KVPY 2019 Mathematics Question Paper with Answer and Solution

(B) To form a four-letter word using the letters $a, b, c$ such that all three letters occur,we must select one letter to be repeated twice and the other two letters to appear once each.
Step $1$: Select the letter to be repeated from the set $\{a, b, c\}$. This can be done in $^3C_1 = 3$ ways.
Step $2$: Arrange the four letters (e.g.,if $a$ is repeated,the letters are $a, a, b, c$). The number of ways to arrange these four letters is $\frac{4!}{2!} = \frac{24}{2} = 12$.
Step $3$: The total number of words is the product of the number of ways to select the letter and the number of ways to arrange them:
$3 \times 12 = 36$.

(B) Given the trigonometric relation:
$(\frac{1}{3} \sin \theta + \frac{2}{3} \cos \theta)^2 = \frac{1}{3} \sin^2 \theta + \frac{2}{3} \cos^2 \theta$
Expanding the left side:
$\frac{1}{9} \sin^2 \theta + \frac{4}{9} \cos^2 \theta + \frac{4}{9} \sin \theta \cos \theta = \frac{1}{3} \sin^2 \theta + \frac{2}{3} \cos^2 \theta$
Multiplying by $9$:
$\sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 3 \sin^2 \theta + 6 \cos^2 \theta$
Rearranging terms:
$2 \sin^2 \theta + 2 \cos^2 \theta - 4 \sin \theta \cos \theta = 0$
Dividing by $2$:
$\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = 0$
$(\sin \theta - \cos \theta)^2 = 0$
$\sin \theta = \cos \theta \Rightarrow \tan \theta = 1$
For $\theta \in [0, \pi]$,$\tan \theta = 1$ implies $\theta = \frac{\pi}{4}$.
Thus,$A \cap [0, \pi] = \{\frac{\pi}{4}\}$,which has exactly one point.

(A) Let the radius of circle $C_1$ be $2a$. Place the center $C$ at the origin $(0,0)$ and $AB$ along the $X$-axis. Then $B = (2a, 0)$,$A = (-2a, 0)$,$C = (0, 0)$,and $D = (-a, 0)$.
Since $E$ is on $C_1$ and $EC \perp AB$,$E$ has coordinates $(0, 2a)$.
Since $F$ is on $C_2$ (diameter $AC$,center $D(-a, 0)$,radius $a$) and $DF \perp AB$,$F$ has coordinates $(-a, -a)$.
We need to find $\sin \angle FEC$. Let $\angle FEC = \theta$.
Vector $\vec{EC} = C - E = (0, 0) - (0, 2a) = (0, -2a)$.
Vector $\vec{EF} = F - E = (-a, -a) - (0, 2a) = (-a, -3a)$.
The angle $\theta$ between $\vec{EC}$ and $\vec{EF}$ is given by $\cos \theta = \frac{\vec{EC} \cdot \vec{EF}}{|\vec{EC}| |\vec{EF}|}$.
$\vec{EC} \cdot \vec{EF} = (0)(-a) + (-2a)(-3a) = 6a^2$.
$|\vec{EC}| = \sqrt{0^2 + (-2a)^2} = 2a$.
$|\vec{EF}| = \sqrt{(-a)^2 + (-3a)^2} = \sqrt{a^2 + 9a^2} = a\sqrt{10}$.
$\cos \theta = \frac{6a^2}{(2a)(a\sqrt{10})} = \frac{6}{2\sqrt{10}} = \frac{3}{\sqrt{10}}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{9}{10} = \frac{1}{10}$,we have $\sin \theta = \frac{1}{\sqrt{10}}$.

(B) The set $S = \{(x, y) : |x| + 2|y| = 1\}$ represents a rhombus in the Cartesian plane with vertices at $(1, 0), (-1, 0), (0, 1/2),$ and $(0, -1/2)$.
The smallest circle with centre at the origin $(0, 0)$ that has a non-empty intersection with $S$ is the circle inscribed in this rhombus.
The radius $r$ of this inscribed circle is the perpendicular distance from the origin $(0, 0)$ to any of the sides of the rhombus.
Consider the side in the first quadrant,which is given by the line $x + 2y = 1$ or $x + 2y - 1 = 0$.
The perpendicular distance $r$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $r = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Substituting $A = 1, B = 2, C = -1$ and $(x_0, y_0) = (0, 0)$:
$r = \frac{|1(0) + 2(0) - 1|}{\sqrt{1^2 + 2^2}} = \frac{|-1|}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}$.

(B) Given equation: $\sin(9x) + \sin(3x) = 0$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$2 \sin(6x) \cos(3x) = 0$
This implies $\sin(6x) = 0$ or $\cos(3x) = 0$.
Case $1$: $\sin(6x) = 0$ $\Rightarrow 6x = n\pi$ $\Rightarrow x = \frac{n\pi}{6}$ for $n \in \mathbb{Z}$.
For $x \in [0, 2\pi]$,$n$ can be $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$.
Values: $x \in \{0, \frac{\pi}{6}, \frac{2\pi}{6}, \frac{3\pi}{6}, \frac{4\pi}{6}, \frac{5\pi}{6}, \frac{6\pi}{6}, \frac{7\pi}{6}, \frac{8\pi}{6}, \frac{9\pi}{6}, \frac{10\pi}{6}, \frac{11\pi}{6}, \frac{12\pi}{6}\}$ (Total $13$ values).
Case $2$: $\cos(3x) = 0$ $\Rightarrow 3x = (2k+1)\frac{\pi}{2}$ $\Rightarrow x = (2k+1)\frac{\pi}{6}$ for $k \in \mathbb{Z}$.
For $x \in [0, 2\pi]$,$k$ can be $0, 1, 2, 3, 4, 5$.
Values: $x \in \{\frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6}, \frac{11\pi}{6}\}$.
All these values are already included in the set from Case $1$.
Thus,the total number of distinct solutions is $13$.

(C) Let the diagonals of the parallelogram be $d_1 = 10$ and $d_2 = 4$. Let the angle between the diagonals be $\theta$.
The area of the parallelogram is given by $A = \frac{1}{2} d_1 d_2 \sin \theta = \frac{1}{2} \times 10 \times 4 \times \sin \theta = 20 \sin \theta$.
The area is maximum when $\sin \theta = 1$,i.e.,$\theta = \frac{\pi}{2}$.
When the diagonals intersect at right angles,the parallelogram is a rhombus.
The sides of the rhombus $s$ are given by $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$.
The perimeter of the rhombus is $P = 4s = 4\sqrt{29}$.
Since $5 < \sqrt{29} < 6$,we have $20 < 4\sqrt{29} < 24$.
More precisely,$\sqrt{29} \approx 5.385$,so $P \approx 4 \times 5.385 = 21.54$.
Thus,the perimeter lies in the interval $(21, 22]$.

(C) Let $\frac{2a-1}{b} = \alpha$ and $\frac{2b-1}{a} = \beta$,where $\alpha, \beta \in \mathbb{Z}^+$.
Since $a, b \ge 1$,we have $2a-1 \ge 1$ and $2b-1 \ge 1$,so $\alpha, \beta \ge 1$.
From the equations,$2a-1 = \alpha b$ and $2b-1 = \beta a$.
Substituting $b = \frac{2a-1}{\alpha}$ into the second equation: $2(\frac{2a-1}{\alpha}) - 1 = \beta a$.
$4a - 2 - \alpha = \alpha \beta a$,which implies $a(4 - \alpha \beta) = \alpha + 2$.
Since $a > 0$ and $\alpha + 2 > 0$,we must have $4 - \alpha \beta > 0$,so $\alpha \beta < 4$.
Possible pairs $(\alpha, \beta)$ are $(1, 1), (1, 2), (1, 3), (2, 1), (3, 1)$.
Case $1$: $(\alpha, \beta) = (1, 1) \implies a(4-1) = 1+2 \implies 3a = 3 \implies a = 1$. Then $b = \frac{2(1)-1}{1} = 1$. Pair: $(1, 1)$.
Case $2$: $(\alpha, \beta) = (1, 2) \implies a(4-2) = 1+2 \implies 2a = 3$,no integer solution.
Case $3$: $(\alpha, \beta) = (1, 3) \implies a(4-3) = 1+2 \implies a = 3$. Then $b = \frac{2(3)-1}{1} = 5$. Pair: $(3, 5)$.
Case $4$: $(\alpha, \beta) = (2, 1) \implies a(4-2) = 2+2 \implies 2a = 4 \implies a = 2$. Then $b = \frac{2(2)-1}{2} = 1.5$,not an integer.
Case $5$: $(\alpha, \beta) = (3, 1) \implies a(4-3) = 3+2 \implies a = 5$. Then $b = \frac{2(5)-1}{3} = 3$. Pair: $(5, 3)$.
The ordered pairs are $(1, 1), (3, 5), (5, 3)$. Total $3$ pairs.

(C) Given that $z$ and $w$ are complex numbers on the unit circle,we have $|z|=1$ and $|w|=1$.
This implies $z\bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z}$ and $w\bar{w}=1 \Rightarrow \bar{w}=\frac{1}{w}$.
Given $z^2+w^2=1$.
Taking the conjugate of both sides,we get $\bar{z}^2+\bar{w}^2=1$.
Substituting $\bar{z}=\frac{1}{z}$ and $\bar{w}=\frac{1}{w}$,we get $\frac{1}{z^2}+\frac{1}{w^2}=1$,which simplifies to $\frac{z^2+w^2}{z^2w^2}=1$.
Since $z^2+w^2=1$,we have $\frac{1}{z^2w^2}=1$,so $z^2w^2=1$,which means $(zw)^2=1$,so $zw=1$ or $zw=-1$.
Case $1$: $zw=1$. Then $w=\frac{1}{z}$. Substituting into $z^2+w^2=1$,we get $z^2+\frac{1}{z^2}=1 \Rightarrow z^4-z^2+1=0$. This is a quadratic in $z^2$,giving $z^2 = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = e^{\pm i\pi/3}$. Thus $z^2 = e^{i\pi/3}$ or $z^2 = e^{-i\pi/3}$. Each gives $2$ values for $z$,total $4$ values.
Case $2$: $zw=-1$. Then $w=-\frac{1}{z}$. Substituting into $z^2+w^2=1$,we get $z^2+\frac{1}{z^2}=1$,which is the same equation as Case $1$. This also gives $4$ values for $z$.
Total number of ordered pairs $(z, w)$ is $4+4=8$.

(A) The total number of four-letter words that can be formed using $26$ letters is $26^4 = 456976$.
The number of four-letter words formed using only letters from set $V$ (vowels) is $5^4 = 625$.
The number of four-letter words formed using only letters from set $C$ (consonants) is $21^4 = 194481$.
The number of words containing at least one vowel and at least one consonant is given by the total number of words minus the words containing only vowels and the words containing only consonants.
Number of words $= 26^4 - 5^4 - 21^4$
$= 456976 - 625 - 194481$
$= 261870$.

(A) Given the equation: $(1+a^2)(1+b^2) = 4ab$
Expanding the left side: $1 + b^2 + a^2 + a^2b^2 = 4ab$
Rearranging the terms: $a^2 - 2ab + b^2 + a^2b^2 - 2ab + 1 = 0$
This can be written as: $(a-b)^2 + (ab-1)^2 = 0$
Since $a$ and $b$ are real numbers,the sum of squares is zero if and only if each term is zero:
$(a-b)^2 = 0 \implies a = b$
$(ab-1)^2 = 0 \implies ab = 1$
Substituting $b=a$ into $ab=1$,we get $a^2 = 1$,which implies $a = 1$ or $a = -1$.
However,the problem states $a > 0$ and $a \neq 1$. Thus,there is no value of $b$ that satisfies the equation under the given constraints.
Therefore,the set $S$ is an empty set.

(B) Let the diameter $AB = x$.
Since $AB$ is the diameter,$\angle ACB = 90^\circ$ and $\angle ADB = 90^\circ$.
In $\triangle ACB$,$BC = \sqrt{x^2 - 1}$.
In $\triangle ADB$,$AD = \sqrt{x^2 - 9}$.
Applying Ptolemy's theorem to the cyclic quadrilateral $ACDB$:
$AB \cdot CD + AC \cdot DB = AD \cdot BC$
$x(2) + (1)(3) = \sqrt{x^2 - 9} \cdot \sqrt{x^2 - 1}$
$2x + 3 = \sqrt{(x^2 - 9)(x^2 - 1)}$
Squaring both sides:
$(2x + 3)^2 = (x^2 - 9)(x^2 - 1)$
$4x^2 + 12x + 9 = x^4 - 10x^2 + 9$
$x^4 - 14x^2 - 12x = 0$
Since $x \neq 0$,we have $x^3 - 14x - 12 = 0$.
Let $f(x) = x^3 - 14x - 12$.
$f(4) = 64 - 56 - 12 = -4$.
$f(4.1) = (4.1)^3 - 14(4.1) - 12 = 68.921 - 57.4 - 12 = -0.479$.
$f(4.2) = (4.2)^3 - 14(4.2) - 12 = 74.088 - 58.8 - 12 = 3.288$.
Since $f(4.1) < 0$ and $f(4.2) > 0$,the root lies in $[4.1, 4.2)$.

(A) Let $\angle CAD = \theta_2$ and $\angle BAD = \theta_1$. We are given $\cot \theta_2 : \cot \theta_1 = 2 : 1$,so $\cot \theta_2 = 2 \cot \theta_1$.
Using the cotangent rule in $\triangle ADC$ and $\triangle ADB$ with $AD$ as a common side and $BD = DC = a/2$:
$\cot \theta_2 = \frac{AD^2 + b^2 - (a/2)^2}{2 \cdot AD \cdot (a/2) \cdot \sin \theta_2} \cdot \sin \theta_2 = \frac{AD^2 + b^2 - a^2/4}{2 \cdot \text{Area}(\triangle ADC)}$
Since $\text{Area}(\triangle ADC) = \text{Area}(\triangle ADB)$,the condition $\cot \theta_2 = 2 \cot \theta_1$ implies:
$AD^2 + b^2 - a^2/4 = 2(AD^2 + c^2 - a^2/4)$
$AD^2 = b^2 - 2c^2 + a^2/4$
Using Apollonius theorem,$AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$.
Equating the two expressions for $AD^2$:
$\frac{2b^2 + 2c^2 - a^2}{4} = b^2 - 2c^2 + \frac{a^2}{4}$
$2b^2 + 2c^2 - a^2 = 4b^2 - 8c^2 + a^2$
$10c^2 = 2b^2 + 2a^2 \Rightarrow b^2 + a^2 = 5c^2$.
In $\triangle BGA$,by the Law of Cosines:
$\cos(\angle BGA) = \frac{AG^2 + BG^2 - AB^2}{2 \cdot AG \cdot BG}$.
Using $AG = \frac{2}{3}m_a$ and $BG = \frac{2}{3}m_b$,where $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$ and $m_b^2 = \frac{2a^2+2c^2-b^2}{4}$,and substituting $a^2+b^2=5c^2$,we find $\cos(\angle BGA) = 0$,so $\angle BGA = 90^{\circ}$.

(B) Given $f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1$ and $g(x) = x^4 - x^3 - x^2 - 1$.
By performing polynomial division of $f(x)$ by $g(x)$,we get:
$f(x) = (x^2 - x)g(x) + (2x^2 - 2x - 1)$.
Since $a, b, c, d$ are roots of $g(x) = 0$,we have $g(a) = g(b) = g(c) = g(d) = 0$.
Thus,$f(a) = 2a^2 - 2a - 1$,$f(b) = 2b^2 - 2b - 1$,$f(c) = 2c^2 - 2c - 1$,and $f(d) = 2d^2 - 2d - 1$.
Summing these,we get $\sum f(a) = 2\sum a^2 - 2\sum a - 4$.
From $g(x) = x^4 - x^3 - x^2 - 1 = 0$,by Vieta's formulas,$\sum a = 1$ and $\sum ab = -1$.
We know $\sum a^2 = (\sum a)^2 - 2\sum ab = (1)^2 - 2(-1) = 1 + 2 = 3$.
Substituting these values,$\sum f(a) = 2(3) - 2(1) - 4 = 6 - 2 - 4 = 0$.

(D) The given equation is $\sin(\pi \sin^2 \theta) + \sin(\pi \cos^2 \theta) = 2 \cos(\frac{\pi}{2} \cos \theta)$.
Using the identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:
$2 \sin(\frac{\pi(\sin^2 \theta + \cos^2 \theta)}{2}) \cos(\frac{\pi(\sin^2 \theta - \cos^2 \theta)}{2}) = 2 \cos(\frac{\pi}{2} \cos \theta)$.
Since $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin^2 \theta - \cos^2 \theta = -\cos 2\theta$,we have:
$2 \sin(\frac{\pi}{2}) \cos(-\frac{\pi}{2} \cos 2\theta) = 2 \cos(\frac{\pi}{2} \cos \theta)$.
Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(-x) = \cos x$,this simplifies to:
$\cos(\frac{\pi}{2} \cos 2\theta) = \cos(\frac{\pi}{2} \cos \theta)$.
This implies $\frac{\pi}{2} \cos 2\theta = 2n\pi \pm \frac{\pi}{2} \cos \theta$,so $\cos 2\theta = 4n \pm \cos \theta$ for $n \in Z$.
Case $I$: $\cos 2\theta - \cos \theta = 4n$. For $n=0$,$2\cos^2 \theta - 1 - \cos \theta = 0$,so $(2\cos \theta + 1)(\cos \theta - 1) = 0$. Thus $\cos \theta = 1$ or $\cos \theta = -1/2$. Solutions in $[0, 2\pi]$ are $\theta = 0, 2\pi, 2\pi/3, 4\pi/3$.
Case $II$: $\cos 2\theta + \cos \theta = 4n$. For $n=0$,$2\cos^2 \theta + \cos \theta - 1 = 0$,so $(2\cos \theta - 1)(\cos \theta + 1) = 0$. Thus $\cos \theta = 1/2$ or $\cos \theta = -1$. Solutions in $[0, 2\pi]$ are $\theta = \pi, \pi/3, 5\pi/3$.
Total solutions are ${0, 2\pi, 2\pi/3, 4\pi/3, \pi, \pi/3, 5\pi/3}$,which is $7$ solutions.

(C) For an equilateral triangle $ABC$ with side length $a$,let $R$ be the circumradius and $r$ be the inradius.
In an equilateral triangle,the circumradius $R$ is given by $R = \frac{a}{2 \sin 60^{\circ}} = \frac{a}{2(\sqrt{3}/2)} = \frac{a}{\sqrt{3}}$.
The inradius $r$ is given by $r = \frac{a}{2 \tan 60^{\circ}} = \frac{a}{2\sqrt{3}}$.
Therefore,the ratio $\frac{R}{r} = \frac{a/\sqrt{3}}{a/(2\sqrt{3})} = \frac{a}{\sqrt{3}} \times \frac{2\sqrt{3}}{a} = 2$.
Since the ratio is $2$,which is independent of $a$,it remains constant.

(C) Given the quadratic equation $2x^2 + bx + \frac{1}{b} = 0$,it has two distinct real roots,so the discriminant $D > 0$.
$D = b^2 - 4(2)(\frac{1}{b}) > 0$
$b^2 - \frac{8}{b} > 0 \Rightarrow \frac{b^3 - 8}{b} > 0$
Using the factorization $b^3 - 8 = (b - 2)(b^2 + 2b + 4)$,and noting that $b^2 + 2b + 4 = (b + 1)^2 + 3 > 0$ for all real $b$,the inequality simplifies to:
$\frac{b - 2}{b} > 0$
This holds when $b \in (-\infty, 0) \cup (2, \infty)$.
Now,check option $(c)$:
$b^2 - 3b > -2 \Rightarrow b^2 - 3b + 2 > 0$
$(b - 2)(b - 1) > 0$
This holds when $b \in (-\infty, 1) \cup (2, \infty)$.
Since the set $(-\infty, 0) \cup (2, \infty)$ is a subset of $(-\infty, 1) \cup (2, \infty)$,the condition $b^2 - 3b > -2$ is satisfied for all $b$ that satisfy the original quadratic equation condition.

(B) Given $p(x) = x^2 + ax + b$ has two distinct real roots,say $r_1$ and $r_2$. Thus,$p(x) = (x - r_1)(x - r_2)$.
Then $g(x) = p(x^3) = (x^3 - r_1)(x^3 - r_2)$.
Since $x^3 - r = 0$ has exactly one real root for any real $r$,the roots of $g(x)$ are $x = \sqrt[3]{r_1}$ and $x = \sqrt[3]{r_2}$.
Since $r_1 \neq r_2$,we have $\sqrt[3]{r_1} \neq \sqrt[3]{r_2}$. Thus,$g(x)$ has exactly two distinct real roots. Statement $I$ is true and $II$ is false.
For statement $III$,$g(x) = (x^3)^2 + a(x^3) + b = x^6 + ax^3 + b$. As $x \to \infty$,$g(x) \to \infty$. Since $g(x)$ is a continuous polynomial of even degree $(6)$,it must have a global minimum value. Thus,there exists a real number $\alpha$ such that $g(x) \geq \alpha$ for all real $x$. Statement $III$ is true.
Therefore,both $I$ and $III$ are true.

(A) The sum of the first $n$ terms of an arithmetic progression with first term $a = 2$ and common difference $d = 4$ is given by:
$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(2) + (n-1)4] = \frac{n}{2}[4 + 4n - 4] = \frac{n}{2}[4n] = 2n^2$.
The average of the first $n$ terms is $M_n = \frac{S_n}{n} = \frac{2n^2}{n} = 2n$.
We need to calculate the sum $\sum_{n=1}^{10} M_n = \sum_{n=1}^{10} 2n$.
$= 2 \sum_{n=1}^{10} n = 2 \times \frac{10(10+1)}{2} = 10 \times 11 = 110$.

(A) Given that in $\triangle ABC$,$\angle BAC = 90^{\circ}$ and $AD$ is the altitude from $A$ onto $BC$.
Since $AB = 15$ and $BC = 25$,by the Pythagorean theorem:
$AC = \sqrt{BC^2 - AB^2} = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20$.
The area of $\triangle ABC$ can be calculated in two ways:
Area $= \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 15 \times 20 = 150$.
Also,Area $= \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 25 \times AD$.
Equating the two areas: $\frac{1}{2} \times 25 \times AD = 150 \implies AD = \frac{300}{25} = 12$.
In quadrilateral $AEDF$,$\angle FAE = 90^{\circ}$,$\angle AFD = 90^{\circ}$,and $\angle AED = 90^{\circ}$. Thus,$AEDF$ is a rectangle.
In a rectangle,the diagonals are equal. Therefore,$EF = AD$.
Since $AD = 12$,the length of $EF$ is $12$.

(B) Given relations for the sides of the triangle are:
$c^2 = 2ab$ $(i)$
$a^2 + c^2 = 3b^2$ $(ii)$
Substituting $(i)$ into $(ii)$:
$a^2 + 2ab = 3b^2$
$a^2 + 2ab - 3b^2 = 0$
$(a + 3b)(a - b) = 0$
Since $a$ and $b$ are side lengths,$a, b > 0$,so $a = b$.
Substituting $a = b$ into $(i)$:
$c^2 = 2(b)(b) = 2b^2$
$c = \sqrt{2}b$
Now,we have sides $a = b$,$b = b$,and $c = \sqrt{2}b$.
Since $a^2 + b^2 = b^2 + b^2 = 2b^2 = c^2$,the triangle is a right-angled triangle with the right angle at $C$ (i.e.,$\angle C = 90^{\circ}$).
Since $a = b$,the triangle is an isosceles right-angled triangle.
Therefore,$\angle A = \angle B = 45^{\circ}$.
The measure of $\angle BAC$ is $45^{\circ}$.

(A) Let the number be $N$. When a non-zero digit $c$ is appended to $N$,the new number is $10N + c$.
Given that $10N + c$ is divisible by $c$ for all $c \in \{1, 2, \dots, 9\}$.
This implies that $10N$ must be divisible by $c$ for all $c \in \{1, 2, \dots, 9\}$.
Therefore,$10N$ must be a multiple of the least common multiple $(LCM)$ of the digits $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
$LCM(1, 2, 3, 4, 5, 6, 7, 8, 9) = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$.
So,$10N$ must be a multiple of $2520$,which means $N$ must be a multiple of $252$.
The least positive integer $N$ is $252$.
The sum of the digits of $N$ is $2 + 5 + 2 = 9$.

(C) Let the $11$ distinct positive integers be arranged in increasing order: $x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10} < x_{11}$.
The median of these $11$ integers is $x_6$.
The other $10$ integers are $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}$.
The median of these $10$ integers is $m = \frac{x_5 + x_6}{2}$.
Since $x_5 < x_6$,it follows that $x_5 < \frac{x_5 + x_6}{2} < x_6$,so $x_5 < m < x_6$.
Replacing the largest integer $x_{11}$ with $m$,the new set of $11$ integers in increasing order is $x_1, x_2, x_3, x_4, x_5, m, x_6, x_7, x_8, x_9, x_{10}$.
The new median is the $6^{th}$ term,which is $m$.
Since $m < x_6$,the median decreases.

(D) two-digit number $\overline{ab}$ is almost prime if it can be transformed into a prime number by changing at most one digit.
Every two-digit number $\overline{ab}$ can be written as $10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.
There are $90$ two-digit numbers in total,ranging from $10$ to $99$.
Any prime number $p$ (where $10 \le p \le 99$) is itself an almost prime number because we can change zero digits to obtain a prime.
For any non-prime two-digit number,we check if it can be changed into a prime by altering one digit.
Since there are many primes ending in $1, 3, 7, 9$,almost every two-digit number can be converted into a prime by changing the units digit $b$.
For example,any number $\overline{a0}, \overline{a2}, \overline{a4}, \overline{a5}, \overline{a6}, \overline{a8}$ can be changed to a prime by modifying the last digit.
Exhaustive checking shows that all $90$ two-digit numbers satisfy this condition.
Thus,the total count is $90$.

(C) Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively.
Let the areas of the triangles be as follows:
$\text{Area}(\triangle AKP) = \text{Area}(\triangle BKP) = x$
$\text{Area}(\triangle BLP) = \text{Area}(\triangle CLP) = y$
$\text{Area}(\triangle CPM) = \text{Area}(\triangle DPM) = z$
$\text{Area}(\triangle DNP) = \text{Area}(\triangle ANP) = w$
Given:
$\text{Area}(PKAN) = x + w = 25$
$\text{Area}(PLBK) = x + y = 36$
$\text{Area}(PMDN) = z + w = 41$
We need to find $\text{Area}(PLCM) = y + z$.
From the equations:
$(x + y) + (z + w) = 36 + 41 = 77$
$(x + w) + (y + z) = 77$
$25 + (y + z) = 77$
$y + z = 77 - 25 = 52$
Thus,$\text{Area}(PLCM) = 52$.

(C) Given equations are:
$6x + 4y + z = 200$ $(i)$
$x + y + z = 100$ $(ii)$
Subtracting equation $(ii)$ from $(i)$:
$(6x - x) + (4y - y) + (z - z) = 200 - 100$
$5x + 3y = 100$
Since $x$ and $y$ must be non-negative integers,we can express $y$ as $y = \frac{100 - 5x}{3} = \frac{5(20 - x)}{3}$.
For $y$ to be an integer,$(20 - x)$ must be a multiple of $3$. Let $20 - x = 3k$,where $k$ is an integer.
Then $x = 20 - 3k$. Since $x \ge 0$,$20 - 3k \ge 0 \implies 3k \le 20 \implies k \le 6.66$.
Also,since $y \ge 0$,$5(20 - x) \ge 0 \implies x \le 20$. Since $y = 5k$,$k$ must be $\ge 0$.
Possible values for $k$ are $0, 1, 2, 3, 4, 5, 6$.
For each $k$,we find $x = 20 - 3k$ and $y = 5k$. Then $z = 100 - x - y = 100 - (20 - 3k) - 5k = 80 - 2k$.
Since $k \in \{0, 1, 2, 3, 4, 5, 6\}$,$z$ will always be non-negative $(80, 78, 76, 74, 72, 70, 68)$.
There are $7$ such solutions.

(D) Given $N_2 = 165 = 3 \times 5 \times 11$ and $N_1 = 2^{55} + 1$.
We know that if $n$ is an odd integer,then $x^n + y^n$ is divisible by $x + y$.
Since $55$ is odd,$N_1 = 2^{55} + 1^{55}$ is divisible by $2 + 1 = 3$.
Also,$N_1 = (2^5)^{11} + 1^{11} = 32^{11} + 1^{11}$,which is divisible by $32 + 1 = 33$.
Since $33 = 3 \times 11$,$N_1$ is divisible by both $3$ and $11$.
$N_2 = 165 = 15 \times 11 = 5 \times 33$.
Since $N_1$ is a multiple of $33$ and $N_2$ is a multiple of $33$,and $N_1$ is not divisible by $5$ (as $2^{55} + 1 \equiv (2^2)^{27} \times 2 + 1 \equiv (-1)^{27} \times 2 + 1 \equiv -2 + 1 \equiv -1 \equiv 4 \pmod{5}$),the $HCF$ of $N_1$ and $N_2$ is $33$.

(A) Given the circumference of circle $C$ is $l$,we have $2 \pi r = l$,which implies $r = \frac{l}{2 \pi}$.
Thus,the area of circle $C$ is $A_C = \pi r^2 = \frac{l^2}{4 \pi}$.
For a triangle $T$ with perimeter $l$,the area $A_T$ can be arbitrarily small by choosing a very thin triangle (e.g.,sides $\frac{l}{2}-\epsilon, \frac{l}{2}-\epsilon, 2\epsilon$).
Since $A_C$ is fixed for a given $l$ and $A_T$ can be made arbitrarily close to $0$,the ratio $\frac{A_C}{A_T}$ can be made arbitrarily large.
Therefore,for any positive real number $\alpha$,we can choose $T$ such that $\frac{\operatorname{Area}(C)}{\operatorname{Area}(T)} > \alpha$.

(B) Let the three-digit number be $\overline{abc}$,where $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$.
The arithmetic mean of $b$ and $c$ is $\frac{b+c}{2}$.
The geometric mean of $b$ and $c$ is $\sqrt{bc}$. The square of the geometric mean is $bc$.
Given that $\frac{b+c}{2} = bc$,which implies $b+c = 2bc$.
Case $1$: If $b=0$,then $c=0$. Since $a$ can be any digit from $1$ to $9$,there are $9$ such numbers $(100, 200, \dots, 900)$.
Case $2$: If $b, c \neq 0$,then $\frac{1}{c} + \frac{1}{b} = 2$. For $b, c \in \{1, 2, \dots, 9\}$,the only solution is $b=1$ and $c=1$.
Since $a$ can be any digit from $1$ to $9$,there are $9$ such numbers $(111, 211, \dots, 911)$.
Total number of such three-digit numbers = $9 + 9 = 18$.

(C) Given that $a, b$ are roots of $x^2-5cx-6d=0$,we have:
$a+b=5c$ $(1)$
$ab=-6d$ $(2)$
Given that $c, d$ are roots of $x^2-5ax-6b=0$,we have:
$c+d=5a$ $(3)$
$cd=-6b$ $(4)$
Subtracting $(3)$ from $(1)$:
$(a+b)-(c+d) = 5c-5a$
$(a-c)+(b-d) = -5(a-c)$
$b-d = -6(a-c) = 6(c-a)$ $(5)$
Adding $(1)$ and $(3)$:
$(a+b)+(c+d) = 5c+5a$
$(b+d) = 4(a+c)$ $(6)$
From $(2)$ and $(4)$,$ab-cd = -6d+6b = 6(b-d)$.
Substituting $(5)$ into this: $ab-cd = 6(6(c-a)) = 36(c-a)$.
Also,$ab-cd = a(5c-a) - c(5a-c) = 5ac - a^2 - 5ac + c^2 = c^2-a^2 = (c-a)(c+a)$.
Since $a, b, c, d$ are distinct,$c-a \neq 0$,so $c+a = 36$.
Substituting this into $(6)$: $b+d = 4(36) = 144$.

(A) The quadratic equation $4x^2+bx+c=0$ has equal roots if its discriminant $D = b^2 - 4(4)(c) = 0$.
This implies $b^2 = 16c$,or $b^2 = (4\sqrt{c})^2$,which means $b = 4\sqrt{c}$.
Since $b$ must be an integer and $b \in S$,$c$ must be a perfect square such that $4\sqrt{c} \in \{1, 2, \ldots, 100\}$.
Let $c = k^2$ for some integer $k$. Then $b = 4k$.
Since $1 \le b \le 100$,we have $1 \le 4k \le 100$,which implies $1 \le k \le 25$.
Also,$c = k^2$ must be in $S$,so $1 \le k^2 \le 100$,which implies $1 \le k \le 10$.
Thus,the possible values for $k$ are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
There are $10$ such pairs $(b, c)$.
The total number of ways to choose $b$ and $c$ from $S$ is $100 \times 100 = 10000$.
The probability is $\frac{10}{10000} = \frac{1}{1000} = 0.001$.

(B) Given $p = p_1 + p_2 = p_3 - p_4$ where $p, p_1, p_2, p_3, p_4$ are primes.
Case $I$: If all $p_1, p_2, p_3, p_4$ are odd,then $p_1 + p_2$ is even,which implies $p$ is even. Since $p$ is a prime,$p = 2$. However,$p_1 + p_2 = 2$ is impossible for primes $p_1, p_2 \ge 2$.
Case $II$: At least one of $p_1, p_2$ must be $2$. Let $p_2 = 2$. Then $p = p_1 + 2$. Since $p$ and $p_1$ are primes,$p_1$ must be odd (unless $p_1=2$,then $p=4$ not prime). If $p_1$ is odd,$p$ is odd.
From $p = p_3 - p_4$,we have $p_3 = p + p_4$. Since $p$ is odd,for $p_3$ to be prime,$p_4$ must be $2$ (if $p_4$ were odd,$p_3$ would be even and $>2$,hence not prime).
Thus,$p = p_1 + 2$ and $p = p_3 - 2$,which means $p_3 = p + 2$ and $p_1 = p - 2$.
We need $p-2, p, p+2$ to be primes. This is a prime triplet of the form $(n, n+2, n+4)$. The only such triplet is $(3, 5, 7)$.
Therefore,$p = 5$ is the only solution.
The number of special primes is $1$.

(B) Given in $\triangle ABC$,$AB=BC$ and $\frac{AX}{XB} = \frac{AB}{AX} = k$.
Since $AB = AX + XB$,we have $\frac{AB}{AX} = \frac{AX+XB}{AX} = 1 + \frac{XB}{AX} = 1 + \frac{1}{k}$.
Thus,$k = 1 + \frac{1}{k} \Rightarrow k^2 - k - 1 = 0$.
Since $k > 0$,$k = \frac{1+\sqrt{5}}{2}$.
Then $\frac{AX}{AB} = \frac{1}{k} = \frac{2}{1+\sqrt{5}} = \frac{\sqrt{5}-1}{2}$.
Let $\angle ABC = \theta$. Since $AB=BC$,$\angle BAC = \angle BCA = \frac{180^{\circ}-\theta}{2} = 90^{\circ} - \frac{\theta}{2}$.
In $\triangle AXC$,$AC=AX$,so $\angle AXC = \angle ACX = \angle BCA = 90^{\circ} - \frac{\theta}{2}$.
Then $\angle XAC = 180^{\circ} - 2(90^{\circ} - \frac{\theta}{2}) = \theta$.
Since $\angle BAC = 90^{\circ} - \frac{\theta}{2}$,we have $\angle BAX = \angle BAC - \angle XAC = (90^{\circ} - \frac{\theta}{2}) - \theta = 90^{\circ} - \frac{3\theta}{2}$.
In $\triangle BAX$,by the Law of Sines: $\frac{AX}{\sin \theta} = \frac{AB}{\sin(90^{\circ}-\frac{\theta}{2})}$.
$\frac{AX}{AB} = \frac{\sin \theta}{\cos(\theta/2)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos(\theta/2)} = 2\sin(\theta/2)$.
Equating the two expressions for $\frac{AX}{AB}$: $2\sin(\theta/2) = \frac{\sqrt{5}-1}{2} \Rightarrow \sin(\theta/2) = \frac{\sqrt{5}-1}{4} = \sin 18^{\circ}$.
Therefore,$\frac{\theta}{2} = 18^{\circ} \Rightarrow \theta = 36^{\circ}$.

(B) Given the curves:
$x(y-e^x)=\sin x \implies y = \frac{\sin x}{x} + e^x$
$2xy = 2\sin x + x^3 \implies y = \frac{\sin x}{x} + \frac{x^2}{2}$
For $x \in [1, 2]$,$e^x > \frac{x^2}{2}$,so the area $A$ is given by:
$A = \int_{1}^{2} \left( \left( \frac{\sin x}{x} + e^x \right) - \left( \frac{\sin x}{x} + \frac{x^2}{2} \right) \right) dx$
$A = \int_{1}^{2} \left( e^x - \frac{x^2}{2} \right) dx$
$A = \left[ e^x - \frac{x^3}{6} \right]_{1}^{2}$
$A = \left( e^2 - \frac{2^3}{6} \right) - \left( e^1 - \frac{1^3}{6} \right)$
$A = e^2 - \frac{8}{6} - e + \frac{1}{6}$
$A = e^2 - e - \frac{7}{6}$

(B) Given the equation: $-3 x^4 + \operatorname{det}\begin{bmatrix} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{bmatrix} = 0$.
First,we evaluate the determinant $D = \operatorname{det}\begin{bmatrix} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{bmatrix}$.
Using the property of Vandermonde-like determinants or direct expansion:
$D = 1(x^2 \cdot x^6 - x^4 \cdot x^3) - x(1 \cdot x^6 - x^4 \cdot 1) + x^2(1 \cdot x^3 - x^2 \cdot 1)$
$D = (x^8 - x^7) - x(x^6 - x^4) + x^2(x^3 - x^2)$
$D = x^8 - x^7 - x^7 + x^5 + x^5 - x^4 = x^8 - 2x^7 + 2x^5 - x^4$.
Substituting this into the original equation:
$-3x^4 + x^8 - 2x^7 + 2x^5 - x^4 = 0$
$x^8 - 2x^7 + 2x^5 - 4x^4 = 0$.
Factoring the expression:
$x^4(x^4 - 2x^3 + 2x - 4) = 0$
$x^4(x^3(x - 2) + 2(x - 2)) = 0$
$x^4(x^3 + 2)(x - 2) = 0$.
Setting each factor to zero:
$x^4 = 0 \implies x = 0$
$x^3 + 2 = 0 \implies x^3 = -2 \implies x = \sqrt[3]{-2}$ (not an integer)
$x - 2 = 0 \implies x = 2$.
Since $x$ must be an integer,the possible values are $x = 0$ and $x = 2$.
Thus,there are $2$ such integers.

(B) Given that $P(x)$ is a non-zero polynomial satisfying $P(1+x)=P(1-x)$ for all real $x$ and $P(1)=0$.
Differentiating both sides with respect to $x$,we get:
$P'(1+x) = -P'(1-x)$.
Setting $x=0$,we obtain:
$P'(1) = -P'(1) \implies 2P'(1) = 0 \implies P'(1) = 0$.
Since $P(1)=0$ and $P'(1)=0$,by the Factor Theorem,$(x-1)^2$ must be a factor of $P(x)$.
To check if $m$ can be larger,consider $P(x) = (x-1)^2$. This satisfies $P(1+x) = (1+x-1)^2 = x^2$ and $P(1-x) = (1-x-1)^2 = (-x)^2 = x^2$. Thus,$P(1+x)=P(1-x)$ holds.
Since $P(x) = (x-1)^2$ is a valid polynomial,the largest integer $m$ such that $(x-1)^m$ divides $P(x)$ for all such $P(x)$ is $2$.

(A) Given the function $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 1 & \text{when } x = 0 \end{cases}$.
We want to find the set $A = \{x \in \mathbb{R} : f(x) = 1\}$.
Case $1$: If $x = 0$,then $f(0) = 1$. Thus,$0 \in A$.
Case $2$: If $x \neq 0$,we solve $x \sin \left(\frac{1}{x}\right) = 1$,which implies $\sin \left(\frac{1}{x}\right) = \frac{1}{x}$.
Let $t = \frac{1}{x}$. Then the equation becomes $\sin(t) = t$.
We know that for all $t \neq 0$,$|\sin(t)| < |t|$.
Specifically,if $t > 0$,$\sin(t) < t$,and if $t < 0$,$\sin(t) > t$.
Therefore,the equation $\sin(t) = t$ has only one solution at $t = 0$.
However,our substitution was $t = \frac{1}{x}$,and $x \neq 0$ implies $t$ cannot be $0$.
Thus,there are no solutions for $x \neq 0$.
Consequently,the only element in set $A$ is ${0}$.
Therefore,$A$ has exactly one element.

(C) The equations of the planes passing through the origin are given by the dot product of the normal vectors with the position vector $(x, y, z) = 0$.
$\sigma_1: x + y + z = 0$
$\sigma_2: ax + by + cz = 0$
$\sigma_3: a^2x + b^2y + c^2z = 0$
For the intersection of these three planes to be a single point (the origin),the determinant of the coefficient matrix must be non-zero:
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \neq 0$
This is a standard Vandermonde determinant. We can simplify it by performing column operations:
Apply $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$\Delta = \begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{vmatrix} = (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^2 & b+a & c+a \end{vmatrix}$
Apply $C_3 \rightarrow C_3 - C_2$:
$\Delta = (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ a^2 & b+a & c-b \end{vmatrix} = (b-a)(c-a)(c-b) = -(a-b)(b-c)(c-a)$
For $\Delta \neq 0$,we must have $a \neq b$,$b \neq c$,and $c \neq a$. Thus,$a, b$,and $c$ must be three distinct positive values.

(A) Initially,Ravi has ${2R, 2B}$ and Rashmi has ${2R, 2B}$. Total cards with each is $4$.
Step $1$: Ravi picks a card from Rashmi. Rashmi now has $3$ cards. Then Rashmi picks a card from Ravi. Ravi now has $4$ cards again.
To have all $4$ cards of the same colour,Ravi must end up with $4$ Red cards and Rashmi with $4$ Black cards,or vice versa.
Case $1$: Ravi ends with $4$ Red cards and Rashmi with $4$ Black cards.
In the first exchange,Ravi must pick a Red card from Rashmi (prob $\frac{2}{4}$) and Rashmi must pick a Black card from Ravi (prob $\frac{2}{4}$). After this,Ravi has ${2R, 1B, 1R} = {3R, 1B}$ and Rashmi has ${1R, 2B, 1B} = {1R, 3B}$.
In the second exchange,Ravi must pick the remaining Red card from Rashmi (prob $\frac{1}{4}$) and Rashmi must pick the remaining Black card from Ravi (prob $\frac{1}{4}$).
Probability of Case $1$ = $(\frac{2}{4} \times \frac{2}{4}) \times (\frac{1}{4} \times \frac{1}{4}) = \frac{4}{16} \times \frac{1}{16} = \frac{4}{256} = \frac{1}{64}$.
Case $2$: Ravi ends with $4$ Black cards and Rashmi with $4$ Red cards.
By symmetry,the probability is also $\frac{1}{64}$.
Total probability $p = \frac{1}{64} + \frac{1}{64} = \frac{2}{64} = \frac{1}{32} = 0.03125 = 3.125 \%$.
Since $3.125 \% \leq 5 \%$,the correct option is $A$.

(C) The region $A_1$ is defined by $x \geq 0, y \geq 0$ and $2x + 2y - x^2 - y^2 > 1 > x + y$.
Rearranging the inequality $2x + 2y - x^2 - y^2 > 1$,we get $x^2 - 2x + y^2 - 2y < -1$,which simplifies to $(x-1)^2 + (y-1)^2 < 1$. This is the interior of a circle centered at $(1, 1)$ with radius $1$. The condition $x + y < 1$ represents the region below the line $x + y = 1$ in the first quadrant.
For $A_2$,the region is $x \geq 0, y \geq 0$ and $x + y > 1 > x^2 + y^2$. This is the region between the line $x + y = 1$ and the circle $x^2 + y^2 = 1$ in the first quadrant.
For $A_3$,the region is $x \geq 0, y \geq 0$ and $x + y > 1 > x^3 + y^3$. This is the region between the line $x + y = 1$ and the curve $x^3 + y^3 = 1$ in the first quadrant.
By symmetry and geometric transformation,$|A_1| = |A_2|$.
Comparing the curves $x^2 + y^2 = 1$ and $x^3 + y^3 = 1$ in the first quadrant,for $0 < x < 1$,$x^3 < x^2$ and $y^3 < y^2$,so $x^3 + y^3 < x^2 + y^2$. Thus,the region bounded by $x^3 + y^3 = 1$ and $x + y = 1$ is larger than the region bounded by $x^2 + y^2 = 1$ and $x + y = 1$.
Therefore,$|A_3| > |A_2| = |A_1|$.
Hence,the correct option is $(c)$.

(D) Given $f(x^2) = f(x^3)$ for all $x \in R$.
For any $x > 0$,let $x = t^6$. Then $f(t^{12}) = f(t^{18})$.
More generally,for any $x > 0$,we can write $x = t^{6^n}$.
By repeating the substitution,$f(x) = f(x^{2/3}) = f(x^{(2/3)^2}) = \dots = f(x^{(2/3)^n})$.
As $n \to \infty$,$(2/3)^n \to 0$,so $f(x) = f(x^0) = f(1)$ for all $x > 0$.
Since $f$ is continuous at $x = 0$,$f(0) = \lim_{x \to 0^+} f(x) = f(1)$.
For $x < 0$,let $x = -t$ where $t > 0$. Then $f((-t)^2) = f((-t)^3) \implies f(t^2) = f(-t^3)$.
Since $f(t^2) = f(1)$,we have $f(-t^3) = f(1)$.
As $t^3$ covers all positive values,$f(x) = f(1)$ for all $x < 0$.
Thus,$f(x) = c$ (a constant) for all $x \in R$.
$A$ constant function is an even function because $f(-x) = c = f(x)$.
$A$ constant function is also differentiable everywhere with $f'(x) = 0$.
Therefore,both $II$ and $III$ are true.

(A) Given the continuous function $f:[0, \infty) \rightarrow R$ satisfying $f(x) = 2 \int_0^x t f(t) d t + 1, \forall x \geq 0$.
Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:
$f'(x) = 2x f(x)$.
Rearranging the terms,we get $\frac{f'(x)}{f(x)} = 2x$.
Integrating both sides with respect to $x$:
$\int \frac{f'(x)}{f(x)} dx = \int 2x dx$.
$\ln |f(x)| = x^2 + C$.
$f(x) = K e^{x^2}$,where $K = e^C$.
To find $K$,we evaluate the original equation at $x=0$:
$f(0) = 2 \int_0^0 t f(t) dt + 1 = 0 + 1 = 1$.
Substituting $x=0$ into $f(x) = K e^{x^2}$,we get $f(0) = K e^0 = K$.
Thus,$K = 1$,which gives $f(x) = e^{x^2}$.
Finally,$f(1) = e^{(1)^2} = e^1 = e$.

(D) Given the function $f(x) = \begin{cases} \frac{\sin(x^2)}{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$.
First,check for continuity at $x=0$:
$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x} = \lim_{x \rightarrow 0} x \cdot \frac{\sin(x^2)}{x^2} = 0 \cdot 1 = 0 = f(0)$.
Since the limit equals the function value,$f(x)$ is continuous at $x=0$.
Next,check for differentiability at $x=0$:
$f'(0) = \lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{\frac{\sin(h^2)}{h} - 0}{h} = \lim_{h \rightarrow 0} \frac{\sin(h^2)}{h^2} = 1$.
Since the limit exists,$f(x)$ is differentiable at $x=0$ and $f'(0) = 1$.
Now,find $f'(x)$ for $x \neq 0$:
$f'(x) = \frac{d}{dx} \left( \frac{\sin(x^2)}{x} \right) = \frac{x \cdot \cos(x^2) \cdot 2x - \sin(x^2) \cdot 1}{x^2} = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}$.
Check if $f'(x)$ is continuous at $x=0$:
$\lim_{x \rightarrow 0} f'(x) = \lim_{x \rightarrow 0} \left( 2\cos(x^2) - \frac{\sin(x^2)}{x^2} \right) = 2(1) - 1 = 1$.
Since $\lim_{x \rightarrow 0} f'(x) = f'(0) = 1$,the derivative is continuous at $x=0$.
Thus,$f$ is differentiable and the derivative is continuous.

(A) The equation $\vec{r} \cdot \vec{a} = 5$ represents a plane $x + y + z = 5$.
The equation $|\vec{r} - \vec{b}| + |\vec{r} - \vec{c}| = 4$ represents an ellipse in space with foci at $\vec{b}$ and $\vec{c}$,provided the plane contains the foci.
First,check if $\vec{b}$ and $\vec{c}$ lie on the plane $x + y + z = 5$:
For $\vec{b}(2, 2, 1)$,$2 + 2 + 1 = 5$ (True).
For $\vec{c}(5, 1, -1)$,$5 + 1 - 1 = 5$ (True).
Since both foci lie on the plane,the intersection is an ellipse in the plane.
The length of the major axis is $2a = 4$,so $a = 2$.
The distance between the foci is $2ae = |\vec{b} - \vec{c}| = |(2-5)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k}| = |-3\hat{i} + \hat{j} + 2\hat{k}| = \sqrt{(-3)^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
Thus,$2ae = \sqrt{14} \Rightarrow 4e = \sqrt{14} \Rightarrow e = \frac{\sqrt{14}}{4}$.
Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 4(1 - \frac{14}{16}) = 4(\frac{2}{16}) = \frac{8}{16} = \frac{1}{2}$,so $b = \frac{1}{\sqrt{2}}$.
The area of the ellipse is $\pi ab = \pi(2)(\frac{1}{\sqrt{2}}) = \sqrt{2}\pi \approx 1.414 \times 3.14159 \approx 4.44$.
The closest integer is $4$.

(A) To evaluate $J = \int_0^1 \frac{x}{1+x^8} dx$,we analyze the assertions.
For assertion $I$:
Since $1+x^8 < 2$ for all $x \in (0, 1)$,we have $\frac{1}{1+x^8} > \frac{1}{2}$.
Multiplying by $x > 0$,we get $\frac{x}{1+x^8} > \frac{x}{2}$.
Integrating both sides from $0$ to $1$:
$J = \int_0^1 \frac{x}{1+x^8} dx > \int_0^1 \frac{x}{2} dx = \left[ \frac{x^2}{4} \right]_0^1 = \frac{1}{4}$.
Thus,assertion $I$ is true.
For assertion $II$:
Let $u = x^4$,then $du = 4x^3 dx$,which is not directly helpful. Instead,consider the integral $J = \int_0^1 \frac{x}{1+x^8} dx$. Let $u = x^4$,then $du = 4x^3 dx$. This doesn't simplify easily. Let's use the substitution $u = x^4$,$du = 4x^3 dx$. Then $J = \int_0^1 \frac{x}{1+(x^4)^2} dx$. Let $u = x^4$,$du = 4x^3 dx$. This is not standard.
Actually,let $u = x^4$,then $du = 4x^3 dx$. The integral is $J = \frac{1}{4} \int_0^1 \frac{4x^3}{x^2(1+x^8)} dx$.
Alternatively,note that for $x \in (0, 1)$,$x^8 < x^4$,so $1+x^8 < 1+x^4$.
Thus $\frac{1}{1+x^8} > \frac{1}{1+x^4}$.
Then $J = \int_0^1 \frac{x}{1+x^8} dx > \int_0^1 \frac{x}{1+x^4} dx$.
Let $u = x^2$,$du = 2x dx$. Then $\int_0^1 \frac{x}{1+x^4} dx = \frac{1}{2} \int_0^1 \frac{du}{1+u^2} = \frac{1}{2} [\tan^{-1}(u)]_0^1 = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}$.
Since $J > \frac{\pi}{8}$,assertion $II$ is false.
Therefore,only $I$ is true.

(D) Given the differential equation $(f^{\prime}(x))^4 = 16(f(x))^2$,we have $(f^{\prime}(x))^2 = \pm 4f(x)$.
Since $(f^{\prime}(x))^2 \geq 0$,we must have $4f(x) \geq 0$,so $f(x) \geq 0$.
Thus,$(f^{\prime}(x))^2 = 4f(x)$,which implies $f^{\prime}(x) = \pm 2\sqrt{f(x)}$.
If $f(x) > 0$,then $\frac{f^{\prime}(x)}{\sqrt{f(x)}} = \pm 2$. Integrating both sides,we get $2\sqrt{f(x)} = \pm 2x + C$. Since $f(0)=0$,$C=0$,so $\sqrt{f(x)} = \pm x$,which gives $f(x) = x^2$.
We can construct solutions by patching $f(x) = x^2$,$f(x) = -x^2$ (only if $f(x) \geq 0$ is not strictly required,but here $f(x) \geq 0$ is forced by the square),and $f(x) = 0$.
Specifically,for any $a, b \in [0, 1)$,we can define $f(x)$ piecewise using $x^2$ and $0$ such that it remains differentiable at $x=0$. Since there are infinitely many such combinations,the number of such functions is more than $4$.

(A) Given $f(x) = |\sin x|$ and $g(x) = \int_0^x |\sin t| \, dt$.
$p(x) = g(x) - \frac{2}{\pi} x$.
We evaluate $p(x + \pi) = g(x + \pi) - \frac{2}{\pi}(x + \pi)$.
$g(x + \pi) = \int_0^{x + \pi} |\sin t| \, dt = \int_0^{\pi} |\sin t| \, dt + \int_{\pi}^{x + \pi} |\sin t| \, dt$.
Since $|\sin t|$ is periodic with period $\pi$,$\int_{\pi}^{x + \pi} |\sin t| \, dt = \int_0^x |\sin t| \, dt = g(x)$.
Also,$\int_0^{\pi} |\sin t| \, dt = 2 \int_0^{\pi/2} \sin t \, dt = 2[-\cos t]_0^{\pi/2} = 2(0 - (-1)) = 2$.
Thus,$g(x + \pi) = 2 + g(x)$.
Substituting this into $p(x + \pi)$:
$p(x + \pi) = (2 + g(x)) - \frac{2}{\pi} x - \frac{2}{\pi} \cdot \pi = 2 + g(x) - \frac{2}{\pi} x - 2 = g(x) - \frac{2}{\pi} x = p(x)$.
Therefore,$p(x + \pi) = p(x)$ for all $x$.

(B) Given the equation: $\left(\frac{a_1}{2} + \frac{a_2}{4} + \frac{a_3}{8}\right)^2 = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.
Expanding the left side: $\frac{a_1^2}{4} + \frac{a_2^2}{16} + \frac{a_3^2}{64} + \frac{a_1 a_2}{4} + \frac{a_2 a_3}{16} + \frac{a_3 a_1}{8} = \frac{a_1^2}{2} + \frac{a_2^2}{4} + \frac{a_3^2}{8}$.
Multiplying by $64$ to clear denominators: $16a_1^2 + 4a_2^2 + a_3^2 + 16a_1 a_2 + 4a_2 a_3 + 8a_3 a_1 = 32a_1^2 + 16a_2^2 + 8a_3^2$.
Rearranging terms: $16a_1^2 + 12a_2^2 + 7a_3^2 - 16a_1 a_2 - 4a_2 a_3 - 8a_3 a_1 = 0$.
This can be rewritten as: $8(a_1 - a_2)^2 + (2a_2 - a_3)^2 + 2(a_3 - 2a_1)^2 + 4a_3^2 = 0$.
Since the sum of squares is zero,each term must be zero: $a_1 - a_2 = 0$,$2a_2 - a_3 = 0$,$a_3 - 2a_1 = 0$,and $a_3 = 0$.
This implies $a_1 = a_2 = a_3 = 0$.
Thus,the set $A$ contains exactly one element,which is the zero vector $(0, 0, 0)$.

(A) Given that $f:[0,1] \rightarrow [0,1]$ is a continuous function satisfying $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$.
This implies $(f(x))^2 \leq 1-x^2$,so $f(x) \leq \sqrt{1-x^2}$ since $f(x) \geq 0$.
Integrating both sides from $0$ to $1$,we get $\int_0^1 f(x) dx \leq \int_0^1 \sqrt{1-x^2} dx$.
The integral $\int_0^1 \sqrt{1-x^2} dx$ represents the area of a quarter circle of radius $1$,which is $\frac{1}{4} \pi (1)^2 = \frac{\pi}{4}$.
Since it is given that $\int_0^1 f(x) dx = \frac{\pi}{4}$,the equality must hold,which implies $f(x) = \sqrt{1-x^2}$ for all $x \in [0,1]$.
Now,we evaluate the integral $\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx = \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2}}{1-x^2} dx = \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{1-x^2}} dx$.
This is equal to $[\sin^{-1} x]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} = \sin^{-1}(\frac{1}{\sqrt{2}}) - \sin^{-1}(\frac{1}{2}) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$.

(A) Let $P(x)$ be a cubic polynomial. We are given $P(x) = 2x$ for $x = 1, 2, 3, 4$.
Define a new polynomial $Q(x) = P(x) - 2x$.
Since $P(x)$ is a cubic polynomial,$Q(x)$ is also a polynomial of degree at most $3$.
From the given conditions,$Q(1) = P(1) - 2(1) = 2 - 2 = 0$,$Q(2) = P(2) - 2(2) = 4 - 4 = 0$,$Q(3) = P(3) - 2(3) = 6 - 6 = 0$,and $Q(4) = P(4) - 2(4) = 8 - 8 = 0$.
Thus,$1, 2, 3, 4$ are the roots of $Q(x)$.
Since $Q(x)$ is a polynomial of degree at most $3$ and it has $4$ distinct roots,$Q(x)$ must be the zero polynomial.
Therefore,$P(x) - 2x = 0$,which implies $P(x) = 2x$.
However,$P(x) = 2x$ is a polynomial of degree $1$,not $3$.
Thus,there is no cubic polynomial $P(x)$ that satisfies the given conditions.
Hence,the number of such cubic polynomials is $0$.

(A) Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,let $a = (n+1)^{1/3}$ and $b = n^{1/3}$.
Then $f_n = a - b = \frac{a^3 - b^3}{a^2 + ab + b^2} = \frac{1}{(n+1)^{2/3} + (n+1)^{1/3}n^{1/3} + n^{2/3}}$.
Since $n < n+1$,we have $n^{2/3} < (n+1)^{2/3}$.
Thus,$3n^{2/3} < (n+1)^{2/3} + (n+1)^{1/3}n^{1/3} + n^{2/3} < 3(n+1)^{2/3}$.
Taking the reciprocal,we get $\frac{1}{3(n+1)^{2/3}} < f_n < \frac{1}{3n^{2/3}}$.
Replacing $n$ with $n+1$,we get $\frac{1}{3(n+2)^{2/3}} < f_{n+1} < \frac{1}{3(n+1)^{2/3}}$.
Combining these inequalities,we have $f_{n+1} < \frac{1}{3(n+1)^{2/3}} < f_n$ for all $n \in N$.
Therefore,$A = N$.