AIEEE 2004 Mathematics Question Paper with Answer and Solution

(B) Given $|z^2 - 1| = |z|^2 + 1$.
Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$.
Substituting $z = x + iy$ into the equation:
$|(x + iy)^2 - 1| = x^2 + y^2 + 1$
$|(x^2 - y^2 - 1) + i(2xy)| = x^2 + y^2 + 1$
Squaring both sides:
$(x^2 - y^2 - 1)^2 + (2xy)^2 = (x^2 + y^2 + 1)^2$
$(x^2 - y^2 - 1)^2 + 4x^2y^2 = (x^2 + y^2 + 1)^2$
Expanding both sides:
$(x^4 + y^4 + 1 - 2x^2y^2 - 2x^2 + 2y^2) + 4x^2y^2 = x^4 + y^4 + 1 + 2x^2y^2 + 2x^2 + 2y^2$
$x^4 + y^4 + 1 + 2x^2y^2 - 2x^2 + 2y^2 = x^4 + y^4 + 1 + 2x^2y^2 + 2x^2 + 2y^2$
Subtracting common terms from both sides:
$-2x^2 = 2x^2$
$4x^2 = 0 \implies x = 0$.
Since $x = 0$,the complex number $z = 0 + iy = iy$ lies on the imaginary axis.

(C) Given that $\text{arg}(zw) = \pi$ $(i)$
$\overline{z} + i\overline{w} = 0$ $\Rightarrow \overline{z} = -i\overline{w}$ $\Rightarrow z = i w$ $\Rightarrow w = -iz$
Substitute $w = -iz$ into $(i)$:
$\text{arg}(z(-iz)) = \pi$
$\text{arg}(-iz^2) = \pi$
$\text{arg}(-i) + \text{arg}(z^2) = \pi$
$\text{arg}(-i) + 2\text{arg}(z) = \pi$
Since $\text{arg}(-i) = -\pi / 2$,we have:
$-\pi / 2 + 2\text{arg}(z) = \pi$
$2\text{arg}(z) = 3\pi / 2$
$\text{arg}(z) = 3\pi / 4$

(C) Given that $4$ is a root of the equation $x^2 + px + 12 = 0$.
Substituting $x = 4$ into the equation: $4^2 + p(4) + 12 = 0$.
$16 + 4p + 12 = 0$ $\Rightarrow 4p = -28$ $\Rightarrow p = -7$.
Now,the second equation is $x^2 + px + q = 0$,which becomes $x^2 - 7x + q = 0$.
Since the roots of this equation are equal,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
$(-7)^2 - 4(1)(q) = 0$.
$49 - 4q = 0$ $\Rightarrow 4q = 49$ $\Rightarrow q = \frac{49}{4}$.

(B) Let the two numbers be $x_1$ and $x_2$.
The arithmetic mean is given by $\frac{x_1 + x_2}{2} = 9$,which implies $x_1 + x_2 = 18$.
The geometric mean is given by $\sqrt{x_1 x_2} = 4$,which implies $x_1 x_2 = 4^2 = 16$.
$A$ quadratic equation with roots $x_1$ and $x_2$ is given by $x^2 - (x_1 + x_2)x + (x_1 x_2) = 0$.
Substituting the values,we get $x^2 - 18x + 16 = 0$.

(B) The expression is $(1 + x)(1 - x)^n = (1 - x)^n + x(1 - x)^n$.
We need to find the coefficient of $x^n$ in this expression.
The coefficient of $x^n$ in $(1 - x)^n$ is given by the general term $T_{r+1} = \binom{n}{r} (1)^{n-r} (-x)^r = \binom{n}{r} (-1)^r x^r$.
For $r = n$,the coefficient is $\binom{n}{n} (-1)^n = (-1)^n$.
The coefficient of $x^n$ in $x(1 - x)^n$ is the coefficient of $x^{n-1}$ in $(1 - x)^n$.
For $r = n-1$,the coefficient is $\binom{n}{n-1} (-1)^{n-1} = n (-1)^{n-1} = -n (-1)^n$.
Adding these,the total coefficient is $(-1)^n - n(-1)^n = (-1)^n(1 - n)$.

(D) We have ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $.
Using the property ${^n{C_r}} = {^n{C_{n - r}}}$,we can write ${t_n} = \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_{n - r}}}}} = \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_r}}}} $.
Thus,${t_n} = n \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} - \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $.
This simplifies to ${t_n} = n \cdot {S_n} - {t_n}$.
Adding ${t_n}$ to both sides,we get $2{t_n} = n \cdot {S_n}$.
Therefore,$\frac{{{t_n}}}{{{S_n}}} = \frac{n}{2}$.

(A) Let the height of the tree be $h$ and the breadth of the river be $b$.
From the given information:
In the first triangle,$\tan 60^\circ = \frac{h}{b} \Rightarrow h = b \tan 60^\circ = b\sqrt{3}$.
In the second triangle,$\tan 30^\circ = \frac{h}{b + 40} \Rightarrow h = (b + 40) \tan 30^\circ = \frac{b + 40}{\sqrt{3}}$.
Equating the two expressions for $h$:
$b\sqrt{3} = \frac{b + 40}{\sqrt{3}}$
$3b = b + 40$
$2b = 40$
$b = 20 \ m$.

(D) Let the third vertex $C$ be $(x, y)$.
The centroid $G$ of triangle $ABC$ is given by:
$G = \left( \frac{2 - 2 + x}{3}, \frac{-3 + 1 + y}{3} \right) = \left( \frac{x}{3}, \frac{y - 2}{3} \right)$.
Given that the centroid moves on the line $2x + 3y = 1$,we substitute the coordinates of $G$ into this equation:
$2\left( \frac{x}{3} \right) + 3\left( \frac{y - 2}{3} \right) = 1$.
Multiplying the entire equation by $3$,we get:
$2x + 3(y - 2) = 3$
$2x + 3y - 6 = 3$
$2x + 3y = 9$.
Thus,the locus of vertex $C$ is $2x + 3y = 9$.

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively.
Given that $a + b = -1$,so $b = -1 - a$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting $b$,we get $\frac{x}{a} + \frac{y}{-(1+a)} = 1$,which is $\frac{x}{a} - \frac{y}{1+a} = 1$.
Since the line passes through $(4, 3)$,we have $\frac{4}{a} - \frac{3}{1+a} = 1$.
Multiplying by $a(1+a)$,we get $4(1+a) - 3a = a(1+a)$.
$4 + 4a - 3a = a + a^2$.
$4 + a = a + a^2$,which simplifies to $a^2 = 4$,so $a = 2$ or $a = -2$.
If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$,or $\frac{x}{2} - \frac{y}{3} = 1$.
If $a = -2$,then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

(A) The given equation of the pair of lines is $6x^2 - xy + 4cy^2 = 0$ $(i)$.
Since $3x + 4y = 0$ is a line represented by this equation,its slope is $m_1 = -\frac{3}{4}$.
For a homogeneous equation $ax^2 + 2hxy + by^2 = 0$,the product of the slopes of the lines is $m_1 m_2 = \frac{a}{b}$.
Here,$a = 6$,$2h = -1$,and $b = 4c$. Thus,$m_1 m_2 = \frac{6}{4c} = \frac{3}{2c}$.
Substituting $m_1 = -\frac{3}{4}$,we get $(-\frac{3}{4}) m_2 = \frac{3}{2c}$,which implies $m_2 = -\frac{2}{c}$.
The sum of the slopes is $m_1 + m_2 = -\frac{2h}{b} = -\frac{-1}{4c} = \frac{1}{4c}$.
Substituting the values of $m_1$ and $m_2$: $-\frac{3}{4} - \frac{2}{c} = \frac{1}{4c}$.
Multiplying by $4c$,we get $-3c - 8 = 1$.
$-3c = 9$,so $c = -3$.

(C) The given equation of the pair of lines is ${x^2} - 2cxy - 7{y^2} = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = -2c$,and $b = -7$.
Let the slopes of the lines be $m_1$ and $m_2$.
The sum of the slopes is $m_1 + m_2 = -\frac{2h}{b} = -\frac{-2c}{-7} = -\frac{2c}{7}$.
The product of the slopes is $m_1m_2 = \frac{a}{b} = \frac{1}{-7} = -\frac{1}{7}$.
According to the problem,the sum of the slopes is four times their product:
$m_1 + m_2 = 4(m_1m_2)$
$-\frac{2c}{7} = 4 \times (-\frac{1}{7})$
$-\frac{2c}{7} = -\frac{4}{7}$
$2c = 4$
$c = 2$.

(D) The center of the circle is the intersection point of its diameters. Solving the system of equations:
$2x + 3y = -1$ $(i)$
$3x - y = 4$ (ii)
Multiplying (ii) by $3$: $9x - 3y = 12$ (iii)
Adding $(i)$ and (iii): $11x = 11 \Rightarrow x = 1$.
Substituting $x = 1$ into (ii): $3(1) - y = 4 \Rightarrow y = -1$.
Thus,the center $(h, k)$ is $(1, -1)$.
Given the circumference $2\pi r = 10\pi$,we find the radius $r = 5$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 5^2$
$x^2 - 2x + 1 + y^2 + 2y + 1 = 25$
$x^2 + y^2 - 2x + 2y - 23 = 0$.

(D) Given,the circle is ${x^2} + {y^2} - 2x = 0$ $(i)$ and the line is $y = x$ $(ii)$.
Substituting $y = x$ in $(i)$,we get:
${x^2} + {x^2} - 2x = 0$
$2{x^2} - 2x = 0$
$2x(x - 1) = 0$
So,$x = 0$ or $x = 1$.
Since $y = x$,the points of intersection are $A = (0, 0)$ and $B = (1, 1)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting $A(0, 0)$ and $B(1, 1)$:
$(x - 0)(x - 1) + (y - 0)(y - 1) = 0$
$x(x - 1) + y(y - 1) = 0$
${x^2} - x + {y^2} - y = 0$
${x^2} + {y^2} - x - y = 0$.

(B) Let the equation of the variable circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$ $... (i)$.
Since circle $(i)$ cuts the circle ${x^2} + {y^2} - 4 = 0$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ gives $2g(0) + 2f(0) = c - 4$,which implies $c = 4$.
Since the circle $(i)$ passes through the point $(a, b)$,we have ${a^2} + {b^2} + 2ga + 2fb + 4 = 0$.
To find the locus of the centre $(-g, -f)$,let $x = -g$ and $y = -f$,so $g = -x$ and $f = -y$.
Substituting these into the equation,we get ${a^2} + {b^2} + 2(-x)a + 2(-y)b + 4 = 0$.
Thus,the locus is $2ax + 2by - ({a^2} + {b^2} + 4) = 0$.

(D) The given parabolas are $y^2 = 4ax$ $(i)$ and $x^2 = 4ay$ $(ii)$.
Substituting $y = \frac{x^2}{4a}$ from $(ii)$ into $(i)$,we get $\frac{x^4}{16a^2} = 4ax$.
This simplifies to $x^4 - 64a^3x = 0$,or $x(x^3 - 64a^3) = 0$.
Thus,$x = 0$ or $x = 4a$.
For $x = 0$,$y = 0$. For $x = 4a$,$y = 4a$.
The points of intersection are $A(0, 0)$ and $B(4a, 4a)$.
The line $2bx + 3cy + 4d = 0$ passes through $(0, 0)$,so $2b(0) + 3c(0) + 4d = 0$,which implies $d = 0$.
The line also passes through $(4a, 4a)$,so $2b(4a) + 3c(4a) + 4(0) = 0$.
Dividing by $4a$ (since $a \ne 0$),we get $2b + 3c = 0$.
Since $d = 0$ and $2b + 3c = 0$,it follows that $d^2 + (2b + 3c)^2 = 0$.

(B) Given that the directrix is $x = 4$,which is parallel to the $y$-axis,the major axis of the ellipse lies along the $x$-axis.
Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.
The eccentricity $e$ is given as $\frac{1}{2}$.
Using the relation $e^2 = 1 - \frac{b^2}{a^2}$,we have $\frac{b^2}{a^2} = 1 - e^2 = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
The equation of the directrix is $x = \frac{a}{e}$. Given $x = 4$,we have $\frac{a}{e} = 4$.
Substituting $e = \frac{1}{2}$,we get $a = 4e = 4 \times \frac{1}{2} = 2$.
Now,$b^2 = \frac{3}{4}a^2 = \frac{3}{4} \times (2)^2 = \frac{3}{4} \times 4 = 3$.
Substituting $a^2 = 4$ and $b^2 = 3$ into the ellipse equation,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.

(B) We know that $\mathop {\lim }\limits_{x \to \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x \to \infty } f(x)g(x)}$ if $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$.
Given $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2}$.
Applying the formula,we get $e^{\mathop {\lim }\limits_{x \to \infty } (\frac{a}{x} + \frac{b}{{{x^2}}}) \cdot 2x} = {e^2}$.
This simplifies to $e^{\mathop {\lim }\limits_{x \to \infty } (2a + \frac{2b}{x})} = {e^2}$.
Evaluating the limit,we get $e^{2a + 0} = {e^2}$.
Thus,$e^{2a} = {e^2}$,which implies $2a = 2$,so $a = 1$.
Since the term $\frac{b}{x^2}$ vanishes as $x \to \infty$ regardless of the value of $b$,$b$ can be any real number.
Therefore,$a = 1$ and $b \in \mathbb{R}$.

(D) For the permutation ${}^{n}P_{r}$ to be defined,we must have $n \ge r \ge 0$ and $n, r \in \mathbb{Z}^+ \cup \{0\}$.
Here,$n = 7 - x$ and $r = x - 3$.
$1$. $n \ge r \implies 7 - x \ge x - 3 \implies 10 \ge 2x \implies x \le 5$.
$2$. $r \ge 0 \implies x - 3 \ge 0 \implies x \ge 3$.
$3$. $n \ge 0 \implies 7 - x \ge 0 \implies x \le 7$.
Combining these,the domain is $x \in \{3, 4, 5\}$.
Now,calculate $f(x)$ for each value in the domain:
For $x = 3$: $f(3) = {}^{7-3}P_{3-3} = {}^{4}P_{0} = 1$.
For $x = 4$: $f(4) = {}^{7-4}P_{4-3} = {}^{3}P_{1} = 3$.
For $x = 5$: $f(5) = {}^{7-5}P_{5-3} = {}^{2}P_{2} = 2$.
The range is the set of values ${f(3), f(4), f(5)} = \{1, 3, 2\}$,which is $\{1, 2, 3\}$.

(A) Given the equation of the parabola: $y^2 = 18x$.
Differentiating both sides with respect to $t$:
$2y \left( \frac{dy}{dt} \right) = 18 \left( \frac{dx}{dt} \right)$.
We are given that the ordinate $(y)$ increases at twice the rate of the abscissa $(x)$,so $\frac{dy}{dt} = 2 \frac{dx}{dt}$.
Substituting this into the differentiated equation:
$2y \left( 2 \frac{dx}{dt} \right) = 18 \left( \frac{dx}{dt} \right)$.
Assuming $\frac{dx}{dt} \neq 0$,we get:
$4y = 18 \implies y = \frac{18}{4} = \frac{9}{2}$.
Now,substitute $y = \frac{9}{2}$ into the original parabola equation to find $x$:
$\left( \frac{9}{2} \right)^2 = 18x$
$\frac{81}{4} = 18x$
$x = \frac{81}{4 \times 18} = \frac{81}{72} = \frac{9}{8}$.
Thus,the required point is $\left( \frac{9}{8}, \frac{9}{2} \right)$.

(C) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.
Given: $P(A) = \frac{4}{5}$ and $P(B) = \frac{3}{4}$.
Then,the probability that $A$ lies is $P(\bar{A}) = 1 - \frac{4}{5} = \frac{1}{5}$.
The probability that $B$ lies is $P(\bar{B}) = 1 - \frac{3}{4} = \frac{1}{4}$.
They contradict each other if ($A$ speaks the truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks the truth).
Required probability $= P(A) \times P(\bar{B}) + P(\bar{A}) \times P(B)$.
$= (\frac{4}{5} \times \frac{1}{4}) + (\frac{1}{5} \times \frac{3}{4})$.
$= \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$.

(D) Statement $(1)$ is correct: The mode of a grouped frequency distribution can be determined graphically using a histogram.
Statement $(2)$ is incorrect: Median is independent of the change of origin,but it is $NOT$ independent of the change of scale. If $y = a + bx$,then $Median(y) = a + b \times Median(x)$. Since it depends on $b$,it is not independent of the change of scale.
Statement $(3)$ is incorrect: Variance is independent of the change of origin,but it is $NOT$ independent of the change of scale. If $y = a + bx$,then $Var(y) = b^2 \times Var(x)$. Since it depends on $b^2$,it is not independent of the change of scale.
Therefore,only statement $(1)$ is correct.

(C) The total number of observations is $2n$.
There are $n$ observations equal to $a$ and $n$ observations equal to $-a$.
The mean $\bar{x} = \frac{n(a) + n(-a)}{2n} = \frac{0}{2n} = 0$.
The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{2n}}$.
Substituting the values: $2 = \sqrt{\frac{n(a - 0)^2 + n(-a - 0)^2}{2n}}$.
$2 = \sqrt{\frac{na^2 + na^2}{2n}} = \sqrt{\frac{2na^2}{2n}} = \sqrt{a^2} = |a|$.
Therefore,$|a| = 2$.

(A) Given $z = x - iy$ and $z^{1/3} = p + iq$.
Taking the cube on both sides,we get $z = (p + iq)^3$.
$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$.
$z = p^3 + 3ip^2q - 3pq^2 - iq^3$.
$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$.
Comparing the real and imaginary parts with $z = x - iy$,we have:
$x = p^3 - 3pq^2 = p(p^2 - 3q^2)$ and $-y = 3p^2q - q^3$,which implies $y = q^2 - 3p^2q = q(q^2 - 3p^2)$.
Now,$\frac{x}{p} = p^2 - 3q^2$ and $\frac{y}{q} = q^2 - 3p^2$.
Adding these two expressions:
$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$.
Therefore,$\frac{\frac{x}{p} + \frac{y}{q}}{p^2 + q^2} = \frac{-2(p^2 + q^2)}{p^2 + q^2} = -2$.

(C) The middle term in the expansion of $(1 + \alpha x)^4$ is the $3^{rd}$ term,given by $T_3 = ^4C_2 (\alpha x)^2 = 6 \alpha^2 x^2$. The coefficient is $6 \alpha^2$.
The middle term in the expansion of $(1 - \alpha x)^6$ is the $4^{th}$ term,given by $T_4 = ^6C_3 (-\alpha x)^3 = -20 \alpha^3 x^3$. The coefficient is $-20 \alpha^3$.
According to the problem,the coefficients are equal:
$6 \alpha^2 = -20 \alpha^3$
Since $\alpha \neq 0$ (otherwise the middle terms would be constants,not involving $x$ in the same way),we can divide by $2 \alpha^2$:
$3 = -10 \alpha$
$\alpha = -3/10$.

(C) Given $S(k) = 1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$.
For $k = 1$,$S(1) \Rightarrow 1 = 3 + 1^2 = 4$,which is false.
For $k = 2$,$S(2) \Rightarrow 1 + 3 = 3 + 2^2 = 7$,which is false.
Now,assume $S(k)$ is true,i.e.,$1 + 3 + 5 + \dots + (2k - 1) = 3 + k^2$.
Adding $(2(k + 1) - 1) = 2k + 1$ to both sides:
$1 + 3 + 5 + \dots + (2k - 1) + (2k + 1) = 3 + k^2 + 2k + 1 = 3 + (k + 1)^2$.
This shows that $S(k) \Rightarrow S(k + 1)$ is true,even though the base case $S(1)$ is false.

(B) We know the expansion of $\cosh(x)$ is given by $\frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots$
Setting $x = 1$,we get:
$\frac{e^1 + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$
$\frac{e + \frac{1}{e}}{2} = 1 + \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right)$
$\frac{e^2 + 1}{2e} = 1 + \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right)$
Therefore,the sum of the series is:
$\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots = \frac{e^2 + 1}{2e} - 1$
$= \frac{e^2 + 1 - 2e}{2e} = \frac{(e - 1)^2}{2e}$.

(D) Given $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$.
Squaring and adding both equations:
$(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$
$2 + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{441 + 729}{4225}$
$2 + 2\cos(\alpha - \beta) = \frac{1170}{4225} = \frac{18}{65}$
$2(1 + \cos(\alpha - \beta)) = \frac{18}{65}$
$4 \cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{65}$
$\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{260} = \frac{9}{130}$
$\cos\left(\frac{\alpha - \beta}{2}\right) = \pm \frac{3}{\sqrt{130}}$
Since $\pi < \alpha - \beta < 3\pi$,dividing by $2$ gives $\frac{\pi}{2} < \frac{\alpha - \beta}{2} < \frac{3\pi}{2}$.
In this interval,the cosine function is negative.
Therefore,$\cos\left(\frac{\alpha - \beta}{2}\right) = -\frac{3}{\sqrt{130}}$.

(C) Let the sides be $a = \sin \alpha$,$b = \cos \alpha$,and $c = \sqrt{1 + \sin \alpha \cos \alpha}$.
Since $0 < \alpha < \frac{\pi}{2}$,both $\sin \alpha$ and $\cos \alpha$ are positive,and $c$ is clearly the longest side because $c^2 = 1 + \sin \alpha \cos \alpha > \sin^2 \alpha + \cos^2 \alpha = 1$.
Using the Law of Cosines for the angle $C$ opposite to side $c$:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos C = \frac{\sin^2 \alpha + \cos^2 \alpha - (1 + \sin \alpha \cos \alpha)}{2 \sin \alpha \cos \alpha}$
$\cos C = \frac{1 - 1 - \sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha}$
$\cos C = \frac{-\sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha} = -\frac{1}{2}$
Since $\cos C = -\frac{1}{2}$,the angle $C = 120^\circ$.

(D) Let the other end of the diameter be $B(\alpha, \beta)$.
The circle has $AB$ as its diameter,so its equation is $(x - p)(x - \alpha) + (y - q)(y - \beta) = 0$.
Expanding this,we get $x^2 - (p + \alpha)x + p\alpha + y^2 - (q + \beta)y + q\beta = 0$,or $x^2 + y^2 - (p + \alpha)x - (q + \beta)y + (p\alpha + q\beta) = 0$.
Since the circle touches the $x$-axis,the $y$-coordinate of the center is equal to the radius,or the discriminant of the equation when $y=0$ is zero.
Setting $y = 0$,we get $x^2 - (p + \alpha)x + (p\alpha + q\beta) = 0$.
For the circle to touch the $x$-axis,the discriminant $D = 0$.
$D = (p + \alpha)^2 - 4(p\alpha + q\beta) = 0$.
$p^2 + 2p\alpha + \alpha^2 - 4p\alpha - 4q\beta = 0$.
$p^2 - 2p\alpha + \alpha^2 - 4q\beta = 0$.
$(p - \alpha)^2 = 4q\beta$.
Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $(x - p)^2 = 4qy$.

(A) Given $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $.
Squaring both sides,we get:
${u^2} = ({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta ) + ({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta ) + 2\sqrt {({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta )({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta )} $
${u^2} = {a^2} + {b^2} + 2\sqrt {({a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta )({a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta )} $
Let $t = {a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta $. Then ${a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta = {a^2} + {b^2} - t$.
So,${u^2} = {a^2} + {b^2} + 2\sqrt {t({a^2} + {b^2} - t)} = {a^2} + {b^2} + 2\sqrt { - {t^2} + ({a^2} + {b^2})t} $.
Let $f(t) = - {t^2} + ({a^2} + {b^2})t$. The maximum value of $f(t)$ occurs at $t = \frac{{{a^2} + {b^2}}}{2}$,which is $f\left( \frac{{{a^2} + {b^2}}}{2} \right) = \frac{{{({a^2} + {b^2})^2}}}{4}$.
Thus,${({u^2})_{\max }} = {a^2} + {b^2} + 2\sqrt {\frac{{{({a^2} + {b^2})^2}}}{4}} = {a^2} + {b^2} + ({a^2} + {b^2}) = 2({a^2} + {b^2})$.
The minimum value of $f(t)$ occurs at the boundaries of $t$,i.e.,$t = {a^2}$ or $t = {b^2}$.
At $t = {a^2}$,$f({a^2}) = - {a^4} + ({a^2} + {b^2}){a^2} = {a^2}{b^2}$.
Thus,${({u^2})_{\min }} = {a^2} + {b^2} + 2\sqrt {{a^2}{b^2}} = {a^2} + {b^2} + 2ab = {(a + b)^2}$.
The difference is ${({u^2})_{\max }} - {({u^2})_{\min }} = 2{a^2} + 2{b^2} - ({a^2} + {b^2} + 2ab) = {a^2} + {b^2} - 2ab = {(a - b)^2}$.

(C) Given $A = \{1, 2, 3, 4\}$ and $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$.
$1$. For $R$ to be symmetric,$(a, b) \in R$ must imply $(b, a) \in R$. Here,$(2, 3) \in R$ but $(3, 2) \notin R$. Hence,$R$ is not symmetric.
$2$. For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Since $(1, 1) \notin R$,$R$ is not reflexive.
$3$. For $R$ to be a function,each element in $A$ must have a unique image. Here,$2$ is mapped to both $4$ and $3$ (i.e.,$(2, 4) \in R$ and $(2, 3) \in R$),so $R$ is not a function.
$4$. For $R$ to be transitive,$(a, b) \in R$ and $(b, c) \in R$ must imply $(a, c) \in R$. Here,$(1, 3) \in R$ and $(3, 1) \in R$,but $(1, 1) \notin R$. Thus,$R$ is not transitive.
Therefore,the correct statement is that $R$ is not symmetric.

(A) Let $f(x) = ax^2 + bx + c$.
Define $F(x) = \int_{0}^{x} f(t) dt = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$.
Clearly,$F(0) = 0$.
Also,$F(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.
Given $2a + 3b + 6c = 0$,we have $F(1) = 0$.
Since $F(0) = F(1) = 0$ and $F(x)$ is a polynomial (and thus continuous and differentiable),by Rolle's Theorem,there exists at least one $x \in (0, 1)$ such that $F'(x) = 0$.
Since $F'(x) = f(x) = ax^2 + bx + c$,there exists at least one root of $ax^2 + bx + c = 0$ in the interval $(0, 1)$.

(A) Given $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
We evaluate the options:
$(i)$ Calculate $A^2$:
$A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Thus,option $A$ is correct.
$(ii)$ $(-1)I = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \neq A$.
$(iii)$ Calculate the determinant $|A|$:
$|A| = 0(0 - 0) - 0(0 - 0) - 1(0 - 1) = 1$.
Since $|A| \neq 0$,$A^{-1}$ exists.
$(iv)$ $A$ is clearly not a zero matrix as it contains non-zero elements.

(A) Given that $B = A^{-1}$,we have $10B = 10A^{-1}$.
Therefore,$10A^{-1} = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$.
Multiplying both sides by $A$ on the right,we get $10I = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$.
This results in $10I = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}$.
To find $\alpha$,we equate the element in the $2^{nd}$ row and $1^{st}$ column of the product matrix to the corresponding element in $10I$ (which is $0$):
$(-5 \times 1) + (0 \times 2) + (\alpha \times 1) = 0$.
$-5 + 0 + \alpha = 0$.
$\alpha = 5$.

(D) Given that $a + 2b$ is collinear with $c$,there exists a scalar $x$ such that $a + 2b = xc$.
Given that $b + 3c$ is collinear with $a$,there exists a scalar $y$ such that $b + 3c = ya$.
From the first equation,$a + 2b = xc$.
From the second equation,$b = ya - 3c$.
Substitute $b$ into the first equation: $a + 2(ya - 3c) = xc$.
$a + 2ya - 6c = xc$.
$(1 + 2y)a = (x + 6)c$.
Since $a$ and $c$ are non-zero and non-collinear,the coefficients must be zero:
$1 + 2y = 0 \implies y = -\frac{1}{2}$.
$x + 6 = 0 \implies x = -6$.
Substituting $x = -6$ into $a + 2b = xc$,we get $a + 2b = -6c$,which implies $a + 2b + 6c = 0$.

(C) The resultant force $\overrightarrow{F}$ is the sum of the individual forces:
$\overrightarrow{F} = (4i + j - 3k) + (3i + j - k) = (4+3)i + (1+1)j + (-3-1)k = 7i + 2j - 4k$.
The displacement vector $\overrightarrow{d}$ is the difference between the final position vector and the initial position vector:
$\overrightarrow{d} = (5i + 4j + k) - (i + 2j + 3k) = (5-1)i + (4-2)j + (1-3)k = 4i + 2j - 2k$.
The work done $W$ is the dot product of the force and displacement vectors:
$W = \overrightarrow{F} \cdot \overrightarrow{d} = (7i + 2j - 4k) \cdot (4i + 2j - 2k)$.
$W = (7 \times 4) + (2 \times 2) + (-4 \times -2) = 28 + 4 + 8 = 40 \text{ units}$.

(C) Since $a, b, c$ are non-coplanar vectors,their scalar triple product $[a, b, c] \neq 0$.
The vectors $a + 2b + 3c, \lambda b + 4c$,and $(2\lambda - 1)c$ are non-coplanar if and only if their scalar triple product is non-zero:
$[(a + 2b + 3c), (\lambda b + 4c), (2\lambda - 1)c] \neq 0$.
Using the properties of the scalar triple product,we can write this as:
$(a + 2b + 3c) \cdot [(\lambda b + 4c) \times (2\lambda - 1)c] \neq 0$
$(a + 2b + 3c) \cdot [\lambda(2\lambda - 1)(b \times c)] \neq 0$
Since $a \cdot (b \times c) = [a, b, c]$,$b \cdot (b \times c) = 0$,and $c \cdot (b \times c) = 0$,the expression simplifies to:
$\lambda(2\lambda - 1)[a, b, c] \neq 0$.
Given $[a, b, c] \neq 0$,the condition for non-coplanarity is $\lambda(2\lambda - 1) \neq 0$.
This implies $\lambda \neq 0$ and $\lambda \neq \frac{1}{2}$.
Thus,the vectors are non-coplanar for all values of $\lambda$ except $\lambda = 0$ and $\lambda = \frac{1}{2}$.

(A) Let the direction cosines of the line be $l, m, n$.
Since the line makes the same angle $\theta$ with the $x$ and $z$-axes,we have $l = \cos \theta$ and $n = \cos \theta$.
The angle with the $y$-axis is $\beta$,so $m = \cos \beta$.
The sum of the squares of the direction cosines is always $1$,so $l^2 + m^2 + n^2 = 1$.
Substituting the values,we get $\cos^2 \theta + \cos^2 \beta + \cos^2 \theta = 1$,which simplifies to $2 \cos^2 \theta + \cos^2 \beta = 1$.
Using the identity $\cos^2 \beta = 1 - \sin^2 \beta$,we have $2 \cos^2 \theta + (1 - \sin^2 \beta) = 1$,which implies $2 \cos^2 \theta = \sin^2 \beta$.
Given the condition $\sin^2 \beta = 3 \sin^2 \theta$,we substitute this into our equation: $2 \cos^2 \theta = 3 \sin^2 \theta$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get $2 \cos^2 \theta = 3(1 - \cos^2 \theta)$.
Expanding this,$2 \cos^2 \theta = 3 - 3 \cos^2 \theta$,which leads to $5 \cos^2 \theta = 3$.
Therefore,$\cos^2 \theta = \frac{3}{5}$.

(C) The given equations of the planes are $2x + y + 2z - 8 = 0$ and $4x + 2y + 4z + 5 = 0$.
To find the distance between parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$,we first make the coefficients of $x, y, z$ identical.
Multiply the first equation by $2$: $4x + 2y + 4z - 16 = 0$.
Now,the planes are $4x + 2y + 4z - 16 = 0$ and $4x + 2y + 4z + 5 = 0$.
The distance $d$ is given by the formula $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A = 4, B = 2, C = 4, D_1 = -16, D_2 = 5$.
$d = \frac{|-16 - 5|}{\sqrt{4^2 + 2^2 + 4^2}} = \frac{|-21|}{\sqrt{16 + 4 + 16}} = \frac{21}{\sqrt{36}} = \frac{21}{6} = \frac{7}{2}$.

(D) The given lines are:
Line $1$: $\frac{x - 1}{1} = \frac{y + 3}{-\lambda} = \frac{z - 1}{\lambda} = s$
Line $2$: $\frac{x - 0}{1/2} = \frac{y - 1}{1} = \frac{z - 2}{-1} = t$
The points on the lines are $P_1(1, -3, 1)$ and $P_2(0, 1, 2)$. The direction vectors are $\vec{v_1} = (1, -\lambda, \lambda)$ and $\vec{v_2} = (1/2, 1, -1)$.
Two lines are coplanar if the scalar triple product of the vector connecting the points and the two direction vectors is zero:
$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$
Substituting the values:
$\begin{vmatrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix} = 0$
Expanding the determinant:
$-1(\lambda - \lambda) - 4(-1 - \lambda/2) + 1(1 + \lambda/2) = 0$
$0 + 4 + 2\lambda + 1 + \lambda/2 = 0$
$5 + 5\lambda/2 = 0$
$5\lambda/2 = -5$
$\lambda = -2$.

(B) Let the two lines be $L_1: x = y + a = z = \lambda$ and $L_2: x + a = 2y = 2z = 2\mu$.
From $L_1$,any point $P$ is $(\lambda, \lambda - a, \lambda)$.
From $L_2$,any point $Q$ is $(2\mu - a, \mu, \mu)$.
The direction ratios of the line $PQ$ are $(2\mu - a - \lambda, \mu - \lambda + a, \mu - \lambda)$.
Since the line $PQ$ is proportional to $2, 1, 2$,we have:
$\frac{2\mu - a - \lambda}{2} = \frac{\mu - \lambda + a}{1} = \frac{\mu - \lambda}{2} = k$.
From $\frac{\mu - \lambda}{2} = k$,we get $\mu - \lambda = 2k$.
Substituting this into the second ratio: $2k + a = k \implies k = -a$.
Then $\mu - \lambda = -2a$.
From the first ratio: $\frac{2\mu - \lambda - a}{2} = -a \implies 2\mu - \lambda = a$.
Solving $\mu - \lambda = -2a$ and $2\mu - \lambda = a$,we get $\mu = 3a$ and $\lambda = 3a$.
For $P$,substituting $\lambda = 3a$: $P = (3a, 3a - a, 3a) = (3a, 2a, 3a)$.
For $Q$,substituting $\mu = a$ (Wait,re-solving: $2\mu - \lambda = a$ and $\mu - \lambda = -2a \implies \mu = 3a, \lambda = 5a$ is incorrect. Let's re-evaluate: $\frac{2\mu - a - \lambda}{2} = \frac{\mu - \lambda + a}{1} = \frac{\mu - \lambda}{2}$.
Let $\mu - \lambda = m$. Then $\frac{2\mu - a - \lambda}{2} = \frac{m+a}{1} = \frac{m}{2}$.
$m = 2m + 2a \implies m = -2a$.
$2\mu - a - \lambda = 2(m+a) = 2(-2a+a) = -2a \implies 2\mu - \lambda = -a$.
With $\mu - \lambda = -2a$ and $2\mu - \lambda = -a$,we get $\mu = a$ and $\lambda = 3a$.
Thus,$P = (3a, 2a, 3a)$ and $Q = (2(a) - a, a, a) = (a, a, a)$.

(B) function $y = f(x)$ is symmetrical about the line $x = a$ if $f(a + x) = f(a - x)$ for all $x$ in the domain.
Given that the graph is symmetrical about the line $x = 2$,we substitute $a = 2$ into the condition.
Therefore,$f(2 + x) = f(2 - x)$.

(A) The function is given by $f(x) = \sin x - \sqrt{3} \cos x + 1$.
We know that the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$a = 1$ and $b = -\sqrt{3}$.
Thus,the range of $\sin x - \sqrt{3} \cos x$ is $[-\sqrt{1^2 + (-\sqrt{3})^2}, \sqrt{1^2 + (-\sqrt{3})^2}] = [-\sqrt{1+3}, \sqrt{1+3}] = [-2, 2]$.
Adding $1$ to the entire inequality,we get:
$-2 + 1 \le \sin x - \sqrt{3} \cos x + 1 \le 2 + 1$.
$-1 \le f(x) \le 3$.
Since the function $f$ is onto,the codomain $S$ must be equal to the range of the function.
Therefore,$S = [-1, 3]$.

(B) To define the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$,we must satisfy two conditions:
$1$. The argument of the square root in the denominator must be strictly positive:
$9 - x^2 > 0$
$x^2 < 9$
$-3 < x < 3$ --- $(i)$
$2$. The argument of the inverse sine function must lie in the interval $[-1, 1]$:
$-1 \le x - 3 \le 1$
Adding $3$ to all parts:
$2 \le x \le 4$ --- $(ii)$
Taking the intersection of conditions $(i)$ and $(ii)$:
$x \in (-3, 3) \cap [2, 4]$
$x \in [2, 3)$
Thus,the domain of the function is $[2, 3)$.

(C) For $f(x)$ to be continuous at $x = \frac{\pi }{4}$,we must have $f(\frac{\pi }{4}) = \lim_{x \to \frac{\pi }{4}} f(x)$.
Evaluating the limit: $\lim_{x \to \frac{\pi }{4}} \frac{1 - \tan x}{4x - \pi }$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'$H$ôpital's rule:
$\lim_{x \to \frac{\pi }{4}} \frac{\frac{d}{dx}(1 - \tan x)}{\frac{d}{dx}(4x - \pi )} = \lim_{x \to \frac{\pi }{4}} \frac{-\sec^2 x}{4}$.
Substituting $x = \frac{\pi }{4}$:
$\frac{-\sec^2(\frac{\pi }{4})}{4} = \frac{-(\sqrt{2})^2}{4} = \frac{-2}{4} = -\frac{1}{2}$.
Therefore,$f(\frac{\pi }{4}) = -\frac{1}{2}$.

(C) Given the equation $x = e^{y + e^{y + \dots \infty}}$.
Since the exponent is an infinite series,we can write it as $x = e^{y + x}$.
Taking the natural logarithm on both sides,we get $\ln(x) = y + x$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(\ln(x)) = \frac{d}{dx}(y + x)$
$\frac{1}{x} = \frac{dy}{dx} + 1$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{1}{x} - 1 = \frac{1 - x}{x}$.

(D) Given the parametric equations of the curve: $x = a(1 + \cos \theta)$ and $y = a \sin \theta$.
First,we find the derivative $\frac{dy}{dx}$:
$\frac{dx}{d\theta} = -a \sin \theta$
$\frac{dy}{d\theta} = a \cos \theta$
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$.
The slope of the normal is the negative reciprocal of the slope of the tangent:
$m_n = -\frac{1}{dy/dx} = -\frac{1}{-\cot \theta} = \tan \theta$.
The equation of the normal at point $(a(1 + \cos \theta), a \sin \theta)$ is:
$y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta))$
$y - a \sin \theta = \frac{\sin \theta}{\cos \theta} (x - a - a \cos \theta)$
$y \cos \theta - a \sin \theta \cos \theta = x \sin \theta - a \sin \theta - a \sin \theta \cos \theta$
$y \cos \theta = x \sin \theta - a \sin \theta$
$y \cos \theta = \sin \theta (x - a)$.
If we substitute the point $(a, 0)$ into the equation:
$0 \cdot \cos \theta = \sin \theta (a - a)$
$0 = 0$.
Thus,the normal always passes through the fixed point $(a, 0)$.

(B) To evaluate the integral $I = \int \frac{\sin x}{\sin (x - \alpha )} dx$,we rewrite the numerator as $\sin(x - \alpha + \alpha)$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get:
$I = \int \frac{\sin(x - \alpha) \cos \alpha + \cos(x - \alpha) \sin \alpha}{\sin(x - \alpha)} dx$
$I = \int \left( \cos \alpha + \sin \alpha \cdot \cot(x - \alpha) \right) dx$
$I = \int \cos \alpha \, dx + \sin \alpha \int \cot(x - \alpha) \, dx$
Since $\int \cos \alpha \, dx = x \cos \alpha$ and $\int \cot(x - \alpha) \, dx = \log |\sin(x - \alpha)|$,we have:
$I = x \cos \alpha + \sin \alpha \log |\sin(x - \alpha)| + c$.

(A) Let $I = \int \frac{dx}{\cos x - \sin x}$.
Multiply and divide by $\sqrt{2}$:
$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}$
Using $\cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:
$I = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\cos(x + \frac{\pi}{4})}$
$I = \frac{1}{\sqrt{2}} \int \sec(x + \frac{\pi}{4}) dx$
Using the formula $\int \sec \theta d\theta = \log |\tan(\frac{\theta}{2} + \frac{\pi}{4})| + c$:
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x + \frac{\pi}{4}}{2} + \frac{\pi}{4} \right) \right| + c$
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} + \frac{\pi}{8} + \frac{2\pi}{8} \right) \right| + c$
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left( \frac{x}{2} + \frac{3\pi}{8} \right) \right| + c$.

(C) Given $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$.
We know that $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$.
Substituting this into the integral,we get:
$I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{(\sin x + \cos x)^2}} dx$.
Since $\sin x + \cos x > 0$ for $x \in [0, \pi/2]$,$\sqrt{(\sin x + \cos x)^2} = \sin x + \cos x$.
Thus,$I = \int_0^{\pi /2} (\sin x + \cos x) dx$.
Evaluating the integral:
$I = [-\cos x + \sin x]_0^{\pi /2}$.
$I = (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos(0) + \sin(0))$.
$I = (0 + 1) - (-1 + 0) = 1 + 1 = 2$.

(D) We need to evaluate the integral $I = \int_{-2}^{3} |1 - x^2| dx$.
The expression $|1 - x^2|$ changes sign at $x = -1$ and $x = 1$.
For $x \in [-2, -1]$,$1 - x^2 \le 0$,so $|1 - x^2| = x^2 - 1$.
For $x \in [-1, 1]$,$1 - x^2 \ge 0$,so $|1 - x^2| = 1 - x^2$.
For $x \in [1, 3]$,$1 - x^2 \le 0$,so $|1 - x^2| = x^2 - 1$.
Thus,$I = \int_{-2}^{-1} (x^2 - 1) dx + \int_{-1}^{1} (1 - x^2) dx + \int_{1}^{3} (x^2 - 1) dx$.
Evaluating each integral:
$\int_{-2}^{-1} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{-2}^{-1} = (-\frac{1}{3} + 1) - (-\frac{8}{3} + 2) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.
$\int_{-1}^{1} (1 - x^2) dx = [x - \frac{x^3}{3}]_{-1}^{1} = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$.
$\int_{1}^{3} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{1}^{3} = (9 - 3) - (\frac{1}{3} - 1) = 6 - (-\frac{2}{3}) = \frac{20}{3}$.
Summing these values: $I = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}$.

(B) Let $I = \int_0^\pi {xf(\sin x)dx}$.
Using the property $\int_0^a f(x)dx = \int_0^a f(a-x)dx$,we have:
$I = \int_0^\pi {(\pi - x)f(\sin(\pi - x))dx} = \int_0^\pi {(\pi - x)f(\sin x)dx}$.
Adding the two expressions for $I$:
$2I = \int_0^\pi {xf(\sin x)dx} + \int_0^\pi {(\pi - x)f(\sin x)dx} = \int_0^\pi {\pi f(\sin x)dx} = \pi \int_0^\pi {f(\sin x)dx}$.
Using the property $\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if $f(2a-x) = f(x)$,we have $\int_0^\pi {f(\sin x)dx} = 2\int_0^{\pi /2} {f(\sin x)dx}$.
Thus,$2I = \pi \times 2 \int_0^{\pi /2} {f(\sin x)dx} = 2\pi \int_0^{\pi /2} {f(\sin x)dx}$.
Therefore,$I = \pi \int_0^{\pi /2} {f(\sin x)dx}$.
Comparing this with $A \int_0^{\pi /2} {f(\sin x)dx}$,we get $A = \pi$.

(B) The given limit is of the form of a definite integral as a limit of a sum.
We know that $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}f\left( \frac{r}{n} \right) = \int_0^1 {f(x)dx} }$.
Here,$f\left( \frac{r}{n} \right) = e^{\frac{r}{n}}$,so $f(x) = e^x$.
Therefore,the expression becomes $\int_0^1 {e^x dx}$.
Evaluating the integral: $\int_0^1 {e^x dx} = [e^x]_0^1 = e^1 - e^0 = e - 1$.
Thus,the correct option is $B$.

(D) The required area is given by the integral $\int_{1}^{3} |x - 2| \, dx$.
Since $|x - 2| = -(x - 2)$ for $x < 2$ and $|x - 2| = (x - 2)$ for $x \geq 2$,we split the integral at $x = 2$:
$\text{Area} = \int_{1}^{2} -(x - 2) \, dx + \int_{2}^{3} (x - 2) \, dx$
$= \int_{1}^{2} (2 - x) \, dx + \int_{2}^{3} (x - 2) \, dx$
$= \left[ 2x - \frac{x^2}{2} \right]_{1}^{2} + \left[ \frac{x^2}{2} - 2x \right]_{2}^{3}$
$= \left( (4 - 2) - (2 - 0.5) \right) + \left( (4.5 - 6) - (2 - 4) \right)$
$= (2 - 1.5) + (-1.5 - (-2))$
$= 0.5 + 0.5 = 1$.
Thus,the area is $1$ square unit.

(D) Given $f(x) = \frac{e^x}{1 + e^x}$. Note that $f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1}$.
Thus,$f(a) + f(-a) = \frac{e^a}{1 + e^a} + \frac{1}{1 + e^a} = 1$.
Let $I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$. Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A + B - x) dx$,where $A + B = 1$:
$I_1 = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)(1 - (1 - x))\} dx = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)x\} dx$.
$I_1 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx - \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$.
$I_1 = I_2 - I_1$.
$2I_1 = I_2$.
Therefore,$\frac{I_2}{I_1} = 2$.

(A) Given the differential equation: $\frac{dy}{dx} = e^{x - y} + x^2 e^{-y}$
We can rewrite the right side as: $\frac{dy}{dx} = e^{-y}(e^x + x^2)$
Separating the variables $x$ and $y$,we get: $e^y dy = (e^x + x^2) dx$
Integrating both sides: $\int e^y dy = \int (e^x + x^2) dx$
Performing the integration: $e^y = e^x + \frac{x^3}{3} + c$
Thus,the correct option is $A$.

(B) Given $f''(x) = 6(x - 1)$.
Integrating with respect to $x$,we get $f'(x) = \int 6(x - 1) dx = 3(x - 1)^2 + c_1$.
Since the tangent at $(2, 1)$ is $y = 3x - 5$,the slope of the tangent at $x = 2$ is $f'(2) = 3$.
Substituting $x = 2$ into the expression for $f'(x)$,we get $3 = 3(2 - 1)^2 + c_1$,which implies $3 = 3 + c_1$,so $c_1 = 0$.
Thus,$f'(x) = 3(x - 1)^2$.
Integrating again with respect to $x$,we get $f(x) = \int 3(x - 1)^2 dx = (x - 1)^3 + c_2$.
Since the graph passes through $(2, 1)$,we have $f(2) = 1$.
Substituting $x = 2$ into the expression for $f(x)$,we get $1 = (2 - 1)^3 + c_2$,which implies $1 = 1 + c_2$,so $c_2 = 0$.
Therefore,the function is $f(x) = (x - 1)^3$.

(B) The event $E$ is defined as $X$ being a prime number. The prime numbers in the given set are $\{2, 3, 5, 7\}$.
$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.
The event $F$ is defined as $X < 4$. The values satisfying this are $\{1, 2, 3\}$.
$P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50$.
The intersection event $E \cap F$ consists of values that are both prime and less than $4$,which are $\{2, 3\}$.
$P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35$.
Using the addition rule for probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.
$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

(A) For a binomial distribution,the mean is given by $\mu = np = 4$ and the variance is given by $\sigma^2 = npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{4}$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n(\frac{1}{2}) = 4$,so $n = 8$.
The probability of $X$ successes is given by $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $k = 2$,we have $P(X = 2) = \binom{8}{2} (\frac{1}{2})^2 (\frac{1}{2})^{8-2} = \binom{8}{2} (\frac{1}{2})^8$.
Calculating the combination,$\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Thus,$P(X = 2) = 28 \times \frac{1}{256} = \frac{28}{256}$.

(D) Given that ${a_1}, {a_2}, {a_3}, \dots$ are in $G.P.$ with common ratio $r$.
Thus,${a_{n+1}} = {a_n} \cdot r$,which implies $\log {a_{n+1}} = \log {a_n} + \log r$.
Similarly,$\log {a_{n+k}} = \log {a_n} + k \log r$.
Let the determinant be $\Delta = \left| \begin{array}{ccc} \log {a_n} & \log {a_{n+1}} & \log {a_{n+2}} \\ \log {a_{n+3}} & \log {a_{n+4}} & \log {a_{n+5}} \\ \log {a_{n+6}} & \log {a_{n+7}} & \log {a_{n+8}} \end{array} \right|$.
Applying column operations ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_2}$:
$\Delta = \left| \begin{array}{ccc} \log {a_n} & \log {a_{n+1}} - \log {a_n} & \log {a_{n+2}} - \log {a_{n+1}} \\ \log {a_{n+3}} & \log {a_{n+4}} - \log {a_{n+3}} & \log {a_{n+5}} - \log {a_{n+4}} \\ \log {a_{n+6}} & \log {a_{n+7}} - \log {a_{n+6}} & \log {a_{n+8}} - \log {a_{n+7}} \end{array} \right|$.
Since $\log {a_{n+k}} - \log {a_{n+k-1}} = \log r$ for any $k$,the determinant becomes:
$\Delta = \left| \begin{array}{ccc} \log {a_n} & \log r & \log r \\ \log {a_{n+3}} & \log r & \log r \\ \log {a_{n+6}} & \log r & \log r \end{array} \right|$.
Since column $2$ and column $3$ are identical,the value of the determinant is $0$.

(C) Given $|u| = 1, |v| = 2, |w| = 3$. Since $v \perp w$,we have $v \cdot w = 0$.
Let the projection of $v$ along $u$ be equal to the projection of $w$ along $u$. This implies $\frac{v \cdot u}{|u|} = \frac{w \cdot u}{|u|}$.
Since $|u| = 1$,we have $v \cdot u = w \cdot u$,which means $(v - w) \cdot u = 0$.
We want to find $|u - v + w|$. Let $X = u - (v - w)$.
Then $|X|^2 = |u - (v - w)|^2 = |u|^2 + |v - w|^2 - 2u \cdot (v - w)$.
Since $u \cdot (v - w) = 0$,we have $|X|^2 = |u|^2 + |v - w|^2$.
$|v - w|^2 = |v|^2 + |w|^2 - 2(v \cdot w) = 2^2 + 3^2 - 2(0) = 4 + 9 = 13$.
Thus,$|u - v + w|^2 = 1^2 + 13 = 14$.
Therefore,$|u - v + w| = \sqrt{14}$.

(A) Given the vector triple product identity: $(a \times b) \times c = (a \cdot c)b - (b \cdot c)a$.
Substituting this into the given equation: $(a \cdot c)b - (b \cdot c)a = \frac{1}{3}|b||c|a$.
Rearranging the terms: $(a \cdot c)b = (b \cdot c + \frac{1}{3}|b||c|)a$.
Since $a$ and $b$ are non-zero and not parallel,the coefficients must be zero:
$a \cdot c = 0$ and $b \cdot c + \frac{1}{3}|b||c| = 0$.
Using the definition of the dot product $b \cdot c = |b||c| \cos \theta$,we get:
$|b||c| \cos \theta + \frac{1}{3}|b||c| = 0$.
Since $b$ and $c$ are non-zero,$\cos \theta = -\frac{1}{3}$.
However,the problem states $\theta$ is the acute angle between $b$ and $c$,implying $\cos \theta > 0$.
Re-evaluating the equation $(a \cdot c)b - (b \cdot c)a = \frac{1}{3}|b||c|a$,if $a$ and $b$ are linearly independent,then $a \cdot c = 0$ and $-(b \cdot c) = \frac{1}{3}|b||c|$.
Thus,$\cos \theta = -\frac{1}{3}$.
Given the constraint that $\theta$ is acute,there might be a sign convention in the problem statement. Assuming the magnitude of the cosine is $1/3$,$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (1/3)^2} = \sqrt{8/9} = \frac{2\sqrt{2}}{3}$.

(A) The equation of the plane passing through the intersection of two spheres ${S_1} = 0$ and ${S_2} = 0$ is given by ${S_1} - {S_2} = 0$.
Given the spheres:
${S_1}: {x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$
${S_2}: {x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$
Subtracting the two equations:
$({x^2} + {y^2} + {z^2} + 7x - 2y - z - 13) - ({x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8) = 0$
$(7x - (-3x)) + (-2y - 3y) + (-z - 4z) + (-13 - (-8)) = 0$
$10x - 5y - 5z - 5 = 0$
Dividing the entire equation by $5$:
$2x - y - z = 1$
Thus,the intersection of the two spheres lies on the plane $2x - y - z = 1$.

(C) Given the family of curves is ${x^2} + {y^2} - 2ay = 0$ ..... $(i)$
Differentiating both sides with respect to $x$,we get:
$2x + 2yy' - 2ay' = 0$
$2ay' = 2x + 2yy'$
$2a = \frac{2x + 2yy'}{y'} = \frac{2x}{y'} + 2y$ ..... $(ii)$
Substituting the value of $2a$ from equation $(ii)$ into equation $(i)$:
${x^2} + {y^2} - (\frac{2x}{y'} + 2y)y = 0$
${x^2} + {y^2} - \frac{2xy}{y'} - 2{y^2} = 0$
${x^2} - {y^2} - \frac{2xy}{y'} = 0$
Multiplying by $y'$,we get:
$({x^2} - {y^2})y' - 2xy = 0$
$({x^2} - {y^2})y' = 2xy$.