AIEEE 2012 Mathematics Question Paper with Answer and Solution

(A) Since $\sec(\theta - \phi)$,$\sec \theta$,and $\sec(\theta + \phi)$ are in $A.P.$,we have:
$2 \sec \theta = \sec(\theta - \phi) + \sec(\theta + \phi)$
$\frac{2}{\cos \theta} = \frac{\cos(\theta + \phi) + \cos(\theta - \phi)}{\cos(\theta - \phi) \cos(\theta + \phi)}$
Using the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$ and $\cos(A-B) \cos(A+B) = \cos^2 A - \sin^2 B$:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \phi}{\cos^2 \theta - \sin^2 \phi}$
$\cos^2 \theta - \sin^2 \phi = \cos^2 \theta \cos \phi$
$\cos^2 \theta (1 - \cos \phi) = \sin^2 \phi$
$\cos^2 \theta (1 - \cos \phi) = 1 - \cos^2 \phi = (1 - \cos \phi)(1 + \cos \phi)$
Since $\phi \neq 0$,$1 - \cos \phi \neq 0$,so:
$\cos^2 \theta = 1 + \cos \phi = 2 \cos^2(\frac{\phi}{2})$
$\cos \theta = \pm \sqrt{2} \cos(\frac{\phi}{2})$
Given $\cos \theta = k \cos(\frac{\phi}{2})$,we find $k = \pm \sqrt{2}$.

(A) Given $n = ^mC_2 = \frac{m(m-1)}{2}$.
We need to find $^nC_2 = \frac{n(n-1)}{2}$.
Substituting $n = \frac{m(m-1)}{2}$:
$^nC_2 = \frac{\frac{m(m-1)}{2} \left( \frac{m(m-1)}{2} - 1 \right)}{2}$
$= \frac{\frac{m(m-1)}{2} \left( \frac{m^2-m-2}{2} \right)}{2}$
$= \frac{m(m-1)(m^2-m-2)}{8}$
$= \frac{m(m-1)(m-2)(m+1)}{8}$
$= \frac{3 \times (m+1)m(m-1)(m-2)}{4 \times 3 \times 2 \times 1}$
$= 3(^{m+1}C_4)$.

(A) The set of letters is $\{a, b, c, d, e, f\}$,which contains $2$ vowels $(\{a, e\})$ and $4$ consonants $(\{b, c, d, f\})$.
We need to form arrangements of $3$ letters containing at least one vowel.
Total arrangements with at least one vowel = (Total arrangements of $3$ letters) - (Arrangements with no vowels).
Total arrangements of $3$ letters from $6$ distinct letters = $^6P_3 = 6 \times 5 \times 4 = 120$.
Arrangements with no vowels (only consonants) = $^4P_3 = 4 \times 3 \times 2 = 24$.
Therefore,the number of arrangements with at least one vowel = $120 - 24 = 96$.

(C) Let $P(0, 1)$ be the given point and $L: 2x - y = 0$ be the line.
The formula for the image $(x_1, y_1)$ of a point $(x_0, y_0)$ in the line $ax + by + c = 0$ is given by $\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
Substituting the values: $\frac{x_1 - 0}{2} = \frac{y_1 - 1}{-1} = -2 \frac{2(0) - 1(1)}{2^2 + (-1)^2} = -2 \frac{-1}{5} = \frac{2}{5}$.
Thus,$x_1 = 2 \times \frac{2}{5} = \frac{4}{5}$ and $y_1 = 1 - \frac{2}{5} = \frac{3}{5}$.
So,Statement $1$ is true.
By definition,the image of a point in a line is the point such that the line is the perpendicular bisector of the segment joining the point and its image. This implies the points lie on opposite sides and are equidistant from the line. Thus,Statement $2$ is true and is the correct explanation for Statement $1$.

(A) The formula for the median of grouped data is:
$M = l + \left( \frac{\frac{N}{2} - F}{f} \right) \times C$
Where:
$l = 20$ (lower limit of the median class)
$N = 100$ (total number of observations)
$F = 45$ (cumulative frequency of the class preceding the median class)
$C = 30 - 20 = 10$ (width of the median class)
$M = 25$ (median)
$f$ is the frequency of the median class.
Substituting the values into the formula:
$25 = 20 + \left( \frac{\frac{100}{2} - 45}{f} \right) \times 10$
$25 - 20 = \left( \frac{50 - 45}{f} \right) \times 10$
$5 = \frac{5}{f} \times 10$
$5 = \frac{50}{f}$
$f = \frac{50}{5} = 10$

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

(D) Given that the $A.M.$ of the $p^{th}$ and $q^{th}$ terms is equal to the $A.M.$ of the $r^{th}$ and $s^{th}$ terms of an $A.P.$:
$\frac{a_p + a_q}{2} = \frac{a_r + a_s}{2}$
Using the formula for the $n^{th}$ term of an $A.P.$,$a_n = a + (n-1)d$:
$a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d$
Simplifying the equation:
$2a + (p + q - 2)d = 2a + (r + s - 2)d$
Subtracting $2a$ from both sides:
$(p + q - 2)d = (r + s - 2)d$
Assuming $d \neq 0$,we can divide by $d$:
$p + q - 2 = r + s - 2$
Therefore:
$p + q = r + s$

(C) Consider $\cos 255^o + \sin 195^o$.
Using the reduction formulas:
$\cos 255^o = \cos(270^o - 15^o) = -\sin 15^o$.
$\sin 195^o = \sin(180^o + 15^o) = -\sin 15^o$.
Therefore,the expression becomes:
$-\sin 15^o - \sin 15^o = -2 \sin 15^o$.
Using the value $\sin 15^o = \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$:
$-2 \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) = -\frac{\sqrt{3} - 1}{\sqrt{2}}$.

(D) The given expression is ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$.
This can be rewritten as ${\left( \frac{x-1}{x} \right)^n} {(1-x)^n} = {\left( \frac{-(1-x)}{x} \right)^n} {(1-x)^n} = {(-1)^n} {x^{-n}} {(1-x)^{2n}}$.
The expansion of ${(1-x)^{2n}}$ has $2n+1$ terms.
The middle term is the ${\left( \frac{2n+1+1}{2} \right)}^{\text{th}} = {(n+1)}^{\text{th}}$ term.
The general term $T_{r+1}$ in the expansion of ${(1-x)^{2n}}$ is given by ${}^{2n}C_r {(-x)^r} = {}^{2n}C_r {(-1)^r} {x^r}$.
For the $(n+1)^{\text{th}}$ term,we set $r=n$,which gives ${}^{2n}C_n {(-1)^n} {x^n}$.
Substituting this back into the expression: ${(-1)^n} {x^{-n}} \cdot {}^{2n}C_n {(-1)^n} {x^n} = {(-1)^{2n}} {}^{2n}C_n = {}^{2n}C_n$ (since $2n$ is even,${(-1)^{2n}} = 1$).

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$
Using the identity $\cos^2 x = 1 - \sin^2 x$,we rewrite the expression:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi (1 - \sin^2 x)} \right)}}{{{x^2}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi - \pi \sin^2 x} \right)}}{{{x^2}}}$
Since $\sin(\pi - \theta) = \sin \theta$,we have:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{{x^2}}}$
Multiply and divide by $\pi \sin^2 x$:
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{\pi \sin^2 x}} \times \frac{{\pi \sin^2 x}}{{{x^2}}} \right)$
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta}}{\theta} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$:
$= 1 \times \pi \times (\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x})^2 = 1 \times \pi \times 1^2 = \pi$

(C) Given lines are:
$ax + 2by + 3b = 0$ $(1)$
$bx - 2ay - 3a = 0$ $(2)$
To find the intersection point,multiply $(1)$ by $a$ and $(2)$ by $b$:
$a^2x + 2aby + 3ab = 0$
$b^2x - 2aby - 3ab = 0$
Adding these equations: $(a^2 + b^2)x = 0$. Since $(a, b) \neq (0, 0)$,$a^2 + b^2 \neq 0$,so $x = 0$.
Substituting $x = 0$ into $(1)$: $2by = -3b$. Since $b$ cannot be $0$ (otherwise the lines would not intersect at a unique point or would be coincident),we get $y = -3/2$.
The point of intersection is $(0, -3/2)$.
$A$ line parallel to the $x-$ axis has the equation $y = k$. Since it passes through $(0, -3/2)$,the equation is $y = -3/2$.
This represents a line below the $x-$ axis at a distance of $3/2$ from it.

(A) The equation of the parabola is $y^2 = x$.
Given that the chord $PQ$ is perpendicular to the axis of the parabola (the $x$-axis) and one end $P$ is $(4, -2)$,the $x$-coordinate of $Q$ must also be $4$.
Since $Q$ lies on the parabola $y^2 = x$,substituting $x = 4$ gives $y^2 = 4$,so $y = \pm 2$. Since $P$ is $(4, -2)$,$Q$ must be $(4, 2)$.
The derivative of the parabola $y^2 = x$ with respect to $x$ is $2y \frac{dy}{dx} = 1$,which gives $\frac{dy}{dx} = \frac{1}{2y}$.
The slope of the tangent at $Q(4, 2)$ is $\frac{dy}{dx} \Big|_{(4, 2)} = \frac{1}{2(2)} = \frac{1}{4}$.
The slope of the normal at $Q$ is the negative reciprocal of the slope of the tangent.
Therefore,the slope of the normal is $-\frac{1}{1/4} = -4$.

(C) Let $p$: The sun is shining.
Let $q$: $I$ shall play tennis in the afternoon.
The given statement is of the form $p \to q$.
The negation of a conditional statement $p \to q$ is given by $\sim(p \to q) \equiv p \wedge \sim q$.
Therefore,the negation is "The sun is shining and $I$ shall not play tennis in the afternoon" which corresponds to $p \wedge \sim q$.

(C) For an odd $n$,the series consists of $n$ terms: $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2 + n^2$.
This can be written as the sum of the first $(n-1)$ terms plus the $n$-th term.
Since $(n-1)$ is even,we use the given formula for even terms: $S_{n-1} = \frac{(n-1)(n-1+1)^2}{2} = \frac{(n-1)n^2}{2}$.
Now,adding the $n$-th term $(n^2)$: $S_n = \frac{(n-1)n^2}{2} + n^2$.
Factoring out $n^2$: $S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.

(C) The given circles are $C_1: x^2 + y^2 - 8x - 2y + 1 = 0$ and $C_2: x^2 + y^2 + 6x + 8y = 0$.
For $C_1$,the center is $(4, 1)$ and the radius $r_1 = \sqrt{4^2 + 1^2 - 1} = \sqrt{16} = 4$.
For $C_2$,the center is $(-3, -4)$ and the radius $r_2 = \sqrt{(-3)^2 + (-4)^2 - 0} = \sqrt{25} = 5$.
The distance between the centers $d = \sqrt{(4 - (-3))^2 + (1 - (-4))^2} = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74}$.
Since $\sqrt{49} < \sqrt{74} < \sqrt{81}$,we have $7 < d < 9$.
Also,$r_1 + r_2 = 4 + 5 = 9$ and $|r_1 - r_2| = |4 - 5| = 1$.
Since $|r_1 - r_2| < d < r_1 + r_2$,the circles intersect at two distinct points.
Therefore,the number of common tangents is $2$.

(B) We use the property of the modulus of complex numbers: $\left| z \right|^2 = z \bar{z}$.
Expanding the terms:
${\left| {{z_1} + {z_2}} \right|^2} = (z_1 + z_2)(\bar{z_1} + \bar{z_2}) = z_1\bar{z_1} + z_1\bar{z_2} + z_2\bar{z_1} + z_2\bar{z_2} = \left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 + z_1\bar{z_2} + z_2\bar{z_1}$.
Similarly:
${\left| {{z_1} - {z_2}} \right|^2} = (z_1 - z_2)(\bar{z_1} - \bar{z_2}) = z_1\bar{z_1} - z_1\bar{z_2} - z_2\bar{z_1} + z_2\bar{z_2} = \left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 - z_1\bar{z_2} - z_2\bar{z_1}$.
Adding these two expressions:
${\left| {{z_1} + {z_2}} \right|^2} + {\left| {{z_1} - {z_2}} \right|^2} = (\left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 + z_1\bar{z_2} + z_2\bar{z_1}) + (\left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2 - z_1\bar{z_2} - z_2\bar{z_1})$
$= 2\left| {{z_1}} \right|^2 + 2\left| {{z_2}} \right|^2 = 2(\left| {{z_1}} \right|^2 + \left| {{z_2}} \right|^2)$.

(A) The given ellipse is $\frac{x^2}{16} + \frac{y^2}{3} = 1$.
Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 3$.
The equation of the normal at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.
Substituting $(x_1, y_1) = \left( 2, \frac{3}{2} \right)$,$a^2 = 16$,and $b^2 = 3$:
$\frac{16x}{2} - \frac{3y}{3/2} = 16 - 3$
$8x - 2y = 13$
$2y = 8x - 13 \Rightarrow y = 4x - \frac{13}{2}$.
$A$ line $y = mx + c$ touches the parabola $y^2 = 4Ax$ if $c = \frac{A}{m}$.
Here $m = 4$ and $c = -\frac{13}{2}$.
$-\frac{13}{2} = \frac{A}{4} \Rightarrow A = -26$.
Thus,the equation of the parabola is $y^2 = 4(-26)x = -104x$.

(C) Given that $1$ is a root of $ax^2 + bx + c = 0$,we have $a(1)^2 + b(1) + c = 0$,which implies $a + b + c = 0$,or $c = -a - b$.
We want to find the intersection of $y = 4ax^2 + 3bx + 2c$ with the $x$-axis,so we set $y = 0$:
$4ax^2 + 3bx + 2c = 0$.
Substitute $c = -a - b$ into the equation:
$4ax^2 + 3bx + 2(-a - b) = 0$
$4ax^2 + 3bx - 2a - 2b = 0$
Rearranging the terms:
$4ax^2 - 2a + 3bx - 2b = 0$
$2a(2x^2 - 1) + b(3x - 2) = 0$.
Alternatively,using the condition $a+b+c=0$,we can write $c = -(a+b)$.
$y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.
At $x=1$,$y = 4a + 3b - 2a - 2b = 2a + b$. This is not necessarily zero.
Let us re-evaluate the intersection $4ax^2 + 3bx + 2c = 0$ with $c = -a-b$:
$4ax^2 + 3bx - 2a - 2b = 0$.
For this to be a quadratic equation in $x$,the discriminant $D = (3b)^2 - 4(4a)(-2a-2b) = 9b^2 + 16a(2a+2b) = 9b^2 + 32a^2 + 32ab$.
This discriminant is not necessarily positive,zero,or negative without further constraints on $a$ and $b$. However,if we test specific values,e.g.,$a=1, b=-2, c=1$ (where $1-2+1=0$),then $y = 4x^2 - 6x + 2 = 2(2x^2 - 3x + 1) = 2(2x-1)(x-1)$. This intersects at $x=1$ and $x=1/2$ (two points).
If $a=1, b=-1, c=0$,then $y = 4x^2 - 3x = x(4x-3)$. This intersects at $x=0$ and $x=3/4$ (two points).
Thus,the curve intersects the $x$-axis at exactly two distinct points.

(D) The slopes of the lines are:
$m_1 = 1$ (for $L_1$)
$m_2 = -1$ (for $L_2$)
$m_3 = -1$ (for $L_3$)
$m_4 = 1$ (for $L_4$)
$1$. Check perpendicularity:
$m_1 \times m_2 = 1 \times (-1) = -1$,so $L_1 \perp L_2$.
$m_1 \times m_3 = 1 \times (-1) = -1$,so $L_1 \perp L_3$.
$2$. Check parallelism:
$m_2 = m_3 = -1$,so $L_2 || L_3$.
$3$. Check intersection:
Since $m_2 \neq m_4$ $(-1 \neq 1)$,$L_2$ and $L_4$ are not parallel and therefore intersect.
Comparing with the options,$L_1 \perp L_2$,$L_1 \perp L_3$,and $L_2$ intersects $L_4$ is the correct set of statements.

(D) The first $n$ odd natural numbers are $1, 3, 5, \dots, (2n-1)$.
The mean $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (2i-1) = \frac{1}{n} \cdot n^2 = n$.
The sum of squares is $\sum_{i=1}^{n} (2i-1)^2 = \sum (4i^2 - 4i + 1) = 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n = \frac{n(4n^2-1)}{3}$.
Variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = \frac{4n^2-1}{3} - n^2 = \frac{4n^2-1-3n^2}{3} = \frac{n^2-1}{3}$.
Thus,Statement $1$ is true and Statement $2$ is true. Statement $2$ provides the correct formulas used to derive Statement $1$.

(A) To ensure that there is a man on either side of every woman,the men and women must alternate around the circular table.
First,arrange the $7$ men around the circular table. The number of ways to arrange $n$ items in a circle is $(n-1)!$. Thus,$7$ men can be arranged in $(7-1)! = 6!$ ways.
Once the men are seated,there are $7$ distinct gaps created between them.
Since there are $7$ women to be seated in these $7$ gaps,they can be arranged in $7!$ ways.
Therefore,the total number of seating arrangements is $6! \times 7!$.

(B) Let the direction ratios of the common line be $\ell, m, n$.
Since the line lies on the plane $x = 5$,its direction vector must be perpendicular to the normal $(1, 0, 0)$. Thus,$\ell = 0$.
Since the line lies on the other two planes,its direction vector must be perpendicular to their normals $(2, -5a, 3)$ and $(3b, 1, -3)$.
For the plane $2x - 5ay + 3z - 2 = 0$,we have $2\ell - 5am + 3n = 0$. Since $\ell = 0$,we get $-5am + 3n = 0$,or $3n = 5am$.
For the plane $3bx + y - 3z = 0$,we have $3b\ell + m - 3n = 0$. Since $\ell = 0$,we get $m - 3n = 0$,or $m = 3n$.
Substituting $m = 3n$ into $3n = 5am$,we get $3n = 5a(3n)$.
Assuming $n \neq 0$,we have $3 = 15a$,which gives $a = \frac{1}{5}$.
Now,the common line must also satisfy the equations of the planes. Since $x = 5$,the other two equations become $-5ay + 3z = 2 - 2(5) = -8$ and $y - 3z = -3b(5) = -15b$.
Substituting $a = \frac{1}{5}$,the first equation becomes $-y + 3z = -8$,or $y - 3z = 8$.
Comparing $y - 3z = 8$ and $y - 3z = -15b$,we get $-15b = 8$,so $b = -\frac{8}{15}$.
Thus,$(a, b) = \left( \frac{1}{5}, -\frac{8}{15} \right)$.

(D) Let $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx.$
Then $g'(x) = ax^2 + bx + c.$
We are given $2a + 3b + 6c = 0.$
Statement $2:$
$g(0) = 0.$
$g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6} = \frac{0}{6} = 0.$
Since $g(0) = g(1) = 0$ and $g(x)$ is a polynomial,it is continuous on $[0, 1]$ and differentiable on $(0, 1).$
By Rolle's theorem,there exists at least one $k \in (0, 1)$ such that $g'(k) = 0.$
Thus,Statement $2$ is true.
Statement $1:$
Since $g'(k) = ak^2 + bk + c = 0$ for some $k \in (0, 1),$ the quadratic equation $ax^2 + bx + c = 0$ has at least one root in $(0, 1).$
Thus,Statement $1$ is true and Statement $2$ is the correct explanation for Statement $1.$

(B) The given lines are $L_1: \frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $L_2: \frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$.
Both lines have the same direction vector $\vec{b} = 2\hat{i} - \hat{j} + 2\hat{k}$,so they are parallel.
The formula for the shortest distance between two parallel lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}$ is $d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$.
Here,$\vec{a}_1 = 0\hat{i} + 0\hat{j} + 0\hat{k}$ and $\vec{a}_2 = 1\hat{i} + 1\hat{j} + 1\hat{k}$.
Thus,$\vec{a}_2 - \vec{a}_1 = \hat{i} + \hat{j} + \hat{k}$.
$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 2 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(2 - 2) + \hat{k}(-1 - 2) = 3\hat{i} + 0\hat{j} - 3\hat{k}$.
The magnitude is $|3\hat{i} - 3\hat{k}| = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.
The magnitude of $\vec{b}$ is $|2\hat{i} - \hat{j} + 2\hat{k}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$.
Therefore,$d = \frac{3\sqrt{2}}{3} = \sqrt{2}$.
Since the calculated distance is $\sqrt{2}$,Statement $1$ is true.
Statement $2$ is the standard definition of the shortest distance between parallel lines,which is also true.
Since Statement $2$ provides the method used to calculate the distance in Statement $1$,it is the correct explanation.

(C) Given the set $S = \{ 1, 2, 3 \}$.
The number of elements in $S$ is $n(S) = 3$.
$P(S)$ is the power set of $S$,which is the set of all subsets of $S$.
The number of elements in the power set is $n(P(S)) = 2^{n(S)} = 2^3 = 8$.
$A$ function $f: S \to P(S)$ is one-to-one (injective) if each element in $S$ maps to a unique element in $P(S)$.
The number of one-to-one functions from a set with $m$ elements to a set with $n$ elements is given by the permutation formula $^nP_m = \frac{n!}{(n-m)!}$.
Here,$m = n(S) = 3$ and $n = n(P(S)) = 8$.
Therefore,the number of one-to-one functions is $^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$.

(C) Given $W = nw$.
Using the product rule for differentiation,we have:
$\frac{dW}{dt} = n \frac{dw}{dt} + w \frac{dn}{dt}$
Given $n = 2t^2 + 3$ and $w = t^2 - t + 2$.
Calculating the derivatives with respect to $t$:
$\frac{dn}{dt} = 4t$
$\frac{dw}{dt} = 2t - 1$
At $t = 1$:
$n = 2(1)^2 + 3 = 5$
$w = (1)^2 - 1 + 2 = 2$
$\frac{dn}{dt} = 4(1) = 4$
$\frac{dw}{dt} = 2(1) - 1 = 1$
Substituting these values into the derivative formula:
$\frac{dW}{dt} = (5)(1) + (2)(4)$
$\frac{dW}{dt} = 5 + 8 = 13$

(B) The region is bounded by the curve $y = x^3$,the line $y = 8$,and the $y$-axis $(x = 0)$.
To find the area with respect to the $y$-axis,we express $x$ in terms of $y$: $x = y^{1/3}$.
The limits for $y$ are from $0$ to $8$.
Required Area $= \int\limits_{0}^{8} x \, dy = \int\limits_{0}^{8} y^{1/3} \, dy$
$= \left[ \frac{y^{1/3 + 1}}{1/3 + 1} \right]_{0}^{8} = \left[ \frac{y^{4/3}}{4/3} \right]_{0}^{8} = \left[ \frac{3}{4} y^{4/3} \right]_{0}^{8}$
$= \frac{3}{4} (8^{4/3} - 0^{4/3}) = \frac{3}{4} ((2^3)^{4/3}) = \frac{3}{4} (2^4) = \frac{3}{4} \times 16 = 12 \text{ sq. units.}$

(C) Given vectors are $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$,and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}$.
Since $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$,the vectors $\vec{a}, \vec{b}$,and $\vec{c}$ are coplanar.
For three vectors to be coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \, \vec{b} \, \vec{c}] = 0$.
This is equivalent to the determinant of their components being zero:
$\begin{vmatrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ r & 1 & 2r - 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((3)(2r - 1) - (-1)(1)) - (-2)((2)(2r - 1) - (-1)(r)) + 3((2)(1) - (3)(r)) = 0$
$1(6r - 3 + 1) + 2(4r - 2 + r) + 3(2 - 3r) = 0$
$(6r - 2) + 2(5r - 2) + (6 - 9r) = 0$
$6r - 2 + 10r - 4 + 6 - 9r = 0$
$7r = 0$
$r = 0$

(C) Given $A = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix}$. The determinant $|A| = 0(bf - ce) - 0(0 - cd) + a(0 - bd) = -abd$.
Since $abd \neq 0$,the matrix is invertible.
The cofactor matrix is calculated as:
$C_{11} = bf - ce, C_{12} = -(-cd) = cd, C_{13} = -bd$
$C_{21} = -(-ae) = ae, C_{22} = -ad, C_{23} = 0$
$C_{31} = -ab, C_{32} = 0, C_{33} = 0$
Thus,$\text{adj}(A) = \begin{bmatrix} bf-ce & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-abd} \begin{bmatrix} bf-ce & ae & -ab \\ cd & -ad & 0 \\ -bd & 0 & 0 \end{bmatrix}$.
Given $A^{-1} = A^T$,where $A^T = \begin{bmatrix} 0 & 0 & d \\ 0 & b & e \\ a & c & f \end{bmatrix}$.
Equating $A^{-1} = A^T$ leads to the condition $A A^T = I$ (since $A^{-1} = A^T \implies A A^T = I$).
Calculating $A A^T = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix} \begin{bmatrix} 0 & 0 & d \\ 0 & b & e \\ a & c & f \end{bmatrix} = \begin{bmatrix} a^2 & ac & af \\ ac & b^2+c^2 & be+cf \\ ad & ae+cf & d^2+e^2+f^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
From $a^2 = 1$,$a = \pm 1$.
From $ac = 0$,$c = 0$.
From $af = 0$,$f = 0$.
From $b^2+c^2 = 1$,$b^2 = 1 \implies b = \pm 1$.
From $be+cf = 0$,$be = 0 \implies e = 0$.
From $d^2+e^2+f^2 = 1$,$d^2 = 1 \implies d = \pm 1$.
Thus,$a, b, d \in \{1, -1\}$ and $c=e=f=0$.
The number of such matrices is $2 \times 2 \times 2 = 2^3 = 8$.

(C) Let the given point be $P(-1, 2, 6)$ and the line $L$ pass through $A(2, 3, -4)$ with direction vector $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.
The vector $\vec{AP} = (-1-2)\hat{i} + (2-3)\hat{j} + (6-(-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.
The distance $d$ of point $P$ from the line is given by $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$.
First,calculate the cross product $\vec{AP} \times \vec{v}$:
$\vec{AP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & 10 \\ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 + 6) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.
The magnitude $|\vec{AP} \times \vec{v}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989}$.
The magnitude $|\vec{v}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Thus,$d = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

(D) The given planes are:
$P: x + y - 2z + 7 = 0$
$Q: x + y + 2z + 2 = 0$
$R: 3x + 3y - 6z - 11 = 0$
To check if planes are parallel,we compare the ratios of the coefficients of $x, y,$ and $z$.
For plane $P$,the normal vector is $\vec{n_1} = (1, 1, -2)$.
For plane $R$,the normal vector is $\vec{n_2} = (3, 3, -6)$.
We observe that $\vec{n_2} = 3(1, 1, -2) = 3\vec{n_1}$.
Since the normal vectors are proportional,the planes $P$ and $R$ are parallel.
Thus,the correct option is $D$.

(A) Given matrices are $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{bmatrix}$.
To find $AB$,we perform matrix multiplication:
$AB = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{bmatrix}$
Calculating each element:
Row $1$: $(1)(1) + (0)(-2) + (0)(7) = 1$,$(1)(0) + (0)(1) + (0)(-2) = 0$,$(1)(0) + (0)(0) + (0)(1) = 0$
Row $2$: $(2)(1) + (1)(-2) + (0)(7) = 2 - 2 = 0$,$(2)(0) + (1)(1) + (0)(-2) = 1$,$(2)(0) + (1)(0) + (0)(1) = 0$
Row $3$: $(-3)(1) + (2)(-2) + (1)(7) = -3 - 4 + 7 = 0$,$(-3)(0) + (2)(1) + (1)(-2) = 2 - 2 = 0$,$(-3)(0) + (2)(0) + (1)(1) = 1$
Thus,$AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.

(B) Given equations are $y^2 = 4x$ and $2x - 3y + 4 = 0$.
From the line equation,$x = \frac{3y - 4}{2}$.
Substituting this into the parabola equation: $y^2 = 4 \left( \frac{3y - 4}{2} \right) = 2(3y - 4) = 6y - 8$.
$y^2 - 6y + 8 = 0 \implies (y - 2)(y - 4) = 0$.
So,$y = 2$ and $y = 4$.
When $y = 2$,$x = \frac{3(2) - 4}{2} = 1$. When $y = 4$,$x = \frac{3(4) - 4}{2} = 4$.
The area is given by $\int_{2}^{4} (x_{line} - x_{parabola}) dy = \int_{2}^{4} \left( \frac{3y - 4}{2} - \frac{y^2}{4} \right) dy$.
$= \left[ \frac{3y^2}{4} - 2y - \frac{y^3}{12} \right]_{2}^{4}$.
$= \left( \frac{3(16)}{4} - 2(4) - \frac{64}{12} \right) - \left( \frac{3(4)}{4} - 2(2) - \frac{8}{12} \right)$.
$= \left( 12 - 8 - \frac{16}{3} \right) - \left( 3 - 4 - \frac{2}{3} \right) = \left( 4 - \frac{16}{3} \right) - \left( -1 - \frac{2}{3} \right) = -\frac{4}{3} + \frac{5}{3} = \frac{1}{3}$ square units.

(D) Given the function $f(x) = x^3 + 1$.
To check for local extrema,we find the derivative: $f'(x) = 3x^2$.
Setting $f'(x) = 0$,we get $3x^2 = 0$,which implies $x = 0$.
Now,we check the sign of $f'(x)$ around $x = 0$:
For $x < 0$,$f'(x) = 3x^2 > 0$.
For $x > 0$,$f'(x) = 3x^2 > 0$.
Since the derivative $f'(x)$ does not change sign as $x$ passes through $0$,the function $f(x)$ does not have a local extremum at $x = 0$. Thus,Statement $1$ is false.
For Statement $2$: The function $f(x) = x^3 + 1$ is a polynomial function,which is continuous and differentiable everywhere on $( -\infty, \infty )$.
Also,$f'(x) = 3x^2$,so $f'(0) = 3(0)^2 = 0$. Thus,Statement $2$ is true.
Therefore,Statement $1$ is false and Statement $2$ is true.

(D) function $f : A \to B$ is bijective if it is both one-to-one (injective) and onto (surjective).
Statement $1$ states that $f$ is an onto function,which is true by the definition of a bijective function.
Statement $2$ states that there exists a function $g : B \to A$ such that $f \circ g = I_B$. Since $f$ is bijective,it is invertible,meaning there exists an inverse function $f^{-1} : B \to A$ such that $f \circ f^{-1} = I_B$. Thus,$g = f^{-1}$ satisfies the condition. Therefore,Statement $2$ is also true.
However,Statement $2$ describes the property of invertibility resulting from bijectivity,while Statement $1$ is simply the definition of one part of bijectivity. Statement $2$ is not the reason why $f$ is onto; rather,both are consequences of $f$ being bijective. Thus,Statement $2$ is not the correct explanation for Statement $1$.

(A) We have $f(x) = \int \csc^6 x \, dx$.
Using the identity $\csc^2 x = 1 + \cot^2 x$,we can write:
$f(x) = \int \csc^4 x \cdot \csc^2 x \, dx = \int (1 + \cot^2 x)^2 \csc^2 x \, dx$.
Let $u = \cot x$,then $du = -\csc^2 x \, dx$,which implies $\csc^2 x \, dx = -du$.
Substituting these into the integral:
$f(x) = \int (1 + u^2)^2 (-du) = -\int (1 + 2u^2 + u^4) \, du$.
Integrating with respect to $u$:
$f(x) = -(u + \frac{2u^3}{3} + \frac{u^5}{5}) + C$.
Substituting $u = \cot x$ back:
$f(x) = -\cot x - \frac{2}{3} \cot^3 x - \frac{1}{5} \cot^5 x + C$.
This is a polynomial in $\cot x$ of degree $5$.

(A) The equation of a plane passing through the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ is given by $a(x + 1) + b(y - 3) + c(z + 2) = 0$,where the normal vector $(a, b, c)$ is perpendicular to the line's direction vector $(-3, 2, 1)$.
Thus,$-3a + 2b + c = 0$ --- $(1)$
Since the plane passes through the point $(0, 7, -7)$,we substitute these coordinates into the plane equation:
$a(0 + 1) + b(7 - 3) + c(-7 + 2) = 0$
$a + 4b - 5c = 0$ --- $(2)$
Solving equations $(1)$ and $(2)$ using cross-multiplication:
$\frac{a}{2(-5) - 1(4)} = \frac{-b}{-3(-5) - 1(1)} = \frac{c}{-3(4) - 2(1)}$
$\frac{a}{-10 - 4} = \frac{-b}{15 - 1} = \frac{c}{-12 - 2}$
$\frac{a}{-14} = \frac{-b}{14} = \frac{c}{-14}$
Dividing by $-14$,we get $a = 1, b = 1, c = 1$.
Substituting these values into the plane equation:
$1(x + 1) + 1(y - 3) + 1(z + 2) = 0$
$x + y + z = 0$.

(D) Statement $1$: The condition for three vectors $\vec{a}, \vec{b}, \vec{c}$ to be coplanar is that their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. Thus,Statement $1$ is true.
Statement $2$: Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product is zero,i.e.,$\vec{u} \cdot \vec{v} = 0$. The cross product $\vec{u} \times \vec{v}$ is indeed a vector perpendicular to the plane containing $\vec{u}$ and $\vec{v}$. Thus,Statement $2$ is true.
However,Statement $2$ provides a general property of perpendicular vectors and cross products,which is not the logical explanation for the coplanarity condition in Statement $1$. Therefore,Statement $2$ is not the correct explanation for Statement $1$.

(A) For a system of homogeneous linear equations to have a nontrivial solution,the determinant of the coefficient matrix must be zero.
The coefficient matrix is $A = \begin{bmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{bmatrix}$.
Setting the determinant to zero: $\left| \begin{matrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{matrix} \right| = 0$.
Expanding along the first row:
$1(-4k - (-6)) - k(-12 - (-4)) + 3(9 - 2k) = 0$
$1(-4k + 6) - k(-8) + 27 - 6k = 0$
$-4k + 6 + 8k + 27 - 6k = 0$
$-2k + 33 = 0$
$2k = 33 \Rightarrow k = \frac{33}{2}$.
Since the calculated value of $k$ is $\frac{33}{2}$ and not $\frac{31}{2}$,Statement $1$ is false.
Statement $2$ is a standard theorem in linear algebra regarding homogeneous systems,so it is true.

(D) Let $I = \int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$.
We can split this into two integrals: $I = \int_{-0.9}^{0.9} [x^2] dx + \int_{-0.9}^{0.9} \log \left( \frac{2-x}{2+x} \right) dx$.
For the first part,since $-0.9 < x < 0.9$,we have $0 \leq x^2 < 0.81$. Thus,$[x^2] = 0$ for all $x \in (-0.9, 0.9)$. Therefore,$\int_{-0.9}^{0.9} [x^2] dx = 0$.
For the second part,let $f(x) = \log \left( \frac{2-x}{2+x} \right)$.
Then $f(-x) = \log \left( \frac{2-(-x)}{2+(-x)} \right) = \log \left( \frac{2+x}{2-x} \right) = \log \left( \left( \frac{2-x}{2+x} \right)^{-1} \right) = -\log \left( \frac{2-x}{2+x} \right) = -f(x)$.
Since $f(x)$ is an odd function and the interval $[-0.9, 0.9]$ is symmetric about the origin,$\int_{-0.9}^{0.9} f(x) dx = 0$.
Thus,$I = 0 + 0 = 0$.

(D) The function $f(x) = a|\sin x| + be^{|x|} + c|x|^3$ is differentiable at $x = 0$ if and only if each component is differentiable or their non-differentiable parts cancel out.
First,consider $|\sin x|$. The left-hand derivative at $x = 0$ is $\lim_{h \to 0^-} \frac{|\sin h| - 0}{h} = \lim_{h \to 0^-} \frac{-\sin h}{h} = -1$. The right-hand derivative is $\lim_{h \to 0^+} \frac{\sin h}{h} = 1$. Since $-1 \neq 1$,$|\sin x|$ is not differentiable at $x = 0$.
Second,consider $e^{|x|}$. The left-hand derivative at $x = 0$ is $\lim_{h \to 0^-} \frac{e^{-h} - 1}{h} = -1$. The right-hand derivative is $\lim_{h \to 0^+} \frac{e^h - 1}{h} = 1$. Since $-1 \neq 1$,$e^{|x|}$ is not differentiable at $x = 0$.
Third,consider $|x|^3$. Since $|x|^3 = x^3$ for $x \ge 0$ and $-x^3$ for $x < 0$,the derivative at $x = 0$ is $\lim_{h \to 0} \frac{|h|^3 - 0}{h} = 0$. Thus,$|x|^3$ is differentiable at $x = 0$.
For $f(x)$ to be differentiable at $x = 0$,the non-differentiable parts must vanish. This requires $a = 0$ and $b = 0$. The constant $c$ can be any real number because $|x|^3$ is already differentiable.
Therefore,the correct option is $D$.

(B) The given differential equation is $(x^2 - 1)\frac{dy}{dx} + 2xy = x$.
Dividing both sides by $(x^2 - 1)$,we get $\frac{dy}{dx} + \frac{2x}{x^2 - 1}y = \frac{x}{x^2 - 1}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{x^2 - 1}$ and $Q = \frac{x}{x^2 - 1}$.
The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{2x}{x^2 - 1} dx}$.
Let $t = x^2 - 1$,then $dt = 2x dx$.
Thus,$IF = e^{\int \frac{dt}{t}} = e^{\ln|t|} = t = x^2 - 1$.
Therefore,the integrating factor is $x^2 - 1$.

(B) Let $A$ be the area of the circular plate and $r$ be its radius. The area is given by $A = \pi r^2$.
Given that the radius $r = 50 \, cm$ and the rate of change of the radius is $\frac{dr}{dt} = 1 \, mm/hour = 0.1 \, cm/hour = \frac{1}{10} \, cm/hour$.
Differentiating the area formula with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Substituting the given values,$\frac{dA}{dt} = 2 \times \pi \times 50 \times \frac{1}{10} = 10\pi \, cm^2/hour$.
Thus,the rate at which the area of the plate increases is $10\pi \, cm^2/hour$.

(B) Let the initial composition of the two balls in the urn be represented by the number of white balls $W$. Since each ball can be white or black,the possible initial states are $0$ white balls $(BB)$,$1$ white ball $(BW)$,or $2$ white balls $(WW)$. Assuming each state is equally likely,the prior probability for each is $P(S_i) = \frac{1}{3}$.
After adding one white ball,the new compositions are:
$1) BB \to BBW$ (Number of white balls = $1$,Total = $3$)
$2) BW \to BWW$ (Number of white balls = $2$,Total = $3$)
$3) WW \to WWW$ (Number of white balls = $3$,Total = $3$)
The probability of drawing a white ball given the state $S_i$ is $P(W|S_i) = \frac{\text{white balls}}{3}$.
$P(W|S_1) = \frac{1}{3}$,$P(W|S_2) = \frac{2}{3}$,$P(W|S_3) = \frac{3}{3} = 1$.
Using the law of total probability: $P(W) = \sum P(W|S_i)P(S_i) = \frac{1}{3} \times (\frac{1}{3} + \frac{2}{3} + 1) = \frac{1}{3} \times (\frac{6}{3}) = \frac{2}{3}$.

(C) Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$.
This implies $\vec{a}+\vec{b}=-\vec{c}$.
Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = |-\vec{c}|^2$.
$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.
Substituting the given values: $3^2 + 5^2 + 2(\vec{a} \cdot \vec{b}) = 7^2$.
$9 + 25 + 2(\vec{a} \cdot \vec{b}) = 49$.
$34 + 2(\vec{a} \cdot \vec{b}) = 49$.
$2(\vec{a} \cdot \vec{b}) = 49 - 34 = 15$.
$\vec{a} \cdot \vec{b} = \frac{15}{2}$.
We know that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$\frac{15}{2} = (3)(5) \cos \theta$.
$\frac{15}{2} = 15 \cos \theta$.
$\cos \theta = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.