WBJEE 2018 Mathematics Question Paper with Answer and Solution

(C) Given equation: $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$
Rearranging the terms:
$\log _{10}\left(1+2^{x}\right) = x \log _{10} 5 - x + \log _{10} 6$
$\log _{10}\left(1+2^{x}\right) = x \log _{10} 5 - x \log _{10} 10 + \log _{10} 6$
$\log _{10}\left(1+2^{x}\right) = \log _{10} 5^{x} - \log _{10} 10^{x} + \log _{10} 6$
$\log _{10}\left(1+2^{x}\right) = \log _{10}\left(\frac{5^{x} \cdot 6}{10^{x}}\right)$
Taking antilog on both sides:
$1+2^{x} = \frac{5^{x} \cdot 6}{2^{x} \cdot 5^{x}}$
$1+2^{x} = \frac{6}{2^{x}}$
Let $2^{x} = t$. Then $1+t = \frac{6}{t}$
$t + t^{2} = 6$
$t^{2} + t - 6 = 0$
$(t+3)(t-2) = 0$
Since $t = 2^{x} > 0$,we have $t = 2$.
$2^{x} = 2^{1} \Rightarrow x = 1$.

(B) Given $0 \leq p \leq 1$ and $a, b > 0$.
Consider the function $f(x) = x^p$.
Since $0 \leq p \leq 1$,the function $f(x) = x^p$ is a concave function for $x > 0$.
By the property of concave functions,for any $a, b > 0$ and $0 < p < 1$,we have $(a+b)^p \leq a^p + b^p$.
This implies $I(p) \leq J(p)$.
For $p=0$,$I(0) = (a+b)^0 = 1$ and $J(0) = a^0 + b^0 = 1 + 1 = 2$,so $1 \leq 2$.
For $p=1$,$I(1) = a+b$ and $J(1) = a+b$,so $a+b \leq a+b$.
Thus,$I(p) \leq J(p)$ holds for all $0 \leq p \leq 1$.

(A) Consider the two quadratic equations:
$x^{2} + b_{1}x + c_{1} = 0$ and $x^{2} + b_{2}x + c_{2} = 0$.
Let $D_{1}$ and $D_{2}$ be the discriminants of these equations respectively.
$D_{1} = b_{1}^{2} - 4c_{1}$
$D_{2} = b_{2}^{2} - 4c_{2}$
Adding the two discriminants,we get:
$D_{1} + D_{2} = b_{1}^{2} + b_{2}^{2} - 4(c_{1} + c_{2})$
Given that $b_{1}b_{2} = 2(c_{1} + c_{2})$,we substitute $4(c_{1} + c_{2}) = 2b_{1}b_{2}$ into the equation:
$D_{1} + D_{2} = b_{1}^{2} + b_{2}^{2} - 2b_{1}b_{2}$
$D_{1} + D_{2} = (b_{1} - b_{2})^{2}$
Since $(b_{1} - b_{2})^{2} \geq 0$ for all real $b_{1}, b_{2}$,it follows that $D_{1} + D_{2} \geq 0$.
If the sum of two real numbers is non-negative,then at least one of them must be non-negative.
Therefore,at least one of $D_{1}$ or $D_{2}$ is $\geq 0$,which implies that at least one of the equations has real roots.

(D) Let $\alpha$ and $\beta$ be the roots of $x^{2}+ax+b=0$ and the roots of $x^{2}-cx+d=0$ be $\alpha^{4}$ and $\beta^{4}$.
From the relations between roots and coefficients:
$\alpha+\beta=-a, \alpha\beta=b$ ...$(i)$
$\alpha^{4}+\beta^{4}=c, \alpha^{4}\beta^{4}=d$ ...$(ii)$
From $(ii),$ $c = \alpha^{4}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2}-2(\alpha\beta)^{2} = ((\alpha+\beta)^{2}-2\alpha\beta)^{2}-2(\alpha\beta)^{2}$.
Substituting $(i)$: $c = (a^{2}-2b)^{2}-2b^{2} = a^{4}+4b^{2}-4a^{2}b-2b^{2} = a^{4}-4a^{2}b+2b^{2}$.
Thus,$2b^{2}-c = 2b^{2}-(a^{4}-4a^{2}b+2b^{2}) = 4a^{2}b-a^{4} = a^{2}(4b-a^{2})$.
Given $a^{2}>4b$,we have $4b-a^{2} < 0$,so $2b^{2}-c < 0$.
For the equation $x^{2}-4bx+(2b^{2}-c)=0$,the product of roots is $2b^{2}-c < 0$.
Since the product of the roots is negative,one root must be positive and the other must be negative.

(D) Let $w = \frac{z_{1}+z_{2}}{z_{1}-z_{2}}$.
To check if $w$ is purely imaginary, we evaluate $w + \bar{w}$.
$w + \bar{w} = \frac{z_{1}+z_{2}}{z_{1}-z_{2}} + \frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{2}}$
$= \frac{(z_{1}+z_{2})(\bar{z}_{1}-\bar{z}_{2}) + (\bar{z}_{1}+\bar{z}_{2})(z_{1}-z_{2})}{(z_{1}-z_{2})(\bar{z}_{1}-\bar{z}_{2})}$
$= \frac{(z_{1}\bar{z}_{1} - z_{1}\bar{z}_{2} + z_{2}\bar{z}_{1} - z_{2}\bar{z}_{2}) + (\bar{z}_{1}z_{1} - \bar{z}_{1}z_{2} + \bar{z}_{2}z_{1} - \bar{z}_{2}z_{2})}{|z_{1}-z_{2}|^2}$
Since $|z_{1}| = |z_{2}|$, we have $z_{1}\bar{z}_{1} = z_{2}\bar{z}_{2} = |z_{1}|^2 = |z_{2}|^2$.
Substituting this, the numerator becomes:
$|z_{1}|^2 - z_{1}\bar{z}_{2} + z_{2}\bar{z}_{1} - |z_{2}|^2 + |z_{1}|^2 - \bar{z}_{1}z_{2} + \bar{z}_{2}z_{1} - |z_{2}|^2$
$= 2|z_{1}|^2 - 2|z_{2}|^2 = 0$.
Since $w + \bar{w} = 0$, $w$ must be purely imaginary.

(B) Given $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11}$.
We can rewrite this as $Z_{r} = -i (\cos \frac{2 \pi r}{11} + i \sin \frac{2 \pi r}{11})$.
Using Euler's formula $e^{i \theta} = \cos \theta + i \sin \theta$,we have $Z_{r} = -i e^{i \frac{2 \pi r}{11}}$.
Now,$\sum_{r=0}^{10} Z_{r} = -i \sum_{r=0}^{10} (e^{i \frac{2 \pi}{11}})^{r}$.
This is a geometric series with $11$ terms where the common ratio is $\omega = e^{i \frac{2 \pi}{11}} \neq 1$.
The sum of the roots of unity $\sum_{r=0}^{n-1} e^{i \frac{2 \pi r}{n}} = 0$ for $n > 1$.
Thus,$\sum_{r=0}^{10} e^{i \frac{2 \pi r}{11}} = 0$.
Therefore,$\sum_{r=0}^{10} Z_{r} = -i \times 0 = 0$.

(C) We know that if $z_{1}, z_{2},$ and $z_{3}$ are the vertices of an equilateral triangle,then $z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}=0$.
Given that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$,we can multiply by $z_{1}z_{2}$ to get $z_{1}^{2}+z_{2}^{2}=z_{1}z_{2}$,which implies $z_{1}^{2}+z_{2}^{2}-z_{1}z_{2}=0$.
If we consider the origin as the third point $z_{3}=0$,the condition becomes $z_{1}^{2}+z_{2}^{2}+0^{2}-z_{1}z_{2}-z_{2}(0)-0(z_{1})=0$,which simplifies to $z_{1}^{2}+z_{2}^{2}-z_{1}z_{2}=0$.
This matches the condition for an equilateral triangle.
Thus,the origin and the points $z_{1}$ and $z_{2}$ form an equilateral triangle.

(B) Each student sends a card to every other student. This means for every pair of students $(A, B)$,student $A$ sends a card to $B$ and student $B$ sends a card to $A$.
This is a permutation problem where we need to arrange $2$ students out of $20$ in a specific order (sender and receiver).
The number of ways to arrange $2$ students out of $20$ is given by the permutation formula ${}^{n}P_{r} = \frac{n!}{(n-r)!}$.
Here $n = 20$ and $r = 2$,so the number of cards is ${}^{20}P_{2} = 20 \times 19 = 380$.
Alternatively,this is equal to ${}^{20}C_{2} \times 2! = \frac{20 \times 19}{2 \times 1} \times 2 = 380$.

(A) The total number of ways to select $4$ numbers from $20$ is given by $^{20}C_{4}$.
The number of ways to select $4$ consecutive numbers from $20$ is $17$ (these are $(1,2,3,4), (2,3,4,5), \ldots, (17,18,19,20)$).
The number of ways to select $4$ non-consecutive numbers is the total number of selections minus the number of consecutive selections.
$\text{Required selections} = {}^{20}C_{4} - 17$
$\text{Required selections} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} - 17$
$\text{Required selections} = 4845 - 17 = 4828$
Calculating the options: $284 \times 17 = 4828$.
Thus,the correct option is $A$.

(A) We need to select $n$ objects from $2n$ objects,where $n$ objects are identical and $n$ objects are distinct.
Let $k$ be the number of distinct objects selected,where $0 \le k \le n$.
Then the remaining $(n-k)$ objects must be selected from the $n$ identical objects.
Since the $n$ identical objects are indistinguishable,there is only $1$ way to select any number of them.
Thus,for each $k$ from $0$ to $n$,the number of ways to select $k$ distinct objects is given by $\binom{n}{k}$.
The total number of ways is the sum of these possibilities:
$\sum_{k=0}^{n} \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^{n}$.

(D) Let $d_1$ be the common difference for the first set $(a, 2b)$. The $(n+2)^{th}$ term is $2b = a + (n+1)d_1$,so $d_1 = \frac{2b-a}{n+1}$.
The $m^{th}$ arithmetic mean is $A_m = a + m d_1 = a + m \left( \frac{2b-a}{n+1} \right)$.
Let $d_2$ be the common difference for the second set $(2a, b)$. The $(n+2)^{th}$ term is $b = 2a + (n+1)d_2$,so $d_2 = \frac{b-2a}{n+1}$.
The $m^{th}$ arithmetic mean is $A'_m = 2a + m d_2 = 2a + m \left( \frac{b-2a}{n+1} \right)$.
Equating the two means: $a + m \left( \frac{2b-a}{n+1} \right) = 2a + m \left( \frac{b-2a}{n+1} \right)$.
Multiplying by $(n+1)$: $a(n+1) + m(2b-a) = 2a(n+1) + m(b-2a)$.
$an + a + 2bm - am = 2an + 2a + bm - 2am$.
Rearranging terms: $2bm - bm - am + 2am = 2an - an + 2a - a$.
$bm + am = an + a$.
$m(b+a) = a(n+1)$.
Thus,$\frac{a}{b} = \frac{m}{n+1-m}$.

(A) For the expansion $(1+x)^n$,the greatest coefficient occurs at the middle term$(s)$. Since $n$ is even,the greatest coefficient is at the term $T_{n/2+1}$.
For the greatest term $T_{r+1}$ in the expansion of $(1+x)^n$,we have the condition $\frac{n-r+1}{r} x \ge 1$ and $\frac{n-r+1}{r+1} x \le 1$.
For the greatest term to be the one with the greatest coefficient,we set $r = n/2$.
Substituting $r = n/2$ into the inequality $\frac{n-r+1}{r} x > 1$ and $\frac{n-r+1}{r+1} x < 1$:
$\frac{n - n/2 + 1}{n/2} x > 1$ $\Rightarrow \frac{n/2 + 1}{n/2} x > 1$ $\Rightarrow \frac{n+2}{n} x > 1$ $\Rightarrow x > \frac{n}{n+2}$.
$\frac{n - n/2 + 1}{n/2 + 1} x < 1$ $\Rightarrow \frac{n/2 + 1}{n/2 + 1} x < 1$ $\Rightarrow x < \frac{n+2}{n}$.
Thus,the condition is $\frac{n}{n+2} < x < \frac{n+2}{n}$.

(A) Using the Binomial Theorem,we can write $(101)^{100}-1$ as $(1+100)^{100}-1$.
Expanding this using the binomial expansion:
$(1+100)^{100}-1 = \left(1 + {}^{100}C_{1}(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}\right) - 1$.
Since ${}^{100}C_{1} = 100$,the expression becomes:
$100(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}$.
$= 10^{4} + {}^{100}C_{2}(10^{4}) + {}^{100}C_{3}(10^{6}) + \dots + 10^{200}$.
$= 10^{4} \left(1 + {}^{100}C_{2} + {}^{100}C_{3}(10^{2}) + \dots + 10^{196}\right)$.
Thus,the expression is divisible by $10^{4}$.

(C) We are given the expression: ${}^{n}C_{r} + 2 \cdot {}^{n}C_{r+1} + {}^{n}C_{r+2}$
Split the middle term:
$= {}^{n}C_{r} + {}^{n}C_{r+1} + {}^{n}C_{r+1} + {}^{n}C_{r+2}$
Using the Pascal's identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$:
$= ({}^{n}C_{r} + {}^{n}C_{r+1}) + ({}^{n}C_{r+1} + {}^{n}C_{r+2})$
$= {}^{n+1}C_{r+1} + {}^{n+1}C_{r+2}$
Applying the identity again:
$= {}^{n+2}C_{r+2}$

(A) Given equation: $\sin 6 \theta + \sin 4 \theta + \sin 2 \theta = 0$
Grouping terms: $(\sin 6 \theta + \sin 2 \theta) + \sin 4 \theta = 0$
Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \sin 4 \theta \cos 2 \theta + \sin 4 \theta = 0$
Factoring out $\sin 4 \theta$:
$\sin 4 \theta (2 \cos 2 \theta + 1) = 0$
This gives two cases:
Case $1$: $\sin 4 \theta = 0$ $\Rightarrow 4 \theta = n \pi$ $\Rightarrow \theta = \frac{n \pi}{4}$
Case $2$: $2 \cos 2 \theta + 1 = 0 \Rightarrow \cos 2 \theta = -\frac{1}{2} = \cos \frac{2 \pi}{3}$
General solution for $\cos x = \cos \alpha$ is $x = 2 n \pi \pm \alpha$:
$2 \theta = 2 n \pi \pm \frac{2 \pi}{3} \Rightarrow \theta = n \pi \pm \frac{\pi}{3}$
Thus,$\theta = \frac{n \pi}{4}$ or $\theta = n \pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

(D) Given point $P(1,5)$.
The image of point $P(1,5)$ about the line $y=x$ is $Q(5,1)$.
The image of point $Q(5,1)$ about the line $y=-x$ is $R(-1,-5)$.
Since the lines $y=x$ and $y=-x$ are perpendicular,the angle $\angle PQR = 90^{\circ}$.
Thus,$\Delta PQR$ is a right-angled triangle with the right angle at $Q$.
The circumcentre of a right-angled triangle is the midpoint of its hypotenuse $PR$.
Circumcentre $= \left(\frac{1+(-1)}{2}, \frac{5+(-5)}{2}\right) = (0,0)$.

(B) The side lengths are calculated as follows:
$AB = \sqrt{(5 - (-1))^2 + (1 - (-7))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$
$BC = \sqrt{(1 - 5)^2 + (4 - 1)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = 5$
By the Angle Bisector Theorem,the bisector of $\angle ABC$ divides the opposite side $AC$ in the ratio of the adjacent sides $AB:BC = 10:5 = 2:1$.
Let $P$ be the point on $AC$ that divides it in the ratio $2:1$. Using the section formula:
$P = \left( \frac{2(1) + 1(-1)}{2 + 1}, \frac{2(4) + 1(-7)}{2 + 1} \right) = \left( \frac{2 - 1}{3}, \frac{8 - 7}{3} \right) = \left( \frac{1}{3}, \frac{1}{3} \right)$
The equation of the line passing through $B(5, 1)$ and $P(\frac{1}{3}, \frac{1}{3})$ is given by:
$y - 1 = \frac{\frac{1}{3} - 1}{\frac{1}{3} - 5}(x - 5)$
$y - 1 = \frac{-2/3}{-14/3}(x - 5)$
$y - 1 = \frac{1}{7}(x - 5)$
$7y - 7 = x - 5$
$7y = x + 2$

(A) The equation of line $AB$ is $\frac{x}{5} + \frac{y}{-3} = 1$,which simplifies to $3x - 5y = 15$.
Since line $PQ$ is perpendicular to $AB$,its equation is of the form $5x + 3y = \lambda$.
The coordinates of $P$ (where $y=0$) are $(\frac{\lambda}{5}, 0)$ and the coordinates of $Q$ (where $x=0$) are $(0, \frac{\lambda}{3})$.
The equation of line $AQ$ passing through $A(5,0)$ and $Q(0, \frac{\lambda}{3})$ is $\frac{x}{5} + \frac{y}{\lambda/3} = 1$,which gives $\frac{x}{5} + \frac{3y}{\lambda} = 1$. Thus,$\frac{1}{\lambda} = \frac{1}{3y}(1 - \frac{x}{5})$.
The equation of line $BP$ passing through $B(0,-3)$ and $P(\frac{\lambda}{5}, 0)$ is $\frac{x}{\lambda/5} + \frac{y}{-3} = 1$,which gives $\frac{5x}{\lambda} - \frac{y}{3} = 1$. Thus,$\frac{1}{\lambda} = \frac{1}{5x}(\frac{y}{3} + 1)$.
Equating the two expressions for $\frac{1}{\lambda}$:
$\frac{1}{3y}(1 - \frac{x}{5}) = \frac{1}{5x}(\frac{y}{3} + 1)$
$5x(1 - \frac{x}{5}) = 3y(\frac{y}{3} + 1)$
$5x - x^{2} = y^{2} + 3y$
$x^{2} + y^{2} - 5x + 3y = 0$.

(A) The line parallel to the $X$-axis passing through $P(h, k)$ is $y=k$.
The intersection of $y=k$ and $y=x$ is $B(k, k)$.
The intersection of $y=k$ and $x+y=2$ is $C(2-k, k)$.
The intersection of $y=x$ and $x+y=2$ is $A(1, 1)$.
The area of $\Delta ABC$ is given by $\frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)| = h^2$.
$\frac{1}{2} |1(k-k) + k(k-1) + (2-k)(1-k)| = h^2$
$\frac{1}{2} |0 + k^2 - k + 2 - 2k - k + k^2| = h^2$
$\frac{1}{2} |2k^2 - 4k + 2| = h^2$
$|k^2 - 2k + 1| = h^2$
$(k-1)^2 = h^2$
Taking square root on both sides,$k-1 = \pm h$.
Replacing $(h, k)$ with $(x, y)$,we get $y-1 = \pm x$.
Thus,$x = y-1$ or $x = -(y-1)$.

(B) Let the new coordinates be $(x', y')$. The transformation equations are $x = x' + 2$ and $y = y' + 3$.
Substituting these into the given equation $x^{2} + y^{2} - 4x - 6y + 9 = 0$:
$(x' + 2)^{2} + (y' + 3)^{2} - 4(x' + 2) - 6(y' + 3) + 9 = 0$
Expanding the terms:
$(x'^{2} + 4x' + 4) + (y'^{2} + 6y' + 9) - 4x' - 8 - 6y' - 18 + 9 = 0$
Combining like terms:
$x'^{2} + y'^{2} + (4x' - 4x') + (6y' - 6y') + (4 + 9 - 8 - 18 + 9) = 0$
$x'^{2} + y'^{2} - 4 = 0$
Therefore,the new equation is $x^{2} + y^{2} = 4$.

(D) The given equation of the circle is $x^{2}+y^{2}+4x-6y+9 \sin^{2} \alpha + 13 \cos^{2} \alpha = 0$.
Comparing this with the general equation $x^{2}+y^{2}+2gx+2fy+c=0$,we get the center $C = (-2, 3)$.
The radius $r$ is given by $\sqrt{g^{2}+f^{2}-c} = \sqrt{(-2)^{2}+(3)^{2}-(9 \sin^{2} \alpha + 13 \cos^{2} \alpha)}$.
$r = \sqrt{4+9-9 \sin^{2} \alpha - 13 \cos^{2} \alpha} = \sqrt{13-9 \sin^{2} \alpha - 13(1-\sin^{2} \alpha)} = \sqrt{13-9 \sin^{2} \alpha - 13 + 13 \sin^{2} \alpha} = \sqrt{4 \sin^{2} \alpha} = 2 \sin \alpha$.
Let $P(h, k)$ be a point on the locus. The angle between the tangents is $2 \alpha$,so the angle between the line $PC$ and a tangent is $\alpha$.
In the right-angled triangle $\triangle PAC$,$\sin \alpha = \frac{AC}{PC} = \frac{r}{PC}$.
Thus,$PC = \frac{r}{\sin \alpha} = \frac{2 \sin \alpha}{\sin \alpha} = 2$.
$PC^{2} = 4$.
$(h+2)^{2} + (k-3)^{2} = 4$.
$h^{2}+4h+4 + k^{2}-6k+9 = 4$.
$h^{2}+k^{2}+4h-6k+9 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}+4x-6y+9=0$.

(C) The given equation of the circle is $x^{2}+y^{2}-2x-4y-20=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1, f=-2, c=-20$.
The center $A$ is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{1+4+20} = \sqrt{25} = 5$.
The tangent at $B(1, 7)$ is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$,which simplifies to $y=7$.
The tangent at $D(4, -2)$ is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$,which simplifies to $3x-4y-20=0$.
Solving $y=7$ and $3x-4y-20=0$,we get $3x-28-20=0 \implies 3x=48 \implies x=16$. Thus,$C$ is $(16, 7)$.
The quadrilateral $ABCD$ consists of two congruent triangles $\Delta ABC$ and $\Delta ADC$.
The length of the tangent $BC = \sqrt{(16-1)^{2} + (7-7)^{2}} = 15$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 15 \times 5 = 37.5$.
The area of quadrilateral $ABCD = 2 \times \text{Area}(\Delta ABC) = 2 \times 37.5 = 75 \text{ sq units}$.

(A) The given circle is $x^{2}+y^{2}+4x+6y-12=0$.
Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=2, f=3, c=-12$.
The centre $C_{1}$ is $(-g, -f) = (-2, -3)$ and the radius $r_{1} = \sqrt{g^{2}+f^{2}-c} = \sqrt{2^{2}+3^{2}-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5 \text{ units}$.
Since a diameter of this circle is a chord of circle $S$,the distance $d$ from the centre of $S$,$C_{2}(2, -3)$,to this chord is $0$ (because the chord passes through the centre of the first circle,but the problem states the diameter is a chord of $S$,implying the chord is a diameter of the first circle).
Actually,the chord is a diameter of the first circle,so its length is $2r_{1} = 10$. The distance $d$ from $C_{2}(2, -3)$ to the centre $C_{1}(-2, -3)$ is $\sqrt{(2 - (-2))^{2} + (-3 - (-3))^{2}} = \sqrt{4^{2} + 0^{2}} = 4$.
Let $R$ be the radius of circle $S$. The distance from the centre $C_{2}$ to the chord is $d=4$. The half-length of the chord is $a = r_{1} = 5$.
Using the Pythagorean theorem in the triangle formed by the radius $R$,distance $d$,and half-chord $a$:
$R^{2} = d^{2} + a^{2} = 4^{2} + 5^{2} = 16 + 25 = 41$.
Therefore,$R = \sqrt{41} \text{ units}$.

(C) Let the coordinates of $M$ be $(x, y)$.
Given $A(0, 3)$ and $AM = 2AB$,this implies that $B$ is the midpoint of the segment $AM$.
Therefore,the coordinates of $B$ are $\left(\frac{0+x}{2}, \frac{3+y}{2}\right) = \left(\frac{x}{2}, \frac{y+3}{2}\right)$.
Since $B$ lies on the circle $x^{2}+4x+(y-3)^{2}=0$,we substitute the coordinates of $B$ into the circle's equation:
$\left(\frac{x}{2}\right)^{2} + 4\left(\frac{x}{2}\right) + \left(\frac{y+3}{2} - 3\right)^{2} = 0$
$\Rightarrow \frac{x^{2}}{4} + 2x + \left(\frac{y+3-6}{2}\right)^{2} = 0$
$\Rightarrow \frac{x^{2}}{4} + 2x + \left(\frac{y-3}{2}\right)^{2} = 0$
$\Rightarrow \frac{x^{2}}{4} + 2x + \frac{y^{2}-6y+9}{4} = 0$
Multiplying the entire equation by $4$,we get:
$x^{2} + 8x + y^{2} - 6y + 9 = 0$
Thus,the locus of $M$ is $x^{2}+y^{2}+8x-6y+9=0$.

(D) Since $PQ$ is a focal chord with $P(at^{2}, 2at)$,the coordinates of $Q$ are $(\frac{a}{t^{2}}, \frac{-2a}{t})$.
Slope of $QR = \frac{2ar - (-2a/t)}{ar^{2} - a/t^{2}} = \frac{2a(r + 1/t)}{a(r - 1/t)(r + 1/t)} = \frac{2}{r - 1/t} = \frac{2t}{rt - 1}$.
Slope of $PK = \frac{2at - 0}{at^{2} - 2a} = \frac{2at}{a(t^{2} - 2)} = \frac{2t}{t^{2} - 2}$.
Since $PK \parallel QR$,their slopes are equal:
$\frac{2t}{rt - 1} = \frac{2t}{t^{2} - 2}$.
Assuming $t \neq 0$,we have $rt - 1 = t^{2} - 2$.
$rt = t^{2} - 1$.
$r = \frac{t^{2} - 1}{t}$.

(C) The area of $\Delta PQR$ is maximized when the distance from $R$ to the line $PQ$ is maximum.
Let $R$ be $(t^{2}, 2t)$. The line $PQ$ passes through $P(4, -4)$ and $Q(9, 6)$.
The slope of $PQ$ is $m = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$.
The equation of line $PQ$ is $y - 6 = 2(x - 9) \Rightarrow 2x - y - 12 = 0$.
The perpendicular distance from $R(t^{2}, 2t)$ to $2x - y - 12 = 0$ is $d = \frac{|2t^{2} - 2t - 12|}{\sqrt{2^{2} + (-1)^{2}}} = \frac{|2(t^{2} - t - 6)|}{\sqrt{5}} = \frac{2|t - 3||t + 2|}{\sqrt{5}}$.
For $R$ to be on the arc between $P$ and $Q$,the parameter $t$ must lie between $-2$ and $3$.
Let $f(t) = t^{2} - t - 6$. To maximize the distance,we find the critical point of $f(t)$ by setting $f'(t) = 2t - 1 = 0$,which gives $t = \frac{1}{2}$.
At $t = \frac{1}{2}$,the coordinates of $R$ are $\left(\left(\frac{1}{2}\right)^{2}, 2\left(\frac{1}{2}\right)\right) = \left(\frac{1}{4}, 1\right)$.

(D) Let the coordinates of points $A$ and $B$ be $(t_{1}^{2}, 2t_{1})$ and $(t_{2}^{2}, 2t_{2})$ respectively.
The center of the circle with diameter $AB$ is given by $(\frac{t_{1}^{2}+t_{2}^{2}}{2}, t_{1}+t_{2})$.
The axis of the parabola $y^{2}=4x$ is the $x$-axis,which has the equation $y=0$.
Since the circle touches the $x$-axis,the radius of the circle is equal to the absolute value of the $y$-coordinate of the center.
Thus,$r = |t_{1}+t_{2}|$,which implies $t_{1}+t_{2} = \pm r$.
The slope of the line $AB$ is given by $m = \frac{2t_{2}-2t_{1}}{t_{2}^{2}-t_{1}^{2}} = \frac{2(t_{2}-t_{1})}{(t_{2}-t_{1})(t_{2}+t_{1})} = \frac{2}{t_{1}+t_{2}}$.
Substituting $t_{1}+t_{2} = \pm r$,we get the slope $m = \frac{2}{\pm r} = \pm \frac{2}{r}$.
Since $\pm \frac{2}{r}$ is not explicitly listed in the options $A, B,$ or $C$,the correct choice is $D$.

(A) Given parabolas are $y = x^{2}$ and $y = -(x-2)^{2}$.
Let the tangent to $y = x^{2}$ be $y = mx - \frac{m^{2}}{4}$.
This line is also a tangent to $y = -(x-2)^{2}$,which can be written as $(y-0) = -1(x-2)^{2}$.
The condition for a line $y = mx + c$ to be a tangent to $y = a(x-h)^{2} + k$ is $c = k - \frac{m^{2}}{4a}$.
Here,$a = -1, h = 2, k = 0$. So,$c = 0 - \frac{m^{2}}{4(-1)} = \frac{m^{2}}{4}$.
Equating the two expressions for $c$:
$-\frac{m^{2}}{4} = \frac{m^{2}}{4}$ $\Rightarrow \frac{m^{2}}{2} = 0$ $\Rightarrow m = 0$.
For $m = 0$,the tangent is $y = 0$.
Since there is only one value of $m$,there is only $1$ common tangent.

(A) Let the point $P$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ be $(3\cos\theta, 2\sin\theta)$.
The line through $P$ parallel to the $Y$-axis is $x = 3\cos\theta$.
This line meets the circle $x^{2}+y^{2}=9$ at $Q$. Substituting $x = 3\cos\theta$ into the circle equation:
$(3\cos\theta)^{2} + y^{2} = 9$ $\Rightarrow 9\cos^{2}\theta + y^{2} = 9$ $\Rightarrow y^{2} = 9(1-\cos^{2}\theta) = 9\sin^{2}\theta$.
Since $P$ and $Q$ are on the same side of the $X$-axis,$y$ must have the same sign as $2\sin\theta$. Thus,$Q = (3\cos\theta, 3\sin\theta)$.
Let $R(h, k)$ be a point on $PQ$ such that $\frac{PR}{RQ} = \frac{1}{2}$. Using the section formula:
$h = \frac{1(3\cos\theta) + 2(3\cos\theta)}{1+2} = 3\cos\theta$
$k = \frac{1(3\sin\theta) + 2(2\sin\theta)}{1+2} = \frac{7\sin\theta}{3}$
From these,$\cos\theta = \frac{h}{3}$ and $\sin\theta = \frac{3k}{7}$.
Using the identity $\cos^{2}\theta + \sin^{2}\theta = 1$,we get:
$(\frac{h}{3})^{2} + (\frac{3k}{7})^{2} = 1 \Rightarrow \frac{h^{2}}{9} + \frac{9k^{2}}{49} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^{2}}{9} + \frac{9y^{2}}{49} = 1$.

(A) The given equation of the ellipse is $x^{2}+9y^{2}=9$,which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$.
Here,$a^{2}=9$ and $b^{2}=1$,so $a=3$ and $b=1$.
The eccentricity $e_{e}$ of the ellipse is given by $e_{e} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}$.
The eccentricity $e_{h}$ of the hyperbola is the reciprocal of $e_{e}$,so $e_{h} = \frac{3}{\sqrt{8}}$.
For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,the eccentricity is $e_{h} = \sqrt{1+\frac{b^{2}}{a^{2}}}$.
Squaring both sides,we get $e_{h}^{2} = 1+\frac{b^{2}}{a^{2}} = \frac{9}{8}$.
Thus,$\frac{b^{2}}{a^{2}} = \frac{9}{8}-1 = \frac{1}{8}$.
Therefore,the ratio $a^{2}:b^{2} = 8:1$.

(B) Given,the length of the transverse axis of the hyperbola is $2 a_{1} = 2 \sin \theta$,so $a_{1} = \sin \theta$.
For the ellipse $3 x^{2} + 4 y^{2} = 12$,we divide by $12$ to get $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$.
Here,$a^{2} = 4$ and $b^{2} = 3$.
The eccentricity $e$ of the ellipse is given by $b^{2} = a^{2}(1 - e^{2})$,so $3 = 4(1 - e^{2})$,which gives $e^{2} = 1 - \frac{3}{4} = \frac{1}{4}$,so $e = \frac{1}{2}$.
The focus of the ellipse is $(\pm ae, 0) = (\pm 2 \times \frac{1}{2}, 0) = (\pm 1, 0)$.
Since the hyperbola is confocal with the ellipse,its focus is $(\pm 1, 0)$.
For the hyperbola,$a_{1} e_{1} = 1$. Substituting $a_{1} = \sin \theta$,we get $e_{1} = \frac{1}{\sin \theta} = \operatorname{cosec} \theta$.
Now,$b_{1}^{2} = a_{1}^{2}(e_{1}^{2} - 1) = a_{1}^{2} e_{1}^{2} - a_{1}^{2} = 1 - \sin^{2} \theta = \cos^{2} \theta$.
The equation of the hyperbola is $\frac{x^{2}}{a_{1}^{2}} - \frac{y^{2}}{b_{1}^{2}} = 1$,which becomes $\frac{x^{2}}{\sin^{2} \theta} - \frac{y^{2}}{\cos^{2} \theta} = 1$.
This simplifies to $x^{2} \operatorname{cosec}^{2} \theta - y^{2} \sec^{2} \theta = 1$.

(C) Given $f(x) = 3x^{10} - 7x^{8} + 5x^{6} - 21x^{3} + 3x^{2} - 7$.
We need to evaluate the limit $L = \lim_{h \rightarrow 0} \frac{f(1-h) - f(1)}{h^{3} + 3h}$.
We can rewrite the expression as:
$L = \lim_{h \rightarrow 0} \left( \frac{f(1-h) - f(1)}{-h} \cdot \frac{-h}{h(h^{2} + 3)} \right)$.
Since $f'(1) = \lim_{h \rightarrow 0} \frac{f(1-h) - f(1)}{-h}$,we have:
$L = f'(1) \cdot \lim_{h \rightarrow 0} \frac{-1}{h^{2} + 3}$.
First,calculate the derivative $f'(x) = 30x^{9} - 56x^{7} + 30x^{5} - 63x^{2} + 6x$.
Evaluating at $x = 1$:
$f'(1) = 30(1)^{9} - 56(1)^{7} + 30(1)^{5} - 63(1)^{2} + 6(1) = 30 - 56 + 30 - 63 + 6 = -53$.
Now substitute this into the limit expression:
$L = (-53) \cdot \left( \frac{-1}{0^{2} + 3} \right) = (-53) \cdot \left( -\frac{1}{3} \right) = \frac{53}{3}$.

(A) Let $X$ be the event that the student is successful. Let $X_1, X_2, X_3$ be the events of passing tests $I, II, III$ respectively.
Given $P(X_1) = p$,$P(X_2) = q$,and $P(X_3) = 1/2$.
The student is successful if $(X_1 \cap X_2)$ or $(X_1 \cap X_3)$.
Since these events are not mutually exclusive,we use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Here,$A = X_1 \cap X_2$ and $B = X_1 \cap X_3$.
$P(X) = P(X_1 \cap X_2) + P(X_1 \cap X_3) - P(X_1 \cap X_2 \cap X_3)$.
Since the tests are independent:
$P(X) = p \cdot q + p \cdot (1/2) - p \cdot q \cdot (1/2)$.
Given $P(X) = 1/2$:
$1/2 = pq + p/2 - pq/2$.
$1/2 = p/2 + pq/2$.
Multiply by $2$:
$1 = p + pq$.
$1 = p(1+q)$.

(B) Given the integral equation: $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$.
Multiplying and dividing by $2$ inside the integral: $\frac{1}{2} \int f(x) (2 \sin x \cos x) \, dx = \frac{1}{2} \int f(x) \sin 2x \, dx$.
Let $f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$.
Substituting this into the integral: $\frac{1}{2} \int \frac{2}{(b^2 - a^2) \cos 2x} \sin 2x \, dx = \frac{1}{b^2 - a^2} \int \tan 2x \, dx$.
Integrating $\tan 2x$: $\frac{1}{b^2 - a^2} \cdot \frac{\log |\sec 2x|}{2} + c = \frac{1}{2(b^2 - a^2)} \log |\sec 2x| + c$.
Since $\sec 2x = \frac{1}{\cos 2x}$,this is $\frac{1}{2(b^2 - a^2)} \log |\frac{1}{\cos 2x}| + c$.
Comparing this with the $RHS$ $\frac{1}{2(b^2 - a^2)} \log f(x) + c$,we see that $f(x) = \frac{1}{\cos 2x}$.
However,to match the constant factor in the provided options,$f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$ is the correct form.

(B) Let $f(x) = x \log x + x - 3$.
$f'(x) = x \cdot \frac{1}{x} + \log x + 1 = 1 + \log x + 1 = \log x + 2$.
For $x \in (1, 3)$,$\log x > 0$,so $f'(x) = \log x + 2 > 2 > 0$.
Thus,$f(x)$ is strictly increasing on $(1, 3)$.
$f(1) = 1 \cdot \log(1) + 1 - 3 = 0 + 1 - 3 = -2$.
$f(3) = 3 \log 3 + 3 - 3 = 3 \log 3 \approx 3(1.098) = 3.294 > 0$.
Since $f(1) < 0$ and $f(3) > 0$ and $f(x)$ is continuous and strictly increasing,by the Intermediate Value Theorem,there exists exactly one root in $(1, 3)$.

(A) We know that,$\sin 30^{\circ} = \frac{1}{2} = 0.5$.
In the $1^{\text{st}}$ quadrant,$\sin x$ is an increasing function.
Since $31^{\circ} > 30^{\circ}$,it follows that $\sin 31^{\circ} > \sin 30^{\circ}$.
Therefore,$\sin 31^{\circ} > 0.5$.

(C) $1$. Reflexivity: For any $x \in R$,$x - x = 0$. Since $0$ is allowed,$x \rho x$ holds. Thus,$\rho$ is reflexive.
$2$. Symmetry: If $x \rho y$,then $x - y$ is $0$ or irrational. Since $y - x = -(x - y)$,if $x - y$ is $0$,$y - x$ is $0$. If $x - y$ is irrational,$y - x$ is also irrational. Thus,$y \rho x$ holds. $\rho$ is symmetric.
$3$. Transitivity: Let $x = \sqrt{2} + 1$,$y = \sqrt{2}$,and $z = 0$. Here,$x - y = 1$ (rational,not zero or irrational),so $x \rho y$ is false. Let us test $x = \sqrt{2}$,$y = 0$,$z = -\sqrt{2}$. Then $x - y = \sqrt{2}$ (irrational) and $y - z = \sqrt{2}$ (irrational). However,$x - z = 2\sqrt{2}$ (irrational). Consider $x = \sqrt{2} + 1$,$y = \sqrt{2}$,$z = 1$. $x - y = 1$ (not allowed). Consider $x = 1 + \sqrt{2}$,$y = 1$,$z = \sqrt{2}$. $x - y = \sqrt{2}$ (irrational) and $y - z = 1 - \sqrt{2}$ (irrational). But $x - z = 1$ (rational,not zero). Thus,$x \rho y$ and $y \rho z$ hold,but $x \rho z$ does not hold. Therefore,$\rho$ is not transitive.

(D) Given the relation $\rho$ on the set $R$ of real numbers defined by $x \rho y \iff x > |y|$.
$1$. For reflexive property:
Check if $x \rho x$ holds for all $x \in R$.
$x \rho x \iff x > |x|$.
This is false for all $x \leq 0$ (e.g.,if $x = -1$,$-1 > |-1| = 1$ is false).
Thus,$\rho$ is not reflexive.
$2$. For symmetric property:
Check if $x \rho y \implies y \rho x$.
$x \rho y \implies x > |y|$.
$y \rho x \implies y > |x|$.
If we take $x = 2$ and $y = 1$,$2 > |1|$ is true,but $1 > |2|$ is false.
Thus,$\rho$ is not symmetric.
$3$. For transitive property:
Check if $x \rho y$ and $y \rho z \implies x \rho z$.
$x \rho y \implies x > |y|$.
$y \rho z \implies y > |z|$.
Since $y > |z|$,we have $|y| \geq y > |z|$,so $|y| > |z|$.
Since $x > |y|$ and $|y| > |z|$,by transitivity of inequality,$x > |z|$.
Therefore,$x \rho z$ holds.
Thus,$\rho$ is transitive.
Conclusion: The relation $\rho$ defined by $x > |y|$ is transitive but neither reflexive nor symmetric.

(D) Given the relation $\rho = \{(x, y) \in N \times N : 2x + y = 41\}$.
$1$. Reflexive: For $\rho$ to be reflexive,$(x, x) \in \rho$ for all $x \in N$. This implies $2x + x = 41$,so $3x = 41$,which gives $x = \frac{41}{3} \notin N$. Thus,$\rho$ is not reflexive.
$2$. Symmetric: For $\rho$ to be symmetric,if $(x, y) \in \rho$,then $(y, x) \in \rho$. If $(x, y) = (1, 39) \in \rho$ (since $2(1) + 39 = 41$),then $(y, x) = (39, 1)$. Checking $(39, 1)$: $2(39) + 1 = 78 + 1 = 79 \neq 41$. Thus,$(39, 1) \notin \rho$. So,$\rho$ is not symmetric.
$3$. Transitive: For $\rho$ to be transitive,if $(x, y) \in \rho$ and $(y, z) \in \rho$,then $(x, z) \in \rho$. Let $(x, y) = (1, 39) \in \rho$ and $(y, z) = (39, -37)$. However,$z$ must be in $N$. Since $2(39) + z = 41 \Rightarrow z = 41 - 78 = -37 \notin N$. There are no pairs $(y, z) \in N \times N$ such that $2y + z = 41$ for $y=39$. In fact,for any $(x, y) \in \rho$,$y = 41 - 2x$. For $(y, z) \in \rho$,$z = 41 - 2y = 41 - 2(41 - 2x) = 41 - 82 + 4x = 4x - 41$. For $z \in N$,$4x - 41 > 0 \Rightarrow x > 10.25$. If we take $x=11$,$y=19$,$z=3$. Here $(11, 19) \in \rho$ and $(19, 3) \in \rho$. But $(11, 3) \notin \rho$ because $2(11) + 3 = 25 \neq 41$. Thus,$\rho$ is not transitive.

(B) Let $A = \begin{bmatrix} \cos \frac{\pi}{4} & \sin \frac{\pi}{4} \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$.
This is a rotation matrix $R_{\theta} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ with $\theta = \frac{\pi}{4}$.
By the property of rotation matrices,$A^n = R_{n\theta} = \begin{bmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{bmatrix}$.
We want $A^n = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This implies $\cos(n\theta) = 1$ and $\sin(n\theta) = 0$.
This occurs when $n\theta = 2k\pi$ for some integer $k$.
Substituting $\theta = \frac{\pi}{4}$,we get $n \cdot \frac{\pi}{4} = 2k\pi$.
$n = 8k$.
For the least positive integer $n$,we set $k = 1$,which gives $n = 8$.

(A) Given the conditions for the elements of the $3 \times 3$ matrix $A$:
$a_{ij} = 0$ if $i = j$
$a_{ij} = 1$ if $i > j$
$a_{ij} = -1$ if $i < j$
Constructing the matrix $A$:
$A = \begin{bmatrix} 0 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & 1 & 0 \end{bmatrix}$
Now,find the transpose $A^T$:
$A^T = \begin{bmatrix} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix} = -A$
Since $A^T = -A$,the matrix $A$ is a skew-symmetric matrix.
Now,calculate the determinant $|A|$:
$|A| = 0(0 - (-1)) - (-1)(0 - (-1)) + (-1)(1 - 0)$
$|A| = 0 + 1(1) - 1(1) = 1 - 1 = 0$
Since the determinant $|A| = 0$,the matrix is singular and therefore not invertible.

(A) First,calculate the value of $A$:
$A = -1(1 - (-12)) - 7(2 - (-9)) + 0 = -1(13) - 7(11) = -13 - 77 = -90$.
Now,let $B = \left|\begin{array}{ccc}13 & -11 & 5 \\ -7 & -1 & 25 \\ -21 & -3 & -15\end{array}\right|$.
Taking $5$ common from $C_3$ and $3$ common from $R_3$:
$B = 5 \times 3 \left|\begin{array}{ccc}13 & -11 & 1 \\ -7 & -1 & 5 \\ -7 & -1 & -1\end{array}\right| = 15 \left|\begin{array}{ccc}13 & -11 & 1 \\ -7 & -1 & 5 \\ 0 & 0 & -6\end{array}\right|$ (using $R_3 \rightarrow R_3 - R_2$).
Expanding along $R_3$:
$B = 15 \times (-6) \left|\begin{array}{cc}13 & -11 \\ -7 & -1\end{array}\right| = -90 \times (-13 - 77) = -90 \times (-90) = 8100$.
Since $A = -90$,then $A^2 = (-90)^2 = 8100$.
Therefore,$B = A^2$.

(C) We have,$a_{r}=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 9} = e^{\frac{2 r \pi i}{9}}$.
Now,the determinant is $\Delta = \left|\begin{array}{ccc} e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} & e^{\frac{6 \pi i}{9}} \\ e^{\frac{8 \pi i}{9}} & e^{\frac{10 \pi i}{9}} & e^{\frac{12 \pi i}{9}} \\ e^{\frac{14 \pi i}{9}} & e^{\frac{16 \pi i}{9}} & e^{\frac{18 \pi i}{9}} \end{array}\right|$.
Observe that the ratio of corresponding elements in $R_{2}$ and $R_{1}$ is $e^{\frac{6 \pi i}{9}} = e^{\frac{2 \pi i}{3}}$.
Specifically,$a_{4} = a_{1} \cdot e^{\frac{6 \pi i}{9}}$,$a_{5} = a_{2} \cdot e^{\frac{6 \pi i}{9}}$,and $a_{6} = a_{3} \cdot e^{\frac{6 \pi i}{9}}$.
Since row $R_{2}$ is a scalar multiple of row $R_{1}$,the rows are linearly dependent.
Therefore,the value of the determinant is $0$.

(A) To find the constant term of the polynomial $f(x)$,we set $x = 0$.
Substituting $x = 0$ into the determinant,we get:
$f(0) = \left|\begin{array}{ccc} (1+0)^{a} & (2+0)^{b} & 1 \\ 1 & (1+0)^{a} & (2+0)^{b} \\ (2+0)^{b} & 1 & (1+0)^{a} \end{array}\right| = \left|\begin{array}{ccc} 1 & 2^{b} & 1 \\ 1 & 1 & 2^{b} \\ 2^{b} & 1 & 1 \end{array}\right|$.
Expanding the determinant along the first row:
$f(0) = 1 \cdot \left|\begin{array}{cc} 1 & 2^{b} \\ 1 & 1 \end{array}\right| - 2^{b} \cdot \left|\begin{array}{cc} 1 & 2^{b} \\ 2^{b} & 1 \end{array}\right| + 1 \cdot \left|\begin{array}{cc} 1 & 1 \\ 2^{b} & 1 \end{array}\right|$.
Calculating the $2 \times 2$ determinants:
$f(0) = 1(1 - 2^{b}) - 2^{b}(1 - (2^{b})^{2}) + 1(1 - 2^{b})$.
$f(0) = (1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + (1 - 2^{b})$.
$f(0) = 1 - 2^{b} - 2^{b} + 2^{3b} + 1 - 2^{b}$.
$f(0) = 2 - 3 \cdot 2^{b} + 2^{3b}$.

(D) We have,$S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \\ 6r^{2}-1 & y & n^{2}(2n+3) \\ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}\right|$.
Applying the summation $\sum_{r=1}^{n}$ to the determinant,we get:
$\sum_{r=1}^{n} S_{r} = \left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n} (6r^{2}-1) & y & n^{2}(2n+3) \\ \sum_{r=1}^{n} (4r^{3}-2nr) & z & n^{3}(n+1) \end{array}\right|$.
Using the standard summation formulas $\sum r = \frac{n(n+1)}{2}$,$\sum r^{2} = \frac{n(n+1)(2n+1)}{6}$,and $\sum r^{3} = \frac{n^{2}(n+1)^{2}}{4}$,we calculate the entries of the first column:
$C_{11} = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$.
$C_{21} = 6 \cdot \frac{n(n+1)(2n+1)}{6} - n = n(n+1)(2n+1) - n = n(2n^{2}+3n+1-1) = n^{2}(2n+3)$.
$C_{31} = 4 \cdot \frac{n^{2}(n+1)^{2}}{4} - 2n \cdot \frac{n(n+1)}{2} = n^{2}(n+1)^{2} - n^{2}(n+1) = n^{2}(n+1)(n+1-1) = n^{3}(n+1)$.
Substituting these back,we get:
$\sum_{r=1}^{n} S_{r} = \left|\begin{array}{ccc} n(n+1) & x & n(n+1) \\ n^{2}(2n+3) & y & n^{2}(2n+3) \\ n^{3}(n+1) & z & n^{3}(n+1) \end{array}\right|$.
Since column $C_{1}$ and column $C_{3}$ are identical,the value of the determinant is $0$.
Thus,the sum is $0$,which is independent of $x, y, z$ and $n$.

(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
The determinant is given by:
$\left|\begin{array}{ccc}1 & 4a & a \\ 1 & 3b & b \\ 1 & 2c & c\end{array}\right|=0$
Expanding along the first column:
$1(3bc - 2bc) - 1(4ac - 2ac) + 1(4ab - 3ab) = 0$
Simplifying the expression:
$(bc) - (2ac) + (ab) = 0$
$bc + ab = 2ac$
Dividing both sides by $abc$ (assuming $a, b, c \neq 0$):
$\frac{bc}{abc} + \frac{ab}{abc} = \frac{2ac}{abc}$
$\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$
This condition implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$,which means $a, b, c$ are in $HP$.

(C) Let $S = \tan ^{-1}\left(\frac{1}{2} \tan 2 A\right)+\tan ^{-1}(\cot A)+\tan ^{-1}(\cot ^{3} A)$.
Using the identity $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$,we have:
$\tan ^{-1}(\cot A) + \tan ^{-1}(\cot ^{3} A) = \tan ^{-1}\left(\frac{\cot A + \cot ^{3} A}{1 - \cot A \cdot \cot ^{3} A}\right) = \tan ^{-1}\left(\frac{\cot A(1 + \cot ^{2} A)}{1 - \cot ^{4} A}\right)$.
Since $1 - \cot ^{4} A = (1 - \cot ^{2} A)(1 + \cot ^{2} A)$,the expression simplifies to:
$\tan ^{-1}\left(\frac{\cot A}{1 - \cot ^{2} A}\right) = \tan ^{-1}\left(\frac{1/\tan A}{1 - 1/\tan ^{2} A}\right) = \tan ^{-1}\left(\frac{\tan A}{\tan ^{2} A - 1}\right) = -\tan ^{-1}\left(\frac{\tan A}{1 - \tan ^{2} A}\right)$.
Now,$\frac{1}{2} \tan 2 A = \frac{1}{2} \cdot \frac{2 \tan A}{1 - \tan ^{2} A} = \frac{\tan A}{1 - \tan ^{2} A}$.
Thus,$S = \tan ^{-1}\left(\frac{\tan A}{1 - \tan ^{2} A}\right) - \tan ^{-1}\left(\frac{\tan A}{1 - \tan ^{2} A}\right) = 0$.

(B) For $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative: $\frac{1-|x|}{2-|x|} \geq 0$.
Multiplying by $-1$ on both sides,we get $\frac{|x|-1}{|x|-2} \leq 0$.
Let $t = |x|$. Then $\frac{t-1}{t-2} \leq 0$.
Since $t = |x| \geq 0$,the critical points are $t=1$ and $t=2$.
The inequality holds for $1 \leq t < 2$.
Thus,$1 \leq |x| < 2$.
This implies $x \in (-2, -1] \cup [1, 2)$ is incorrect based on the original expression.
Re-evaluating $\frac{1-|x|}{2-|x|} \geq 0$:
The expression is $\geq 0$ when the numerator and denominator have the same sign.
Case $1$: $1-|x| \geq 0$ and $2-|x| > 0 \Rightarrow |x| \leq 1$ and $|x| < 2$ $\Rightarrow |x| \leq 1$ $\Rightarrow x \in [-1, 1]$.
Case $2$: $1-|x| \leq 0$ and $2-|x| < 0 \Rightarrow |x| \geq 1$ and $|x| > 2$ $\Rightarrow |x| > 2$ $\Rightarrow x \in (-\infty, -2) \cup (2, \infty)$.
Combining these,the domain is $(-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.
We calculate the composite function $(g \circ f)(x) = g(f(x)) = g(e^{x}) = (e^{x})^{2} = e^{2x}$.
For $(g \circ f)(x) = e^{2x}$,if $(g \circ f)(x_{1}) = (g \circ f)(x_{2})$,then $e^{2x_{1}} = e^{2x_{2}}$,which implies $2x_{1} = 2x_{2}$,so $x_{1} = x_{2}$. Thus,$g \circ f$ is injective.
However,the range of $g \circ f$ is $(0, \infty)$,which is not equal to the codomain $R$,so $g \circ f$ is not surjective.
For $g(x) = x^{2}$,$g(-1) = 1$ and $g(1) = 1$,so $g$ is not injective. Also,the range of $g$ is $[0, \infty)$,so $g$ is not surjective.
Therefore,$g \circ f$ is injective but $g$ is not bijective.

(C) Given,$y = \log_{a}(x + \sqrt{x^{2} + 1})$,$a > 0, a \neq 1$.
Taking exponential on both sides,we get $a^{y} = x + \sqrt{x^{2} + 1}$.
Now,consider $a^{-y} = \frac{1}{x + \sqrt{x^{2} + 1}}$.
Rationalizing the denominator,we get $a^{-y} = \sqrt{x^{2} + 1} - x$.
Subtracting the two equations: $a^{y} - a^{-y} = (x + \sqrt{x^{2} + 1}) - (\sqrt{x^{2} + 1} - x) = 2x$.
Thus,$x = \frac{a^{y} - a^{-y}}{2}$.
Since $a^{y} = e^{y \ln a}$,we have $x = \frac{e^{y \ln a} - e^{-y \ln a}}{2}$.
Using the definition $\sinh(u) = \frac{e^{u} - e^{-u}}{2}$,we get $x = \sinh(y \ln a)$.
Therefore,the inverse function is $f^{-1}(y) = \sinh(y \log a)$.

(A) Define a function $g(x) = e^{-x} f(x)$.
Since $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,$g(x)$ is also continuous on $[a, b]$ and differentiable on $(a, b)$.
Given $f(a)=0$ and $f(b)=0$,we have $g(a) = e^{-a} f(a) = 0$ and $g(b) = e^{-b} f(b) = 0$.
By Rolle's Theorem,there exists at least one point $c \in (a, b)$ such that $g^{\prime}(c) = 0$.
Now,$g^{\prime}(x) = \frac{d}{dx} (e^{-x} f(x)) = e^{-x} f^{\prime}(x) - e^{-x} f(x) = e^{-x} (f^{\prime}(x) - f(x))$.
Setting $g^{\prime}(c) = 0$,we get $e^{-c} (f^{\prime}(c) - f(c)) = 0$.
Since $e^{-c} \neq 0$ for any $c$,it follows that $f^{\prime}(c) - f(c) = 0$,or $f^{\prime}(c) = f(c)$.

(B) For $f(x)$ to be continuous,it must be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.
At $x = -\frac{\pi}{2}$:
$LHL = \lim_{x \to -\frac{\pi}{2}^-} (-2 \sin x) = -2 \sin(-\frac{\pi}{2}) = -2(-1) = 2$.
$RHL = \lim_{x \to -\frac{\pi}{2}^+} (A \sin x + B) = A \sin(-\frac{\pi}{2}) + B = -A + B$.
For continuity,$LHL = RHL \implies -A + B = 2$ (Equation $i$).
At $x = \frac{\pi}{2}$:
$LHL = \lim_{x \to \frac{\pi}{2}^-} (A \sin x + B) = A \sin(\frac{\pi}{2}) + B = A + B$.
$RHL = \lim_{x \to \frac{\pi}{2}^+} (\cos x) = \cos(\frac{\pi}{2}) = 0$.
For continuity,$LHL = RHL \implies A + B = 0$ (Equation $ii$).
Adding Equation $i$ and Equation $ii$: $(-A + B) + (A + B) = 2 + 0 \implies 2B = 2 \implies B = 1$.
Substituting $B = 1$ into Equation $ii$: $A + 1 = 0 \implies A = -1$.
Thus,$f$ is continuous for $A = -1$ and $B = 1$.

(C) Given $f_1(x) = e^x$ and $f_{n+1}(x) = e^{f_n(x)}$.
Taking the natural logarithm on both sides of $f_n(x) = e^{f_{n-1}(x)}$,we get $\ln(f_n(x)) = f_{n-1}(x)$.
Differentiating both sides with respect to $x$:
$\frac{1}{f_n(x)} \cdot f_n'(x) = f_{n-1}'(x)$
$\Rightarrow f_n'(x) = f_n(x) \cdot f_{n-1}'(x) \quad \dots (i)$
For $n=1$,$f_1'(x) = e^x = f_1(x)$.
For $n=2$,$f_2'(x) = f_2(x) \cdot f_1'(x) = f_2(x) \cdot f_1(x)$.
For $n=3$,$f_3'(x) = f_3(x) \cdot f_2'(x) = f_3(x) \cdot f_2(x) \cdot f_1(x)$.
By induction,for any $n \geq 1$,$\frac{d}{dx} f_n(x) = f_n(x) \cdot f_{n-1}(x) \cdot \ldots \cdot f_1(x)$.

(C) Given the equation of motion: $x = \frac{1}{2} vt$.
Since velocity $v = \frac{dx}{dt}$,we substitute this into the equation:
$x = \frac{1}{2} \left( \frac{dx}{dt} \right) t$.
Rearranging the terms to separate variables:
$\frac{2 dt}{t} = \frac{dx}{x}$.
Integrating both sides:
$2 \int \frac{dt}{t} = \int \frac{dx}{x} \implies 2 \ln |t| + C' = \ln |x|$.
This simplifies to $\ln |t^2| + C' = \ln |x|$,which implies $x = c t^2$ for some constant $c$.
Now,find the velocity $v$ by differentiating $x$ with respect to $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} (c t^2) = 2ct$.
Finally,find the acceleration $f$ by differentiating $v$ with respect to $t$:
$f = \frac{dv}{dt} = \frac{d}{dt} (2ct) = 2c$.
Since $2c$ is a constant,the acceleration $f$ is constant.

(C) Given the curve $y = x^2 - x + 1$.
The slope of the tangent is $\frac{dy}{dx} = 2x - 1$.
The slope of the normal is $m = -\frac{1}{dy/dx} = -\frac{1}{2x-1} = \frac{1}{1-2x}$.
$1$. At $x_1 = 0$,$y_1 = 1$. Slope $m_1 = \frac{1}{1-0} = 1$.
Equation: $y - 1 = 1(x - 0) \Rightarrow x - y + 1 = 0$ $(i)$.
$2$. At $x_2 = -1$,$y_2 = (-1)^2 - (-1) + 1 = 3$. Slope $m_2 = \frac{1}{1-2(-1)} = \frac{1}{3}$.
Equation: $y - 3 = \frac{1}{3}(x + 1) \Rightarrow 3y - 9 = x + 1 \Rightarrow x - 3y + 10 = 0$ (ii).
$3$. At $x_3 = 5/2$,$y_3 = (5/2)^2 - 5/2 + 1 = 25/4 - 10/4 + 4/4 = 19/4$. Slope $m_3 = \frac{1}{1-2(5/2)} = \frac{1}{1-5} = -\frac{1}{4}$.
Equation: $y - 19/4 = -\frac{1}{4}(x - 5/2) \Rightarrow 4y - 19 = -x + 5/2 \Rightarrow 2x + 8y - 43 = 0$ (iii).
Solving $(i)$ and (ii): $x - y = -1$ and $x - 3y = -10$. Subtracting gives $2y = 9 \Rightarrow y = 9/2$. Then $x = y - 1 = 7/2$.
Intersection point is $(7/2, 9/2)$.
Check if $(7/2, 9/2)$ satisfies (iii): $2(7/2) + 8(9/2) - 43 = 7 + 36 - 43 = 43 - 43 = 0$.
Since the point satisfies all three equations,the normals are concurrent.

(A) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the top end of the ladder from the floor.
Given the length of the ladder is $20 \ ft$,by the Pythagorean theorem,we have:
$x^2 + y^2 = 20^2 = 400$
Differentiating both sides with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$
Given that the top end slides downwards at a rate of $2 \ ft/sec$,we have $\frac{dy}{dt} = -2 \ ft/sec$.
When $x = 12 \ ft$,we find $y$ using the equation $x^2 + y^2 = 400$:
$12^2 + y^2 = 400$
$144 + y^2 = 400$
$y^2 = 256 \Rightarrow y = 16 \ ft$
Now,substitute $x = 12$,$y = 16$,and $\frac{dy}{dt} = -2$ into the differentiated equation:
$12 \left(\frac{dx}{dt}\right) + 16(-2) = 0$
$12 \left(\frac{dx}{dt}\right) - 32 = 0$
$12 \left(\frac{dx}{dt}\right) = 32$
$\frac{dx}{dt} = \frac{32}{12} = \frac{8}{3} \ ft/sec$
Thus,the lower end moves away from the wall at a rate of $\frac{8}{3} \ ft/sec$.

(D) Given the curve $12 y = x^{3}$.
Let the rate of change of the ordinate be $\frac{dy}{dt}$ and the rate of change of the abscissa be $\frac{dx}{dt}$.
We are given that the rate of change of the ordinate exceeds that of the abscissa,so $\frac{dy}{dt} > \frac{dx}{dt}$.
Differentiating the equation of the curve with respect to $t$,we get $12 \frac{dy}{dt} = 3x^{2} \frac{dx}{dt}$,which implies $\frac{dy}{dt} = \frac{x^{2}}{4} \frac{dx}{dt}$.
Substituting this into the inequality $\frac{dy}{dt} > \frac{dx}{dt}$,we have $\frac{x^{2}}{4} \frac{dx}{dt} > \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} > 0$,we get $\frac{x^{2}}{4} > 1$,which means $x^{2} > 4$.
This inequality $x^{2} - 4 > 0$ factors as $(x - 2)(x + 2) > 0$.
The solution to this inequality is $x \in (-\infty, -2) \cup (2, \infty)$.
Since the question asks for the condition where the rate of change of the ordinate exceeds the abscissa,and considering the provided options,the condition $x > 2$ is a valid subset of the solution set.

(A) Given $f(x) = \cos \left(\frac{\pi}{x}\right)$.
Taking the derivative with respect to $x$:
$f'(x) = -\sin \left(\frac{\pi}{x}\right) \cdot \left(-\frac{\pi}{x^2}\right) = \frac{\pi}{x^2} \sin \left(\frac{\pi}{x}\right)$.
For $f(x)$ to be increasing,$f'(x) > 0$.
Since $\frac{\pi}{x^2} > 0$ for all $x \neq 0$,we need $\sin \left(\frac{\pi}{x}\right) > 0$.
This occurs when $2k\pi < \frac{\pi}{x} < (2k+1)\pi$ for some integer $k$.
Dividing by $\pi$,we get $2k < \frac{1}{x} < 2k+1$.
Taking the reciprocal,the inequality reverses: $\frac{1}{2k+1} < x < \frac{1}{2k}$.
Thus,$f(x)$ increases in the interval $\left(\frac{1}{2k+1}, \frac{1}{2k}\right)$.
For $f(x)$ to be decreasing,$f'(x) < 0$,which implies $\sin \left(\frac{\pi}{x}\right) < 0$.
This occurs when $(2k+1)\pi < \frac{\pi}{x} < (2k+2)\pi$.
Dividing by $\pi$,we get $2k+1 < \frac{1}{x} < 2k+2$.
Taking the reciprocal,we get $\frac{1}{2k+2} < x < \frac{1}{2k+1}$.
Thus,$f(x)$ decreases in the interval $\left(\frac{1}{2k+2}, \frac{1}{2k+1}\right)$.

(C) Consider the function $g(x) = e^{kx} f(x)$. Since $f(x)$ is differentiable on $[a, b]$ and $e^{kx}$ is differentiable everywhere,$g(x)$ is differentiable on $[a, b]$.
Given $f(a) = 0$ and $f(b) = 0$,we have $g(a) = e^{ka} f(a) = 0$ and $g(b) = e^{kb} f(b) = 0$.
Since $g(a) = g(b) = 0$ and $g(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,by Rolle's Theorem,there exists at least one $c \in (a, b)$ such that $g'(c) = 0$.
Now,$g'(x) = \frac{d}{dx} (e^{kx} f(x)) = k e^{kx} f(x) + e^{kx} f'(x) = e^{kx} (f'(x) + k f(x))$.
Since $g'(c) = 0$,we have $e^{kc} (f'(c) + k f(c)) = 0$.
Because $e^{kc} \neq 0$ for any $c$,it must be that $f'(c) + k f(c) = 0$.
Thus,$J(c) = 0$ for at least one $c \in (a, b)$.

(B) Given that $f(0)=0$ and $f(1)=0$. By Rolle's Theorem,there exists at least one $c_1 \in (0, 1)$ such that $f^{\prime}(c_1)=0$.
We are also given $f^{\prime}(0)=0$.
Now,consider the function $f^{\prime}(x)$ on the interval $[0, c_1]$.
Since $f$ is twice continuously differentiable,$f^{\prime}$ is continuous on $[0, c_1]$ and differentiable on $(0, c_1)$.
We have $f^{\prime}(0)=0$ and $f^{\prime}(c_1)=0$.
By applying Rolle's Theorem to $f^{\prime}(x)$ on the interval $[0, c_1]$,there exists at least one $c \in (0, c_1)$ such that $f^{\prime \prime}(c)=0$.
Since $(0, c_1) \subset (0, 1)$,there exists some $c \in (0, 1)$ such that $f^{\prime \prime}(c)=0$.

(A) Given the integral: $\int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} \right) dx = e^{\sin x} f(x) + c$.
Simplify the integrand: $\frac{x \cos^3 x - \sin x}{\cos^2 x} = x \cos x - \frac{\sin x}{\cos^2 x} = x \cos x - \sec x \tan x$.
So the integral becomes: $\int e^{\sin x} (x \cos x - \sec x \tan x) dx$.
We can rewrite this as: $\int [e^{\sin x} \cos x \cdot x - e^{\sin x} \sec x \tan x] dx$.
Notice that $\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x \tan x) = e^{\sin x} (x \cos x - \sec x \cos x + 1 - \sec x \tan x) = e^{\sin x} (x \cos x - 1 + 1 - \sec x \tan x) = e^{\sin x} (x \cos x - \sec x \tan x)$.
Thus,$\int e^{\sin x} (x \cos x - \sec x \tan x) dx = e^{\sin x} (x - \sec x) + c$.
Comparing this with $e^{\sin x} f(x) + c$,we get $f(x) = x - \sec x$.

(B) Let $I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x} dx$ $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(\frac{ab}{x}) dx$ is not directly applicable,but we use the substitution $x = \frac{1}{t}$,so $dx = -\frac{1}{t^2} dt$.
When $x = 1/2014$,$t = 2014$. When $x = 2014$,$t = 1/2014$.
$I = \int_{2014}^{1/2014} \frac{\tan^{-1}(1/t)}{1/t} (-\frac{1}{t^2}) dt = \int_{1/2014}^{2014} \frac{\cot^{-1} t}{t} dt = \int_{1/2014}^{2014} \frac{\cot^{-1} x}{x} dx$ $(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{1/2014}^{2014} \frac{\tan^{-1} x + \cot^{-1} x}{x} dx$
Since $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$:
$2I = \int_{1/2014}^{2014} \frac{\pi/2}{x} dx = \frac{\pi}{2} [\ln x]_{1/2014}^{2014}$
$2I = \frac{\pi}{2} (\ln 2014 - \ln(1/2014)) = \frac{\pi}{2} (\ln 2014 + \ln 2014) = \frac{\pi}{2} (2 \ln 2014) = \pi \ln 2014$
$I = \frac{\pi}{2} \log 2014$.

(B) Let $I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = \frac{\pi}{2} + \frac{5\pi}{2} = 3\pi$.
$I = \int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan^{-1}(\sin(3\pi - x))}}{e^{\tan^{-1}(\sin(3\pi - x))} + e^{\tan^{-1}(\cos(3\pi - x))}} dx$.
Since $\sin(3\pi - x) = \sin x$ and $\cos(3\pi - x) = -\cos x$,this does not simplify directly.
Let us use the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ on the interval $[0, 2\pi]$.
The integrand $f(x) = \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}}$ has a period of $\frac{\pi}{2}$.
$I = \int_{\pi/2}^{5\pi/2} f(x) dx = \int_{0}^{2\pi} f(x) dx$.
Since $f(x)$ is periodic with period $\frac{\pi}{2}$,$\int_{0}^{2\pi} f(x) dx = 4 \int_{0}^{\pi/2} f(x) dx$.
Let $J = \int_{0}^{\pi/2} \frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)} + e^{\tan^{-1}(\cos x)}} dx$.
Using $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,$J = \int_{0}^{\pi/2} \frac{e^{\tan^{-1}(\cos x)}}{e^{\tan^{-1}(\cos x)} + e^{\tan^{-1}(\sin x)}} dx$.
$2J = \int_{0}^{\pi/2} 1 dx = \frac{\pi}{2} \implies J = \frac{\pi}{4}$.
Thus,$I = 4 \times \frac{\pi}{4} = \pi$.

(D) Given,$M = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} dx$.
Using the identity $\sin x \cos x = \frac{1}{2} \sin 2x$,we have $N = \int_{0}^{\frac{\pi}{4}} \frac{\sin 2x}{2(x+1)^{2}} dx$.
Let $2x = t$,then $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=\frac{\pi}{2}$.
Substituting these into $N$,we get $N = \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{2(\frac{t}{2}+1)^{2}} \cdot \frac{dt}{2} = \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{(t+2)^{2}} dt$.
Now,$M - N = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{x+2} dx - \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} dx$.
Using integration by parts on the first integral,let $u = \frac{1}{x+2}$ and $dv = \cos x dx$. Then $du = -\frac{1}{(x+2)^{2}} dx$ and $v = \sin x$.
$M = \left[ \frac{\sin x}{x+2} \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \sin x \left( -\frac{1}{(x+2)^{2}} \right) dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(x+2)^{2}} dx$.
Thus,$M - N = \left[ \frac{\sin x}{x+2} \right]_{0}^{\frac{\pi}{2}} = \frac{\sin(\pi/2)}{\pi/2 + 2} - \frac{\sin(0)}{0+2} = \frac{1}{\frac{\pi+4}{2}} = \frac{2}{\pi+4}$.

(C) We have $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$.
Since $f(x) = \frac{\sin x}{x}$ is a decreasing function on the interval $[\frac{\pi}{4}, \frac{\pi}{3}]$,the minimum value occurs at $x = \frac{\pi}{3}$ and the maximum value occurs at $x = \frac{\pi}{4}$.
The length of the interval is $\Delta x = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$.
Using the property of definite integrals for a monotonic function,we have:
$(\text{length of interval}) \times f(\text{upper limit}) \leq I \leq (\text{length of interval}) \times f(\text{lower limit})$.
Substituting the values:
$\frac{\pi}{12} \times \frac{\sin(\pi/3)}{\pi/3} \leq I \leq \frac{\pi}{12} \times \frac{\sin(\pi/4)}{\pi/4}$.
$\frac{\pi}{12} \times \frac{\sqrt{3}/2}{\pi/3} \leq I \leq \frac{\pi}{12} \times \frac{1/\sqrt{2}}{\pi/4}$.
$\frac{\pi}{12} \times \frac{3\sqrt{3}}{2\pi} \leq I \leq \frac{\pi}{12} \times \frac{4}{\pi\sqrt{2}}$.
$\frac{3\sqrt{3}}{24} \leq I \leq \frac{4}{12\sqrt{2}}$.
$\frac{\sqrt{3}}{8} \leq I \leq \frac{1}{3\sqrt{2}} = \frac{\sqrt{2}}{6}$.

(A) Given $I = \int_{0}^{1} \frac{x^{3} \cos 3x}{2+x^{2}} dx$.
Since $-1 \leq \cos 3x \leq 1$,we have $-x^{3} \leq x^{3} \cos 3x \leq x^{3}$.
Dividing by $2+x^{2} > 0$,we get $\frac{-x^{3}}{2+x^{2}} \leq \frac{x^{3} \cos 3x}{2+x^{2}} \leq \frac{x^{3}}{2+x^{2}}$.
Since $2+x^{2} > x^{2}$ for $x \in [0, 1]$,it follows that $\frac{x^{3}}{2+x^{2}} < \frac{x^{3}}{x^{2}} = x$.
Thus,$-\int_{0}^{1} x dx < I < \int_{0}^{1} x dx$.
Evaluating the integrals,$-\left[ \frac{x^{2}}{2} \right]_{0}^{1} < I < \left[ \frac{x^{2}}{2} \right]_{0}^{1}$.
This gives $-\frac{1}{2} < I < \frac{1}{2}$.

(C) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sec ^{2} \left( \frac{r \pi}{4 n} \right)$.
By the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f \left( \frac{r}{n} \right) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = \sec ^{2} \left( \frac{\pi x}{4} \right)$.
So,the integral becomes $\int_{0}^{1} \sec ^{2} \left( \frac{\pi x}{4} \right) dx$.
Integrating $\sec ^{2} \left( \frac{\pi x}{4} \right)$,we get $\frac{4}{\pi} \tan \left( \frac{\pi x}{4} \right)$.
Evaluating the definite integral from $0$ to $1$: $\left[ \frac{4}{\pi} \tan \left( \frac{\pi x}{4} \right) \right]_{0}^{1} = \frac{4}{\pi} \left( \tan \frac{\pi}{4} - \tan 0 \right)$.
Since $\tan \frac{\pi}{4} = 1$ and $\tan 0 = 0$,the value is $\frac{4}{\pi} (1 - 0) = \frac{4}{\pi}$.

(B) Given,the equation of the circle is $x^{2}+y^{2}=2ax$,which can be written as $(x-a)^{2}+y^{2}=a^{2}$.
The equation of the parabola is $y^{2}=ax$,where $a>0$.
To find the intersection points,substitute $y^{2}=ax$ into the circle equation:
$x^{2}+ax=2ax$
$x^{2}-ax=0$
$x(x-a)=0$
Thus,$x=0$ or $x=a$.
The intersection points are $(0,0)$ and $(a, a)$ (considering the region above the $X$-axis).
The required area is the area of the quarter circle in the first quadrant minus the area under the parabola from $x=0$ to $x=a$.
Area $= \int_{0}^{a} \sqrt{2ax-x^{2}} dx - \int_{0}^{a} \sqrt{ax} dx$
$= \frac{\pi a^{2}}{4} - \sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{a}$
$= \frac{\pi a^{2}}{4} - \frac{2}{3} \sqrt{a} (a^{3/2})$
$= \frac{\pi a^{2}}{4} - \frac{2a^{2}}{3}$
$= a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$

(C) Given,$y^{2}=2 d(x+\sqrt{d})$ $(i)$
Differentiating with respect to $x$,we get:
$2y y_{1} = 2d \Rightarrow d = y y_{1}$
Substituting $d = y y_{1}$ into equation $(i)$:
$y^{2} = 2(y y_{1})(x + \sqrt{y y_{1}})$
Rearranging the terms to isolate the radical:
$y^{2} - 2y y_{1} x = 2y y_{1} \sqrt{y y_{1}}$
Squaring both sides:
$(y^{2} - 2y y_{1} x)^{2} = (2y y_{1})^{2} (y y_{1})$
$(y^{2} - 2y y_{1} x)^{2} = 4(y y_{1})^{3}$
The highest derivative present is $y_{1}$ (order $1$),and the highest power of the highest derivative is $3$. Therefore,the degree of the differential equation is $3$.

(C) The given differential equation is $(1+x^{2}) \frac{dy}{dx} + 2xy = 4x^{2}$.
Dividing by $(1+x^{2})$,we get $\frac{dy}{dx} + \left(\frac{2x}{1+x^{2}}\right)y = \frac{4x^{2}}{1+x^{2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x}{1+x^{2}}$ and $Q(x) = \frac{4x^{2}}{1+x^{2}}$.
The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^{2}} dx} = e^{\ln(1+x^{2})} = 1+x^{2}$.
The general solution is given by $y(IF) = \int Q(x)(IF) dx + C$.
$y(1+x^{2}) = \int \left(\frac{4x^{2}}{1+x^{2}}\right)(1+x^{2}) dx + C$.
$y(1+x^{2}) = \int 4x^{2} dx + C = \frac{4x^{3}}{3} + C$.
Using the initial condition $y(0) = -1$,we substitute $x=0$ and $y=-1$:
$-1(1+0^{2}) = \frac{4(0)^{3}}{3} + C \Rightarrow C = -1$.
Thus,$y(1+x^{2}) = \frac{4x^{3}}{3} - 1$.
For $x=1$,$y(1+1^{2}) = \frac{4(1)^{3}}{3} - 1$.
$2y = \frac{4}{3} - 1 = \frac{1}{3}$.
Therefore,$y(1) = \frac{1}{6}$.

(C) Since $\vec{\delta}$ lies in the plane of $\vec{\alpha}$ and $\vec{\beta}$,we can write $\vec{\delta} = \vec{\alpha} + \lambda \vec{\beta}$ for some scalar $\lambda$.
Substituting the given vectors,we get $\vec{\delta} = (\hat{i}+\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}-\hat{k}) = (1+\lambda)\hat{i} + (1-\lambda)\hat{j} + (1-\lambda)\hat{k}$.
The projection of $\vec{\delta}$ on $\vec{\gamma}$ is given by $\frac{\vec{\delta} \cdot \vec{\gamma}}{|\vec{\gamma}|} = \frac{1}{\sqrt{3}}$.
Calculating the dot product: $\vec{\delta} \cdot \vec{\gamma} = (1+\lambda)(-1) + (1-\lambda)(1) + (1-\lambda)(-1) = -1 - \lambda + 1 - \lambda - 1 + \lambda = -1 - \lambda$.
The magnitude $|\vec{\gamma}| = \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{3}$.
Thus,$\frac{-1-\lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow -1-\lambda = 1 \Rightarrow \lambda = -2$.
Substituting $\lambda = -2$ into the expression for $\vec{\delta}$,we get $\vec{\delta} = (1-2)\hat{i} + (1-(-2))\hat{j} + (1-(-2))\hat{k} = -\hat{i} + 3\hat{j} + 3\hat{k}$.

(C) Since $\vec{\alpha} \cdot \vec{\beta} = 0$ and $\vec{\alpha} \cdot \vec{\gamma} = 0$,the vector $\vec{\alpha}$ is perpendicular to both $\vec{\beta}$ and $\vec{\gamma}$.
Thus,$\vec{\alpha}$ must be parallel to the cross product $\vec{\beta} \times \vec{\gamma}$.
Let $\vec{\alpha} = \lambda(\vec{\beta} \times \vec{\gamma})$ for some scalar $\lambda$.
Since $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are unit vectors,we have $|\vec{\alpha}| = 1, |\vec{\beta}| = 1, |\vec{\gamma}| = 1$.
Taking the magnitude on both sides: $|\vec{\alpha}| = |\lambda| |\vec{\beta} \times \vec{\gamma}|$.
We know that $|\vec{\beta} \times \vec{\gamma}| = |\vec{\beta}| |\vec{\gamma}| \sin(30^{\circ}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$.
Substituting the values: $1 = |\lambda| \cdot \frac{1}{2} \Rightarrow |\lambda| = 2 \Rightarrow \lambda = \pm 2$.
Therefore,$\vec{\alpha} = \pm 2(\vec{\beta} \times \vec{\gamma})$.

(D) Let the points be $A(0, -11, 4)$ and $B(2, -3, 1)$. The equation of the line passing through $A$ and $B$ is given by $\frac{x-0}{2-0} = \frac{y-(-11)}{-3-(-11)} = \frac{z-4}{1-4} = \lambda$.
This simplifies to $\frac{x}{2} = \frac{y+11}{8} = \frac{z-4}{-3} = \lambda$.
So,any point $P$ on this line is $(2\lambda, 8\lambda-11, -3\lambda+4)$.
Let $Q$ be the point $(1, 8, 4)$. The direction ratios of the line $PQ$ are $(2\lambda-1, 8\lambda-11-8, -3\lambda+4-4) = (2\lambda-1, 8\lambda-19, -3\lambda)$.
Since $PQ$ is perpendicular to the line,the dot product of the direction ratios of $PQ$ and the line $(2, 8, -3)$ must be zero:
$2(2\lambda-1) + 8(8\lambda-19) - 3(-3\lambda) = 0$.
$4\lambda - 2 + 64\lambda - 152 + 9\lambda = 0$.
$77\lambda - 154 = 0 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ into the coordinates of $P$:
$x = 2(2) = 4$,$y = 8(2)-11 = 5$,$z = -3(2)+4 = -2$.
Thus,the foot of the perpendicular is $(4, 5, -2)$.

(A) The equation of the line passing through $Q(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$.
Since point $P$ lies on this line,its coordinates can be represented as $P(\lambda+1, 4\lambda-2, 5\lambda+3)$.
Since $P$ also lies on the plane $2x + 3y - 4z + 22 = 0$,we substitute the coordinates of $P$ into the plane equation:
$2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$.
Expanding this,we get $2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$.
Simplifying,we get $-6\lambda + 6 = 0$,which implies $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $P$,we get $P(1+1, 4(1)-2, 5(1)+3) = P(2, 2, 8)$.
The distance $PQ$ is calculated as $\sqrt{(2-1)^2 + (2 - (-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$ units.

(B) Let $n$ be the number of times the coin is tossed.
For an unbiased coin,the probability of getting a head is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
The probability of getting at least one head is given by $P(X \geq 1) = 1 - P(X = 0)$.
Here,$P(X = 0)$ is the probability of getting no heads (i.e.,all tails),which is $q^n = (\frac{1}{2})^n$.
We are given that $P(X \geq 1) \geq 0.9$.
So,$1 - (\frac{1}{2})^n \geq 0.9$.
$1 - 0.9 \geq (\frac{1}{2})^n$.
$0.1 \geq \frac{1}{2^n}$.
$\frac{1}{10} \geq \frac{1}{2^n}$.
$2^n \geq 10$.
For $n = 3$,$2^3 = 8 < 10$.
For $n = 4$,$2^4 = 16 \geq 10$.
Thus,the minimum number of tosses required is $4$.