IIT JEE 2014 Mathematics Question Paper with Answer and Solution

(B) Let the circle be $x^2+y^2+2gx+2fy+c=0$ $\quad\quad(1)$
The given circles are:
$C_1: x^2+y^2-2x-15=0$ $\quad\quad(2)$
$C_2: x^2+y^2-1=0$ $\quad\quad(3)$
Since $(1)$ is orthogonal to $(2)$,we have $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here $g_1=g, f_1=f, c_1=c$ and for $(2)$,$g_2=-1, f_2=0, c_2=-15$:
$2(g)(-1) + 2(f)(0) = c-15 \Rightarrow -2g = c-15$ $\quad\quad(4)$
Since $(1)$ is orthogonal to $(3)$,where $g_3=0, f_3=0, c_3=-1$:
$2(g)(0) + 2(f)(0) = c-1 \Rightarrow c=1$
Substituting $c=1$ into $(4)$:
$-2g = 1-15 = -14 \Rightarrow g=7$
Since the circle passes through $(0,1)$:
$0^2+1^2+2g(0)+2f(1)+c=0$ $\Rightarrow 1+2f+1=0$ $\Rightarrow 2f=-2$ $\Rightarrow f=-1$
The circle equation is $x^2+y^2+14x-2y+1=0$.
Centre is $(-g, -f) = (-7, 1)$.
Radius is $\sqrt{g^2+f^2-c} = \sqrt{7^2+(-1)^2-1} = \sqrt{49+1-1} = \sqrt{49} = 7$.
Thus,$(B)$ and $(C)$ are correct.

(A) Let $L = \lim _{x \rightarrow 1}\left\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\}^{\frac{1-x}{1-\sqrt{x}}}$.
Since $\frac{1-x}{1-\sqrt{x}} = \frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}} = 1+\sqrt{x}$,the limit becomes $\lim _{x \rightarrow 1}\left\{\frac{-a(x-1)+\sin(x-1)}{(x-1)+\sin(x-1)}\right\}^{1+\sqrt{x}} = \frac{1}{4}$.
Let $x-1 = h$. As $x \rightarrow 1$,$h \rightarrow 0$. The expression becomes $\lim _{h \rightarrow 0}\left\{\frac{-ah+\sin h}{h+\sin h}\right\}^{1+\sqrt{1+h}} = \frac{1}{4}$.
Dividing numerator and denominator by $h$,we get $\lim _{h \rightarrow 0}\left\{\frac{-a+\frac{\sin h}{h}}{1+\frac{\sin h}{h}}\right\}^{1+\sqrt{1+h}} = \frac{1}{4}$.
Substituting $h=0$,we get $\left(\frac{-a+1}{1+1}\right)^{1+1} = \frac{1}{4}$,which simplifies to $\left(\frac{1-a}{2}\right)^2 = \frac{1}{4}$.
This implies $\frac{1-a}{2} = \frac{1}{2}$ or $\frac{1-a}{2} = -\frac{1}{2}$.
If $\frac{1-a}{2} = \frac{1}{2}$,then $1-a = 1$,so $a=0$.
If $\frac{1-a}{2} = -\frac{1}{2}$,then $1-a = -1$,so $a=2$.
For $a=2$,the base $\frac{-2h+\sin h}{h+\sin h} \approx \frac{-h}{2h} = -\frac{1}{2}$ as $h \rightarrow 0$. Since the base is negative,the power $1+\sqrt{1+h} \approx 2$ makes the result positive,but the limit of the base itself is negative,which is generally undefined in real-valued limits of the form $f(x)^{g(x)}$. Thus,$a=0$ is the valid non-negative integer.

(C) Let $P(x, y)$ be a point in the first quadrant,so $x > 0$ and $y > 0$.
The distances are $d_1(P) = \frac{|x-y|}{\sqrt{2}}$ and $d_2(P) = \frac{|x+y|}{\sqrt{2}}$.
The given condition is $2 \leq \frac{|x-y|}{\sqrt{2}} + \frac{|x+y|}{\sqrt{2}} \leq 4$,which simplifies to $2\sqrt{2} \leq |x-y| + |x+y| \leq 4\sqrt{2}$.
Case $1$: $x \geq y$. Then $|x-y| + |x+y| = (x-y) + (x+y) = 2x$.
So,$2\sqrt{2} \leq 2x \leq 4\sqrt{2} \implies \sqrt{2} \leq x \leq 2\sqrt{2}$. Since $x \geq y > 0$,this region is a rectangle in the first quadrant.
Case $2$: $y > x$. Then $|x-y| + |x+y| = (y-x) + (x+y) = 2y$.
So,$2\sqrt{2} \leq 2y \leq 4\sqrt{2} \implies \sqrt{2} \leq y \leq 2\sqrt{2}$. Since $y > x > 0$,this region is also a rectangle.
The region $R$ is the union of these two rectangles,which forms an $L$-shaped region.
The area is the difference between the area of a large square of side $2\sqrt{2}$ and a small square of side $\sqrt{2}$.
Area $= (2\sqrt{2})^2 - (\sqrt{2})^2 = 8 - 2 = 6$.

(D) We are given $n_1 < n_2 < n_3 < n_4 < n_5$ where $n_i \in \mathbb{Z}^+$ and $\sum_{i=1}^5 n_i = 20$.
Let $n_1 = a_1$,$n_2 = a_1 + a_2 + 1$,$n_3 = a_1 + a_2 + a_3 + 2$,$n_4 = a_1 + a_2 + a_3 + a_4 + 3$,$n_5 = a_1 + a_2 + a_3 + a_4 + a_5 + 4$,where $a_1 \geq 1$ and $a_2, a_3, a_4, a_5 \geq 0$.
Substituting these into the sum:
$5a_1 + 4a_2 + 3a_3 + 2a_4 + a_5 + 10 = 20 \implies 5a_1 + 4a_2 + 3a_3 + 2a_4 + a_5 = 10$.
Since $a_1 \geq 1$,let $a_1 = 1 + k$ where $k \geq 0$. Then $5(1+k) + 4a_2 + 3a_3 + 2a_4 + a_5 = 10 \implies 5k + 4a_2 + 3a_3 + 2a_4 + a_5 = 5$.
If $k=1$,then $a_2=a_3=a_4=a_5=0$,which gives $(n_1, n_2, n_3, n_4, n_5) = (2, 3, 4, 5, 6)$.
If $k=0$,then $4a_2 + 3a_3 + 2a_4 + a_5 = 5$. The possible non-negative integer solutions $(a_2, a_3, a_4, a_5)$ are:
$1$) $(1, 0, 0, 1) \implies (1, 3, 4, 5, 7)$
$2$) $(1, 0, 1, -1)$ (Invalid)
$3$) $(0, 1, 1, 0) \implies (1, 2, 4, 6, 7)$
$4$) $(0, 1, 0, 2) \implies (1, 2, 3, 6, 8)$
$5$) $(0, 0, 2, 1) \implies (1, 2, 3, 5, 9)$
$6$) $(0, 0, 1, 3) \implies (1, 2, 3, 4, 10)$
$7$) $(0, 0, 0, 5) \implies (1, 2, 3, 4, 10)$ (Wait,checking manually: $(1, 2, 4, 5, 8)$ is another solution).
Listing all valid partitions of $20$ into $5$ distinct parts: $(1, 2, 3, 4, 10), (1, 2, 3, 5, 9), (1, 2, 3, 6, 8), (1, 2, 4, 5, 8), (1, 2, 4, 6, 7), (1, 3, 4, 5, 7), (2, 3, 4, 5, 6)$.
There are exactly $7$ such arrangements.

(D) Given that $a, b, c$ are in geometric progression,let $b = ar$ and $c = ar^2$,where $r$ is an integer since $\frac{b}{a} = r$ is an integer.
The arithmetic mean of $a, b, c$ is given by $\frac{a+b+c}{3} = b+2$.
Substituting $b = ar$ and $c = ar^2$,we get $\frac{a + ar + ar^2}{3} = ar + 2$.
Multiplying by $3$,we have $a + ar + ar^2 = 3ar + 6$,which simplifies to $a + ar^2 = 2ar + 6$.
Rearranging the terms,we get $a(1 - 2r + r^2) = 6$,or $a(r-1)^2 = 6$.
Since $a$ and $r$ are integers,$(r-1)^2$ must be a factor of $6$. The perfect square factors of $6$ are $1$. Thus,$(r-1)^2 = 1$,which implies $r-1 = 1$ (since $r$ must be positive for $a, b, c$ to be positive integers),so $r = 2$.
Substituting $r = 2$ into $a(r-1)^2 = 6$,we get $a(1)^2 = 6$,so $a = 6$.
Now,we calculate the value of $\frac{a^2+a-14}{a+1}$ for $a = 6$:
$\frac{6^2 + 6 - 14}{6 + 1} = \frac{36 + 6 - 14}{7} = \frac{28}{7} = 4$.

(C) The total number of line segments formed by joining $n$ points is given by $\binom{n}{2} = \frac{n(n-1)}{2}$.
The number of line segments joining adjacent points (which form the sides of an $n$-sided polygon) is $n$.
The number of line segments joining non-adjacent points (which are the diagonals of the polygon) is $\binom{n}{2} - n$.
According to the problem,the number of blue segments (adjacent) equals the number of red segments (non-adjacent):
$n = \binom{n}{2} - n$
$2n = \frac{n(n-1)}{2}$
$4n = n^2 - n$
$n^2 - 5n = 0$
$n(n - 5) = 0$
Since $n \geq 2$,we have $n = 5$.

(D) Let $p(x) = ax^2 + c$ where $a, c \in \mathbb{R}$. Since the roots are purely imaginary,let them be $\pm i k$ $(k \neq 0)$.
Then $p(ik) = a(ik)^2 + c = -ak^2 + c = 0$,which implies $c = ak^2$.
Thus,$p(x) = a(x^2 + k^2)$.
Now,consider $p(p(x)) = 0$,which implies $p(x) = \pm ik$.
$a(x^2 + k^2) = ik$ or $a(x^2 + k^2) = -ik$.
$x^2 + k^2 = \pm \frac{ik}{a}$.
$x^2 = -k^2 \pm \frac{ik}{a}$.
Since $k^2$ is real and $\pm \frac{ik}{a}$ is purely imaginary,$x^2$ is a complex number with a non-zero imaginary part.
Therefore,$x$ cannot be purely real (as $x^2$ would be real) and $x$ cannot be purely imaginary (as $x^2$ would be real).
Thus,the roots are neither real nor purely imaginary. The correct option is $(D)$.

(A) Let $B$ represent a boy and $G$ represent a girl. There are $3$ boys and $2$ girls. The total number of ways to arrange $5$ people is $5! = 120$.
Let $b_i$ be the number of boys ahead of the $i$-th girl and $g_i$ be the number of girls ahead of the $i$-th girl. The condition is $b_i \ge g_i + 1$ for both girls.
Let the positions of the girls be $x_1$ and $x_2$ where $1 \le x_1 < x_2 \le 5$.
For the first girl at $x_1$,the number of boys ahead is $x_1 - 1$ and girls ahead is $0$. So,$x_1 - 1 \ge 0 + 1 \implies x_1 \ge 2$.
For the second girl at $x_2$,the number of boys ahead is $x_2 - 2$ and girls ahead is $1$. So,$x_2 - 2 \ge 1 + 1 \implies x_2 \ge 4$.
Possible positions $(x_1, x_2)$ are $(2, 4), (2, 5), (3, 4), (3, 5), (4, 5)$.
Checking the condition $b_i \ge g_i + 1$:
$1$. $(2, 4): b_1=1, g_1=0 (1 \ge 1); b_2=2, g_2=1 (2 \ge 2)$. Valid.
$2$. $(2, 5): b_1=1, g_1=0 (1 \ge 1); b_2=3, g_2=1 (3 \ge 2)$. Valid.
$3$. $(3, 4): b_1=2, g_1=0 (2 \ge 1); b_2=2, g_2=1 (2 \ge 2)$. Valid.
$4$. $(3, 5): b_1=2, g_1=0 (2 \ge 1); b_2=3, g_2=1 (3 \ge 2)$. Valid.
$5$. $(4, 5): b_1=3, g_1=0 (3 \ge 1); b_2=3, g_2=1 (3 \ge 2)$. Valid.
Total valid arrangements = $5 \times (3! \times 2!) = 5 \times 12 = 60$.
Probability = $\frac{60}{120} = \frac{1}{2}$.

(C) Let the cards be $C_1, C_2, C_3, C_4, C_5, C_6$ and envelopes be $E_1, E_2, E_3, E_4, E_5, E_6$.
Given $C_1$ is in $E_2$.
We need to place $C_2, C_3, C_4, C_5, C_6$ in $E_1, E_3, E_4, E_5, E_6$ such that $C_i$ is not in $E_i$ for $i \in \{2, 3, 4, 5, 6\}$.
Case $1$: $C_2$ is in $E_1$.
Then we need to place $C_3, C_4, C_5, C_6$ in $E_3, E_4, E_5, E_6$ such that no card $C_i$ is in $E_i$. This is a derangement of $4$ objects,$D_4 = 4!(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 24(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 12 - 4 + 1 = 9$.
Case $2$: $C_2$ is not in $E_1$.
We have $5$ cards $C_2, C_3, C_4, C_5, C_6$ to be placed in $E_1, E_3, E_4, E_5, E_6$ such that $C_2 \neq E_1, C_3 \neq E_3, C_4 \neq E_4, C_5 \neq E_5, C_6 \neq E_6$. This is a derangement of $5$ objects,$D_5 = 5!(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}) = 120(\frac{60-20+5-1}{120}) = 44$.
Total ways = $9 + 44 = 53$.

(B) Let the two sides be $a$ and $b$. Given $a+b=x$ and $ab=y$.
Given $x^2-c^2=y$,substituting $x=a+b$ and $y=ab$,we get $(a+b)^2-c^2=ab$.
$a^2+b^2+2ab-c^2=ab \implies a^2+b^2-c^2=-ab$.
Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{-ab}{2ab} = -\frac{1}{2}$.
Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$.
The ratio of in-radius $r$ to circum-radius $R$ is given by $\frac{r}{R} = 4 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}$.
Alternatively,$\frac{r}{R} = \frac{\Delta/s}{abc/4\Delta} = \frac{4\Delta^2}{sabc} = \frac{4 \cdot (\frac{1}{2}ab \sin C)^2}{(\frac{a+b+c}{2})abc} = \frac{a^2b^2 \sin^2(120^\circ)}{(\frac{x+c}{2})abc} = \frac{y^2 \cdot (\frac{\sqrt{3}}{2})^2}{(\frac{x+c}{2})abc} = \frac{3y^2}{2(x+c)abc}$.
Since $ab=y$,this simplifies to $\frac{3y}{2c(x+c)}$.
Wait,re-evaluating: $\frac{r}{R} = \frac{4\Delta^2}{sabc} = \frac{a^2b^2 \sin^2 C}{sabc} = \frac{y^2 \cdot (3/4)}{(\frac{x+c}{2})abc} = \frac{3y^2}{2(x+c)abc} = \frac{3y^2}{2(x+c)yc} = \frac{3y}{2c(x+c)}$.

(D) The equation of a tangent to the parabola $y^2=8x$ is $y=mx+\frac{2}{m}$.
For this to be a tangent to the circle $x^2+y^2=2$,the perpendicular distance from the center $(0,0)$ to the line $mx-y+\frac{2}{m}=0$ must be equal to the radius $\sqrt{2}$.
$\left|\frac{2/m}{\sqrt{m^2+1}}\right|=\sqrt{2}$ $\Rightarrow \frac{4}{m^2(m^2+1)}=2$ $\Rightarrow m^4+m^2-2=0$.
Let $m^2=t$,then $t^2+t-2=0 \Rightarrow (t+2)(t-1)=0$. Since $t=m^2 > 0$,we have $m^2=1$,so $m=\pm 1$.
The common tangents are $y=x+2$ and $y=-x-2$.
The points of contact $P, Q$ on the circle $x^2+y^2=2$ are found by the chord of contact formula $xx_1+yy_1=r^2$. For the tangent $x-y+2=0$,the point of contact is $(-1, 1)$. For $x+y+2=0$,it is $(-1, -1)$. Thus $P=(-1, 1)$ and $Q=(-1, -1)$.
The points of contact $R, S$ on the parabola $y^2=8x$ are found using $yy_1=4(x+x_1)$. For $y=x+2$,$y_1y=4(x+x_1)$ $\Rightarrow 1(y)=4(x+1)$ $\Rightarrow y=4x+4$. Comparing with $y=x+2$,we get $x=2, y=4$. For $y=-x-2$,we get $x=2, y=-4$. Thus $R=(2, 4)$ and $S=(2, -4)$.
The quadrilateral $PQRS$ is a trapezium with parallel sides $PQ$ and $RS$.
Length $PQ = |1 - (-1)| = 2$.
Length $RS = |4 - (-4)| = 8$.
Height $h = |2 - (-1)| = 3$.
Area $= \frac{1}{2}(PQ+RS) \times h = \frac{1}{2}(2+8) \times 3 = \frac{1}{2}(10) \times 3 = 15$.

(C) We need to find the coefficient of $x^{11}$ in the expansion of $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$.
This is equivalent to finding the number of non-negative integer solutions to $2a + 3b + 4c = 11$,where $0 \le a \le 4$,$0 \le b \le 7$,and $0 \le c \le 12$.
We list the possible combinations of $(a, b, c)$:
$1$. If $b=1$,then $2a + 4c = 11 - 3 = 8 \implies a + 2c = 4$. Possible $(a, c)$ are $(4, 0), (2, 1), (0, 2)$.
$2$. If $b=3$,then $2a + 4c = 11 - 9 = 2 \implies a + 2c = 1$. Possible $(a, c)$ is $(1, 0)$.
Now calculate the coefficients using the binomial theorem $\binom{n}{r}$:
- For $(a, b, c) = (4, 1, 0)$: $\binom{4}{4} \times \binom{7}{1} \times \binom{12}{0} = 1 \times 7 \times 1 = 7$.
- For $(a, b, c) = (2, 1, 1)$: $\binom{4}{2} \times \binom{7}{1} \times \binom{12}{1} = 6 \times 7 \times 12 = 504$.
- For $(a, b, c) = (0, 1, 2)$: $\binom{4}{0} \times \binom{7}{1} \times \binom{12}{2} = 1 \times 7 \times 66 = 462$.
- For $(a, b, c) = (1, 3, 0)$: $\binom{4}{1} \times \binom{7}{3} \times \binom{12}{0} = 4 \times 35 \times 1 = 140$.
Summing these values: $7 + 504 + 462 + 140 = 1113$.

(D) Given the equation: $\sin x + 2 \sin 2x - \sin 3x = 3$.
Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we substitute:
$\sin x + 2(2 \sin x \cos x) - (3 \sin x - 4 \sin^3 x) = 3$.
$-2 \sin x + 4 \sin x \cos x + 4 \sin^3 x = 3$.
Since $\sin x \in (0, 1]$ for $x \in (0, \pi)$,we analyze the maximum value of the expression $f(x) = \sin x + 2 \sin 2x - \sin 3x$.
We know that $\sin x \leq 1$,$\sin 2x \leq 1$,and $-\sin 3x \leq 1$.
However,these cannot all be $1$ simultaneously for the same $x$.
Alternatively,consider the range of $f(x) = \sin x + 4 \sin x \cos x - 3 \sin x + 4 \sin^3 x = 4 \sin x \cos x + 4 \sin^3 x - 2 \sin x$.
$f(x) = 2 \sin x (2 \cos x + 2 \sin^2 x - 1) = 2 \sin x (2 \cos x + 2(1 - \cos^2 x) - 1) = 2 \sin x (-2 \cos^2 x + 2 \cos x + 1)$.
Let $u = \cos x$,where $u \in (-1, 1)$. The function $g(u) = 2 \sqrt{1-u^2} (-2u^2 + 2u + 1)$.
By checking the maximum value,it is found that $f(x) < 3$ for all $x \in (0, \pi)$.
Thus,there is no solution.

(D,B) $1.$ Since $PQ$ is a focal chord,$t \cdot t' = -1$,so $t' = -\frac{1}{t}$.
The slope of $PK$ is $m_{PK} = \frac{2at - 0}{at^2 - 2a} = \frac{2at}{a(t^2-2)} = \frac{2t}{t^2-2}$.
The slope of $QR$ is $m_{QR} = \frac{2ar - 2at'}{ar^2 - at'^2} = \frac{2a(r-t')}{a(r-t')(r+t')} = \frac{2}{r+t'}$.
Since $QR \parallel PK$,$m_{QR} = m_{PK} \implies \frac{2}{r+t'} = \frac{2t}{t^2-2}$.
$r+t' = \frac{t^2-2}{t} = t - \frac{2}{t}$.
Substituting $t' = -\frac{1}{t}$,we get $r - \frac{1}{t} = t - \frac{2}{t} \implies r = t - \frac{1}{t} = \frac{t^2-1}{t}$.
Thus,the correct option for $1$ is $(D)$.
$2.$ The tangent at $P(at^2, 2at)$ is $ty = x + at^2$.
The normal at $S(as^2, 2as)$ is $y = -sx + 2as + as^3$,or $y + sx = 2as + as^3$.
Given $st = 1$,so $s = \frac{1}{t}$.
The normal equation becomes $y + \frac{1}{t}x = 2a(\frac{1}{t}) + a(\frac{1}{t^3}) = \frac{2at^2+a}{t^3}$.
$ty + x = \frac{2at^2+a}{t^2}$.
We have the system:
$ty - x = at^2$
$ty + x = \frac{2at^2+a}{t^2}$
Adding the two equations: $2ty = at^2 + \frac{2at^2+a}{t^2} = \frac{at^4 + 2at^2 + a}{t^2} = \frac{a(t^2+1)^2}{t^2}$.
$y = \frac{a(t^2+1)^2}{2t^3}$.
Thus,the correct option for $2$ is $(B)$.

(C) $(P)$ Since $z_k = e^{i(2k\pi/10)}$,we have $z_k \cdot z_j = e^{i(2(k+j)\pi/10)} = 1$ if $k+j = 10$. For any $k \in \{1, \ldots, 9\}$,we can choose $j = 10-k \in \{1, \ldots, 9\}$. Thus,the statement is True $(1)$.
$(Q)$ The equation $z_1 \cdot z = z_k$ is a linear equation in complex numbers,which always has a solution $z = z_k / z_1$. Thus,the statement is False $(2)$.
$(R)$ The values $z_1, z_2, \ldots, z_9$ are the roots of $\frac{z^{10}-1}{z-1} = 0$. Thus,$z^{10}-1 = (z-1)(z-z_1)(z-z_2)\ldots(z-z_9)$. Dividing by $(z-1)$,we get $1+z+z^2+\ldots+z^9 = (z-z_1)(z-z_2)\ldots(z-z_9)$. Setting $z=1$,we get $10 = (1-z_1)(1-z_2)\ldots(1-z_9)$. Taking the modulus,$|1-z_1||1-z_2|\ldots|1-z_9| = 10$. Thus,the expression equals $10/10 = 1$ $(3)$.
$(S)$ The sum of all roots of $z^{10}-1=0$ is $1 + z_1 + z_2 + \ldots + z_9 = 0$. Thus,$\sum_{k=1}^9 z_k = -1$. Taking the real part,$\sum_{k=1}^9 \cos(2k\pi/10) = -1$. Then $1 - (-1) = 2$ $(4)$.
Therefore,the correct matching is $P-1, Q-2, R-3, S-4$.

(C) First,we check the continuity of $g(x)$ at $x = a$ and $x = b$.
At $x = a$: $\lim_{x \rightarrow a^-} g(x) = 0$ and $\lim_{x \rightarrow a^+} g(x) = \int_a^a f(t) dt = 0$. Since $g(a) = 0$,$g(x)$ is continuous at $x = a$.
At $x = b$: $\lim_{x \rightarrow b^-} g(x) = \int_a^b f(t) dt$ and $\lim_{x \rightarrow b^+} g(x) = \int_a^b f(t) dt$. Since $g(b) = \int_a^b f(t) dt$,$g(x)$ is continuous at $x = b$.
Thus,$g(x)$ is continuous for all $x \in \mathbb{R}$.
Next,we check differentiability using $g'(x) = \begin{cases} 0 & x < a \\ f(x) & a < x < b \\ 0 & x > b \end{cases}$.
At $x = a$: $g'(a^-) = 0$ and $g'(a^+) = f(a)$. Since $f(a) \in [1, \infty)$,$f(a) \neq 0$,so $g'(a^-) \neq g'(a^+)$. Thus,$g(x)$ is not differentiable at $x = a$.
At $x = b$: $g'(b^-) = f(b)$ and $g'(b^+) = 0$. Since $f(b) \in [1, \infty)$,$f(b) \neq 0$,so $g'(b^-) \neq g'(b^+)$. Thus,$g(x)$ is not differentiable at $x = b$.
Therefore,$g(x)$ is continuous but not differentiable at $a$ and $b$.

(D) Let $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$.
Since $f$ and $g$ are continuous on $[0, 1]$,there exist $a, b \in [0, 1]$ such that $f(a) = \lambda$ and $g(b) = \lambda$.
Define $h(x) = f(x) - g(x)$.
Then $h(a) = f(a) - g(a) = \lambda - g(a) \ge 0$ and $h(b) = f(b) - g(b) = f(b) - \lambda \le 0$.
By the Intermediate Value Theorem,there exists $c \in [0, 1]$ such that $h(c) = 0$,which implies $f(c) = g(c)$.
For option $(A)$: $(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$. Since $f(c) = g(c)$,this holds true.
For option $(D)$: $(f(c))^2 = (g(c))^2$. Since $f(c) = g(c)$,this holds true.
For options $(B)$ and $(C)$,consider $f(x) = g(x) = \lambda$ where $\lambda \neq 0$. Then $(B)$ becomes $\lambda^2 + \lambda = \lambda^2 + 3\lambda$,which implies $\lambda = 3\lambda$,or $\lambda = 0$,which contradicts $\lambda \neq 0$. Similarly for $(C)$.
Thus,$(A)$ and $(D)$ are correct.

(B) Let $M = \begin{bmatrix} a & b \\ b & c \end{bmatrix}$ where $a, b, c \in \mathbb{Z}$.
$(A)$ The first column is $\begin{bmatrix} a \\ b \end{bmatrix}$ and the transpose of the second row is $\begin{bmatrix} b \\ c \end{bmatrix}$. If they are equal,then $a=b$ and $b=c$,so $a=b=c$. Then $M = \begin{bmatrix} a & a \\ a & a \end{bmatrix}$,which has determinant $|M| = a^2 - a^2 = 0$. Thus,$M$ is not invertible. $(A)$ is incorrect.
$(B)$ The second row is $[b, c]$ and the transpose of the first column is $\begin{bmatrix} a \\ b \end{bmatrix}$. If they are equal,then $b=a$ and $c=b$,so $a=b=c$. This leads to the same matrix as $(A)$,which is not invertible. $(B)$ is incorrect.
$(C)$ If $M$ is a diagonal matrix,$M = \begin{bmatrix} a & 0 \\ 0 & c \end{bmatrix}$. The determinant is $|M| = ac$. Since $a, c \neq 0$,$|M| \neq 0$. Thus,$M$ is invertible. $(C)$ is correct.
$(D)$ The determinant of $M$ is $|M| = ac - b^2$. For $M$ to be invertible,$|M| \neq 0$,which means $ac - b^2 \neq 0$,or $ac \neq b^2$. Since $b$ is an integer,$b^2$ is a perfect square. Thus,if $ac$ is not a perfect square of an integer,then $ac \neq b^2$ is guaranteed. $(D)$ is correct.
Therefore,$(C, D)$ are correct.

(A, B, C) Given $|\vec{x}| = |\vec{y}| = |\vec{z}| = \sqrt{2}$ and the angle between each pair is $\frac{\pi}{3}$.
Thus,$\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{z} = \vec{z} \cdot \vec{x} = |\vec{x}||\vec{y}| \cos(\frac{\pi}{3}) = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1$.
Since $\vec{a}$ is perpendicular to $\vec{x}$ and $\vec{y} \times \vec{z}$,$\vec{a}$ is parallel to $\vec{x} \times (\vec{y} \times \vec{z})$.
Using the vector triple product formula,$\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} = 1\vec{y} - 1\vec{z} = \vec{y} - \vec{z}$.
So,$\vec{a} = \lambda(\vec{y} - \vec{z})$.
Then $\vec{a} \cdot \vec{y} = \lambda(\vec{y} \cdot \vec{y} - \vec{z} \cdot \vec{y}) = \lambda(2 - 1) = \lambda$. Thus,$\vec{a} = (\vec{a} \cdot \vec{y})(\vec{y} - \vec{z})$,which is $(B)$.
Similarly,$\vec{b}$ is perpendicular to $\vec{y}$ and $\vec{z} \times \vec{x}$,so $\vec{b}$ is parallel to $\vec{y} \times (\vec{z} \times \vec{x}) = (\vec{y} \cdot \vec{x})\vec{z} - (\vec{y} \cdot \vec{z})\vec{x} = 1\vec{z} - 1\vec{x} = \vec{z} - \vec{x}$.
So,$\vec{b} = \mu(\vec{z} - \vec{x})$.
Then $\vec{b} \cdot \vec{z} = \mu(\vec{z} \cdot \vec{z} - \vec{x} \cdot \vec{z}) = \mu(2 - 1) = \mu$. Thus,$\vec{b} = (\vec{b} \cdot \vec{z})(\vec{z} - \vec{x})$,which is $(A)$.
Now,$\vec{a} \cdot \vec{b} = \lambda \mu (\vec{y} - \vec{z}) \cdot (\vec{z} - \vec{x}) = \lambda \mu (\vec{y} \cdot \vec{z} - \vec{y} \cdot \vec{x} - \vec{z} \cdot \vec{z} + \vec{z} \cdot \vec{x}) = \lambda \mu (1 - 1 - 2 + 1) = -\lambda \mu = -(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$,which is $(C)$.

(C) The first line is $L_1: \frac{x}{1} = \frac{y}{1}, z=1$. Let any point $Q$ on $L_1$ be $(\alpha, \alpha, 1)$.
Direction ratios of $PQ$ are $(\alpha-\lambda, \alpha-\lambda, 1-\lambda)$.
Since $PQ$ is perpendicular to $L_1$ (direction vector $(1, 1, 0)$),we have $(\alpha-\lambda)(1) + (\alpha-\lambda)(1) + (1-\lambda)(0) = 0$,which gives $2(\alpha-\lambda) = 0$,so $\alpha = \lambda$.
Thus,$Q = (\lambda, \lambda, 1)$ and the vector $\vec{PQ} = (0, 0, 1-\lambda)$.
The second line is $L_2: \frac{x}{-1} = \frac{y}{1}, z=-1$. Let any point $R$ on $L_2$ be $(-\beta, \beta, -1)$.
Direction ratios of $PR$ are $(-\beta-\lambda, \beta-\lambda, -1-\lambda)$.
Since $PR$ is perpendicular to $L_2$ (direction vector $(-1, 1, 0)$),we have $(-\beta-\lambda)(-1) + (\beta-\lambda)(1) + (-1-\lambda)(0) = 0$,which gives $\beta+\lambda+\beta-\lambda = 0$,so $2\beta = 0$,hence $\beta = 0$.
Thus,$R = (0, 0, -1)$ and the vector $\vec{PR} = (-\lambda, -\lambda, -1-\lambda)$.
Given $\angle QPR = 90^\circ$,so $\vec{PQ} \cdot \vec{PR} = 0$.
$(0)(-\lambda) + (0)(-\lambda) + (1-\lambda)(-1-\lambda) = 0$.
$-(1-\lambda)(1+\lambda) = 0 \Rightarrow \lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1$.
If $\lambda = 1$,$P = (1, 1, 1)$,which lies on $L_1$,so $PQ$ is not uniquely defined. Thus,$\lambda = -1$ is the only solution.

(C) Given $MN = NM$ and $M^2 = N^4$.
This implies $M^2 - N^4 = 0$,which can be factored as $(M - N^2)(M + N^2) = 0$ because $M$ and $N$ commute.
Since $M \neq N^2$,the matrix $(M - N^2)$ is not necessarily the zero matrix,but the product $(M - N^2)(M + N^2) = 0$ implies that the determinant of the product is zero:
$|M - N^2| \cdot |M + N^2| = 0$.
From the provided logic,in any case,$|M + N^2| = 0$.
Now,consider the expression $M^2 + MN^2 = M(M + N^2)$.
The determinant is $|M^2 + MN^2| = |M| \cdot |M + N^2| = |M| \cdot 0 = 0$.
Thus,$(A)$ is correct.
Since the determinant of $(M^2 + MN^2)$ is $0$,the matrix $(M^2 + MN^2)$ is singular.
Therefore,there exists a non-zero matrix $U$ such that $(M^2 + MN^2)U = 0$ (as the system of linear equations $AX = 0$ with $|A| = 0$ has non-trivial solutions).
Thus,$(B)$ is correct.
$(C)$ is incorrect because the determinant is $0$.
$(D)$ is incorrect because for a singular matrix $A$,$AU = 0$ does not imply $U = 0$ (non-trivial solutions exist).
Therefore,the correct options are $(A)$ and $(B)$.

(A) Given $f(x)=\int_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$.
Using Leibniz's rule for differentiation under the integral sign:
$f^{\prime}(x) = e^{-\left(x+\frac{1}{x}\right)} \cdot \frac{1}{x} - e^{-\left(\frac{1}{x}+x\right)} \cdot \left(-\frac{1}{x^2}\right) = \frac{e^{-\left(x+\frac{1}{x}\right)}}{x} + \frac{e^{-\left(x+\frac{1}{x}\right)}}{x} = \frac{2 e^{-\left(x+\frac{1}{x}\right)}}{x}$.
$(A)$ For $x \in [1, \infty)$,$f^{\prime}(x) > 0$,so $f(x)$ is monotonically increasing. Thus,$(A)$ is correct.
$(B)$ For $x \in (0, 1)$,$f^{\prime}(x) > 0$,so $f(x)$ is monotonically increasing on $(0, 1)$. Thus,$(B)$ is incorrect.
$(C)$ $f(1/x) = \int_{x}^{1/x} e^{-\left(t+\frac{1}{t}\right)} \frac{dt}{t}$. Let $t = 1/u$,then $dt = -1/u^2 du$.
$f(1/x) = \int_{1/x}^{x} e^{-\left(1/u+u\right)} \cdot u \cdot (-1/u^2) du = -\int_{1/x}^{x} e^{-\left(u+\frac{1}{u}\right)} \frac{du}{u} = -f(x)$.
Therefore,$f(x) + f(1/x) = 0$. Thus,$(C)$ is correct.
$(D)$ Let $g(x) = f(2^x)$. Then $g(-x) = f(2^{-x}) = f(1/2^x) = -f(2^x) = -g(x)$.
Thus,$f(2^x)$ is an odd function. Thus,$(D)$ is correct.
The correct options are $(A, C, D)$.

(A, C) Step $1$: Check for odd/even function.
$f(-x) = (\log(\sec(-x) + \tan(-x)))^3 = (\log(\sec x - \tan x))^3$.
Since $\sec x - \tan x = \frac{1}{\sec x + \tan x}$,we have $\log(\sec x - \tan x) = \log((\sec x + \tan x)^{-1}) = -\log(\sec x + \tan x)$.
Thus,$f(-x) = (-\log(\sec x + \tan x))^3 = -(\log(\sec x + \tan x))^3 = -f(x)$.
Therefore,$f(x)$ is an odd function.
Step $2$: Check for one-one function.
$f'(x) = 3(\log(\sec x + \tan x))^2 \cdot \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) = 3(\log(\sec x + \tan x))^2 \cdot \sec x$.
Since $\sec x > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $(\log(\sec x + \tan x))^2 \ge 0$,$f'(x) \ge 0$. The function is strictly increasing,so it is a one-one function.
Step $3$: Check for onto function.
As $x \rightarrow \frac{\pi}{2}^-$,$\sec x + \tan x \rightarrow \infty$,so $f(x) \rightarrow \infty$.
As $x \rightarrow -\frac{\pi}{2}^+$,$\sec x + \tan x \rightarrow 0^+$,so $\log(\sec x + \tan x) \rightarrow -\infty$,and $f(x) \rightarrow -\infty$.
Since the range is $(-\infty, \infty) = \mathbb{R}$,the function is onto.
Conclusion: $f(x)$ is an odd function and an onto function.

(A) Given $f(x) = x^5 - 5x + a$.
To find the number of real roots,we analyze the function $g(x) = x^5 - 5x$,where $f(x) = g(x) + a = 0$,which implies $g(x) = -a$.
First,find the critical points of $g(x)$ by setting $g'(x) = 0$:
$g'(x) = 5x^4 - 5 = 5(x^4 - 1) = 5(x^2 - 1)(x^2 + 1) = 5(x - 1)(x + 1)(x^2 + 1) = 0$.
The critical points are $x = 1$ and $x = -1$.
The local maximum value is $g(-1) = (-1)^5 - 5(-1) = -1 + 5 = 4$.
The local minimum value is $g(1) = (1)^5 - 5(1) = 1 - 5 = -4$.
For $f(x) = 0$,we need $g(x) = -a$.
$1$. If $-a > 4$ (i.e.,$a < -4$),the line $y = -a$ is above the local maximum,so there is only $1$ real root.
$2$. If $-a < -4$ (i.e.,$a > 4$),the line $y = -a$ is below the local minimum,so there is only $1$ real root.
$3$. If $-4 < -a < 4$ (i.e.,$-4 < a < 4$),the line $y = -a$ intersects the graph at $3$ points,so there are $3$ real roots.
Thus,$(B)$ is correct ($a > 4$ implies $1$ root) and $(D)$ is correct ($-4 < a < 4$ implies $3$ roots).

(D) Given the curve equation: $(y-x^5)^2=x(1+x^2)^2$
Differentiating both sides with respect to $x$:
$2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + x \cdot 2(1+x^2)(2x)$
$2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + 4x^2(1+x^2)$
Substitute the point $(1,3)$ into the equation:
$2(3-1^5)(\frac{dy}{dx}-5(1)^4) = (1+1^2)^2 + 4(1)^2(1+1^2)$
$2(3-1)(\frac{dy}{dx}-5) = (2)^2 + 4(2)$
$2(2)(\frac{dy}{dx}-5) = 4 + 8$
$4(\frac{dy}{dx}-5) = 12$
$\frac{dy}{dx}-5 = 3$
$\frac{dy}{dx} = 8$
Thus,the slope of the tangent at $(1,3)$ is $8$.

(B) The function $f(x) = \cos^{-1}(\cos x)$ is a periodic function with period $2\pi$.
In the interval $[0, 4\pi]$,the graph of $f(x)$ consists of two triangular waves.
Specifically,$f(x) = x$ for $x \in [0, \pi]$,$f(x) = 2\pi - x$ for $x \in [\pi, 2\pi]$,$f(x) = x - 2\pi$ for $x \in [2\pi, 3\pi]$,and $f(x) = 4\pi - x$ for $x \in [3\pi, 4\pi]$.
We need to find the number of intersection points of $f(x)$ and the line $y = 1 - \frac{x}{10}$.
At $x = 0$,$f(0) = 0$ and $y = 1$.
At $x = \pi \approx 3.14$,$f(\pi) = \pi \approx 3.14$ and $y = 1 - 0.314 = 0.686$.
At $x = 2\pi \approx 6.28$,$f(2\pi) = 0$ and $y = 1 - 0.628 = 0.372$.
At $x = 3\pi \approx 9.42$,$f(3\pi) = \pi \approx 3.14$ and $y = 1 - 0.942 = 0.058$.
At $x = 4\pi \approx 12.56$,$f(4\pi) = 0$ and $y = 1 - 1.256 = -0.256$.
By observing the graph,the line $y = 1 - \frac{x}{10}$ intersects the graph of $f(x)$ at $3$ distinct points in the interval $[0, 4\pi]$.
Thus,the number of points is $3$.

(B) Given $f(x) = |x| + 1$ and $g(x) = x^2 + 1$.
For $x \leq 0$,$f(x) = -x + 1$ and $g(x) = x^2 + 1$. We define $h(x) = \max\{-x+1, x^2+1\}$.
Since $x^2+1 \geq -x+1$ for $x \in [-1, 0]$ (as $x^2+x \geq 0$),$h(x) = x^2+1$ for $x \in [-1, 0]$ and $h(x) = -x+1$ for $x < -1$.
For $x > 0$,$f(x) = x + 1$ and $g(x) = x^2 + 1$. We define $h(x) = \min\{x+1, x^2+1\}$.
Since $x^2+1 \leq x+1$ for $x \in [0, 1]$ (as $x^2-x \leq 0$),$h(x) = x^2+1$ for $x \in [0, 1]$ and $h(x) = x+1$ for $x > 1$.
Thus,$h(x) = \begin{cases} -x+1 & x < -1 \\ x^2+1 & -1 \leq x \leq 1 \\ x+1 & x > 1 \end{cases}$.
Checking differentiability:
At $x = -1$: $h(-1) = 2$. Left derivative is $-1$,right derivative is $2(-1) = -2$. Not differentiable.
At $x = 1$: $h(1) = 2$. Left derivative is $2(1) = 2$,right derivative is $1$. Not differentiable.
At $x = 0$: $h(0) = 1$. Left derivative is $2(0) = 0$,right derivative is $2(0) = 0$. Differentiable.
There are $2$ points of non-differentiability.

(B) Let $I = \int_0^1 4 x^3 \frac{d^2}{d x^2} (1-x^2)^5 d x$.
Using integration by parts,let $u = 4x^3$ and $dv = \frac{d^2}{dx^2}(1-x^2)^5 dx$.
Then $du = 12x^2 dx$ and $v = \frac{d}{dx}(1-x^2)^5 = 5(1-x^2)^4(-2x) = -10x(1-x^2)^4$.
$I = [4x^3 \cdot (-10x(1-x^2)^4)]_0^1 - \int_0^1 (-10x(1-x^2)^4) \cdot 12x^2 dx$.
The boundary term $[ -40x^4(1-x^2)^4 ]_0^1 = 0 - 0 = 0$.
So,$I = 120 \int_0^1 x^3(1-x^2)^4 dx$.
Let $t = 1-x^2$,then $dt = -2x dx$,so $x^2 = 1-t$ and $x dx = -\frac{1}{2} dt$.
$I = 120 \int_1^0 (1-t) t^4 (-\frac{1}{2} dt) = 60 \int_0^1 (t^4 - t^5) dt$.
$I = 60 [\frac{t^5}{5} - \frac{t^6}{6}]_0^1 = 60 (\frac{1}{5} - \frac{1}{6}) = 60 (\frac{6-5}{30}) = 60 (\frac{1}{30}) = 2$.

(D) Given $\vec{a}, \vec{b}, \vec{c}$ are unit vectors with angle $\frac{\pi}{3}$ between any pair,so $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
Given $p \vec{a} + q \vec{b} + r \vec{c} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c}$.
Taking the dot product with $\vec{a}, \vec{b}, \vec{c}$ respectively:
$1$) $\vec{a} \cdot (p \vec{a} + q \vec{b} + r \vec{c}) = \vec{a} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) \Rightarrow p + \frac{q}{2} + \frac{r}{2} = [\vec{a} \vec{b} \vec{c}]$.
$2$) $\vec{b} \cdot (p \vec{a} + q \vec{b} + r \vec{c}) = \vec{b} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) \Rightarrow \frac{p}{2} + q + \frac{r}{2} = 0$.
$3$) $\vec{c} \cdot (p \vec{a} + q \vec{b} + r \vec{c}) = \vec{c} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c}) \Rightarrow \frac{p}{2} + \frac{q}{2} + r = [\vec{a} \vec{b} \vec{c}]$.
From $(1)$ and $(3)$,$p + \frac{q}{2} + \frac{r}{2} = \frac{p}{2} + \frac{q}{2} + r \Rightarrow p = r$.
Substituting $r = p$ into $(2)$: $\frac{p}{2} + q + \frac{p}{2} = 0 \Rightarrow p + q = 0 \Rightarrow q = -p$.
Now,$\frac{p^2 + 2q^2 + r^2}{q^2} = \frac{p^2 + 2(-p)^2 + p^2}{(-p)^2} = \frac{p^2 + 2p^2 + p^2}{p^2} = \frac{4p^2}{p^2} = 4$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q(x) = \frac{x^4+2x}{\sqrt{1-x^2}}$.
Integrating factor ($I$.$F$.) $= e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = e^{\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$.
The general solution is $y \cdot \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + c = \int (x^4+2x) dx + c = \frac{x^5}{5} + x^2 + c$.
Given $f(0)=0$,we have $0 \cdot 1 = 0 + 0 + c$,so $c=0$.
Thus,$f(x) = \frac{x^5/5 + x^2}{\sqrt{1-x^2}}$.
We need to evaluate $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$. Since $f(x) = \frac{x^5/5}{\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}}$,the first part is an odd function,so its integral over the symmetric interval is $0$.
Thus,$I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$.
Let $x = \sin \theta$,then $dx = \cos \theta d\theta$. When $x=0, \theta=0$; when $x=\frac{\sqrt{3}}{2}, \theta=\frac{\pi}{3}$.
$I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 \theta d\theta = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2\theta) d\theta$.
$I = [\theta - \frac{1}{2} \sin 2\theta]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{1}{2} \sin(\frac{2\pi}{3}) = \frac{\pi}{3} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

(B) Given $F(x) = \int_0^{x^2} f(\sqrt{t}) dt$. Using the Leibniz rule for differentiation under the integral sign,we get $F'(x) = f(\sqrt{x^2}) \cdot \frac{d}{dx}(x^2) = f(x) \cdot 2x$.
Given $F'(x) = f'(x)$,we have $f'(x) = 2x f(x)$.
This is a first-order linear differential equation: $\frac{f'(x)}{f(x)} = 2x$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x^2 + C$.
Since $f(0) = 1$,we have $\ln(1) = 0^2 + C$,which implies $C = 0$.
Thus,$\ln(f(x)) = x^2$,which gives $f(x) = e^{x^2}$.
Now,$F(x) = \int_0^{x^2} e^{(\sqrt{t})^2} dt = \int_0^{x^2} e^t dt = [e^t]_0^{x^2} = e^{x^2} - e^0 = e^{x^2} - 1$.
Therefore,$F(2) = e^{2^2} - 1 = e^4 - 1$.

(A) Let $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 \operatorname{cosec} x)^{17} dx$.
Substitute $\tan(\frac{x}{2}) = e^u$,then $\frac{1}{2} \sec^2(\frac{x}{2}) dx = e^u du$,so $dx = \frac{2 e^u}{1+e^{2u}} du$.
Also,$\operatorname{cosec} x = \frac{1}{\sin x} = \frac{1+e^{2u}}{2e^u} = \frac{e^u + e^{-u}}{2}$.
When $x = \frac{\pi}{4}$,$e^u = \tan(\frac{\pi}{8}) = \sqrt{2}-1$,so $u = \ln(\sqrt{2}-1) = -\ln(\sqrt{2}+1)$.
When $x = \frac{\pi}{2}$,$e^u = \tan(\frac{\pi}{4}) = 1$,so $u = 0$.
Substituting these into the integral:
$I = \int_{-\ln(\sqrt{2}+1)}^{0} (e^u + e^{-u})^{17} \cdot \frac{2}{e^u + e^{-u}} du = \int_{-\ln(\sqrt{2}+1)}^{0} 2(e^u + e^{-u})^{16} du$.
Since $f(u) = (e^u + e^{-u})^{16}$ is an even function,$\int_{-a}^{0} f(u) du = \int_{0}^{a} f(u) du$.
Thus,$I = \int_{0}^{\ln(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$.

(A-D) $1.$ The sum $x_1 + x_2 + x_3$ is odd if all three are odd or two are even and one is odd.
Case $1$: $(O, O, O) \rightarrow P = \frac{2}{3} \times \frac{3}{5} \times \frac{4}{7} = \frac{24}{105}$
Case $2$: $(O, E, E) \rightarrow P = \frac{2}{3} \times \frac{2}{5} \times \frac{3}{7} = \frac{12}{105}$
Case $3$: $(E, O, E) \rightarrow P = \frac{1}{3} \times \frac{3}{5} \times \frac{3}{7} = \frac{9}{105}$
Case $4$: $(E, E, O) \rightarrow P = \frac{1}{3} \times \frac{2}{5} \times \frac{4}{7} = \frac{8}{105}$
Total probability $= \frac{24+12+9+8}{105} = \frac{53}{105}$.
$2.$ For $x_1, x_2, x_3$ to be in arithmetic progression,$2x_2 = x_1 + x_3$.
This implies $x_1 + x_3$ must be even,meaning $x_1$ and $x_3$ must have the same parity.
If $x_1, x_3$ are both odd: $(1,1), (1,3), (1,5), (1,7), (3,1), (3,3), (3,5), (3,7)$ (Total $8$ pairs).
If $x_1, x_3$ are both even: $(2,2), (2,4), (2,6)$ (Total $3$ pairs).
Total favorable outcomes $= 8 + 3 = 11$.
Total possible outcomes $= 3 \times 5 \times 7 = 105$.
Probability $= \frac{11}{105}$.

(C) $1.$ To find $g(a)$,we evaluate the integral $g(a) = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} t^{-a}(1-t)^{a-1} dt$.
For $a = \frac{1}{2}$,$g\left(\frac{1}{2}\right) = \int_0^1 t^{-1/2}(1-t)^{-1/2} dt = \int_0^1 \frac{dt}{\sqrt{t(1-t)}}$.
Using the substitution $t = \sin^2 \theta$,$dt = 2 \sin \theta \cos \theta d\theta$,the integral becomes $\int_0^{\pi/2} \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta} d\theta = \int_0^{\pi/2} 2 d\theta = \pi$.
Thus,$g\left(\frac{1}{2}\right) = \pi$.
$2.$ We observe that $g(1-a) = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} t^{-(1-a)}(1-t)^{(1-a)-1} dt = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} t^{a-1}(1-t)^{-a} dt$.
Using the property $\int_n^{1-n} f(t) dt = \int_n^{1-n} f(1-t) dt$,we get $g(1-a) = \lim_{n \rightarrow 0^{+}} \int_n^{1-n} (1-t)^{a-1} t^{-a} dt = g(a)$.
Since $g(1-a) = g(a)$,differentiating both sides with respect to $a$ using the chain rule gives $-g'(1-a) = g'(a)$.
At $a = \frac{1}{2}$,$-g'\left(\frac{1}{2}\right) = g'\left(\frac{1}{2}\right)$,which implies $2g'\left(\frac{1}{2}\right) = 0$,so $g'\left(\frac{1}{2}\right) = 0$.
Therefore,the correct pair is $(A, D)$.

(D) $(P)$ Let $f(x) = ax^2 + bx$,where $a, b \in \mathbb{W}$ (since $f(0)=0$).
$\int_0^1 (ax^2 + bx) dx = \frac{a}{3} + \frac{b}{2} = 1 \implies 2a + 3b = 6$.
Possible non-negative integer solutions $(a, b)$ are $(3, 0)$ and $(0, 2)$.
Thus,the number of such polynomials is $2$.
$(Q)$ $f(x) = \sqrt{2} \sin(x^2 + \frac{\pi}{4})$.
$f(x)$ is maximum when $x^2 + \frac{\pi}{4} = 2n\pi + \frac{\pi}{2} \implies x^2 = 2n\pi + \frac{\pi}{4}$.
For $n=0$,$x^2 = \frac{\pi}{4} \approx 0.785 \in [0, 13)$.
For $n=1$,$x^2 = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4} \approx 7.068 \in [0, 13)$.
For $n=2$,$x^2 = 4\pi + \frac{\pi}{4} = \frac{17\pi}{4} \approx 13.35 \notin [0, 13)$.
Since $x^2$ can take $2$ values for each $n$,and $x$ can be $\pm \sqrt{x^2}$,we check the interval $(-\sqrt{13}, \sqrt{13})$.
For $n=0$,$x = \pm \sqrt{\pi/4} = \pm \sqrt{\pi}/2$ ($2$ points).
For $n=1$,$x = \pm \sqrt{9\pi/4} = \pm 3\sqrt{\pi}/2$ ($2$ points).
Total points $= 4$. Wait,checking the options,$Q$ matches $3$ ($4$ points).
$(R)$ $\int_{-2}^2 \frac{3x^2}{1+e^x} dx = \int_0^2 3x^2 (\frac{1}{1+e^x} + \frac{1}{1+e^{-x}}) dx = \int_0^2 3x^2 (\frac{1+e^x}{1+e^x}) dx = \int_0^2 3x^2 dx = [x^3]_0^2 = 8$.
$(S)$ The numerator is $\int_{-1/2}^{1/2} \cos 2x \log(\frac{1+x}{1-x}) dx$. Since $\cos 2x$ is even and $\log(\frac{1+x}{1-x})$ is odd,the product is odd. Thus,the integral is $0$.

(C) $(P)$ Given $y(x) = \cos(3 \cos^{-1} x) = 4x^3 - 3x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 12x^2 - 3$ and $\frac{d^2y}{dx^2} = 24x$.
Substituting these into the expression: $(x^2-1)(24x) + x(12x^2-3) = 24x^3 - 24x + 12x^3 - 3x = 36x^3 - 27x = 9(4x^3 - 3x) = 9y$.
Thus,$\frac{1}{y} \{9y\} = 9$. So $P \to 4$.
$(Q)$ The magnitude of the cross product is $\sum |\vec{a}_k| |\vec{a}_{k+1}| \sin(\frac{2\pi}{n}) = (n-1) \lambda^2 \sin(\frac{2\pi}{n})$.
The magnitude of the dot product is $\sum |\vec{a}_k| |\vec{a}_{k+1}| \cos(\frac{2\pi}{n}) = (n-1) \lambda^2 \cos(\frac{2\pi}{n})$.
Equating them gives $\tan(\frac{2\pi}{n}) = 1$,so $\frac{2\pi}{n} = \frac{\pi}{4}$,which implies $n = 8$. So $Q \to 3$.
$(R)$ The normal to $\frac{x^2}{6} + \frac{y^2}{3} = 1$ at $(x_1, y_1)$ is $\frac{6x}{x_1} - \frac{3y}{y_1} = 3$. Given $P(h, 1)$ lies on the ellipse,$\frac{h^2}{6} + \frac{1}{3} = 1 \implies h^2 = 4 \implies h = 2$ (for positive $h$). The slope of the normal is $\frac{6/x_1}{3/y_1} = \frac{2y_1}{x_1}$. Since it is perpendicular to $x+y=8$ (slope $-1$),the normal slope is $1$. Thus $2y_1 = x_1$. Substituting into the ellipse equation: $\frac{(2y_1)^2}{6} + \frac{y_1^2}{3} = 1 \implies \frac{4y_1^2}{6} + \frac{2y_1^2}{6} = 1 \implies y_1^2 = 1 \implies y_1 = 1$. Then $x_1 = 2$. The normal equation is $\frac{6x}{2} - \frac{3y}{1} = 3 \implies 3x - 3y = 3 \implies x - y = 1$. Since $P(h, 1)$ is on the normal,$h - 1 = 1 \implies h = 2$. So $R \to 2$.
$(S)$ Using $\tan^{-1} A + \tan^{-1} B = \tan^{-1}(\frac{A+B}{1-AB})$,we get $\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{(2x+1)(4x+1)}} = \frac{2}{x^2} \implies \frac{6x+2}{8x^2+6x} = \frac{2}{x^2} \implies \frac{3x+1}{4x^2+3x} = \frac{2}{x^2} \implies 3x^3 + x^2 = 8x^2 + 6x \implies 3x^3 - 7x^2 - 6x = 0$. Since $x > 0$,$3x^2 - 7x - 6 = 0 \implies (3x+2)(x-3) = 0$. Only $x=3$ is a positive solution. Thus,there is $1$ positive solution. So $S \to 1$.

(D) $1$. Analyze $f_4(x)$:
$f_2(f_1(x)) = (f_1(x))^2 = \begin{cases} x^2 & x < 0 \\ e^{2x} & x \geq 0 \end{cases}$
$f_4(x) = \begin{cases} x^2 & x < 0 \\ e^{2x} - 1 & x \geq 0 \end{cases}$
At $x=0$,$f_4(0) = e^0 - 1 = 0$. $\lim_{x \to 0^-} f_4(x) = 0$. So $f_4$ is continuous. It is many-one (e.g.,$f_4(-1) = 1, f_4(\frac{1}{2}\ln 2) = 1$). Range is $[0, \infty)$,so it is onto. Thus $P \to 1$.
$2$. Analyze $f_3(x)$:
$f_3(x) = \begin{cases} \sin x & x < 0 \\ x & x \geq 0 \end{cases}$
At $x=0$,$f_3(0) = 0$. $\lim_{x \to 0^-} \sin x = 0$. Continuous. $f_3'(0^-) = \cos(0) = 1$,$f_3'(0^+) = 1$. Differentiable. It is many-one (e.g.,$f_3(-\pi) = 0, f_3(0) = 0$). Thus $Q \to 3$.
$3$. Analyze $f_2 \circ f_1(x)$:
$(f_2 \circ f_1)(x) = \begin{cases} x^2 & x < 0 \\ e^{2x} & x \geq 0 \end{cases}$
At $x=0$,$LHL = 0, RHL = 1$. Discontinuous. Thus $R \to 2$.
$4$. Analyze $f_2(x) = x^2$ on $[0, \infty)$:
It is strictly increasing,so one-one. Continuous. Thus $S \to 4$.
Matching: $P-1, Q-3, R-2, S-4$. Correct option is $(D)$.