AIEEE 2009 Mathematics Question Paper with Answer and Solution

(B) Given $A \cap B = A \cap C$ and $A \cup B = A \cup C$.
Consider $B = B \cap (A \cup B)$.
Since $A \cup B = A \cup C$,we have $B = B \cap (A \cup C)$.
Using the distributive law,$B = (B \cap A) \cup (B \cap C)$.
Since $A \cap B = A \cap C$,we have $B = (A \cap C) \cup (B \cap C)$.
$B = (A \cup B) \cap C$.
Since $A \cup B = A \cup C$,we have $B = (A \cup C) \cap C$.
Since $(A \cup C) \cap C = C$,it follows that $B = C$.

(B) Given $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$.
Multiplying by $2$,we get $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] = -3$.
Adding $3$ to both sides: $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] + 3 = 0$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we can write $3 = (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + (\sin^2 \gamma + \cos^2 \gamma)$.
Substituting this into the equation:
$(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha) + (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha) = 0$.
This simplifies to $(\sin \alpha + \sin \beta + \sin \gamma)^2 + (\cos \alpha + \cos \beta + \cos \gamma)^2 = 0$.
Since the sum of squares is zero,each term must be zero:
$\sin \alpha + \sin \beta + \sin \gamma = 0$ and $\cos \alpha + \cos \beta + \cos \gamma = 0$.
Therefore,both statements $A$ and $B$ are true.

(C) Step $1$: Select $4$ novels from $6$ different novels in $^6C_4$ ways.
$^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$ ways.
Step $2$: Select $1$ dictionary from $3$ different dictionaries in $^3C_1$ ways.
$^3C_1 = 3$ ways.
Step $3$: Arrange the $4$ selected novels and $1$ dictionary in a row such that the dictionary is always in the middle.
Since the dictionary is fixed in the middle,we only need to arrange the $4$ novels in the remaining $4$ positions.
The number of ways to arrange $4$ novels is $4! = 24$.
Step $4$: Total number of arrangements = $^6C_4 \times ^3C_1 \times 4! = 15 \times 3 \times 24 = 1080$.

(A) We need to find the remainder when $8^{2n} - 62^{2n+1}$ is divided by $9$.
Note that $8 \equiv -1 \pmod{9}$ and $62 \equiv 8 \equiv -1 \pmod{9}$.
Substituting these into the expression:
$8^{2n} - 62^{2n+1} \equiv (-1)^{2n} - (-1)^{2n+1} \pmod{9}$.
Since $2n$ is an even integer,$(-1)^{2n} = 1$.
Since $2n+1$ is an odd integer,$(-1)^{2n+1} = -1$.
Therefore,$1 - (-1) = 1 + 1 = 2$.
Thus,the remainder is $2$.

(A) Let the sum be $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ $(1)$
Multiply by $\frac{1}{3}$: $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots$ $(2)$
Subtracting $(2)$ from $(1)$:
$S - \frac{1}{3}S = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{3^2} - \frac{2}{3^2}) + (\frac{10}{3^3} - \frac{6}{3^3}) + (\frac{14}{3^4} - \frac{10}{3^4}) + \dots$
$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots$
$\frac{2}{3}S = 1 + \frac{1}{3} + [\frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots]$
The term in the bracket is an infinite geometric series with $a = \frac{4}{9}$ and $r = \frac{1}{3}$.
Sum $= \frac{a}{1-r} = \frac{4/9}{1 - 1/3} = \frac{4/9}{2/3} = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3}$.
So,$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{2}{3} = 1 + 1 = 2$.
$S = 2 \times \frac{3}{2} = 3$.

(A) If two lines are perpendicular to a common line,they must be parallel to each other.
Let the slopes of the lines be $m_1$ and $m_2$.
For the first line $p(p^2 + 1)x - y + q = 0$,the slope $m_1 = \frac{p(p^2 + 1)}{1} = p(p^2 + 1)$.
For the second line $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$,the slope $m_2 = -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1)$.
Since the lines are parallel,$m_1 = m_2$.
$p(p^2 + 1) = -(p^2 + 1)$.
Since $p^2 + 1 \neq 0$ for any real $p$,we can divide by $(p^2 + 1)$.
$p = -1$.
Thus,there is exactly one value of $p$.

(A) The equation of the common chord (radical axis) of the two circles $S_1: x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $S_2: x^2 + y^2 + 2x + 2y - p^2 = 0$ is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 + 3x + 7y + 2p - 5) - (x^2 + y^2 + 2x + 2y - p^2) = 0$
$x + 5y + 2p - 5 + p^2 = 0 \quad \dots(i)$
For a circle to pass through the intersection points $P$ and $Q$ and the point $(1, 1)$,the point $(1, 1)$ must not lie on the common chord $PQ$.
Substituting $(1, 1)$ into equation $(i)$:
$1 + 5(1) + 2p - 5 + p^2 \neq 0$
$p^2 + 2p + 1 \neq 0$
$(p + 1)^2 \neq 0$
$p \neq -1$
Thus,there exists a circle passing through $P, Q$ and $(1, 1)$ for all values of $p$ except $p = -1$.

(A) The given ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.
This ellipse has semi-axes $a = 2$ and $b = 1$.
The rectangle inscribed around this ellipse has vertices at $(\pm 2, \pm 1)$.
The outer ellipse is inscribed in this rectangle,meaning it passes through the points $(\pm 2, \pm 1)$.
Let the equation of the outer ellipse be $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.
Since it passes through $(4,0)$,we have $\frac{4^2}{A^2} + \frac{0^2}{B^2} = 1$,which gives $A^2 = 16$.
Since it also passes through $(2,1)$,we have $\frac{2^2}{16} + \frac{1^2}{B^2} = 1$.
$\frac{4}{16} + \frac{1}{B^2} = 1 \implies \frac{1}{4} + \frac{1}{B^2} = 1 \implies \frac{1}{B^2} = \frac{3}{4} \implies B^2 = \frac{4}{3}$.
Substituting $A^2$ and $B^2$ into the equation: $\frac{x^2}{16} + \frac{y^2}{4/3} = 1$.
$\frac{x^2}{16} + \frac{3y^2}{4} = 1$.
Multiplying by $16$,we get $x^2 + 12y^2 = 16$.

(B) The given series $1, 1+d, 1+2d, \dots, 1+100d$ is an $A.P.$ with $n = 101$ terms.
The mean of this series is $\bar{x} = \frac{\sum_{r=0}^{100} (1+rd)}{101} = \frac{101 + d \frac{100 \times 101}{2}}{101} = 1 + 50d$.
The mean deviation from the mean is given by $\frac{1}{101} \sum_{r=0}^{100} |(1+rd) - (1+50d)| = \frac{1}{101} \sum_{r=0}^{100} |(r-50)d|$.
Assuming $d > 0$,this becomes $\frac{d}{101} [\sum_{r=0}^{50} (50-r) + \sum_{r=51}^{100} (r-50)]$.
$= \frac{d}{101} [ (50+49+\dots+0) + (1+2+\dots+50) ] = \frac{d}{101} [ 2 \times \frac{50 \times 51}{2} ] = \frac{50 \times 51 \times d}{101}$.
Given the mean deviation is $255$,we have $\frac{50 \times 51 \times d}{101} = 255$.
$d = \frac{255 \times 101}{50 \times 51} = \frac{5 \times 101}{50} = \frac{101}{10} = 10.1$.

(C) Given that the equation $bx^2 + cx + a = 0$ has imaginary roots,the discriminant $D < 0$.
Therefore,$c^2 - 4ab < 0$,which implies $c^2 < 4ab$.
Multiplying by $-1$,we get $-c^2 > -4ab$.
Now,consider the expression $E = 3b^2x^2 + 6bcx + 2c^2$.
$E = 3(b^2x^2 + 2bcx + c^2) - c^2$.
$E = 3(bx + c)^2 - c^2$.
Since $(bx + c)^2 \geq 0$,it follows that $3(bx + c)^2 \geq 0$.
Thus,$E \geq -c^2$.
Since $-c^2 > -4ab$,we have $E > -4ab$.

(D) To evaluate Statement $-1$,we analyze the truth table for $\sim (p \leftrightarrow \sim q)$.
Recall that $(p \leftrightarrow \sim q)$ is true when $p$ and $\sim q$ have the same truth value,which means $p$ and $q$ have opposite truth values.
Thus,$(p \leftrightarrow \sim q)$ is equivalent to $\sim (p \leftrightarrow q)$.
Therefore,$\sim (p \leftrightarrow \sim q)$ is equivalent to $\sim (\sim (p \leftrightarrow q))$,which simplifies to $(p \leftrightarrow q)$.
So,Statement $-1$ is true.
To evaluate Statement $-2$,we check if $\sim (p \leftrightarrow \sim q)$ is a tautology.
Since $\sim (p \leftrightarrow \sim q) \equiv (p \leftrightarrow q)$,and $(p \leftrightarrow q)$ is not true for all values of $p$ and $q$ (it is false when $p$ and $q$ have different truth values),the expression is not a tautology.
Thus,Statement $-2$ is false.

(D) Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$
Since $(\sim q \rightarrow \sim p)$ is the contrapositive of $(p \rightarrow q)$,they have the same truth values.
Thus,$(p \rightarrow q) \leftrightarrow (p \rightarrow q)$ is always true,which means it is a tautology.
So,Statement-$2$ is true.
Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$
Using the associative and commutative laws:
$= (p \wedge \sim p) \wedge (\sim q \wedge q)$
$= F \wedge F = F$
Since the result is always false,it is a fallacy.
So,Statement-$1$ is true.
Both statements are true,and Statement-$2$ is not an explanation for Statement-$1$.

(C) Let the common root be $\alpha$. The equations are $x^2-6x+a=0$ and $x^2-cx+6=0$.
Let the other roots be $\beta_1$ and $\beta_2$ respectively. Given $\beta_1 : \beta_2 = 4:3$,so $\beta_1 = 4k$ and $\beta_2 = 3k$ for some constant $k$.
From the properties of roots:
For the first equation: $\alpha + 4k = 6$ and $\alpha \cdot 4k = a$.
For the second equation: $\alpha + 3k = c$ and $\alpha \cdot 3k = 6$.
From $\alpha \cdot 3k = 6$,we have $k = \frac{2}{\alpha}$.
Substitute $k$ into the first equation: $\alpha + 4(\frac{2}{\alpha}) = 6 \Rightarrow \alpha + \frac{8}{\alpha} = 6$.
Multiplying by $\alpha$: $\alpha^2 - 6\alpha + 8 = 0$.
Factoring: $(\alpha - 4)(\alpha - 2) = 0$,so $\alpha = 4$ or $\alpha = 2$.
If $\alpha = 4$,then $4k = 6 - 4 = 2 \Rightarrow k = 0.5$. Then $\beta_1 = 4(0.5) = 2$ and $\beta_2 = 3(0.5) = 1.5$. Since $\beta_2$ must be an integer,this is rejected.
If $\alpha = 2$,then $4k = 6 - 2 = 4 \Rightarrow k = 1$. Then $\beta_1 = 4(1) = 4$ and $\beta_2 = 3(1) = 3$. Both are integers.
Thus,the common root is $2$.

(A) The line $\frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$ lies on the plane $x + 3y - \alpha z + \beta = 0$.
Since the line is parallel to the plane,the normal vector of the plane $(1, 3, -\alpha)$ must be perpendicular to the direction vector of the line $(3, -5, 2)$.
Thus,$1(3) + 3(-5) + (-\alpha)(2) = 0$.
$3 - 15 - 2\alpha = 0$.
$-12 - 2\alpha = 0 \implies \alpha = -6$.
Now,the plane equation is $x + 3y + 6z + \beta = 0$.
Since the line lies on the plane,any point on the line must satisfy the plane equation. The point $(2, 1, -2)$ lies on the line.
Substituting $(2, 1, -2)$ into the plane equation: $2 + 3(1) + 6(-2) + \beta = 0$.
$2 + 3 - 12 + \beta = 0$.
$-7 + \beta = 0 \implies \beta = 7$.
Therefore,$(\alpha, \beta) = (-6, 7)$.

(B) The equation of the parabola is $(y-2)^2 = x-1$.
The tangent at the point $(2,3)$ is found using the formula $y-y_1 = m(x-x_1)$.
First,differentiate the parabola equation with respect to $x$: $2(y-2) \frac{dy}{dx} = 1$,so $\frac{dy}{dx} = \frac{1}{2(y-2)}$.
At $(2,3)$,the slope $m = \frac{1}{2(3-2)} = \frac{1}{2}$.
The equation of the tangent is $y-3 = \frac{1}{2}(x-2)$,which simplifies to $2y-6 = x-2$,or $x = 2y-4$.
The region is bounded by the parabola $x = (y-2)^2 + 1$,the tangent $x = 2y-4$,and the $x$-axis $(y=0)$.
The intersection of the tangent with the $x$-axis occurs at $y=0$,which gives $x = 2(0)-4 = -4$.
The intersection of the parabola with the $x$-axis occurs at $y=0$,which gives $x = (0-2)^2 + 1 = 5$.
However,the region is bounded by the tangent and the parabola up to the point $(2,3)$. Integrating with respect to $y$ from $y=0$ to $y=3$:
Area $A = \int_{0}^{3} [x_{parabola} - x_{tangent}] dy = \int_{0}^{3} [((y-2)^2 + 1) - (2y-4)] dy$
$A = \int_{0}^{3} [y^2 - 4y + 4 + 1 - 2y + 4] dy = \int_{0}^{3} [y^2 - 6y + 9] dy$
$A = \int_{0}^{3} (y-3)^2 dy = \left[ \frac{(y-3)^3}{3} \right]_{0}^{3} = 0 - \left( \frac{-27}{3} \right) = 9$.

(C) Let $I = \int_{0}^{\pi} [\cot x] dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi} [\cot(\pi - x)] dx = \int_{0}^{\pi} [-\cot x] dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi} ([\cot x] + [-\cot x]) dx$.
We know that for any $y \notin \mathbb{Z}$,$[y] + [-y] = -1$,and for $y \in \mathbb{Z}$,$[y] + [-y] = 0$.
The function $\cot x$ is not defined at $x=0, \pi/2, \pi$. At $x=\pi/2$,$\cot x = 0$,so $[0] + [0] = 0$.
For all other $x \in (0, \pi)$,$\cot x$ is not an integer.
Thus,$[\cot x] + [-\cot x] = -1$ almost everywhere in $(0, \pi)$.
$2I = \int_{0}^{\pi} (-1) dx = -\pi$.
Therefore,$I = -\frac{\pi}{2}$.

(C) For any $n \times n$ matrix $A$,the property of the adjoint is $adj(adj A) = |A|^{n-2} A$.
Given $n = 2$,we have $adj(adj A) = |A|^{2-2} A = |A|^0 A = I \cdot A = A$. Thus,$Statement-1$ is true.
For $Statement-2$,we know that $|adj A| = |A|^{n-1}$.
Given $n = 2$,we have $|adj A| = |A|^{2-1} = |A|^1 = |A|$. Thus,$Statement-2$ is true.
Since $adj(adj A) = |A|^{n-2} A$,the result $adj(adj A) = A$ depends on the property $|A| = 1$ or the general identity for $n=2$. However,the standard property $adj(adj A) = |A|^{n-2} A$ is the fundamental reason why $adj(adj A) = A$ for $n=2$. Therefore,$Statement-2$ provides the necessary context for the identity in $Statement-1$.

(C) Let the given equation be $D_1 + D_2 = 0$.
First,we observe that the second determinant $D_2$ can be transposed without changing its value.
$D_2 = \left| \begin{array}{ccc} a+1 & a-1 & (-1)^{n+2}a \\ b+1 & b-1 & (-1)^{n+1}b \\ c-1 & c+1 & (-1)^n c \end{array} \right|$.
By adding the two determinants,we find that for the sum to be zero for arbitrary $a, b, c$,the columns must cancel out.
Specifically,if $n$ is an odd integer,then $(-1)^{n+2} = -1$,$(-1)^{n+1} = 1$,and $(-1)^n = -1$.
Substituting these into the sum of determinants,the terms cancel out to zero.
Thus,$n$ must be any odd integer.

(D) Given the equation: ${x^{2x}} - 2{x^x}\cot y - 1 = 0$ ........$(i)$
At $x = 1$,we have:
$1^{2(1)} - 2(1^1)\cot y - 1 = 0$
$1 - 2\cot y - 1 = 0$
$\Rightarrow \cot y = 0 \quad \therefore \quad y = \frac{\pi}{2}$
Differentiating $(i)$ with respect to $x$:
$\frac{d}{dx}(x^{2x}) - 2 \frac{d}{dx}(x^x \cot y) = 0$
$x^{2x} \cdot \frac{d}{dx}(2x \ln x) - 2 \left[ x^x \ln x \cdot \cot y + x^x \cdot (-\csc^2 y) \frac{dy}{dx} + \cot y \cdot x^x(1 + \ln x) \right] = 0$
Using $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$:
$2x^{2x}(1 + \ln x) - 2 \left[ x^x(1 + \ln x) \cot y - x^x \csc^2 y \frac{dy}{dx} \right] = 0$
At $P(1, \frac{\pi}{2})$,we have $x = 1, y = \frac{\pi}{2}, \cot y = 0, \csc^2 y = 1, \ln 1 = 0$:
$2(1)^{2}(1 + 0) - 2 \left[ 1(1 + 0)(0) - 1(1) \left(\frac{dy}{dx}\right)_{P} \right] = 0$
$2 - 2 \left[ -\left(\frac{dy}{dx}\right)_{P} \right] = 0$
$2 + 2 \left(\frac{dy}{dx}\right)_{P} = 0$
$\left(\frac{dy}{dx}\right)_{P} = -1$

(D) We have $f(x) = x|x|$ and $g(x) = \sin x$.
Therefore,$(gof)(x) = \sin(x|x|) = \begin{cases} -\sin(x^2), & x < 0 \\ \sin(x^2), & x \ge 0 \end{cases}$.
$LHD$ of $gof$ at $x=0$: $\lim_{x \to 0^-} \frac{-\sin(x^2) - 0}{x - 0} = \lim_{x \to 0^-} -x \left(\frac{\sin(x^2)}{x^2}\right) = 0$.
$RHD$ of $gof$ at $x=0$: $\lim_{x \to 0^+} \frac{\sin(x^2) - 0}{x - 0} = \lim_{x \to 0^+} x \left(\frac{\sin(x^2)}{x^2}\right) = 0$.
Since $LHD = RHD = 0$,$gof$ is differentiable at $x=0$ and $(gof)'(0) = 0$.
The derivative is $(gof)'(x) = \begin{cases} -2x\cos(x^2), & x < 0 \\ 2x\cos(x^2), & x \ge 0 \end{cases}$.
Since $\lim_{x \to 0^-} (gof)'(x) = 0$ and $\lim_{x \to 0^+} (gof)'(x) = 0$,the derivative is continuous at $x=0$. Thus,Statement-$1$ is true.
Now,check for second differentiability at $x=0$:
$LHD$ of $(gof)'$ at $x=0$: $\lim_{x \to 0^-} \frac{-2x\cos(x^2) - 0}{x - 0} = -2\cos(0) = -2$.
$RHD$ of $(gof)'$ at $x=0$: $\lim_{x \to 0^+} \frac{2x\cos(x^2) - 0}{x - 0} = 2\cos(0) = 2$.
Since $LHD \neq RHD$,$gof$ is not twice differentiable at $x=0$. Thus,Statement-$2$ is false.

(B) Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$.
$P'(x) = 4x^3 + 3ax^2 + 2bx + c$.
Since $x=0$ is the only real root of $P'(x)=0$,we have $P'(0) = 0$,which implies $c=0$.
Thus,$P'(x) = x(4x^2 + 3ax + 2b)$.
Since $x=0$ is the only real root,the quadratic factor $4x^2 + 3ax + 2b$ must have no real roots,meaning its discriminant $D < 0$.
$D = (3a)^2 - 4(4)(2b) = 9a^2 - 32b < 0$.
Since the leading coefficient $4 > 0$,the quadratic $4x^2 + 3ax + 2b > 0$ for all $x \in \mathbb{R}$.
Therefore,the sign of $P'(x)$ is determined by the sign of $x$.
For $x \in [-1, 0)$,$P'(x) < 0$,so $P(x)$ is strictly decreasing.
For $x \in (0, 1]$,$P'(x) > 0$,so $P(x)$ is strictly increasing.
Since $P(x)$ decreases on $[-1, 0]$ and increases on $[0, 1]$,the global minimum on $[-1, 1]$ occurs at $x=0$.
Thus,$P(-1)$ is not the minimum.
Since $P(x)$ is increasing on $(0, 1]$ and $P(-1) < P(1)$,the maximum value on $[-1, 1]$ must be $P(1)$.
Therefore,$P(-1)$ is not the minimum,but $P(1)$ is the maximum.

(C) Given the family of curves: $y = c_1 e^{c_2 x} \dots (i)$
Differentiating with respect to $x$,we get:
$y' = c_1 c_2 e^{c_2 x} = c_2 y \dots (ii)$
Again,differentiating with respect to $x$,we get:
$y'' = c_2 y' \dots (iii)$
From equation $(ii)$,we have $c_2 = \frac{y'}{y}$.
Substituting this value of $c_2$ into equation $(iii)$:
$y'' = \left( \frac{y'}{y} \right) y'$
Multiplying both sides by $y$:
$y y'' = (y')^2$
This is the required differential equation.

(D) Let $S = \{00, 01, 02, \ldots, 49\}$ be the sample space. The total number of tickets is $50$.
Let $A$ be the event that the sum of the digits on the selected ticket is $8$.
$A = \{08, 17, 26, 35, 44\}$.
Let $B$ be the event that the product of the digits is zero.
This occurs if at least one digit is $0$. The tickets are $\{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$.
There are $14$ such tickets,so $n(B) = 14$.
We need to find the conditional probability $P(A|B) = \frac{n(A \cap B)}{n(B)}$.
$A \cap B$ is the set of tickets where the sum of digits is $8$ $AND$ the product of digits is $0$.
Looking at set $A = \{08, 17, 26, 35, 44\}$,only $08$ has a product of digits equal to $0$ $(0 \times 8 = 0)$.
Thus,$A \cap B = \{08\}$,so $n(A \cap B) = 1$.
Therefore,$P(A|B) = \frac{1}{14}$.

(A) The probability of at least one success is given by $P(X \geq 1) = 1 - P(X = 0)$.
In a binomial distribution $B(n, p)$,$P(X = 0) = q^n$,where $q = 1 - p$.
Given $p = \frac{1}{4}$,we have $q = 1 - \frac{1}{4} = \frac{3}{4}$.
The condition is $1 - (\frac{3}{4})^n \geq \frac{9}{10}$.
Rearranging the inequality,we get $1 - \frac{9}{10} \geq (\frac{3}{4})^n$,which simplifies to $\frac{1}{10} \geq (\frac{3}{4})^n$.
Taking the logarithm with base $10$ on both sides: $\log_{10}(\frac{1}{10}) \geq \log_{10}((\frac{3}{4})^n)$.
$-1 \geq n(\log_{10} 3 - \log_{10} 4)$.
Multiplying by $-1$ reverses the inequality sign: $1 \leq n(\log_{10} 4 - \log_{10} 3)$.
Therefore,$n \geq \frac{1}{\log_{10} 4 - \log_{10} 3}$.

(C) Given $f(x) = x^3 + 5x + 1.$
First,we find the derivative: $f'(x) = 3x^2 + 5.$
Since $x^2 \ge 0$ for all $x \in R,$ it follows that $3x^2 + 5 \ge 5 > 0$ for all $x \in R.$
Because $f'(x) > 0$ for all $x,$ the function $f(x)$ is strictly increasing on $R.$
$A$ strictly increasing function is always one-one.
Next,we check if the function is onto. Since $f(x)$ is a polynomial of odd degree,its range is $(-\infty, \infty).$
Specifically,$\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty.$
Since $f(x)$ is continuous and its range is the set of all real numbers $R,$ the function is onto.
Therefore,$f$ is both one-one and onto.

(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$.
To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Since $x \geq -1$,$y \geq -1$.
Solving for $x$: $(x+1)^2 = y+1 \Rightarrow x+1 = \sqrt{y+1} \Rightarrow x = \sqrt{y+1} - 1$.
Thus,$f^{-1}(x) = \sqrt{x+1} - 1$. Since $f$ is strictly increasing on $[-1, \infty)$,it is a bijection (one-one and onto). So,Statement-$2$ is true.
To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$ because $f$ is increasing.
$(x+1)^2 - 1 = x \Rightarrow x^2 + 2x + 1 - 1 = x \Rightarrow x^2 + x = 0 \Rightarrow x(x+1) = 0$.
Thus,$x = 0$ or $x = -1$. So,$S = \{0, -1\}$. Statement-$1$ is true.
Since $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for increasing functions,Statement-$2$ explains Statement-$1$.

(B) Let the vector be $\vec{v} = (a, b, c) = (6, -3, 2)$.
The magnitude of the vector is given by $|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$.
$|\vec{v}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
The direction cosines $(l, m, n)$ of a vector are given by dividing the components by the magnitude of the vector:
$l = \frac{a}{|\vec{v}|} = \frac{6}{7}$,
$m = \frac{b}{|\vec{v}|} = \frac{-3}{7}$,
$n = \frac{c}{|\vec{v}|} = \frac{2}{7}$.
Thus,the direction cosines are $\left(\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}\right)$.

(D) Given the equation: $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$
Using the property of scalar triple product $[k\vec{a}, l\vec{b}, m\vec{c}] = klm[\vec{a}, \vec{b}, \vec{c}]$,we get:
$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{v}, \vec{w}, \vec{u}] - 2q^2[\vec{w}, \vec{v}, \vec{u}] = 0$
Since $[\vec{v}, \vec{w}, \vec{u}] = [\vec{u}, \vec{v}, \vec{w}]$ and $[\vec{w}, \vec{v}, \vec{u}] = -[\vec{u}, \vec{v}, \vec{w}]$,the equation becomes:
$3p^2[\vec{u}, \vec{v}, \vec{w}] - pq[\vec{u}, \vec{v}, \vec{w}] + 2q^2[\vec{u}, \vec{v}, \vec{w}] = 0$
$(3p^2 - pq + 2q^2)[\vec{u}, \vec{v}, \vec{w}] = 0$
Since $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar,$[\vec{u}, \vec{v}, \vec{w}] \neq 0$. Therefore:
$3p^2 - pq + 2q^2 = 0$
This is a quadratic equation in $p$. For $p$ to be a real number,the discriminant $D$ must be $\geq 0$:
$D = (-q)^2 - 4(3)(2q^2) = q^2 - 24q^2 = -23q^2$
For $D \geq 0$,we must have $-23q^2 \geq 0$,which implies $q^2 \leq 0$. Since $q$ is a real number,this is only possible if $q = 0$.
Substituting $q = 0$ into $3p^2 - pq + 2q^2 = 0$,we get $3p^2 = 0$,so $p = 0$.
Thus,the only solution is $(p, q) = (0, 0)$.