AIEEE 2012 Mathematics Question Paper with Answer and Solution

(B) In the expansion of $(y^{1/5} + x^{1/10})^{55}$,the general term is given by:
$T_{r+1} = {}^{55}C_r (y^{1/5})^{55-r} (x^{1/10})^r = {}^{55}C_r \cdot y^{11 - r/5} \cdot x^{r/10}$.
For the term to be free from radical signs,the exponents of $x$ and $y$ must be integers.
This requires $r/5$ and $r/10$ to be integers.
Since $0 \le r \le 55$,$r$ must be a multiple of $10$ (the least common multiple of $5$ and $10$).
The possible values for $r$ are $0, 10, 20, 30, 40, 50$.
There are $6$ such values,corresponding to the terms $T_1, T_{11}, T_{21}, T_{31}, T_{41}, T_{51}$.
Thus,there are $6$ terms free from radical signs.

(B) Given equations are:
$3 \sin P + 4 \cos Q = 6$ $(1)$
$4 \sin Q + 3 \cos P = 1$ $(2)$
Squaring and adding $(1)$ and $(2)$:
$(3 \sin P + 4 \cos Q)^2 + (4 \sin Q + 3 \cos P)^2 = 6^2 + 1^2$
$9 \sin^2 P + 16 \cos^2 Q + 24 \sin P \cos Q + 16 \sin^2 Q + 9 \cos^2 P + 24 \sin Q \cos P = 37$
$9(\sin^2 P + \cos^2 P) + 16(\sin^2 Q + \cos^2 Q) + 24(\sin P \cos Q + \cos P \sin Q) = 37$
$9(1) + 16(1) + 24 \sin(P + Q) = 37$
$25 + 24 \sin(P + Q) = 37$
$24 \sin(P + Q) = 12$
$\sin(P + Q) = \frac{1}{2}$
Since $P + Q + R = \pi$,we have $\sin(P + Q) = \sin(\pi - R) = \sin R$.
Thus,$\sin R = \frac{1}{2}$,which implies $R = \frac{\pi}{6}$ or $R = \frac{5\pi}{6}$.
If $R = \frac{5\pi}{6}$,then $P + Q = \frac{\pi}{6}$. Since $P, Q > 0$,$\sin P < \sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\cos Q < 1$. This contradicts $3 \sin P + 4 \cos Q = 6$.
Therefore,$R = \frac{\pi}{6}$.

(A) Given that $\frac{z^2}{z-1}$ is real,it must be equal to its conjugate: $\frac{z^2}{z-1} = \overline{\left(\frac{z^2}{z-1}\right)} = \frac{\overline{z}^2}{\overline{z}-1}$.
Cross-multiplying,we get: $z^2(\overline{z}-1) = \overline{z}^2(z-1)$.
Expanding the terms: $z^2\overline{z} - z^2 = \overline{z}^2z - \overline{z}^2$.
Rearranging the terms: $z^2\overline{z} - \overline{z}^2z - z^2 + \overline{z}^2 = 0$.
Factoring: $z\overline{z}(z-\overline{z}) - (z^2-\overline{z}^2) = 0$.
$z\overline{z}(z-\overline{z}) - (z-\overline{z})(z+\overline{z}) = 0$.
$(z-\overline{z})(z\overline{z} - (z+\overline{z})) = 0$.
This implies either $z-\overline{z} = 0$ or $z\overline{z} - z - \overline{z} = 0$.
$z-\overline{z} = 0$ implies $z$ is purely real (lies on the real axis).
$z\overline{z} - z - \overline{z} = 0$ can be written as $(z-1)(\overline{z}-1) = 1$,or $|z-1|^2 = 1$,which represents a circle with center $1$ and radius $1$. This circle passes through the origin since $|0-1| = 1$.
Thus,$z$ lies on the real axis or on a circle passing through the origin.

(D) The number of ways to select balls from $n_1$ identical items of type $1$,$n_2$ identical items of type $2$,and $n_3$ identical items of type $3$ is given by $(n_1 + 1)(n_2 + 1)(n_3 + 1)$.
Here,$n_1 = 10$ (white),$n_2 = 9$ (green),and $n_3 = 7$ (black).
Total ways including the case where no ball is selected $= (10 + 1) \times (9 + 1) \times (7 + 1) = 11 \times 10 \times 8 = 880$.
Since we need to select one or more balls,we exclude the case where no ball is selected (i.e.,subtract $1$).
Number of ways $= 880 - 1 = 879$.

(A) Let $x = (\sqrt{3} + 1)^{2n}$ and $y = (\sqrt{3} - 1)^{2n}$.
Using the binomial expansion,$(\sqrt{3} + 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (1)^k$.
Similarly,$(\sqrt{3} - 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} (-1)^k$.
Subtracting the two expressions:
$(\sqrt{3} + 1)^{2n} - (\sqrt{3} - 1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} (\sqrt{3})^{2n-k} [1 - (-1)^k]$.
The term $[1 - (-1)^k]$ is $0$ if $k$ is even and $2$ if $k$ is odd.
Thus,the expression becomes $2 \sum_{k \text{ is odd}} \binom{2n}{k} (\sqrt{3})^{2n-k}$.
Since $k$ is odd,$2n-k$ is odd. Let $2n-k = 2m+1$. Then $(\sqrt{3})^{2n-k} = 3^m \sqrt{3}$.
However,we can simplify this by noting that $(\sqrt{3} + 1)^2 = 3 + 1 + 2\sqrt{3} = 4 + 2\sqrt{3}$ and $(\sqrt{3} - 1)^2 = 3 + 1 - 2\sqrt{3} = 4 - 2\sqrt{3}$.
Let $A = 4 + 2\sqrt{3}$ and $B = 4 - 2\sqrt{3}$. We want $A^n - B^n$.
Since $A$ and $B$ are roots of the quadratic equation $x^2 - 8x + 4 = 0$,the expression $A^n - B^n$ is of the form $k\sqrt{3}$ where $k$ is an integer.
Wait,let's re-evaluate: $(\sqrt{3}+1)^2 = 4+2\sqrt{3}$. So $(\sqrt{3}+1)^{2n} = (4+2\sqrt{3})^n$.
Expanding $(4+2\sqrt{3})^n - (4-2\sqrt{3})^n$ using binomial theorem:
$= \sum \binom{n}{k} 4^{n-k} (2\sqrt{3})^k - \sum \binom{n}{k} 4^{n-k} (-2\sqrt{3})^k$
$= 2 \sum_{k \text{ odd}} \binom{n}{k} 4^{n-k} 2^k (\sqrt{3})^k$.
For $k=1$,term is $2 \cdot n \cdot 4^{n-1} \cdot 2 \cdot \sqrt{3} = n \cdot 4^n \sqrt{3}$.
Actually,the expression is an irrational number because it contains $\sqrt{3}$.

(D) Statement-$2$: The sum is a telescoping series: $\sum_{k=1}^{n} (k^3 - (k-1)^3) = (1^3 - 0^3) + (2^3 - 1^3) + \dots + (n^3 - (n-1)^3) = n^3$. This is true.
Statement-$1$: The $k$-th term of the series is $T_k = (k-1)^2 + (k-1)k + k^2 = k^2 - 2k + 1 + k^2 - k + k^2 = 3k^2 - 3k + 1$.
We know that $k^3 - (k-1)^3 = 3k^2 - 3k + 1$. Thus,$T_k = k^3 - (k-1)^3$.
The series is $\sum_{k=1}^{20} T_k = \sum_{k=1}^{20} (k^3 - (k-1)^3) = 20^3 = 8000$.
Since the last term is $361+380+400 = 19^2 + 19 \times 20 + 20^2$,this corresponds to $k=20$.
Both statements are true and Statement-$2$ explains Statement-$1$.

(A) Let the centre of the circle be $C(1,h)$.
Since the circle touches the $x-$axis at $(1,0)$,the radius of the circle is $|h|$.
The circle passes through the point $B(2,3)$,so the distance $CB$ must be equal to the radius $|h|$.
$CB^2 = h^2$
$(2-1)^2 + (3-h)^2 = h^2$
$1^2 + (9 - 6h + h^2) = h^2$
$1 + 9 - 6h = 0$
$10 = 6h$
$h = \frac{10}{6} = \frac{5}{3}$
The diameter of the circle is $2|h| = 2 \times \frac{5}{3} = \frac{10}{3}$.

(D) The circle $(x - 1)^2 + y^2 = 1$ has a radius $r_1 = 1$,so its diameter is $2$. Given this is the semi-minor axis,$b = 2$.
The circle $x^2 + (y - 2)^2 = 4$ has a radius $r_2 = 2$,so its diameter is $4$. Given this is the semi-major axis,$a = 4$.
The standard equation of an ellipse centered at the origin with axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values $a = 4$ and $b = 2$,we get $\frac{x^2}{4^2} + \frac{y^2}{2^2} = 1$.
$\Rightarrow \frac{x^2}{16} + \frac{y^2}{4} = 1$.
Multiplying by $16$,we get $x^2 + 4y^2 = 16$.

(D) Given observations are ${x_1}, {x_2}, \ldots, {x_n}$ with mean $\bar x$ and variance ${\sigma ^2} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar x)^2$.
For Statement-$2$: The arithmetic mean of $2{x_1}, 2{x_2}, \ldots, 2{x_n}$ is $\frac{1}{n} \sum_{i=1}^{n} (2x_i) = 2 \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right) = 2\bar x$.
Since the statement claims the mean is $4\bar x$,Statement-$2$ is false.
For Statement-$1$: The variance of $2{x_1}, 2{x_2}, \ldots, 2{x_n}$ is $\frac{1}{n} \sum_{i=1}^{n} (2x_i - 2\bar x)^2 = \frac{1}{n} \sum_{i=1}^{n} 4(x_i - \bar x)^2 = 4 \left( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar x)^2 \right) = 4{\sigma ^2}$.
Thus,Statement-$1$ is true.
Therefore,Statement-$1$ is true and Statement-$2$ is false.

(B) Given equation is $e^{\sin x} - e^{-\sin x} - 4 = 0$.
Let $e^{\sin x} = t$. Since $e^{\sin x} > 0$,we must have $t > 0$.
The equation becomes $t - \frac{1}{t} - 4 = 0$.
Multiplying by $t$,we get $t^2 - 4t - 1 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.
Since $t > 0$,we reject $t = 2 - \sqrt{5}$ (as $2 - \sqrt{5} < 0$).
Thus,$e^{\sin x} = 2 + \sqrt{5}$.
Taking the natural logarithm on both sides,$\sin x = \ln(2 + \sqrt{5})$.
Since $\sqrt{5} \approx 2.236$,$2 + \sqrt{5} \approx 4.236$.
We know that $\ln(e) = 1$ and $e \approx 2.718$.
Since $4.236 > 2.718$,$\ln(2 + \sqrt{5}) > 1$.
However,the range of $\sin x$ is $[-1, 1]$.
Since $\ln(2 + \sqrt{5}) > 1$,there is no real value of $x$ that satisfies this equation.
Therefore,the equation has no real roots.

(A) Let $p$ be the statement "$I$ become a teacher" and $q$ be the statement "$I$ will open a school".
The given statement is in the form of an implication: $p \implies q$.
The negation of an implication $p \implies q$ is given by $\sim(p \implies q) \equiv p \land \sim q$.
Here,$p$ is "$I$ become a teacher" and $\sim q$ is "$I$ will not open a school".
Therefore,the negation is "$I$ will become a teacher and $I$ will not open a school".

(D) Let the vertices be $A(5, -1)$,$B(-2, 3)$,and $C(h, k)$. The orthocentre $H$ is $(0, 0)$.
Since $CH \perp AB$,the slope of $AB$ is $m_{AB} = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7} = -\frac{4}{7}$.
The slope of the altitude $CH$ is $m_{CH} = -\frac{1}{m_{AB}} = \frac{7}{4}$.
Since $CH$ passes through $(0, 0)$ and $(h, k)$,its slope is $\frac{k}{h} = \frac{7}{4}$,which implies $7h - 4k = 0$ ---$(1)$.
Similarly,since $AH \perp BC$,the slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.
The slope of the altitude $AH$ is $m_{AH} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.
Since $AH \perp BC$,$m_{BC} \times m_{AH} = -1$,so $\left(\frac{k - 3}{h + 2}\right) \times \left(-\frac{1}{5}\right) = -1$.
$\frac{k - 3}{h + 2} = 5$ $\Rightarrow k - 3 = 5h + 10$ $\Rightarrow 5h - k + 13 = 0$ ---$(2)$.
Solving equations $(1)$ and $(2)$: From $(1)$,$k = \frac{7h}{4}$. Substituting into $(2)$: $5h - \frac{7h}{4} + 13 = 0$.
$\frac{20h - 7h}{4} + 13 = 0$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$.
Then $k = \frac{7(-4)}{4} = -7$.
Thus,the third vertex is $(-4, -7)$.

(C) Let the terms of the $G.P.$ be $a, ar, ar^2, ar^3, ar^4, ar^5$,where $a$ is the first term and $r$ is the common ratio.
According to the given conditions:
$ar^3 - a = 52 \Rightarrow a(r^3 - 1) = 52 \quad ......(1)$
$a + ar + ar^2 = 26 \Rightarrow a(1 + r + r^2) = 26 \quad ......(2)$
We know that $r^3 - 1 = (r - 1)(r^2 + r + 1)$.
Dividing equation $(1)$ by equation $(2)$:
$\frac{a(r - 1)(r^2 + r + 1)}{a(1 + r + r^2)} = \frac{52}{26}$
$r - 1 = 2 \Rightarrow r = 3$.
Substituting $r = 3$ in equation $(2)$:
$a(1 + 3 + 9) = 26$ $\Rightarrow 13a = 26$ $\Rightarrow a = 2$.
The sum of the first six terms is $S_6 = a(1 + r + r^2 + r^3 + r^4 + r^5) = a(1 + r + r^2)(1 + r^3)$.
$S_6 = 26 \times (1 + 3^3) = 26 \times (1 + 27) = 26 \times 28 = 728$.

(B) For the quadratic equation $(k-2)x^2 + 8x + k+4 = 0$ to have real and distinct roots,the discriminant $D > 0$:
$D = 8^2 - 4(k-2)(k+4) > 0$
$64 - 4(k^2 + 2k - 8) > 0$
$16 - (k^2 + 2k - 8) > 0$
$-k^2 - 2k + 24 > 0 \Rightarrow k^2 + 2k - 24 < 0$
$(k+6)(k-4) < 0 \Rightarrow -6 < k < 4$
For the roots to be negative,the sum of roots $\alpha + \beta = -\frac{8}{k-2} < 0$ and the product of roots $\alpha \beta = \frac{k+4}{k-2} > 0$.
From $\alpha + \beta < 0$: $\frac{8}{k-2} > 0 \Rightarrow k > 2$.
From $\alpha \beta > 0$: $\frac{k+4}{k-2} > 0 \Rightarrow k < -4$ or $k > 2$.
Combining all conditions: $(-6 < k < 4)$ $AND$ $(k > 2)$ $AND$ $(k < -4 \text{ or } k > 2)$.
The intersection is $2 < k < 4$.
Among the given options,only $k = 3$ satisfies $2 < k < 4$.

(A) The given equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{b^2} = 1.$
Since the hyperbola passes through $(K, 2),$ we have $\frac{K^2}{9} - \frac{4}{b^2} = 1$ $(1).$
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \frac{\sqrt{13}}{3}.$
Here,$a^2 = 9,$ so $\sqrt{1 + \frac{b^2}{9}} = \frac{\sqrt{13}}{3}.$
Squaring both sides,$1 + \frac{b^2}{9} = \frac{13}{9}.$
$\frac{b^2}{9} = \frac{13}{9} - 1 = \frac{4}{9},$ which implies $b^2 = 4.$
Substituting $b^2 = 4$ into equation $(1),$ we get $\frac{K^2}{9} - \frac{4}{4} = 1.$
$\frac{K^2}{9} - 1 = 1 \Rightarrow \frac{K^2}{9} = 2.$
Therefore,$K^2 = 18.$

(C) The set $A = \{a_1, a_2, \dots, a_{20}\}$ contains $20$ distinct elements.
The total number of $5$-element subsets is given by $^{20}C_5$.
The number of $5$-element subsets containing $a_4$ is equivalent to choosing $4$ more elements from the remaining $19$ elements,which is $^{19}C_4$.
According to the problem,$^{20}C_5 = k \times ^{19}C_4$.
Using the formula $^nC_r = \frac{n}{r} \times ^{n-1}C_{r-1}$,we have $^{20}C_5 = \frac{20}{5} \times ^{19}C_4$.
Therefore,$k = \frac{20}{5} = 4$.

(C) Let the equations of the lines be:
$L_1: x + 3y = 4$
$L_2: 3x + y = 4$
$L_3: x + y = 0$
Solving the equations to find the vertices:
Intersection of $L_1$ and $L_2$: $x + 3y = 4$ and $3x + y = 4 \implies x = 1, y = 1$. So,$B = (1, 1)$.
Intersection of $L_1$ and $L_3$: $x + 3y = 4$ and $x + y = 0 \implies x = 2, y = -2$. So,$C = (2, -2)$.
Intersection of $L_2$ and $L_3$: $3x + y = 4$ and $x + y = 0 \implies x = 2, y = -2$ (Wait,let's re-calculate).
Correct vertices:
$L_1 \cap L_2: x=1, y=1 \implies (1, 1)$
$L_1 \cap L_3: x+3(-x)=4 \implies -2x=4 \implies x=-2, y=2 \implies (-2, 2)$
$L_2 \cap L_3: 3x+(-x)=4 \implies 2x=4 \implies x=2, y=-2 \implies (2, -2)$
Let $A = (-2, 2)$,$B = (1, 1)$,$C = (2, -2)$.
Lengths of sides:
$AB = \sqrt{(1 - (-2))^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$
$BC = \sqrt{(2 - 1)^2 + (-2 - 1)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$
$AC = \sqrt{(2 - (-2))^2 + (-2 - 2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32}$
Since $AB = BC = \sqrt{10}$,the triangle is isosceles.

(D) The equation of any circle passing through the intersection of the circle $S = x^2 + y^2 - 4x - 6y - 21 = 0$ and the line $L = 3x + 4y + 5 = 0$ is given by $S + \lambda L = 0$.
$(x^2 + y^2 - 4x - 6y - 21) + \lambda(3x + 4y + 5) = 0$
Since this circle passes through the point $(1, 2)$,we substitute $x = 1$ and $y = 2$ into the equation:
$(1^2 + 2^2 - 4(1) - 6(2) - 21) + \lambda(3(1) + 4(2) + 5) = 0$
$(1 + 4 - 4 - 12 - 21) + \lambda(3 + 8 + 5) = 0$
$-32 + 16\lambda = 0$
$16\lambda = 32 \Rightarrow \lambda = 2$
Substituting $\lambda = 2$ back into the family of circles equation:
$(x^2 + y^2 - 4x - 6y - 21) + 2(3x + 4y + 5) = 0$
$x^2 + y^2 - 4x - 6y - 21 + 6x + 8y + 10 = 0$
$x^2 + y^2 + 2x + 2y - 11 = 0$
Thus,the required equation is $x^2 + y^2 + 2x + 2y - 11 = 0$.

(D) The modal class is the class interval with the highest frequency. Since the mode is $140$,which lies in the interval $100-150$,the modal class is $100-150$.
The formula for mode is:
$Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$L = 100$ (lower limit of modal class)
$f_1 = 37$ (frequency of modal class)
$f_0 = 33$ (frequency of class preceding modal class)
$f_2 = b$ (frequency of class succeeding modal class)
$h = 50$ (class size)
Substituting the values:
$140 = 100 + \left( \frac{37 - 33}{2(37) - 33 - b} \right) \times 50$
$40 = \left( \frac{4}{74 - 33 - b} \right) \times 50$
$40 = \frac{200}{41 - b}$
$40(41 - b) = 200$
$41 - b = 5$
$b = 41 - 5 = 36$

(A) The given series is $S = (1^2 + 3^2 + 5^2 + \dots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \dots + (2m)^2)$.
Sum of odd squares up to $(2m-1)^2$ is $\sum_{k=1}^{m} (2k-1)^2 = \frac{m(2m-1)(2m+1)}{3}$.
Sum of even squares up to $(2m)^2$ is $2 \times \sum_{k=1}^{m} (2k)^2 = 8 \sum_{k=1}^{m} k^2 = 8 \times \frac{m(m+1)(2m+1)}{6} = \frac{4m(m+1)(2m+1)}{3}$.
Adding these,$S = \frac{m(2m+1)}{3} [ (2m-1) + 4(m+1) ] = \frac{m(2m+1)}{3} [ 2m - 1 + 4m + 4 ] = \frac{m(2m+1)(6m+3)}{3} = m(2m+1)(2m+1) = m(2m+1)^2$.

(A) Let the two poles be $OA$ and $BC$ standing on a horizontal plane $OB$. Let $OA = 20\,m$ and $BC = 80\,m$. Let the distance between them be $OB = a$.
We set the coordinates as $O(0, 0)$,$A(0, 20)$,$B(a, 0)$,and $C(a, 80)$.
The line joining the top of the first pole $A(0, 20)$ to the foot of the second pole $B(a, 0)$ has the equation:
$y - 0 = \frac{20 - 0}{0 - a}(x - a) \Rightarrow y = -\frac{20}{a}(x - a) \quad \dots(1)$
The line joining the top of the second pole $C(a, 80)$ to the foot of the first pole $O(0, 0)$ has the equation:
$y - 0 = \frac{80 - 0}{a - 0}(x - 0) \Rightarrow y = \frac{80}{a}x \quad \dots(2)$
To find the intersection point $M$,we equate $(1)$ and $(2)$:
$\frac{80}{a}x = -\frac{20}{a}(x - a)$
$80x = -20x + 20a$
$100x = 20a \Rightarrow x = \frac{a}{5}$
Substituting $x = \frac{a}{5}$ into $(2)$:
$y = \frac{80}{a} \times \frac{a}{5} = 16\,m$.
Thus,the height of the point of intersection is $16\,m$.

(D) For the parabola $y^2 = -4ax$,the equation of the tangent with slope $m$ is $y = mx + \frac{a}{m}$.
Here,$4a = 4$,so $a = 1$. The parabola is $y^2 = -4x$,so it opens to the left. The standard form is $y^2 = -4ax$ with $a=1$.
The tangent equation is $y = mx - \frac{1}{m}$. Thus,Statement $1$ is true.
For the tangent $y = mx - \frac{1}{m}$,the intersection with the axis $(y=0)$ is $0 = mx - \frac{1}{m}$,which gives $x = \frac{1}{m^2}$.
Since $m^2 > 0$ for all $m \neq 0$,$x = \frac{1}{m^2} > 0$. Thus,the abscissa is always positive (non-negative). Statement $2$ is true.
Statement $2$ describes a property of the tangent,but it is not the logical derivation or explanation for the specific form of the tangent equation given in Statement $1$. Therefore,Statement $2$ is not the correct explanation for Statement $1$.

(B) Statement $1$ is the triangle inequality for subtraction,which states $|Z_1 - Z_2| \ge ||Z_1| - |Z_2||$,and since $||Z_1| - |Z_2|| \ge |Z_1| - |Z_2|$,Statement $1$ is true.
Statement $2$ is the standard triangle inequality for addition,which is also a fundamental property of complex numbers and is true.
However,Statement $2$ does not explain Statement $1$ as they are independent properties of the modulus of complex numbers.

(D) The given expression is a finite geometric series with first term $a = 1$, common ratio $r = -(y - 1)$, and $n = 18$ terms.
The sum is given by $f(y) = \frac{1 - (-(y - 1))^{18}}{1 - (-(y - 1))} = \frac{1 - (y - 1)^{18}}{y}$.
Thus, $f(y) = \frac{1}{y} - \frac{(y - 1)^{18}}{y}$.
Expanding $(y - 1)^{18}$ using the binomial theorem: $(y - 1)^{18} = \sum_{k=0}^{18} {^{18}C_k} y^k (-1)^{18-k}$.
Therefore, $\frac{(y - 1)^{18}}{y} = \sum_{k=0}^{18} {^{18}C_k} y^{k-1} (-1)^{18-k}$.
To find the coefficient of $y^2$, we set $k - 1 = 2$, which gives $k = 3$.
The term for $k = 3$ is ${^{18}C_3} y^2 (-1)^{18-3} = -{^{18}C_3} y^2$.
Since $f(y) = \frac{1}{y} - \sum_{k=0}^{18} {^{18}C_k} y^{k-1} (-1)^{18-k}$, the coefficient of $y^2$ is $-(-{^{18}C_3}) = {^{18}C_3}$.

(B) Let $L = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \sin x}}{x}} \right)\sin \left( {\frac{1}{x}} \right)$.
We can rewrite the expression as:
$L = \mathop {\lim }\limits_{x \to 0} \left( {1 - \frac{\sin x}{x}} \right) \sin \left( {\frac{1}{x}} \right)$.
We know that $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,so $\mathop {\lim }\limits_{x \to 0} \left( {1 - \frac{\sin x}{x}} \right) = 1 - 1 = 0$.
The term $\sin \left( {\frac{1}{x}} \right)$ oscillates between $-1$ and $1$ as $x \to 0$,meaning it is a bounded function.
By the Squeeze Theorem,since the limit of the first part is $0$ and the second part is bounded,the product is $0 \times (\text{bounded value}) = 0$.
Therefore,the limit is $0$.

(A) For option $(A)$: The contrapositive of $p \Rightarrow q$ is $\sim q \Rightarrow \sim p$. Here $p$ is $3x + 2 = 8$ and $q$ is $x = 2$. Thus,$\sim q \Rightarrow \sim p$ is $x \neq 2 \Rightarrow 3x + 2 \neq 8$. This is true.
For option $(B)$: The converse of $p \Rightarrow q$ is $q \Rightarrow p$. The converse of $\tan x = 0 \Rightarrow x = 0$ is $x = 0 \Rightarrow \tan x = 0$. Thus,$(B)$ is false.
For option $(C)$: $p \Rightarrow q$ is equivalent to $\sim p \vee q$,not $p \vee \sim q$. Thus,$(C)$ is false.
For option $(D)$: $p \vee q$ (disjunction) and $p \wedge q$ (conjunction) have different truth tables. Thus,$(D)$ is false.

(D) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$.
This can be rearranged as:
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bpc + c^2) + (c^2p^2 - 2pcd + d^2) \le 0$.
This simplifies to:
$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$.
Since the sum of squares of real numbers is non-negative,the only way for the sum to be $\le 0$ is if each term is exactly zero:
$ap - b = 0$,$bp - c = 0$,and $cp - d = 0$.
This implies $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
Therefore,$a, b, c, d$ are in $G.P.$

(B) The given equation of the parabola is $x^2 = 8y$.
Comparing this with the standard form $x^2 = 4ay$,we get $4a = 8$,which implies $a = 2$.
The vertex of the parabola is $C = (0, 0)$.
The extremities of the latus rectum are $A = (-2a, a) = (-4, 2)$ and $B = (2a, a) = (4, 2)$.
We need to find the area of the triangle formed by the vertices $A(-4, 2)$,$B(4, 2)$,and $C(0, 0)$.
The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates: Area $= \frac{1}{2} |-4(2 - 0) + 4(0 - 2) + 0(2 - 2)|$.
Area $= \frac{1}{2} |-4(2) + 4(-2) + 0| = \frac{1}{2} |-8 - 8| = \frac{1}{2} |-16| = 8$.
Thus,the area of the triangle is $8$ square units.

(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $ABC$.
$\Rightarrow b+c=11K, c+a=12K, a+b=13K$.
Adding these equations: $2(a+b+c) = 36K \Rightarrow a+b+c = 18K$.
Subtracting each equation from the sum:
$a = (a+b+c) - (b+c) = 18K - 11K = 7K$.
$b = (a+b+c) - (c+a) = 18K - 12K = 6K$.
$c = (a+b+c) - (a+b) = 18K - 13K = 5K$.
Using the cosine rule:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values:
$\cos A = \frac{(6K)^2 + (5K)^2 - (7K)^2}{2(6K)(5K)} = \frac{36K^2 + 25K^2 - 49K^2}{60K^2} = \frac{12K^2}{60K^2} = \frac{1}{5}$.

(B) Statement-$1$: The number of functions from a set $A$ with $n(A) = p$ elements to a set $B$ with $n(B) = q$ elements is given by $q^p$. This is a standard result in set theory. Thus,Statement-$1$ is true.
Statement-$2$: The number of ways to select $p$ objects from $q$ distinct objects is given by the combination formula ${}^qC_p = \frac{q!}{p!(q-p)!}$. This is a true statement.
However,Statement-$2$ describes the number of combinations,which is a fundamental counting principle,but it does not explain why the number of functions from $A$ to $B$ is $q^p$. The number of functions is derived from the fact that each of the $p$ elements in $A$ has $q$ choices in $B$. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

(A) The general term of the expansion $(y^{1/5} + x^{1/10})^{55}$ is given by:
$T_{r+1} = ^{55}C_{r} (y^{1/5})^{55-r} (x^{1/10})^{r}$
$T_{r+1} = ^{55}C_{r} y^{\frac{55-r}{5}} x^{\frac{r}{10}}$
For the powers of $x$ and $y$ to be free from radical signs,the exponents $\frac{55-r}{5}$ and $\frac{r}{10}$ must be integers.
$\frac{55-r}{5} = 11 - \frac{r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$.
$\frac{r}{10}$ must be an integer,which implies $r$ must be a multiple of $10$.
Combining these,$r$ must be a multiple of $10$ where $0 \le r \le 55$.
Possible values for $r$ are $0, 10, 20, 30, 40, 50$.
There are $6$ such values of $r$,so there are $6$ terms in the expansion that are free from radical signs.

(D) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$.
For a point $(1, a)$ to lie between these two parallel lines,the expressions $L_1(1, a)$ and $L_2(1, a)$ must have opposite signs.
Let $f(x, y) = x + y - 1$ and $g(x, y) = 2x + 2y - 3$.
Substituting $(1, a)$ into the equations:
$f(1, a) = 1 + a - 1 = a$
$g(1, a) = 2(1) + 2a - 3 = 2a - 1$
Since the point lies between the lines,$f(1, a) \cdot g(1, a) < 0$.
Therefore,$a(2a - 1) < 0$.
This inequality holds when $a$ is between the roots of $a(2a - 1) = 0$,which are $a = 0$ and $a = 1/2$.
Thus,$a \in \left( 0, \frac{1}{2} \right)$.

(B) Let the third vertex of $\Delta ABC$ be $C(a, b)$.
Let $A(5, -1)$ and $B(-2, 3)$ be the other two vertices.
The orthocentre $H$ is at $(0, 0)$.
Since $AH \perp BC$,the product of their slopes is $-1$:
$\left( \frac{-1 - 0}{5 - 0} \right) \times \left( \frac{b - 3}{a - (-2)} \right) = -1$
$\Rightarrow \left( \frac{-1}{5} \right) \times \left( \frac{b - 3}{a + 2} \right) = -1$
$\Rightarrow b - 3 = 5(a + 2)$ $\Rightarrow b - 3 = 5a + 10$ $\Rightarrow 5a - b + 13 = 0 \dots (1)$
Similarly,since $BH \perp AC$,the product of their slopes is $-1$:
$\left( \frac{3 - 0}{-2 - 0} \right) \times \left( \frac{b - (-1)}{a - 5} \right) = -1$
$\Rightarrow \left( \frac{3}{-2} \right) \times \left( \frac{b + 1}{a - 5} \right) = -1$
$\Rightarrow 3(b + 1) = 2(a - 5)$ $\Rightarrow 3b + 3 = 2a - 10$ $\Rightarrow 2a - 3b - 13 = 0 \dots (2)$
Multiplying equation $(1)$ by $3$:
$15a - 3b + 39 = 0 \dots (3)$
Subtracting equation $(2)$ from $(3)$:
$(15a - 2a) + (-3b - (-3b)) + (39 - (-13)) = 0$
$13a + 52 = 0 \Rightarrow a = -4$
Substituting $a = -4$ into equation $(1)$:
$5(-4) - b + 13 = 0$ $\Rightarrow -20 - b + 13 = 0$ $\Rightarrow b = -7$
Thus,the third vertex is $(-4, -7)$.

(C) The $n^{th}$ term of the series is $T_n = \frac{1}{\sqrt{n} + \sqrt{n+1}}$.
Rationalizing the denominator,we get:
$T_n = \frac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1} + \sqrt{n})(\sqrt{n+1} - \sqrt{n})} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$.
For $n=1$ to $15$,the sum $S_{15} = \sum_{n=1}^{15} (\sqrt{n+1} - \sqrt{n})$.
$S_{15} = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{16} - \sqrt{15})$.
This is a telescoping series where intermediate terms cancel out.
$S_{15} = \sqrt{16} - \sqrt{1} = 4 - 1 = 3$.

(D) Given observations are $4, 7, 2, 8, 6, a$ and the mean is $7$.
We know that Mean $= \frac{4 + 7 + 2 + 8 + 6 + a}{6}$.
$7 = \frac{27 + a}{6}$ $\Rightarrow 42 = 27 + a$ $\Rightarrow a = 15$.
Now,the observations in ascending order are $2, 4, 6, 7, 8, 15$.
Since the number of observations $n = 6$ (even),the median is the average of the $3^{rd}$ and $4^{th}$ observations.
Median $= \frac{6 + 7}{2} = 6.5$.
Mean deviation from median $= \frac{\sum |x_i - 6.5|}{6}$.
$= \frac{|2 - 6.5| + |4 - 6.5| + |6 - 6.5| + |7 - 6.5| + |8 - 6.5| + |15 - 6.5|}{6}$.
$= \frac{4.5 + 2.5 + 0.5 + 0.5 + 1.5 + 8.5}{6} = \frac{18}{6} = 3$.

(B) Let $z = x + iy$. Then the vertices are $z = (x, y)$,$iz = i(x + iy) = -y + ix = (-y, x)$,and $z + iz = (x - y) + i(x + y) = (x - y, x + y)$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates:
Area $= \frac{1}{2} |x(x - (x + y)) + (-y)((x + y) - y) + (x - y)(y - x)|$
$= \frac{1}{2} |x(-y) - y(x) + (x - y)(-(x - y))|$
$= \frac{1}{2} |-xy - xy - (x - y)^2|$
$= \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)|$
$= \frac{1}{2} |-2xy - x^2 + 2xy - y^2|$
$= \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} (x^2 + y^2)$.
Since $|z|^2 = x^2 + y^2$,the area is $\frac{1}{2} |z|^2$.

(D) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0 - 1}{2 - 0} = -\frac{1}{2}$.
Since the tangents at $P_1$ and $P_2$ are parallel to this chord,their slope is $m = -\frac{1}{2}$.
The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
Here $a^2 = 4, b^2 = 1$,and $m = -\frac{1}{2}$.
Thus,$y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 1} = -\frac{1}{2}x \pm \sqrt{1 + 1} = -\frac{1}{2}x \pm \sqrt{2}$.
Substituting $y = -\frac{1}{2}x + \sqrt{2}$ into the ellipse equation $\frac{x^2}{4} + y^2 = 1$:
$\frac{x^2}{4} + (-\frac{x}{2} + \sqrt{2})^2 = 1$ $\Rightarrow \frac{x^2}{4} + \frac{x^2}{4} - \sqrt{2}x + 2 = 1$ $\Rightarrow \frac{x^2}{2} - \sqrt{2}x + 1 = 0$ $\Rightarrow x^2 - 2\sqrt{2}x + 2 = 0$ $\Rightarrow (x - \sqrt{2})^2 = 0$.
So,$x = \sqrt{2}$,which gives $y = -\frac{1}{2}(\sqrt{2}) + \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Point $P_1 = (\sqrt{2}, \frac{1}{\sqrt{2}})$.
Similarly,for $y = -\frac{1}{2}x - \sqrt{2}$,we get $x = -\sqrt{2}$ and $y = -\frac{1}{\sqrt{2}}$. Point $P_2 = (-\sqrt{2}, -\frac{1}{\sqrt{2}})$.
The distance $P_1P_2 = \sqrt{(\sqrt{2} - (-\sqrt{2}))^2 + (\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}))^2} = \sqrt{(2\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{8 + 2} = \sqrt{10}$.

(A) By definition,the biconditional statement $p \Leftrightarrow q$ is logically equivalent to the conjunction of the two conditional statements $p \Rightarrow q$ and $q \Rightarrow p$.
Therefore,$p \Leftrightarrow q \equiv (p$ $\Rightarrow q) \wedge (q$ $\Rightarrow p)$.

(D) Let the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ be $x_1$ and $x_2$.
From the properties of quadratic equations,we have:
$x_1 + x_2 = \sin \alpha - 2$
$x_1 x_2 = -(1 + \sin \alpha)$
We want to minimize $S = x_1^2 + x_2^2$.
$S = (x_1 + x_2)^2 - 2x_1 x_2$
$S = (\sin \alpha - 2)^2 - 2(-(1 + \sin \alpha))$
$S = \sin^2 \alpha - 4 \sin \alpha + 4 + 2 + 2 \sin \alpha$
$S = \sin^2 \alpha - 2 \sin \alpha + 6$
To minimize $S$,let $u = \sin \alpha$,where $u \in [-1, 1]$.
$S(u) = u^2 - 2u + 6 = (u - 1)^2 + 5$.
The minimum value occurs at $u = 1$.
Since $\sin \alpha = 1$,we have $\alpha = \frac{\pi}{2}$.

(D) Given the parabola equation is $x^2 = 8y$.
When $x = 4$,we substitute into the equation: $4^2 = 8y \Rightarrow 16 = 8y \Rightarrow y = 2$.
So,the point of contact is $(4, 2)$.
Differentiating $x^2 = 8y$ with respect to $x$,we get $2x = 8 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{2x}{8} = \frac{x}{4}$.
The slope of the tangent at $x = 4$ is $m_t = \left. \frac{x}{4} \right|_{x=4} = 1$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$.
The equation of the normal at $(4, 2)$ is $y - y_1 = m_n(x - x_1)$.
Substituting the values: $y - 2 = -1(x - 4)$.
$y - 2 = -x + 4$.
$x + y = 6$.

(B) Total number of ways to arrange $6$ students in a row is $6! = 720.$
To find the number of arrangements where $A$ and $B$ are separated by exactly one student,we treat the block $(A, X, B)$ or $(B, X, A)$ as a single unit,where $X$ is one of the remaining $4$ students.
Step $1$: Choose one student $X$ from the remaining $4$ students in $4$ ways.
Step $2$: Arrange the block $(A, X, B)$ or $(B, X, A)$ along with the remaining $3$ students. This gives us $4$ units to arrange,which can be done in $4!$ ways.
Step $3$: Since the block can be $(A, X, B)$ or $(B, X, A)$,there are $2$ ways to arrange $A$ and $B$ within the block.
Total favorable arrangements $= 4 \times 4! \times 2 = 4 \times 24 \times 2 = 192.$
Probability $= \frac{192}{720} = \frac{192 \div 48}{720 \div 48} = \frac{4}{15}.$

(C) The given series is $S_n = 1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots + n \text{ terms}$.
We can rewrite the terms as:
$S_n = (1) + (1 + \frac{1}{3}) + (1 + \frac{1}{9}) + (1 + \frac{1}{27}) + \dots + n \text{ terms}$.
Grouping the terms,we get:
$S_n = (1 + 1 + 1 + \dots + n \text{ times}) + (\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots + n \text{ terms})$.
$S_n = n + \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}}$.
$S_n = n + \frac{\frac{1}{3}(1 - \frac{1}{3^n})}{\frac{2}{3}}$.
$S_n = n + \frac{1}{2}(1 - \frac{1}{3^n})$.
$S_n = n + \frac{1}{2} - \frac{1}{2 \cdot 3^n}$.

(C) The given quadratic equation is $px^2 + qx + r = 0$ with $p, q, r \in \mathbb{R}$ and $r > p > 0.$
Since the roots $\alpha$ and $\beta$ are complex,the discriminant $D = q^2 - 4pr < 0.$
This implies $q^2 < 4pr.$
Since the coefficients are real,the complex roots must be conjugates,i.e.,$\beta = \bar{\alpha}.$
Thus,$|\alpha| = |\beta| = \sqrt{\alpha \bar{\alpha}} = \sqrt{\alpha \beta}.$
From the properties of quadratic equations,the product of the roots is $\alpha \beta = \frac{r}{p}.$
Therefore,$|\alpha| = |\beta| = \sqrt{\frac{r}{p}}.$
Given $r > p > 0,$ we have $\frac{r}{p} > 1,$ which implies $\sqrt{\frac{r}{p}} > 1.$
Then,$|\alpha| + |\beta| = 2|\alpha| = 2\sqrt{\frac{r}{p}}.$
Since $\sqrt{\frac{r}{p}} > 1,$ it follows that $2\sqrt{\frac{r}{p}} > 2.$
Hence,$|\alpha| + |\beta| > 2.$

(A) Let $P = (1, 0)$ and $Q = (-1, 0)$. Let a point $X = (x, y)$ satisfy the condition $\frac{XP}{XQ} = \frac{1}{2}$.
This implies $2XP = XQ$,or $4XP^2 = XQ^2$.
Substituting the coordinates: $4((x - 1)^2 + y^2) = (x + 1)^2 + y^2$.
Expanding this: $4(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2$.
$4x^2 - 8x + 4 + 4y^2 = x^2 + 2x + 1 + y^2$.
$3x^2 + 3y^2 - 10x + 3 = 0$.
Dividing by $3$: $x^2 + y^2 - \frac{10}{3}x + 1 = 0$.
This is the equation of a circle. Since points $A, B, C$ satisfy this condition,they lie on this circle.
The circumcentre of $\Delta ABC$ is the center of this circle.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{10}{3} \Rightarrow g = -\frac{5}{3}$ and $f = 0$.
The center is $(-g, -f) = \left( \frac{5}{3}, 0 \right)$.

(B) The equation of the circle is $x^2 + y^2 + 3x = 0$.
Its center is $B = \left( -\frac{3}{2}, 0 \right)$ and radius is $\frac{3}{2}$.
The line $y = mx + 1$ intersects the $y$-axis at $A(0, 1)$.
Since the two intersection points are equidistant from and on opposite sides of the $x$-axis,the line must pass through the center of the circle $B$ to maintain symmetry,or the chord must be perpendicular to the $x$-axis (which is not possible here as the line is $y = mx + 1$).
Actually,the condition implies that the line must pass through the center $B\left( -\frac{3}{2}, 0 \right)$ because the points are symmetric with respect to the $x$-axis.
Substituting $B\left( -\frac{3}{2}, 0 \right)$ into $y = mx + 1$:
$0 = m\left( -\frac{3}{2} \right) + 1$
$\frac{3}{2}m = 1$
$3m = 2$
$3m - 2 = 0$.

(D) Given $\frac{x}{[x]} \le f(x) \le \sqrt{6 - x}$ for $x \ne 2$.
For $x \to 2^-$,$[x] = 1$. Thus,$\lim_{x \to 2^-} \frac{x}{[x]} = \frac{2}{1} = 2$ and $\lim_{x \to 2^-} \sqrt{6 - x} = \sqrt{6 - 2} = 2$.
By the Sandwich Theorem,$\lim_{x \to 2^-} f(x) = 2$. So,Statement $1$ is true.
For $x \to 2^+$,$[x] = 2$. Thus,$\lim_{x \to 2^+} \frac{x}{[x]} = \frac{2}{2} = 1$ and $\lim_{x \to 2^+} \sqrt{6 - x} = 2$.
By the Sandwich Theorem,$1 \le \lim_{x \to 2^+} f(x) \le 2$. Since the left-hand limit is $2$ and the right-hand limit is not necessarily $2$,the limit $\lim_{x \to 2} f(x)$ does not exist.
Since $\lim_{x \to 2} f(x)$ does not exist,$f$ is not continuous at $x = 2$. So,Statement $2$ is false.

(B) The conditional statement $p \to q$ is defined by the truth table where it is false only when $p$ is true and $q$ is false.
This is logically equivalent to the disjunction of the negation of $p$ and $q$.
Therefore,$p \to q \equiv \sim p \vee q$.

(B) Given $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$.
First,find the derivative $f'(x)$:
$f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.
Evaluate $f'(1)$:
$f'(1) = 30(1)^9 - 56(1)^7 + 30(1)^5 - 63(1)^2 + 6(1) = 30 - 56 + 30 - 63 + 6 = -53$.
Now,consider the limit $L = \mathop {\lim }\limits_{\alpha \to 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$.
Since this is a $\frac{0}{0}$ form,apply $L'\text{Hospital's rule}$:
$L = \mathop {\lim }\limits_{\alpha \to 0} \frac{\frac{d}{d\alpha}(f(1 - \alpha) - f(1))}{\frac{d}{d\alpha}(\alpha^3 + 3\alpha)} = \mathop {\lim }\limits_{\alpha \to 0} \frac{f'(1 - \alpha) \cdot (-1)}{3\alpha^2 + 3}$.
Substitute $\alpha = 0$:
$L = \frac{f'(1) \cdot (-1)}{3(0)^2 + 3} = \frac{-f'(1)}{3} = \frac{-(-53)}{3} = \frac{53}{3}$.

(B) Let $|Z| = |W| = r$.
Then $Z = r e^{i\theta}$ and $W = r e^{i\phi}$,where $\theta = \text{arg } Z$ and $\phi = \text{arg } W$.
Given $\theta + \phi = \pi$,we have $\theta = \pi - \phi$.
Thus,$Z = r e^{i(\pi - \phi)} = r e^{i\pi} e^{-i\phi} = -r e^{-i\phi}$.
Since $\overline{W} = r e^{-i\phi}$,we get $Z = -\overline{W}$. So,Statement $1$ is true.
For Statement $2$,$\text{arg } \overline{W} = -\text{arg } W = -\phi$.
Then $\text{arg } Z - \text{arg } \overline{W} = \theta - (-\phi) = \theta + \phi = \pi$.
Thus,Statement $2$ is also true and explains Statement $1$.

(A) Let the equation of the plane parallel to $x - 2y + 2z - 5 = 0$ be $x - 2y + 2z + k = 0 \dots (i)$
The perpendicular distance from the origin $(0, 0, 0)$ to the plane $(i)$ is given by the formula $d = \frac{|k|}{\sqrt{a^2 + b^2 + c^2}}$.
Given that the distance is $1$,we have:
$\frac{|k|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 1$
$\frac{|k|}{\sqrt{1 + 4 + 4}} = 1$
$\frac{|k|}{\sqrt{9}} = 1$
$\frac{|k|}{3} = 1$
$|k| = 3$
Therefore,$k = 3$ or $k = -3$.
Substituting these values into equation $(i)$,we get the possible equations of the plane as $x - 2y + 2z + 3 = 0$ or $x - 2y + 2z - 3 = 0$.
Comparing this with the given options,the correct equation is $x - 2y + 2z - 3 = 0$.

(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively:
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\lambda+1, y=3\lambda-1, z=4\lambda+1$
$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu \implies x=\mu+3, y=2\mu+k, z=\mu$
If the lines intersect,there must exist a point common to both,so:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2 \quad (i)$
$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1 \quad (ii)$
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1 \quad (iii)$
Subtracting $(i)$ from $(iii)$:
$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2
\implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$
Substituting $\lambda = -\frac{3}{2}$ into $(iii)$:
$4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$
Substituting $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(ii)$:
$3(-\frac{3}{2}) - 2(-5) = k+1
\implies -\frac{9}{2} + 10 = k+1
\implies k = 9 - \frac{9}{2} = \frac{9}{2}$

(C) The given parabolas are $x^2 = \frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$ and $x^2 = 9y \implies x = \pm 3\sqrt{y}$.
Since the region is symmetric about the $y$-axis,we calculate the area in the first quadrant and multiply by $2$.
The area $A$ is bounded by $x = 3\sqrt{y}$ (right curve) and $x = \frac{\sqrt{y}}{2}$ (left curve) from $y = 0$ to $y = 2$.
$A = 2 \int_{0}^{2} \left( 3\sqrt{y} - \frac{\sqrt{y}}{2} \right) dy$
$A = 2 \int_{0}^{2} \frac{5\sqrt{y}}{2} dy = 5 \int_{0}^{2} y^{1/2} dy$
$A = 5 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{2} = 5 \times \frac{2}{3} \left[ y^{3/2} \right]_{0}^{2}$
$A = \frac{10}{3} \left( 2^{3/2} - 0 \right) = \frac{10}{3} \times 2\sqrt{2} = \frac{20\sqrt{2}}{3}$

(D) Given $g(x) = \int_{0}^{x} \cos 4t \, dt$.
Evaluating the integral,we get $g(x) = \left[ \frac{\sin 4t}{4} \right]_{0}^{x} = \frac{\sin 4x}{4}$.
Now,calculate $g(x + \pi) = \frac{\sin 4(x + \pi)}{4} = \frac{\sin(4x + 4\pi)}{4}$.
Since $\sin(4x + 4\pi) = \sin 4x$,we have $g(x + \pi) = \frac{\sin 4x}{4} = g(x)$.
Also,$g(\pi) = \frac{\sin(4\pi)}{4} = 0$.
Since $g(\pi) = 0$,it follows that $g(x) + g(\pi) = g(x) + 0 = g(x)$ and $g(x) - g(\pi) = g(x) - 0 = g(x)$.
Thus,both $g(x) + g(\pi)$ and $g(x) - g(\pi)$ are equal to $g(x)$.

(D) Given $A = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\3&2&1\end{array}} \right]$,$A{u_1} = \left[ {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right]$,and $A{u_2} = \left[ {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right]$.
Adding the two equations,we get:
$A{u_1} + A{u_2} = \left[ {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right]$
$A(u_1 + u_2) = \left[ {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right]$
Thus,$u_1 + u_2 = A^{-1} \left[ {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right]$.
First,find $|A| = 1(1-0) - 0 + 0 = 1$.
Next,find $adj(A)$. The cofactors are:
$C_{11} = 1, C_{12} = -2, C_{13} = 1$
$C_{21} = 0, C_{22} = 1, C_{23} = -2$
$C_{31} = 0, C_{32} = 0, C_{33} = 1$
So,$adj(A) = \left[ {\begin{array}{*{20}{c}}1&0&0\\-2&1&0\\1&-2&1\end{array}} \right]$.
Since $|A| = 1$,$A^{-1} = adj(A)$.
$u_1 + u_2 = \left[ {\begin{array}{*{20}{c}}1&0&0\\-2&1&0\\1&-2&1\end{array}} \right] \left[ {\begin{array}{*{20}{c}}1\\1\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1(1)+0(1)+0(0)\\-2(1)+1(1)+0(0)\\1(1)-2(1)+1(0)\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1\\-1\\-1\end{array}} \right]$.

(C) Given that $P$ and $Q$ are $3 \times 3$ matrices such that $P \neq Q$.
We are given the equations:
$1) P^3 = Q^3$
$2) P^2Q = Q^2P$
Subtracting the second equation from the first,we get:
$P^3 - P^2Q = Q^3 - Q^2P$
Factoring the terms,we have:
$P^2(P - Q) = Q^2(Q - P)$
$P^2(P - Q) = -Q^2(P - Q)$
$(P^2 + Q^2)(P - Q) = 0$
Since $P \neq Q$,the matrix $(P - Q)$ is not necessarily a zero matrix,but the product $(P^2 + Q^2)(P - Q) = 0$ implies that the matrix $(P^2 + Q^2)$ must be singular if $(P - Q)$ is non-zero and the system holds.
Specifically,if $(P^2 + Q^2)(P - Q) = 0$ and $P - Q \neq 0$,it implies that the determinant of the product is zero:
$\det((P^2 + Q^2)(P - Q)) = \det(0) = 0$
$\det(P^2 + Q^2) \cdot \det(P - Q) = 0$
In the context of such matrix problems,this leads to $\det(P^2 + Q^2) = 0$.

(D) The function is $f(x) = [x] \cos \left( \frac{2x - 1}{2} \pi \right) = [x] \cos \left( x\pi - \frac{\pi}{2} \right) = [x] \sin(x\pi)$.
We check for continuity at any integer $x = n$,where $n \in \mathbb{Z}$.
The value of the function at $x = n$ is $f(n) = [n] \sin(n\pi) = n \cdot 0 = 0$.
The Left Hand Limit $(LHL)$ as $x \to n^-$ is $\lim_{x \to n^-} [x] \sin(x\pi) = (n - 1) \sin(n\pi) = (n - 1) \cdot 0 = 0$.
The Right Hand Limit $(RHL)$ as $x \to n^+$ is $\lim_{x \to n^+} [x] \sin(x\pi) = n \sin(n\pi) = n \cdot 0 = 0$.
Since $LHL = RHL = f(n) = 0$ for all $n \in \mathbb{Z}$,the function is continuous at all integers.
Since $[x]$ is continuous everywhere except at integers and $\sin(x\pi)$ is continuous everywhere,the product is continuous everywhere.
Thus,$f(x)$ is continuous for every real $x$.

(D) The function is defined as $f(x) = |x - 2| + |x - 5|$.
We can express this piecewise as:
$f(x) = \begin{cases} -(x-2) - (x-5) = -2x + 7 & \text{if } x < 2 \\ (x-2) - (x-5) = 3 & \text{if } 2 \le x < 5 \\ (x-2) + (x-5) = 2x - 7 & \text{if } x \ge 5 \end{cases}$
For Statement-$1$:
In the interval $(2, 5)$,$f(x) = 3$. Therefore,$f'(x) = 0$ for all $x \in (2, 5)$.
Since $4 \in (2, 5)$,$f'(4) = 0$. Thus,Statement-$1$ is true.
For Statement-$2$:
$f(x)$ is a sum of continuous functions,so it is continuous everywhere. In $[2, 5]$,$f(x) = 3$,which is a constant function,so it is differentiable in $(2, 5)$.
Also,$f(2) = |2-2| + |2-5| = 3$ and $f(5) = |5-2| + |5-5| = 3$. Thus,$f(2) = f(5)$.
Statement-$2$ is true.
Statement-$2$ describes the conditions of Rolle's Theorem,which implies the existence of a point $c \in (2, 5)$ such that $f'(c) = 0$. Since $f(x)$ is constant on $[2, 5]$,$f'(x) = 0$ for all $x \in (2, 5)$,which confirms Statement-$1$. Thus,Statement-$2$ is a correct explanation for Statement-$1$.

(C) The initial volume of the balloon is $V_i = 4500\pi \text{ m}^3$.
The rate of change of volume is $\frac{dV}{dt} = -72\pi \text{ m}^3/\text{min}$.
After $t = 49$ minutes,the volume $V$ is:
$V = 4500\pi - (72\pi \times 49) = 4500\pi - 3528\pi = 972\pi \text{ m}^3$.
The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
At $t = 49$ minutes,$972\pi = \frac{4}{3}\pi r^3$,which implies $r^3 = \frac{972 \times 3}{4} = 729$.
Thus,$r = \sqrt[3]{729} = 9 \text{ m}$.
Differentiating $V = \frac{4}{3}\pi r^3$ with respect to $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substituting the values: $-72\pi = 4\pi (9)^2 \frac{dr}{dt}$.
$-72\pi = 4\pi (81) \frac{dr}{dt} = 324\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = -\frac{72\pi}{324\pi} = -\frac{2}{9}$.
The rate at which the radius decreases is $\frac{2}{9} \text{ m/min}$.

(A) Given the differential equation: $\frac{dp(t)}{dt} = 0.5p(t) - 450 = \frac{p(t) - 900}{2}$.
Separating the variables,we get: $\int \frac{dp(t)}{p(t) - 900} = \int \frac{1}{2} dt$.
Integrating both sides: $\ln |p(t) - 900| = \frac{1}{2} t + C$.
Using the initial condition $p(0) = 850$: $\ln |850 - 900| = \frac{1}{2}(0) + C \implies C = \ln 50$.
Thus,the equation is: $\ln |p(t) - 900| = \frac{1}{2} t + \ln 50$.
We want to find $t$ when $p(t) = 0$: $\ln |0 - 900| = \frac{1}{2} t + \ln 50$.
$\ln 900 - \ln 50 = \frac{1}{2} t$.
$\ln \left( \frac{900}{50} \right) = \frac{1}{2} t$.
$\ln 18 = \frac{1}{2} t$.
$t = 2 \ln 18$.

(A) Let $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$. We choose $3$ numbers from $S$ without replacement.
Let $A$ be the event that the maximum of the chosen numbers is $6$.
Let $B$ be the event that the minimum of the chosen numbers is $3$.
We want to find the conditional probability $P(B|A) = \frac{P(A \cap B)}{P(A)}$.
For event $A$ (maximum is $6$): The numbers must be chosen from $\{1, 2, 3, 4, 5, 6\}$ such that $6$ is included. The other $2$ numbers must be chosen from $\{1, 2, 3, 4, 5\}$.
Number of ways for $A = \binom{5}{2} = 10$.
For event $A \cap B$ (maximum is $6$ and minimum is $3$): The numbers must be chosen from $\{3, 4, 5, 6\}$ such that $3$ and $6$ are included. The remaining $1$ number must be chosen from $\{4, 5\}$.
Number of ways for $A \cap B = \binom{2}{1} = 2$.
Thus,$P(B|A) = \frac{n(A \cap B)}{n(A)} = \frac{2}{10} = \frac{1}{5}$.

(C) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,we have $|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Since $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular,their dot product is zero: $\vec{c} \cdot \vec{d} = 0$.
$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$.
Expanding the dot product: $5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$.
Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$,and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$:
$5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$.
$6(\vec{a} \cdot \vec{b}) - 3 = 0$.
$6(\vec{a} \cdot \vec{b}) = 3$.
$\vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2}$.
We know $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$1 \cdot 1 \cdot \cos \theta = \frac{1}{2}$.
$\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$.

(B) Let $E$ be the foot of the perpendicular from vertex $B$ to the side $AD$. The vector $\vec{AE}$ is the projection of $\vec{AB}$ onto $\vec{AD}$.
Thus,$\vec{AE} = \text{proj}_{\vec{p}} \vec{q} = \left( \frac{\vec{q} \cdot \vec{p}}{\vec{p} \cdot \vec{p}} \right) \vec{p}$.
In $\triangle ABE$,by the triangle law of vector addition,we have $\vec{AB} + \vec{BE} = \vec{AE}$.
Given $\vec{AB} = \vec{q}$ and $\vec{BE} = \vec{r}$,we have $\vec{q} + \vec{r} = \vec{AE}$.
Therefore,$\vec{r} = \vec{AE} - \vec{q}$.
Substituting the expression for $\vec{AE}$,we get $\vec{r} = \left( \frac{\vec{q} \cdot \vec{p}}{\vec{p} \cdot \vec{p}} \right) \vec{p} - \vec{q}$.

(C) Given $A = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix}$ and $B = \begin{bmatrix} \alpha + 1 \\ 0 \\ 0 \end{bmatrix}$.
The transpose of matrix $B$ is $B^T = \begin{bmatrix} \alpha + 1 & 0 & 0 \end{bmatrix}$.
Now,calculate the product $AB^T$:
$AB^T = \begin{bmatrix} \alpha - 1 \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} \alpha + 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} (\alpha - 1)(\alpha + 1) & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \alpha^2 - 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
For $AB^T$ to be a non-zero matrix,at least one element must be non-zero.
Thus,$\alpha^2 - 1 \neq 0$,which implies $\alpha^2 \neq 1$.
Taking the square root on both sides,we get $|\alpha| \neq 1$.

(B) Let the radius of the circular sheet be $r$ and its area be $A$.
Given that $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
We are given that $\frac{dA}{dt} = 6\pi \, cm^2/hr$ and $r = 30 \, cm$.
Substituting these values into the equation:
$6\pi = 2\pi (30) \frac{dr}{dt}$.
$6\pi = 60\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{6\pi}{60\pi} = \frac{1}{10} = 0.1 \, cm/hr$.
Thus,the rate at which the radius of the circular sheet increases is $0.1 \, cm/hr$.

(C) The given differential equation is $\frac{(2 + \sin x)}{(1 + y)} \frac{dy}{dx} = \cos x.$
Separating the variables,we get $\frac{dy}{1 + y} = \frac{\cos x}{2 + \sin x} dx.$
Integrating both sides,we have $\int \frac{dy}{1 + y} = \int \frac{\cos x}{2 + \sin x} dx.$
This yields $\ln(1 + y) = \ln(2 + \sin x) + \ln C.$
Exponentiating both sides,we get $1 + y = C(2 + \sin x).$
Given $y(0) = 2,$ we substitute $x = 0$ and $y = 2$ to find $C$: $1 + 2 = C(2 + \sin 0) \Rightarrow 3 = 2C \Rightarrow C = \frac{3}{2}.$
Now,we find $y\left( \frac{\pi}{2} \right)$ by substituting $x = \frac{\pi}{2}$ and $C = \frac{3}{2}$ into the equation $1 + y = \frac{3}{2}(2 + \sin x).$
$1 + y\left( \frac{\pi}{2} \right) = \frac{3}{2}(2 + \sin \frac{\pi}{2}) = \frac{3}{2}(2 + 1) = \frac{3}{2}(3) = \frac{9}{2}.$
Therefore,$y\left( \frac{\pi}{2} \right) = \frac{9}{2} - 1 = \frac{7}{2}.$

(D) Given the equation $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$.
Applying the Leibniz integral rule,we differentiate both sides with respect to $x$:
$\frac{d}{{dx}}\left[ {\int\limits_e^x {t\,f(t)\,dt} } \right] = \frac{d}{{dx}}\left[ {\sin x - x\cos x - \frac{{{x^2}}}{2}} \right]$
Using the product rule on the right side:
$x\,f(x) = \cos x - (\cos x - x\sin x) - x$
$x\,f(x) = \cos x - \cos x + x\sin x - x$
$x\,f(x) = x\sin x - x$
Dividing by $x$ (since $x \neq 0$):
$f(x) = \sin x - 1$
Now,substitute $x = \frac{\pi}{6}$:
$f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) - 1$
$f(\frac{\pi}{6}) = \frac{1}{2} - 1 = -\frac{1}{2}$.

(D) The given system of equations is:
$x + y + z = 6$
$x + 2y + 3z = 10$
$x + 2y + \lambda z = 0$
For a system of linear equations to have a unique solution,the determinant of the coefficient matrix must be non-zero.
Let $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \neq 0$
Expanding the determinant along the first row:
$D = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) \neq 0$
$D = 2\lambda - 6 - \lambda + 3 + 0 \neq 0$
$D = \lambda - 3 \neq 0$
Therefore,$\lambda \neq 3$.

(D) Statement-$2$: $\cos^3 x$ is a periodic function. This is a true statement because $\cos x$ is periodic with period $2\pi$,and any power of a periodic function is also periodic.
Given $f(x) = \int \cos^3 x \, dx$.
Using the identity $\cos 3x = 4\cos^3 x - 3\cos x$,we have $\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$.
$f(x) = \int \left(\frac{\cos 3x}{4} + \frac{3\cos x}{4}\right) dx = \frac{1}{4} \cdot \frac{\sin 3x}{3} + \frac{3}{4} \sin x + C = \frac{1}{12} \sin 3x + \frac{3}{4} \sin x + C$.
The period of $\sin 3x$ is $\frac{2\pi}{3}$ and the period of $\sin x$ is $2\pi$.
The period of the sum of two periodic functions is the least common multiple $(LCM)$ of their individual periods.
Period of $f(x) = \text{LCM}\left(\frac{2\pi}{3}, 2\pi\right) = 2\pi$.
Since the period is $2\pi$ and not $\pi$,Statement-$1$ is false.

(B) The intersection points of $y^2 = x$ and $x^2 + y^2 = 2$ are found by substituting $y^2 = x$ into the circle equation: $x^2 + x - 2 = 0$,which gives $(x+2)(x-1) = 0$. Since $x \ge 0$,we have $x = 1$. Thus,the intersection points are $(1, 1)$ and $(1, -1)$.
The area of the circle is $A_{total} = \pi r^2 = 2\pi$.
The area of the region bounded by the parabola and the circle to the right of the $y$-axis is given by:
$A_1 = 2 \int_{0}^{1} \sqrt{x} \, dx + 2 \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx$
Calculating the first integral:
$2 \int_{0}^{1} x^{1/2} \, dx = 2 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{4}{3}$.
Calculating the second integral:
$2 \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx = 2 \left[ \frac{x}{2} \sqrt{2 - x^2} + \frac{2}{2} \sin^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_{1}^{\sqrt{2}}$
$= 2 \left[ (0 + \sin^{-1}(1)) - (\frac{1}{2} \sqrt{1} + \sin^{-1} \left( \frac{1}{\sqrt{2}} \right)) \right]$
$= 2 \left[ \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} \right] = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \right] = \frac{\pi}{2} - 1$.
So,$A_1 = \frac{4}{3} + \frac{\pi}{2} - 1 = \frac{\pi}{2} + \frac{1}{3} = \frac{3\pi + 2}{6}$.
The other area is $A_2 = A_{total} - A_1 = 2\pi - \frac{3\pi + 2}{6} = \frac{12\pi - 3\pi - 2}{6} = \frac{9\pi - 2}{6}$.
The ratio of the areas is $A_2 : A_1 = \frac{9\pi - 2}{6} : \frac{3\pi + 2}{6} = 9\pi - 2 : 3\pi + 2$.

(D) In a parallelogram,the diagonals bisect each other. Therefore,the midpoint $M$ of the diagonal $DB$ is the same as the midpoint of the diagonal $AC$.
The position vector of $M$ is given by $\vec{OM} = \frac{\vec{OA} + \vec{OC}}{2} = \frac{(3\hat{i} + 3\hat{j} + 5\hat{k}) + (\hat{i} - 5\hat{j} - 5\hat{k})}{2} = \frac{4\hat{i} - 2\hat{j} + 0\hat{k}}{2} = 2\hat{i} - \hat{j}$.
The projection of vector $\vec{OM}$ on $\vec{OC}$ is given by the formula $\frac{\vec{OM} \cdot \vec{OC}}{|\vec{OC}|}$.
We have $\vec{OM} = 2\hat{i} - \hat{j}$ and $\vec{OC} = \hat{i} - 5\hat{j} - 5\hat{k}$.
The dot product $\vec{OM} \cdot \vec{OC} = (2)(1) + (-1)(-5) + (0)(-5) = 2 + 5 + 0 = 7$.
The magnitude of $\vec{OC}$ is $|\vec{OC}| = \sqrt{1^2 + (-5)^2 + (-5)^2} = \sqrt{1 + 25 + 25} = \sqrt{51}$.
Therefore,the magnitude of the projection is $\frac{\vec{OM} \cdot \vec{OC}}{|\vec{OC}|} = \frac{7}{\sqrt{51}}$.

(B) Given the function $f(x) = \frac{x}{1 + |x|}$ for $x \in R$.
Case $1$: If $x \ge 0$,then $|x| = x$.
$f(x) = \frac{x}{1 + x} = \frac{x + 1 - 1}{1 + x} = 1 - \frac{1}{1 + x}$.
As $x$ increases from $0$ to $\infty$,$1 + x$ increases from $1$ to $\infty$,so $\frac{1}{1 + x}$ decreases from $1$ to $0$.
Thus,$f(x)$ increases from $0$ to $1$. So,the range for $x \ge 0$ is $[0, 1)$.
Case $2$: If $x < 0$,then $|x| = -x$.
$f(x) = \frac{x}{1 - x} = \frac{-(1 - x) + 1}{1 - x} = -1 + \frac{1}{1 - x}$.
As $x$ decreases from $0$ to $-\infty$,$1 - x$ increases from $1$ to $\infty$,so $\frac{1}{1 - x}$ decreases from $1$ to $0$.
Thus,$f(x)$ decreases from $0$ to $-1$. So,the range for $x < 0$ is $(-1, 0)$.
Combining both cases,the range of the function is $(-1, 0) \cup [0, 1) = (-1, 1)$.

(A) Three vectors $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
The scalar triple product is given by the determinant of the components of the vectors:
$\begin{vmatrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ \lambda & 1 & 2\lambda - 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1[3(2\lambda - 1) - (-1)(1)] - (-2)[2(2\lambda - 1) - (-1)(\lambda)] + 3[2(1) - 3(\lambda)] = 0$
$1[6\lambda - 3 + 1] + 2[4\lambda - 2 + \lambda] + 3[2 - 3\lambda] = 0$
$(6\lambda - 2) + 2(5\lambda - 2) + (6 - 9\lambda) = 0$
$6\lambda - 2 + 10\lambda - 4 + 6 - 9\lambda = 0$
$(6\lambda + 10\lambda - 9\lambda) + (-2 - 4 + 6) = 0$
$7\lambda + 0 = 0$
$\lambda = 0$

(D) Given $P(X \cup Y) = P(X \cap Y)$.
We know that $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$.
Substituting the given condition,we get $P(X \cap Y) = P(X) + P(Y) - P(X \cap Y)$,which implies $P(X) + P(Y) = 2P(X \cap Y)$. Thus,Statement $2$ is true.
Now,$P(X \cap Y') = P(X) - P(X \cap Y)$ and $P(X' \cap Y) = P(Y) - P(X \cap Y)$.
Since $P(X \cup Y) = P(X \cap Y)$,it implies $P(X \cup Y) - P(X \cap Y) = 0$,which is $P(X \Delta Y) = 0$,where $X \Delta Y$ is the symmetric difference.
This means $P(X \cap Y') = 0$ and $P(X' \cap Y) = 0$. Thus,Statement $1$ is true.
Since $P(X \cap Y') = P(X) - P(X \cap Y) = 0$ and $P(X' \cap Y) = P(Y) - P(X \cap Y) = 0$,we have $P(X) = P(X \cap Y)$ and $P(Y) = P(X \cap Y)$,which leads to $P(X) + P(Y) = 2P(X \cap Y)$. Therefore,Statement $2$ is the correct explanation of Statement $1$.

(C) Let the plane be $f(x, y, z) = 3x + 4y - 12z + 13 = 0$.
Two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie on opposite sides of the plane if $f(x_1, y_1, z_1) \cdot f(x_2, y_2, z_2) < 0$.
For point $P_1 = (1, a, 1)$,$f(1, a, 1) = 3(1) + 4(a) - 12(1) + 13 = 4a + 4$.
For point $P_2 = (-3, 0, a)$,$f(-3, 0, a) = 3(-3) + 4(0) - 12(a) + 13 = 4 - 12a$.
We require $(4a + 4)(4 - 12a) < 0$.
Dividing by $16$,we get $(a + 1)(1 - 3a) < 0$.
Multiplying by $-1$ reverses the inequality: $(a + 1)(3a - 1) > 0$.
The roots are $a = -1$ and $a = \frac{1}{3}$.
The inequality holds for $a < -1$ or $a > \frac{1}{3}$.

(C) The line makes equal angles with the coordinate axes,so its direction cosines are equal. Let the direction cosines be $(l, l, l)$. Since $l^2 + l^2 + l^2 = 1,$ we have $3l^2 = 1,$ so $l = \frac{1}{\sqrt{3}}$ (as direction cosines are positive).
The direction ratios of the line are proportional to $(1, 1, 1).$
The equation of the line passing through $P(2, -1, 2)$ with direction ratios $(1, 1, 1)$ is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = r.$
Any point $Q$ on this line is given by $(r+2, r-1, r+2).$
Since $Q$ lies on the plane $2x + y + z = 9,$ we substitute the coordinates of $Q$ into the plane equation:
$2(r+2) + (r-1) + (r+2) = 9$
$2r + 4 + r - 1 + r + 2 = 9$
$4r + 5 = 9$
$4r = 4 \Rightarrow r = 1.$
Thus,the point $Q$ is $(1+2, 1-1, 1+2) = (3, 0, 3).$
The distance $PQ$ is $\sqrt{(3-2)^2 + (0 - (-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.$

(C) Consider the function $h(x) = g(x) - f(x) = x - \sin x$ for $x \in (0, \infty)$.
Taking the derivative,$h'(x) = 1 - \cos x$.
Since $-1 \le \cos x \le 1$,we have $h'(x) = 1 - \cos x \ge 0$ for all $x$.
Since $h'(x) \ge 0$ and $h(0) = 0 - \sin(0) = 0$,the function $h(x)$ is non-decreasing and $h(x) \ge 0$ for all $x \in (0, \infty)$.
Thus,$x \ge \sin x$,which means $g(x) \ge f(x)$ is true. So,Statement $1$ is true.
For Statement $2$,we know that $\sin x \le 1$ for all $x \in (0, \infty)$ and $\lim_{x \to \infty} x = \infty$. Both parts of Statement $2$ are true.
However,the fact that $\sin x \le 1$ and $x \to \infty$ does not directly prove $\sin x \le x$ for all $x > 0$ without considering the behavior of the difference $x - \sin x$. Therefore,Statement $2$ is not the correct explanation for Statement $1$.

(B) Given the equation $x + |y| = 2y$.
Case $1$: If $y \ge 0,$ then $|y| = y.$
$x + y = 2y \Rightarrow y = x.$
Case $2$: If $y < 0,$ then $|y| = -y.$
$x - y = 2y \Rightarrow x = 3y \Rightarrow y = x/3.$
Thus,the function is defined as $f(x) = \begin{cases} x, & x \ge 0 \\ x/3, & x < 0 \end{cases}$.
Continuity at $x = 0$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x/3 = 0$.
Since $f(0) = 0,$ the function is continuous at $x = 0$.
Differentiability at $x = 0$:
$LHD = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{(0-h)/3 - 0}{-h} = 1/3$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1$.
Since $LHD \neq RHD,$ the function is not differentiable at $x = 0$.

(C) Given $A = \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix}$ and $B = \begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix}$.
Calculating $AB$:
$AB = \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix} \begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix} = \begin{bmatrix} 0 & \alpha \gamma \\ \beta \delta & 0 \end{bmatrix}$.
Calculating $BA$:
$BA = \begin{bmatrix} 0 & \gamma \\ \delta & 0 \end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix} = \begin{bmatrix} 0 & \gamma \beta \\ \delta \alpha & 0 \end{bmatrix}$.
Now,$AB - BA = \begin{bmatrix} 0 & \alpha \gamma - \beta \gamma \\ \beta \delta - \alpha \delta & 0 \end{bmatrix} = \begin{bmatrix} 0 & \gamma(\alpha - \beta) \\ \delta(\beta - \alpha) & 0 \end{bmatrix}$.
The determinant $|AB - BA| = 0 - (\gamma(\alpha - \beta) \cdot \delta(\beta - \alpha)) = \gamma \delta (\alpha - \beta)^2$.
For $AB - BA$ to be invertible,the determinant must be non-zero. This depends on $\alpha \neq \beta$ and $\gamma \delta \neq 0$. If these conditions are not met,it is not always invertible. However,assuming the standard context of such problems where $\alpha \neq \beta$ and $\gamma, \delta \neq 0$,Statement $1$ is considered true.
For Statement $2$: $AB - BA = \begin{bmatrix} 0 & \gamma(\alpha - \beta) \\ -\delta(\alpha - \beta) & 0 \end{bmatrix}$. For this to be the identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,the diagonal elements must be $1$. Since the diagonal elements are $0$,it can never be the identity matrix. Thus,Statement $2$ is true.

(D) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ be the required unit vector.
Since $\vec{v}$ is perpendicular to $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$,we have $\vec{v} \cdot \vec{a} = 0$.
$2x - y + 2z = 0$ ...... $(i)$
Since $\vec{v}$ is coplanar with $\vec{b} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + 2\hat{j} - \hat{k}$,$\vec{v}$ can be written as a linear combination of $\vec{b}$ and $\vec{c}$.
$\vec{v} = p(\hat{i} + \hat{j} - \hat{k}) + q(2\hat{i} + 2\hat{j} - \hat{k}) = (p + 2q)\hat{i} + (p + 2q)\hat{j} - (p + q)\hat{k}$.
Comparing components,$x = p + 2q$,$y = p + 2q$,$z = -(p + q)$.
Substitute into $(i)$: $2(p + 2q) - (p + 2q) + 2(-p - q) = 0$.
$2p + 4q - p - 2q - 2p - 2q = 0 \Rightarrow -p = 0 \Rightarrow p = 0$.
Thus,$x = 2q, y = 2q, z = -q$.
Since $\vec{v}$ is a unit vector,$x^2 + y^2 + z^2 = 1$.
$(2q)^2 + (2q)^2 + (-q)^2 = 1 \Rightarrow 4q^2 + 4q^2 + q^2 = 1 \Rightarrow 9q^2 = 1 \Rightarrow q = \pm \frac{1}{3}$.
For $q = \frac{1}{3}$,$\vec{v} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} = \frac{2\hat{i} + 2\hat{j} - \hat{k}}{3}$.
This matches option $D$.

(D) Statement $1$ is true. This is the standard definition of continuity at a point $x_0$.
Statement $2$ is false. $A$ function $f$ is discontinuous at $x_0$ if it is not continuous at $x_0$. This occurs if $\lim_{x \to x_0} f(x)$ does not exist,$OR$ if $\lim_{x \to x_0} f(x)$ exists but $\lim_{x \to x_0} f(x) \neq f(x_0)$. Statement $2$ only describes one specific case of discontinuity (removable discontinuity) and ignores other cases like jump discontinuity or infinite discontinuity where the limit might not exist at all.

(D) Let the given line be $\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8} = k$.
Any point $L$ on the line is given by $L = (2k + 1, -3k - 1, 8k - 10)$.
Let $P = (1, 0, 0)$. The direction ratios of the line $PL$ are $(2k + 1 - 1, -3k - 1 - 0, 8k - 10 - 0) = (2k, -3k - 1, 8k - 10)$.
Since $PL$ is perpendicular to the given line with direction ratios $(2, -3, 8)$,the dot product of their direction ratios must be zero:
$2(2k) - 3(-3k - 1) + 8(8k - 10) = 0$.
$4k + 9k + 3 + 64k - 80 = 0$.
$77k - 77 = 0$,which gives $k = 1$.
Substituting $k = 1$ into the coordinates of $L$:
$L = (2(1) + 1, -3(1) - 1, 8(1) - 10) = (3, -4, -2)$.

(B) Given vectors are $\vec{u} = \hat{j} + 4\hat{k}$,$\vec{v} = \hat{i} + 3\hat{k}$,and $\vec{w} = \cos \theta \hat{i} + \sin \theta \hat{j}$.
First,calculate the cross product $\vec{u} \times \vec{v}$:
$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 4 \\ 1 & 0 & 3 \end{vmatrix}$
$= \hat{i}(1 \times 3 - 4 \times 0) - \hat{j}(0 \times 3 - 4 \times 1) + \hat{k}(0 \times 0 - 1 \times 1)$
$= 3\hat{i} + 4\hat{j} - \hat{k}$.
Now,calculate the scalar triple product $(\vec{u} \times \vec{v}) \cdot \vec{w}$:
$(\vec{u} \times \vec{v}) \cdot \vec{w} = (3\hat{i} + 4\hat{j} - \hat{k}) \cdot (\cos \theta \hat{i} + \sin \theta \hat{j})$
$= 3 \cos \theta + 4 \sin \theta$.
The expression is of the form $a \cos \theta + b \sin \theta$,whose maximum value is $\sqrt{a^2 + b^2}$.
Here,$a = 3$ and $b = 4$.
Maximum value $= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

(D) Let $A$ be the area,$b$ be the breadth,and $\ell$ be the length of the rectangle.
Given: $\frac{dA}{dt} = -5$,$\frac{d\ell}{dt} = 2$,and $\frac{db}{dt} = -3$.
We know that the area of a rectangle is $A = \ell \times b$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \ell \cdot \frac{db}{dt} + b \cdot \frac{d\ell}{dt}$.
Substituting the given values:
$-5 = \ell(-3) + b(2)$.
$-5 = -3\ell + 2b$.
When the breadth $b = 2 \, m$,we substitute this into the equation:
$-5 = -3\ell + 2(2)$.
$-5 = -3\ell + 4$.
$3\ell = 4 + 5$.
$3\ell = 9$.
$\ell = 3 \, m$.
Thus,the length of the rectangle is $3 \, m$.

(C) Given $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$.
Using the chain rule,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = f'\left(\frac{2x + 3}{3 - 2x}\right) \cdot \frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right)$.
First,calculate the derivative of the inner function using the quotient rule:
$\frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right) = \frac{(3 - 2x)(2) - (2x + 3)(-2)}{(3 - 2x)^2} = \frac{6 - 4x + 4x + 6}{(3 - 2x)^2} = \frac{12}{(3 - 2x)^2}$.
Now,substitute this back into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right] \cdot \frac{12}{(3 - 2x)^2}$.
Thus,$\frac{dy}{dx} = \frac{12}{(3 - 2x)^2} \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right]$.

(D) For Statement $1$:
Consider the first differential equation: $\frac{dy}{dx} + y^2 = x$. The highest order derivative is $\frac{dy}{dx}$ (order $1$),and its power is $1$. Thus,the degree is $1$.
Consider the second differential equation: $\frac{d^2y}{dx^2} + y = \sin x$. The highest order derivative is $\frac{d^2y}{dx^2}$ (order $2$),and its power is $1$. Thus,the degree is $1$.
Since both equations have a degree of $1$,Statement $1$ is true.
For Statement $2$:
This is the standard definition of the degree of a differential equation. It is a polynomial in derivatives,and the degree is the highest power of the highest order derivative. Thus,Statement $2$ is true.
Conclusion:
Statement $2$ provides the definition used to determine the degrees in Statement $1$,making it the correct explanation. Therefore,the correct option is $D$.

(A) Let the points be $A(1, 2, 2), B(2, 1, 2), C(2, 2, z)$ and $D(1, 1, 1)$.
The points are coplanar if the scalar triple product of vectors $\vec{AB}, \vec{AC}$ and $\vec{AD}$ is $0$.
$\vec{AB} = (2-1, 1-2, 2-2) = (1, -1, 0)$
$\vec{AC} = (2-1, 2-2, z-2) = (1, 0, z-2)$
$\vec{AD} = (1-1, 1-2, 1-2) = (0, -1, -1)$
The condition for coplanarity is $\begin{vmatrix} 1 & -1 & 0 \\ 1 & 0 & z-2 \\ 0 & -1 & -1 \end{vmatrix} = 0$.
Expanding the determinant: $1(0 - (-(z-2))) - (-1)(-1 - 0) + 0 = 0$
$1(z-2) - 1 = 0 \Rightarrow z-3 = 0 \Rightarrow z = 3$.
Since $z=3 \neq 2$,Statement $1$ is false.
Statement $2$ is a standard geometric property: if four points are coplanar,they do not form a tetrahedron of non-zero volume,hence the volume is $0$. Thus,Statement $2$ is true.

(D) The given curves are $y = x^2$ and $y = x^3$.
The intersection point of $y = x^2$ and $y = x^3$ is found by setting $x^2 = x^3$,which gives $x^2(x - 1) = 0$,so $x = 0$ or $x = 1$.
For $0 < x < 1$,$x^2 > x^3$,and for $x > 1$,$x^3 > x^2$.
Given $p > 1$,the area is the sum of the areas in the intervals $[0, 1]$ and $[1, p]$:
Area $= \int_{0}^{1} (x^2 - x^3) \, dx + \int_{1}^{p} (x^3 - x^2) \, dx = \frac{1}{6}$.
Calculating the first integral:
$\int_{0}^{1} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$.
Calculating the second integral:
$\int_{1}^{p} (x^3 - x^2) \, dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} \right]_{1}^{p} = \left( \frac{p^4}{4} - \frac{p^3}{3} \right) - \left( \frac{1}{4} - \frac{1}{3} \right) = \frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{12}$.
Summing the areas:
$\frac{1}{12} + \frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{12} = \frac{1}{6}$.
$\frac{p^4}{4} - \frac{p^3}{3} + \frac{2}{12} = \frac{1}{6}$.
$\frac{p^4}{4} - \frac{p^3}{3} + \frac{1}{6} = \frac{1}{6}$.
$\frac{p^4}{4} - \frac{p^3}{3} = 0$.
Multiplying by $12$:
$3p^4 - 4p^3 = 0$.
$p^3(3p - 4) = 0$.
Since $p > 1$,we have $p = 4/3$.

(C) Given $f(x) = x e^{x(1-x)}$.
Applying the product rule and chain rule,we find the derivative:
$f'(x) = 1 \cdot e^{x(1-x)} + x \cdot e^{x(1-x)} \cdot (1-2x)$
$f'(x) = e^{x(1-x)} [1 + x - 2x^2]$
$f'(x) = -e^{x(1-x)} [2x^2 - x - 1]$
Factoring the quadratic expression:
$f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$
$f'(x) = -2 e^{x(1-x)} (x + 1/2)(x - 1)$
Since $e^{x(1-x)} > 0$ for all $x \in R$,the sign of $f'(x)$ depends on the expression $-(x + 1/2)(x - 1)$.
For $x \in [-1/2, 1]$,the product $(x + 1/2)(x - 1) \leq 0$.
Therefore,$-(x + 1/2)(x - 1) \geq 0$.
Thus,$f'(x) \geq 0$ for $x \in [-1/2, 1]$.
Hence,$f(x)$ is increasing on $[-1/2, 1]$.

(A) Let $I = \int \frac{x^2 - x}{x^3 - x^2 + x - 1} dx$.
Factor the denominator: $x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1)$.
Substitute this into the integral: $I = \int \frac{x(x - 1)}{(x^2 + 1)(x - 1)} dx$.
Cancel the common term $(x - 1)$: $I = \int \frac{x}{x^2 + 1} dx$.
Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2x}{x^2 + 1} dx$.
Let $u = x^2 + 1$,then $du = 2x dx$.
Substituting these into the integral: $I = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log |u| + c$.
Since $x^2 + 1 > 0$ for all real $x$,we can write: $I = \frac{1}{2} \log (x^2 + 1) + c$.

(C) Let $\Delta = \left| \begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \end{array} \right|$.
Applying $C_1 \to C_1 + C_3$ and $C_2 \to C_2 + C_3$:
$\Delta = \left| \begin{array}{ccc} -a+c & 2a+b+c & a+c \\ 2b+a+c & -b+c & b+c \\ a-c & b-c & -2c \end{array} \right|$.
Applying $R_1 \to R_1 + R_3$ and $R_2 \to R_2 + R_3$:
$\Delta = \left| \begin{array}{ccc} 0 & 2(a+b) & a-c \\ 2(a+b) & 0 & b-c \\ a-c & b-c & -2c \end{array} \right|$.
Expanding along the first row:
$\Delta = 0 - 2(a+b) \left[ -4c(a+b) - (a-c)(b-c) \right] + (a-c) \left[ 2(a+b)(b-c) - 0 \right]$.
$\Delta = 8c(a+b)^2 + 2(a+b)(a-c)(b-c) + 2(a+b)(a-c)(b-c)$.
$\Delta = 8c(a+b)^2 + 4(a+b)(a-c)(b-c)$.
$\Delta = 4(a+b) \left[ 2c(a+b) + (a-c)(b-c) \right]$.
$\Delta = 4(a+b) \left[ 2ac + 2bc + ab - ac - bc + c^2 \right]$.
$\Delta = 4(a+b) \left[ ac + bc + ab + c^2 \right]$.
$\Delta = 4(a+b) \left[ c(a+c) + b(a+c) \right]$.
$\Delta = 4(a+b)(b+c)(c+a)$.
Comparing with $\alpha(a+b)(b+c)(c+a)$,we get $\alpha = 4$.

(B) relation $R$ on a set $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.
First,we identify the elements of set $A$: $A = \{3, 4, 6, 8, 9\}$. The number of elements $n = 5$.
The total number of ordered pairs in $A \times A$ is $n^2 = 5^2 = 25$.
These $25$ pairs can be categorized into:
$1$. Pairs of the form $(a, a)$: There are $n = 5$ such pairs: $(3, 3), (4, 4), (6, 6), (8, 8), (9, 9)$.
$2$. Pairs of the form $(a, b)$ where $a \neq b$: There are $n^2 - n = 25 - 5 = 20$ such pairs.
For a symmetric relation,if $(a, b)$ is included,$(b, a)$ must also be included. These $20$ pairs form $10$ pairs of the form $\{(a, b), (b, a)\}$.
For each of the $5$ diagonal pairs $(a, a)$,we have $2$ choices (include or exclude).
For each of the $10$ pairs of the form $\{(a, b), (b, a)\}$,we have $2$ choices (include both or exclude both).
Thus,the total number of symmetric relations is $2^5 \times 2^{10} = 2^{5+10} = 2^{15}$.

(A) Given $\frac{d}{dx} G(x) = \frac{e^{\tan x}}{x}$ for $x \in (0, \pi/2)$.
Let $I = \int_{1/4}^{1/2} \frac{2}{x} e^{\tan(\pi x^2)} dx$.
Multiply and divide by $\pi$ inside the integral: $I = \int_{1/4}^{1/2} \frac{2\pi x}{\pi x^2} e^{\tan(\pi x^2)} dx$.
Let $t = \pi x^2$. Then $dt = 2\pi x dx$.
When $x = 1/4$,$t = \pi/16$. When $x = 1/2$,$t = \pi/4$.
Substituting these into the integral,we get $I = \int_{\pi/16}^{\pi/4} \frac{e^{\tan t}}{t} dt$.
Since $\frac{d}{dt} G(t) = \frac{e^{\tan t}}{t}$,the integral becomes $[G(t)]_{\pi/16}^{\pi/4} = G(\pi/4) - G(\pi/16)$.

(C) The given expression is $\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i} = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{2n+1}{3}$.
For the expression to be an integer,$2n+1$ must be divisible by $3$.
This means $2n+1 \equiv 0 \pmod{3}$,which implies $2n \equiv -1 \equiv 2 \pmod{3}$,so $n \equiv 1 \pmod{3}$.
We need to find the number of values of $n$ in the set $\{1, 2, 3, \dots, 1000\}$ such that $n = 3k+1$ for some integer $k \ge 0$.
For $k=0$,$n=1$. For $k=333$,$n=3(333)+1 = 1000$.
Thus,$k$ can take values from $0$ to $333$,which gives a total of $333 - 0 + 1 = 334$ values.
The total number of possible values for $n$ is $1000$.
Therefore,the required probability is $\frac{334}{1000} = 0.334$.

(D) The given region is a circle $x^2 + y^2 = 4$ with radius $r = 2$ and center at $(0, 0)$. The line $y = x + 2$ passes through $(-2, 0)$ and $(0, 2)$.
Let $I$ be the smaller portion and $II$ be the greater portion of the circle.
The area of the smaller portion $I$ is given by the integral:
Area of $I = \int_{-2}^{0} [\sqrt{4 - x^2} - (x + 2)] dx$
$= [\frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2})]_{-2}^{0} - [\frac{x^2}{2} + 2x]_{-2}^{0}$
$= [0 + 2 \sin^{-1}(0)] - [(-1) \sqrt{0} + 2 \sin^{-1}(-1)] - [0 - (\frac{4}{2} - 4)]$
$= [0 - 2(-\frac{\pi}{2})] - [0 - (-2)]$
$= \pi - 2$
Now,the area of the greater portion $II$ is:
Area of $II = \text{Area of circle} - \text{Area of } I$
$= 4\pi - (\pi - 2) = 3\pi + 2$
Therefore,the required ratio is:
$\frac{\text{Area of } I}{\text{Area of } II} = \frac{\pi - 2}{3\pi + 2}$

(D) Let $I = \int_{0}^{0.9} [x - 2[x]] \, dx$.
Since $0 \le x < 0.9$,the greatest integer function $[x] = 0$.
Substituting $[x] = 0$ into the integral,we get:
$I = \int_{0}^{0.9} [x - 2(0)] \, dx = \int_{0}^{0.9} [x] \, dx$.
Since $0 \le x < 0.9$,$[x] = 0$ for all $x$ in the interval $[0, 0.9)$.
Therefore,$I = \int_{0}^{0.9} 0 \, dx = 0$.

(C) Let $\Delta = \left| \begin{array}{ccc} b^2 + c^2 & ab & ac \\ ab & c^2 + a^2 & bc \\ ac & bc & a^2 + b^2 \end{array} \right|$.
Multiply $C_1$ by $a$,$C_2$ by $b$,and $C_3$ by $c$,and divide the determinant by $abc$:
$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} a(b^2 + c^2) & ab^2 & ac^2 \\ a^2b & b(c^2 + a^2) & bc^2 \\ a^2c & b^2c & c(a^2 + b^2) \end{array} \right|$.
Take out $a, b, c$ common from $R_1, R_2, R_3$ respectively:
$\Delta = \frac{abc}{abc} \left| \begin{array}{ccc} b^2 + c^2 & b^2 & c^2 \\ a^2 & c^2 + a^2 & c^2 \\ a^2 & b^2 & a^2 + b^2 \end{array} \right| = \left| \begin{array}{ccc} b^2 + c^2 & b^2 & c^2 \\ a^2 & c^2 + a^2 & c^2 \\ a^2 & b^2 & a^2 + b^2 \end{array} \right|$.
Apply $C_1 \to C_1 - C_2 - C_3$:
$\Delta = \left| \begin{array}{ccc} 0 & b^2 & c^2 \\ -2c^2 & c^2 + a^2 & c^2 \\ -2b^2 & b^2 & a^2 + b^2 \end{array} \right| = -2 \left| \begin{array}{ccc} 0 & b^2 & c^2 \\ c^2 & c^2 + a^2 & c^2 \\ b^2 & b^2 & a^2 + b^2 \end{array} \right|$.
Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = -2 \left| \begin{array}{ccc} 0 & b^2 & c^2 \\ c^2 & a^2 & 0 \\ b^2 & 0 & a^2 \end{array} \right| = -2 [ -b^2(c^2 a^2) + c^2(-a^2 b^2) ] = -2 [-a^2 b^2 c^2 - a^2 b^2 c^2] = 4a^2 b^2 c^2$.
Since $\Delta = k a^2 b^2 c^2$,we have $k = 4$.

(A) Given $f(x) = \int \left( \frac{x^2 + \sin^2 x}{1 + x^2} \right) \sec^2 x \, dx$.
We can rewrite the integrand as:
$f(x) = \int \frac{x^2 \sec^2 x + \sin^2 x \sec^2 x}{1 + x^2} \, dx$
Since $\sin^2 x \sec^2 x = \tan^2 x$,we have:
$f(x) = \int \frac{x^2 \sec^2 x + \tan^2 x}{1 + x^2} \, dx$
Using $\sec^2 x = 1 + \tan^2 x$,we get:
$f(x) = \int \frac{x^2(1 + \tan^2 x) + \tan^2 x}{1 + x^2} \, dx$
$f(x) = \int \frac{x^2 + x^2 \tan^2 x + \tan^2 x}{1 + x^2} \, dx = \int \frac{x^2 + \tan^2 x(1 + x^2)}{1 + x^2} \, dx$
$f(x) = \int \frac{x^2}{1 + x^2} \, dx + \int \tan^2 x \, dx$
$f(x) = \int \frac{x^2 + 1 - 1}{1 + x^2} \, dx + \int (\sec^2 x - 1) \, dx$
$f(x) = \int (1 - \frac{1}{1 + x^2}) \, dx + \int \sec^2 x \, dx - \int 1 \, dx$
$f(x) = x - \tan^{-1} x + \tan x - x + C = \tan x - \tan^{-1} x + C$
Given $f(0) = 0$,we have $0 = \tan 0 - \tan^{-1} 0 + C$,so $C = 0$.
Thus,$f(x) = \tan x - \tan^{-1} x$.
Therefore,$f(1) = \tan 1 - \tan^{-1}(1) = \tan 1 - \frac{\pi}{4}$.

(D) The given differential equation is $\frac{dy}{dx} + \frac{2}{x}y = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x^2$.
First,we calculate the integrating factor ($I$.$F$.):
$I.F. = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values,we get $y \cdot x^2 = \int (x^2 \cdot x^2) dx + c$.
$y \cdot x^2 = \int x^4 dx + c$.
$y \cdot x^2 = \frac{x^5}{5} + c$.
Dividing both sides by $x^2$,we get $y = \frac{x^3}{5} + cx^{-2}$.

(D) Let $\theta = \cos^{-1} \sqrt{\frac{2}{3}}$.
Then $\cos \theta = \sqrt{\frac{2}{3}}$.
We know that $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
Substituting the value of $\cos \theta$,we get $\sin \theta = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.
Now,the expression becomes $\tan^{-1} (\sin \theta) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$.
Since $\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}$,we have $\tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$.