$.
Using the external section formula for $Q(\alpha, \beta, \gamma)$:
$\alpha = \frac{2(-1) - 1(2)}{2-1} = -2 - 2 = -4$.
$\beta = \frac{2(-8) - 1(-5)}{2-1} = -16 + 5 = -11$.
$\gamma = \frac{2(0) - 1(3)}{2-1} = 0 - 3 = -3$.
Thus,$Q = (-4, -11, -3)$.
We need to find $\alpha - \beta + \gamma = -4 - (-11) + (-3) = -4 + 11 - 3 = 4$.

(D) To determine the nature of the triangle formed by points $A(-1, 2, 3)$,$B(2, -3, 1)$,and $C(3, 1, -2)$,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$1$. Length of $AB$:
$AB = \sqrt{(2 - (-1))^2 + (-3 - 2)^2 + (1 - 3)^2} = \sqrt{3^2 + (-5)^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.
$2$. Length of $BC$:
$BC = \sqrt{(3 - 2)^2 + (1 - (-3))^2 + (-2 - 1)^2} = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.
$3$. Length of $AC$:
$AC = \sqrt{(3 - (-1))^2 + (1 - 2)^2 + (-2 - 3)^2} = \sqrt{4^2 + (-1)^2 + (-5)^2} = \sqrt{16 + 1 + 25} = \sqrt{42}$.
Since all sides $AB = \sqrt{38}$,$BC = \sqrt{26}$,and $AC = \sqrt{42}$ are unequal,the triangle is a scalene triangle.

(D) Let the vertices be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.
First,calculate the side lengths:
$AB = \sqrt{(1-2)^2 + (-3-(-1))^2 + (-5-1)^2} = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.
$BC = \sqrt{(3-1)^2 + (-4-(-3))^2 + (-4-(-5))^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$AC = \sqrt{(3-2)^2 + (-4-(-1))^2 + (-4-1)^2} = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
Observe that $AB^2 = 41$ and $BC^2 + AC^2 = 6 + 35 = 41$.
Since $AB^2 = BC^2 + AC^2$,the triangle is a right-angled triangle with the right angle at $C$.
For a right-angled triangle,the circumradius $R$ is half of the hypotenuse.
Here,the hypotenuse is $AB = \sqrt{41}$.
Therefore,$R = \frac{AB}{2} = \frac{\sqrt{41}}{2}$.

(D) The length of the sides of $\triangle ABC$ are calculated using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$c = AB = \sqrt{(2-2)^2 + (5-(-1))^2 + (1-1)^2} = \sqrt{0^2 + 6^2 + 0^2} = 6$.
$b = AC = \sqrt{(0-2)^2 + (-2-(-1))^2 + (3-1)^2} = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = 3$.
$a = BC = \sqrt{(0-2)^2 + (-2-5)^2 + (3-1)^2} = \sqrt{(-2)^2 + (-7)^2 + 2^2} = \sqrt{4+49+4} = \sqrt{57}$.
By the Angle Bisector Theorem,point $D$ divides $BC$ in the ratio $c:b$,which is $6:3 = 2:1$.
Using the section formula,the coordinates of $D$ are $\left(\frac{2(0)+1(2)}{2+1}, \frac{2(-2)+1(5)}{2+1}, \frac{2(3)+1(1)}{2+1}\right) = \left(\frac{2}{3}, \frac{1}{3}, \frac{7}{3}\right)$.
Now,$AD = \sqrt{(\frac{2}{3}-2)^2 + (\frac{1}{3}-(-1))^2 + (\frac{7}{3}-1)^2} = \sqrt{(-\frac{4}{3})^2 + (\frac{4}{3})^2 + (\frac{4}{3})^2} = \sqrt{\frac{16}{9} + \frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{48}{9}} = \frac{4\sqrt{3}}{3} = \frac{4}{\sqrt{3}}$.

(B) The total number of ways to draw $3$ balls from $12$ is given by $C(12, 3) = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
We want the probability that exactly $2$ balls have the same colour. This means $2$ balls are of one colour and $1$ ball is of a different colour.
Case $1$: $2$ red and $1$ non-red (green or white). Number of ways = $C(4, 2) \times C(8, 1) = 6 \times 8 = 48$.
Case $2$: $2$ green and $1$ non-green (red or white). Number of ways = $C(5, 2) \times C(7, 1) = 10 \times 7 = 70$.
Case $3$: $2$ white and $1$ non-white (red or green). Number of ways = $C(3, 2) \times C(9, 1) = 3 \times 9 = 27$.
Total favorable outcomes = $48 + 70 + 27 = 145$.
The probability is $\frac{145}{220} = \frac{29}{44}$.

(C) The total number of ways to choose $3$ squares from $64$ is $\binom{64}{3} = \frac{64 \times 63 \times 62}{3 \times 2 \times 1} = 41664$.
On a chess board,the diagonals have lengths $1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1$ in both directions.
For a diagonal of length $k$,the number of ways to choose $3$ squares is $\binom{k}{3}$.
We only consider diagonals with length $k \ge 3$.
The lengths are $3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3$.
There are two sets of these diagonals (one for each direction).
Sum of $\binom{k}{3}$ for one direction: $\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3} + \binom{7}{3} + \binom{8}{3} + \binom{7}{3} + \binom{6}{3} + \binom{5}{3} + \binom{4}{3} + \binom{3}{3} = 1 + 4 + 10 + 20 + 35 + 56 + 35 + 20 + 10 + 4 + 1 = 196$.
Since there are two directions,the total number of favorable outcomes is $196 \times 2 = 392$.
The probability is $\frac{392}{41664} = \frac{7}{744}$.

(A) The word $VARIABLE$ consists of $8$ letters: $V, A, R, I, A, B, L, E$. There are $4$ consonants $(V, R, B, L)$ and $4$ vowels $(A, A, I, E)$.
Total number of ways to form a $3$-letter word from $8$ letters is $P(8, 3) = 8 \times 7 \times 6 = 336$.
To find the number of words with a consonant in the middle,we fix the middle position with one of the $4$ consonants. This can be done in $4$ ways.
The remaining $2$ positions can be filled by the remaining $7$ letters in $P(7, 2) = 7 \times 6 = 42$ ways.
Total favorable words = $4 \times 42 = 168$.
Probability = $\frac{168}{336} = \frac{1}{2}$.
Note: The calculated probability is $\frac{1}{2}$. Given the options provided,there is a discrepancy in the question's options as $\frac{1}{2}$ is not listed.

(C) The set of possible values for $p$ and $q$ is $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of pairs $(p, q)$ is $6 \times 6 = 36$.
We need to find the number of distinct rational numbers $p/q$.
The distinct values are:
$1/1=1, 1/2, 1/3, 1/4, 1/5, 1/6$
$2/1=2, 2/2=1, 2/3, 2/4=1/2, 2/5, 2/6=1/3$
$3/1=3, 3/2, 3/3=1, 3/4, 3/5, 3/6=1/2$
$4/1=4, 4/2=2, 4/3, 4/4=1, 4/5, 4/6=2/3$
$5/1=5, 5/2, 5/3, 5/4, 5/5=1, 5/6$
$6/1=6, 6/2=3, 6/3=2, 6/4=3/2, 6/5, 6/6=1$
Listing the distinct values: $\{1, 2, 3, 4, 5, 6, 1/2, 1/3, 1/4, 1/5, 1/6, 2/3, 2/5, 3/2, 3/4, 3/5, 4/3, 4/5, 5/2, 5/3, 5/4, 5/6, 6/5\}$.
Counting these,we find there are $23$ distinct rational numbers.
$A$ proper fraction is a fraction where $p < q$.
The proper fractions are: $\{1/2, 1/3, 1/4, 1/5, 1/6, 2/3, 2/5, 3/4, 3/5, 4/5, 5/6\}$.
There are $11$ such distinct proper fractions.
Thus,the probability is $11/23$.

(C) Total number of children = $8 + 7 = 15$.
Total ways to choose $3$ slips from $15$ is $^{15}C_3 = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$.
Case $1$: One boy and two girls.
Ways = $^8C_1 \times ^7C_2 = 8 \times \frac{7 \times 6}{2} = 8 \times 21 = 168$.
Case $2$: One girl and two boys.
Ways = $^7C_1 \times ^8C_2 = 7 \times \frac{8 \times 7}{2} = 7 \times 28 = 196$.
Total favorable ways = $168 + 196 = 364$.
Probability = $\frac{364}{455}$.
Dividing both by $91$,we get $\frac{364 \div 91}{455 \div 91} = \frac{4}{5}$.

(D) The total number of ways to choose $3$ numbers from $30$ is given by $C(30, 3) = \frac{30 \times 29 \times 28}{3 \times 2 \times 1} = 4060$.
Next,we find the number of ways to choose $3$ consecutive numbers. These are $(1, 2, 3), (2, 3, 4), \dots, (28, 29, 30)$.
There are $28$ such sets of consecutive numbers.
The probability of choosing $3$ consecutive numbers is $P(E) = \frac{28}{4060} = \frac{1}{145}$.
The probability that they are not three consecutive numbers is $1 - P(E) = 1 - \frac{1}{145} = \frac{144}{145}$.

(B) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
Let $S$ be the sum of the numbers on the two dice. The possible values for $S$ are $\{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$.
Event $A$ is the event that the sum is a prime number: $A = \{2, 3, 5, 7, 11\}$.
Event $B$ is the event that the sum is greater than $8$: $B = \{9, 10, 11, 12\}$.
We need to find $P(A \cap \overline{B})$,which is the probability of the event $A$ occurring and $B$ not occurring.
This is equivalent to the event that the sum is a prime number $AND$ the sum is less than or equal to $8$.
$A \cap \overline{B} = \{2, 3, 5, 7\}$.
Now,we count the number of outcomes for each sum:
Sum $= 2: (1,1) \rightarrow 1$ outcome
Sum $= 3: (1,2), (2,1) \rightarrow 2$ outcomes
Sum $= 5: (1,4), (2,3), (3,2), (4,1) \rightarrow 4$ outcomes
Sum $= 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \rightarrow 6$ outcomes
Total outcomes for $A \cap \overline{B} = 1 + 2 + 4 + 6 = 13$.
Therefore,$P(A \cap \overline{B}) = \frac{13}{36}$.

(B) Given that $P(\bar{A}) = \frac{2}{3}$,we can find $P(A)$ using the complement rule: $P(A) = 1 - P(\bar{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.
We know the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.
Rearranging to solve for $P(B)$: $P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find $P(\bar{A} \cap B)$. Using the set theory property,$P(\bar{A} \cap B) = P(B) - P(A \cap B)$.
Substituting the values: $P(\bar{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.
Thus,the correct option is $B$.

(B) Total number of ways to choose $4$ shoes out of $8$ is $\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
Let $E$ be the event of getting at least one correct pair.
It is easier to calculate the probability of the complement event $E^c$,which is the event of getting no correct pairs.
To have no correct pairs,we must choose $4$ shoes such that no two shoes form a pair.
There are $4$ pairs. We must select $4$ pairs out of $4$ and then select $1$ shoe from each of these $4$ pairs.
Number of ways to select $4$ pairs out of $4$ is $\binom{4}{4} = 1$.
Number of ways to select $1$ shoe from each of the $4$ pairs is $2^4 = 16$.
So,the number of ways to choose $4$ shoes with no correct pair is $1 \times 16 = 16$.
Probability of no correct pair $P(E^c) = \frac{16}{70} = \frac{8}{35}$.
Probability of at least one correct pair $P(E) = 1 - P(E^c) = 1 - \frac{8}{35} = \frac{27}{35}$.

(B) Let $P(A)$ be the probability that candidate $A$ gets the job and $P(B)$ be the probability that candidate $B$ gets the job.
Given: $P(A) = 0.8$ and $P(B) = 0.7$.
Since the events are independent,the probability that both get the job is $P(A \cap B) = P(A) \times P(B) = 0.8 \times 0.7 = 0.56$.
The probability that at least one of them gets the job is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = 0.8 + 0.7 - 0.56 = 1.5 - 0.56 = 0.94$.

(D) Let $A$ be the event that the first student qualifies and $B$ be the event that the second student qualifies.
Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{2}{5}$.
Since the events are independent,the probability that neither qualifies is $P(A^c \cap B^c) = P(A^c) \times P(B^c)$.
$P(A^c) = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
$P(B^c) = 1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5}$.
Probability that neither qualifies = $\frac{3}{4} \times \frac{3}{5} = \frac{9}{20}$.
Probability that at least one qualifies = $1 - P(\text{neither qualifies}) = 1 - \frac{9}{20} = \frac{11}{20}$.

(A) Let $P(A)$ be the probability that student $A$ solves the problem,so $P(A) = \frac{2}{5}$.
Let $P(B)$ be the probability that student $B$ solves the problem,so $P(B) = \frac{3}{4}$.
The probability that student $A$ does not solve the problem is $P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$.
The probability that student $B$ does not solve the problem is $P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$.
The problem is solved if at least one of them solves it. This is the complement of the event that neither of them solves the problem.
Probability that neither solves the problem = $P(A') \times P(B') = \frac{3}{5} \times \frac{1}{4} = \frac{3}{20}$.
Probability that the problem is solved = $1 - P(\text{neither solves}) = 1 - \frac{3}{20} = \frac{17}{20}$.

(A) Let $P(A)$ be the probability that person $A$ completes the work,$P(A) = \frac{2}{3}$.
Let $P(B)$ be the probability that person $B$ completes the work,$P(B) = \frac{3}{4}$.
The work is completed if at least one of them completes the work.
This is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \times P(B) = \frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$.
Therefore,$P(A \cup B) = \frac{2}{3} + \frac{3}{4} - \frac{1}{2}$.
Finding a common denominator of $12$:
$P(A \cup B) = \frac{8}{12} + \frac{9}{12} - \frac{6}{12} = \frac{11}{12}$.
Thus,the probability that the work is completed is $\frac{11}{12}$.

(A) The total number of people is $9 + 5 = 14$.
We need to form a committee of $4$ members.
The total number of ways to select $4$ members from $14$ is given by $^{14}C_4 = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$.
We want the probability that the committee contains at least one woman.
It is easier to calculate the complement: the probability that the committee contains no women (i.e.,all $4$ members are men).
The number of ways to select $4$ men from $9$ men is $^{9}C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
The probability of selecting no women is $P(\text{No women}) = \frac{126}{1001} = \frac{18}{143}$.
The probability of selecting at least one woman is $1 - P(\text{No women}) = 1 - \frac{18}{143} = \frac{125}{143}$.

(D) Total number of girls $= 3 + 8 = 11$.
Number of ways to arrange $11$ girls around a circle is $(11 - 1)! = 10!$.
To find the number of ways where the $3$ sisters are $NOT$ seated together,we use the complement method: Total arrangements $-$ Arrangements where $3$ sisters are together.
Treating the $3$ sisters as a single unit,we have $8 + 1 = 9$ units to arrange in a circle,which can be done in $(9 - 1)! = 8!$ ways.
The $3$ sisters can be arranged among themselves in $3!$ ways.
So,arrangements where $3$ sisters are together $= 8! \times 3!$.
Required number of ways $= 10! - (8! \times 3!) = 10! - (8! \times 6)$.
$= 8! \times (10 \times 9 - 6) = 8! \times (90 - 6) = 8! \times 84$.

(C) The number of diagonals of a regular polygon having $n$ sides is given by $\frac{n(n-3)}{2}$.
$\therefore \frac{n(n-3)}{2} = 35$
$\Rightarrow n(n-3) = 70$
$\Rightarrow n^2 - 3n - 70 = 0$
$\Rightarrow n^2 - 10n + 7n - 70 = 0$
$\Rightarrow n(n - 10) + 7(n - 10) = 0$
$\Rightarrow (n - 10)(n + 7) = 0$
Since the number of sides $n$ must be positive,we have $n = 10$.

(C) Given,$S_n = \sum_{k=1}^{n} k^3$ and $T_n = \sum_{k=1}^{n} k$.
We know that the sum of the first $n$ natural numbers is $T_n = \frac{n(n+1)}{2}$.
The sum of the cubes of the first $n$ natural numbers is $S_n = \left[\frac{n(n+1)}{2}\right]^2$.
Substituting $T_n$ into the expression for $S_n$,we get $S_n = (T_n)^2$.
Therefore,the correct relation is $S_n = T_n^2$.

(B) Given the series $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots \infty$.
Adding $1$ to both sides,we get $y + 1 = 1 + \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots \infty$.
This is of the form $(1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \ldots$.
Here,$nx = \frac{3}{4}$ and $\frac{n(n+1)}{2!}x^2 = \frac{15}{32}$.
Solving for $n$ and $x$,we find $n = \frac{3}{2}$ and $x = \frac{1}{2}$.
Thus,$y + 1 = (1 - \frac{1}{2})^{-\frac{3}{2}} = (\frac{1}{2})^{-\frac{3}{2}} = 2^{\frac{3}{2}}$.
Squaring both sides,$(y + 1)^2 = 2^3 = 8$.
$y^2 + 2y + 1 = 8 \Rightarrow y^2 + 2y - 7 = 0$.

(D) Given equations are:
$1) \ 3 \cos^2 A + 2 \cos^2 B = 4$
$2) \ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$ $\Rightarrow 3 \sin A \cos A = 2 \sin B \cos B$ $\Rightarrow \frac{3}{2} \sin 2A = \sin 2B$
From $(1)$:
$3(1 - \sin^2 A) + 2(1 - \sin^2 B) = 4$
$5 - 3 \sin^2 A - 2 \sin^2 B = 4$
$3 \sin^2 A + 2 \sin^2 B = 1$
$3 \sin^2 A = 1 - 2 \sin^2 B = \cos 2B$
Consider $\cos(A + 2B) = \cos A \cos 2B - \sin A \sin 2B$
Substitute $\cos 2B = 3 \sin^2 A$ and $\sin 2B = \frac{3}{2} \sin 2A = 3 \sin A \cos A$:
$\cos(A + 2B) = \cos A (3 \sin^2 A) - \sin A (3 \sin A \cos A)$
$\cos(A + 2B) = 3 \sin^2 A \cos A - 3 \sin^2 A \cos A = 0$
Since $A, B$ are acute,$A + 2B$ must be $90^{\circ}$.

(D) We know that $\cosh 2x = \frac{1 + \tanh^2 x}{1 - \tanh^2 x}$.
Given $\cosh 2x = 199$,we have:
$\frac{1 + \tanh^2 x}{1 - \tanh^2 x} = 199$
$1 + \tanh^2 x = 199 - 199 \tanh^2 x$
$200 \tanh^2 x = 198$
$\tanh^2 x = \frac{198}{200} = \frac{99}{100}$
$\tanh x = \sqrt{\frac{99}{100}} = \frac{3 \sqrt{11}}{10}$
Since $\coth x = \frac{1}{\tanh x}$,we get:
$\coth x = \frac{10}{3 \sqrt{11}}$

(B) Given the equation $\cos A \cos B + \sin A \sin B \sin C = 1$ in $\triangle ABC$.
Since $\sin C \le 1$,we have $\cos A \cos B + \sin A \sin B \sin C \le \cos A \cos B + \sin A \sin B = \cos(A - B)$.
Since $\cos(A - B) \le 1$,the equality holds only if $\cos(A - B) = 1$ and $\sin C = 1$.
This implies $A = B$ and $C = 90^{\circ}$.
Since $A + B + C = 180^{\circ}$,we have $2A + 90^{\circ} = 180^{\circ}$,so $A = 45^{\circ}$ and $B = 45^{\circ}$.
Thus,$\sin A + \sin B + \sin C = \sin 45^{\circ} + \sin 45^{\circ} + \sin 90^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1 = \sqrt{2} + 1$.

(B) The parabola is given by $y^2=4x$,which is of the form $y^2=4ax$ with $a=1$. The focus is $S(a,0) = (1,0)$.
Since $PQ$ is a focal chord,the segments $PS$ and $SQ$ are related to the semi-latus rectum $l=2a=2$ by the harmonic mean property: $\frac{1}{PS} + \frac{1}{SQ} = \frac{1}{a} = \frac{1}{1} = 1$.
Given $P=(4,4)$ and $S=(1,0)$,the distance $PS = \sqrt{(4-1)^2 + (4-0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
Substituting $PS=5$ into the relation: $\frac{1}{5} + \frac{1}{SQ} = 1$.
$\frac{1}{SQ} = 1 - \frac{1}{5} = \frac{4}{5}$.
Therefore,$SQ = \frac{5}{4}$.

(A) The given equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$. Here $a^2=2$ and $b^2=1$,so $a=\sqrt{2}$ and $b=1$.
The equation of the tangent to the ellipse at point $P(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$.
Substituting the values,we get $\frac{x \cos \theta}{\sqrt{2}} + y \sin \theta = 1$.
For the $x$-intercept $(A)$,set $y=0$: $\frac{x \cos \theta}{\sqrt{2}} = 1 \Rightarrow x = \sqrt{2} \sec \theta$. So,$A = (\sqrt{2} \sec \theta, 0)$.
For the $y$-intercept $(B)$,set $x=0$: $y \sin \theta = 1 \Rightarrow y = \operatorname{cosec} \theta$. So,$B = (0, \operatorname{cosec} \theta)$.
Let $M(h, k)$ be the midpoint of $AB$. Then:
$h = \frac{\sqrt{2} \sec \theta + 0}{2} = \frac{\sec \theta}{\sqrt{2}} \Rightarrow \sec \theta = \sqrt{2}h$
$k = \frac{0 + \operatorname{cosec} \theta}{2} = \frac{\operatorname{cosec} \theta}{2} \Rightarrow \operatorname{cosec} \theta = 2k$
We know that $\cos^2 \theta + \sin^2 \theta = 1$,which implies $\frac{1}{\sec^2 \theta} + \frac{1}{\operatorname{cosec}^2 \theta} = 1$.
Substituting the values of $\sec \theta$ and $\operatorname{cosec} \theta$:
$\frac{1}{(\sqrt{2}h)^2} + \frac{1}{(2k)^2} = 1$
$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(C) The equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$.
Any point $P$ on the ellipse is $(\sqrt{2}\cos\theta, \sin\theta)$.
The equation of the tangent at $P$ is $\frac{x(\sqrt{2}\cos\theta)}{2} + \frac{y(\sin\theta)}{1} = 1$,which simplifies to $\frac{x\cos\theta}{\sqrt{2}} + y\sin\theta = 1$.
The intercepts made by the tangent on the coordinate axes are $A\left(\frac{\sqrt{2}}{\cos\theta}, 0\right)$ and $B\left(0, \frac{1}{\sin\theta}\right)$.
Let $(h, k)$ be the mid-point of $AB$. Then $h = \frac{\sqrt{2}}{2\cos\theta}$ and $k = \frac{1}{2\sin\theta}$.
This implies $\cos\theta = \frac{1}{\sqrt{2}h}$ and $\sin\theta = \frac{1}{2k}$.
Using the identity $\cos^2\theta + \sin^2\theta = 1$,we get $\left(\frac{1}{\sqrt{2}h}\right)^2 + \left(\frac{1}{2k}\right)^2 = 1$.
Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

(D) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The equation of the normal at a point $P(a \sec \theta, b \tan \theta)$ is $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2+b^2$,which simplifies to $ax \cos \theta + by \cot \theta = a^2+b^2$.
Similarly,the equation of the normal at $Q(a \sec \phi, b \tan \phi)$ is $ax \cos \phi + by \cot \phi = a^2+b^2$.
Given $\theta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - \theta$. Thus,$\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
The two equations are:
$(1) \quad ax \cos \theta + by \cot \theta = a^2+b^2$
$(2) \quad ax \sin \theta + by \tan \theta = a^2+b^2$
To find $k$ (the $y$-coordinate),we eliminate $x$ by multiplying $(1)$ by $\sin \theta$ and $(2)$ by $\cos \theta$:
$ax \cos \theta \sin \theta + by \cot \theta \sin \theta = (a^2+b^2) \sin \theta$
$ax \sin \theta \cos \theta + by \tan \theta \cos \theta = (a^2+b^2) \cos \theta$
Subtracting the two equations:
$by(\cot \theta \sin \theta - \tan \theta \cos \theta) = (a^2+b^2)(\sin \theta - \cos \theta)$
$by(\cos \theta - \sin \theta) = (a^2+b^2)(\sin \theta - \cos \theta)$
$by = -(a^2+b^2)$
$y = -\frac{a^2+b^2}{b}$
Therefore,$k = -\frac{a^2+b^2}{b}$.

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a) \cdot s(s-c)}{(s-b)(s-c) \cdot (s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.
Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.
Thus,$\frac{s}{s-b} = \frac{2s}{2s-2b} = \frac{a+b+c}{a+b+c-2b} = \frac{(a+c)+b}{(a+c)-b}$.
Given $a+c=5b$,substituting this into the expression gives $\frac{5b+b}{5b-b} = \frac{6b}{4b} = \frac{3}{2}$.

(D) Given,$r_1 = 2r_2 = 3r_3$.
Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have:
$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{3\Delta}{s-c}$.
Dividing by $\Delta$,we get $\frac{1}{s-a} = \frac{2}{s-b} = \frac{3}{s-c} = k$ (let).
Then $s-a = \frac{1}{k}$,$s-b = \frac{2}{k}$,and $s-c = \frac{3}{k}$.
Adding these,$(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.
So,$s = \frac{1+2+3}{k} = \frac{6}{k}$,which implies $k = \frac{6}{s}$.
Now,$s-a = \frac{1}{6/s} = \frac{s}{6}$ $\Rightarrow 6s - 6a = s$ $\Rightarrow 5s = 6a$ $\Rightarrow a = \frac{5s}{6}$.
And $s-b = \frac{2}{6/s} = \frac{2s}{6} = \frac{s}{3}$ $\Rightarrow 3s - 3b = s$ $\Rightarrow 2s = 3b$ $\Rightarrow b = \frac{2s}{3} = \frac{4s}{6}$.
Therefore,$a : b = \frac{5s}{6} : \frac{4s}{6} = 5 : 4$.