KVPY 2021 Mathematics Question Paper with Answer and Solution

(B) The set of digits is $\{0, 1, 2, 3, 4, 5, 6\}$.
The total number of $7$-digit numbers that can be formed is $7! - 6! = 6 \times 6! = 4320$.
$A$ number is divisible by $4$ if its last two digits form a number divisible by $4$.
Case $1$: The last two digits do not contain $0$. The possible pairs are $\{12, 16, 24, 32, 36, 52, 56, 64\}$. There are $8$ such pairs.
For each pair,the remaining $5$ digits can be arranged in $5! - 4!$ ways (excluding $0$ at the first position).
So,$8 \times (5! - 4!) = 8 \times (120 - 24) = 8 \times 96 = 768$.
Case $2$: The last two digits contain $0$. The possible pairs are $\{04, 20, 40, 60\}$. There are $4$ such pairs.
For each pair,the remaining $5$ digits can be arranged in $5!$ ways.
So,$4 \times 5! = 4 \times 120 = 480$.
Total favorable outcomes $= 768 + 480 = 1248$.
The probability $P = \frac{1248}{4320} = \frac{13}{45}$.

(C) Given $\log_{a} b = 10$,which implies $\log b = 10 \log a$.
The expression is $\frac{\log_{a} x \cdot \log_{x}(\frac{b}{a})}{\log_{x} b \cdot \log_{ab} x}$.
Using the change of base formula $\log_{m} n = \frac{\log n}{\log m}$,we have:
$\log_{a} x = \frac{\log x}{\log a}$,$\log_{x}(\frac{b}{a}) = \frac{\log b - \log a}{\log x}$,$\log_{x} b = \frac{\log b}{\log x}$,and $\log_{ab} x = \frac{\log x}{\log a + \log b}$.
Substituting these into the expression:
$\frac{p}{q} = \frac{(\frac{\log x}{\log a}) \cdot (\frac{\log b - \log a}{\log x})}{(\frac{\log b}{\log x}) \cdot (\frac{\log x}{\log a + \log b})} = \frac{\frac{\log b - \log a}{\log a}}{\frac{\log b}{\log a + \log b}} = \frac{(\log b - \log a)(\log a + \log b)}{\log a \cdot \log b} = \frac{(\log b)^2 - (\log a)^2}{\log a \cdot \log b}$.
Since $\log b = 10 \log a$,we substitute this into the expression:
$\frac{p}{q} = \frac{(10 \log a)^2 - (\log a)^2}{\log a \cdot (10 \log a)} = \frac{100(\log a)^2 - (\log a)^2}{10(\log a)^2} = \frac{99(\log a)^2}{10(\log a)^2} = \frac{99}{10}$.
Since $p=99$ and $q=10$ are coprime,$p+q = 99+10 = 109$.

(A) Let $P = \sqrt{|x-y|} + \sqrt{|y-z|} + \sqrt{|z-x|}$.
Without loss of generality,assume $0 \leq x \leq y \leq z \leq 1$.
Then $P = \sqrt{y-x} + \sqrt{z-y} + \sqrt{z-x}$.
To maximize $P$,we set $x=0$ and $z=1$,which gives $P = \sqrt{y} + \sqrt{1-y} + 1$.
Let $f(y) = \sqrt{y} + \sqrt{1-y} + 1$ for $y \in [0, 1]$.
Using the substitution $y = \sin^2 \theta$,where $\theta \in [0, \pi/2]$,we get $f(\theta) = \sin \theta + \cos \theta + 1$.
We know that $\sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4)$,which has a maximum value of $\sqrt{2}$ at $\theta = \pi/4$.
Thus,the maximum value of $P$ is $\sqrt{2} + 1$.

(A) Let $n$ be an integer. When a nonzero digit $d$ is appended to the right of $n$,the new number is $10n + d$.
We are given that $10n + d$ is divisible by $d$ for all $d \in \{1, 2, \dots, 9\}$.
This implies $\frac{10n + d}{d} = \frac{10n}{d} + 1$ must be an integer for all $d \in \{1, 2, \dots, 9\}$.
Thus,$10n$ must be divisible by every $d \in \{1, 2, \dots, 9\}$.
This means $10n$ must be a multiple of the least common multiple of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
$\text{LCM}(1, 2, 3, 4, 5, 6, 7, 8, 9) = 2^3 \times 3^2 \times 5 \times 7 = 2520$.
So,$10n$ must be a multiple of $2520$,which means $n$ must be a multiple of $252$.
The smallest positive integer $n$ is $252$.
The product of the digits of $N = 252$ is $2 \times 5 \times 2 = 20$.

(C) The points of intersection of the parabola $y = x^2 + 2ax + 2021$ and the $x$-axis $(y = 0)$ are given by the roots of the quadratic equation $x^2 + 2ax + 2021 = 0$.
For the roots to be rational,the discriminant $D$ must be a perfect square.
$D = (2a)^2 - 4(1)(2021) = 4a^2 - 8084 = 4(a^2 - 2021)$.
Thus,$a^2 - 2021$ must be a perfect square,say $\lambda^2$ for some non-negative integer $\lambda$.
$a^2 - \lambda^2 = 2021 \Rightarrow (a - \lambda)(a + \lambda) = 2021$.
Since $2021 = 43 \times 47$,we consider the factors of $2021$ as $(1, 2021)$ and $(43, 47)$.
For the largest value of $a$,we set $a + \lambda = 2021$ and $a - \lambda = 1$.
Adding these equations: $2a = 2022 \Rightarrow a = 1011$.
Checking the other case: $a + \lambda = 47$ and $a - \lambda = 43$ $\Rightarrow 2a = 90$ $\Rightarrow a = 45$.
The largest element of $E$ is $1011$.

(C) The region is bounded by the lines $y = mx + 3$,$y = nx + 3$,and the $x$-axis $(y = 0)$.
For $y = 0$,the $x$-intercepts are $x_1 = -\frac{3}{m}$ (for $m > 0$) and $x_2 = -\frac{3}{n}$ (for $n < 0$).
The intersection point of the two lines $y = mx + 3$ and $y = nx + 3$ is $(0, 3)$.
The base of the triangle on the $x$-axis has length $b = |x_2 - x_1| = |-\frac{3}{n} - (-\frac{3}{m})| = 3 |\frac{1}{m} - \frac{1}{n}|$.
The height of the triangle is $h = 3$.
Area $A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 3 |\frac{1}{m} - \frac{1}{n}| \times 3 = \frac{9}{2} (\frac{1}{m} - \frac{1}{n})$.
To minimize the area,we maximize $m$ and $n$. Given $m \leq \sqrt{3}$ and $n \geq -\sqrt{3}$,we take $m = \sqrt{3}$ and $n = -\sqrt{3}$.
Area $= \frac{9}{2} (\frac{1}{\sqrt{3}} - \frac{1}{-\sqrt{3}}) = \frac{9}{2} (\frac{2}{\sqrt{3}}) = \frac{9}{\sqrt{3}} = 3 \sqrt{3}$.

(C) Let $R$ be the radius of the semicircle $S$ and $r$ be the radius of the small circle with center $C$.
Let $O$ be the center of the semicircle $S$.
The distance from $C$ to the diameter $AB$ is $r$.
The distance from $C$ to the center $O$ is $R - r$ (since the small circle is tangent to the semicircle $S$).
Let $T$ be the projection of $C$ onto the diameter $AB$. Then $CT = r$.
Thus,the distance of $C$ from the point $O$ is equal to the distance of $C$ from the line $AB$ plus some constant,or more precisely,$CO + CT = R$.
Since the distance from $C$ to the point $O$ is equal to the distance from $C$ to a line parallel to $AB$ at a distance $r$ below it,this definition fits the geometric property of a parabola where the distance to a fixed point (focus) equals the distance to a fixed line (directrix).
Therefore,the locus of $C$ is an arc of a parabola.

(B) For the quadratic equation $x^2 + 4x \cos \theta + \cot \theta = 0$ to have equal roots,its discriminant $D$ must be zero.
$D = b^2 - 4ac = (4 \cos \theta)^2 - 4(1)(\cot \theta) = 0$
$16 \cos^2 \theta - 4 \cot \theta = 0$
$4 \cos^2 \theta = \cot \theta$
$4 \cos^2 \theta = \frac{\cos \theta}{\sin \theta}$
Since $0 < \theta < \pi / 2$,$\cos \theta \neq 0$,so we can divide by $\cos \theta$:
$4 \cos \theta \sin \theta = 1$
$2 \sin 2 \theta = 1$
$\sin 2 \theta = \frac{1}{2}$
Since $0 < \theta < \pi / 2$,we have $0 < 2 \theta < \pi$. The solutions for $2 \theta$ are $\frac{\pi}{6}$ and $\frac{5 \pi}{6}$.
Therefore,$\theta = \frac{\pi}{12}$ or $\theta = \frac{5 \pi}{12}$.

(A) Consider the equation $x^2 + y^2 = a^2 + b^2 + c^2$ where $x, y, a, b, c$ are primes.
Case $1$: If all primes are odd,then $x^2, y^2, a^2, b^2, c^2 \equiv 1 \pmod{4}$ or $1 \pmod{8}$.
Specifically,for any odd prime $p$,$p^2 \equiv 1 \pmod{8}$.
Then $x^2 + y^2 \equiv 1 + 1 = 2 \pmod{8}$ and $a^2 + b^2 + c^2 \equiv 1 + 1 + 1 = 3 \pmod{8}$.
Since $2 \not\equiv 3 \pmod{8}$,there are no solutions where all variables are odd primes.
Case $2$: If at least one variable is $2$.
If $x=2$,then $4 + y^2 = a^2 + b^2 + c^2$. If $y$ is also $2$,then $8 = a^2 + b^2 + c^2$. The only primes whose squares are less than $8$ are $2$. If $a=b=c=2$,then $a^2 + b^2 + c^2 = 4 + 4 + 4 = 12 \neq 8$. If any are $2$,we cannot satisfy the sum.
Testing small primes,we find no combination of prime numbers satisfies the equation.
Thus,the number of solutions is $0$.

(B) Let the points in the complex plane be $O(0,0)$,$A(1,0)$,$B(1,1)$,and $C(0,1)$.
The expression represents the sum of distances from a point $z$ to the vertices of a unit square $OABC$.
By the triangle inequality,the sum of distances from any point $z$ to the vertices of a convex quadrilateral is minimized at the intersection of its diagonals.
The diagonals are $OB$ (connecting $(0,0)$ and $(1,1)$) and $AC$ (connecting $(1,0)$ and $(0,1)$).
The intersection point is $z = \frac{1}{2} + \frac{1}{2}i$.
The minimum value is the sum of the lengths of the diagonals:
$|z-0| + |z-(1+i)| = |\frac{1}{2} + \frac{1}{2}i| + |-\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
$|z-1| + |z-i| = |-\frac{1}{2} + \frac{1}{2}i| + |\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
Total minimum value = $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

(D) Let the angles of the triangle be $\alpha, \beta, \gamma$. We are given two angles as $A = 2x + 7$ and $B = 7x + 10$.
For an isosceles triangle,at least two angles must be equal. We consider three cases:
Case $1$: $A = B$
$2x + 7 = 7x + 10$ $\Rightarrow 5x = -3$ $\Rightarrow x = -0.6$.
The angles are $A = 5.8^\circ, B = 5.8^\circ, C = 180 - 11.6 = 168.4^\circ$. This is a valid triangle.
Case $2$: $A = C$
Since $A + B + C = 180^\circ$,we have $2A + B = 180^\circ$.
$2(2x + 7) + (7x + 10) = 180$ $\Rightarrow 4x + 14 + 7x + 10 = 180$ $\Rightarrow 11x = 156$ $\Rightarrow x = \frac{156}{11} \approx 14.18$.
The angles are $A = 35.36^\circ, B = 109.26^\circ, C = 35.36^\circ$. This is a valid triangle.
Case $3$: $B = C$
Since $A + B + C = 180^\circ$,we have $A + 2B = 180^\circ$.
$(2x + 7) + 2(7x + 10) = 180$ $\Rightarrow 2x + 7 + 14x + 20 = 180$ $\Rightarrow 16x = 153$ $\Rightarrow x = \frac{153}{16} = 9.5625$.
The angles are $A = 26.125^\circ, B = 76.9375^\circ, C = 76.9375^\circ$. This is a valid triangle.
Since all three cases yield valid real values for $x$,the total number of such real numbers is $3$.

(C) Since the ellipse is tangent to both axes in the first quadrant,its center is $(a, b)$. The foci are $(a \pm ae, b)$ or $(a, b \pm ae)$. Given the geometry,the center is $(a, b)$ and the distance from the origin to the center is $\sqrt{a^2 + b^2}$.
For an ellipse tangent to the axes,$x_0 = a$ and $y_0 = b$. The foci are $(a \pm ae, b)$.
Given $\triangle OF_1F_2$ is isosceles with $\angle OF_1F_2 = 120^{\circ}$,we have $OF_1 = F_1F_2 = 2ae$.
Using the law of cosines in $\triangle OF_1F_2$ or coordinate geometry properties,we find $e = \frac{1}{2}$.

(D) Let the sides of the triangle be $a=1$,$b$,and $c$. Let the altitudes corresponding to these sides be $h_a$,$h_b$,and $h_c$. The area $\Delta = \frac{1}{2} a h_a = \frac{1}{2} b h_b = \frac{1}{2} c h_c$.
Given $h_a = 1 \cdot h_a$ and $h_b = 2 h_a$ (or vice versa). Since $h_a = \frac{2\Delta}{1}$ and $h_b = \frac{2\Delta}{b}$,the condition that the longest altitude is twice the shortest implies $b=2$ or $c=2$.
For a triangle with sides $1, 2, c$,the triangle inequality requires $2-1 < c < 2+1$,so $1 < c < 3$. Since $c$ is an integer,$c=2$.
The sides are $1, 2, 2$. This is an isosceles triangle.
The semi-perimeter $s = \frac{1+2+2}{2} = \frac{5}{2}$.
The area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} = \frac{\sqrt{15}}{4}$.
The inradius $r = \frac{\Delta}{s} = \frac{\sqrt{15}/4}{5/2} = \frac{\sqrt{15}}{10}$.
The circumradius $R = \frac{abc}{4\Delta} = \frac{1 \cdot 2 \cdot 2}{4 \cdot \sqrt{15}/4} = \frac{4}{\sqrt{15}}$.
Then $\frac{R}{r} = \frac{4/\sqrt{15}}{\sqrt{15}/10} = \frac{40}{15} = \frac{8}{3}$.
Thus $m=8$ and $n=3$. Since $m, n$ are coprime,$m+n = 8+3 = 11$.

(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$.
Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$,the expression simplifies to $1 - 4 \cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 1$.
This implies $\cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 0$.
Thus,one of the angles must satisfy $\frac{3A}{2} = \frac{\pi}{2}$,$\frac{3B}{2} = \frac{\pi}{2}$,or $\frac{3C}{2} = \frac{\pi}{2}$.
This gives $A = \frac{\pi}{3}$,$B = \frac{\pi}{3}$,or $C = \frac{\pi}{3}$.
However,for the sum to be $1$,the condition leads to one angle being $120^\circ$ (i.e.,$\frac{2\pi}{3}$).
Let $A = \frac{2\pi}{3}$. Given circumradius $R = \sqrt{3}$.
Using the sine rule,the side $a = 2R \sin A = 2(\sqrt{3}) \sin \frac{2\pi}{3} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$.
Since $A$ is the largest angle $(120^\circ)$,$a$ is the longest side.
Therefore,the length of the longest side is $3$.

(C) Given the expression $E = \frac{\sin A \cot C + \cos A}{\sin B \cot C + \cos B}$.
Substituting $\cot C = \frac{\cos C}{\sin C}$,we get $E = \frac{\frac{\sin A \cos C + \cos A \sin C}{\sin C}}{\frac{\sin B \cos C + \cos B \sin C}{\sin C}} = \frac{\sin(A+C)}{\sin(B+C)}$.
Since $A+B+C = \pi$,we have $A+C = \pi - B$ and $B+C = \pi - A$.
Thus,$E = \frac{\sin(\pi - B)}{\sin(\pi - A)} = \frac{\sin B}{\sin A} = \frac{b}{a}$.
Given $b^2 = ac$,the sides $a, b, c$ are in a Geometric Progression $(G.P.)$. Let $b = ar$ and $c = ar^2$.
Then $E = \frac{b}{a} = \frac{ar}{a} = r$.
For a triangle to exist,the sum of any two sides must be greater than the third side:
$1) a+b > c$ $\Rightarrow a+ar > ar^2$ $\Rightarrow 1+r > r^2$ $\Rightarrow r^2-r-1 < 0$. Solving this gives $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Since $r > 0$,$r \in \left(0, \frac{1+\sqrt{5}}{2}\right)$.
$2) a+c > b$ $\Rightarrow a+ar^2 > ar$ $\Rightarrow r^2-r+1 > 0$. This is true for all $r$.
$3) b+c > a$ $\Rightarrow ar+ar^2 > a$ $\Rightarrow r^2+r-1 > 0$. Solving this gives $r > \frac{-1+\sqrt{5}}{2}$ (since $r > 0$).
Combining these,$r \in \left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$.

(B) We know that $\sum_{k=0}^{n} \binom{n}{k} k^2 = n(n+1) 2^{n-2}$.
For $n=10$,the sum is $\sum_{k=0}^{10} \binom{10}{k} k^2 = 10(11) 2^{10-2} = 110 \times 2^8$.
Now,we calculate the given expression:
$\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2 = \frac{110 \times 2^8}{2^{10}} = \frac{110}{2^2} = \frac{110}{4} = 27.5$.
Since the question asks for the interval,and $27.5$ is not in the original options provided,we note that the calculation yields $27.5$,which lies in the interval $(27, 28)$.

(C) The total number of ways to draw $5$ tickets from $10$ is $^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
$p_1$ is the probability of obtaining exactly $1$ winning ticket:
$p_1 = \frac{^2C_1 \cdot ^8C_4}{^{10}C_5} = \frac{2 \times 70}{252} = \frac{140}{252} = \frac{5}{9}$.
$p_2$ is the probability of obtaining exactly $2$ winning tickets:
$p_2 = \frac{^2C_2 \cdot ^8C_3}{^{10}C_5} = \frac{1 \times 56}{252} = \frac{56}{252} = \frac{2}{9}$.
Therefore,$p_1 + p_2 = \frac{5}{9} + \frac{2}{9} = \frac{7}{9}$.
Since $\frac{3}{4} = 0.75$ and $\frac{7}{9} \approx 0.777$,the value $\frac{7}{9}$ lies in the interval $\left(\frac{3}{4}, 1\right]$.

(C) If $B, I, O, C$ are concyclic,then $\angle BIC = \angle BOC$ (Angles in the same segment).
We know that $\angle BIC = 90^{\circ} + \frac{A}{2}$ and $\angle BOC = 2A$.
Equating these,we get $90^{\circ} + \frac{A}{2} = 2A$.
$\frac{3A}{2} = 90^{\circ} \implies A = 60^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$\angle B + \angle C = 180^{\circ} - A = 180^{\circ} - 60^{\circ} = 120^{\circ}$.

(C) Given that $AC$ is the bisector of $\angle DAB$,we have $\angle DAC = \angle CAB$.
Since $DC \parallel AB$,the alternate interior angles are equal,so $\angle CAB = \angle ACD$.
Therefore,$\angle DAC = \angle ACD$,which implies that $\triangle ADC$ is an isosceles triangle with $AD = DC$.
Similarly,since $BD$ is the bisector of $\angle CBA$,we have $\angle CBD = \angle DBA$.
Since $DC \parallel AB$,the alternate interior angles are equal,so $\angle DBA = \angle BDC$.
Therefore,$\angle CBD = \angle BDC$,which implies that $\triangle BCD$ is an isosceles triangle with $BC = DC$.
Since $AD = DC$ and $BC = DC$,we have $AD = BC = DC$.
Thus,exactly three sides of the trapezium are equal.

(B) Let $\angle A = \theta$.
The area of $\triangle ABC$ is given by $\text{ar}(\triangle ABC) = \frac{1}{2} \cdot AB \cdot AC \cdot \sin \theta$.
The area of $\triangle ADE$ is given by $\text{ar}(\triangle ADE) = \frac{1}{2} \cdot AD \cdot AE \cdot \sin \theta$.
Therefore,the ratio of the areas is:
$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle ADE)} = \frac{\frac{1}{2} \cdot AB \cdot AC \cdot \sin \theta}{\frac{1}{2} \cdot AD \cdot AE \cdot \sin \theta} = \frac{AB}{AD} \times \frac{AC}{AE}$.
Given $AD : AB = 3 : 5$,we have $\frac{AB}{AD} = \frac{5}{3}$.
Given $AE : AC = 2 : 3$,we have $\frac{AC}{AE} = \frac{3}{2}$.
Thus,the ratio is $\frac{5}{3} \times \frac{3}{2} = \frac{5}{2} = 2.5$.
The value $2.5$ lies in the interval $\left(2, \frac{5}{2}\right]$.
Hence,the correct option is $B$.

(C) In $\triangle ABC$ and $\triangle DCB$,
$AB = DC$ (Given)
$BC = CB$ (Common side)
$AC = DB$ (Given)
By $SSS$ congruence criterion,$\triangle ABC \cong \triangle DCB$.
Therefore,$\angle BAC = \angle CDB = 70^{\circ}$ and $\angle ABC = \angle DCB = 60^{\circ}$.
In $\triangle ABC$,$\angle ACB = 180^{\circ} - \angle BAC - \angle ABC = 180^{\circ} - 70^{\circ} - 60^{\circ} = 50^{\circ}$.
Now,$\angle DCO = \angle BCD - \angle ACB = 60^{\circ} - 50^{\circ} = 10^{\circ}$.
In $\triangle DOC$,$\angle DOC = 180^{\circ} - \angle ODC - \angle DCO = 180^{\circ} - 70^{\circ} - 10^{\circ} = 100^{\circ}$.
The acute angle between $AC$ and $BD$ is $180^{\circ} - 100^{\circ} = 80^{\circ}$.

(C) The sum of the first $n$ integers is given by $S_n = \frac{n(n+1)}{2}$.
After erasing the integer $k$,the sum of the remaining $(n-1)$ integers is $\frac{n(n+1)}{2} - k$.
The average of the remaining numbers is given as $16$,so:
$\frac{\frac{n(n+1)}{2} - k}{n-1} = 16$
$n(n+1) - 2k = 32(n-1)$
$n^2 + n - 2k = 32n - 32$
$2k = n^2 - 31n + 32$
$k = \frac{n^2 - 31n + 32}{2}$
Since $1 < k < n$,we have:
$1 < \frac{n^2 - 31n + 32}{2} < n$
From $k < n$: $n^2 - 31n + 32 < 2n \implies n^2 - 33n + 32 < 0 \implies (n-32)(n-1) < 0$. Since $n \geq 3$,we have $n < 32$.
From $k > 1$: $n^2 - 31n + 32 > 2 \implies n^2 - 31n + 30 > 0 \implies (n-30)(n-1) > 0$. Since $n \geq 3$,we have $n > 30$.
Thus,$n$ must be $31$.
Substituting $n = 31$ into the equation for $k$:
$k = \frac{31^2 - 31(31) + 32}{2} = \frac{32}{2} = 16$.
Therefore,$n + k = 31 + 16 = 47$.

(A) Given $p = 99$ and $q = 101$.
$p_1 = \log_{10} \left(\frac{99+101}{2}\right) = \log_{10} 100 = 2$.
$q_1 = \frac{1}{2}(\log_{10} 99 + \log_{10} 101) = \log_{10} \sqrt{99 \times 101} = \log_{10} \sqrt{9999}$.
Since $9999 < 10000$,$\sqrt{9999} < 100$,so $q_1 < \log_{10} 100 = 2$.
Thus,$p_1 > q_1$.
By the Arithmetic Mean-Geometric Mean inequality,for any two positive numbers $a$ and $b$,$\frac{a+b}{2} > \sqrt{ab}$ if $a \neq b$.
Since $p_1 > q_1$,we have $\frac{p_1+q_1}{2} > \sqrt{p_1 q_1}$.
Taking $\log_{10}$ on both sides,$\log_{10} \left(\frac{p_1+q_1}{2}\right) > \log_{10} \sqrt{p_1 q_1} = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$.
This implies $p_2 > q_2$.
Also,since $p_1 > q_1$,it follows that $\log_{10} p_1 > \log_{10} q_1$.
Since $p_1 > \frac{p_1+q_1}{2} > q_1$,applying the logarithm function (which is strictly increasing) gives $\log_{10} p_1 > p_2 > \log_{10} q_1$.
Combining these,we get $\log_{10} p_1 > p_2 > q_2 > \log_{10} q_1$.

(B) The given polynomial equation is $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$.
This expression follows the binomial expansion of $(x-1)^6$.
Thus,the equation can be written as $(x-1)^6=0$.
This implies that all roots of the equation are equal to $1$.
Therefore,the largest real root $a = 1$ and the smallest real root $b = 1$.
Substituting these values into the expression,we get $\frac{a^2+b^2}{a+b+1} = \frac{1^2+1^2}{1+1+1} = \frac{1+1}{3} = \frac{2}{3}$.

(B) Let $\alpha$ be the common real root of the equations $x^2 - ax + b = 0$ and $x^3 - ax^2 + bx + a - b = 0$.
Since $\alpha$ is a root of the first equation,we have $\alpha^2 - a\alpha + b = 0$.
Substituting this into the second equation: $\alpha(\alpha^2 - a\alpha + b) + a - b = 0$.
Since $\alpha^2 - a\alpha + b = 0$,this simplifies to $a - b = 0$,which means $a = b$.
Substituting $a = b$ into the first equation,we get $x^2 - ax + a = 0$.
For the roots to be real,the discriminant $D \geq 0$,so $a^2 - 4a \geq 0$.
This implies $a(a - 4) \geq 0$,which gives $a \leq 0$ or $a \geq 4$.
Given $1 \leq a, b \leq 2021$ and $a = b$,we must have $4 \leq a \leq 2021$.
The number of such integers $a$ is $2021 - 4 + 1 = 2018$.

(B) Given equation: $\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x+2} = \frac{13}{12}$.
Combining the fractions on the left side:
$\frac{(x+1)(x+2) + x(x+2) + x(x+1)}{x(x+1)(x+2)} = \frac{13}{12}$.
Expanding the numerator:
$\frac{(x^2 + 3x + 2) + (x^2 + 2x) + (x^2 + x)}{x(x+1)(x+2)} = \frac{13}{12}$.
$\frac{3x^2 + 6x + 2}{x^3 + 3x^2 + 2x} = \frac{13}{12}$.
Cross-multiplying:
$12(3x^2 + 6x + 2) = 13(x^3 + 3x^2 + 2x)$.
$36x^2 + 72x + 24 = 13x^3 + 39x^2 + 26x$.
Rearranging into a cubic equation:
$13x^3 + 3x^2 - 46x - 24 = 0$.
Testing for integer roots using the Rational Root Theorem,we find $x=2$ is a root:
$13(8) + 3(4) - 46(2) - 24 = 104 + 12 - 92 - 24 = 0$.
Dividing the cubic by $(x-2)$ gives $(x-2)(13x^2 + 29x + 12) = 0$.
The quadratic $13x^2 + 29x + 12 = 0$ has discriminant $D = 29^2 - 4(13)(12) = 841 - 624 = 217$,which is not a perfect square,so the roots are not integers.
Thus,the only positive integer solution is $x=2$.

(B) Work done by Team $A$ in one day $= \frac{1}{12}$.
Work done by Team $B$ in one day $= \frac{1}{36}$.
Work done by Team $A$ in first $4$ days $= 4 \times \frac{1}{12} = \frac{1}{3}$.
Work done by Team $A$ and $B$ together in the next $2$ days $= 2 \times (\frac{1}{12} + \frac{1}{36}) = 2 \times (\frac{3+1}{36}) = 2 \times \frac{4}{36} = \frac{2}{9}$.
Total work done in $6$ days $= \frac{1}{3} + \frac{2}{9} = \frac{3+2}{9} = \frac{5}{9}$.
Remaining work $= 1 - \frac{5}{9} = \frac{4}{9}$.
Team $B$ doubles its efficiency,so new efficiency $= 2 \times \frac{1}{36} = \frac{1}{18}$.
Additional days required for Team $B$ to complete the remaining work $= \frac{4/9}{1/18} = \frac{4}{9} \times 18 = 8$ days.

(C) We are given that $(n+3)$ divides $(n^3-3)$.
Using polynomial division,we can write:
$\frac{n^3-3}{n+3} = \frac{n^3+27-30}{n+3} = \frac{(n+3)(n^2-3n+9)-30}{n+3} = (n^2-3n+9) - \frac{30}{n+3}$.
For the expression to be an integer,$(n+3)$ must be a divisor of $30$.
Since $n$ is a positive integer,$n \ge 1$,so $n+3 \ge 4$.
The divisors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$.
Considering the condition $n+3 \ge 4$,the possible values for $(n+3)$ are $5, 6, 10, 15, 30$.
Calculating $n$ for each case:
$n+3 = 5 \Rightarrow n = 2$
$n+3 = 6 \Rightarrow n = 3$
$n+3 = 10 \Rightarrow n = 7$
$n+3 = 15 \Rightarrow n = 12$
$n+3 = 30 \Rightarrow n = 27$
Thus,there are $5$ such positive integers $n$.

(A) Given the arithmetic progression $a_1, a_2, \ldots, a_n$ with $a_1 = 1$ and common difference $d = a_2 - a_1 = 5$.
The $n$-th term is given by $a_n = a_1 + (n - 1)d = 1 + (n - 1)5 = 5n - 4$.
We are given $a_k \leq 2021$,so $5k - 4 \leq 2021$,which implies $5k \leq 2025$,or $k \leq 405$.
Since $a_{k+1} > 2021$,the sequence is $a_1, a_2, \ldots, a_{405}$.
The number of terms is $405$,which is odd. The median of a sequence with $n$ terms (where $n$ is odd) is the term at position $\frac{n+1}{2}$.
Here,the median is the term at position $\frac{405+1}{2} = 203$.
The $203$-rd term is $a_{203} = a_1 + (203 - 1)d = 1 + 202 \times 5 = 1 + 1010 = 1011$.

(A) The fifth root of a number $x$ is given by $x^{1/5}$.
Given the expression: $\left(10^{10^{10}}\right)^{\frac{1}{5}}$.
Using the power rule $(a^m)^n = a^{m \times n}$,we get:
$= 10^{10^{10} \times \frac{1}{5}}$.
We can rewrite $10^{10}$ as $10 \times 10^9$:
$= 10^{\frac{10}{5} \times 10^9}$.
$= 10^{2 \times 10^9}$.

(C) Let the $2$-digit number be represented as $10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.
According to the problem,$10a + b = 4(a! + b!)$.
Since $10a + b \leq 99$,we have $4(a! + b!) \leq 99$,which implies $a! + b! \leq 24.75$.
This restricts $a$ and $b$ to values $\leq 4$ because $5! = 120 > 24$.
Testing possible values for $a \in \{1, 2, 3, 4\}$ and $b \in \{0, 1, 2, 3, 4\}$:
If $a=1$: $10(1) + b = 4(1! + b!)$ $\Rightarrow 10 + b = 4(1 + b!)$ $\Rightarrow 6 + b = 4b!$. For $b=2$,$6+2 = 4(2!) = 8$. So,$12$ is a solution.
If $a=2$: $20 + b = 4(2! + b!)$ $\Rightarrow 20 + b = 8 + 4b!$ $\Rightarrow 12 + b = 4b!$. No integer solution for $b \in \{0, 1, 2, 3, 4\}$.
If $a=3$: $30 + b = 4(3! + b!)$ $\Rightarrow 30 + b = 24 + 4b!$ $\Rightarrow 6 + b = 4b!$. For $b=2$,$6+2 = 4(2!) = 8$. So,$32$ is a solution.
If $a=4$: $40 + b = 4(4! + b!) \Rightarrow 40 + b = 96 + 4b!$. No solution as $40+b < 96$.
The set $A = \{12, 32\}$.
The sum of the numbers in $A$ is $12 + 32 = 44$.

(B) Let $x$ be the number of students who chose all three subjects.
Using the Venn diagram approach,the total number of students is the sum of all disjoint regions:
$100 = 15 + 3 + 45 + (23 - x) + (20 - x) + (12 - x) + x$
$100 = 63 + 55 - 2x$
$100 = 118 - 2x$
$2x = 118 - 100$
$2x = 18$
$x = 9$
Thus,the number of students who chose all three subjects is $9$.

(C) Let the sides of the right-angled triangle be $b$,$c$ (legs) and $x$ (hypotenuse).
Given $b + c + x = 42$ --- $(1)$
In a right-angled triangle,the median to the hypotenuse is half the hypotenuse,so $M = \frac{x}{2}$.
The altitude $h$ to the hypotenuse is given by $h = \frac{bc}{x}$.
Given $M - h = 2$,so $\frac{x}{2} - \frac{bc}{x} = 2 \Rightarrow x^2 - 2bc = 4x$ --- $(2)$
From $(1)$,$b + c = 42 - x$. Squaring both sides: $b^2 + c^2 + 2bc = (42 - x)^2$.
Since $b^2 + c^2 = x^2$,we have $x^2 + 2bc = (42 - x)^2$ --- $(3)$
Adding $(2)$ and $(3)$: $2x^2 = (42 - x)^2 + 4x$.
$2x^2 = 1764 - 84x + x^2 + 4x \Rightarrow x^2 + 80x - 1764 = 0$.
Solving the quadratic equation: $(x + 98)(x - 18) = 0$. Since $x > 0$,$x = 18$.
Substitute $x = 18$ into $(2)$: $18^2 - 2bc = 4(18)$ $\Rightarrow 324 - 2bc = 72$ $\Rightarrow 2bc = 252$ $\Rightarrow bc = 126$.
The area of the triangle is $\frac{1}{2}bc = \frac{1}{2} \times 126 = 63$.

(B) Given the quadratic equation $x^2+ax+b=0$. Since $a-b$ is a root,it must satisfy the equation:
$(a-b)^2 + a(a-b) + b = 0$
$a^2 - 2ab + b^2 + a^2 - ab + b = 0$
$b^2 - 3ab + b + 2a^2 = 0$
$b^2 + b(1-3a) + 2a^2 = 0$
Solving for $b$ using the quadratic formula:
$b = \frac{-(1-3a) \pm \sqrt{(3a-1)^2 - 8a^2}}{2} = \frac{(3a-1) \pm \sqrt{9a^2 - 6a + 1 - 8a^2}}{2} = \frac{(3a-1) \pm \sqrt{a^2 - 6a + 1}}{2}$
For $b$ to be an integer,the discriminant $D = a^2 - 6a + 1$ must be a perfect square,say $k^2$.
$a^2 - 6a + 1 = k^2$
$(a-3)^2 - 8 = k^2$
$(a-3)^2 - k^2 = 8$
$(a-3-k)(a-3+k) = 8$
Since the product is $8$ (even),both factors must be even. Let $X = a-3-k$ and $Y = a-3+k$. Then $XY = 8$ and $X+Y = 2(a-3)$.
Possible pairs $(X, Y)$ such that $XY=8$ and $X, Y$ are even:
$1$) $X=2, Y=4 \implies 2(a-3)=6 \implies a-3=3 \implies a=6$. Then $k=1$,$b = \frac{(3(6)-1) \pm 1}{2} = \frac{17 \pm 1}{2} \implies b=9, 8$.
$2$) $X=-4, Y=-2 \implies 2(a-3)=-6 \implies a-3=-3 \implies a=0$. Then $k=1$,$b = \frac{(3(0)-1) \pm 1}{2} = \frac{-1 \pm 1}{2} \implies b=0, -1$.
Thus,the pairs $(a, b)$ are $(6, 9), (6, 8), (0, 0), (0, -1)$.
There are $4$ such ordered pairs.

(B) For any integer $n$,$n^3 \equiv 0, 1, \text{ or } 8 \pmod{9}$,which is equivalent to $n^3 \equiv 0, 1, -1 \pmod{9}$.
Statement $I$: We want to check if $a^3+b^3+c^3 \equiv 0 \pmod{9}$ implies $abc \equiv 0 \pmod{3}$.
If $abc$ is not divisible by $3$,then $a, b, c \not\equiv 0 \pmod{3}$. Thus $a, b, c \equiv 1, 2 \pmod{3}$.
Then $a^3, b^3, c^3 \equiv 1, 8 \pmod{9}$ (since $1^3=1$ and $2^3=8 \equiv -1 \pmod{9}$).
To have $a^3+b^3+c^3 \equiv 0 \pmod{9}$,we need the sum of three values from ${1, -1}$ to be $0 \pmod{9}$,which is impossible as the possible sums are $\{3, 1, -1, -3\}$.
Thus,at least one of $a, b, c$ must be a multiple of $3$,so $abc$ is divisible by $3$. Statement $I$ is true.
Statement $II$: Consider $a=1, b=2, c=1, d=2$.
Then $a^3+b^3+c^3+d^3 = 1^3+2^3+1^3+2^3 = 1+8+1+8 = 18$,which is divisible by $9$.
However,$abcd = 1 \times 2 \times 1 \times 2 = 4$,which is not divisible by $3$.
Thus,Statement $II$ is false.

(A) The given equation is $x^2-x-1=0$. The roots are $x = \frac{1 \pm \sqrt{5}}{2}$.
Given $\lambda = \frac{1+\sqrt{5}}{2}$,then $1-\lambda = \frac{1-\sqrt{5}}{2}$.
Note that $a_n$ is the $n$-th Fibonacci number $F_n$,defined by the Binet formula.
$a_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)$.
Expanding using the binomial theorem:
$(1+\sqrt{5})^n = \sum_{k=0}^n \binom{n}{k} (\sqrt{5})^k$ and $(1-\sqrt{5})^n = \sum_{k=0}^n \binom{n}{k} (-\sqrt{5})^k$.
Then $(1+\sqrt{5})^n - (1-\sqrt{5})^n = 2 \sum_{k \text{ odd}} \binom{n}{k} 5^{(k-1)/2} \cdot 5^{1/2} = 2\sqrt{5} \sum_{j=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2j+1} 5^j$.
Thus,$a_n = \frac{1}{\sqrt{5}} \cdot \frac{1}{2^n} \cdot 2\sqrt{5} \sum_{j=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2j+1} 5^j = \frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2j+1} 5^j$.
Since $a_n$ is the $n$-th Fibonacci number,$a_n$ is an integer for all $n \in N$.
Therefore,$a_n$ is never a non-integer rational number,so $A = \emptyset$.
Also,$a_n$ is never an irrational number,so $B = \emptyset$.
Hence,both sets $A$ and $B$ are empty.

(B) For $\frac{1}{q}$ to have a terminating decimal expansion,$q$ must be of the form $2^m 5^n$ where $m, n \in \mathbb{W}$.
For $\sqrt{q}$ to be rational,$q$ must be a perfect square.
If $q = 2^m 5^n$ is a perfect square,then both $m$ and $n$ must be even.
Let $m = 2a$ and $n = 2b$ where $a, b \in \mathbb{W}$.
Then $q = 2^{2a} 5^{2b} = (2^a 5^b)^2$.
We are given $1 \leq q \leq 2021$,so $1 \leq (2^a 5^b)^2 \leq 2021$,which implies $1 \leq 2^a 5^b \leq \sqrt{2021} \approx 44.95$.
We list the possible values for $2^a 5^b \leq 44$:
If $b=0$: $2^a \leq 44 \Rightarrow a \in \{0, 1, 2, 3, 4, 5\}$ (Values: $1, 2, 4, 8, 16, 32$)
If $b=1$: $2^a \cdot 5 \leq 44$ $\Rightarrow 2^a \leq 8.8$ $\Rightarrow a \in \{0, 1, 2, 3\}$ (Values: $5, 10, 20, 40$)
If $b=2$: $2^a \cdot 25 \leq 44$ $\Rightarrow 2^a \leq 1.76$ $\Rightarrow a = 0$ (Value: $25$)
Total values for $2^a 5^b$ are $6 + 4 + 1 = 11$.
Thus,there are $11$ such integers $q$.

(D) For $x \neq 0$,the function is given by the sum of a geometric series:
$f(x) = |x|^\alpha \sum \limits_{n=0}^{\infty} \left(\frac{1}{1+x^2}\right)^n$.
Since $|\frac{1}{1+x^2}| < 1$ for $x \neq 0$,the sum of the infinite geometric series is $\frac{1}{1 - \frac{1}{1+x^2}} = \frac{1+x^2}{x^2}$.
Thus,$f(x) = |x|^\alpha \cdot \frac{1+x^2}{x^2} = |x|^\alpha \cdot |x|^{-2} (1+x^2) = |x|^{\alpha-2} (1+x^2)$.
For $f$ to be continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0) = 0$.
$\lim_{x \to 0} |x|^{\alpha-2} (1+x^2) = 0$ holds if and only if $\alpha - 2 > 0$,which means $\alpha > 2$.
The set of such real numbers $\alpha$ is the interval $(2, \infty)$.
Since this interval contains infinitely many real numbers,the set has more than $4$ elements.

(B) Given the functional equation $f(x) + f(2x) = 0$ for all $x \in R$.
Substituting $x = 0$,we get $f(0) + f(0) = 0$,which implies $2f(0) = 0$,so $f(0) = 0$.
From the given equation,$f(x) = -f(x/2)$.
By iterating this relation,we get $f(x) = -f(x/2) = f(x/4) = -f(x/8) = \dots = (-1)^n f(x/2^n)$.
Since $f$ is continuous at $x = 0$,we have $\lim_{n \to \infty} f(x/2^n) = f(\lim_{n \to \infty} x/2^n) = f(0) = 0$.
Thus,$f(x) = \lim_{n \to \infty} (-1)^n f(x/2^n) = 0$.
Therefore,the only continuous function satisfying the condition is the zero function $f(x) = 0$.
Hence,there is only $1$ such function.

(C) Given $f(x) = \frac{\cos x}{\cos 2x}$.
Using the quotient rule,$f'(x) = \frac{-\sin x \cos 2x - \cos x (-2 \sin 2x)}{(\cos 2x)^2} = \frac{-\sin x \cos 2x + 2 \sin 2x \cos x}{(\cos 2x)^2}$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have $f'(x) = \frac{\sin(2x-x) + \sin 2x \cos x}{(\cos 2x)^2} = \frac{\sin x + 2 \sin x \cos^2 x}{(\cos 2x)^2}$.
$f'(x) = \frac{\sin x (1 + 2 \cos^2 x)}{(\cos 2x)^2}$.
Since $(1 + 2 \cos^2 x) > 0$ and $(\cos 2x)^2 > 0$ for all $x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$,the sign of $f'(x)$ depends only on $\sin x$.
For $x \in \left(-\frac{\pi}{4}, 0\right)$,$\sin x < 0$,so $f'(x) < 0$,meaning $f(x)$ is decreasing.
For $x \in \left(0, \frac{\pi}{4}\right)$,$\sin x > 0$,so $f'(x) > 0$,meaning $f(x)$ is increasing.
Thus,the correct option is $C$.

(D) Given the differential equation $\frac{dy}{dx} = 2 \sqrt{y}$ with the initial condition $y(0) = 0$.
Case $I$: The trivial solution $y(x) = 0$ for all $x \in \mathbb{R}$ satisfies the equation and the initial condition.
Case $II$: For any constant $a \geq 0$,we can define a family of functions:
$y(x) = \begin{cases} 0 & x < a \\ (x-a)^2 & x \geq a \end{cases}$
Let us check the differentiability at $x = a$:
Left-hand derivative: $\lim_{h \to 0^-} \frac{y(a+h) - y(a)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$.
Right-hand derivative: $\lim_{h \to 0^+} \frac{y(a+h) - y(a)}{h} = \lim_{h \to 0^+} \frac{(a+h-a)^2 - 0}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = 0$.
Since the left-hand derivative equals the right-hand derivative,the function is differentiable at $x = a$.
For $x > a$,$y'(x) = 2(x-a) = 2\sqrt{(x-a)^2} = 2\sqrt{y(x)}$.
Since $a$ can be any non-negative real number,there are infinitely many such functions.

(D) Given the condition: $\int_0^1 (f(x))^2 dx = 2 \int_0^1 f(x) dx$.
Rearranging the terms,we get: $\int_0^1 (f(x))^2 dx - 2 \int_0^1 f(x) dx = 0$.
Adding and subtracting $1$ inside the integral: $\int_0^1 ((f(x))^2 - 2f(x) + 1 - 1) dx = 0$.
This simplifies to: $\int_0^1 (f(x) - 1)^2 dx - \int_0^1 1 dx = 0$.
Therefore,$\int_0^1 (f(x) - 1)^2 dx = 1$.
We are looking for continuous functions $f(x)$ such that the integral of $(f(x) - 1)^2$ over $[0, 1]$ is $1$.
Let $g(x) = f(x) - 1$. Then we need $\int_0^1 (g(x))^2 dx = 1$.
There are infinitely many continuous functions $g(x)$ that satisfy this condition (e.g.,$g(x) = 1$,$g(x) = -1$,$g(x) = \sqrt{3}x$,$g(x) = \sqrt{5}x^2$,etc.).
Since there are infinitely many such functions $g(x)$,there are infinitely many such functions $f(x) = g(x) + 1$.
Thus,the number of such functions is more than $4$.

(A) Let $I = \int \limits_0^{\pi / 2} \frac{\sin x \cos x}{1+\cos ^4 x} d x$.
Substitute $\cos ^2 x = t$. Then,differentiating both sides with respect to $x$,we get $2 \cos x (-\sin x) d x = d t$,which implies $\sin x \cos x d x = -\frac{1}{2} d t$.
Change the limits of integration:
When $x = 0$,$t = \cos ^2(0) = 1$.
When $x = \frac{\pi}{2}$,$t = \cos ^2(\frac{\pi}{2}) = 0$.
Substituting these into the integral:
$I = \int \limits_1^0 \frac{-1/2}{1+t^2} d t = \frac{1}{2} \int \limits_0^1 \frac{1}{1+t^2} d t$.
Evaluating the integral:
$I = \frac{1}{2} [\tan ^{-1}(t)]_0^1 = \frac{1}{2} (\tan ^{-1}(1) - \tan ^{-1}(0)) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$.

(A) We use the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.
First,calculate the inner cross product: $(3\hat{i}+4\hat{j}) \times (\hat{j}+\hat{k}) = 3(\hat{i} \times \hat{j}) + 3(\hat{i} \times \hat{k}) + 4(\hat{j} \times \hat{j}) + 4(\hat{j} \times \hat{k}) = 3\hat{k} - 3\hat{j} + 0 + 4\hat{i} = 4\hat{i} - 3\hat{j} + 3\hat{k}$.
Now,calculate the next cross product: $(\hat{i}-\hat{k}) \times (4\hat{i}-3\hat{j}+3\hat{k}) = \hat{i} \times (4\hat{i}-3\hat{j}+3\hat{k}) - \hat{k} \times (4\hat{i}-3\hat{j}+3\hat{k}) = (0 - 3\hat{k} - 3\hat{j}) - (4\hat{j} + 3\hat{i} + 0) = -3\hat{i} - 7\hat{j} - 3\hat{k}$.
Given $\vec{v} \times (-3\hat{i} - 7\hat{j} - 3\hat{k}) = \vec{0}$,this implies $\vec{v}$ is parallel to $3\hat{i} + 7\hat{j} + 3\hat{k}$.
Let $\vec{v} = \lambda(3\hat{i} + 7\hat{j} + 3\hat{k})$.
Given $\vec{v} \cdot \hat{j} = -7$,we have $7\lambda = -7$,so $\lambda = -1$.
Thus,$\vec{v} = -3\hat{i} - 7\hat{j} - 3\hat{k}$.
Therefore,$\vec{v} \cdot \hat{i} = -3$.

(D) The total number of questions is $n = 8$.
Each question has $4$ options,meaning there is $1$ correct option and $3$ incorrect options.
The student needs to choose exactly $5$ correct answers out of $8$.
The number of ways to select $5$ questions to be correct is given by the combination formula $\binom{8}{5}$.
For the $5$ correct questions,there is only $1$ way to choose the correct option.
For the remaining $8 - 5 = 3$ incorrect questions,each can be answered in $3$ different ways (since there are $4$ options and $1$ is correct,$4 - 1 = 3$ are incorrect).
Thus,the total number of ways is $\binom{8}{5} \times (1)^5 \times (3)^3$.
Calculating this: $\binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Total ways $= 56 \times 1 \times 27 = 1512$.

(D) Let $E$ be the event that the person has the disease,and $E^c$ be the event that the person does not have the disease. Let $A$ be the event that the test result is positive.
Given:
$P(E) = \frac{2}{3}$
$P(E^c) = 1 - \frac{2}{3} = \frac{1}{3}$
$P(A|E) = \frac{2}{3}$ (Probability of testing positive given the person has the disease)
$P(A|E^c) = 1 - \frac{2}{3} = \frac{1}{3}$ (Probability of testing positive given the person does not have the disease)
Using the Law of Total Probability,the probability that a person tests positive is:
$P(A) = P(E) \times P(A|E) + P(E^c) \times P(A|E^c)$
$P(A) = \left(\frac{2}{3} \times \frac{2}{3}\right) + \left(\frac{1}{3} \times \frac{1}{3}\right) = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}$
By Bayes' Theorem,the probability that the person has the disease given that they tested positive is:
$P(E|A) = \frac{P(E) \times P(A|E)}{P(A)}$
$P(E|A) = \frac{\frac{2}{3} \times \frac{2}{3}}{\frac{5}{9}} = \frac{\frac{4}{9}}{\frac{5}{9}} = \frac{4}{5}$

(B) Let $I = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x)^2}$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
When $x = 0$,$\theta = 0$,and when $x \to \infty$,$\theta = \frac{\pi}{2}$.
$I = \int_0^{\pi/2} \frac{\sec^2 \theta \, d\theta}{(1+\tan^2 \theta)(1+\tan \theta)^2} = \int_0^{\pi/2} \frac{\sec^2 \theta \, d\theta}{\sec^2 \theta (1+\tan \theta)^2} = \int_0^{\pi/2} \frac{d\theta}{(1+\tan \theta)^2}$.
Using the property $\int_0^a f(\theta) \, d\theta = \int_0^a f(a-\theta) \, d\theta$,we have:
$I = \int_0^{\pi/2} \frac{d\theta}{(1+\tan(\frac{\pi}{2}-\theta))^2} = \int_0^{\pi/2} \frac{d\theta}{(1+\cot \theta)^2} = \int_0^{\pi/2} \frac{\tan^2 \theta \, d\theta}{(\tan \theta + 1)^2}$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi/2} \frac{1 + \tan^2 \theta}{(1+\tan \theta)^2} \, d\theta = \int_0^{\pi/2} \frac{\sec^2 \theta}{(1+\tan \theta)^2} \, d\theta$.
Let $u = 1 + \tan \theta$,then $du = \sec^2 \theta \, d\theta$.
$2I = \int_1^{\infty} \frac{du}{u^2} = \left[ -\frac{1}{u} \right]_1^{\infty} = 0 - (-1) = 1$.
Therefore,$I = \frac{1}{2}$.

(B) Let $f(x) + x^2 = g^2(x)$ where $g(x) > 0$.
Substituting $f(x) = g^2(x) - x^2$ into the equation:
$4 \int_0^{3/2} (g^2(x) - x^2) dx + 125 \int_0^{3/2} \frac{dx}{g(x)} = 108$
$4 \int_0^{3/2} g^2(x) dx + 125 \int_0^{3/2} \frac{dx}{g(x)} = 108 + 4 \int_0^{3/2} x^2 dx$
$4 \int_0^{3/2} g^2(x) dx + 125 \int_0^{3/2} \frac{dx}{g(x)} = 108 + 4 [\frac{x^3}{3}]_0^{3/2} = 108 + 4(\frac{27/8}{3}) = 108 + 4.5 = 112.5$.
Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$ for the integrand $4g^2(x) + \frac{125}{g(x)} = 4g^2(x) + \frac{62.5}{g(x)} + \frac{62.5}{g(x)}$:
$4g^2(x) + \frac{62.5}{g(x)} + \frac{62.5}{g(x)} \geq 3 \sqrt[3]{4g^2(x) \cdot \frac{62.5}{g(x)} \cdot \frac{62.5}{g(x)}} = 3 \sqrt[3]{4 \cdot 3906.25} = 3 \sqrt[3]{15625} = 3 \cdot 25 = 75$.
Integrating both sides from $0$ to $3/2$:
$\int_0^{3/2} (4g^2(x) + \frac{125}{g(x)}) dx \geq \int_0^{3/2} 75 dx = 75 \cdot \frac{3}{2} = 112.5$.
Since the integral equals $112.5$,the equality must hold for all $x \in [0, 3/2]$.
Thus,$4g^2(x) = \frac{62.5}{g(x)} \Rightarrow g^3(x) = \frac{62.5}{4} = 15.625 = (2.5)^3$.
So,$g(x) = 2.5 = 5/2$.
Therefore,$f(x) = g^2(x) - x^2 = \frac{25}{4} - x^2$.
There is exactly $1$ such function.

(C) We are given the integral $I = \int_1^M \{x\}^{[x]} dx$. Since $[x]$ is constant on intervals $[n, n+1)$,we can write the integral as a sum of integrals over unit intervals:
$I = \sum_{n=1}^{M-1} \int_n^{n+1} (x-n)^n dx$.
Let $u = x-n$,then $du = dx$. The integral becomes $\int_0^1 u^n du = \left[ \frac{u^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1}$.
Thus,$I = \sum_{n=1}^{M-1} \frac{1}{n+1} = \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{M}$.
For $M=2$,$I = \frac{1}{2} = 0.5 < 1$.
For $M=3$,$I = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \approx 0.833 < 1$.
For $M=4$,$I = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6+4+3}{12} = \frac{13}{12} > 1$.
Therefore,the smallest positive integer $M$ is $4$.

(C) Expanding the determinant $f(x)$ along the first column:
$f(x) = 1 \cdot (|x-1|(x-2)^2 - |x-2|(x-1)^2) - |x| \cdot ((x-2)^2 - (x-1)^2) + x^2 \cdot (|x-2| - |x-1|)$
Simplifying the expression:
$f(x) = |x-1|(x^2-4x+4) - |x-2|(x^2-2x+1) - |x|(x^2-4x+4 - x^2+2x-1) + x^2(|x-2|-|x-1|)$
$f(x) = |x-1|(x^2-4x+4-x^2) + |x-2|(x^2-x^2+2x-1) - |x|(-2x+3)$
$f(x) = |x-1|(4-4x) + |x-2|(2x-1) - |x|(3-2x)$
We check for non-differentiability at the critical points of the absolute values: $x = 0, 1, 2$.
At $x=0$: The term $|x|$ is non-differentiable,but the coefficient $(3-2x)$ is $3 \neq 0$. Thus,$f(x)$ is non-differentiable at $x=0$.
At $x=1$: The term $|x-1|$ is non-differentiable,but the coefficient $(4-4x)$ is $0$. Thus,the non-differentiability is removed.
At $x=2$: The term $|x-2|$ is non-differentiable,but the coefficient $(2x-1)$ is $3 \neq 0$. Thus,$f(x)$ is non-differentiable at $x=2$.
Therefore,the function is non-differentiable at $x=0$ and $x=2$. The number of such values is $2$.