IIT JEE 2020 Mathematics Question Paper with Answer and Solution

(D) Given $x^2+20x-2020=0$ has roots $a, b$. By Vieta's formulas,$a+b = -20$ and $ab = -2020$.
Given $x^2-20x+2020=0$ has roots $c, d$. By Vieta's formulas,$c+d = 20$ and $cd = 2020$.
We need to evaluate $E = ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$.
Expanding the expression: $E = a^2c - ac^2 + a^2d - ad^2 + b^2c - bc^2 + b^2d - bd^2$.
Grouping terms: $E = a^2(c+d) + b^2(c+d) - c^2(a+b) - d^2(a+b)$.
$E = (c+d)(a^2+b^2) - (a+b)(c^2+d^2)$.
Using $a^2+b^2 = (a+b)^2 - 2ab$ and $c^2+d^2 = (c+d)^2 - 2cd$:
$a^2+b^2 = (-20)^2 - 2(-2020) = 400 + 4040 = 4440$.
$c^2+d^2 = (20)^2 - 2(2020) = 400 - 4040 = -3640$.
Substituting these values: $E = (20)(4440) - (-20)(-3640)$.
$E = 88800 - 72800 = 16000$.

(A) The parabola is $y^2 = 4 \lambda x$. The end point of the latus rectum is $P(\lambda, 2 \lambda)$.
The slope of the tangent to the parabola at $P$ is given by differentiating $y^2 = 4 \lambda x$ with respect to $x$: $2y \frac{dy}{dx} = 4 \lambda \Rightarrow \frac{dy}{dx} = \frac{2 \lambda}{y}$.
At $P(\lambda, 2 \lambda)$,the slope $m_1 = \frac{2 \lambda}{2 \lambda} = 1$.
The slope of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $P$ is found by differentiating: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.
At $P(\lambda, 2 \lambda)$,the slope $m_2 = -\frac{b^2 \lambda}{a^2 (2 \lambda)} = -\frac{b^2}{2a^2}$.
Since the tangents are perpendicular,$m_1 \times m_2 = -1$.
$1 \times (-\frac{b^2}{2a^2}) = -1$ $\Rightarrow \frac{b^2}{2a^2} = 1$ $\Rightarrow \frac{b^2}{a^2} = 2$.
Since the ellipse passes through $P(\lambda, 2 \lambda)$,we have $\frac{\lambda^2}{a^2} + \frac{4 \lambda^2}{b^2} = 1$.
Substituting $b^2 = 2a^2$: $\frac{\lambda^2}{a^2} + \frac{4 \lambda^2}{2a^2} = 1$ $\Rightarrow \frac{\lambda^2}{a^2} + \frac{2 \lambda^2}{a^2} = 1$ $\Rightarrow \frac{3 \lambda^2}{a^2} = 1$ $\Rightarrow a^2 = 3 \lambda^2$.
Then $b^2 = 2a^2 = 6 \lambda^2$.
The eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2}$ (if $a < b$) or $e^2 = 1 - \frac{a^2}{b^2}$ (if $a > b$).
Here $\frac{b^2}{a^2} = 2$,so $a^2 < b^2$. Thus $e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$e = \frac{1}{\sqrt{2}}$.

(B) Given $|z^2+z+1|=1$.
We can write this as $|(z+\frac{1}{2})^2 + \frac{3}{4}| = 1$.
Using the triangle inequality $|a+b| \leq |a| + |b|$,we have $1 = |(z+\frac{1}{2})^2 + \frac{3}{4}| \leq |z+\frac{1}{2}|^2 + \frac{3}{4}$,which implies $|z+\frac{1}{2}|^2 \geq \frac{1}{4}$,so $|z+\frac{1}{2}| \geq \frac{1}{2}$. Thus,$(C)$ is true.
Also,$|z^2+z| = |(z^2+z+1)-1| \leq |z^2+z+1| + |-1| = 1+1 = 2$.
Since $|z^2+z| = |z||z+1| \leq 2$,for large $|z|$,$|z|^2 \approx |z^2+z| \leq 2$,so $|z| \leq 2$. Thus,$(B)$ is true.
Since the equation $|z^2+z+1|=1$ represents a curve in the complex plane,the set $S$ is infinite,so $(D)$ is false.
Therefore,the correct statements are $(B)$ and $(C)$.

(B) Given $\tan \frac{X}{2} + \tan \frac{Z}{2} = \frac{2y}{x+y+z}$.
Using the formula $\tan \frac{X}{2} = \frac{\Delta}{S(S-x)}$ where $S = \frac{x+y+z}{2}$ and $\Delta$ is the area of the triangle,we have:
$\frac{\Delta}{S(S-x)} + \frac{\Delta}{S(S-z)} = \frac{2y}{2S} = \frac{y}{S}$
$\frac{\Delta}{S} \left( \frac{S-z + S-x}{(S-x)(S-z)} \right) = \frac{y}{S}$
$\Delta \left( \frac{2S - (x+z)}{(S-x)(S-z)} \right) = y$
Since $2S = x+y+z$,$2S - (x+z) = y$.
$\Delta \left( \frac{y}{(S-x)(S-z)} \right) = y \implies \Delta = (S-x)(S-z)$
Squaring both sides: $\Delta^2 = (S-x)^2(S-z)^2$
$S(S-x)(S-y)(S-z) = (S-x)^2(S-z)^2$
$S(S-y) = (S-x)(S-z)$
Substituting $S = \frac{x+y+z}{2}$,we get $y^2 = x^2 + z^2$.
This implies $\angle Y = 90^\circ$.
Since $X+Y+Z = 180^\circ$,$X+Z = 90^\circ$,so $Y = X+Z$. Thus $(B)$ is true.
Also,for a right-angled triangle at $Y$,$\tan \frac{X}{2} = \sqrt{\frac{(S-y)(S-z)}{S(S-x)}} = \frac{x}{y+z}$. Thus $(C)$ is true.
Therefore,$(B)$ and $(C)$ are true.

(B) By the $AM$-$GM$ inequality,$\frac{3^{y_1}+3^{y_2}+3^{y_3}}{3} \geq \sqrt[3]{3^{y_1} \cdot 3^{y_2} \cdot 3^{y_3}} = \sqrt[3]{3^{y_1+y_2+y_3}}$.
Given $y_1+y_2+y_3=9$,we have $\frac{3^{y_1}+3^{y_2}+3^{y_3}}{3} \geq \sqrt[3]{3^9} = 3^3 = 27$.
So,$3^{y_1}+3^{y_2}+3^{y_3} \geq 81 = 3^4$.
Taking $\log_3$ on both sides,$\log_3(3^{y_1}+3^{y_2}+3^{y_3}) \geq 4$,so $m=4$.
For $M$,by the $AM$-$GM$ inequality,$\frac{x_1+x_2+x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3}$.
Given $x_1+x_2+x_3=9$,we have $3 \geq \sqrt[3]{x_1 x_2 x_3}$,so $x_1 x_2 x_3 \leq 27$.
Then $\log_3(x_1 x_2 x_3) = \log_3 x_1 + \log_3 x_2 + \log_3 x_3 \leq \log_3(27) = 3$,so $M=3$.
Finally,$\log_2(m^3) + \log_3(M^2) = \log_2(4^3) + \log_3(3^2) = \log_2(64) + 2 = 6 + 2 = 8$.

(A) The sum of the arithmetic progression is $S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(2c + (n-1)2) = n(c + n - 1)$.
Given $2(S_n) = b_1 + b_2 + \ldots + b_n$,where $b_n$ is a geometric progression with $b_1 = c$ and $r = 2$.
The sum of the geometric progression is $T_n = c(2^n - 1) / (2 - 1) = c(2^n - 1)$.
Substituting these into the given equation: $2n(c + n - 1) = c(2^n - 1)$.
$2nc + 2n^2 - 2n = c(2^n - 1)$.
$2n^2 - 2n = c(2^n - 1 - 2n)$.
$c = \frac{2n^2 - 2n}{2^n - 2n - 1}$.
For $c$ to be a positive integer,$2n^2 - 2n > 0$ and $2^n - 2n - 1 > 0$.
For $n=1$,$c = 0 / (2-2-1) = 0$ (not a positive integer).
For $n=2$,$c = (8-4) / (4-4-1) = -4$ (not positive).
For $n=3$,$c = (18-6) / (8-6-1) = 12 / 1 = 12$.
For $n=4$,$c = (32-8) / (16-8-1) = 24 / 7$ (not an integer).
For $n=5$,$c = (50-10) / (32-10-1) = 40 / 21$ (not an integer).
For $n=6$,$c = (72-12) / (64-12-1) = 60 / 51$ (not an integer).
For $n \geq 7$,$2^n - 2n - 1 > 2n^2 - 2n$,so $c < 1$.
Thus,the only positive integer value for $c$ is $12$,which corresponds to $n=3$.
Therefore,the number of possible values of $c$ is $1$.

(B) Let $L = \lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$.
We know that $(1-x)^{\frac{1}{x}} = e^{\frac{1}{x} \ln(1-x)}$.
Using the Taylor series expansion for $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$,we have:
$\frac{1}{x} \ln(1-x) = -1 - \frac{x}{2} - \frac{x^2}{3} - \dots$
Thus,$(1-x)^{\frac{1}{x}} = e^{-1 - \frac{x}{2} - \frac{x^2}{3} - \dots} = e^{-1} \cdot e^{-\frac{x}{2} - \frac{x^2}{3} - \dots}$.
Using $e^u = 1 + u + \frac{u^2}{2!} + \dots$,we get:
$(1-x)^{\frac{1}{x}} = e^{-1} \left( 1 + (-\frac{x}{2} - \frac{x^2}{3} - \dots) + \frac{(-\frac{x}{2} - \dots)^2}{2} + \dots \right) = e^{-1} \left( 1 - \frac{x}{2} - \frac{x^2}{3} + \frac{x^2}{8} + \dots \right) = e^{-1} \left( 1 - \frac{x}{2} - \frac{5x^2}{24} + \dots \right)$.
Substituting this into the limit:
$L = \lim_{x \rightarrow 0^{+}} \frac{e^{-1} (1 - \frac{x}{2} - \frac{5x^2}{24} + \dots) - e^{-1}}{x^a} = e^{-1} \lim_{x \rightarrow 0^{+}} \frac{-\frac{x}{2} - \frac{5x^2}{24} - \dots}{x^a}$.
For the limit to be a nonzero real number,the power of $x$ in the numerator must match the denominator $x^a$. Thus,$a = 1$.

(D) Let $z = x + iy$. The given equation is $z^4 - |z|^4 = 4iz^2$.
Since $|z|^2 = z\bar{z}$, we have $|z|^4 = (z\bar{z})^2 = z^2\bar{z}^2$.
Substituting this into the equation: $z^4 - z^2\bar{z}^2 = 4iz^2$.
This implies $z^2(z^2 - \bar{z}^2) = 4iz^2$.
Case $1$: $z^2 = 0 \Rightarrow z = 0$. However, $z = 0$ does not satisfy the condition $\operatorname{Re}(z_1) > 0$ or $\operatorname{Re}(z_2) < 0$.
Case $2$: $z^2 - \bar{z}^2 = 4i$.
Since $z = x + iy$, $z^2 = x^2 - y^2 + 2ixy$ and $\bar{z}^2 = x^2 - y^2 - 2ixy$.
Thus, $z^2 - \bar{z}^2 = (x^2 - y^2 + 2ixy) - (x^2 - y^2 - 2ixy) = 4ixy$.
Equating to $4i$, we get $4ixy = 4i$, which simplifies to $xy = 1$.
We need to minimize $|z_1 - z_2|^2$ where $z_1 = x_1 + iy_1$ with $x_1 > 0$ and $z_2 = x_2 + iy_2$ with $x_2 < 0$.
Since $xy = 1$, $y = 1/x$. Thus $z = x + i/x$.
$|z_1 - z_2|^2 = |(x_1 - x_2) + i(1/x_1 - 1/x_2)|^2 = (x_1 - x_2)^2 + (\frac{x_2 - x_1}{x_1x_2})^2 = (x_1 - x_2)^2 (1 + \frac{1}{x_1^2x_2^2})$.
Let $x_1 = a > 0$ and $x_2 = -b$ where $b > 0$. Then $x_1x_2 = -ab$.
$|z_1 - z_2|^2 = (a + b)^2 (1 + \frac{1}{a^2b^2})$.
By $AM-GM$ inequality, $(a + b)^2 \ge 4ab$. Also $1 + \frac{1}{a^2b^2} \ge 2\sqrt{1 \cdot \frac{1}{a^2b^2}} = \frac{2}{ab}$.
So $|z_1 - z_2|^2 \ge 4ab \cdot \frac{2}{ab} = 8$.

(B) Let the centre of the circumcircle of $\triangle OPQ$ be $C(h, k)$.
Since $O(0, 0)$ is a vertex of the triangle,the circumcircle passes through the origin.
The equation of the chord $PQ$ is $2x + 4y = 5$.
The equation of the chord of the circle $x^2 + y^2 = r^2$ with midpoint $(h, k)$ is $hx + ky = h^2 + k^2$.
Comparing $2x + 4y = 5$ with $hx + ky = r^2$ (since the chord $PQ$ is the polar of the centre $C$ with respect to the circle $x^2 + y^2 = r^2$ is not directly applicable,we use the property that the line $OC$ is perpendicular to $PQ$ and the midpoint of $PQ$ lies on $OC$):
The slope of $PQ$ is $m = -\frac{2}{4} = -\frac{1}{2}$.
Since $OC \perp PQ$,the slope of $OC$ is $2$.
The line $OC$ passes through $(0, 0)$,so its equation is $y = 2x$.
The centre $C(h, k)$ lies on $y = 2x$,so $k = 2h$.
Also,$C(h, k)$ lies on the given line $x + 2y = 4$.
Substituting $k = 2h$ into $x + 2y = 4$,we get $h + 2(2h) = 4 \Rightarrow 5h = 4 \Rightarrow h = \frac{4}{5}$.
Thus,$k = 2(\frac{4}{5}) = \frac{8}{5}$,so $C = (\frac{4}{5}, \frac{8}{5})$.
Since $C$ is the circumcentre of $\triangle OPQ$,$CO = CP = CQ = r_{circum}$.
$CO^2 = (\frac{4}{5})^2 + (\frac{8}{5})^2 = \frac{16+64}{25} = \frac{80}{25} = \frac{16}{5}$.
The distance from $C(\frac{4}{5}, \frac{8}{5})$ to the line $2x + 4y - 5 = 0$ is $d = \frac{|2(\frac{4}{5}) + 4(\frac{8}{5}) - 5|}{\sqrt{2^2 + 4^2}} = \frac{|\frac{8+32-25}{5}|}{\sqrt{20}} = \frac{15/5}{\sqrt{20}} = \frac{3}{\sqrt{20}}$.
In $\triangle CPQ$,$CP^2 = d^2 + (PQ/2)^2$. Let $M$ be the midpoint of $PQ$. $CM = d = \frac{3}{\sqrt{20}}$.
$PM^2 = r^2 - OM^2$. $OM$ is the distance from origin to $2x+4y=5$,$OM = \frac{|-5|}{\sqrt{20}} = \frac{5}{\sqrt{20}}$.
$PM^2 = r^2 - \frac{25}{20} = r^2 - \frac{5}{4}$.
$CP^2 = CM^2 + PM^2 \Rightarrow \frac{16}{5} = \frac{9}{20} + r^2 - \frac{5}{4} = r^2 + \frac{9-25}{20} = r^2 - \frac{16}{20} = r^2 - \frac{4}{5}$.
$r^2 = \frac{16}{5} + \frac{4}{5} = \frac{20}{5} = 4 \Rightarrow r = 2$.

(C) Let the given limit be $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$.
Using the identity $\sin 3x + \sin x = 2 \sin 2x \cos x$,the numerator becomes $8 \sqrt{2} \sin 2x \cos x = 16 \sqrt{2} \sin x \cos^2 x$.
For the denominator,we use $\cos \frac{5x}{2} - \cos \frac{3x}{2} = -2 \sin 2x \sin \frac{x}{2}$ and $\sqrt{2} + \sqrt{2} \cos 2x = \sqrt{2}(1 + \cos 2x) = 2 \sqrt{2} \cos^2 x$.
The denominator simplifies to $2 \sin 2x \sin \frac{3x}{2} - 2 \sin 2x \sin \frac{x}{2} - 2 \sqrt{2} \cos^2 x = 2 \sin 2x (\sin \frac{3x}{2} - \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x$.
Using $\sin \frac{3x}{2} - \sin \frac{x}{2} = 2 \cos x \sin \frac{x}{2}$,the denominator becomes $4 \sin x \cos x (2 \cos x \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x = 2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})$.
Thus,$L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{16 \sqrt{2} \sin x \cos^2 x}{2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})} = \lim_{x \rightarrow \frac{\pi}{2}} \frac{8 \sqrt{2} \sin x}{4 \sin x \sin \frac{x}{2} - \sqrt{2}}$.
Substituting $x = \frac{\pi}{2}$,we get $L = \frac{8 \sqrt{2} (1)}{4 (1) \sin \frac{\pi}{4} - \sqrt{2}} = \frac{8 \sqrt{2}}{4 \cdot \frac{1}{\sqrt{2}} - \sqrt{2}} = \frac{8 \sqrt{2}}{2 \sqrt{2} - \sqrt{2}} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8$.

(A, D) The normal at $P$ makes equal intercepts on the coordinate axes,so its slope is $-1$. Thus,the slope of the tangent at $P$ is $1$.
The tangent passes through $(1, 0)$ with slope $1$,so its equation is $y - 0 = 1(x - 1)$,or $x - y = 1$.
The equation of the tangent at $P(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. Comparing this with $x - y = 1$,we get $\frac{x_1}{a^2} = 1$ and $\frac{y_1}{b^2} = 1$,so $x_1 = a^2$ and $y_1 = b^2$.
Since $P(a^2, b^2)$ lies on the hyperbola,$\frac{(a^2)^2}{a^2} - \frac{(b^2)^2}{b^2} = 1$,which simplifies to $a^2 - b^2 = 1$.
The normal at $P(a^2, b^2)$ has slope $-1$,so its equation is $y - b^2 = -1(x - a^2)$,or $x + y = a^2 + b^2$.
The $x$-intercept of the normal is $x = a^2 + b^2$. The tangent $x - y = 1$ has $x$-intercept $x = 1$.
The triangle is formed by the tangent,the normal,and the $x$-axis. The base is the distance between the $x$-intercepts: $(a^2 + b^2) - 1 = (a^2 - 1) + b^2 = b^2 + b^2 = 2b^2$. The height is the $y$-coordinate of $P$,which is $b^2$.
Thus,$\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2b^2) \times b^2 = b^4$. So $(D)$ is true.
For eccentricity $e$,$e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{a^2 - 1}{a^2} = 2 - \frac{1}{a^2}$.
Since $a > 1$,$0 < \frac{1}{a^2} < 1$,so $1 < 2 - \frac{1}{a^2} < 2$,which means $1 < e^2 < 2$,so $1 < e < \sqrt{2}$. Thus $(A)$ is true.

(A) Given $f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}$.
Using the identity $\binom{n+i}{p} \binom{p+n}{p-i} = \binom{n+p}{p} \binom{n+i}{i}$,we have:
$f(m, n, p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+p}{p} \binom{n+i}{i} = \binom{n+p}{p} \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{i}$.
Using Vandermonde's Identity or combinatorial simplification,$\sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{i} = \binom{m+n+1}{p}$ is not directly applicable,but rather $\sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p-i}$ simplifies to $\binom{m+n+p}{p}$.
Actually,$f(m, n, p) = \binom{n+p}{p} \binom{m+n}{p}$.
Thus,$\frac{f(m, n, p)}{\binom{n+p}{p}} = \binom{m+n}{p}$.
Then $g(m, n) = \sum_{p=0}^{m+n} \binom{m+n}{p} = 2^{m+n}$.
Checking the options:
$(A)$ $g(m, n) = 2^{m+n}$ and $g(n, m) = 2^{n+m}$,so $g(m, n) = g(n, m)$ is $TRUE$.
$(B)$ $g(m, n+1) = 2^{m+n+1}$ and $g(m+1, n) = 2^{m+1+n}$,so $g(m, n+1) = g(m+1, n)$ is $TRUE$.
$(C)$ $g(2m, 2n) = 2^{2m+2n}$ and $2g(m, n) = 2 \cdot 2^{m+n} = 2^{m+n+1}$,so $g(2m, 2n) \neq 2g(m, n)$.
$(D)$ $g(2m, 2n) = 2^{2m+2n} = (2^{m+n})^2 = (g(m, n))^2$,so $g(2m, 2n) = (g(m, n))^2$ is $TRUE$.
Therefore,the correct statements are $A, B, D$.

(B) To select $r$ items from $n$ items such that no two are consecutive,the formula is given by $^{n-r+1}C_r$.
Here,$n = 15$ and $r = 4$.
Number of ways $= ^{15-4+1}C_4 = ^{12}C_4$.
Calculating the value:
$^{12}C_4 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.

(C) Let the number of persons in the four rooms be $n_1, n_2, n_3, n_4$. Since each room has at least $1$ and at most $2$ persons,and the total number of persons is $6$,we have $n_1 + n_2 + n_3 + n_4 = 6$ where $1 \le n_i \le 2$.
This implies two rooms must have $2$ persons and two rooms must have $1$ person (since $2+2+1+1 = 6$).
The number of ways to distribute $6$ distinct persons into these groups is given by the multinomial coefficient $\frac{6!}{2!2!1!1!}$.
Since the rooms are distinct,we must multiply by the number of ways to assign these group sizes to the $4$ rooms,which is $\frac{4!}{2!2!} = 6$.
Thus,the total number of ways is $\frac{6!}{2!2!1!1!} \times 6 = \frac{720}{4} \times 6 = 180 \times 6 = 1080$.

(B) Given the trajectory $y = \frac{x^2}{2}$.
At $t=0$,the particle is at $(0, 0)$ with speed $v = 1 \text{ m/s}$.
Differentiating $y = \frac{x^2}{2}$ with respect to $t$,we get $\frac{dy}{dt} = x \frac{dx}{dt}$,which means $v_y = x v_x$.
Differentiating again with respect to $t$,we get $a_y = v_x^2 + x a_x$.
$(A)$ If $a_x = 1 \text{ m/s}^2$ and the particle is at the origin $(x=0)$,then $a_y = v_x^2 + (0)(1) = v_x^2$. Since the speed is $1 \text{ m/s}$ and at the origin $v_y = x v_x = 0$,we have $v_x = 1 \text{ m/s}$. Thus,$a_y = 1^2 = 1 \text{ m/s}^2$. This is correct.
$(B)$ If $a_x = 0$,then $v_x$ is constant. Since $v_x(0) = 1 \text{ m/s}$,$v_x = 1 \text{ m/s}$ for all $t$. Then $a_y = v_x^2 + x a_x = 1^2 + x(0) = 1 \text{ m/s}^2$ for all $t$. This is correct.
$(C)$ At $t=0$,$x=0$. Since $v_y = x v_x$,we have $v_y = 0 \cdot v_x = 0$. Thus,the velocity vector is $\vec{v} = v_x \hat{i} + 0 \hat{j}$,which points in the $x$-direction. This is correct.
$(D)$ If $a_x = 0$,then $v_x = 1 \text{ m/s}$ and $a_x = 0$. From $a_y = v_x^2 + x a_x$,we get $a_y = 1^2 + 0 = 1 \text{ m/s}^2$. Since $a_y$ is constant,$v_y = a_y t = 1 \cdot t = t$. At $t=1 \text{ s}$,$v_y = 1 \text{ m/s}$. The angle $\theta$ with the $x$-axis is given by $\tan \theta = \frac{v_y}{v_x} = \frac{1}{1} = 1$,so $\theta = 45^{\circ}$. This is correct.
Therefore,all statements $(A), (B), (C), (D)$ are correct.

(C) Given $f(x) = |x|(x - \sin x)$.
Since $f(-x) = |-x|(-x - \sin(-x)) = |x|(-x + \sin x) = -|x|(x - \sin x) = -f(x)$,the function is an odd function.
For $x \geq 0$,$f(x) = x^2 - x \sin x$. For $x < 0$,$f(x) = -x^2 + x \sin x$.
As $x \rightarrow \infty$,$f(x) = x^2(1 - \frac{\sin x}{x}) \rightarrow \infty$. As $x \rightarrow -\infty$,$f(x) \rightarrow -\infty$. Since $f$ is continuous,the range is $R$,so $f$ is onto.
For $x > 0$,$f'(x) = 2x - \sin x - x \cos x = x(1 - \cos x) + (x - \sin x)$. Since $x > \sin x$ and $1 - \cos x \geq 0$ for $x > 0$,$f'(x) > 0$.
For $x < 0$,$f'(x) = -2x + \sin x + x \cos x = -[2x - \sin x - x \cos x] > 0$ (by symmetry of the odd function).
Since $f'(x) > 0$ for all $x \neq 0$ and $f$ is continuous,$f$ is strictly increasing on $R$.
Thus,$f$ is one-one.
Therefore,$f$ is both one-one and onto.

(A) Given $f(x) = e^{x-1} - e^{-|x-1|}$ and $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$.
For $x \leq 1$,$|x-1| = 1-x$,so $f(x) = e^{x-1} - e^{x-1} = 0$.
For $x \geq 1$,$|x-1| = x-1$,so $f(x) = e^{x-1} - e^{1-x}$.
To find the intersection point of $f(x)$ and $g(x)$ for $x \geq 1$:
$e^{x-1} - e^{1-x} = \frac{1}{2}(e^{x-1} + e^{1-x})$
$2e^{x-1} - 2e^{1-x} = e^{x-1} + e^{1-x}$
$e^{x-1} = 3e^{1-x} \Rightarrow e^{2(x-1)} = 3 \Rightarrow 2(x-1) = \ln 3 \Rightarrow x = 1 + \frac{1}{2}\ln 3 = 1 + \ln \sqrt{3}$.
The area $A$ in the first quadrant bounded by $y=f(x)$,$y=g(x)$,and $x=0$ is:
$A = \int_0^1 (g(x) - 0) dx + \int_1^{1+\ln \sqrt{3}} (g(x) - f(x)) dx$
$A = \int_0^1 \frac{1}{2}(e^{x-1} + e^{1-x}) dx + \int_1^{1+\ln \sqrt{3}} (\frac{1}{2}(e^{x-1} + e^{1-x}) - (e^{x-1} - e^{1-x})) dx$
$A = \frac{1}{2}[e^{x-1} - e^{1-x}]_0^1 + \int_1^{1+\ln \sqrt{3}} (\frac{3}{2}e^{1-x} - \frac{1}{2}e^{x-1}) dx$
$A = \frac{1}{2}[(e^0 - e^0) - (e^{-1} - e^1)] + [-\frac{3}{2}e^{1-x} - \frac{1}{2}e^{x-1}]_1^{1+\ln \sqrt{3}}$
$A = \frac{1}{2}(e - e^{-1}) + [(-\frac{3}{2}e^{-\ln \sqrt{3}} - \frac{1}{2}e^{\ln \sqrt{3}}) - (-\frac{3}{2} - \frac{1}{2})]$
$A = \frac{1}{2}(e - e^{-1}) + [-\frac{3}{2}(\frac{1}{\sqrt{3}}) - \frac{1}{2}(\sqrt{3}) + 2]$
$A = \frac{1}{2}(e - e^{-1}) + [-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 2] = 2 - \sqrt{3} + \frac{1}{2}(e - e^{-1})$.

(B) For $C_1$,$P(H) = \frac{2}{3}$. The number of heads $\alpha$ follows a binomial distribution $B(2, \frac{2}{3})$.
$P(\alpha = 0) = (\frac{1}{3})^2 = \frac{1}{9}$
$P(\alpha = 1) = 2 \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{9}$
$P(\alpha = 2) = (\frac{2}{3})^2 = \frac{4}{9}$
For $C_2$,$P(H) = \frac{1}{3}$. The number of heads $\beta$ follows a binomial distribution $B(2, \frac{1}{3})$.
$P(\beta = 0) = (\frac{2}{3})^2 = \frac{4}{9}$
$P(\beta = 1) = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}$
$P(\beta = 2) = (\frac{1}{3})^2 = \frac{1}{9}$
The roots of $x^2 - \alpha x + \beta = 0$ are real and equal if the discriminant $D = \alpha^2 - 4\beta = 0$,i.e.,$\alpha^2 = 4\beta$.
Possible pairs $(\alpha, \beta)$ satisfying this are $(0, 0)$ and $(2, 1)$.
Probability $= P(\alpha=0)P(\beta=0) + P(\alpha=2)P(\beta=1)$
$= (\frac{1}{9} \times \frac{4}{9}) + (\frac{4}{9} \times \frac{4}{9}) = \frac{4}{81} + \frac{16}{81} = \frac{20}{81}$.

(C) Let the rectangle be symmetric about the line $x = \frac{\pi}{4}$. Let the width of the rectangle be $2\alpha$,where the $x$-coordinates of the vertical sides are $\frac{\pi}{4} - \alpha$ and $\frac{\pi}{4} + \alpha$.
The height of the rectangle is $y = 2 \sin(2(\frac{\pi}{4} + \alpha)) = 2 \sin(\frac{\pi}{2} + 2\alpha) = 2 \cos(2\alpha)$.
The perimeter $P$ of the rectangle is given by $P = 2(\text{width} + \text{height}) = 2(2\alpha + 2 \cos(2\alpha)) = 4(\alpha + \cos(2\alpha))$.
To find the maximum perimeter,we differentiate $P$ with respect to $\alpha$:
$\frac{dP}{d\alpha} = 4(1 - 2 \sin(2\alpha)) = 0$.
This gives $\sin(2\alpha) = \frac{1}{2}$,so $2\alpha = \frac{\pi}{6}$ (since $0 < 2\alpha < \frac{\pi}{2}$),which means $\alpha = \frac{\pi}{12}$.
Checking the second derivative: $\frac{d^2P}{d\alpha^2} = -8 \cos(2\alpha)$. At $\alpha = \frac{\pi}{12}$,$\frac{d^2P}{d\alpha^2} = -8 \cos(\frac{\pi}{6}) = -4\sqrt{3} < 0$,so the perimeter is maximum at $\alpha = \frac{\pi}{12}$.
The area of this rectangle is $\text{Area} = \text{width} \times \text{height} = (2\alpha) \times (2 \cos(2\alpha)) = 2(\frac{\pi}{12}) \times 2 \cos(\frac{\pi}{6}) = \frac{\pi}{6} \times 2 \times \frac{\sqrt{3}}{2} = \frac{\pi}{2\sqrt{3}}$.

(D) Given $f(x) = x - x^2 + (x - 1) \sin x = -(x - 1)^2 + (x - 1) \sin x = (x - 1) [-(x - 1) + \sin x]$.
Note that $f(1) = 0$.
Also,$f'(x) = 1 - 2x + \sin x + (x - 1) \cos x$. Thus,$f'(1) = 1 - 2 + \sin 1 + 0 = \sin 1 - 1$.
Let $h(x) = (fg)(x) = f(x)g(x)$.
For differentiability at $x=1$,we check $h'(1) = \lim_{k \to 0} \frac{f(1+k)g(1+k) - f(1)g(1)}{k} = \lim_{k \to 0} \frac{f(1+k)g(1+k)}{k}$.
Since $f(1+k) = k(-k + \sin(1+k))$,the limit becomes $\lim_{k \to 0} \frac{k(-k + \sin(1+k))g(1+k)}{k} = \lim_{k \to 0} (-k + \sin(1+k))g(1+k) = \sin(1) \cdot g(1)$.
If $g$ is continuous at $x=1$,$g(1+k) \to g(1)$,so the limit exists. Thus,$(A)$ is true.
If $g$ is differentiable at $x=1$,it is continuous,so $(C)$ is true.
If $fg$ is differentiable,it does not imply $g$ is continuous or differentiable (e.g.,if $f(x)=0$ at $x=1$,$fg$ could be differentiable even if $g$ is not). Thus,$(B)$ and $(D)$ are false.
Therefore,the correct statements are $(A)$ and $(C)$.

(A) Given $M^{-1} = \operatorname{adj}(\operatorname{adj} M)$.
We know that for a $3 \times 3$ matrix $M$,$\operatorname{adj}(\operatorname{adj} M) = (\operatorname{det} M)^{3-2} M = (\operatorname{det} M) M$.
Thus,$M^{-1} = (\operatorname{det} M) M$.
Multiplying both sides by $M$,we get $M^{-1} M = (\operatorname{det} M) M^2$,which implies $I = (\operatorname{det} M) M^2$.
Taking the determinant on both sides,$\operatorname{det}(I) = \operatorname{det}((\operatorname{det} M) M^2)$.
$1 = (\operatorname{det} M)^3 \cdot (\operatorname{det} M)^2 = (\operatorname{det} M)^5$.
Since $M$ has real entries,$\operatorname{det} M = 1$.
Substituting $\operatorname{det} M = 1$ into $I = (\operatorname{det} M) M^2$,we get $I = M^2$.
Since $M^2 = I$,then $(\operatorname{adj} M)^2 = \operatorname{adj}(M^2) = \operatorname{adj}(I) = I$.
Thus,statements $B, C,$ and $D$ are always true.

(D) The point of intersection of $L_1$ and $L_2$ is $(1, 0, 1)$.
Since line $L$ passes through $(1, 0, 1)$,we have $\frac{1-\alpha}{l} = \frac{0-1}{m} = \frac{1-\gamma}{-2}$,which implies $\frac{1-\alpha}{l} = -\frac{1}{m} = \frac{1-\gamma}{-2} = k$ (say). Thus,$1-\alpha = kl$,$m = 1/k$,and $1-\gamma = -2k$.
Direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{v_2} = -3\hat{i} - \hat{j} + \hat{k}$.
The unit vectors are $\hat{u_1} = \frac{\vec{v_1}}{\sqrt{11}}$ and $\hat{u_2} = \frac{\vec{v_2}}{\sqrt{11}}$.
The direction of the angle bisector is $\vec{v} = \hat{u_1} + \hat{u_2} = \frac{1}{\sqrt{11}} ((\hat{i} - \hat{j} + 3\hat{k}) + (-3\hat{i} - \hat{j} + \hat{k})) = \frac{1}{\sqrt{11}} (-2\hat{i} - 2\hat{j} + 4\hat{k}) \propto \hat{i} + \hat{j} - 2\hat{k}$.
Comparing with the direction ratios $(l, m, -2)$ of $L$,we get $\frac{l}{1} = \frac{m}{1} = \frac{-2}{-2} = 1$. Thus,$l=1$ and $m=1$.
Then $l+m = 1+1 = 2$,so $(B)$ is true.
Using $k=1$ in the intersection equations: $1-\alpha = 1 \Rightarrow \alpha=0$ and $1-\gamma = -2 \Rightarrow \gamma=3$.
Then $\alpha-\gamma = 0-3 = -3$. Checking the options,we re-evaluate the bisector direction. The acute angle bisector direction is $\vec{v_1} - \vec{v_2} = 4\hat{i} + 2\hat{k} \propto 2\hat{i} + \hat{k}$ or $\vec{v_1} + \vec{v_2} = -2\hat{i} - 2\hat{j} + 4\hat{k} \propto \hat{i} + \hat{j} - 2\hat{k}$.
Given the denominator of $L$ is $-2$,we use $\vec{v} = \hat{i} + \hat{j} - 2\hat{k}$,so $l=1, m=1$. $\alpha=0, \gamma=3$. $\alpha-\gamma = -3$. Wait,if we use the other bisector,$l=-1, m=-1$,then $l+m=-2$. Re-checking the intersection: $1-\alpha = l/(-1) = -l$,$1-\gamma = -2/(-1) = 2$. $\alpha = 1+l, \gamma = -1$. $\alpha-\gamma = 2+l$. With $l=1, \alpha=2, \gamma=-1, \alpha-\gamma=3$. Thus $(A)$ and $(B)$ are true.

(C) Using the Taylor series expansion,$\cos x \geq 1 - \frac{x^2}{2}$ for $x \in [0, 1]$.
$\int_0^1 x \cos x \, dx \geq \int_0^1 x(1 - \frac{x^2}{2}) \, dx = [\frac{x^2}{2} - \frac{x^4}{8}]_0^1 = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$. Thus,$(A)$ is $TRUE$.
$(B)$ Using $\sin x \geq x - \frac{x^3}{6}$,we have:
$\int_0^1 x \sin x \, dx \geq \int_0^1 x(x - \frac{x^3}{6}) \, dx = [\frac{x^3}{3} - \frac{x^5}{30}]_0^1 = \frac{1}{3} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10}$. Thus,$(B)$ is $TRUE$.
$(C)$ Since $\cos x < 1$ for $x \in (0, 1]$,$\int_0^1 x^2 \cos x \, dx < \int_0^1 x^2 \, dx = \frac{1}{3}$. Since $\frac{1}{3} < \frac{1}{2}$,$(C)$ is $FALSE$.
$(D)$ Using $\sin x \geq x - \frac{x^3}{6}$,we have:
$\int_0^1 x^2 \sin x \, dx \geq \int_0^1 x^2(x - \frac{x^3}{6}) \, dx = [\frac{x^4}{4} - \frac{x^6}{36}]_0^1 = \frac{1}{4} - \frac{1}{36} = \frac{9-1}{36} = \frac{8}{36} = \frac{2}{9}$. Thus,$(D)$ is $TRUE$.
Therefore,the correct options are $(A), (B), (D)$.

(B) Let $\theta = \pi x - \frac{\pi}{4}$. Since $x \in [0, 2]$,$\theta \in [-\frac{\pi}{4}, \frac{7\pi}{4}]$.
Then $2\pi x = 2\theta + \frac{\pi}{2}$,so $\sin(2\pi x) = \cos(2\theta)$.
Also $3\pi x + \frac{\pi}{4} = 3(\theta + \frac{\pi}{4}) + \frac{\pi}{4} = 3\theta + \pi$,so $\sin(3\pi x + \frac{\pi}{4}) = \sin(3\theta + \pi) = -\sin(3\theta)$.
The inequality $f(x) \geq 0$ becomes $(3 - \cos(2\theta)) \sin \theta - (-\sin(3\theta)) \geq 0$.
$(3 - (1 - 2\sin^2 \theta)) \sin \theta + \sin(3\theta) \geq 0$.
$(2 + 2\sin^2 \theta) \sin \theta + (3\sin \theta - 4\sin^3 \theta) \geq 0$.
$2\sin \theta + 2\sin^3 \theta + 3\sin \theta - 4\sin^3 \theta \geq 0$.
$5\sin \theta - 2\sin^3 \theta \geq 0 \Rightarrow \sin \theta (5 - 2\sin^2 \theta) \geq 0$.
Since $5 - 2\sin^2 \theta > 0$ for all $\theta$,we must have $\sin \theta \geq 0$.
Thus,$\theta \in [0, \pi]$.
$0 \leq \pi x - \frac{\pi}{4} \leq \pi$ $\Rightarrow \frac{\pi}{4} \leq \pi x \leq \frac{5\pi}{4}$ $\Rightarrow x \in [\frac{1}{4}, \frac{5}{4}]$.
Therefore,$\alpha = \frac{1}{4}$ and $\beta = \frac{5}{4}$,so $\beta - \alpha = 1$.

(C) In triangle $PQR$,we have $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.
Since $\vec{a} = \vec{QR}$,$\vec{b} = \vec{RP}$,and $\vec{c} = \vec{PQ}$,we have $\vec{c} = -(\vec{a} + \vec{b})$.
Substituting $\vec{c}$ into the given expression:
$\frac{\vec{a} \cdot (-(\vec{a} + \vec{b}) - \vec{b})}{(-(\vec{a} + \vec{b})) \cdot (\vec{a} - \vec{b})} = \frac{|\vec{a}|}{|\vec{a}| + |\vec{b}|}$
$\frac{\vec{a} \cdot (-\vec{a} - 2\vec{b})}{-(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})} = \frac{3}{3 + 4}$
$\frac{-|\vec{a}|^2 - 2(\vec{a} \cdot \vec{b})}{-(\vec{a}^2 - \vec{b}^2)} = \frac{3}{7}$
Given $|\vec{a}| = 3$ and $|\vec{b}| = 4$,we have $|\vec{a}|^2 = 9$ and $|\vec{b}|^2 = 16$.
$\frac{-9 - 2(\vec{a} \cdot \vec{b})}{-(9 - 16)} = \frac{3}{7}$
$\frac{-9 - 2(\vec{a} \cdot \vec{b})}{7} = \frac{3}{7}$
$-9 - 2(\vec{a} \cdot \vec{b}) = 3$
$-2(\vec{a} \cdot \vec{b}) = 12$
$\vec{a} \cdot \vec{b} = -6$
Now,using the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$:
$|\vec{a} \times \vec{b}|^2 = (9)(16) - (-6)^2$
$|\vec{a} \times \vec{b}|^2 = 144 - 36 = 108$.

(A) Let $f(x) = (x^2-1)^2 h(x)$,where $h(x) = a_0+a_1x+a_2x^2+a_3x^3$.
Since $(x^2-1)^2 = (x-1)^2(x+1)^2$,$f(x)$ has roots at $x=1$ and $x=-1$ with multiplicity at least $2$.
Thus,$f(1)=0, f(-1)=0$ and $f'(1)=0, f'(-1)=0$.
By Rolle's Theorem,there exists $\alpha \in (-1, 1)$ such that $f'(\alpha)=0$.
Since $f'(x)$ has roots at $-1, \alpha, 1$,we have $m_{f'} \ge 3$.
For $f(x) = (x^2-1)^2$,the roots are $1, -1$,so $m_f = 2$.
Then $m_f + m_{f'} = 2 + 3 = 5$.
Thus,the minimum value is $5$.

(A) Let $X$ be the number of successful hits,where $X \sim B(n, p)$ with $p = 0.75 = \frac{3}{4}$ and $q = 1 - p = 0.25 = \frac{1}{4}$.
We require at least $3$ successful hits to destroy the target,so $P(X \geq 3) \geq 0.95$.
This is equivalent to $1 - P(X < 3) \geq 0.95$,or $P(X=0) + P(X=1) + P(X=2) \leq 0.05$.
The probability mass function is $P(X=r) = {}^{n}C_{r} (\frac{3}{4})^r (\frac{1}{4})^{n-r}$.
Substituting the values: ${}^{n}C_{0} (\frac{1}{4})^n + {}^{n}C_{1} (\frac{3}{4}) (\frac{1}{4})^{n-1} + {}^{n}C_{2} (\frac{3}{4})^2 (\frac{1}{4})^{n-2} \leq 0.05$.
$\frac{1}{4^n} [1 + 3n + \frac{9n(n-1)}{2}] \leq 0.05$.
$1 + 3n + 4.5n^2 - 4.5n \leq 0.05 \times 4^n$.
$4.5n^2 - 1.5n + 1 \leq 0.05 \times 4^n$.
For $n=5$: $4.5(25) - 1.5(5) + 1 = 112.5 - 7.5 + 1 = 106 \leq 0.05(1024) = 51.2$ (False).
For $n=6$: $4.5(36) - 1.5(6) + 1 = 162 - 9 + 1 = 154 \leq 0.05(4096) = 204.8$ (True).
Thus,the minimum number of missiles required is $6$.

(C) Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The characteristic equation of $A$ is given by $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$.
Given $\text{tr}(A) = a+d = 3$ and let $\Delta = \det(A) = ad-bc$.
So,the characteristic equation is $\lambda^2 - 3\lambda + \Delta = 0$.
By the Cayley-Hamilton theorem,$A^2 - 3A + \Delta I = 0$,which implies $A^2 = 3A - \Delta I$.
Multiplying by $A$,we get $A^3 = 3A^2 - \Delta A$.
Substituting $A^2 = 3A - \Delta I$ into the equation:
$A^3 = 3(3A - \Delta I) - \Delta A = 9A - 3\Delta I - \Delta A = (9 - \Delta)A - 3\Delta I$.
Taking the trace on both sides:
$\text{tr}(A^3) = (9 - \Delta)\text{tr}(A) - 3\Delta \text{tr}(I)$.
Since $\text{tr}(A^3) = -18$,$\text{tr}(A) = 3$,and $\text{tr}(I) = 2$ (for a $2 \times 2$ matrix):
$-18 = (9 - \Delta)(3) - 3\Delta(2)$.
$-18 = 27 - 3\Delta - 6\Delta$.
$-18 = 27 - 9\Delta$.
$9\Delta = 27 + 18 = 45$.
$\Delta = 5$.
Thus,the determinant of $A$ is $5$.

(B) Given $f(x) = |2x-1| + |2x+1|$ and $g(x) = x - [x] = \{x\}$.
For $x \in (-1, 1)$,the fractional part function $g(x) = \{x\}$ is defined as:
$g(x) = x+1$ for $x \in (-1, 0)$ and $g(x) = x$ for $x \in [0, 1)$.
Thus,the composite function $h(x) = f(g(x)) = |2\{x\}-1| + |2\{x\}+1|$.
Since $\{x\} \in [0, 1)$,we have $2\{x\}+1 \geq 1$,so $|2\{x\}+1| = 2\{x\}+1$.
Then $h(x) = |2\{x\}-1| + 2\{x\} + 1$.
If $0 \leq \{x\} \leq 1/2$,then $h(x) = -(2\{x\}-1) + 2\{x\} + 1 = 2$.
If $1/2 < \{x\} < 1$,then $h(x) = (2\{x\}-1) + 2\{x\} + 1 = 4\{x\}$.
Analyzing the interval $(-1, 1)$:
For $x \in (-1, 0)$,$\{x\} = x+1$. So $h(x) = 2$ if $x+1 \leq 1/2$ (i.e.,$x \leq -1/2$) and $h(x) = 4(x+1)$ if $x > -1/2$.
For $x \in [0, 1)$,$\{x\} = x$. So $h(x) = 2$ if $x \leq 1/2$ and $h(x) = 4x$ if $x > 1/2$.
Discontinuity: The function $h(x)$ is discontinuous at $x=0$ because $\lim_{x \to 0^-} h(x) = 4(0+1) = 4$ and $h(0) = 2$. Thus,$c=1$.
Non-differentiability: The function is non-differentiable at $x = -1/2$ (corner point),$x = 0$ (discontinuity),and $x = 1/2$ (corner point). Thus,$d=3$.
Therefore,$c+d = 1+3 = 4$.

(D) Given the differential equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$.
Separating variables,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{b^2+x^2}$.
Integrating both sides with respect to $x$,we have $\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{b^2+x^2} dx$.
This yields $\ln|f(x)| = \frac{1}{b} \tan^{-1}\left(\frac{x}{b}\right) + C$.
Using the initial condition $f(0)=1$,we get $\ln(1) = \frac{1}{b} \tan^{-1}(0) + C$,which implies $C=0$.
Thus,$f(x) = e^{\frac{1}{b} \tan^{-1}(\frac{x}{b})}$.
Now,consider $f(x)f(-x) = e^{\frac{1}{b} \tan^{-1}(\frac{x}{b})} \cdot e^{\frac{1}{b} \tan^{-1}(\frac{-x}{b})} = e^{\frac{1}{b} (\tan^{-1}(\frac{x}{b}) - \tan^{-1}(\frac{x}{b}))} = e^0 = 1$. So,$(C)$ is true.
For $b>0$,$f^{\prime}(x) = \frac{f(x)}{b^2+x^2} > 0$ since $f(x) = e^{\dots} > 0$ and $b^2+x^2 > 0$. Thus,$f$ is an increasing function. So,$(A)$ is true.
Therefore,the correct statements are $(A)$ and $(C)$.

(B) Given $f(x+y) = f(x) + f(y) + f(x)f(y)$. Adding $1$ to both sides gives $1 + f(x+y) = (1 + f(x))(1 + f(y))$.
Let $h(x) = 1 + f(x)$. Then $h(x+y) = h(x)h(y)$.
Since $f(x) = xg(x)$ and $\lim_{x \to 0} g(x) = 1$,we have $\lim_{x \to 0} f(x) = 0$,so $h(0) = 1 + f(0) = 1$.
For $h(x+y) = h(x)h(y)$,the solution is $h(x) = e^{cx}$.
Thus $1 + f(x) = e^{cx}$,or $f(x) = e^{cx} - 1$.
Given $f(x) = xg(x)$,we have $g(x) = \frac{e^{cx}-1}{x}$.
$\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{e^{cx}-1}{x} = c$.
Since $\lim_{x \to 0} g(x) = 1$,we have $c = 1$.
Therefore,$f(x) = e^x - 1$.
$f'(x) = e^x$,which is defined for all $x \in R$,so $(A)$ is $TRUE$.
$f'(0) = e^0 = 1$,so $(D)$ is $TRUE$.
$f'(1) = e^1 = e \neq 1$,so $(C)$ is $FALSE$.
For $g(x) = \frac{e^x-1}{x}$ for $x \neq 0$ and $g(0)=1$,$g$ is differentiable at all $x \neq 0$. At $x=0$,$g'(0) = \lim_{h \to 0} \frac{\frac{e^h-1}{h}-1}{h} = \lim_{h \to 0} \frac{e^h-1-h}{h^2} = \frac{1}{2}$. Thus $g$ is differentiable everywhere,so $(B)$ is $TRUE$.

(C) Let the points be $P(1, 0, -1)$ and $Q(3, 2, -1)$. The mirror image of $P$ with respect to the plane is $Q$.
The midpoint $R$ of the line segment $PQ$ lies on the plane.
$R = \left( \frac{1+3}{2}, \frac{0+2}{2}, \frac{-1-1}{2} \right) = (2, 1, -1)$.
Since $R$ lies on the plane $\alpha x + \beta y + \gamma z = \delta$,we have:
$2\alpha + \beta - \gamma = \delta$ --- $(1)$
The normal vector to the plane is $\vec{n} = (\alpha, \beta, \gamma)$.
The vector $\vec{PQ} = (3-1, 2-0, -1-(-1)) = (2, 2, 0)$ is parallel to the normal vector $\vec{n}$.
Thus,$\frac{\alpha}{2} = \frac{\beta}{2} = \frac{\gamma}{0} = k$ (where $k \neq 0$).
This gives $\alpha = 2k$,$\beta = 2k$,and $\gamma = 0$.
Given $\alpha + \gamma = 1$,we have $2k + 0 = 1$,so $k = \frac{1}{2}$.
Therefore,$\alpha = 1$,$\beta = 1$,and $\gamma = 0$.
Substituting these into $(1)$:
$2(1) + 1(1) - 0 = \delta \implies \delta = 3$.
Now,check the options:
$(A)$ $\alpha + \beta = 1 + 1 = 2$ (True)
$(B)$ $\delta - \gamma = 3 - 0 = 3$ (True)
$(C)$ $\delta + \beta = 3 + 1 = 4$ (True)
$(D)$ $\alpha + \beta + \gamma = 1 + 1 + 0 = 2 \neq \delta$ (False)
Thus,statements $A, B, C$ are true.

(D) The projection vector of $\vec{w}$ along $\vec{PQ}$ is $\vec{u} = \left( \frac{\vec{w} \cdot \vec{PQ}}{|\vec{PQ}|^2} \right) \vec{PQ}$.
The magnitude is $|\vec{u}| = \frac{|\vec{w} \cdot \vec{PQ}|}{|\vec{PQ}|} = \frac{|(i+j) \cdot (ai+bj)|}{\sqrt{a^2+b^2}} = \frac{a+b}{\sqrt{a^2+b^2}}$.
Similarly,the magnitude of the projection vector $\vec{v}$ along $\vec{PS}$ is $|\vec{v}| = \frac{|\vec{w} \cdot \vec{PS}|}{|\vec{PS}|} = \frac{|(i+j) \cdot (ai-bj)|}{\sqrt{a^2+b^2}} = \frac{|a-b|}{\sqrt{a^2+b^2}}$.
Given $|\vec{u}| + |\vec{v}| = |\vec{w}| = \sqrt{1^2+1^2} = \sqrt{2}$,we have $\frac{a+b + |a-b|}{\sqrt{a^2+b^2}} = \sqrt{2}$.
If $a \ge b$,then $\frac{2a}{\sqrt{a^2+b^2}} = \sqrt{2} \implies 4a^2 = 2(a^2+b^2) \implies 2a^2 = 2b^2 \implies a=b$.
If $b > a$,then $\frac{2b}{\sqrt{a^2+b^2}} = \sqrt{2} \implies a=b$.
Thus,$a=b$. The area of the parallelogram is $|\vec{PQ} \times \vec{PS}| = |(ai+bj) \times (ai-bj)| = |(-abk - abk)| = |-2abk| = 2ab = 8$.
Since $a=b$,$2a^2 = 8 \implies a^2 = 4 \implies a=2$ (as $a>0$). Thus $a=2, b=2$.
$(A)$ $a+b = 2+2 = 4$ (True).
$(B)$ $a-b = 2-2 = 0 \neq 2$ (False).
$(C)$ The diagonal $\vec{PR} = \vec{PQ} + \vec{PS} = (ai+bj) + (ai-bj) = 2ai = 4i$. Length is $|4i| = 4$ (True).
$(D)$ $\vec{PQ} = 2i+2j$ and $\vec{PS} = 2i-2j$. The angle bisector direction is $\frac{\vec{PQ}}{|\vec{PQ}|} + \frac{\vec{PS}}{|\vec{PS}|} = \frac{2i+2j}{2\sqrt{2}} + \frac{2i-2j}{2\sqrt{2}} = \frac{4i}{2\sqrt{2}} = \sqrt{2}i$. This is along the $x$-axis,while $\vec{w} = i+j$ is at $45^\circ$. (False).

(D) Let $S$ be the sum of the numbers on the two dice. The possible values of $S$ range from $2$ to $12$.
The set of prime sums is $P = \{2, 3, 5, 7, 11\}$. The number of outcomes for these sums are: $P(2)=1, P(3)=2, P(5)=4, P(7)=6, P(11)=2$. Total outcomes for prime = $1+2+4+6+2 = 15$. So,$P(\text{Prime}) = \frac{15}{36}$.
The set of perfect square sums is $Q = \{4, 9\}$. The number of outcomes for these sums are: $P(4)=3, P(9)=4$. Total outcomes for perfect square = $3+4 = 7$. So,$P(\text{Perfect Square}) = \frac{7}{36}$.
The probability of getting neither a prime nor a perfect square is $1 - (\frac{15}{36} + \frac{7}{36}) = 1 - \frac{22}{36} = \frac{14}{36}$.
Let $E$ be the event that a perfect square occurs before a prime. The probability $P(E) = \frac{7/36}{1 - 14/36} = \frac{7}{22}$.
We are given that a perfect square occurs before a prime. We want the probability that this perfect square is odd. The odd perfect square in our set is $9$,which has $4$ outcomes. The even perfect square is $4$,which has $3$ outcomes.
Given the condition,the probability that the perfect square is $9$ (odd) is $\frac{P(9)}{P(4) + P(9)} = \frac{4}{3+4} = \frac{4}{7}$.
Thus,$p = \frac{4}{7}$.
Therefore,$14p = 14 \times \frac{4}{7} = 8$.

(A) Consider the property $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2}$.
$= \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x + 2} = \frac{4^x}{4^x+2} + \frac{4}{4 + 2 \cdot 4^x} = \frac{4^x}{4^x+2} + \frac{2}{2 + 4^x} = \frac{4^x+2}{4^x+2} = 1$.
Let $S = f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + \dots + f\left(\frac{39}{40}\right)$.
Pairing terms $f\left(\frac{k}{40}\right) + f\left(1 - \frac{k}{40}\right) = f\left(\frac{k}{40}\right) + f\left(\frac{40-k}{40}\right) = 1$.
There are $39$ terms in the sum. The middle term is $f\left(\frac{20}{40}\right) = f\left(\frac{1}{2}\right)$.
There are $19$ pairs that sum to $1$,plus the middle term $f\left(\frac{1}{2}\right)$.
So,$S = 19 + f\left(\frac{1}{2}\right)$.
Therefore,$S - f\left(\frac{1}{2}\right) = 19 + f\left(\frac{1}{2}\right) - f\left(\frac{1}{2}\right) = 19$.

(A) Given $F(x) = \int_0^x f(t) dt$,by the Fundamental Theorem of Calculus,$F'(x) = f(x)$ and $F(0) = 0$.
We are given the integral equation:
$\int_0^\pi (f'(x) + F(x)) \cos x dx = 2$
Split the integral into two parts:
$\int_0^\pi f'(x) \cos x dx + \int_0^\pi F(x) \cos x dx = 2$
Apply integration by parts to the first integral $I_1 = \int_0^\pi f'(x) \cos x dx$:
$I_1 = [f(x) \cos x]_0^\pi - \int_0^\pi f(x) (-\sin x) dx$
$I_1 = f(\pi) \cos(\pi) - f(0) \cos(0) + \int_0^\pi f(x) \sin x dx$
Since $f(\pi) = -6$,$\cos(\pi) = -1$,and $\cos(0) = 1$:
$I_1 = (-6)(-1) - f(0)(1) + \int_0^\pi f(x) \sin x dx = 6 - f(0) + \int_0^\pi f(x) \sin x dx$
Now,consider the second integral $I_2 = \int_0^\pi F(x) \cos x dx$. Apply integration by parts:
$I_2 = [F(x) \sin x]_0^\pi - \int_0^\pi F'(x) \sin x dx$
Since $F(0) = 0$ and $\sin(0) = \sin(\pi) = 0$,the boundary term is $0$.
$I_2 = - \int_0^\pi f(x) \sin x dx$
Substitute $I_1$ and $I_2$ back into the original equation:
$(6 - f(0) + \int_0^\pi f(x) \sin x dx) + (- \int_0^\pi f(x) \sin x dx) = 2$
$6 - f(0) = 2$
$f(0) = 4$.

(B) Given $f(\theta) = (\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^4$.
Using $(\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta$ and $(\sin \theta - \cos \theta)^2 = 1 - \sin 2\theta$,we have:
$f(\theta) = (1 + \sin 2\theta) + (1 - \sin 2\theta)^2$
$f(\theta) = 1 + \sin 2\theta + 1 - 2\sin 2\theta + \sin^2 2\theta = \sin^2 2\theta - \sin 2\theta + 2$.
Let $x = \sin 2\theta$. Then $g(x) = x^2 - x + 2$. The derivative $g'(x) = 2x - 1$. Setting $g'(x) = 0$ gives $x = 1/2$.
Thus,$\sin 2\theta = 1/2$. For $\theta \in (0, \pi)$,$2\theta \in (0, 2\pi)$.
$2\theta = \pi/6$ or $2\theta = 5\pi/6$.
$\theta = \pi/12$ or $\theta = 5\pi/12$.
These are the points of local minima. Thus $\lambda_1 = 1/12$ and $\lambda_2 = 5/12$.
The sum $\lambda_1 + \lambda_2 = 1/12 + 5/12 = 6/12 = 0.50$.