AP EAMCET 2003 Mathematics Question Paper with Answer and Solution

(B) Let the remainder when $p(x)$ is divided by $x^2-a^2$ be $R(x) = mx + c$,where $x^2-a^2 = (x-a)(x+a)$.
By the Remainder Theorem,$p(a) = -a$ and $p(-a) = a$.
Since $p(x) = (x^2-a^2)q(x) + (mx+c)$,we have:
$p(a) = m(a) + c = -a$ ... $(i)$
$p(-a) = m(-a) + c = a$ ... (ii)
Subtracting (ii) from $(i)$: $2ma = -2a \Rightarrow m = -1$.
Substituting $m = -1$ into $(i)$: $-a + c = -a \Rightarrow c = 0$.
Thus,the remainder $R(x) = -1(x) + 0 = -x$.

(A) The $n$-th term of the series is $T_n = \frac{1+2+3+\ldots+n}{(n+1)!}$.
Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(n+1)!} = \frac{n(n+1)}{2(n+1)n(n-1)!} = \frac{1}{2(n-1)!}$.
Summing from $n=1$ to $\infty$:
$S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{2(n-1)!} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$.
Let $k = n-1$,then as $n=1, k=0$ and as $n \to \infty, k \to \infty$.
$S = \frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{2} \left( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots \right) = \frac{1}{2} e$.
Thus,the sum is $\frac{e}{2}$.

(A) Let $x = \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$.
Note that $\frac{14 \pi}{15} = \pi - \frac{\pi}{15}$,so $\cos \frac{14 \pi}{15} = -\cos \frac{\pi}{15}$.
Also,$\frac{2 \pi}{15} = 24^{\circ}$,$\frac{4 \pi}{15} = 48^{\circ}$,$\frac{8 \pi}{15} = 96^{\circ}$,and $\frac{\pi}{15} = 12^{\circ}$.
Thus,$x = -\cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}$.
Using the formula $\cos \theta \cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}$:
$x = -\left( \frac{\sin(2^4 \times 12^{\circ})}{2^4 \sin 12^{\circ}} \right) = -\frac{\sin 192^{\circ}}{16 \sin 12^{\circ}}$.
Since $\sin 192^{\circ} = \sin(180^{\circ} + 12^{\circ}) = -\sin 12^{\circ}$,
$x = -\frac{-\sin 12^{\circ}}{16 \sin 12^{\circ}} = \frac{1}{16}$.

(C) Given the equation: $(5+4 \cos \theta)(2 \cos \theta+1)=0$
Since $5+4 \cos \theta$ can never be $0$ (because $-1 \le \cos \theta \le 1$,so $1 \le 5+4 \cos \theta \le 9$),we must have:
$2 \cos \theta + 1 = 0$
$\cos \theta = -\frac{1}{2}$
In the interval $[0, 2\pi]$,the values of $\theta$ for which $\cos \theta = -\frac{1}{2}$ are:
$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$
Thus,the solution set is $\left\{\frac{2\pi}{3}, \frac{4\pi}{3}\right\}$.

(D) We know that $\sin 6 \theta = \sin 3(2 \theta)$.
Using the identity $\sin 3A = 3 \sin A - 4 \sin^3 A$,we get:
$\sin 6 \theta = 3 \sin 2 \theta - 4 \sin^3 2 \theta$.
Substitute $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$\sin 6 \theta = 3(2 \sin \theta \cos \theta) - 4(2 \sin \theta \cos \theta)^3$
$= 6 \sin \theta \cos \theta - 4(8 \sin^3 \theta \cos^3 \theta)$
$= 6 \sin \theta \cos \theta - 32 \sin^3 \theta \cos^3 \theta$
$= 6 \sin \theta \cos \theta - 32 \sin \theta \cos^3 \theta (1 - \cos^2 \theta)$
$= 6 \sin \theta \cos \theta - 32 \sin \theta \cos^3 \theta + 32 \sin \theta \cos^5 \theta$
$= 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3(2 \sin \theta \cos \theta)$
$= 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3 \sin 2 \theta$.
Comparing this with the given equation $\sin 6 \theta = 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3x$,we find $3x = 3 \sin 2 \theta$,which implies $x = \sin 2 \theta$.

(A) Expand the expression using the formula $\sin (A-B) = \sin A \cos B - \cos A \sin B$:
$\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$
Distribute the terms:
$\cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$
Grouping the terms,we see that each term cancels out:
$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$

(C) Given $A+B+C=270^{\circ}$,so $A+B=270^{\circ}-C$.
Using the formula $\cos 2A+\cos 2B=2 \cos(A+B) \cos(A-B)$,we have:
$\cos 2A+\cos 2B+\cos 2C = 2 \cos(A+B) \cos(A-B) + (2 \cos^2 C - 1)$
$= 2 \cos(270^{\circ}-C) \cos(A-B) + 2 \cos^2 C - 1$
$= 2(-\sin C) \cos(A-B) + 2 \cos^2 C - 1$
$= -2 \sin C \cos(A-B) + 2(1-\sin^2 C) - 1$
$= 1 - 2 \sin C [\cos(A-B) + \sin C]$
Since $C = 270^{\circ}-(A+B)$,$\sin C = \sin(270^{\circ}-(A+B)) = -\cos(A+B)$.
$= 1 - 2 \sin C [\cos(A-B) - \cos(A+B)]$
Using $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$:
$= 1 - 2 \sin C [2 \sin A \sin B] = 1 - 4 \sin A \sin B \sin C$.

(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $\theta = 135^{\circ}$.
The transformation equations are:
$x' = x \cos \theta + y \sin \theta$
$y' = -x \sin \theta + y \cos \theta$
Given $(x', y') = (4, -3)$ and $\theta = 135^{\circ}$,we have:
$4 = x \cos 135^{\circ} + y \sin 135^{\circ} = -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}$
$\Rightarrow -x + y = 4\sqrt{2} \quad (i)$
$-3 = -x \sin 135^{\circ} + y \cos 135^{\circ} = -\frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}}$
$\Rightarrow x + y = 3\sqrt{2} \quad (ii)$
Adding $(i)$ and $(ii)$ gives $2y = 7\sqrt{2} \Rightarrow y = \frac{7}{\sqrt{2}}$.
Subtracting $(i)$ from $(ii)$ gives $2x = -\sqrt{2} \Rightarrow x = -\frac{1}{\sqrt{2}}$.
Thus,the coordinates in the original system are $(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$.

(D) Let the coordinates of $P$ be $(x, y)$.
Since $P$ is equidistant from $A, B, C$,we have $PA^2 = PB^2$ and $PB^2 = PC^2$.
From $PA^2 = PB^2$:
$(x-1)^2 + (y-3)^2 = (x+3)^2 + (y-5)^2$
$x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 6x + 9 + y^2 - 10y + 25$
$-2x - 6y + 10 = 6x - 10y + 34$
$8x - 4y + 24 = 0 \Rightarrow 2x - y + 6 = 0$ ... $(i)$
From $PB^2 = PC^2$:
$(x+3)^2 + (y-5)^2 = (x-5)^2 + (y+1)^2$
$x^2 + 6x + 9 + y^2 - 10y + 25 = x^2 - 10x + 25 + y^2 + 2y + 1$
$6x - 10y + 34 = -10x + 2y + 26$
$16x - 12y + 8 = 0 \Rightarrow 4x - 3y + 2 = 0$ ... (ii)
Solving $(i)$ and (ii):
Multiply $(i)$ by $2$: $4x - 2y + 12 = 0$ ... (iii)
Subtract (ii) from (iii): $(-2y - (-3y)) + (12 - 2) = 0$ $\Rightarrow y + 10 = 0$ $\Rightarrow y = -10$.
Substitute $y = -10$ into $(i)$: $2x - (-10) + 6 = 0$ $\Rightarrow 2x + 16 = 0$ $\Rightarrow x = -8$.
Thus,$P$ is $(-8, -10)$.
$PA = \sqrt{(-8-1)^2 + (-10-3)^2} = \sqrt{(-9)^2 + (-13)^2} = \sqrt{81 + 169} = \sqrt{250} = 5 \sqrt{10}$.

(A) Given equation of the line is $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$,which can be written as $\sqrt{3} r \sin \theta + 2 r \cos \theta = 4$.
Converting to Cartesian coordinates using $x = r \cos \theta$ and $y = r \sin \theta$,we get $\sqrt{3} y + 2 x = 4$,or $2x + \sqrt{3} y - 4 = 0$.
The slope of this line is $m_1 = -\frac{2}{\sqrt{3}}$.
The slope of a line perpendicular to this is $m_2 = -\frac{1}{m_1} = \frac{\sqrt{3}}{2}$.
The point $\left(-1, \frac{\pi}{2}\right)$ in polar coordinates corresponds to Cartesian coordinates $(x_1, y_1) = (-1 \cos \frac{\pi}{2}, -1 \sin \frac{\pi}{2}) = (0, -1)$.
The equation of the line passing through $(0, -1)$ with slope $\frac{\sqrt{3}}{2}$ is $y - (-1) = \frac{\sqrt{3}}{2}(x - 0)$.
$y + 1 = \frac{\sqrt{3}}{2} x$ $\Rightarrow 2y + 2 = \sqrt{3} x$ $\Rightarrow \sqrt{3} x - 2y = 2$.
Substituting $x = r \cos \theta$ and $y = r \sin \theta$,we get $\sqrt{3} r \cos \theta - 2 r \sin \theta = 2$.

(D) The pair of lines is given by $Ax^2+2Hxy+By^2=0$.
Let the angle between these lines be $2\theta$. Since the triangle formed with the line $ax+by+c=0$ is equilateral,the angle between each pair of lines is $60^{\circ}$.
The angle $\theta$ between the lines $y=m_1x$ and $y=m_2x$ is given by $\tan \theta = \left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$.
For an equilateral triangle,the angle between the lines is $60^{\circ}$,so $\theta = 30^{\circ}$.
Thus,$\tan 30^{\circ} = \frac{\sqrt{H^2-AB}}{|A+B|}$.
Squaring both sides,we get $\frac{1}{3} = \frac{H^2-AB}{(A+B)^2}$.
$(A+B)^2 = 3(H^2-AB)$.
$A^2+B^2+2AB = 3H^2-3AB$.
$A^2+B^2+5AB = 3H^2$.
We need to evaluate $(A+3B)(3A+B) = 3A^2+AB+9AB+3B^2 = 3A^2+10AB+3B^2$.
From the condition of the angle between the lines $Ax^2+2Hxy+By^2=0$ being $60^{\circ}$,we use the formula $\cos 60^{\circ} = \frac{|A+B|}{\sqrt{(A-B)^2+4H^2}}$.
$\frac{1}{2} = \frac{|A+B|}{\sqrt{(A-B)^2+4H^2}} \implies (A-B)^2+4H^2 = 4(A+B)^2$.
$A^2+B^2-2AB+4H^2 = 4(A^2+B^2+2AB)$.
$4H^2 = 3A^2+3B^2+10AB$.
Therefore,$(A+3B)(3A+B) = 3A^2+10AB+3B^2 = 4H^2$.

(A) The given equations can be factored as follows:
First pair: $(\lambda x - my)(\lambda x + my) - n(\lambda x + my) = 0 \implies (\lambda x + my)(\lambda x - my - n) = 0$.
This represents the lines $L_1: \lambda x + my = 0$ and $L_2: \lambda x - my = n$.
Second pair: $(\lambda x - my)(\lambda x + my) + n(\lambda x + my) = 0 \implies (\lambda x + my)(\lambda x - my + n) = 0$.
This represents the lines $L_3: \lambda x + my = -n$ and $L_4: \lambda x - my = 0$.
The area of the parallelogram formed by lines $a_1 x + b_1 y + c_1 = 0, a_1 x + b_1 y + c_2 = 0, a_2 x + b_2 y + d_1 = 0, a_2 x + b_2 y + d_2 = 0$ is given by $\left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1 b_2 - a_2 b_1} \right|$.
Here,the lines are $\lambda x + my + 0 = 0, \lambda x + my + n = 0$ and $\lambda x - my + 0 = 0, \lambda x - my - n = 0$.
Area $= \left| \frac{(0 - n)(0 - (-n))}{\lambda(-m) - m(\lambda)} \right| = \left| \frac{-n^2}{-2\lambda m} \right| = \frac{n^2}{2|\lambda m|}$ sq units.

(A) Let the roots of the equation $x^3+p x^2-q x+r=0$ be $\alpha, \beta, \gamma$.
From the relation between roots and coefficients:
$\alpha+\beta+\gamma = -p$
$\alpha \beta + \beta \gamma + \gamma \alpha = -q$
$\alpha \beta \gamma = -r$
Given that the sum of two roots is zero,let $\alpha + \beta = 0$,which implies $\beta = -\alpha$.
Substituting $\alpha + \beta = 0$ into the first relation:
$0 + \gamma = -p \implies \gamma = -p$.
Substituting $\gamma = -p$ into the third relation:
$\alpha \beta (-p) = -r \implies \alpha \beta p = r$.
Now,substitute $\beta = -\alpha$ into the second relation:
$\alpha(-\alpha) + \gamma(\alpha + \beta) = -q$
$-\alpha^2 + \gamma(0) = -q
-\alpha^2 = -q
\alpha^2 = q$.
Since $\alpha \beta = \alpha(-\alpha) = -\alpha^2 = -q$,we have $\alpha \beta = -q$.
From $\alpha \beta p = r$,we get $(-q)p = r$,which means $-pq = r$,or $pq = -r$.

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+0x^2+4x+1=0$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 0$
$\alpha\beta+\beta\gamma+\gamma\alpha = 4$
$\alpha\beta\gamma = -1$
Since $\alpha+\beta+\gamma = 0$,we can write:
$\alpha+\beta = -\gamma$
$\beta+\gamma = -\alpha$
$\gamma+\alpha = -\beta$
Substituting these into the expression:
$(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta}$
$= -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) = -\left(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\right)$
$= -\left(\frac{4}{-1}\right) = 4$

(D) Given expression: $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$
Expanding the squares: $225 + (9 \omega^2 + 64 \omega^4 + 48 \omega^3) + (9 \omega^4 + 64 \omega^2 + 48 \omega^3)$
Using the properties $\omega^3 = 1$ and $\omega^4 = \omega$: $225 + 9 \omega^2 + 64 \omega + 48 + 9 \omega + 64 \omega^2 + 48$
Grouping terms: $225 + 73(\omega^2 + \omega) + 96$
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$: $225 + 73(-1) + 96$
Calculating the final value: $225 - 73 + 96 = 152 + 96 = 248$

(D) Given,$\arg(z-2-3i) = \frac{\pi}{4}$.
Let $z = x+iy$.
Then $z-2-3i = (x-2) + i(y-3)$.
Since $\arg(z-2-3i) = \frac{\pi}{4}$,we have $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{\pi}{4}$.
This implies $\frac{y-3}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Therefore,$y-3 = x-2$,which simplifies to $x-y+1 = 0$.
Thus,the locus of $z$ is $x-y+1 = 0$.

(B) Given $z = x+iy$,we have $z-2-3i = (x-2) + i(y-3)$.
Since the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,we have $\arg((x-2) + i(y-3)) = \frac{\pi}{4}$.
This implies $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{\pi}{4}$.
Taking the tangent on both sides,we get $\frac{y-3}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Therefore,$y-3 = x-2$,which simplifies to $x-y+1=0$.

(C) The given lines are $y=x+r$ and $y=-x+r$ for $r \in \{0, 1, 2, 3, 4, 5, 6\}$.
These lines form a grid of squares.
Two lines $y=x+r_1$ and $y=x+r_2$ are parallel,and the distance between them is $\frac{|r_1-r_2|}{\sqrt{1^2+(-1)^2}} = \frac{|r_1-r_2|}{\sqrt{2}}$.
For the side length of the square to be $\sqrt{2}$,we require $\frac{|r_1-r_2|}{\sqrt{2}} = \sqrt{2}$,which implies $|r_1-r_2| = 2$.
For the set $r \in \{0, 1, 2, 3, 4, 5, 6\}$,the pairs $(r_1, r_2)$ with a difference of $2$ are $(0, 2), (1, 3), (2, 4), (3, 5), (4, 6)$. There are $5$ such pairs.
Similarly,for the lines $y=-x+r$,the distance between $y=-x+r_3$ and $y=-x+r_4$ is $\frac{|r_3-r_4|}{\sqrt{1^2+1^2}} = \frac{|r_3-r_4|}{\sqrt{2}}$.
Setting this to $\sqrt{2}$ gives $|r_3-r_4| = 2$,which also yields $5$ pairs.
The total number of squares formed is the product of the number of intervals of length $2$ in each direction,which is $5 \times 5 = 25$.

(D) The total number of points is $8$ (excluding $P$). Including $P$,we have $9$ points in total.
To form a triangle,we need to select $3$ non-collinear points.
The total number of ways to select $3$ points from $9$ points is ${^9C_3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
However,points on the same line cannot form a triangle.
Points on $l_1$ (including $P$) are $4$ $(A_1, B_1, C_1, P)$. The number of ways to choose $3$ collinear points from these is ${^4C_3} = 4$.
Points on $l_2$ (including $P$) are $6$ $(A_2, B_2, C_2, D_2, E_2, P)$. The number of ways to choose $3$ collinear points from these is ${^6C_3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Total triangles = (Total ways to choose $3$ points) - (Collinear sets on $l_1$) - (Collinear sets on $l_2$)
Total triangles = $84 - 4 - 20 = 60$.

(A) The $n$th term of the series is given by $T_n = \frac{1+2+3+\ldots+n}{(n+1) !}$.
Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(n+1)!} = \frac{n}{2(n)!} = \frac{1}{2(n-1)!}$.
For $n=1, 2, 3, \ldots$,the terms are $T_1 = \frac{1}{2(0!)}, T_2 = \frac{1}{2(1!)}, T_3 = \frac{1}{2(2!)}, \ldots$.
The sum $S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \left[ \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots \right]$.
Since $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots$,we have $S = \frac{1}{2} e$.

(A) We have the expression $(1+x^2)^5(1+x)^4$.
Using the binomial expansion formula $(1+a)^n = \sum_{k=0}^{n} {^nC_k} a^k$:
$(1+x^2)^5 = {^5C_0} + {^5C_1}x^2 + {^5C_2}x^4 + {^5C_3}x^6 + \dots = 1 + 5x^2 + 10x^4 + 10x^6 + \dots$
$(1+x)^4 = {^4C_0} + {^4C_1}x + {^4C_2}x^2 + {^4C_3}x^3 + {^4C_4}x^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$
To find the coefficient of $x^5$,we multiply terms from the two expansions such that the sum of their powers is $5$:
$(5x^2) \cdot (4x^3) + (10x^4) \cdot (4x) = 20x^5 + 40x^5 = 60x^5$
Thus,the coefficient of $x^5$ is $60$.

(B) The general term in the expansion of $(1+x)^n$ is given by $T_{k+1} = {}^{n}C_k x^k$.
For the $(2r+1)$-th term,$k = (2r+1)-1 = 2r$. The coefficient is ${}^{43}C_{2r}$.
For the $(r+2)$-th term,$k = (r+2)-1 = r+1$. The coefficient is ${}^{43}C_{r+1}$.
Given that the coefficients are equal:
${}^{43}C_{2r} = {}^{43}C_{r+1}$.
Using the property ${}^{n}C_a = {}^{n}C_b$,we have either $a = b$ or $a+b = n$.
Case $1$: $2r = r+1 \Rightarrow r = 1$.
Case $2$: $2r + (r+1) = 43$ $\Rightarrow 3r + 1 = 43$ $\Rightarrow 3r = 42$ $\Rightarrow r = 14$.
Since $14$ is one of the options,$r = 14$ is the correct value.

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.
Let $k = \frac{1}{x} = y^3 - 1$. Note that $x = \frac{1}{k}$.
The series is $S = 2(\frac{1}{x}) + \frac{2}{3}(\frac{1}{x})^3 + \frac{2}{5}(\frac{1}{x})^5 + \dots = 2k + \frac{2}{3}k^3 + \frac{2}{5}k^5 + \dots$
We know the logarithmic expansion $\log \left( \frac{1+k}{1-k} \right) = 2(k + \frac{k^3}{3} + \frac{k^5}{5} + \dots) = 2k + \frac{2}{3}k^3 + \frac{2}{5}k^5 + \dots$
Substituting $k = y^3 - 1$:
$S = \log \left( \frac{1 + (y^3 - 1)}{1 - (y^3 - 1)} \right) = \log \left( \frac{y^3}{1 - y^3 + 1} \right) = \log \left( \frac{y^3}{2 - y^3} \right)$.

(D) The period of $\sin(a\theta)$ is $\frac{2\pi}{|a|}$ and the period of $\cos(b\theta)$ is $\frac{2\pi}{|b|}$.
For $f(\theta) = \sin \frac{\theta}{3} + \cos \frac{\theta}{2}$, the period of $\sin \frac{\theta}{3}$ is $T_1 = \frac{2\pi}{1/3} = 6\pi$.
The period of $\cos \frac{\theta}{2}$ is $T_2 = \frac{2\pi}{1/2} = 4\pi$.
The period of the sum of two periodic functions is the Least Common Multiple $(LCM)$ of their individual periods.
We need to find the $LCM(6\pi, 4\pi)$.
$LCM(6, 4) = 12$.
Therefore, the period of $f(\theta)$ is $12\pi$.

(A) Given the two tangent lines to the circle are:
$5x - 12y + 10 = 0$ $\dots$ $(i)$
$-5x + 12y + 16 = 0$ $\dots$ $(ii)$
Since the slopes of both lines are equal to $\frac{5}{12}$,the lines are parallel.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 5, B = -12, C_1 = 10, C_2 = 16$.
The distance between the tangents is the diameter of the circle:
$d = \frac{|10 - 16|}{\sqrt{5^2 + (-12)^2}} = \frac{|-6|}{\sqrt{25 + 144}} = \frac{6}{13}$.
Wait,re-evaluating the constant terms:
Equation $(i): 5x - 12y + 10 = 0$
Equation $(ii): -5x + 12y + 16 = 0 \implies 5x - 12y - 16 = 0$
Distance $d = \frac{|10 - (-16)|}{\sqrt{5^2 + (-12)^2}} = \frac{26}{13} = 2$.
Since the distance between parallel tangents is the diameter,$2r = 2$.
Therefore,the radius $r = 1$.

(D) Given that,the circle $S_1 \equiv x^2+y^2+6x-2y+k=0$ and $S_2 \equiv x^2+y^2+2x-6y-15=0$.
Since $S_1$ bisects the circumference of $S_2$,the common chord of $S_1$ and $S_2$ must be the diameter of $S_2$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$.
$4x + 4y + k + 15 = 0$.
Since this chord is the diameter of $S_2$,it must pass through the center of $S_2$.
The center of $S_2$ is $(-g, -f) = (-1, 3)$.
Substituting $(-1, 3)$ into the chord equation:
$4(-1) + 4(3) + k + 15 = 0$.
$-4 + 12 + k + 15 = 0$.
$8 + k + 15 = 0$.
$k + 23 = 0$.
$k = -23$.

(B) Let the coordinates of $P$ be $(x, y)$.
Given the equations of the circles:
$S_1: x^2+y^2+2x-4y-20=0$
$S_2: x^2+y^2-4x+2y-44=0$
The square of the length of the tangent from $P(x, y)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $x^2+y^2+2gx+2fy+c$.
Thus,the ratio of the squares of the lengths of the tangents is:
$\frac{x^2+y^2+2x-4y-20}{x^2+y^2-4x+2y-44} = \frac{2}{3}$
Cross-multiplying gives:
$3(x^2+y^2+2x-4y-20) = 2(x^2+y^2-4x+2y-44)$
$3x^2+3y^2+6x-12y-60 = 2x^2+2y^2-8x+4y-88$
Rearranging the terms to one side:
$x^2+y^2+14x-16y+28 = 0$
This is the equation of a circle in the form $x^2+y^2+2gx+2fy+c=0$,where $2g=14$ and $2f=-16$.
Thus,$g=7$ and $f=-8$.
The centre of the circle is $(-g, -f) = (-7, 8)$.

(A) Given focus of parabola is $S(0,0)$.
Equation of directrix is $x+y-4=0$.
Let $P(x, y)$ be any point on the parabola.
By definition,the distance from $P$ to the focus equals the distance from $P$ to the directrix,so $SP^2 = PM^2$.
$SP^2 = (x-0)^2 + (y-0)^2 = x^2 + y^2$.
The perpendicular distance $PM$ from $P(x, y)$ to the line $x+y-4=0$ is $\frac{|x+y-4|}{\sqrt{1^2+1^2}} = \frac{|x+y-4|}{\sqrt{2}}$.
Thus,$x^2 + y^2 = \left(\frac{x+y-4}{\sqrt{2}}\right)^2$.
$x^2 + y^2 = \frac{(x+y-4)^2}{2}$.
$2(x^2 + y^2) = x^2 + y^2 + 16 + 2xy - 8x - 8y$.
$x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$.

(B) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$.
Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$.
Completing the square,we have $9(x^2 - 2x + 1) + 5(y^2 - 4y + 4) = 16 + 9 + 20$.
This simplifies to $9(x - 1)^2 + 5(y - 2)^2 = 45$.
Dividing by $45$,we obtain the standard form $\frac{(x - 1)^2}{5} + \frac{(y - 2)^2}{9} = 1$.
Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.
Here,$a^2 = 2$ and $b^2 = 1$.
The equations of the asymptotes are given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$,which simplifies to $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$.
For any point $P(x_1, y_1)$ on the hyperbola,the product of the lengths of the perpendiculars to the asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$.
Substituting the values $a^2 = 2$ and $b^2 = 1$,we get:
Product $= \frac{2 \times 1}{2 + 1} = \frac{2}{3}$.

(B) Let $L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x - \sqrt{3} \cos x}{6x - \pi}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:
$L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{\frac{d}{dx}(3 \sin x - \sqrt{3} \cos x)}{\frac{d}{dx}(6x - \pi)}$
$L = \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x + \sqrt{3} \sin x}{6}$
Substituting $x = \frac{\pi}{6}$:
$L = \frac{3 \cos(\frac{\pi}{6}) + \sqrt{3} \sin(\frac{\pi}{6})}{6}$
$L = \frac{3(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2})}{6}$
$L = \frac{\frac{3\sqrt{3} + \sqrt{3}}{2}}{6} = \frac{4\sqrt{3}}{12} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.

(B) Given the limit: $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$.
Applying $L^{\prime}$Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx}(a^x - x^a) = a^x \ln a - a x^{a-1}$.
Denominator derivative: $\frac{d}{dx}(x^x - a^a) = x^x(1 + \ln x)$.
Substituting $x = a$:
$\frac{a^a \ln a - a \cdot a^{a-1}}{a^a(1 + \ln a)} = -1$.
$\frac{a^a \ln a - a^a}{a^a(1 + \ln a)} = -1$.
Dividing numerator and denominator by $a^a$:
$\frac{\ln a - 1}{1 + \ln a} = -1$.
$\ln a - 1 = -(1 + \ln a)$.
$\ln a - 1 = -1 - \ln a$.
$2 \ln a = 0$.
$\ln a = 0 \Rightarrow a = e^0 = 1$.

(B) Given: $b=20, c=21$ and $\sin A=\frac{3}{5}$.
Using the identity $\cos^2 A = 1 - \sin^2 A$,we get:
$\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Thus,$\cos A = \frac{4}{5}$ (assuming $A$ is an acute angle).
Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Substituting the values: $\frac{4}{5} = \frac{20^2+21^2-a^2}{2 \times 20 \times 21}$.
$\frac{4}{5} = \frac{400+441-a^2}{840}$.
$840 \times \frac{4}{5} = 841 - a^2$.
$168 \times 4 = 841 - a^2$.
$672 = 841 - a^2$.
$a^2 = 841 - 672 = 169$.
Therefore,$a = \sqrt{169} = 13$.

(B) We know the formula for the cotangent of half-angles in a triangle: $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get: $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $3a = b + c$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.
Substituting $s = 2a$ into the expression: $\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.

(D) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{B}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $b+c=3a$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.
Substituting $s = 2a$ into the expression,we get $\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.

(A) We have the formulas for the exradii of a triangle:
$r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$
Given that $r_1 < r_2 < r_3$,we substitute the expressions:
$\frac{\Delta}{s-a} < \frac{\Delta}{s-b} < \frac{\Delta}{s-c}$
Since $\Delta$ is the area of the triangle and is positive,taking the reciprocal reverses the inequality signs:
$s-a > s-b > s-c$
Subtracting $s$ from all parts:
$-a > -b > -c$
Multiplying by $-1$ reverses the inequality signs again:
$a < b < c$

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh^{-1}(x) = \log\{x + \sqrt{x^2 + 1}\}$.
Substituting $x = 2^{3/2}$ into the formula:
$\sinh^{-1}(2^{3/2}) = \log\{2^{3/2} + \sqrt{(2^{3/2})^2 + 1\}}$.
Since $2^{3/2} = \sqrt{2^3} = \sqrt{8}$,we have:
$\sinh^{-1}(2^{3/2}) = \log\{\sqrt{8} + \sqrt{8 + 1\}}$.
$= \log\{\sqrt{8} + \sqrt{9\}}$.
$= \log\{3 + \sqrt{8\}}$.
Thus,the correct option is $A$.

(B) Given the inequation: $3^x + 3^{1-x} - 4 < 0$.
Substitute $3^{1-x} = \frac{3}{3^x}$ into the expression: $3^x + \frac{3}{3^x} - 4 < 0$.
Multiply the entire inequality by $3^x$ (since $3^x > 0$ for all $x \in R$): $(3^x)^2 - 4(3^x) + 3 < 0$.
Let $y = 3^x$. The inequality becomes $y^2 - 4y + 3 < 0$.
Factor the quadratic expression: $(y - 1)(y - 3) < 0$.
This implies $1 < y < 3$.
Substituting back $y = 3^x$,we get $1 < 3^x < 3$.
Since $3^0 = 1$ and $3^1 = 3$,we have $3^0 < 3^x < 3^1$.
Comparing the exponents,we get $0 < x < 1$.
Thus,the solution set is $(0, 1)$.

(A) Given the partial fraction decomposition: $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$.
Multiplying both sides by $(1-x)(1-2x)(1-3x)$,we get:
$1 = a(1-2x)(1-3x) + b(1-x)(1-3x) + c(1-x)(1-2x)$.
To find $a$,set $x = 1$: $1 = a(1-2)(1-3) = a(-1)(-2) = 2a \Rightarrow a = \frac{1}{2}$.
To find $b$,set $x = \frac{1}{2}$: $1 = b(1-\frac{1}{2})(1-\frac{3}{2}) = b(\frac{1}{2})(-\frac{1}{2}) = -\frac{1}{4}b \Rightarrow b = -4$.
To find $c$,set $x = \frac{1}{3}$: $1 = c(1-\frac{1}{3})(1-\frac{2}{3}) = c(\frac{2}{3})(\frac{1}{3}) = \frac{2}{9}c \Rightarrow c = \frac{9}{2}$.
Now,calculate $\frac{a}{1} + \frac{b}{3} + \frac{c}{5} = \frac{1/2}{1} + \frac{-4}{3} + \frac{9/2}{5} = \frac{1}{2} - \frac{4}{3} + \frac{9}{10}$.
Finding a common denominator of $30$: $\frac{15}{30} - \frac{40}{30} + \frac{27}{30} = \frac{15 - 40 + 27}{30} = \frac{2}{30} = \frac{1}{15}$.

(C) We have,$f(x) = \frac{1}{\sqrt{x+2 \sqrt{2x-4}}} + \frac{1}{\sqrt{x-2 \sqrt{2x-4}}}$
Substituting $x = 11$:
$f(11) = \frac{1}{\sqrt{11+2 \sqrt{2(11)-4}}} + \frac{1}{\sqrt{11-2 \sqrt{2(11)-4}}}$
$f(11) = \frac{1}{\sqrt{11+2 \sqrt{18}}} + \frac{1}{\sqrt{11-2 \sqrt{18}}}$
$f(11) = \frac{1}{\sqrt{11+6 \sqrt{2}}} + \frac{1}{\sqrt{11-6 \sqrt{2}}}$
Since $11+6 \sqrt{2} = 9+2+2(3)(\sqrt{2}) = (3+\sqrt{2})^2$ and $11-6 \sqrt{2} = (3-\sqrt{2})^2$,
$f(11) = \frac{1}{3+\sqrt{2}} + \frac{1}{3-\sqrt{2}}$
$f(11) = \frac{(3-\sqrt{2}) + (3+\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$
$f(11) = \frac{6}{9-2} = \frac{6}{7}$

(D) Given $t_n = \frac{1}{4}(n+2)(n+3)$.
Then,$\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}$.
Using partial fractions,$\frac{1}{t_n} = 4 \left[ \frac{1}{n+2} - \frac{1}{n+3} \right]$.
Let $S = \sum_{n=1}^{2003} \frac{1}{t_n} = 4 \sum_{n=1}^{2003} \left( \frac{1}{n+2} - \frac{1}{n+3} \right)$.
This is a telescoping series:
$S = 4 \left[ \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots + \left( \frac{1}{2005} - \frac{1}{2006} \right) \right]$.
$S = 4 \left( \frac{1}{3} - \frac{1}{2006} \right)$.
$S = 4 \left( \frac{2006 - 3}{3 \times 2006} \right) = 4 \left( \frac{2003}{6018} \right) = \frac{8012}{6018} = \frac{4006}{3009}$.

(B) In $\triangle ECD$,$\tan 3 \alpha = \frac{h}{CD} \Rightarrow CD = h \cot 3 \alpha \quad \dots(i)$
In $\triangle EBD$,$\tan 2 \alpha = \frac{h}{BD} \Rightarrow BD = h \cot 2 \alpha \quad \dots(ii)$
In $\triangle EAD$,$\tan \alpha = \frac{h}{AD} \Rightarrow AD = h \cot \alpha \quad \dots(iii)$
From Eqs. $(ii)$ and $(iii)$,$AB = AD - BD = h(\cot \alpha - \cot 2 \alpha) \quad \dots(iv)$
From Eqs. $(i)$ and $(ii)$,$BC = BD - CD = h(\cot 2 \alpha - \cot 3 \alpha) \quad \dots(v)$
Dividing $(iv)$ by $(v)$:
$\frac{AB}{BC} = \frac{\cot \alpha - \cot 2 \alpha}{\cot 2 \alpha - \cot 3 \alpha} = \frac{\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha} - \frac{\cos 3 \alpha}{\sin 3 \alpha}}$
$= \frac{\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin(3 \alpha - 2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}} = \frac{\sin \alpha}{\sin \alpha \sin 2 \alpha} \times \frac{\sin 2 \alpha \sin 3 \alpha}{\sin \alpha} = \frac{\sin 3 \alpha}{\sin \alpha}$
$= \frac{3 \sin \alpha - 4 \sin^3 \alpha}{\sin \alpha} = 3 - 4 \sin^2 \alpha = 3 - 2(1 - \cos 2 \alpha) = 1 + 2 \cos 2 \alpha$.

(C) Let the $XOZ$-plane divide the line segment joining the points $A(2, 3, 1)$ and $B(6, 7, 1)$ in the ratio $m: n$.
Using the section formula,the coordinates of the point of division are given by:
$\left(\frac{m(6)+n(2)}{m+n}, \frac{m(7)+n(3)}{m+n}, \frac{m(1)+n(1)}{m+n}\right)$
Since the point lies on the $XOZ$-plane,its $y$-coordinate must be zero.
Therefore,$\frac{7m + 3n}{m+n} = 0$.
$7m + 3n = 0$
$7m = -3n$
$\frac{m}{n} = -\frac{3}{7}$
Thus,the ratio is $-3: 7$.

(D) We know that for any two events $A$ and $B$,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Therefore,$P(A) + P(B) = P(A \cup B) + P(A \cap B)$.
Substituting the given values,$P(A) + P(B) = 0.8 + 0.3 = 1.1$.
We know that $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B)$.
Thus,$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting $P(A) + P(B) = 1.1$,we get $P(\bar{A}) + P(\bar{B}) = 2 - 1.1 = 0.9$.

(B) We know that for any $x \in [-1, 1]$,the identity $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}$ holds true.
Given the expression $\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$.
Substituting $x = -\frac{1}{7}$ into the identity,we get:
$\cos \left[\frac{\pi}{2}\right]$
Since $\cos \frac{\pi}{2} = 0$,the value of the expression is $0$.

(B) To find the number of solutions,we write the system in matrix form $AX = B$,where $A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B = \begin{bmatrix} 7 \\ 1 \\ 5 \end{bmatrix}$.
First,calculate the determinant of $A$ $(|A|)$:
$|A| = 2((-3)(-3) - (2)(4)) - 1((1)(-3) - (2)(1)) - 1((1)(4) - (-3)(1))$
$|A| = 2(9 - 8) - 1(-3 - 2) - 1(4 + 3)$
$|A| = 2(1) - 1(-5) - 1(7) = 2 + 5 - 7 = 0$.
Since $|A| = 0$,the system is either inconsistent (no solution) or has infinitely many solutions.
We check the consistency using the augmented matrix $[A|B]$:
$\begin{bmatrix} 2 & 1 & -1 & | & 7 \\ 1 & -3 & 2 & | & 1 \\ 1 & 4 & -3 & | & 5 \end{bmatrix}$
Performing row operations: $R_1 \leftrightarrow R_2$ gives $\begin{bmatrix} 1 & -3 & 2 & | & 1 \\ 2 & 1 & -1 & | & 7 \\ 1 & 4 & -3 & | & 5 \end{bmatrix}$.
$R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ gives $\begin{bmatrix} 1 & -3 & 2 & | & 1 \\ 0 & 7 & -5 & | & 5 \\ 0 & 7 & -5 & | & 4 \end{bmatrix}$.
$R_3 \to R_3 - R_2$ gives $\begin{bmatrix} 1 & -3 & 2 & | & 1 \\ 0 & 7 & -5 & | & 5 \\ 0 & 0 & 0 & | & -1 \end{bmatrix}$.
The last row implies $0 = -1$,which is a contradiction.
Therefore,the system has no solution.

(C) Given the function $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$.
We need to find the set of $x$ such that $x \leq 0$ and $f(|x|) = x$.
Since $x \leq 0$,we have $|x| = -x$. Since $x \in [-2, 0]$,then $|x| \in [0, 2]$.
For $0 < |x| \leq 2$,$f(|x|) = |x| - 1 = -x - 1$.
Setting $f(|x|) = x$,we get $-x - 1 = x$,which implies $2x = -1$,so $x = -\frac{1}{2}$.
Checking the condition $x \leq 0$,we see that $x = -\frac{1}{2}$ satisfies this.
If $x = 0$,then $f(|0|) = f(0) = -1$. But $x = 0$,so $f(0) \neq 0$.
Thus,the set is $\{-\frac{1}{2}\}$.

(B) Since $f(x)$ is continuous at $x=0$,the left-hand limit must equal the right-hand limit,i.e.,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the right-hand limit: $\lim_{x \to 0^+} (2x^2+3x-2) = 2(0)^2+3(0)-2 = -2$.
Now,calculate the left-hand limit: $\lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$.
Rationalizing the numerator: $\lim_{x \to 0^-} \frac{(\sqrt{1+kx}-\sqrt{1-kx})(\sqrt{1+kx}+\sqrt{1-kx})}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \lim_{x \to 0^-} \frac{(1+kx)-(1-kx)}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}} = \frac{2k}{1+1} = k$.
Equating the limits: $k = -2$.

(B) Given $f(x) = \frac{x-1}{2x^2-7x+5}$ for $x \neq 1$.
Factoring the denominator: $2x^2-7x+5 = 2x^2-2x-5x+5 = 2x(x-1)-5(x-1) = (2x-5)(x-1)$.
Thus,for $x \neq 1$,$f(x) = \frac{x-1}{(2x-5)(x-1)} = \frac{1}{2x-5}$.
The function is $f(x) = \begin{cases} \frac{1}{2x-5}, & x \neq 1 \\ -\frac{1}{3}, & x=1 \end{cases}$.
By the definition of the derivative,$f^{\prime}(1) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{\frac{1}{2(1+h)-5} - (-\frac{1}{3})}{h} = \lim_{h \to 0} \frac{\frac{1}{2h-3} + \frac{1}{3}}{h}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{3 + (2h-3)}{3h(2h-3)} = \lim_{h \to 0} \frac{2h}{3h(2h-3)}$.
$f^{\prime}(1) = \lim_{h \to 0} \frac{2}{3(2h-3)} = \frac{2}{3(-3)} = -\frac{2}{9}$.

(B) Given $f(x) = \frac{x}{1+|x|}$.
To find $f^{\prime}(0)$,we use the definition of the derivative:
$f^{\prime}(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$.
Since $f(0) = \frac{0}{1+|0|} = 0$,we have:
$f^{\prime}(0) = \lim_{h \to 0} \frac{\frac{h}{1+|h|} - 0}{h} = \lim_{h \to 0} \frac{1}{1+|h|}$.
Now,we evaluate the left-hand and right-hand limits:
Left-hand limit $(h \to 0^-)$: $\lim_{h \to 0^-} \frac{1}{1-h} = 1$.
Right-hand limit $(h \to 0^+)$: $\lim_{h \to 0^+} \frac{1}{1+h} = 1$.
Since both limits are equal,$f^{\prime}(0) = 1$.

(C) Given,$u(x, y)=y \log x+x \log y$.
First,we find the partial derivatives $u_x$ and $u_y$:
$u_x = \frac{\partial}{\partial x}(y \log x + x \log y) = \frac{y}{x} + \log y$.
$u_y = \frac{\partial}{\partial y}(y \log x + x \log y) = \log x + \frac{x}{y}$.
Now,consider the expression $E = u_x u_y - u_x \log x - u_y \log y + \log x \log y$.
We can factor this expression as:
$E = u_x(u_y - \log x) - \log y(u_y - \log x)$.
$E = (u_x - \log y)(u_y - \log x)$.
Substitute the values of $u_x$ and $u_y$:
$u_x - \log y = (\frac{y}{x} + \log y) - \log y = \frac{y}{x}$.
$u_y - \log x = (\log x + \frac{x}{y}) - \log x = \frac{x}{y}$.
Therefore,$E = (\frac{y}{x}) \times (\frac{x}{y}) = 1$.

(A) Given that,the rate of change of volume is $\frac{dV}{dt} = 30 \ ft^3 / \text{min}$ and the radius $r = 15 \ ft$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values,$30 = 4 \pi (15)^2 \frac{dr}{dt}$.
$30 = 4 \pi (225) \frac{dr}{dt} = 900 \pi \frac{dr}{dt}$.
Therefore,$\frac{dr}{dt} = \frac{30}{900 \pi} = \frac{1}{30 \pi} \ ft / \text{min}$.

(B) We are given the integral $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$.
First,we rewrite the numerator as $(1+x) + \sqrt{x(1+x)}$.
This can be factored as $\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})$.
Substituting this back into the integral,we get:
$I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x} + \sqrt{1+x}} d x$.
The term $(\sqrt{1+x} + \sqrt{x})$ cancels out from the numerator and denominator.
$I = \int \sqrt{1+x} d x$.
Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$,where $u = 1+x$ and $du = dx$:
$I = \frac{(1+x)^{3/2}}{3/2} + C = \frac{2}{3}(1+x)^{3/2} + C$.

(C) Given the interval $[2,6]$ is divided into $n=4$ sub-intervals of equal length.
$h = \frac{6-2}{4} = 1$.
Let $f(x) = \frac{1}{x^2-x}$.
The values of $x$ are $x_0=2, x_1=3, x_2=4, x_3=5, x_4=6$.
The corresponding values of $y = f(x)$ are:
$y_0 = f(2) = \frac{1}{4-2} = \frac{1}{2} = 0.5$
$y_1 = f(3) = \frac{1}{9-3} = \frac{1}{6} \approx 0.1667$
$y_2 = f(4) = \frac{1}{16-4} = \frac{1}{12} \approx 0.0833$
$y_3 = f(5) = \frac{1}{25-5} = \frac{1}{20} = 0.05$
$y_4 = f(6) = \frac{1}{36-6} = \frac{1}{30} \approx 0.0333$
Using Simpson's rule: $\int_2^6 f(x) dx \approx \frac{h}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2y_2]$
$= \frac{1}{3} [\frac{1}{2} + \frac{1}{30} + 4(\frac{1}{6} + \frac{1}{20}) + 2(\frac{1}{12})]$
$= \frac{1}{3} [\frac{16}{30} + 4(\frac{10+3}{60}) + \frac{1}{6}]$
$= \frac{1}{3} [\frac{16}{30} + \frac{52}{60} + \frac{5}{30}] = \frac{1}{3} [\frac{32+52+10}{60}] = \frac{94}{180} = \frac{47}{90} \approx 0.5222$.

(A) Given differential equation is $y^2 dx + (x^2 - xy + y^2) dy = 0$.
Rearranging,we get $\frac{dy}{dx} = \frac{-y^2}{x^2 - xy + y^2}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{-(vx)^2}{x^2 - x(vx) + (vx)^2} = \frac{-v^2}{1 - v + v^2}$.
Then $x \frac{dv}{dx} = \frac{-v^2}{1 - v + v^2} - v = \frac{-v^2 - v + v^2 - v^3}{1 - v + v^2} = \frac{-(v^3 + v)}{v^2 - v + 1}$.
Separating variables: $\frac{v^2 - v + 1}{v^3 + v} dv = -\frac{1}{x} dx$.
Using partial fractions: $\frac{v^2 - v + 1}{v(v^2 + 1)} = \frac{A}{v} + \frac{Bv + C}{v^2 + 1}$. Solving gives $A=1, B=0, C=-1$.
So,$(\frac{1}{v} - \frac{1}{v^2 + 1}) dv = -\frac{1}{x} dx$.
Integrating both sides: $\log|v| - \tan^{-1}(v) = -\log|x| + C$.
$\log|vx| = \tan^{-1}(v) + C$.
Since $y = vx$,we have $\log|y| = \tan^{-1}(\frac{y}{x}) + C$,or $\tan^{-1}(\frac{y}{x}) = \log y + C$.

(C) Let $E_1$ be the event of selecting bag $X$ and $E_2$ be the event of selecting bag $Y$. Since one bag is selected at random,$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $W$ be the event of drawing a white ball.
The probability of drawing a white ball from bag $X$ is $P(W|E_1) = \frac{2}{2+3} = \frac{2}{5}$.
The probability of drawing a white ball from bag $Y$ is $P(W|E_2) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.
Using the law of total probability,the probability of drawing a white ball is $P(W) = P(E_1) \cdot P(W|E_1) + P(E_2) \cdot P(W|E_2)$.
$P(W) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{3}\right) = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda$ is the mean of the distribution.
Given the condition $P(X=2) = 3 P(X=3)$.
Substituting the formula:
$\frac{e^{-\lambda} \lambda^2}{2!} = 3 \cdot \frac{e^{-\lambda} \lambda^3}{3!}$
Dividing both sides by $e^{-\lambda} \lambda^2$ (assuming $\lambda \neq 0$):
$\frac{1}{2} = 3 \cdot \frac{\lambda}{3 \cdot 2 \cdot 1}$
$\frac{1}{2} = \frac{3 \lambda}{6}$
$\frac{1}{2} = \frac{\lambda}{2}$
$\lambda = 1$.
Thus,the mean of the Poisson distribution is $1$.

(D) Given that the mean of the random variable $X$ is $1.3$.
The formula for the mean is $\Sigma x_i P(X=x_i) = 1.3$.
Substituting the values: $0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) = 1.3$.
Given $P(X=2) = 0.3$ and $P(X=3) = 2 P(X=1)$,we substitute these into the equation:
$0 + P(X=1) + 2(0.3) + 3(2 P(X=1)) = 1.3$.
$P(X=1) + 0.6 + 6 P(X=1) = 1.3$.
$7 P(X=1) = 0.7$,which gives $P(X=1) = 0.1$.
Now,$P(X=3) = 2 P(X=1) = 2(0.1) = 0.2$.
Since the sum of all probabilities is $1$:
$P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$.
$P(X=0) + 0.1 + 0.3 + 0.2 = 1$.
$P(X=0) + 0.6 = 1$.
Therefore,$P(X=0) = 0.4$.

(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$.
Applying the property of determinants to split the second row:
$\left|\begin{array}{ccc}p & b & c \\ p & q & c \\ a & b & r\end{array}\right| + \left|\begin{array}{ccc}p & b & c \\ a & b & c \\ a & b & r\end{array}\right| = 0$.
Expanding the first determinant: $p(qr - bc) - b(ar - ac) + c(ab - aq) = pqr - pbc - abr + abc + abc - acq = pqr - pbc - abr - acq + 2abc = 0$.
Thus,$pqr - pbc - abr - acq = -2abc$.
Now,consider the expression $E = \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$.
This can be rewritten as $E = (1 + \frac{a}{p-a}) + (1 + \frac{b}{q-b}) + (1 + \frac{c}{r-c}) = 3 + \frac{a}{p-a} + \frac{b}{q-b} + \frac{c}{r-c}$.
Alternatively,dividing the equation $pqr - pbc - abr - acq = -2abc$ by $(p-a)(q-b)(r-c)$ leads to the result $2$.

(A) Given the determinant equation:
$\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\\sin A & \cos A & \sin B \\-\cos A & \sin A & \cos B\end{array}\right|=0$
Expanding along the first row:
$\cos (A+B)[\cos A \cos B - \sin A \sin B] + \sin (A+B)[\sin A \cos B + \cos A \sin B] + \cos 2 B[\sin^2 A + \cos^2 A] = 0$
Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$\cos (A+B) \cos (A+B) + \sin (A+B) \sin (A+B) + \cos 2 B(1) = 0$
$\cos^2 (A+B) + \sin^2 (A+B) + \cos 2 B = 0$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$1 + \cos 2 B = 0$
$\cos 2 B = -1$
$2 B = (2 n+1) \pi$
$B = \frac{(2 n+1) \pi}{2} = (2 n+1) \frac{\pi}{2}$

(A) Given $e^{f(x)}=\frac{10+x}{10-x}$.
Taking the natural logarithm on both sides,we get $f(x)=\log \left(\frac{10+x}{10-x}\right)$.
We are given the relation $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$.
Substituting the expression for $f(x)$ into the right side:
$f\left(\frac{200 x}{100+x^2}\right) = \log \left( \frac{10 + \frac{200x}{100+x^2}}{10 - \frac{200x}{100+x^2}} \right)$
$= \log \left( \frac{10(100+x^2) + 200x}{10(100+x^2) - 200x} \right)$
$= \log \left( \frac{1000 + 10x^2 + 200x}{1000 + 10x^2 - 200x} \right)$
$= \log \left( \frac{10(x^2 + 20x + 100)}{10(x^2 - 20x + 100)} \right)$
$= \log \left( \frac{(x+10)^2}{(10-x)^2} \right)$
$= 2 \log \left( \frac{10+x}{10-x} \right) = 2f(x)$.
Substituting this back into the given equation: $f(x) = k \cdot 2f(x)$.
Since $f(x) \neq 0$ for $x \in (-10, 10)$,we have $1 = 2k$,which implies $k = 0.5$.

(D) Given $f(x) = |x|$ and $g(x) = [x]$.
We need to find the set of $x \in R$ such that $g(f(x)) \leq f(g(x))$.
Substituting the functions,we get $[|x|] \leq |[x]|$.
Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$. The inequality becomes $[x] \leq |[x]|$. Since $[x]$ is an integer and $[x] \geq 0$,$|[x]| = [x]$. Thus,$[x] \leq [x]$,which is true for all $x \geq 0$.
Case $2$: If $x < 0$,let $x = -n - f$,where $n \geq 0$ is an integer and $0 \leq f < 1$. If $f=0$,$x = -n$ (an integer),then $[|x|] = [n] = n$ and $|[x]| = |-n| = n$. So $n \leq n$ is true.
If $0 < f < 1$,then $x = -(n+f)$. $|x| = n+f$,so $[|x|] = [n+f] = n$. Also $[x] = [-(n+f)] = -(n+1)$. Then $|[x]| = |-(n+1)| = n+1$. The inequality becomes $n \leq n+1$,which is true.
Since the inequality holds for all $x \in R$,the solution set is $R$.

(C) Given functions are $f(x) = 2x + 3$ and $g(x) = x^2 + 7$.
We need to find $x$ such that $g(f(x)) = 8$.
First,calculate the composite function $g(f(x))$:
$g(f(x)) = g(2x + 3) = (2x + 3)^2 + 7$.
Set this equal to $8$:
$(2x + 3)^2 + 7 = 8$.
Subtract $7$ from both sides:
$(2x + 3)^2 = 1$.
Taking the square root of both sides:
$2x + 3 = 1$ or $2x + 3 = -1$.
Case $1$: $2x + 3 = 1 \Rightarrow 2x = -2 \Rightarrow x = -1$.
Case $2$: $2x + 3 = -1 \Rightarrow 2x = -4 \Rightarrow x = -2$.
Thus,the values of $x$ are $-1$ and $-2$.

(A) Given the curves $y=\sin x$ and $y=\cos x$.
To find the intersection point,set $\sin x = \cos x$,which implies $\tan x = 1$.
Thus,$x = \frac{\pi}{4}$.
Now,find the slopes of the tangents at $x = \frac{\pi}{4}$.
For $y = \sin x$,the slope $m_1 = \frac{dy}{dx} = \cos x$. At $x = \frac{\pi}{4}$,$m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
For $y = \cos x$,the slope $m_2 = \frac{dy}{dx} = -\sin x$. At $x = \frac{\pi}{4}$,$m_2 = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting the values: $\tan \theta = |\frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})}| = |\frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}}| = |\frac{\sqrt{2}}{\frac{1}{2}}| = 2\sqrt{2}$.
Therefore,$\theta = \tan^{-1}(2\sqrt{2})$.

(C) Let $y = 2x^2 + x - 1$.
To find the minimum value,we find the derivative $y' = 4x + 1$.
Setting $y' = 0$ for critical points,we get $4x + 1 = 0$,which implies $x = -\frac{1}{4}$.
The second derivative is $y'' = 4$,which is positive $(> 0)$,confirming that the function has a minimum at $x = -\frac{1}{4}$.
Substituting $x = -\frac{1}{4}$ into the original expression:
$y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) - 1$
$y = 2(\frac{1}{16}) - \frac{1}{4} - 1$
$y = \frac{1}{8} - \frac{2}{8} - \frac{8}{8} = -\frac{9}{8}$.
Thus,the minimum value is $-\frac{9}{8}$.

(A) Let the two numbers be $x$ and $y$.
Given that $x + y = 20$,so $y = 20 - x$.
Let the product be $P = x^2 y^3$.
Substituting $y$,we get $P(x) = x^2(20 - x)^3$.
To find the maximum,differentiate $P$ with respect to $x$:
$\frac{dP}{dx} = 2x(20 - x)^3 + x^2 \cdot 3(20 - x)^2(-1) = x(20 - x)^2 [2(20 - x) - 3x] = x(20 - x)^2 (40 - 5x)$.
Setting $\frac{dP}{dx} = 0$,we get $x = 0$,$x = 20$,or $x = 8$.
Since $x$ and $y$ must be positive,we test $x = 8$.
If $x = 8$,then $y = 20 - 8 = 12$.
Thus,the numbers are $8$ and $12$.

(D) Let $I = \int (1 + x - x^{-1}) e^{x + x^{-1}} dx$.
We can rewrite the integral as:
$I = \int e^{x + x^{-1}} dx + \int (x - x^{-1}) e^{x + x^{-1}} dx$.
Using integration by parts on the first term $\int e^{x + x^{-1}} dx$:
Let $u = x$ and $dv = e^{x + x^{-1}} dx$. This is not straightforward.
Instead,consider the derivative of $x e^{x + x^{-1}}$:
$\frac{d}{dx} (x e^{x + x^{-1}}) = 1 \cdot e^{x + x^{-1}} + x \cdot e^{x + x^{-1}} \cdot (1 - x^{-2}) = e^{x + x^{-1}} + x e^{x + x^{-1}} - x^{-1} e^{x + x^{-1}} = (1 + x - x^{-1}) e^{x + x^{-1}}$.
Therefore,$\int (1 + x - x^{-1}) e^{x + x^{-1}} dx = x e^{x + x^{-1}} + C$.

(A) We need to evaluate the integral $I = \int_0^3 \frac{3x+1}{x^2+9} dx$.
Split the integral into two parts:
$I = \int_0^3 \frac{3x}{x^2+9} dx + \int_0^3 \frac{1}{x^2+9} dx$
For the first part,let $u = x^2+9$,then $du = 2x dx$,so $x dx = \frac{du}{2}$.
$\int_0^3 \frac{3x}{x^2+9} dx = \frac{3}{2} \int_9^{18} \frac{du}{u} = \frac{3}{2} [\log |u|]_9^{18} = \frac{3}{2} (\log 18 - \log 9) = \frac{3}{2} \log 2 = \log (2^{3/2}) = \log (2\sqrt{2})$.
For the second part,use the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$\int_0^3 \frac{1}{x^2+3^2} dx = [\frac{1}{3} \tan^{-1}(\frac{x}{3})]_0^3 = \frac{1}{3} (\tan^{-1}(1) - \tan^{-1}(0)) = \frac{1}{3} (\frac{\pi}{4} - 0) = \frac{\pi}{12}$.
Adding both parts,we get $I = \log (2\sqrt{2}) + \frac{\pi}{12}$.

(B) Let $I = \int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$.
Put $x = \cos \theta$,then $d x = -\sin \theta d \theta$.
When $x = 0$,$\theta = \frac{\pi}{2}$ and when $x = 1$,$\theta = 0$.
Substituting these into the integral:
$I = \int_{\pi/2}^0 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right) (-\sin \theta) d \theta$.
Using the identity $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \cot(\theta/2) = \tan(\frac{\pi}{2} - \frac{\theta}{2})$.
$I = \int_0^{\pi/2} \sin \left(2 \tan ^{-1} \tan \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) \sin \theta d \theta$.
$I = \int_0^{\pi/2} \sin \left(\pi - \theta\right) \sin \theta d \theta = \int_0^{\pi/2} \sin^2 \theta d \theta$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$I = \int_0^{\pi/2} \frac{1 - \cos 2\theta}{2} d \theta = \frac{1}{2} \left[ \theta - \frac{\sin 2\theta}{2} \right]_0^{\pi/2}$.
$I = \frac{1}{2} \left[ (\frac{\pi}{2} - 0) - (0 - 0) \right] = \frac{\pi}{4}$.

(A) Let $f(x) = \int_0^x t e^{t^2} dt$.
To evaluate the integral,substitute $u = t^2$,so $du = 2t dt$,which implies $t dt = \frac{1}{2} du$.
When $t = 0$,$u = 0$. When $t = x$,$u = x^2$.
Thus,$f(x) = \int_0^{x^2} \frac{1}{2} e^u du = \frac{1}{2} [e^u]_0^{x^2} = \frac{1}{2} (e^{x^2} - 1)$.
To find the minimum,differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{1}{2} (e^{x^2} \cdot 2x) = x e^{x^2}$.
Setting $f'(x) = 0$ gives $x = 0$.
Now,find the second derivative:
$f''(x) = e^{x^2} + x(e^{x^2} \cdot 2x) = e^{x^2} (1 + 2x^2)$.
At $x = 0$,$f''(0) = e^0 (1 + 0) = 1$.
Since $f''(0) > 0$,the function has a local minimum at $x = 0$.
The minimum value is $f(0) = \frac{1}{2} (e^0 - 1) = \frac{1}{2} (1 - 1) = 0$.

(D) We have $\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} |[x]| \, dx + \int_{-1}^0 |[x]| \, dx + \int_0^1 |[x]| \, dx + \int_1^2 |[x]| \, dx$.
Since $[x]$ is the greatest integer function:
For $x \in [-2, -1)$,$[x] = -2$,so $|[x]| = |-2| = 2$.
For $x \in [-1, 0)$,$[x] = -1$,so $|[x]| = |-1| = 1$.
For $x \in [0, 1)$,$[x] = 0$,so $|[x]| = |0| = 0$.
For $x \in [1, 2)$,$[x] = 1$,so $|[x]| = |1| = 1$.
Substituting these values:
$\int_{-2}^2 |[x]| \, dx = \int_{-2}^{-1} 2 \, dx + \int_{-1}^0 1 \, dx + \int_0^1 0 \, dx + \int_1^2 1 \, dx$.
$= 2[x]_{-2}^{-1} + [x]_{-1}^0 + 0 + [x]_1^2$.
$= 2(-1 - (-2)) + (0 - (-1)) + (2 - 1)$.
$= 2(1) + 1 + 1 = 4$.

(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is given by $(x+a)^2 + y^2 = (x+2a)^2$,which simplifies to $y^2 = 4a(x+a)$. However,for a parabola with focus at the origin and axis as the $X$-axis,the equation is $y^2 = 4a(x+a)$.
Let the parabola be $y^2 = 2a(x + a/2)$,or more simply $y^2 = 2ax + a^2$ where $a$ is the parameter.
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = 2a \Rightarrow a = y \frac{dy}{dx}$.
Substituting $a$ into the original equation $y^2 = 2ax + a^2$:
$y^2 = 2x(y \frac{dy}{dx}) + (y \frac{dy}{dx})^2$.
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}\right)^2$.
Rearranging the terms:
$y \left(\frac{dy}{dx}\right)^2 = y - 2x \frac{dy}{dx}$,which is equivalent to $-y \left(\frac{dy}{dx}\right)^2 = 2x \frac{dy}{dx} - y$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$.
By separating the variables,we get: $(\sin y + y \cos y) dy = (x \log x^2 + x) dx$.
Integrating both sides: $\int (\sin y + y \cos y) dy = \int (x \log x^2 + x) dx$.
For the left side,using integration by parts on $\int y \cos y dy$: $\int y \cos y dy = y \sin y - \int \sin y dy = y \sin y + \cos y$.
Thus,$\int (\sin y + y \cos y) dy = -\cos y + y \sin y + \cos y = y \sin y$.
For the right side,$\int (x \log x^2 + x) dx = \int (2x \log x + x) dx$.
Using integration by parts for $\int 2x \log x dx$: Let $u = \log x$,$dv = 2x dx$,then $du = \frac{1}{x} dx$,$v = x^2$.
$\int 2x \log x dx = x^2 \log x - \int x^2 \cdot \frac{1}{x} dx = x^2 \log x - \int x dx = x^2 \log x - \frac{x^2}{2}$.
Adding the integral of $x$: $\int (2x \log x + x) dx = x^2 \log x - \frac{x^2}{2} + \frac{x^2}{2} + C = x^2 \log x + C$.
Equating both sides: $y \sin y = x^2 \log x + C$.

(A) Given that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$.
Taking the dot product of both sides with themselves:
$\overrightarrow{a} \cdot \overrightarrow{a} = (\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{b}+\overrightarrow{c})$
$|\overrightarrow{a}|^2 = |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{b} \cdot \overrightarrow{c})$
Since the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$,their dot product $\overrightarrow{b} \cdot \overrightarrow{c} = |\overrightarrow{b}| |\overrightarrow{c}| \cos(\frac{\pi}{2}) = 0$.
Therefore,$a^2 = b^2 + c^2 + 0$
$a^2 = b^2 + c^2$.

(B) The given vector equation is $\overrightarrow{r}=(1-p-q) \overrightarrow{a}+p \overrightarrow{b}+q \overrightarrow{c}$.
We can rewrite this as $\overrightarrow{r} = \overrightarrow{a} - p\overrightarrow{a} - q\overrightarrow{a} + p\overrightarrow{b} + q\overrightarrow{c}$.
Rearranging the terms,we get $\overrightarrow{r} - \overrightarrow{a} = p(\overrightarrow{b} - \overrightarrow{a}) + q(\overrightarrow{c} - \overrightarrow{a})$.
This is the parametric form of the equation of a plane passing through the point with position vector $\overrightarrow{a}$ and parallel to the vectors $(\overrightarrow{b} - \overrightarrow{a})$ and $(\overrightarrow{c} - \overrightarrow{a})$.
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-coplanar,the vectors $(\overrightarrow{b} - \overrightarrow{a})$ and $(\overrightarrow{c} - \overrightarrow{a})$ are linearly independent,thus defining a unique plane.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $D, E, F$ are mid-points of $AB, AC, BC$ respectively,their position vectors are:
$\vec{d} = \frac{\vec{a} + \vec{b}}{2}$,$\vec{e} = \frac{\vec{a} + \vec{c}}{2}$,$\vec{f} = \frac{\vec{b} + \vec{c}}{2}$.
Now,$\overrightarrow{BE} = \vec{e} - \vec{b} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$.
And $\overrightarrow{AF} = \vec{f} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.
Adding these two vectors:
$\overrightarrow{BE} + \overrightarrow{AF} = \frac{\vec{a} + \vec{c} - 2\vec{b} + \vec{b} + \vec{c} - 2\vec{a}}{2} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$.
Since $\vec{d} = \frac{\vec{a} + \vec{b}}{2}$,we have $\overrightarrow{BE} + \overrightarrow{AF} = \vec{c} - \vec{d} = \overrightarrow{DC}$.

(A) Given $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$.
First,calculate the cross product $\overrightarrow{a} \times \overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0-1) + \hat{k}(1-1) = -\hat{i} + \hat{j}$.
Now,calculate $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}$:
$(-\hat{i} + \hat{j}) \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} = \hat{k}(0-1) = -\hat{k}$.
We are given $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = \lambda \overrightarrow{a} + \mu \overrightarrow{b}$.
Substituting the vectors: $-\hat{k} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j})$.
$-\hat{k} = (\lambda + \mu)\hat{i} + (\lambda + \mu)\hat{j} + \lambda\hat{k}$.
Comparing the coefficients on both sides:
For $\hat{i}$: $\lambda + \mu = 0$.
For $\hat{j}$: $\lambda + \mu = 0$.
For $\hat{k}$: $\lambda = -1$.
Since $\lambda + \mu = 0$,the value of $\lambda + \mu$ is $0$.

(B) The vector area of a triangle with vertices having position vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ is given by the formula:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{AB} \times \overrightarrow{AC})$
Since $\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a}$ and $\overrightarrow{AC} = \overrightarrow{c} - \overrightarrow{a}$,we have:
$\text{Vector Area} = \frac{1}{2} ((\overrightarrow{b} - \overrightarrow{a}) \times (\overrightarrow{c} - \overrightarrow{a}))$
Expanding the cross product:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{b} \times \overrightarrow{c} - \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{a} \times \overrightarrow{c} + \overrightarrow{a} \times \overrightarrow{a})$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,$-\overrightarrow{b} \times \overrightarrow{a} = \overrightarrow{a} \times \overrightarrow{b}$,and $-\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{c} \times \overrightarrow{a}$,we get:
$\text{Vector Area} = \frac{1}{2} (\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a})$

(B) Let the equation of the plane passing through $(1, 1, 1)$ be $a(x - 1) + b(y - 1) + c(z - 1) = 0$ $\dots (i)$.
Since it passes through $(1, -1, -1)$,we have $a(1 - 1) + b(-1 - 1) + c(-1 - 1) = 0$,which simplifies to $-2b - 2c = 0$,or $b + c = 0$ $\dots (ii)$.
The plane is perpendicular to $2x - y + z + 5 = 0$,so the normal vectors are perpendicular. Thus,$2a - b + c = 0$ $\dots (iii)$.
From $(ii)$,$c = -b$. Substituting into $(iii)$,we get $2a - b - b = 0$,which means $2a = 2b$,or $a = b$.
Let $a = 1$,then $b = 1$ and $c = -1$.
Substituting these into $(i)$,we get $1(x - 1) + 1(y - 1) - 1(z - 1) = 0$.
$x - 1 + y - 1 - z + 1 = 0$,which simplifies to $x + y - z - 1 = 0$.

(A) The intercept form of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $X, Y, Z$ axes respectively.
Given that the plane makes an intercept of $4$ on the $X$-axis $(a = 4)$ and an intercept of $3$ on the $Z$-axis $(c = 3)$.
Since the plane is parallel to the $Y$-axis,it does not intersect the $Y$-axis at any finite distance,which implies the intercept $b$ on the $Y$-axis is at infinity $(b \to \infty)$.
Therefore,the term $\frac{y}{b}$ becomes $\frac{y}{\infty} = 0$.
The equation of the plane becomes $\frac{x}{4} + \frac{z}{3} = 1$.
Multiplying the entire equation by $12$,we get $3x + 4z = 12$.

(B) The probability of getting a head in a single toss is $p = \frac{1}{2}$,and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
The probability of getting no heads in $n$ tosses is $P(\text{no heads}) = q^n = (\frac{1}{2})^n$.
The probability of getting at least one head is $P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - (\frac{1}{2})^n$.
Given that $1 - (\frac{1}{2})^n > 0.8$,we have:
$1 - 0.8 > (\frac{1}{2})^n$
$0.2 > (\frac{1}{2})^n$
$\frac{1}{5} > \frac{1}{2^n}$
$2^n > 5$.
For $n = 2$,$2^2 = 4 < 5$.
For $n = 3$,$2^3 = 8 > 5$.
Thus,the least value of $n$ is $3$.