AIEEE 2003 Mathematics Question Paper with Answer and Solution

(D) Given the equation $|x|^2 - 3|x| + 2 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 - 3t + 2 = 0$.
Factoring the quadratic equation: $(t - 1)(t - 2) = 0$.
This gives $t = 1$ or $t = 2$.
Since $t = |x|$,we have $|x| = 1$ or $|x| = 2$.
For $|x| = 1$,the solutions are $x = 1$ and $x = -1$.
For $|x| = 2$,the solutions are $x = 2$ and $x = -2$.
Thus,the real solutions are $x \in \{1, -1, 2, -2\}$.
The total number of real solutions is $4$.

(C) Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$.
Then,$\alpha + \beta = -b/a$ and $\alpha\beta = c/a$.
According to the problem,$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$.
This implies $\alpha + \beta = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$.
Substituting the values,we get $-b/a = \frac{(-b/a)^2 - 2(c/a)}{(c/a)^2} = \frac{b^2/a^2 - 2c/a}{c^2/a^2} = \frac{b^2 - 2ac}{c^2}$.
Thus,$-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$.
Rearranging,$2a^2c = ab^2 + bc^2$.
Dividing both sides by $abc$,we get $\frac{2a}{b} = \frac{b}{c} + \frac{c}{a}$.
This shows that $\frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in $A.P.$
Therefore,their reciprocals $\frac{a}{c}, \frac{b}{a}, \frac{c}{b}$ are in $H.P.$

(A) Let the roots be $\alpha$ and $2\alpha$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + 2\alpha = 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3} = \frac{1 - 3a}{a^2 - 5a + 3} \implies \alpha = \frac{1 - 3a}{3(a^2 - 5a + 3)}$.
Product of roots: $\alpha \cdot 2\alpha = 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \implies \alpha^2 = \frac{1}{a^2 - 5a + 3}$.
Substituting $\alpha$ from the sum equation into the product equation:
$\left[\frac{1 - 3a}{3(a^2 - 5a + 3)}\right]^2 = \frac{1}{a^2 - 5a + 3}$.
$\frac{(1 - 3a)^2}{9(a^2 - 5a + 3)^2} = \frac{1}{a^2 - 5a + 3}$.
$(1 - 3a)^2 = 9(a^2 - 5a + 3)$.
$1 - 6a + 9a^2 = 9a^2 - 45a + 27$.
$39a = 26$.
$a = \frac{26}{39} = \frac{2}{3}$.

(A) First,arrange the $6$ men at a round table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$6$ men can be arranged in $(6-1)! = 5!$ ways.
After arranging the men,there are $6$ gaps created between them.
Since no two women should sit together,we need to place $5$ women in these $6$ gaps.
The number of ways to choose $5$ gaps out of $6$ is $^6C_5$,and the $5$ women can be arranged in these gaps in $5!$ ways.
Alternatively,we can place $5$ women in $6$ gaps in $P(6, 5)$ ways.
Total number of ways $= 5! \times P(6, 5) = 5! \times \frac{6!}{(6-5)!} = 5! \times 6!$.

(B) The student needs to select $10$ questions out of $13$ with the condition that at least $4$ must be chosen from the first $5$ questions.
Case $I$: Selecting $4$ questions from the first $5$ and $6$ questions from the remaining $8$.
Number of ways $= {^5C_4} \times {^8C_6} = 5 \times 28 = 140$.
Case $II$: Selecting $5$ questions from the first $5$ and $5$ questions from the remaining $8$.
Number of ways $= {^5C_5} \times {^8C_5} = 1 \times 56 = 56$.
Total number of choices $= 140 + 56 = 196$.

(C) The general term $T_{r+1}$ in the expansion of $(1+x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$.
For the term to be negative,the product $n(n-1)(n-2)\dots(n-r+1)$ must be negative since $x^r$ and $r!$ are positive for $x > 0$.
Here $n = \frac{27}{5} = 5.4$.
The terms become negative when the factor $(n-r+1) < 0$.
$5.4 - r + 1 < 0 \implies 6.4 < r$.
Since $r$ must be an integer,the smallest integer $r$ is $7$.
Thus,the first negative term is $T_{7+1} = T_8$,which is the $8^{th}$ term.

(B) The general term in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (\sqrt{3})^{256-r} (\sqrt[8]{5})^r$.
This simplifies to $T_{r+1} = {}^{256}C_r (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}}$.
For the term to be an integer,both exponents $\frac{256-r}{2}$ and $\frac{r}{8}$ must be non-negative integers.
Since $0 \leq r \leq 256$,for $\frac{r}{8}$ to be an integer,$r$ must be a multiple of $8$,i.e.,$r \in \{0, 8, 16, \dots, 256\}$.
For these values of $r$,$\frac{256-r}{2} = 128 - \frac{r}{2}$ is also an integer because $r$ is a multiple of $8$ (and thus a multiple of $2$).
The number of such values of $r$ is given by the arithmetic progression $0, 8, 16, \dots, 256$.
Using the formula $a_n = a + (n-1)d$,we have $256 = 0 + (n-1)8$,which gives $n-1 = 32$,so $n = 33$.
Thus,there are $33$ integral terms.

(A) We know the logarithmic series expansions:
${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$
However,the given series is $S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)}$.
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
So,$S = \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
$S = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{1}{4} - \frac{1}{5} \right) + \dots$
$S = 1 - 2\left( \frac{1}{2} \right) + 2\left( \frac{1}{3} \right) - 2\left( \frac{1}{4} \right) + \dots$
$S = 1 - 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots \right)$.
Since ${\log_e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,we have $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - {\log_e} 2$.
Substituting this into $S$:
$S = 1 - 2(1 - {\log_e} 2) = 1 - 2 + 2{\log_e} 2 = 2{\log_e} 2 - 1$.
$S = {\log_e} 4 - {\log_e} e = {\log_e} \left( \frac{4}{e} \right)$.

(A) Given: $a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$
Using the half-angle formula $\cos^2\frac{A}{2} = \frac{s(s-a)}{bc}$ and $\cos^2\frac{C}{2} = \frac{s(s-c)}{ab}$,where $s = \frac{a+b+c}{2}$ is the semi-perimeter:
$a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s-c + s-a) = \frac{3b}{2}$
Since $s-c+s-a = 2s - (a+c) = (a+b+c) - (a+c) = b$:
$\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$
Substituting $s = \frac{a+b+c}{2}$:
$\frac{a+b+c}{2} = \frac{3b}{2}$ $\Rightarrow a+b+c = 3b$ $\Rightarrow a+c = 2b$
Thus,$a, b, c$ are in $A.P.$

(B) For a regular $n$-sided polygon with side length $a$,the inradius $r$ and circumradius $R$ are given by:
$r = \frac{a}{2} \cot \left( \frac{\pi}{n} \right)$
$R = \frac{a}{2} \csc \left( \frac{\pi}{n} \right)$
Sum $= r + R = \frac{a}{2} \left( \cot \frac{\pi}{n} + \csc \frac{\pi}{n} \right)$
Using the identity $\cot \theta + \csc \theta = \cot \left( \frac{\theta}{2} \right)$:
Sum $= \frac{a}{2} \cot \left( \frac{\pi}{2n} \right)$.

(B) Let the centroid of the triangle be $(x, y)$.
Given vertices are $(x_1, y_1) = (a \cos t, a \sin t)$,$(x_2, y_2) = (b \sin t, -b \cos t)$,and $(x_3, y_3) = (1, 0)$.
The coordinates of the centroid are given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{y_1 + y_2 + y_3}{3}$.
Thus,$3x = a \cos t + b \sin t + 1$ and $3y = a \sin t - b \cos t$.
Rearranging,we get $3x - 1 = a \cos t + b \sin t$ and $3y = a \sin t - b \cos t$.
Squaring and adding both equations:
$(3x - 1)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$.
Expanding the right side:
$= a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$.
$= a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t) = a^2 + b^2$.
Therefore,the locus is $(3x - 1)^2 + (3y)^2 = a^2 + b^2$.

(A) Let the common ratio be $r$. Since ${x_1}, {x_2}, {x_3}$ are in $G$.$P$.,we have ${x_2} = {x_1}r$ and ${x_3} = {x_1}r^2$.
Similarly,since ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have ${y_2} = {y_1}r$ and ${y_3} = {y_1}r^2$.
The points are $P_1 = ({x_1}, {y_1})$,$P_2 = ({x_1}r, {y_1}r)$,and $P_3 = ({x_1}r^2, {y_1}r^2)$.
Notice that for all points,the ratio of the $y$-coordinate to the $x$-coordinate is constant: $\frac{{y_1}}{{x_1}} = \frac{{y_1}r}{{x_1}r} = \frac{{y_1}r^2}{{x_1}r^2} = m$ (where $m = \frac{{y_1}}{{x_1}}$).
This implies that all three points satisfy the linear equation $y = mx$,which represents a straight line passing through the origin.

(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
For the given equation $x^2 - 2pxy - y^2 = 0$,we have $a = 1, h = -p, b = -1$.
Substituting these values,the equation of the bisectors is $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-p}$.
$\frac{x^2 - y^2}{2} = \frac{xy}{-p}$ $\Rightarrow -p(x^2 - y^2) = 2xy$ $\Rightarrow -px^2 + py^2 = 2xy$ $\Rightarrow px^2 + 2xy - py^2 = 0$.
Comparing this with the given bisector equation $x^2 - 2qxy - y^2 = 0$,we have the ratio of coefficients:
$\frac{p}{1} = \frac{2}{-2q} = \frac{-p}{-1}$.
From $\frac{p}{1} = \frac{2}{-2q}$,we get $p = -\frac{1}{q}$,which implies $pq = -1$ or $pq + 1 = 0$.

(B) The center of the circle is the point of intersection of the diameters $2x - 3y = 5$ and $3x - 4y = 7$.
Solving these equations: $2x - 3y = 5$ (multiplied by $3$) $\Rightarrow 6x - 9y = 15$ and $3x - 4y = 7$ (multiplied by $2$) $\Rightarrow 6x - 8y = 14$.
Subtracting the first from the second: $(6x - 8y) - (6x - 9y) = 14 - 15 \Rightarrow y = -1$.
Substituting $y = -1$ into $2x - 3y = 5$: $2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow 2x = 2$ $\Rightarrow x = 1$.
So,the center $(h, k) = (1, -1)$.
Given area $= 154$,we have $\pi r^2 = 154$ $\Rightarrow \frac{22}{7} r^2 = 154$ $\Rightarrow r^2 = 154 \times \frac{7}{22} = 49$ $\Rightarrow r = 7$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 7^2$.
$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.
$x^2 + y^2 - 2x + 2y + 2 = 49$.
$x^2 + y^2 - 2x + 2y = 47$.

(A) The first circle is $(x - 1)^2 + (y - 3)^2 = r^2$,with center $C_1 = (1, 3)$ and radius $r_1 = r$.
The second circle is $x^2 + y^2 - 8x + 2y + 8 = 0$. Rewriting in standard form: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$,which is $(x - 4)^2 + (y + 1)^2 = 9 = 3^2$. Thus,center $C_2 = (4, -1)$ and radius $r_2 = 3$.
The distance between the centers $d = \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$.
For two circles to intersect at two distinct points,the condition is $|r_1 - r_2| < d < r_1 + r_2$.
$1$) $r_1 + r_2 > d$ $\Rightarrow r + 3 > 5$ $\Rightarrow r > 2$.
$2$) $|r_1 - r_2| < d$ $\Rightarrow |r - 3| < 5$ $\Rightarrow -5 < r - 3 < 5$ $\Rightarrow -2 < r < 8$. Since $r$ must be positive,$0 < r < 8$.
Combining $r > 2$ and $r < 8$,we get $2 < r < 8$.

(A) The equation of the normal to the parabola $y^2 = 4bx$ at the point $P(bt_1^2, 2bt_1)$ is given by $y + t_1x = 2bt_1 + bt_1^3$.
Since this normal meets the parabola again at $Q(bt_2^2, 2bt_2)$,the point $Q$ must satisfy the equation of the normal.
Substituting $x = bt_2^2$ and $y = 2bt_2$ into the normal equation:
$2bt_2 + t_1(bt_2^2) = 2bt_1 + bt_1^3$.
Dividing by $b$ (assuming $b \neq 0$):
$2t_2 + t_1t_2^2 = 2t_1 + t_1^3$.
Rearranging the terms:
$t_1(t_2^2 - t_1^2) + 2(t_2 - t_1) = 0$.
$t_1(t_2 - t_1)(t_2 + t_1) + 2(t_2 - t_1) = 0$.
Since $t_1 \neq t_2$ (as the points are distinct),we can divide by $(t_2 - t_1)$:
$t_1(t_2 + t_1) + 2 = 0$.
$t_1t_2 + t_1^2 + 2 = 0$.
$t_1t_2 = -t_1^2 - 2$.
$t_2 = -t_1 - \frac{2}{t_1}$.

(C) The given hyperbola is $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,which can be written as $\frac{x^2}{(12/5)^2} - \frac{y^2}{(9/5)^2} = 1$.
Here,$a^2 = \frac{144}{25}$ and $b^2 = \frac{81}{25}$.
The eccentricity $e_1$ of the hyperbola is given by $e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4}$.
The foci of the hyperbola are $(\pm a e_1, 0) = (\pm \frac{12}{5} \times \frac{5}{4}, 0) = (\pm 3, 0)$.
For the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$,the foci are $(\pm ae, 0)$,where $a^2 = 16$,so $a = 4$.
Since the foci coincide,$4e = 3$,which gives $e = \frac{3}{4}$.
For an ellipse,$b^2 = a^2(1 - e^2) = 16(1 - (\frac{3}{4})^2) = 16(1 - \frac{9}{16}) = 16(\frac{7}{16}) = 7$.

(C) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\log (3 + x) - \log (3 - x)}}{x} = k$.
Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(\log(3+x) - \log(3-x))}{\frac{d}{dx}(x)} = k$.
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{3+x} - (\frac{1}{3-x} \times -1)}{1} = k$.
$\mathop {\lim }\limits_{x \to 0} (\frac{1}{3+x} + \frac{1}{3-x}) = k$.
Substituting $x = 0$:
$\frac{1}{3+0} + \frac{1}{3-0} = k$.
$\frac{1}{3} + \frac{1}{3} = k$.
$k = \frac{2}{3}$.

(D) Total number of ways to select $2$ horses out of $5$ is given by $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
There is only $1$ winning horse. The number of ways to select $2$ horses such that the winning horse is included is equivalent to choosing the winning horse and one other horse from the remaining $4$ horses,which is $^4C_1 = 4$.
Therefore,the probability that $Mr. A$ selected the winning horse is $\frac{4}{10} = \frac{2}{5}$.

(A) Since $\frac{1 + 3p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2p}{2}$ are the probabilities of three events,each probability must lie in the interval $[0, 1]$.
$1$) $0 \le \frac{1 + 3p}{3} \le 1 \Rightarrow 0 \le 1 + 3p \le 3 \Rightarrow -1 \le 3p \le 2 \Rightarrow -\frac{1}{3} \le p \le \frac{2}{3}$
$2$) $0 \le \frac{1 - p}{4} \le 1 \Rightarrow 0 \le 1 - p \le 4 \Rightarrow -1 \le -p \le 3 \Rightarrow -3 \le p \le 1$
$3$) $0 \le \frac{1 - 2p}{2} \le 1 \Rightarrow 0 \le 1 - 2p \le 2 \Rightarrow -1 \le -2p \le 1 \Rightarrow -\frac{1}{2} \le p \le \frac{1}{2}$
For mutually exclusive events,the sum of probabilities must be $\le 1$:
$\frac{1 + 3p}{3} + \frac{1 - p}{4} + \frac{1 - 2p}{2} \le 1$
Multiply by $12$: $4(1 + 3p) + 3(1 - p) + 6(1 - 2p) \le 12$
$4 + 12p + 3 - 3p + 6 - 12p \le 12$
$13 - 3p \le 12 \Rightarrow -3p \le -1 \Rightarrow p \ge \frac{1}{3}$
Taking the intersection of all conditions:
$p \in [-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] \cap [\frac{1}{3}, \infty)$
Intersection is $\frac{1}{3} \le p \le \frac{1}{2}$.

(D) Given that the number of observations $n = 9$.
The median of $9$ observations is the $\left( \frac{9+1}{2} \right)^{th} = 5^{th}$ observation.
Let the ordered observations be $x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9$.
The median is $x_5 = 20.5$.
If the largest $4$ observations $(x_6, x_7, x_8, x_9)$ are increased by $2$,the new set of observations is $x_1, x_2, x_3, x_4, x_5, (x_6+2), (x_7+2), (x_8+2), (x_9+2)$.
Since $x_5$ is smaller than $x_6$,and $x_6 < x_6+2$,the order of the first $5$ observations remains unchanged.
Therefore,the $5^{th}$ observation remains $x_5$,which is $20.5$.
Thus,the median remains the same as that of the original set.

(A) Given the equation: ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1$.
First,simplify the term inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:
$\frac{{1 + i}}{{1 - i}} \times \frac{{1 + i}}{{1 + i}} = \frac{{(1 + i)^2}}{{1^2 - i^2}} = \frac{{1 + i^2 + 2i}}{{1 - (-1)}} = \frac{{1 - 1 + 2i}}{2} = \frac{{2i}}{2} = i$.
Substituting this back into the equation,we get: $i^x = 1$.
We know that $i^k = 1$ if and only if $k$ is a multiple of $4$.
Therefore,$x = 4n$,where $n$ is any positive integer.

(C) Given that ${z_1}$ and ${z_2}$ are the roots of the equation ${z^2 + az + b = 0}$.
By Vieta's formulas,we have ${z_1 + z_2 = -a}$ and ${z_1 z_2 = b}$.
Since the origin $({z_3 = 0})$,${z_1}$,and ${z_2}$ form an equilateral triangle,the condition for the vertices ${z_1, z_2, z_3}$ to form an equilateral triangle is ${z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1}$.
Substituting ${z_3 = 0}$,we get ${z_1^2 + z_2^2 = z_1 z_2}$.
Adding ${2 z_1 z_2}$ to both sides,we get ${z_1^2 + z_2^2 + 2 z_1 z_2 = 3 z_1 z_2}$.
This simplifies to ${(z_1 + z_2)^2 = 3 z_1 z_2}$.
Substituting the values from Vieta's formulas,we get ${(-a)^2 = 3b}$,which implies ${a^2 = 3b}$.

(B) Let the total height of the pole be $h$. The pole is divided into two parts: the upper part of length $\frac{3h}{4}$ and the lower part of length $\frac{h}{4}$.
Let the point on the ground be $C$ and the foot of the pole be $B$. Given $BC = 40 \ m$.
Let $\theta$ be the angle of elevation of the top of the pole $(A)$ and $\alpha$ be the angle of elevation of the point $D$ (where the pole is divided).
Then $\tan \theta = \frac{h}{40}$ and $\tan \alpha = \frac{h/4}{40} = \frac{h}{160}$.
The angle subtended by the upper part $AD$ at $C$ is $\beta = \theta - \alpha = \tan^{-1}\left(\frac{3}{5}\right)$.
Thus,$\tan \beta = \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha} = \frac{3}{5}$.
Substituting the values: $\frac{\frac{h}{40} - \frac{h}{160}}{1 + \left(\frac{h}{40}\right)\left(\frac{h}{160}\right)} = \frac{3}{5}$.
$\frac{\frac{3h}{160}}{1 + \frac{h^2}{6400}} = \frac{3}{5}$ $\Rightarrow \frac{3h}{160} \times \frac{6400}{6400 + h^2} = \frac{3}{5}$.
$\frac{120h}{6400 + h^2} = \frac{3}{5}$ $\Rightarrow 600h = 3(6400 + h^2)$ $\Rightarrow 200h = 6400 + h^2$.
$h^2 - 200h + 6400 = 0 \Rightarrow (h - 160)(h - 40) = 0$.
So,$h = 40 \ m$ or $h = 160 \ m$.

(B) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B$,and $C$.
Since the side $OA$ makes an angle $\alpha$ with the $x$-axis,its coordinates are $A(a \cos \alpha, a \sin \alpha)$.
The diagonal $OB$ makes an angle of $\alpha + \frac{\pi}{4}$ with the $x$-axis.
The slope of $OB$ is $\tan(\alpha + \frac{\pi}{4})$.
The diagonal $AC$ is perpendicular to $OB$.
The slope of $AC$ is $-\cot(\alpha + \frac{\pi}{4}) = -\frac{1}{\tan(\alpha + \frac{\pi}{4})} = -\frac{1 - \tan \alpha \tan(\frac{\pi}{4})}{\tan \alpha + \tan(\frac{\pi}{4})} = -\frac{1 - \tan \alpha}{1 + \tan \alpha} = \frac{\tan \alpha - 1}{\tan \alpha + 1} = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.
The equation of the line $AC$ passing through $A(a \cos \alpha, a \sin \alpha)$ is:
$y - a \sin \alpha = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} (x - a \cos \alpha)$
$y(\sin \alpha + \cos \alpha) - a \sin \alpha(\sin \alpha + \cos \alpha) = x(\sin \alpha - \cos \alpha) - a \cos \alpha(\sin \alpha - \cos \alpha)$
$y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a \sin^2 \alpha + a \sin \alpha \cos \alpha - a \sin \alpha \cos \alpha + a \cos^2 \alpha$
$y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a(\sin^2 \alpha + \cos^2 \alpha) = a$.
Thus,the equation is $y(\cos \alpha + \sin \alpha) - x(\sin \alpha - \cos \alpha) = a$.

(C) We know that $\frac{1 - \tan(x/2)}{1 + \tan(x/2)} = \tan(\frac{\pi}{4} - \frac{x}{2})$.
So,the limit becomes $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)\,(1 - \sin x)}}{{{{(\pi - 2x)}^3}}}$.
Let $x = \frac{\pi }{2} + y$,where $y \to 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + y) = -2y$.
Also,$1 - \sin x = 1 - \sin(\frac{\pi}{2} + y) = 1 - \cos y = 2\sin^2(\frac{y}{2})$.
Substituting these,we get $\mathop {\lim }\limits_{y \to 0} \frac{{\tan \left( {\frac{\pi}{4} - (\frac{\pi}{4} + \frac{y}{2})} \right)\,(2\sin^2 \frac{y}{2})}}{{{{( - 2y)}^3}}}$.
$= \mathop {\lim }\limits_{y \to 0} \frac{{-\tan(\frac{y}{2}) \cdot 2\sin^2(\frac{y}{2})}}{-8y^3} = \mathop {\lim }\limits_{y \to 0} \frac{1}{4} \cdot \frac{\tan(y/2)}{y} \cdot \left(\frac{\sin(y/2)}{y}\right)^2$.
$= \frac{1}{4} \cdot \frac{1}{2} \cdot (\frac{1}{2})^2 = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{32}$.

(A) Given the limit: $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$.
Since $f(1) = g(1) = k$,the expression becomes $\lim _{x \rightarrow 1} \frac{k g(x) - k - k f(x) + k}{g(x) - f(x)} = \lim _{x \rightarrow 1} \frac{k(g(x) - f(x))}{g(x) - f(x)}$.
As $x \rightarrow 1$,$g(x) - f(x) \rightarrow g(1) - f(1) = k - k = 0$,which is an indeterminate form of type $\frac{0}{0}$.
Applying $L'\text{Hospital's Rule}$: $\lim _{x \rightarrow 1} \frac{k g'(x) - k f'(x)}{g'(x) - f'(x)} = 4$.
$\lim _{x \rightarrow 1} \frac{k(g'(x) - f'(x))}{g'(x) - f'(x)} = 4$.
Since $g'(1) \neq f'(1)$ (as given that derivatives are not equal),we can cancel the term $(g'(x) - f'(x))$.
Thus,$k = 4$.

(C) Given $f(x) = x^n$.
We know that the $k$-th derivative of $f(x)$ at $x=1$ is $f^k(1) = n(n-1)(n-2)...(n-k+1) = \frac{n!}{(n-k)!}$.
Substituting this into the expression:
$f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!} = \sum_{k=0}^{n} (-1)^k \frac{f^k(1)}{k!}$.
Since $\frac{f^k(1)}{k!} = \frac{n!}{k!(n-k)!} = \binom{n}{k}$,the expression becomes:
$\sum_{k=0}^{n} (-1)^k \binom{n}{k} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ...... + (-1)^n \binom{n}{n}$.
Using the binomial theorem,$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$.
Setting $x=1$,we get $(1-1)^n = 0^n = 0$ for $n \ge 1$.
Thus,the value is $0$.

(D) Given $n = 15$,$\sum x^2 = 2830$,and $\sum x = 170$.
The incorrect observation is $20$ and the correct observation is $30$.
Corrected $\sum x = 170 - 20 + 30 = 180$.
Corrected $\sum x^2 = 2830 - (20)^2 + (30)^2 = 2830 - 400 + 900 = 3330$.
Corrected Mean $\bar{x} = \frac{180}{15} = 12$.
Corrected Variance $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 = \frac{3330}{15} - (12)^2$.
$\sigma^2 = 222 - 144 = 78$.

(D) Given that $|z \omega| = 1$,we have $|z| |\omega| = 1$.
Also,$\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,which implies $\operatorname{Arg}(\frac{z}{\omega}) = \frac{\pi}{2}$.
Let $z = r_1 e^{i \theta_1}$ and $\omega = r_2 e^{i \theta_2}$.
Then $|z| = r_1$ and $|\omega| = r_2$,so $r_1 r_2 = 1$.
$\operatorname{Arg}(z) = \theta_1$ and $\operatorname{Arg}(\omega) = \theta_2$,so $\theta_1 - \theta_2 = \frac{\pi}{2}$.
We need to find $\bar{z} \omega$.
$\bar{z} = r_1 e^{-i \theta_1}$.
$\bar{z} \omega = (r_1 e^{-i \theta_1}) (r_2 e^{i \theta_2}) = (r_1 r_2) e^{i(\theta_2 - \theta_1)}$.
Since $r_1 r_2 = 1$ and $\theta_2 - \theta_1 = -\frac{\pi}{2}$,we have:
$\bar{z} \omega = 1 \cdot e^{-i \frac{\pi}{2}} = \cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2}) = 0 - i = -i$.

(B) We use the identity ${}^n C_r + {}^n C_{r-1} = {}^{n+1} C_r$.
The given expression is ${}^n C_{r+1} + {}^n C_{r-1} + 2{}^n C_r$.
This can be rewritten as $({}^n C_{r+1} + {}^n C_r) + ({}^n C_r + {}^n C_{r-1})$.
Applying the identity,we get ${}^{n+1} C_{r+1} + {}^{n+1} C_r$.
Applying the identity again,we get ${}^{n+2} C_{r+1}$.
Hence,option $B$ is correct.

(A) Expanding the determinant along the first row:
$\Delta = 1(\omega^{2n} \cdot \omega^n - 1 \cdot 1) - \omega^n(\omega^n \cdot \omega^n - 1 \cdot \omega^{2n}) + \omega^{2n}(\omega^n \cdot 1 - \omega^{2n} \cdot \omega^{2n})$
$\Delta = (\omega^{3n} - 1) - \omega^n(\omega^{2n} - \omega^{2n}) + \omega^{2n}(\omega^n - \omega^{4n})$
Since $\omega^3 = 1$,we have $\omega^{3n} = 1$ and $\omega^{4n} = \omega^n$.
$\Delta = (1 - 1) - \omega^n(0) + \omega^{2n}(\omega^n - \omega^n)$
$\Delta = 0 - 0 + 0 = 0$.

(C) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.
$\Delta = \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Applying column operation $C_2 \to C_2 - 2C_3$:
$\begin{vmatrix} 1 & 0 & a \\ 1 & b & b \\ 1 & 2c & c \end{vmatrix} = 0$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\begin{vmatrix} 1 & 0 & a \\ 0 & b & b-a \\ 0 & 2c-b & c-b \end{vmatrix} = 0$
Expanding along the first column:
$1 \cdot [b(c - b) - (b - a)(2c - b)] = 0$
$bc - b^2 - (2bc - b^2 - 2ac + ab) = 0$
$bc - b^2 - 2bc + b^2 + 2ac - ab = 0$
$2ac - ab - bc = 0$
$2ac = ab + bc$
Dividing both sides by $abc$:
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This is the condition for $a, b, c$ to be in Harmonic Progression ($H$.$P$.).

(B) Given $A = \begin{bmatrix} a & b \\ b & a \end{bmatrix}$.
We need to find $A^2 = A \times A$.
$A^2 = \begin{bmatrix} a & b \\ b & a \end{bmatrix} \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ab + ba \\ ba + ab & b^2 + a^2 \end{bmatrix}$.
$A^2 = \begin{bmatrix} a^2 + b^2 & 2ab \\ 2ab & a^2 + b^2 \end{bmatrix}$.
Comparing this with $A^2 = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix}$,we get $\alpha = a^2 + b^2$ and $\beta = 2ab$.

(B) Given that $a + b + c = 0$.
Taking the dot product of the sum with itself: $(a + b + c) \cdot (a + b + c) = 0 \cdot 0 = 0$.
Expanding the dot product,we get: $|a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$.
Substitute the given magnitudes $|a| = 1, |b| = 2, |c| = 3$:
$(1)^2 + (2)^2 + (3)^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$.
$1 + 4 + 9 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$.
$14 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$.
$2(a \cdot b + b \cdot c + c \cdot a) = -14$.
$a \cdot b + b \cdot c + c \cdot a = -7$.

(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:
$W = \overrightarrow{F} \cdot \overrightarrow{d} = (7i + 2j - 4k) \cdot (4i + 2j - 2k)$.
$W = (7 \times 4) + (2 \times 2) + (-4 \times -2) = 28 + 4 + 8 = 40 \text{ units}$.

(B) We need to evaluate the scalar triple product $(u + v - w) \cdot [(u - v) \times (v - w)]$.
First,expand the cross product: $(u - v) \times (v - w) = u \times v - u \times w - v \times v + v \times w$.
Since $v \times v = 0$,this simplifies to $u \times v - u \times w + v \times w$.
Now,compute the dot product with $(u + v - w)$:
$(u + v - w) \cdot (u \times v - u \times w + v \times w) = u \cdot (u \times v) - u \cdot (u \times w) + u \cdot (v \times w) + v \cdot (u \times v) - v \cdot (u \times w) + v \cdot (v \times w) - w \cdot (u \times v) + w \cdot (u \times w) - w \cdot (v \times w)$.
Using the property that the scalar triple product is zero if any two vectors are identical,we have:
$u \cdot (u \times v) = 0, u \cdot (u \times w) = 0, v \cdot (u \times v) = 0, v \cdot (v \times w) = 0, w \cdot (u \times w) = 0, w \cdot (v \times w) = 0$.
Thus,the expression simplifies to:
$u \cdot (v \times w) - v \cdot (u \times w) - w \cdot (u \times v)$.
Using the cyclic property of scalar triple products $[u, v, w] = [v, w, u] = [w, u, v]$ and $[v, u, w] = -[u, v, w]$:
$= [u, v, w] - (-[u, v, w]) - [w, u, v] = [u, v, w] + [u, v, w] - [u, v, w] = [u, v, w] = u \cdot (v \times w)$.

(A) The angle between two faces of a tetrahedron is the angle between the normals to those faces.
First,find the normal vector $\vec{n_1}$ to the face $OAB$ using the cross product of vectors $\vec{OA}$ and $\vec{OB}$:
$\vec{OA} = \hat{i} + 2\hat{j} + \hat{k}$
$\vec{OB} = 2\hat{i} + \hat{j} + 3\hat{k}$
$\vec{n_1} = \vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.
Next,find the normal vector $\vec{n_2}$ to the face $ABC$ using the cross product of vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (2-1)\hat{i} + (1-2)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}$
$\vec{AC} = (-1-1)\hat{i} + (1-2)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$
$\vec{n_2} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1+2) - \hat{j}(1+4) + \hat{k}(-1-2) = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the faces is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$:
$\vec{n_1} \cdot \vec{n_2} = (5)(1) + (-1)(-5) + (-3)(-3) = 5 + 5 + 9 = 19$.
$|\vec{n_1}| = \sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$.
$|\vec{n_2}| = \sqrt{1^2 + (-5)^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.
$\cos \theta = \frac{19}{\sqrt{35} \cdot \sqrt{35}} = \frac{19}{35}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

(C) Two lines $\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1}$ and $\frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2}$ are coplanar if $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$.
Here,$(x_1, y_1, z_1) = (2, 3, 4)$ and $(x_2, y_2, z_2) = (1, 4, 5)$.
Thus,$x_2 - x_1 = -1$,$y_2 - y_1 = 1$,$z_2 - z_1 = 1$.
The condition becomes $\begin{vmatrix} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{vmatrix} = 0$.
Expanding the determinant:
$-1(1 + 2k) - 1(1 + k^2) + 1(2 - k) = 0$
$-1 - 2k - 1 - k^2 + 2 - k = 0$
$-k^2 - 3k = 0$
$k(k + 3) = 0$
Therefore,$k = 0$ or $k = -3$.

(D) Let the equation of the plane in the first system of rectangular axes be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
The perpendicular distance $p$ from the origin $(0, 0, 0)$ to this plane is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.
Similarly,for the second system of rectangular axes,the equation of the same plane is $\frac{x}{a'} + \frac{y}{b'} + \frac{z}{c'} = 1$.
The perpendicular distance $p$ from the origin to this plane is the same,so $p = \frac{|-1|}{\sqrt{\frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.
Equating the two expressions for $\frac{1}{p^2}$,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.
Therefore,$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$.

(C) The equation of the sphere is $x^2 + y^2 + z^2 + 4x - 2y - 6z - 155 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2gx + 2fy + 2hz + c = 0$,we get the center $C = (-g, -f, -h) = (-2, 1, 3)$.
The radius $r$ is given by $\sqrt{g^2 + f^2 + h^2 - c} = \sqrt{(-2)^2 + 1^2 + (-3)^2 - (-155)} = \sqrt{4 + 1 + 9 + 155} = \sqrt{169} = 13$.
The perpendicular distance $d$ from the center $C(-2, 1, 3)$ to the plane $12x + 4y + 3z - 327 = 0$ is calculated as:
$d = \frac{|12(-2) + 4(1) + 3(3) - 327|}{\sqrt{12^2 + 4^2 + 3^2}} = \frac{|-24 + 4 + 9 - 327|}{\sqrt{144 + 16 + 9}} = \frac{|-338|}{\sqrt{169}} = \frac{338}{13} = 26$.
The shortest distance from the plane to the sphere is $d - r = 26 - 13 = 13$.

(C) The equation of the sphere is $x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0$.
Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get the center $C = (-u, -v, -w) = (-1, 1, 2)$ and the radius $R = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1 + 1 + 4 + 19} = \sqrt{25} = 5$.
The perpendicular distance $p$ from the center $C(-1, 1, 2)$ to the plane $x + 2y + 2z + 7 = 0$ is given by $p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1 + 4 + 4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.
The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^2 - p^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.

(C) Given the function $f: \mathbb{N} \to \mathbb{Z}$ defined as:
$f(n) = \begin{cases} \frac{n-1}{2}, & \text{if } n \text{ is odd} \\ -\frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$
Let us calculate the images of the first few natural numbers:
$f(1) = \frac{1-1}{2} = 0$
$f(2) = -\frac{2}{2} = -1$
$f(3) = \frac{3-1}{2} = 1$
$f(4) = -\frac{4}{2} = -2$
$f(5) = \frac{5-1}{2} = 2$
$f(6) = -\frac{6}{2} = -3$
$1$. One-one check: For every distinct $n_1, n_2 \in \mathbb{N}$,we get distinct images in $\mathbb{Z}$. Since each input maps to a unique output,the function is one-one.
$2$. Onto check: The range of the function is $\{0, -1, 1, -2, 2, -3, 3, \dots\}$,which is the set of all integers $\mathbb{Z}$. Since the range equals the codomain,the function is onto.
Therefore,the function is both one-one and onto (bijective).

(D) For the function $f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x)$ to be defined:
$1$. The denominator must not be zero: $4 - x^2 \neq 0$ $\Rightarrow x^2 \neq 4$ $\Rightarrow x \neq \pm 2$.
$2$. The argument of the logarithm must be positive: $x^3 - x > 0$.
Factoring the inequality: $x(x^2 - 1) > 0 \Rightarrow x(x - 1)(x + 1) > 0$.
Using the wavy curve method (sign scheme) on the number line with critical points $-1, 0, 1$,the expression is positive in the intervals $(-1, 0) \cup (1, \infty)$.
Combining the conditions: $x \in (-1, 0) \cup (1, \infty)$ and $x \neq \pm 2$.
Since $2$ lies in the interval $(1, \infty)$,we must exclude it.
Thus,the domain is $D = (-1, 0) \cup (1, 2) \cup (2, \infty)$.

(B) To determine if the function $f(x) = \log (x + \sqrt {{x^2} + 1} )$ is even or odd,we evaluate $f(-x)$.
$f(-x) = \log (-x + \sqrt {(-x)^2 + 1} )$
$f(-x) = \log (-x + \sqrt {x^2 + 1} )$
We can rewrite the argument of the logarithm by multiplying and dividing by the conjugate $(\sqrt {x^2 + 1} + x)$:
$f(-x) = \log \left( \frac{(\sqrt {x^2 + 1} - x)(\sqrt {x^2 + 1} + x)}{\sqrt {x^2 + 1} + x} \right)$
$f(-x) = \log \left( \frac{(x^2 + 1) - x^2}{\sqrt {x^2 + 1} + x} \right)$
$f(-x) = \log \left( \frac{1}{\sqrt {x^2 + 1} + x} \right)$
Using the property $\log(1/a) = -\log(a)$:
$f(-x) = -\log (x + \sqrt {x^2 + 1} )$
$f(-x) = -f(x)$
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.

(D) Given the functional equation $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$.
Since $f(1) = 7$,we have $c(1) = 7$,so $c = 7$.
Thus,$f(x) = 7x$.
Now,we need to calculate the sum $\sum_{r = 1}^n f(r) = \sum_{r = 1}^n 7r$.
This simplifies to $7 \sum_{r = 1}^n r$.
Using the formula for the sum of the first $n$ natural numbers,$\sum_{r = 1}^n r = \frac{n(n + 1)}{2}$.
Therefore,$\sum_{r = 1}^n f(r) = 7 \times \frac{n(n + 1)}{2} = \frac{7n(n + 1)}{2}$.

(B) Given $f(x) = \begin{cases} x e^{-\left( \frac{1}{|x|} + \frac{1}{x} \right)}, & x \ne 0 \\ 0, & x = 0 \end{cases}$.
For continuity at $x = 0$:
$R.H.L. = \lim_{h \to 0^+} f(0+h) = \lim_{h \to 0^+} h e^{-\left( \frac{1}{h} + \frac{1}{h} \right)} = \lim_{h \to 0^+} h e^{-2/h} = 0$.
$L.H.L. = \lim_{h \to 0^+} f(0-h) = \lim_{h \to 0^+} (-h) e^{-\left( \frac{1}{h} - \frac{1}{h} \right)} = \lim_{h \to 0^+} (-h) e^0 = 0$.
Since $R.H.L. = L.H.L. = f(0) = 0$,$f(x)$ is continuous at $x = 0$.
For differentiability at $x = 0$:
$Rf'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h e^{-2/h} - 0}{h} = \lim_{h \to 0^+} e^{-2/h} = 0$.
$Lf'(0) = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{-h e^0 - 0}{-h} = 1$.
Since $Rf'(0) \ne Lf'(0)$,$f(x)$ is not differentiable at $x = 0$.

(B) Let the function be $f(x) = x + \frac{1}{x}$ for $x > 0$.
To find the minimum value,we find the derivative $f'(x) = 1 - \frac{1}{x^2}$.
Setting $f'(x) = 0$,we get $1 - \frac{1}{x^2} = 0$,which implies $x^2 = 1$. Since $x > 0$,we have $x = 1$.
Now,we check the second derivative $f''(x) = \frac{2}{x^3}$.
At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.
Since the second derivative is positive,the function $f(x)$ has a local minimum at $x = 1$.

(C) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
Find the first derivative: $f'(x) = 6x^2 - 18ax + 12a^2$.
Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0 \Rightarrow 6(x - a)(x - 2a) = 0$.
The critical points are $x = a$ and $x = 2a$.
Find the second derivative: $f''(x) = 12x - 18a$.
At $x = a$,$f''(a) = 12a - 18a = -6a < 0$ (since $a > 0$),so $x = a$ is a point of local maximum. Thus,$p = a$.
At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a > 0$,so $x = 2a$ is a point of local minimum. Thus,$q = 2a$.
Given the condition $p^2 = q$,we have $a^2 = 2a$.
Since $a > 0$,we can divide by $a$ to get $a = 2$.

(C) Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:
$I = \int_{0}^{1} (1 - x)(1 - (1 - x))^n dx$
$I = \int_{0}^{1} (1 - x)x^n dx$
$I = \int_{0}^{1} (x^n - x^{n+1}) dx$
$I = \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_{0}^{1}$
$I = \left( \frac{1}{n+1} - \frac{1}{n+2} \right) - (0 - 0)$
$I = \frac{1}{n+1} - \frac{1}{n+2}$

(B) Let $I = \int_a^b x f(x) dx$.
Using the property $\int_a^b g(x) dx = \int_a^b g(a + b - x) dx$,we get:
$I = \int_a^b (a + b - x) f(a + b - x) dx$.
Since it is given that $f(a + b - x) = f(x)$,we substitute this into the integral:
$I = \int_a^b (a + b - x) f(x) dx$.
$I = (a + b) \int_a^b f(x) dx - \int_a^b x f(x) dx$.
$I = (a + b) \int_a^b f(x) dx - I$.
$2I = (a + b) \int_a^b f(x) dx$.
$I = \frac{a + b}{2} \int_a^b f(x) dx$.

(D) Given that $\frac{d}{dx}F(x) = \frac{e^{\sin x}}{x}$.
We need to evaluate the integral $I = \int_{1}^{4} \frac{3}{x} e^{\sin(x^3)} dx$.
Multiply the numerator and denominator by $x^2$ to facilitate substitution:
$I = \int_{1}^{4} \frac{3x^2}{x^3} e^{\sin(x^3)} dx$.
Let $t = x^3$,then $dt = 3x^2 dx$.
When $x = 1$,$t = 1^3 = 1$.
When $x = 4$,$t = 4^3 = 64$.
Substituting these into the integral:
$I = \int_{1}^{64} \frac{e^{\sin t}}{t} dt$.
Since $\frac{d}{dt}F(t) = \frac{e^{\sin t}}{t}$,the integral becomes:
$I = [F(t)]_{1}^{64} = F(64) - F(1)$.
Comparing this with the given expression $F(k) - F(1)$,we get $k = 64$.

(D) Given $f'(x) = f(x)$,we have $\frac{f'(x)}{f(x)} = 1$.
Integrating both sides,we get $\ln|f(x)| = x + C$,which implies $f(x) = ce^x$.
Since $f(0) = 1$,we have $1 = ce^0$,so $c = 1$. Thus,$f(x) = e^x$.
Given $f(x) + g(x) = x^2$,we have $g(x) = x^2 - e^x$.
Now,we calculate the integral:
$\int_0^1 f(x)g(x) dx = \int_0^1 e^x(x^2 - e^x) dx = \int_0^1 (x^2 e^x - e^{2x}) dx$.
Using integration by parts for $\int x^2 e^x dx$:
$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(x e^x - e^x) = e^x(x^2 - 2x + 2)$.
Evaluating the definite integral:
$\int_0^1 x^2 e^x dx = [e^x(x^2 - 2x + 2)]_0^1 = e(1 - 2 + 2) - e^0(0 - 0 + 2) = e - 2$.
Evaluating $\int_0^1 e^{2x} dx$:
$\int_0^1 e^{2x} dx = [\frac{1}{2} e^{2x}]_0^1 = \frac{1}{2}(e^2 - 1)$.
Thus,the integral is $(e - 2) - \frac{1}{2}(e^2 - 1) = e - 2 - \frac{1}{2}e^2 + \frac{1}{2} = e - \frac{1}{2}e^2 - \frac{3}{2}$.

(D) We use the definition of the definite integral as the limit of a sum: $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$.
For the first term: $\mathop {\lim }\limits_{n \to \infty } \frac{\sum_{r=1}^{n} r^4}{n^5} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^4 = \int_0^1 x^4 dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5}$.
For the second term: $\mathop {\lim }\limits_{n \to \infty } \frac{\sum_{r=1}^{n} r^3}{n^5} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n^2} \left( \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^3 \right) = 0 \times \int_0^1 x^3 dx = 0 \times \frac{1}{4} = 0$.
Thus,the result is $\frac{1}{5} - 0 = \frac{1}{5}$.

(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{4}$,which simplifies to $q = \frac{1}{2}$.
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 mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
For $X = 1$,we have $P(X = 1) = \binom{8}{1} (\frac{1}{2})^1 (\frac{1}{2})^{8-1} = 8 \times (\frac{1}{2})^8 = 8 \times \frac{1}{256} = \frac{1}{32}$.

(A) We are given the determinant $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}$.
Expanding the determinant along the first row:
$\Delta = 1(\omega^n \cdot \omega^n - 1 \cdot 1) - \omega^n(\omega^n \cdot \omega^n - \omega^{2n} \cdot 1) + \omega^{2n}(\omega^n \cdot 1 - \omega^{2n} \cdot \omega^{2n})$
$\Delta = 1(\omega^{2n} - 1) - \omega^n(\omega^{2n} - \omega^{2n}) + \omega^{2n}(\omega^n - \omega^{4n})$
Since $\omega^3 = 1$,we have $\omega^{3n} = 1$ and $\omega^{4n} = \omega^n$.
$\Delta = (\omega^{2n} - 1) - 0 + \omega^{2n}(\omega^n - \omega^n)$
$\Delta = \omega^{2n} - 1 + 0 = \omega^{2n} - 1$.
However,if we look at the properties of the cube roots of unity,the sum of the columns $C_1 + C_2 + C_3$ gives:
$1 + \omega^n + \omega^{2n}$.
If $n$ is not a multiple of $3$,$1 + \omega^n + \omega^{2n} = 0$,making the determinant $0$.
If $n$ is a multiple of $3$,then $\omega^n = 1$,so the determinant becomes $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0$.
Thus,in all cases,$\Delta = 0$.

(A) Given that the vectors $(1, a, a^2)$,$(1, b, b^2)$,and $(1, c, c^2)$ are non-coplanar,the determinant of the matrix formed by these vectors is non-zero:
$\Delta = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| \neq 0$.
We are given the equation:
$\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0$.
Using the property of determinants,we can split the third column:
$\left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| = 0$.
In the second determinant,factor out $a, b, c$ from the rows:
$\left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| + abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = 0$.
Note that $\left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| = \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = \Delta$ (after two column swaps).
Thus,$\Delta + abc \Delta = 0 \Rightarrow \Delta(1 + abc) = 0$.
Since $\Delta \neq 0$,we have $1 + abc = 0$,which implies $abc = -1$.

(B) The given lines can be written in symmetric form as follows:
For the first line $x = ay + b$ and $z = cy + d$,we have $\frac{x - b}{a} = y = \frac{z - d}{c}$.
Thus,the direction ratios of the first line are $(a, 1, c)$.
For the second line $x = a'y + b'$ and $z = c'y + d'$,we have $\frac{x - b'}{a'} = y = \frac{z - d'}{c'}$.
Thus,the direction ratios of the second line are $(a', 1, c')$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Substituting the direction ratios,we get $(a)(a') + (1)(1) + (c)(c') = 0$.
Therefore,$aa' + 1 + cc' = 0$,which implies $aa' + cc' = -1$.

(C) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x}$.
Since this is a $\frac{0}{0}$ form,we apply $L'\text{H\^opital's rule}$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(\int_0^{x^2} \sec^2 t \, dt)}{\frac{d}{dx}(x \sin x)}$.
Using the $Leibniz$ rule for differentiation under the integral sign:
$L = \mathop {\lim }\limits_{x \to 0} \frac{\sec^2(x^2) \cdot 2x}{\sin x + x \cos x}$.
Divide numerator and denominator by $x$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{2 \sec^2(x^2)}{\frac{\sin x}{x} + \cos x}$.
Substituting $x = 0$:
$L = \frac{2 \cdot \sec^2(0)}{1 + \cos(0)} = \frac{2 \cdot 1}{1 + 1} = \frac{2}{2} = 1$.

(B) Given $F(t) = \int_{0}^{t} f(t - y) g(y) dy$.
Substituting the given functions $f(y) = e^y$ and $g(y) = y$:
$F(t) = \int_{0}^{t} e^{t - y} y dy = e^t \int_{0}^{t} y e^{-y} dy$.
Using integration by parts for $\int y e^{-y} dy$:
Let $u = y$ and $dv = e^{-y} dy$. Then $du = dy$ and $v = -e^{-y}$.
$\int y e^{-y} dy = -y e^{-y} - \int (-e^{-y}) dy = -y e^{-y} - e^{-y}$.
Evaluating the definite integral from $0$ to $t$:
$F(t) = e^t [ -y e^{-y} - e^{-y} ]_{0}^{t} = e^t [ (-t e^{-t} - e^{-t}) - (0 - 1) ]$.
$F(t) = e^t [ -t e^{-t} - e^{-t} + 1 ] = -t - 1 + e^t$.
Thus,$F(t) = e^t - (1 + t)$.

(C) To find the area bounded by $y = |x - 1|$ and $y = 3 - |x|$,we first find the intersection points of the curves.
For $x \ge 1$,$y = x - 1$ and $y = 3 - |x|$. If $x \ge 1$,then $x - 1 = 3 - x \Rightarrow 2x = 4 \Rightarrow x = 2$. At $x = 2$,$y = 1$.
For $x < 0$,$y = 1 - x$ and $y = 3 + x$. Then $1 - x = 3 + x \Rightarrow 2x = -2 \Rightarrow x = -1$. At $x = -1$,$y = 2$.
For $0 \le x < 1$,$y = 1 - x$ and $y = 3 - x$. These lines are parallel and do not intersect.
The required area is given by the integral:
$A = \int_{-1}^{2} (3 - |x| - |x - 1|) dx$
We split the integral based on the definitions of the absolute values:
$A = \int_{-1}^{0} ((3 + x) - (1 - x)) dx + \int_{0}^{1} ((3 - x) - (1 - x)) dx + \int_{1}^{2} ((3 - x) - (x - 1)) dx$
$A = \int_{-1}^{0} (2 + 2x) dx + \int_{0}^{1} (2) dx + \int_{1}^{2} (4 - 2x) dx$
$A = [2x + x^2]_{-1}^{0} + [2x]_{0}^{1} + [4x - x^2]_{1}^{2}$
$A = (0 - (-2 + 1)) + (2 - 0) + ((8 - 4) - (4 - 1))$
$A = 1 + 2 + (4 - 3) = 1 + 2 + 1 = 4 \text{ sq. units}$.
Thus,the correct option is $C$.

(B) Given the differential equation: $(1 + y^2) + (x - e^{\tan^{-1}y}) \frac{dy}{dx} = 0$.
Rearranging the equation to solve for $\frac{dx}{dy}$:
$(1 + y^2) \frac{dx}{dy} + x = e^{\tan^{-1}y}$.
Dividing by $(1 + y^2)$:
$\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{\tan^{-1}y}}{1 + y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1 + y^2}$ and $Q(y) = \frac{e^{\tan^{-1}y}}{1 + y^2}$.
The integrating factor $(I.F.)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1}y}$.
The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + k$.
$x e^{\tan^{-1}y} = \int \frac{e^{\tan^{-1}y}}{1 + y^2} \cdot e^{\tan^{-1}y} dy + k$.
Let $u = \tan^{-1}y$,then $du = \frac{1}{1 + y^2} dy$.
$x e^{\tan^{-1}y} = \int e^{2u} du + k = \frac{e^{2u}}{2} + k = \frac{e^{2\tan^{-1}y}}{2} + k$.
Multiplying by $2$:
$2xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + k$.

(C) Let $A$ be the origin $(0, 0, 0)$.
Then $\vec{AB}$ and $\vec{AC}$ represent the position vectors of vertices $B$ and $C$ respectively.
Let $M$ be the midpoint of side $BC$.
The position vector of $M$ is given by $\vec{AM} = \frac{\vec{AB} + \vec{AC}}{2}$.
Substituting the given vectors:
$\vec{AM} = \frac{(3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k})}{2}$
$\vec{AM} = \frac{(3+5)\hat{i} + (0-2)\hat{j} + (4+4)\hat{k}}{2}$
$\vec{AM} = \frac{8\hat{i} - 2\hat{j} + 8\hat{k}}{2} = 4\hat{i} - \hat{j} + 4\hat{k}$.
The length of the median $AM$ is the magnitude of vector $\vec{AM}$:
$|\vec{AM}| = \sqrt{4^2 + (-1)^2 + 4^2}$
$|\vec{AM}| = \sqrt{16 + 1 + 16} = \sqrt{33}$.

(C) The equation of the family of parabolas with the $X$-axis as their axis is given by $y^2 = 4a(x - b)$,where $a$ and $b$ are arbitrary constants.
To eliminate these two constants,we differentiate with respect to $x$:
$2y \frac{dy}{dx} = 4a \implies y \frac{dy}{dx} = 2a$.
Differentiating again with respect to $x$:
$y \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power of the highest order derivative is $1$,so the degree is $1$.
Thus,the degree and order are $1$ and $2$ respectively.