AP EAMCET 2021 Mathematics Question Paper with Answer and Solution

(A) The variance measures the spread of data points around the mean. The formula for sample variance is $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$.
For option $A$ $(1, 2, 3, 4, 5)$: Mean $\bar{x} = 3$. Variance $= \frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5-1} = \frac{4+1+0+1+4}{4} = \frac{10}{4} = 2.5$.
For option $B$ $(1, 1, 2, 3, 6)$: Mean $\bar{x} = 2.6$. Variance $= \frac{(1-2.6)^2 + (1-2.6)^2 + (2-2.6)^2 + (3-2.6)^2 + (6-2.6)^2}{4} = \frac{2.56 + 2.56 + 0.36 + 0.16 + 11.56}{4} = \frac{17.2}{4} = 4.3$.
For option $C$ $(1, 1, 2, 3, 5)$: Mean $\bar{x} = 2.4$. Variance $= \frac{(1-2.4)^2 + (1-2.4)^2 + (2-2.4)^2 + (3-2.4)^2 + (5-2.4)^2}{4} = \frac{1.96 + 1.96 + 0.16 + 0.36 + 6.76}{4} = \frac{11.2}{4} = 2.8$.
For option $D$ $(1, 1, 2, 2, 5)$: Mean $\bar{x} = 2.2$. Variance $= \frac{(1-2.2)^2 + (1-2.2)^2 + (2-2.2)^2 + (2-2.2)^2 + (5-2.2)^2}{4} = \frac{1.44 + 1.44 + 0.04 + 0.04 + 7.84}{4} = \frac{10.8}{4} = 2.7$.
Comparing the variances $(2.5, 4.3, 2.8, 2.7)$,the minimum variance is $2.5$ for option $A$.

(C) For set $A$: $\text{Mean} = \frac{\sum x_i}{5} = 2 \Rightarrow \sum x_i = 10$.
Variance $= \frac{1}{5} \sum x_i^2 - (\text{Mean})^2 = 4$ $\Rightarrow \frac{1}{5} \sum x_i^2 - 4 = 4$ $\Rightarrow \sum x_i^2 = 40$.
For set $B$: $\text{Mean} = \frac{\sum y_i}{5} = 4 \Rightarrow \sum y_i = 20$.
Variance $= \frac{1}{5} \sum y_i^2 - (\text{Mean})^2 = 5$ $\Rightarrow \frac{1}{5} \sum y_i^2 - 16 = 5$ $\Rightarrow \sum y_i^2 = 105$.
For $A \cup B$: Total elements $N = 10$.
Combined Mean $\bar{X} = \frac{\sum x_i + \sum y_i}{10} = \frac{10 + 20}{10} = 3$.
Combined Variance $= \frac{\sum x_i^2 + \sum y_i^2}{10} - (\bar{X})^2 = \frac{40 + 105}{10} - (3)^2 = 14.5 - 9 = 5.5$.

(D) Given data: $a, b, 8, 5, 10$.
Mean $= \frac{a+b+8+5+10}{5} = 6$.
$\Rightarrow a+b+23 = 30$ $\Rightarrow a+b = 7$ ...$(i)$
Variance $= \frac{\sum x_i^2}{n} - (\text{Mean})^2 = 6.80$.
$\frac{a^2+b^2+8^2+5^2+10^2}{5} - 6^2 = 6.80$.
$\frac{a^2+b^2+64+25+100}{5} - 36 = 6.80$.
$\frac{a^2+b^2+189}{5} = 42.80$.
$a^2+b^2+189 = 214 \Rightarrow a^2+b^2 = 25$ ...(ii)
From $(i)$,$b = 7-a$. Substituting in (ii):
$a^2 + (7-a)^2 = 25$.
$a^2 + 49 - 14a + a^2 = 25$.
$2a^2 - 14a + 24 = 0 \Rightarrow a^2 - 7a + 12 = 0$.
$(a-3)(a-4) = 0$.
So,$a=3, b=4$ or $a=4, b=3$.

(B) Given,$n=10$,$\Sigma x_i=60$,and $\Sigma x_i^2=1000$.
The mean $\bar{x} = \frac{\Sigma x_i}{n} = \frac{60}{10} = 6$.
The formula for variance is $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$.
Substituting the values:
$\sigma^2 = \frac{1000}{10} - (6)^2$
$\sigma^2 = 100 - 36$
$\sigma^2 = 64$.

(C) Let $d_i = x_i - 5$. We are given $\Sigma d_i = 3$ and $\Sigma d_i^2 = 43$ with $n = 18$.
The variance of a distribution is invariant under change of origin.
Therefore,the variance of $x_i$ is the same as the variance of $d_i$.
The formula for variance is $\sigma^2 = \frac{\Sigma d_i^2}{n} - \left( \frac{\Sigma d_i}{n} \right)^2$.
Substituting the values: $\sigma^2 = \frac{43}{18} - \left( \frac{3}{18} \right)^2$.
$\sigma^2 = \frac{43}{18} - \left( \frac{1}{6} \right)^2 = \frac{43}{18} - \frac{1}{36}$.
$\sigma^2 = \frac{86 - 1}{36} = \frac{85}{36} \approx 2.3611$.
Thus,the variance is approximately $2.36$.

(C) Given the set of numbers: $2, 3, 2x, 11$. The number of observations $n = 4$. The standard deviation $(SD)$ is $3.5 = \frac{7}{2}$.
First,calculate the mean $(\bar{x})$:
$\bar{x} = \frac{2 + 3 + 2x + 11}{4} = \frac{16 + 2x}{4} = \frac{8 + x}{2}$.
The variance $(V)$ is given by $V = (SD)^2 = (3.5)^2 = 12.25$.
The formula for variance is $V = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 3^2 + (2x)^2 + 11^2 = 4 + 9 + 4x^2 + 121 = 134 + 4x^2$.
$12.25 = \frac{134 + 4x^2}{4} - \left(\frac{8 + x}{2}\right)^2$.
$12.25 = \frac{134 + 4x^2}{4} - \frac{64 + 16x + x^2}{4}$.
$12.25 = \frac{134 + 4x^2 - 64 - 16x - x^2}{4}$.
$49 = 3x^2 - 16x + 70$.
$3x^2 - 16x + 21 = 0$.
Solving the quadratic equation $3x^2 - 9x - 7x + 21 = 0$:
$3x(x - 3) - 7(x - 3) = 0$.
$(3x - 7)(x - 3) = 0$.
Therefore,$x = 3$ or $x = \frac{7}{3}$.

(D) The standard deviation measures the dispersion of a data set relative to its mean. $A$ smaller range or smaller differences between consecutive values indicate a smaller standard deviation.
For the given sets:
$A: 10, 20, 30, 40$ (Range $= 30$)
$B: 2, 4, 6, 8$ (Range $= 6$)
$C: 3, 6, 9, 12$ (Range $= 9$)
$D: 1, 2, 3, 4$ (Range $= 3$)
Since the set $1, 2, 3, 4$ has the smallest range and the values are closest to each other,it has the least standard deviation.

(D) The variance of $n$ observations is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^n x_i^2 - (\bar{x})^2$,where $\bar{x}$ is the mean of the observations.
Given that the mean $\bar{x} = 5$,variance $\sigma^2 = 0$,and $\sum_{i=1}^n x_i^2 = 400$.
Substituting these values into the formula:
$0 = \frac{1}{n}(400) - (5)^2$
$0 = \frac{400}{n} - 25$
$\frac{400}{n} = 25$
$n = \frac{400}{25} = 16$
Thus,the value of $n$ is $16$.

(B) Given,$3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$.
Squaring and adding both equations,we get:
$(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$
$9 \sin^2 A + 16 \cos^2 B + 24 \sin A \cos B + 16 \sin^2 B + 9 \cos^2 A + 24 \sin B \cos A = 37$
$9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$
Using the identities $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$9(1) + 16(1) + 24 \sin(A + B) = 37$
$25 + 24 \sin(A + B) = 37$
$24 \sin(A + B) = 12$
$\sin(A + B) = \frac{12}{24} = \frac{1}{2}$

(C) Given,$\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$.
Since $BC=a, AC=b, AB=c$,we have $\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$.
$\Rightarrow \frac{a+c+b}{a+c}+\frac{b+c+a}{b+c}=3$ $\Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3$.
$\Rightarrow \frac{b}{a+c}+\frac{a}{b+c}=1$ $\Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)$.
$\Rightarrow b^2+bc+a^2+ac=ab+ac+bc+c^2$.
$\Rightarrow a^2+b^2-c^2=ab$.
Dividing by $2ab$,we get $\frac{a^2+b^2-c^2}{2ab}=\frac{1}{2}$.
By the Cosine Rule,$\cos C = \frac{1}{2}$,so $C = 60^\circ = \frac{\pi}{3}$.
We need to find $\tan \frac{C}{8} = \tan \frac{\pi}{24}$.
Using $\tan \frac{\theta}{2} = \frac{1-\cos \theta}{\sin \theta}$,we have $\tan \frac{\pi}{24} = \frac{1-\cos(\pi/12)}{\sin(\pi/12)}$.
Since $\cos(\pi/12) = \frac{\sqrt{6}+\sqrt{2}}{4}$ and $\sin(\pi/12) = \frac{\sqrt{6}-\sqrt{2}}{4}$,
$\tan \frac{\pi}{24} = \frac{1-\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}} = \frac{4-\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} \times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}$.
$= \frac{4\sqrt{6}+4\sqrt{2}-6-\sqrt{12}-\sqrt{12}-2}{6-2} = \frac{4\sqrt{6}+4\sqrt{2}-8-2\sqrt{3}}{4} = \sqrt{6}+\sqrt{2}-2-\frac{\sqrt{3}}{2}$.
Wait,re-evaluating: $\tan \frac{\pi}{24} = \sqrt{6}-\sqrt{3}+\sqrt{2}-2$.

(B) Given that $AD$ is the altitude of $\triangle ABC$,so $\angle ADB = 90^{\circ}$.
In $\triangle BDP$,$\angle BPD = 180^{\circ} - 90^{\circ} - \frac{B}{3} = 90^{\circ} - \frac{B}{3}$.
Therefore,$\angle APB = 180^{\circ} - \angle BPD = 180^{\circ} - (90^{\circ} - \frac{B}{3}) = 90^{\circ} + \frac{B}{3}$.
In $\triangle ABP$,using the sine rule:
$\frac{AP}{\sin(\angle ABP)} = \frac{AB}{\sin(\angle APB)}$
$\frac{AP}{\sin(\frac{2B}{3})} = \frac{c}{\sin(90^{\circ} + \frac{B}{3})}$
$AP = \frac{c \sin(\frac{2B}{3})}{\cos(\frac{B}{3})}$
$AP = \frac{c \cdot 2 \sin(\frac{B}{3}) \cos(\frac{B}{3})}{\cos(\frac{B}{3})}$
$AP = 2c \sin \frac{B}{3}$

(C) Given expression: $\operatorname{cosec} A(\sin B \cos C + \cos B \sin C)$
Using the trigonometric identity $\sin(B + C) = \sin B \cos C + \cos B \sin C$,the expression becomes:
$\operatorname{cosec} A \cdot \sin(B + C)$
In a $\triangle ABC$,$A + B + C = \pi$,so $B + C = \pi - A$.
Therefore,$\sin(B + C) = \sin(\pi - A) = \sin A$.
Substituting this into the expression:
$\operatorname{cosec} A \cdot \sin A = \frac{1}{\sin A} \cdot \sin A = 1$.

(B) Given $BC = a, CA = b$ and $AB = c$.
To find $a: b: c$.
Since the sum of angles in a triangle is $180^{\circ}$,we have:
$\angle ACB = 180^{\circ} - (\angle BAC + \angle ABC)$
$\angle ACB = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
By using the sine rule,we have:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$
Substituting the values of the sine functions:
$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$
Multiplying by $1/2$,we get:
$a : b : c = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$
Multiplying the ratio by $2$,we get:
$a : b : c = 1 : \sqrt{3} : 2$.

(C) Given in $\triangle ABC$: $a=3, b=4$ and $\sin A = \frac{3}{4}$.
Using the Sine Rule: $\frac{\sin A}{a} = \frac{\sin B}{b}$.
Substituting the values: $\frac{\frac{3}{4}}{3} = \frac{\sin B}{4}$.
$\Rightarrow \frac{3}{4 \times 3} = \frac{\sin B}{4}$.
$\Rightarrow \frac{1}{4} = \frac{\sin B}{4}$.
$\Rightarrow \sin B = 1$.
Since $\sin B = 1$, we have $B = 90^{\circ}$.
Therefore, $\angle CBA = 90^{\circ}$.

(C) Given in $\triangle ABC$,$A=75^{\circ}$ and $B=45^{\circ}$.
Using the angle sum property of a triangle,$C = 180^{\circ} - (75^{\circ} + 45^{\circ}) = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$b = k \sin 45^{\circ} = k \frac{1}{\sqrt{2}}$ and $c = k \sin 60^{\circ} = k \frac{\sqrt{3}}{2}$.
Also,$a = k \sin 75^{\circ} = k \sin(45^{\circ}+30^{\circ}) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = k \frac{\sqrt{3}+1}{2\sqrt{2}}$.
We need to find $b + c\sqrt{2}$.
$b + c\sqrt{2} = k \frac{1}{\sqrt{2}} + k \frac{\sqrt{3}}{2} \cdot \sqrt{2} = k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} \right) = k \frac{\sqrt{3}+1}{\sqrt{2}}$.
Since $a = k \frac{\sqrt{3}+1}{2\sqrt{2}}$,we have $k = \frac{2\sqrt{2}a}{\sqrt{3}+1}$.
Substituting $k$ back,$b + c\sqrt{2} = \left( \frac{2\sqrt{2}a}{\sqrt{3}+1} \right) \left( \frac{\sqrt{3}+1}{\sqrt{2}} \right) = 2a$.

(D) Given: $a=6 \text{ cm}$,$b=10 \text{ cm}$,$c=14 \text{ cm}$.
Using the Law of Cosines:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6^2 + 10^2 - 14^2}{2 \times 6 \times 10} = \frac{36 + 100 - 196}{120} = \frac{-60}{120} = -\frac{1}{2}$.
Since $\cos C = -\frac{1}{2}$,$C = 120^{\circ}$,which is an obtuse angle.
The sum of all angles in a triangle is $180^{\circ}$.
Therefore,the sum of the remaining two angles (which must be acute) is $A + B = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

(A) Let the sides of the triangle be $a = x^2+x+1$,$b = 2x+1$,and $c = x^2-1$.
Since $x^2+x+1$ is the greatest side,the greatest angle $A$ is opposite to side $a$.
Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Substituting the values: $\cos A = \frac{(2x+1)^2 + (x^2-1)^2 - (x^2+x+1)^2}{2(2x+1)(x^2-1)}$.
Expanding the terms: $(2x+1)^2 = 4x^2+4x+1$,$(x^2-1)^2 = x^4-2x^2+1$,and $(x^2+x+1)^2 = x^4+x^2+1+2x^3+2x^2+2x = x^4+2x^3+3x^2+2x+1$.
Numerator: $(4x^2+4x+1) + (x^4-2x^2+1) - (x^4+2x^3+3x^2+2x+1) = -2x^3-x^2+2x+1$.
Factoring the numerator: $-(2x^3+x^2-2x-1) = -[x^2(2x+1) - 1(2x+1)] = -(x^2-1)(2x+1)$.
Thus,$\cos A = \frac{-(x^2-1)(2x+1)}{2(2x+1)(x^2-1)} = -\frac{1}{2}$.
Therefore,$A = 120^{\circ}$.

(D) We are given the expression $b^2 \sin 2C + c^2 \sin 2B$.
Using the double angle formula $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$= b^2 (2 \sin C \cos C) + c^2 (2 \sin B \cos B)$
Using the Sine Rule,$\sin C = \frac{c}{2R}$ and $\sin B = \frac{b}{2R}$:
$= 2b^2 \left(\frac{c}{2R}\right) \cos C + 2c^2 \left(\frac{b}{2R}\right) \cos B$
$= \frac{b^2 c \cos C}{R} + \frac{c^2 b \cos B}{R} = \frac{bc}{R} (b \cos C + c \cos B)$
By the projection formula,$b \cos C + c \cos B = a$:
$= \frac{bc}{R} (a) = \frac{abc}{R}$
Since $R = \frac{abc}{4\Delta}$,it follows that $\frac{abc}{R} = 4\Delta$.

(C) Given,$b = 2$,$c = 3$,and $\angle B = \frac{\pi}{6}$.
Using the cosine rule in $\triangle ABC$:
$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
Substituting the given values:
$\cos \frac{\pi}{6} = \frac{a^2 + 3^2 - 2^2}{2 \cdot a \cdot 3}$
$\frac{\sqrt{3}}{2} = \frac{a^2 + 9 - 4}{6a}$
$\frac{\sqrt{3}}{2} = \frac{a^2 + 5}{6a}$
Multiplying both sides by $6a$:
$3\sqrt{3} a = a^2 + 5$
Rearranging the terms,we get:
$a^2 - 3\sqrt{3} a + 5 = 0$

(C) Given $2 \Delta^2 = \frac{a^2 b^2 c^2}{a^2+b^2+c^2}$.
Using the relation $\Delta = \frac{abc}{4R}$,we have $a^2 b^2 c^2 = (4R \Delta)^2 = 16 R^2 \Delta^2$.
Substituting this into the given equation: $2 \Delta^2 = \frac{16 R^2 \Delta^2}{a^2+b^2+c^2}$.
Dividing by $2 \Delta^2$,we get $a^2+b^2+c^2 = 8 R^2$.
Using $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,we get $4R^2(\sin^2 A + \sin^2 B + \sin^2 C) = 8R^2$.
Thus,$\sin^2 A + \sin^2 B + \sin^2 C = 2$.
Using the identity $\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C$,we have $2 + 2 \cos A \cos B \cos C = 2$.
This implies $2 \cos A \cos B \cos C = 0$,so $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.
Therefore,one of the angles must be $90^{\circ}$,which means the triangle is right-angled.

(A) Given expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$
$= (a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \frac{C}{2}$
$= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) + 2ab(\sin^2 \frac{C}{2} - \cos^2 \frac{C}{2})$
Since $\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2} = 1$ and $\cos C = \cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}$,we have:
$= (a^2 + b^2)(1) - 2ab(\cos C)$
Using the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$:
$= a^2 + b^2 - 2ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right)$
$= a^2 + b^2 - (a^2 + b^2 - c^2)$
$= c^2$

(A) Given that: $a \tan A + b \tan B = (a + b) \tan \left(\frac{A+B}{2}\right)$
Rearranging the terms: $a \left[ \tan A - \tan \left(\frac{A+B}{2}\right) \right] = b \left[ \tan \left(\frac{A+B}{2}\right) - \tan B \right]$
Using $\tan x - \tan y = \frac{\sin(x-y)}{\cos x \cos y}$:
$a \frac{\sin(A - \frac{A+B}{2})}{\cos A \cos \frac{A+B}{2}} = b \frac{\sin(\frac{A+B}{2} - B)}{\cos B \cos \frac{A+B}{2}}$
$a \frac{\sin(\frac{A-B}{2})}{\cos A} = b \frac{\sin(\frac{A-B}{2})}{\cos B}$
$\sin \left(\frac{A-B}{2}\right) \left( \frac{a}{\cos A} - \frac{b}{\cos B} \right) = 0$
Since $\frac{a}{\sin A} = \frac{b}{\sin B} = k$,we have $a = k \sin A$ and $b = k \sin B$:
$\sin \left(\frac{A-B}{2}\right) \left( \frac{k \sin A}{\cos A} - \frac{k \sin B}{\cos B} \right) = 0$
$\sin \left(\frac{A-B}{2}\right) (\tan A - \tan B) = 0$
This implies $\sin \left(\frac{A-B}{2}\right) = 0$ or $\tan A = \tan B$.
In both cases,$A = B$.

(B) Let $\triangle ABC$ be a triangle such that $\angle A: \angle B: \angle C = 1: 2: 3$.
Let the ratio constant be $x$,then $\angle A = x, \angle B = 2x, \angle C = 3x$.
Using the angle sum property of a triangle:
$\angle A + \angle B + \angle C = 180^{\circ}$
$x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
Thus,$\angle A = 30^{\circ}, \angle B = 60^{\circ}, \angle C = 90^{\circ}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$:
$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$
$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$
Multiplying by $1/2$,we get $a: b: c = 1: \sqrt{3}: 2$.

(C) Given,$s(s-a) = (s-b)(s-c)$.
We know that $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$ and $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$.
Squaring both,we get $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$.
Since $s(s-a) = (s-b)(s-c)$,it follows that $\sin^2 \frac{A}{2} = \cos^2 \frac{A}{2}$.
Dividing by $\cos^2 \frac{A}{2}$,we get $\tan^2 \frac{A}{2} = 1$.
Since $\frac{A}{2}$ is an angle of a triangle,$\frac{A}{2} = \frac{\pi}{4}$,which implies $A = \frac{\pi}{2}$.

(A) Given that in a $\triangle ABC$,$\tan(A/2)$,$\tan(B/2)$,and $\tan(C/2)$ are in Arithmetic Progression $(AP)$.
Therefore,$2 \tan(B/2) = \tan(A/2) + \tan(C/2)$.
Using the identity $\tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this can be solved,but a more direct trigonometric approach is:
$\tan(B/2) - \tan(A/2) = \tan(C/2) - \tan(B/2)$
$\frac{\sin(B/2)}{\cos(B/2)} - \frac{\sin(A/2)}{\cos(A/2)} = \frac{\sin(C/2)}{\cos(C/2)} - \frac{\sin(B/2)}{\cos(B/2)}$
$\frac{\sin(B/2-A/2)}{\cos(B/2)\cos(A/2)} = \frac{\sin(C/2-B/2)}{\cos(C/2)\cos(B/2)}$
$\frac{\sin((B-A)/2)}{\cos(A/2)} = \frac{\sin((C-B)/2)}{\cos(C/2)}$
Using $A/2 = 90^{\circ} - (B+C)/2$ and $C/2 = 90^{\circ} - (A+B)/2$,we get:
$\frac{\sin((B-A)/2)}{\sin((B+C)/2)} = \frac{\sin((C-B)/2)}{\sin((A+B)/2)}$
$\sin((A+B)/2)\sin((B-A)/2) = \sin((B+C)/2)\sin((C-B)/2)$
Using $2\sin x \sin y = \cos(x-y) - \cos(x+y)$:
$\cos A - \cos B = \cos B - \cos C$
$\cos A + \cos C = 2\cos B$
Thus,$\cos A$,$\cos B$,and $\cos C$ are in Arithmetic Progression.

(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-perimeter of the triangle.
Then,$ab - r_1 r_2 = ab - \frac{\Delta^2}{(s-a)(s-b)}$.
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $ab - r_1 r_2 = ab - s(s-c)$.
Substituting $s = \frac{a+b+c}{2}$,we get $ab - \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} = ab - \frac{(a+b)^2 - c^2}{4} = \frac{4ab - (a^2 + 2ab + b^2) + c^2}{4} = \frac{c^2 - (a-b)^2}{4} = \frac{c-(a-b)}{2} \cdot \frac{c+(a-b)}{2} = (s-a)(s-b)$.
Now,$\frac{ab - r_1 r_2}{r_3} = \frac{(s-a)(s-b)}{\frac{\Delta}{s-c}} = \frac{(s-a)(s-b)(s-c)}{\Delta} = \frac{\Delta^2}{s \Delta} = \frac{\Delta}{s} = r$.

(A) For $\triangle ABC$,the exradii are given by $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$ and the inradius is $r=\frac{\Delta}{s}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter.
We know the identity $r_1+r_2+r_3 = 4R+r$.
Rearranging this,we get $r_1+r_2+r_3-r = 4R$.
Given $r_1=2, r_2=3, r_3=6$,we can find $r$ using the relation $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.
$\frac{1}{r} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$.
Thus,$r=1$.
Substituting the values,$r_1+r_2+r_3-r = 2+3+6-1 = 10$.
Since $r_1+r_2+r_3-r = 4R$,we have $10 = 4R$,so $R = 2.5$.
The expression $r_1+r_2+r_3-r$ is equal to $4R$.

(A) Given,in $\triangle ABC$,$r_1=2, r_2=3, r_3=6$.
We know the formulas for exradii: $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$.
Thus,$s-a=\frac{\Delta}{2}, s-b=\frac{\Delta}{3}, s-c=\frac{\Delta}{6}$.
Adding these equations: $(s-a)+(s-b)+(s-c) = \Delta(\frac{1}{2}+\frac{1}{3}+\frac{1}{6})$.
$3s-(a+b+c) = \Delta(\frac{3+2+1}{6}) = \Delta$.
Since $a+b+c=2s$,we have $3s-2s = \Delta$,so $s=\Delta$.
Using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.
Substituting $s=\Delta$ and the expressions for $(s-a), (s-b), (s-c)$:
$\Delta = \sqrt{\Delta \cdot \frac{\Delta}{2} \cdot \frac{\Delta}{3} \cdot \frac{\Delta}{6}} = \sqrt{\frac{\Delta^4}{36}} = \frac{\Delta^2}{6}$.
Since $\Delta \neq 0$,we get $1 = \frac{\Delta}{6}$,so $\Delta = 6$.
Thus,$s = 6$.
Now,$s-a = \frac{6}{2} = 3 \Rightarrow a = 6-3 = 3$.
$s-b = \frac{6}{3} = 2 \Rightarrow b = 6-2 = 4$.
$s-c = \frac{6}{6} = 1 \Rightarrow c = 6-1 = 5$.
Therefore,$(a, b, c) = (3, 4, 5)$.

(A) Given: $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these values:
$\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-b}}\right)\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-c}}\right) = 2$
$\left(1-\frac{s-b}{s-a}\right)\left(1-\frac{s-c}{s-a}\right) = 2$
$\left(\frac{s-a-s+b}{s-a}\right)\left(\frac{s-a-s+c}{s-a}\right) = 2$
$\left(\frac{b-a}{s-a}\right)\left(\frac{c-a}{s-a}\right) = 2$
$(b-a)(c-a) = 2(s-a)^2$
$bc - ab - ac + a^2 = 2\left(\frac{b+c-a}{2}\right)^2$
$bc - ab - ac + a^2 = 2\frac{(b+c-a)^2}{4}$
$2(bc - ab - ac + a^2) = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac$
$2bc - 2ab - 2ac + 2a^2 = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac$
$2a^2 = b^2 + c^2 + a^2$
$a^2 = b^2 + c^2$
Since $a^2 = b^2 + c^2$,the triangle is a right-angled triangle.

(D) We know that in $\triangle ABC$,the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Also,$\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,$\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$,and $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$.
Substituting these into the expression:
$E = \frac{16 R s \Delta}{s-c} \cdot \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-a)(s-c)}{ac}} \cdot \sqrt{\frac{s(s-c)}{ab}}$
$E = \frac{16 R s \Delta}{s-c} \cdot \frac{(s-c) \sqrt{s(s-a)(s-b)(s-c)}}{abc}$
Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ and $abc = 4R\Delta$,we have:
$E = \frac{16 R s \Delta}{s-c} \cdot \frac{(s-c) \Delta}{4R\Delta} = \frac{16 R s \Delta}{4R} = 4s\Delta$.
Now,$r_1 r_2 r_3 = \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} = \frac{\Delta^3}{(s-a)(s-b)(s-c)}$.
Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.
Thus,$r_1 r_2 r_3 = \frac{\Delta^3}{\Delta^2/s} = s\Delta$.
Therefore,$4s\Delta = 4 r_1 r_2 r_3$.

(A) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{4k+5k+6k}{2} = \frac{15k}{2}$.
Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} \cdot \frac{7k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2}} = \sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{16} k^4} = \sqrt{\frac{1575}{16} k^4} = \frac{15\sqrt{7}}{4} k^2$.
Circumradius $R = \frac{abc}{4\Delta} = \frac{(4k)(5k)(6k)}{4 \cdot \frac{15\sqrt{7}}{4} k^2} = \frac{120k^3}{15\sqrt{7}k^2} = \frac{8}{\sqrt{7}} k$.
Inradius $r = \frac{\Delta}{s} = \frac{\frac{15\sqrt{7}}{4} k^2}{\frac{15k}{2}} = \frac{15\sqrt{7}}{4} \cdot \frac{2}{15} k = \frac{\sqrt{7}}{2} k$.
Ratio $R: r = \frac{8}{\sqrt{7}} k : \frac{\sqrt{7}}{2} k = \frac{8}{\sqrt{7}} : \frac{\sqrt{7}}{2} = \frac{16}{7}$.

(B) Given $a = |Z_1|, b = |Z_2|, c = |Z_3|$. Since $Z_1, Z_2, Z_3$ are non-zero,$a, b, c > 0$.
The determinant is given by $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$.
Expanding the determinant: $a(bc - a^2) - b(b^2 - ac) + c(ab - c^2) = 0$.
$abc - a^3 - b^3 + abc + abc - c^3 = 0$.
$3abc - (a^3 + b^3 + c^3) = 0$,which implies $a^3 + b^3 + c^3 - 3abc = 0$.
Using the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.
This can be written as $\frac{1}{2}(a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2) = 0$.
Since $a, b, c > 0$,$a + b + c \neq 0$.
Therefore,$(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$,which implies $a = b = c$.
However,looking at the options provided,the question implies a condition derived from the determinant. If $a=b=c$,then $|Z_1| = |Z_2| = |Z_3|$. Given the options,if we assume the question implies the sum condition or a specific property,we select the most logical outcome. Note: $a+b+c=0$ is impossible for non-zero complex numbers. The condition $a=b=c$ is the correct mathematical result.

(D) Given the system of equations $x+2y=3$ and $3x+6y=a-2$.
For a system of linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Comparing the given equations with the standard form,we have $a_1=1, b_1=2, c_1=3$ and $a_2=3, b_2=6, c_2=a-2$.
Substituting these values into the condition:
$\frac{1}{3} = \frac{2}{6} \neq \frac{3}{a-2}$.
Since $\frac{1}{3} = \frac{2}{6}$ is always true,we focus on the inequality $\frac{1}{3} \neq \frac{3}{a-2}$.
Cross-multiplying gives $a-2 \neq 3 \times 3$,which simplifies to $a-2 \neq 9$.
Therefore,$a \neq 11$.

(D) The identity is given by $\tanh^{-1} x = \frac{1}{2} \log_e \left( \frac{1+x}{1-x} \right)$.
For the logarithmic function $\log_e(u)$ to be defined,we must have $u > 0$.
Therefore,we require $\frac{1+x}{1-x} > 0$.
Let $f(x) = \frac{1+x}{1-x}$. The critical points are $x = -1$ and $x = 1$.
Testing the intervals:
For $x < -1$,$f(x) < 0$.
For $-1 < x < 1$,$f(x) > 0$.
For $x > 1$,$f(x) < 0$.
At $x = 1$ and $x = -1$,the expression is undefined.
Thus,the identity is valid only for $x \in (-1, 1)$.

(D) Given,square $ABCD$ with vertices $A(0,0), B(2,0), C(2,2), D(0,2)$.
Applying $f_1(x, y) \longrightarrow (y, x)$:
$A(0,0)$ $\longrightarrow A'(0,0), B(2,0)$ $\longrightarrow B'(0,2), C(2,2)$ $\longrightarrow C'(2,2), D(0,2)$ $\longrightarrow D'(2,0)$.
Applying $f_2(x, y) \longrightarrow (x+3y, y)$:
$A'(0,0)$ $\longrightarrow A''(0,0), B'(0,2)$ $\longrightarrow B''(6,2), C'(2,2)$ $\longrightarrow C''(8,2), D'(2,0)$ $\longrightarrow D''(2,0)$.
Applying $f_3(x, y) \longrightarrow \left(\frac{x-y}{2}, \frac{x+y}{2}\right)$:
$A''(0,0)$ $\longrightarrow A'''(0,0), B''(6,2)$ $\longrightarrow B'''(2,4), C''(8,2)$ $\longrightarrow C'''(3,5), D''(2,0)$ $\longrightarrow D'''(1,1)$.
Final vertices: $A(0,0), B(2,4), C(3,5), D(1,1)$.
Calculating side lengths:
$AB = \sqrt{(2-0)^2 + (4-0)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
$BC = \sqrt{(3-2)^2 + (5-4)^2} = \sqrt{1+1} = \sqrt{2}$.
$CD = \sqrt{(1-3)^2 + (1-5)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.
$DA = \sqrt{(0-1)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.
Since opposite sides are equal ($AB=CD$ and $BC=DA$) and adjacent sides are not equal,the figure is a parallelogram.

(C) We have,$1^4+2^4+3^4+\ldots+n^4 = f(n) \left(1^2+2^2+3^2+\ldots+n^2\right)$.
$f(n) = \frac{1^4+2^4+3^4+\ldots+n^4}{1^2+2^2+3^2+\ldots+n^2}$.
For $n=4$,we have $f(4) = \frac{1^4+2^4+3^4+4^4}{1^2+2^2+3^2+4^2}$.
$f(4) = \frac{1+16+81+256}{1+4+9+16}$.
$f(4) = \frac{354}{30}$.
Dividing both numerator and denominator by $6$,we get $f(4) = \frac{59}{5}$.

(B) Given equation is $\frac{(x^2+1)^3}{x^3} + \frac{x^2+1}{3x} = 0$ for $x \neq 0$.
Let $t = \frac{x^2+1}{x}$. Then the equation becomes $t^3 + \frac{t}{3} = 0$.
Factoring out $t$,we get $t(t^2 + \frac{1}{3}) = 0$.
This implies $t = 0$ or $t^2 = -\frac{1}{3}$.
Since $t$ must be a real number,$t^2 = -\frac{1}{3}$ has no real solutions.
Thus,we must have $t = 0$,which means $\frac{x^2+1}{x} = 0$.
This implies $x^2 + 1 = 0$,or $x^2 = -1$.
Since $x^2 = -1$ has no real solutions for $x$,there are no real roots for the given equation.
Therefore,the number of real roots is $0$.

(C) Given the limit $L = \lim_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.
Substituting $x = 0$ gives the $\frac{0}{0}$ indeterminate form.
Applying $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$L = \lim_{x \to 0} \frac{2f'(x) - 3 \cdot 2f'(2x) + 4f'(4x)}{2x} = \lim_{x \to 0} \frac{2f'(x) - 6f'(2x) + 4f'(4x)}{2x}$.
Applying $L$'Hospital's rule again:
$L = \lim_{x \to 0} \frac{2f''(x) - 6 \cdot 2f''(2x) + 4 \cdot 4f''(4x)}{2} = \lim_{x \to 0} \frac{2f''(x) - 12f''(2x) + 16f''(4x)}{2}$.
Since $f''(x)$ is continuous at $x = 0$,we have $\lim_{x \to 0} f''(x) = f''(0) = 4$.
Substituting $x = 0$ into the expression:
$L = \frac{2f''(0) - 12f''(0) + 16f''(0)}{2} = \frac{6f''(0)}{2} = 3f''(0)$.
Given $f''(0) = 4$,we get $L = 3 \times 4 = 12$.

(C) Given curves are $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$.
Let the curves intersect at point $(x_1, y_1)$.
For the first curve $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$,differentiating with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{a^2y}$.
So,$m_1 = -\frac{4x_1}{a^2y_1}$.
For the second curve $y^3 = 16x$,differentiating with respect to $x$:
$3y^2 \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{16}{3y^2}$.
So,$m_2 = \frac{16}{3y_1^2}$.
Since the curves intersect at right angles,$m_1 \times m_2 = -1$.
$(-\frac{4x_1}{a^2y_1}) \times (\frac{16}{3y_1^2}) = -1$.
$\frac{64x_1}{3a^2y_1^3} = 1$.
Since $(x_1, y_1)$ lies on $y^3 = 16x$,we have $y_1^3 = 16x_1$.
Substituting this into the equation:
$\frac{64x_1}{3a^2(16x_1)} = 1
\implies \frac{64x_1}{48a^2x_1} = 1
\implies \frac{4}{3a^2} = 1
\implies a^2 = \frac{4}{3}$.

(C) Given the ellipse equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \dots (i)$
To find the equation of the tangent and normal at the point $A\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$,we first find the derivative $\frac{dy}{dx}$.
Differentiating $(i)$ with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.
At point $A$,the slope of the tangent $m_T = -\frac{b^2}{a^2} \left( \frac{a/\sqrt{2}}{b/\sqrt{2}} \right) = -\frac{b}{a}$.
The slope of the normal $m_N = -\frac{1}{m_T} = \frac{a}{b}$.
Equation of tangent at $A$: $y - \frac{b}{\sqrt{2}} = -\frac{b}{a} \left( x - \frac{a}{\sqrt{2}} \right) \dots (ii)$.
Equation of normal at $A$: $y - \frac{b}{\sqrt{2}} = \frac{a}{b} \left( x - \frac{a}{\sqrt{2}} \right) \dots (iii)$.
To find the triangle vertices on the $X$-axis,set $y=0$ in $(ii)$ and $(iii)$:
For tangent,$0 - \frac{b}{\sqrt{2}} = -\frac{b}{a} (x - \frac{a}{\sqrt{2}}) \implies \frac{a}{\sqrt{2}} = x - \frac{a}{\sqrt{2}} \implies x = \sqrt{2}a$. Point $B = (\sqrt{2}a, 0)$.
For normal,$0 - \frac{b}{\sqrt{2}} = \frac{a}{b} (x - \frac{a}{\sqrt{2}}) \implies -\frac{b^2}{\sqrt{2}a} = x - \frac{a}{\sqrt{2}} \implies x = \frac{a}{\sqrt{2}} - \frac{b^2}{\sqrt{2}a} = \frac{a^2-b^2}{\sqrt{2}a}$. Point $C = (\frac{a^2-b^2}{\sqrt{2}a}, 0)$.
The height of the triangle is the $y$-coordinate of point $A$,which is $h = \frac{b}{\sqrt{2}}$.
The base $BC = \sqrt{2}a - \frac{a^2-b^2}{\sqrt{2}a} = \frac{2a^2 - a^2 + b^2}{\sqrt{2}a} = \frac{a^2+b^2}{\sqrt{2}a}$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left( \frac{a^2+b^2}{\sqrt{2}a} \right) \times \left( \frac{b}{\sqrt{2}} \right) = \frac{b(a^2+b^2)}{4a}$.

(C) Given the curve equation: $x^2+y^2=a^2$.
Differentiating both sides with respect to $x$:
$2x + 2y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{x}{y}$.
Since the tangent is parallel to the $x$-axis,the slope of the tangent must be zero:
$\frac{dy}{dx} = 0$
$-\frac{x}{y} = 0 \implies x = 0$.
Substituting $x = 0$ into the curve equation:
$0^2 + y^2 = a^2 \implies y^2 = a^2$.
Since $y \geq 0$,we get $y = a$.
Thus,the point on the curve is $(0, a)$.

(A) Let the equation of the line passing through $(4, 3)$ be $y - 3 = m(x - 4)$,where $m < 0$ for the line to cut the first quadrant.
$y = mx - 4m + 3$.
The $x$-intercept is found by setting $y = 0$: $0 = mx - 4m + 3 \implies x = 4 - \frac{3}{m}$.
The $y$-intercept is found by setting $x = 0$: $y = 3 - 4m$.
The area $A$ of the triangle formed in the first quadrant is $A = \frac{1}{2} \times (x\text{-intercept}) \times (y\text{-intercept}) = \frac{1}{2} (4 - \frac{3}{m})(3 - 4m) = \frac{1}{2} (12 - 16m - \frac{9}{m} + 12) = 12 - 8m - \frac{9}{2m}$.
To minimize the area,differentiate $A$ with respect to $m$: $\frac{dA}{dm} = -8 + \frac{9}{2m^2} = 0$.
$8 = \frac{9}{2m^2} \implies m^2 = \frac{9}{16} \implies m = -\frac{3}{4}$ (since $m < 0$).
Substituting $m = -\frac{3}{4}$ into the line equation: $y - 3 = -\frac{3}{4}(x - 4)$.
$4y - 12 = -3x + 12 \implies 3x + 4y = 24$.

(C) Let the intercepts on the $X$ and $Y$ axes be $m$ and $n$ respectively. The equation of the line is $\frac{x}{m} + \frac{y}{n} = 1$.
Since the line passes through $(4, 5)$,we have $\frac{4}{m} + \frac{5}{n} = 1$,which implies $\frac{4}{m} = 1 - \frac{5}{n} = \frac{n-5}{n}$,so $m = \frac{4n}{n-5}$.
The area of the triangle formed with the coordinate axes is $A = \frac{1}{2}mn = \frac{1}{2} \left( \frac{4n}{n-5} \right) n = \frac{2n^2}{n-5}$.
To find the minimum area,we differentiate $A$ with respect to $n$:
$\frac{dA}{dn} = 2 \left[ \frac{(n-5)(2n) - n^2(1)}{(n-5)^2} \right] = 2 \left[ \frac{2n^2 - 10n - n^2}{(n-5)^2} \right] = \frac{2n^2 - 20n}{(n-5)^2}$.
Setting $\frac{dA}{dn} = 0$,we get $2n(n-10) = 0$. Since $n > 5$ for positive intercepts,we have $n = 10$.
Then $m = \frac{4(10)}{10-5} = \frac{40}{5} = 8$.
The ratio of the intercepts $m : n = 8 : 10 = 4 : 5$.

(B) Given $\frac{x^{4}}{(x - 1)(x - 2)} = f(x) + \frac{A}{x - 1} + \frac{B}{x - 2}$.
By performing polynomial long division,we divide $x^{4}$ by $(x - 1)(x - 2) = x^{2} - 3x + 2$.
$x^{4} = (x^{2} - 3x + 2)(x^{2} + 3x + 7) + (15x - 14)$.
Thus,$\frac{x^{4}}{(x - 1)(x - 2)} = x^{2} + 3x + 7 + \frac{15x - 14}{(x - 1)(x - 2)}$.
We express $\frac{15x - 14}{(x - 1)(x - 2)}$ as partial fractions: $\frac{15x - 14}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2}$.
$15x - 14 = A(x - 2) + B(x - 1)$.
For $x = 1$: $15(1) - 14 = A(1 - 2) \Rightarrow 1 = -A \Rightarrow A = -1$.
For $x = 2$: $15(2) - 14 = B(2 - 1) \Rightarrow 16 = B$.
Comparing this with the given equation,$f(x) = x^{2} + 3x + 7$,$A = -1$,and $B = 16$.
Checking the options: $f(x) = x^{2} + 3x + 7$ is option $B$,and $A + B = -1 + 16 = 15$ (not $17$),$A - B = -1 - 16 = -17$ (not $-18$).
Thus,the correct option is $B$.

(B) Given expression is $\frac{-x^{2} + 6x + 13}{(3x + 5)(x^{2} + 4x + 4)} = \frac{-x^{2} + 6x + 13}{(3x + 5)(x + 2)^{2}}$.
Using partial fraction decomposition,we write:
$\frac{-x^{2} + 6x + 13}{(3x + 5)(x + 2)^{2}} = \frac{A}{3x + 5} + \frac{B}{x + 2} + \frac{C}{(x + 2)^{2}}$.
Multiplying both sides by $(3x + 5)(x + 2)^{2}$,we get:
$-x^{2} + 6x + 13 = A(x + 2)^{2} + B(x + 2)(3x + 5) + C(3x + 5)$.
To find $C$,put $x = -2$:
$-(-2)^{2} + 6(-2) + 13 = C(3(-2) + 5) \Rightarrow -4 - 12 + 13 = C(-1) \Rightarrow -3 = -C \Rightarrow C = 3$.
To find $A$,put $x = -\frac{5}{3}$:
$-(-\frac{5}{3})^{2} + 6(-\frac{5}{3}) + 13 = A(-\frac{5}{3} + 2)^{2} \Rightarrow -\frac{25}{9} - 10 + 13 = A(\frac{1}{3})^{2} \Rightarrow \frac{2}{9} = A(\frac{1}{9}) \Rightarrow A = 2$.
To find $B$,compare the coefficients of $x^{2}$ on both sides:
$-1 = A + 3B \Rightarrow -1 = 2 + 3B \Rightarrow -3 = 3B \Rightarrow B = -1$.
Thus,the partial fraction is $\frac{2}{3x + 5} - \frac{1}{x + 2} + \frac{3}{(x + 2)^{2}}$.

(A) Given the expression: $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A + \frac{B}{2 x-1} + \frac{C}{x+2} + \frac{D}{x-3}$
Multiply both sides by $(2x-1)(x+2)(x-3)$:
$x^3 = A(2 x-1)(x+2)(x-3) + B(x+2)(x-3) + C(x-3)(2 x-1) + D(2 x-1)(x+2)$
To find $A$,we observe the coefficient of $x^3$ on both sides.
The denominator is $(2x-1)(x+2)(x-3) = (2x^2+3x-2)(x-3) = 2x^3 - 6x^2 + 3x^2 - 9x - 2x + 6 = 2x^3 - 3x^2 - 11x + 6$.
Since the degree of the numerator and denominator is the same $(3)$,$A$ is the ratio of the leading coefficients:
$A = \frac{1}{2}$
Alternatively,by substituting values:
For $x=3$: $27 = D(5)(5) \Rightarrow D = \frac{27}{25}$
For $x=-2$: $-8 = C(-5)(-5) \Rightarrow C = -8/25$
For $x=1/2$: $1/8 = B(5/2)(-5/2) \Rightarrow B = -1/50$
For $x=0$: $0 = A(-1)(2)(-3) + B(2)(-3) + C(-3)(-1) + D(-1)(2)$
$0 = 6A - 6B + 3C - 2D$
$6A = 6(-1/50) - 3(-8/25) + 2(27/25) = -6/50 + 24/25 + 54/25 = -3/25 + 78/25 = 75/25 = 3$
$6A = 3 \Rightarrow A = 1/2$

(B) Given,$\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$
Equating the numerators:
$1 = A(2+3x) + B(3-5x)$
$1 = (2A + 3B) + x(3A - 5B)$
Comparing the coefficients of $x$ and the constant terms on both sides:
$3A - 5B = 0$ ...$(i)$
$2A + 3B = 1$ ...$(ii)$
Multiplying equation $(i)$ by $3$ and equation $(ii)$ by $5$:
$9A - 15B = 0$
$10A + 15B = 5$
Adding these two equations:
$19A = 5 \implies A = \frac{5}{19}$
Substituting $A$ in equation $(i)$:
$3(\frac{5}{19}) - 5B = 0 \implies 5B = \frac{15}{19} \implies B = \frac{3}{19}$
Therefore,$A+B = \frac{5}{19} + \frac{3}{19} = \frac{8}{19}$

(D) Given the partial fraction decomposition: $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+3} + \frac{Cx+D}{x^2+1}$.
Multiplying both sides by the denominator $(x-1)(x+3)(x^2+1)$,we get:
$6x^3 + 7x^2 + 6x - 3 = A(x+3)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+3)$.
For $x=1$: $6(1)^3 + 7(1)^2 + 6(1) - 3 = A(1+3)(1^2+1) \Rightarrow 16 = 8A \Rightarrow A=2$.
For $x=-3$: $6(-3)^3 + 7(-3)^2 + 6(-3) - 3 = B(-3-1)((-3)^2+1) \Rightarrow -162 + 63 - 18 - 3 = B(-4)(10) \Rightarrow -120 = -40B \Rightarrow B=3$.
Comparing the constant terms (setting $x=0$): $-3 = A(3)(1) + B(-1)(1) + D(-1)(3) \Rightarrow -3 = 3A - B - 3D$.
Substituting $A=2$ and $B=3$: $-3 = 3(2) - 3 - 3D \Rightarrow -3 = 3 - 3D \Rightarrow 3D = 6 \Rightarrow D=2$.
Comparing the coefficients of $x^3$: $6 = A + B + C \Rightarrow 6 = 2 + 3 + C \Rightarrow C=1$.
Thus,$n = A+B+C+D = 2+3+1+2 = 8$.
Given ${}^{50}C_n = {}^{50}C_r$,we know that either $r=n$ or $r=50-n$.
Since $r=n=8$ is not an option,we take $r = 50-8 = 42$.

(D) Given the partial fraction decomposition: $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$.
Multiplying both sides by $(1+x^2)(3-2x)$,we get:
$x = (Bx+C)(3-2x) + A(1+x^2)$
$x = 3Bx - 2Bx^2 + 3C - 2Cx + A + Ax^2$
$x = (A-2B)x^2 + (3B-2C)x + (A+3C)$
Comparing the coefficients on both sides:
$1$) $A - 2B = 0 \Rightarrow A = 2B$
$2$) $3B - 2C = 1$
$3$) $A + 3C = 0 \Rightarrow A = -3C$
From $A = 2B$ and $A = -3C$,we have $2B = -3C \Rightarrow B = -\frac{3}{2}C$.
Substitute $B$ into equation $(2)$:
$3(-\frac{3}{2}C) - 2C = 1$
$-\frac{9}{2}C - 2C = 1$
$-\frac{13}{2}C = 1$
$C = -\frac{2}{13}$.

(A) Given the expression $\frac{x^2}{x^2+3x-4}$.
Since the degree of the numerator is equal to the degree of the denominator,we perform long division:
$\frac{x^2}{x^2+3x-4} = 1 - \frac{3x-4}{x^2+3x-4} = 1 + \frac{-3x+4}{(x-1)(x+4)}$.
Let $\frac{-3x+4}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}$.
Then $-3x+4 = A(x+4) + B(x-1)$.
Comparing coefficients:
$A+B = -3$ and $4A-B = 4$.
Adding the two equations: $5A = 1 \Rightarrow A = \frac{1}{5}$.
Substituting $A$ into $A+B = -3$: $\frac{1}{5} + B = -3 \Rightarrow B = -3 - \frac{1}{5} = -\frac{16}{5}$.
Thus,$\frac{x^2}{x^2+3x-4} = 1 + \frac{1}{5(x-1)} - \frac{16}{5(x+4)}$.

(A) Let $H$ denote the event of getting a head and $T$ denote the event of getting a tail. The experiment stops when a head appears or after three tosses.
The sample space of the experiment is $S = \{H, TH, TTH, TTT\}$.
Let $A$ be the event that a head does not appear on the first toss. This means the first toss is a tail $(T)$.
The outcomes corresponding to event $A$ are $\{TH, TTH, TTT\}$.
The probability $P(A) = P(T) = \frac{1}{2}$.
Let $B$ be the event that the coin is tossed thrice. The outcomes corresponding to event $B$ are $\{TTH, TTT\}$.
The intersection $A \cap B$ represents the event that the first toss is a tail $AND$ the coin is tossed thrice. This corresponds to the outcomes $\{TTH, TTT\}$.
$P(A \cap B) = P(TTH) + P(TTT) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$.
We need to find the conditional probability $P(B|A)$:
$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$.

(B) Let $F$ be the event that a fair coin is drawn and $B$ be the event that a biased coin is drawn. Let $E$ be the event that the first toss is a head and the second toss is a tail.
Given $P(F) = \frac{m}{n}$ and $P(B) = \frac{n-m}{n}$.
For a fair coin,$P(H) = \frac{1}{2}$ and $P(T) = \frac{1}{2}$. Thus,$P(E|F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
For a biased coin,$P(H) = 2P(T)$. Since $P(H) + P(T) = 1$,we have $3P(T) = 1$,so $P(T) = \frac{1}{3}$ and $P(H) = \frac{2}{3}$. Thus,$P(E|B) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$.
Using Bayes' theorem,the probability that the coin is fair given the event $E$ is:
$P(F|E) = \frac{P(F)P(E|F)}{P(F)P(E|F) + P(B)P(E|B)}$
$P(F|E) = \frac{(\frac{m}{n})(\frac{1}{4})}{(\frac{m}{n})(\frac{1}{4}) + (\frac{n-m}{n})(\frac{2}{9})}$
$P(F|E) = \frac{\frac{m}{4}}{\frac{m}{4} + \frac{2(n-m)}{9}} = \frac{9m}{9m + 8(n-m)} = \frac{9m}{9m + 8n - 8m} = \frac{9m}{8n + m}$.

(B) Let $A$ be the event that a prime number occurs on the die.
Let $B$ be the event that a prime number does not occur on the die.
Let $E$ be the event that the man reports that a prime number occurred.
There are $25$ prime numbers between $1$ and $100$.
Therefore,$P(A) = \frac{25}{100} = \frac{1}{4}$ and $P(B) = 1 - P(A) = \frac{75}{100} = \frac{3}{4}$.
The probability that the man speaks the truth is $P(E|A) = \frac{7}{10}$.
The probability that the man lies (reports a prime when it is not) is $P(E|B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Using Bayes' Theorem,the probability that it is actually a prime given that he reported a prime is:
$P(A|E) = \frac{P(A) \times P(E|A)}{P(A) \times P(E|A) + P(B) \times P(E|B)}$
$P(A|E) = \frac{\frac{1}{4} \times \frac{7}{10}}{(\frac{1}{4} \times \frac{7}{10}) + (\frac{3}{4} \times \frac{3}{10})}$
$P(A|E) = \frac{\frac{7}{40}}{\frac{7}{40} + \frac{9}{40}} = \frac{7}{16}$.

(D) Let $X$ denote the number of successes in $3$ trials. Since the die is tossed $3$ times,$n=3$.
The probability of getting an even number (success) in a single toss is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
$X$ follows a binomial distribution $B(n, p)$,where $P(X=r) = { }^n C_r p^r q^{n-r}$.
We need to find the probability of getting at least $2$ successes,which is $P(X \geq 2) = P(X=2) + P(X=3)$.
$P(X=2) = { }^3 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{3-2} = 3 \times \left(\frac{1}{2}\right)^3 = \frac{3}{8}$.
$P(X=3) = { }^3 C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{3-3} = 1 \times \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Therefore,$P(X \geq 2) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$.

(C) The outcome of each coin toss is an independent event.
For a fair coin,the probability of getting a head in any single toss is always $P(H) = \frac{1}{2}$.
Since the tosses are independent,the result of the $1947^{\text{th}}$ toss does not depend on the results of any other tosses.
Therefore,the probability of getting a head on the $1947^{\text{th}}$ toss remains $\frac{1}{2}$.

(A) For a binomial distribution,the mean is given by $np = 5$ and the variance is given by $npq = 4$.
Substituting $np = 5$ into the variance formula,we get $5q = 4$,which implies $q = \frac{4}{5}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{4}{5} = \frac{1}{5}$.
Using $np = 5$ and $p = \frac{1}{5}$,we find $n = \frac{5}{p} = 5 \times 5 = 25$.
The probability mass function for a binomial distribution is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.
For $X=1$,we have $P(X=1) = {}^{25}C_{1} \left(\frac{1}{5}\right)^{1} \left(\frac{4}{5}\right)^{25-1}$.
$P(X=1) = 25 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^{24} = 5 \times \frac{4^{24}}{5^{24}} = \frac{4^{24}}{5^{23}}$.

(A) Let success be selecting a smart phone used for office purposes.
Here,the probability of success $p = 50 \% = \frac{1}{2}$.
The probability of failure $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
We have $n = 10$ trials and we want exactly $r = 2$ successes.
Using the Binomial distribution formula $P(X = r) = { }^n C_r p^r q^{n-r}$:
$P(X = 2) = { }^{10} C_2 \cdot (\frac{1}{2})^2 \cdot (\frac{1}{2})^{10-2}$
$P(X = 2) = { }^{10} C_2 \cdot (\frac{1}{2})^2 \cdot (\frac{1}{2})^8$
$P(X = 2) = { }^{10} C_2 \cdot (\frac{1}{2})^{10} = { }^{10} C_2 \frac{1}{2^{10}}$.

(C) In a binomial distribution $B(n, p)$,the mean is given by $\mu = np$ and the standard deviation is given by $\sigma = \sqrt{npq}$.
Given,$np = 200$ and $\sqrt{npq} = 10$.
Squaring the standard deviation equation,we get $npq = 100$.
Substituting $np = 200$ into $npq = 100$,we get $200q = 100$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Now,$np = 200 \implies n(\frac{1}{2}) = 200 \implies n = 400$.
We need to calculate $n^2 + \frac{1}{p^2} + \frac{1}{q^2}$.
Substituting the values: $(400)^2 + \frac{1}{(1/2)^2} + \frac{1}{(1/2)^2} = 160000 + 4 + 4 = 160008$.

(B) Let $n$ be the number of times he fires at the target.
Let $X$ be the number of times he hits the target.
Since hitting the target is a Bernoulli trial,$X$ follows a binomial distribution.
Here,$p = \frac{2}{3}$ (probability of hitting) and $q = 1 - p = \frac{1}{3}$ (probability of missing).
The probability of hitting the target at least once is given by $P(X \geq 1) = 1 - P(X = 0)$.
We are given that $P(X \geq 1) > 90 \%$,which means $P(X \geq 1) > 0.9$.
Therefore,$1 - P(X = 0) > 0.9$.
Since $P(X = 0) = {}^nC_0 q^n p^0 = (\frac{1}{3})^n$,we have:
$1 - (\frac{1}{3})^n > 0.9$
$0.1 > (\frac{1}{3})^n$
$(\frac{1}{3})^n < \frac{1}{10}$
$3^n > 10$.
For $n = 1$,$3^1 = 3 < 10$.
For $n = 2$,$3^2 = 9 < 10$.
For $n = 3$,$3^3 = 27 > 10$.
Thus,the minimum number of times he must fire is $3$.

(B) Given Mean = $np = 6$ ... $(i)$
Variance = $npq = 2$ where $q = 1 - p$ ... $(ii)$
Substituting $(i)$ into $(ii)$:
$6q = 2 \implies q = \frac{1}{3}$
Since $p = 1 - q$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$
From $(i)$,$n = \frac{6}{p} = \frac{6}{2/3} = 9$
The probability mass function is $P(X = r) = {}^nC_r p^r q^{n-r}$
For $X = 8$:
$P(X = 8) = {}^9C_8 \left(\frac{2}{3}\right)^8 \left(\frac{1}{3}\right)^{9-8}$
$P(X = 8) = 9 \times \frac{2^8}{3^8} \times \frac{1}{3}$
$P(X = 8) = 9 \times \frac{2^8}{3^9} = \frac{3^2 \times 2^8}{3^9} = \frac{2^8}{3^7}$

(C) The expected value $E(X)$ of a discrete random variable $X$ is calculated using the formula $E(X) = \sum x_i P(x_i)$.
Given values of $X$ are $10, 20, 30, 40$ and their respective probabilities are $0.3, 0.3, 0.2, 0.2$.
Substituting these values into the formula:
$E(X) = (10 \times 0.3) + (20 \times 0.3) + (30 \times 0.2) + (40 \times 0.2)$
$E(X) = 3 + 6 + 6 + 8$
$E(X) = 23$
Therefore,the expected value of $X$ is $23$.

(C) The sum of probabilities for a random variable is $1$.
Given $P(X=r) = K r^3$ for $r \in \{1, 2, 3, 4\}$.
Summing these: $K(1^3) + K(2^3) + K(3^3) + K(4^3) = 1$.
$K(1 + 8 + 27 + 64) = 1 \Rightarrow 100K = 1 \Rightarrow K = \frac{1}{100}$.
We need to find $P\left(\left.\frac{1}{2} < X < \frac{5}{2} \right\rvert X > 1\right)$.
This is equivalent to $P(X=2 \mid X \in \{2, 3, 4\})$.
By the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Here $A = \{X=2\}$ and $B = \{X=2, 3, 4\}$.
$P(A \cap B) = P(X=2) = 8K$.
$P(B) = P(X=2) + P(X=3) + P(X=4) = 8K + 27K + 64K = 99K$.
Thus,$P(A \mid B) = \frac{8K}{99K} = \frac{8}{99}$.
Therefore,$K = \frac{1}{100}$ and the conditional probability is $\frac{8}{99}$.

(C) Given,mean $np = 2$ $(i)$
And variance $npq = 1$ $(ii)$
Dividing $(ii)$ by $(i)$,we get $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $(i)$,we get $n \times \frac{1}{2} = 2$,so $n = 4$.
The binomial distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k} = {}^4C_k (\frac{1}{2})^k (\frac{1}{2})^{4-k} = {}^4C_k (\frac{1}{2})^4$.
We need to find $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$.
$P(X > 1) = {}^4C_2 (\frac{1}{2})^4 + {}^4C_3 (\frac{1}{2})^4 + {}^4C_4 (\frac{1}{2})^4$.
$P(X > 1) = (6 + 4 + 1) \times \frac{1}{16} = \frac{11}{16}$.

(D) Given the probability distribution table:

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

Since the sum of all probabilities must be $1$,we have:
$K + (K + \frac{1}{7}) + 2K + \frac{2}{5} = 1$
$4K + \frac{5 + 14}{35} = 1$
$4K + \frac{19}{35} = 1$
$4K = 1 - \frac{19}{35} = \frac{16}{35}$
$K = \frac{4}{35}$
Now,substituting $K$ back into the table:
$P(X=0) = \frac{4}{35}$
$P(X=1) = \frac{4}{35} + \frac{5}{35} = \frac{9}{35}$
$P(X=2) = 2 \times \frac{4}{35} = \frac{8}{35}$
$P(X=3) = \frac{2}{5} = \frac{14}{35}$
The mean $\mu = E(X) = \sum x_i P(x_i)$:
$\mu = 0 \times \frac{4}{35} + 1 \times \frac{9}{35} + 2 \times \frac{8}{35} + 3 \times \frac{14}{35}$
$\mu = 0 + \frac{9}{35} + \frac{16}{35} + \frac{42}{35}$
$\mu = \frac{67}{35}$

(C) We know that for any probability distribution,the sum of all probabilities must be equal to $1$.
Given $P(X=x) = \frac{c^x}{x!}$ for $x = 1, 2, 3, \ldots$.
Therefore,$\sum_{x=1}^{\infty} P(X=x) = 1$.
Substituting the given expression: $\sum_{x=1}^{\infty} \frac{c^x}{x!} = 1$.
We know the Taylor series expansion for the exponential function is $e^c = \sum_{x=0}^{\infty} \frac{c^x}{x!} = 1 + \frac{c}{1!} + \frac{c^2}{2!} + \frac{c^3}{3!} + \ldots$.
Thus,$\sum_{x=1}^{\infty} \frac{c^x}{x!} = e^c - 1$.
Setting this equal to $1$: $e^c - 1 = 1$.
$e^c = 2$.
Taking the natural logarithm on both sides: $c = \ln(2)$.

(B) Given that $X$ follows a Poisson distribution with parameter $\lambda$. The probability mass function is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$.
Given the equation: $3 P(X=4) = \frac{1}{2} P(X=2) + P(X=0)$.
Substituting the formula: $3 \frac{e^{-\lambda} \lambda^4}{4!} = \frac{1}{2} \frac{e^{-\lambda} \lambda^2}{2!} + \frac{e^{-\lambda} \lambda^0}{0!}$.
Dividing both sides by $e^{-\lambda}$: $\frac{3 \lambda^4}{24} = \frac{\lambda^2}{4} + 1$.
Simplifying: $\frac{\lambda^4}{8} = \frac{\lambda^2}{4} + 1$.
Multiplying by $8$: $\lambda^4 = 2 \lambda^2 + 8$,which gives $\lambda^4 - 2 \lambda^2 - 8 = 0$.
Let $u = \lambda^2$. Then $u^2 - 2u - 8 = 0$.
Factoring the quadratic: $(u - 4)(u + 2) = 0$.
This gives $u = 4$ or $u = -2$.
Since $\lambda^2 = u$ and $\lambda^2$ must be non-negative,we have $\lambda^2 = 4$,which implies $\lambda = 2$ (as $\lambda > 0$).
The mean of a Poisson distribution is $\lambda$,so the mean is $2$.

(D) Given,the mean of the Poisson distribution is $\lambda = \frac{1}{3}$.
For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.
For $k=1$,$P(X=1) = \frac{e^{-1/3} (1/3)^1}{1!} = \frac{1}{3} e^{-1/3}$.
For $k=2$,$P(X=2) = \frac{e^{-1/3} (1/3)^2}{2!} = \frac{1}{2} \times \frac{1}{9} e^{-1/3} = \frac{1}{18} e^{-1/3}$.
Now,the ratio $P(X=1) : P(X=2) = \frac{\frac{1}{3} e^{-1/3}}{\frac{1}{18} e^{-1/3}} = \frac{1/3}{1/18} = \frac{18}{3} = 6$.
Thus,the ratio is $6 : 1$.

(C) Given,$P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$.
Since the sum of all probabilities in a probability distribution is $1$,we have $\sum_{x=0}^{\infty} P(X=x) = 1$.
Substituting the given expression:
$\sum_{x=0}^{\infty} K(x+1)\left(\frac{1}{5}\right)^x = 1$
$K \left[ 1 + 2\left(\frac{1}{5}\right) + 3\left(\frac{1}{5}\right)^2 + 4\left(\frac{1}{5}\right)^3 + \ldots \right] = 1$.
This is an arithmetico-geometric series of the form $\sum_{n=1}^{\infty} n r^{n-1} = (1-r)^{-2}$ where $r = \frac{1}{5}$.
Thus,$K(1 - \frac{1}{5})^{-2} = 1$.
$K(\frac{4}{5})^{-2} = 1 \Rightarrow K(\frac{5}{4})^2 = 1$.
$K(\frac{25}{16}) = 1 \Rightarrow K = \frac{16}{25}$.
Now,$P(X=0) = K(0+1)\left(\frac{1}{5}\right)^0 = K(1)(1) = K$.
Therefore,$P(X=0) = \frac{16}{25}$.

(A) $X$ is a Poisson variate.
Given $P(X=2) = P(X=3)$.
Using the formula $P(X=r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!}$,we have:
$\frac{e^{-\lambda} \cdot \lambda^2}{2!} = \frac{e^{-\lambda} \cdot \lambda^3}{3!}$
$\Rightarrow \frac{\lambda^2}{2} = \frac{\lambda^3}{6}$
$\Rightarrow \lambda = \frac{6}{2} = 3$.
Now,we calculate $P(X=4)$:
$P(X=4) = \frac{e^{-\lambda} \cdot \lambda^4}{4!} = \frac{e^{-3} \cdot 3^4}{4!}$.
Finally,we find $e^3 P(X=4)$:
$e^3 P(X=4) = e^3 \cdot \frac{e^{-3} \cdot 3^4}{4!} = \frac{3^4}{4!} = \frac{81}{24} = \frac{27}{8} = \left(\frac{3}{2}\right)^3$.

(C) Given the probability distribution function $P(X=x)=K\left(\frac{2}{5}\right)^x$ for $x=1, 2, 3, \ldots$.
We know that the sum of all probabilities in a distribution must be equal to $1$,i.e.,$\sum_{x=1}^{\infty} P(X=x) = 1$.
Substituting the given function,we get $K \sum_{x=1}^{\infty} \left(\frac{2}{5}\right)^x = 1$.
This is an infinite geometric series with the first term $a = \frac{2}{5}$ and common ratio $r = \frac{2}{5}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.
Therefore,$K \left( \frac{2/5}{1 - 2/5} \right) = 1$.
$K \left( \frac{2/5}{3/5} \right) = 1$.
$K \left( \frac{2}{3} \right) = 1$.
Thus,$K = \frac{3}{2}$.

(B) Given that,$x=a\left(\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right)$ and $y=a \sin \theta$.
On differentiating $x$ and $y$ with respect to $\theta$,we get:
$\frac{dx}{d\theta} = a \left( -\sin \theta + \frac{1}{\tan(\theta/2)} \cdot \sec^2(\theta/2) \cdot \frac{1}{2} \right)$
$= a \left( -\sin \theta + \frac{\cos(\theta/2)}{\sin(\theta/2)} \cdot \frac{1}{\cos^2(\theta/2)} \cdot \frac{1}{2} \right)$
$= a \left( -\sin \theta + \frac{1}{2 \sin(\theta/2) \cos(\theta/2)} \right) = a \left( -\sin \theta + \frac{1}{\sin \theta} \right)$
$= a \left( \frac{1 - \sin^2 \theta}{\sin \theta} \right) = \frac{a \cos^2 \theta}{\sin \theta}$.
Also,$\frac{dy}{d\theta} = a \cos \theta$.
Therefore,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{a \cos^2 \theta / \sin \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.

(B) Given equations for direction ratios $(l, m, n)$ are $l+m+n=0$ and $l^2=m^2+n^2$.
From $l+m+n=0$,we have $l=-(m+n)$.
Substituting this into $l^2=m^2+n^2$,we get $(-(m+n))^2 = m^2+n^2$.
$m^2+n^2+2mn = m^2+n^2$,which implies $2mn=0$,so $mn=0$.
This gives two cases:
Case $I$: $m=0$. Then $l=-n$. Let the direction ratios be $(k, 0, -k)$. The unit vector is $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
Case $II$: $n=0$. Then $l=-m$. Let the direction ratios be $(k, -k, 0)$. The unit vector is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
Let $\vec{a} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
The cosine of the angle $\theta$ between them is given by $\cos \theta = |\vec{a} \cdot \vec{b}|$.
$\cos \theta = |(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0)| = |\frac{1}{2} + 0 + 0| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0$ --- $(i)$
$l^2+m^2-n^2=0$ --- $(ii)$
From $(i)$,$n = -(l+m)$.
Substituting this into $(ii)$:
$l^2 + m^2 - (-(l+m))^2 = 0$
$l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$
$-2lm = 0 \Rightarrow lm = 0$.
This implies either $l=0$ or $m=0$.
Case $1$: If $l=0$,then from $(i)$,$m+n=0 \Rightarrow m=-n$. Let $m=1$,then $n=-1$. The direction ratios are $(0, 1, -1)$.
Case $2$: If $m=0$,then from $(i)$,$l+n=0 \Rightarrow l=-n$. Let $l=1$,then $n=-1$. The direction ratios are $(1, 0, -1)$.
Let the two vectors be $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$.
$|\vec{a}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(D) Given the equations for direction cosines $l, m, n$:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2=0$
$l^2-5m^2+l^2+2lm+m^2=0$
$2l^2+2lm-4m^2=0$
$l^2+lm-2m^2=0$
$(l+2m)(l-m)=0$
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. Direction ratios are $(l, l, -2l)$,i.e.,$(1, 1, -2)$.
Normalized direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. Direction ratios are $(-2m, m, m)$,i.e.,$(-2, 1, 1)$.
Normalized direction cosines $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) Given equations are $l+m+n=0$ and $l^2+m^2-n^2=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation: $(-m-n)^2 + m^2 - n^2 = 0$.
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$.
$2m^2 + 2mn = 0 \Rightarrow 2m(m+n) = 0$.
This gives two cases:
Case $1$: $m=0$. Then $l = -n$. The direction ratios are $(-1, 0, 1)$.
Case $2$: $m = -n$. Then $l = 0$. The direction ratios are $(0, -1, 1)$.
Let the two lines be represented by vectors $\vec{a} = -\hat{i} + \hat{k}$ and $\vec{b} = -\hat{j} + \hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{a}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.