JEE Main 2013 Mathematics Question Paper with Answer and Solution

(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + \dots + 10(20)^2$.
The $n^{th}$ term is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.
The sum of $10$ terms is $S_{10} = \sum_{n=1}^{10} 4n^3$.
$S_{10} = 4 \sum_{n=1}^{10} n^3$.
Using the formula $\sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} \right)^2$,we get:
$S_{10} = 4 \left( \frac{10 \times 11}{2} \right)^2$.
$S_{10} = 4 \times (55)^2 = 4 \times 3025 = 12100$.

(B) The given equation of the circle is $x^2 + y^2 - 6x - 8y + (25 - a^2) = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$,$f = -4$,and $c = 25 - a^2$.
The center of the circle is $(-g, -f) = (3, 4)$ and the radius $r$ is given by $\sqrt{g^2 + f^2 - c}$.
$r = \sqrt{(-3)^2 + (-4)^2 - (25 - a^2)} = \sqrt{9 + 16 - 25 + a^2} = \sqrt{a^2} = |a|$.
Since the circle touches the $x$-axis,the radius must be equal to the absolute value of the $y$-coordinate of the center.
Thus,$|a| = |4|$,which implies $a = \pm 4$.

(D) Given the inequality: $\frac{x - 5}{(x + 7)(x - 2)} > 0$.
We use the wavy curve method (sign scheme) to find the intervals.
The critical points are $x = -7, 2, 5$.
Testing the intervals:
For $x > 5$,the expression is positive.
For $2 < x < 5$,the expression is negative.
For $-7 < x < 2$,the expression is positive.
For $x < -7$,the expression is negative.
The solution set is $x \in (-7, 2) \cup (5, \infty)$.
The integral values in the interval $(-7, 2)$ are $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.
The integral values in the interval $(5, \infty)$ are $\{6, 7, 8, \dots\}$.
The least integral value $\alpha$ is $-6$.
Checking the options for $\alpha = -6$:
$(A)$ $(-6)^2 + 3(-6) - 4 = 36 - 18 - 4 = 14 \neq 0$.
$(B)$ $(-6)^2 - 5(-6) + 4 = 36 + 30 + 4 = 70 \neq 0$.
$(C)$ $(-6)^2 - 7(-6) + 6 = 36 + 42 + 6 = 84 \neq 0$.
$(D)$ $(-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0$.
Thus,$\alpha = -6$ satisfies option $(D)$.

(A) Given equation: $\sin 2x - 2 \cos x + 4 \sin x = 4$
Using $\sin 2x = 2 \sin x \cos x$,we get:
$2 \sin x \cos x - 2 \cos x + 4 \sin x - 4 = 0$
Factor by grouping:
$2 \cos x (\sin x - 1) + 4 (\sin x - 1) = 0$
$(2 \cos x + 4)(\sin x - 1) = 0$
$2(\cos x + 2)(\sin x - 1) = 0$
Since $\cos x + 2 \neq 0$ for any real $x$ (as $-1 \le \cos x \le 1$),we must have:
$\sin x = 1$
In the interval $[0, 5\pi]$,the values of $x$ for which $\sin x = 1$ are:
$x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}$
Thus,there are $3$ solutions.

(A) Let $z = x + iy$. Since $|z| = 1$,we have $x^2 + y^2 = 1$.
Consider the expression $\frac{1 + z^2}{2iz} = \frac{1 + (x + iy)^2}{2i(x + iy)} = \frac{1 + x^2 - y^2 + 2ixy}{2ix - 2y} = \frac{(1 + x^2 - y^2) + 2ixy}{-2y + 2ix}$.
Since $x^2 + y^2 = 1$,we have $1 - y^2 = x^2$. Substituting this into the numerator:
$\frac{(x^2 + x^2) + 2ixy}{-2y + 2ix} = \frac{2x^2 + 2ixy}{2i(x + iy)} = \frac{2x(x + iy)}{2i(x + iy)} = \frac{x}{i} = -ix$.
Thus,$a = \text{Im}(-ix) = -x$.
Since $|z| = 1$,$x$ ranges from $-1$ to $1$. Given $z \ne \pm 1$,$x \ne \pm 1$.
Therefore,$x \in (-1, 1)$,which implies $a = -x \in (-1, 1)$.
Hence,the set $A = (-1, 1)$.

(A) The angle $\theta$ between two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by $\tan \theta = \left| \frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2} \right|$.
For the lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_2 = 0$,the angle $\theta_1$ satisfies $\tan \theta_1 = \left| \frac{(2)(5) - (3)(-1)}{(2)(-1) + (3)(5)} \right| = \left| \frac{10 + 3}{-2 + 15} \right| = \frac{13}{13} = 1$.
Similarly,for the lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_3 = 0$,the angle $\theta_2$ satisfies $\tan \theta_2 = \left| \frac{(2)(5) - (3)(-1)}{(2)(-1) + (3)(5)} \right| = 1$.
Since $\tan \theta_1 = \tan \theta_2 = 1$,we have $\theta_1 = \theta_2$ for any real values of $c_2$ and $c_3$.
Thus,Statement-$2$ is true.
Since Statement-$2$ is true,Statement-$1$ is also true,and Statement-$2$ provides the logical basis for Statement-$1$.

(C) Let the first term be $a$ and the common difference be $d$.
Given $a_4 - a_7 + a_{10} = m$.
Using the formula $a_n = a + (n-1)d$:
$(a + 3d) - (a + 6d) + (a + 9d) = m$
$a + 6d = m$
Note that $a_7 = a + 6d$,so $a_7 = m$.
The sum of the first $13$ terms is given by $S_{13} = \frac{13}{2} [2a + (13-1)d] = \frac{13}{2} [2a + 12d] = 13(a + 6d)$.
Substituting $a + 6d = m$,we get $S_{13} = 13m$.

(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec \theta, b \tan \theta)$ is $\frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1$.
Setting $y=0$,we get $x = a \cos \theta$,so $P = (a \cos \theta, 0)$.
Setting $x=0$,we get $y = -b \cot \theta$,so $Q = (0, -b \cot \theta)$.
Since $OPRQ$ is a rectangle with $O(0,0)$,the coordinates of $R$ are $(h, k) = (a \cos \theta, -b \cot \theta)$.
From this,$\cos \theta = \frac{h}{a}$ and $\cot \theta = -\frac{k}{b}$.
Using the identity $\csc^2 \theta - \cot^2 \theta = 1$,we have $\frac{1}{\sin^2 \theta} - \cot^2 \theta = 1$,which is $\frac{1}{1 - \cos^2 \theta} - \cot^2 \theta = 1$.
Alternatively,$\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta$ is not helpful here. Let's use $\csc^2 \theta = 1 + \cot^2 \theta$.
$\frac{1}{\sin^2 \theta} = 1 + \frac{k^2}{b^2} = \frac{b^2 + k^2}{b^2}$.
Since $\cos^2 \theta = \frac{h^2}{a^2}$,then $\sin^2 \theta = 1 - \frac{h^2}{a^2} = \frac{a^2 - h^2}{a^2}$.
Substituting these into $\frac{1}{\sin^2 \theta} = 1 + \frac{k^2}{b^2}$ gives $\frac{a^2}{a^2 - h^2} = \frac{b^2 + k^2}{b^2}$.
This simplifies to $a^2 b^2 = (a^2 - h^2)(b^2 + k^2) = a^2 b^2 + a^2 k^2 - b^2 h^2 - h^2 k^2$.
$a^2 k^2 - b^2 h^2 - h^2 k^2 = 0 \Rightarrow \frac{a^2}{h^2} - \frac{b^2}{k^2} = 1$.
Given $a^2 = 4$ and $b^2 = 2$,the locus of $R(x, y)$ is $\frac{4}{x^2} - \frac{2}{y^2} = 1$.

(B) Total number of points $= 3 + 4 + 5 = 12$.
Number of ways to select $3$ points out of $12$ is $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
However,points on the same side are collinear and cannot form a triangle.
Number of triangles formed by $3$ collinear points on side $AB = ^3C_3 = 1$.
Number of triangles formed by $3$ collinear points on side $BC = ^4C_3 = 4$.
Number of triangles formed by $3$ collinear points on side $CA = ^5C_3 = 10$.
Required number of triangles $= 220 - (1 + 4 + 10) = 220 - 15 = 205$.

(D) Given the condition $0 < y < x < 2y$,we can order the numbers as follows:
Since $y < x$ and $x < 2y$,we have $x - y < y < x < 2x + y$.
The median of the four numbers is the average of the two middle terms:
$\text{Median} = \frac{y + x}{2} = 10 \Rightarrow x + y = 20 \quad (i)$
The range is the difference between the largest and smallest values:
$\text{Range} = (2x + y) - (x - y) = x + 2y = 28 \quad (ii)$
Subtracting equation $(i)$ from $(ii)$:
$(x + 2y) - (x + y) = 28 - 20 \Rightarrow y = 8$.
Substituting $y = 8$ into $(i)$:
$x + 8 = 20 \Rightarrow x = 12$.
The four numbers are $(12 - 8), 8, 12, (2(12) + 8)$,which are $4, 8, 12, 32$.
The mean is $\frac{4 + 8 + 12 + 32}{4} = \frac{56}{4} = 14$.

(B) The equation of the parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$,we get $a = 1$.
The ends of the latus rectum of the parabola $y^2 = 4ax$ are $(a, 2a)$ and $(a, -2a)$.
For $a = 1$,the ends are $(1, 2)$ and $(1, -2)$.
The normals to the parabola at the ends of the latus rectum $(a, 2a)$ and $(a, -2a)$ intersect at the point $(3a, 0)$.
Substituting $a = 1$,the point of intersection is $(3(1), 0) = (3, 0)$.

(B) Let the sequence be $a, b, c, d$.
Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.
Since $b, c, d$ are in $A.P.$ with common difference $6$,we have $c - b = 6$ and $d - c = 6$.
Thus,$c = b + 6$ and $d = c + 6 = b + 12$.
Given that the first and last terms are equal,$a = d$,so $a = b + 12$,which implies $b = a - 12$.
Substituting $b = a - 12$ into $c = b + 6$,we get $c = (a - 12) + 6 = a - 6$.
Now,substitute $b$ and $c$ into the $G.P.$ condition $b^2 = ac$:
$(a - 12)^2 = a(a - 6)$
$a^2 - 24a + 144 = a^2 - 6a$
$144 = 18a$
$a = 8$.
Since $d = a$,the last term is $8$.

(A) For the circle $x^2 + y^2 - 6x + 2y = 0$,the center is $(3, -1)$ and the radius is $r = \sqrt{3^2 + (-1)^2 - 0} = \sqrt{10}$.
Check if the center $(3, -1)$ lies on the line $2x + y = 5$: $2(3) + (-1) = 6 - 1 = 5$. Since the center lies on the line,the line is a diameter (and thus a normal) to the circle. So,Statement $2$ is true.
For Statement $1$,there are infinitely many circles with radius $\sqrt{10}$ whose centers lie on the line $2x + y = 5$. Any circle with center $(h, k)$ such that $2h + k = 5$ and radius $\sqrt{10}$ satisfies the condition. Thus,Statement $1$ is false.

(D) Let the first circle be $C_1$ with radius $r_1 = 1$ and center at the origin $(0,0)$.
Let the second circle be $C_2$ with radius $r_2$ and center at $(h,k)$.
The arc of $C_2$ subtends an angle of $60^o$ at the circumference of $C_1$.
By the inscribed angle theorem,the angle subtended by the same arc at the center of $C_1$ would be $120^o$ if the arc is part of a circle passing through the center,or related to the geometry of the intersection.
However,the position of the center of the second circle is not fixed relative to the first circle.
Without knowing the distance between the centers or the specific chord length formed by the intersection,the radius $r_2$ cannot be uniquely determined.
Thus,the information provided is insufficient to solve for $r_2$.

(D) Mid-point of $P(2, 3)$ and $Q(4, 5)$ is $(\frac{2+4}{2}, \frac{3+5}{2}) = (3, 4)$.
Slope of $PQ = \frac{5-3}{4-2} = \frac{2}{2} = 1$.
Since the line $L$ is the perpendicular bisector of $PQ$,the slope of line $L$ is $m = -1$.
The equation of line $L$ passing through $(3, 4)$ with slope $-1$ is $y - 4 = -1(x - 3) \Rightarrow x + y - 7 = 0$.
Let the image of point $R(0, 0)$ be $S(x_1, y_1)$.
The mid-point of $RS$ is $(\frac{x_1}{2}, \frac{y_1}{2})$,which lies on line $L$,so $\frac{x_1}{2} + \frac{y_1}{2} - 7 = 0 \Rightarrow x_1 + y_1 = 14$.
The slope of $RS$ is $\frac{y_1}{x_1}$. Since $RS \perp L$,the slope of $RS$ must be $1$ (negative reciprocal of $-1$).
Thus,$\frac{y_1}{x_1} = 1 \Rightarrow x_1 = y_1$.
Substituting $x_1 = y_1$ into $x_1 + y_1 = 14$,we get $2x_1 = 14 \Rightarrow x_1 = 7$ and $y_1 = 7$.
Therefore,the image of $R$ is $(7, 7)$.

(A) Given conics are $x^2 = 6y$ $(i)$ and $2x^2 - 4y^2 = 9$ $(ii)$.
Consider the line $x - y = \frac{3}{2}$ $(iii)$,which implies $x = y + \frac{3}{2}$.
Substitute $x = y + \frac{3}{2}$ into $(i)$: $(y + \frac{3}{2})^2 = 6y \implies y^2 + 3y + \frac{9}{4} = 6y \implies y^2 - 3y + \frac{9}{4} = 0 \implies (y - \frac{3}{2})^2 = 0$. Thus,$y = \frac{3}{2}$ and $x = 3$.
Since there is only one point of intersection $(3, \frac{3}{2})$,the line is tangent to the parabola.
Substitute $x = y + \frac{3}{2}$ into $(ii)$: $2(y + \frac{3}{2})^2 - 4y^2 = 9 \implies 2(y^2 + 3y + \frac{9}{4}) - 4y^2 = 9 \implies 2y^2 + 6y + \frac{9}{2} - 4y^2 = 9 \implies -2y^2 + 6y - \frac{9}{2} = 0 \implies 4y^2 - 12y + 9 = 0 \implies (2y - 3)^2 = 0$. Thus,$y = \frac{3}{2}$ and $x = 3$.
Since there is only one point of intersection $(3, \frac{3}{2})$,the line is tangent to the hyperbola.
Therefore,$x - y = \frac{3}{2}$ is the common tangent.

(B) Let the vertex $C$ be $(x_1, y_1)$.
The centroid $G$ of triangle $ABC$ with vertices $A(-3, 2)$,$B(-2, 1)$,and $C(x_1, y_1)$ is given by:
$G = \left( \frac{-3 - 2 + x_1}{3}, \frac{2 + 1 + y_1}{3} \right) = \left( \frac{x_1 - 5}{3}, \frac{y_1 + 3}{3} \right)$
Since the centroid lies on the line $3x + 4y + 2 = 0$,we substitute the coordinates of $G$ into the equation:
$3 \left( \frac{x_1 - 5}{3} \right) + 4 \left( \frac{y_1 + 3}{3} \right) + 2 = 0$
Multiply the entire equation by $3$ to clear the denominators:
$3(x_1 - 5) + 4(y_1 + 3) + 6 = 0$
$3x_1 - 15 + 4y_1 + 12 + 6 = 0$
$3x_1 + 4y_1 + 3 = 0$
Thus,the vertex $C(x_1, y_1)$ lies on the line $3x + 4y + 3 = 0$.

(C) The logical statement $r: p \to (\sim p \vee q)$ has a truth value $F$.
An implication $A \to B$ is false if and only if $A$ is true and $B$ is false.
Therefore,$p$ must be $T$ and $(\sim p \vee q)$ must be $F$.
For the disjunction $(\sim p \vee q)$ to be $F$,both $\sim p$ and $q$ must be $F$.
Since $\sim p$ is $F$,it implies $p$ is $T$.
Thus,$p$ is $T$ and $q$ is $F$.

(C) For any event $E$,the probability $P(E)$ must satisfy $0 \le P(E) \le 1$.
For event $A$: $0 \le \frac{3x+1}{3} \le 1 \Rightarrow 0 \le 3x+1 \le 3 \Rightarrow -1 \le 3x \le 2 \Rightarrow -\frac{1}{3} \le x \le \frac{2}{3}$.
For event $B$: $0 \le \frac{1-x}{4} \le 1 \Rightarrow 0 \le 1-x \le 4 \Rightarrow -1 \le -x \le 3 \Rightarrow -3 \le x \le 1$.
Since $A$ and $B$ are mutually exclusive,$P(A \cup B) = P(A) + P(B) \le 1$.
$\frac{3x+1}{3} + \frac{1-x}{4} \le 1 \Rightarrow \frac{4(3x+1) + 3(1-x)}{12} \le 1 \Rightarrow 12x + 4 + 3 - 3x \le 12 \Rightarrow 9x + 7 \le 12 \Rightarrow 9x \le 5 \Rightarrow x \le \frac{5}{9}$.
Combining all conditions: $x \ge -\frac{1}{3}$,$x \le \frac{2}{3}$,$x \ge -3$,$x \le 1$,and $x \le \frac{5}{9}$.
The intersection of these intervals is $[-\frac{1}{3}, \frac{5}{9}]$.

(A) The general term in the expansion of $(1 + x)^{2n}$ is given by $T_{k+1} = ^{2n}C_k x^k$.
The coefficient of $x^{3r}$ is $^{2n}C_{3r}$ and the coefficient of $x^{r+2}$ is $^{2n}C_{r+2}$.
Given that the coefficients are equal,we have $^{2n}C_{3r} = ^{2n}C_{r+2}$.
Using the property $^{n}C_a = ^{n}C_b \Rightarrow a = b$ or $a + b = n$,we have two cases:
Case $1$: $3r = r + 2$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. However,the problem states $r > 1$,so this case is rejected.
Case $2$: $3r + (r + 2) = 2n$ $\Rightarrow 4r + 2 = 2n$ $\Rightarrow n = 2r + 1$.
Thus,$n = 2r + 1$.

(B) Given $\alpha^3 + \beta^3 = -p$ and $\alpha \beta = q$.
Let the roots of the required quadratic equation be $x_1 = \frac{\alpha^2}{\beta}$ and $x_2 = \frac{\beta^2}{\alpha}$.
The sum of the roots is $x_1 + x_2 = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} = \frac{-p}{q}$.
The product of the roots is $x_1 \times x_2 = \frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha} = \alpha \beta = q$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (\frac{-p}{q})x + q = 0$.
$x^2 + \frac{p}{q}x + q = 0$.
Multiplying by $q$,we get $qx^2 + px + q^2 = 0$.

(B) For set $A$,we have $\sin(\theta) = \tan(\theta)$.
This implies $\sin(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$,which gives $\sin(\theta)(1 - \frac{1}{\cos(\theta)}) = 0$.
Thus,$\sin(\theta) = 0$ or $\cos(\theta) = 1$.
If $\sin(\theta) = 0$,then $\theta = n\pi$ for $n \in \mathbb{Z}$.
If $\cos(\theta) = 1$,then $\theta = 2n\pi$ for $n \in \mathbb{Z}$.
Combining these,$A = \{ n\pi : n \in \mathbb{Z} \} = \{ 0, \pm\pi, \pm 2\pi, \dots \}$.
For set $B$,we have $\cos(\theta) = 1$,which implies $\theta = 2n\pi$ for $n \in \mathbb{Z}$.
Thus,$B = \{ 2n\pi : n \in \mathbb{Z} \} = \{ 0, \pm 2\pi, \pm 4\pi, \dots \}$.
Comparing the two sets,every element of $B$ is in $A$,so $B \subset A$.
However,$\pi \in A$ but $\pi \notin B$,so $A \not\subset B$.

(B) Let $z = x + iy$,then $\bar{z} = x - iy$.
Given $z = 1 - \bar{z}$,we have $x + iy = 1 - (x - iy) = 1 - x + iy$.
Equating real parts,$x = 1 - x$,which gives $2x = 1$,so $x = \frac{1}{2}$.
Since $|z| = 1$,we have $x^2 + y^2 = 1$.
Substituting $x = \frac{1}{2}$,we get $\frac{1}{4} + y^2 = 1$,so $y^2 = \frac{3}{4}$,which gives $y = \pm \frac{\sqrt{3}}{2}$.
Thus,$z = \frac{1}{2} \pm i\frac{\sqrt{3}}{2}$.
Since $z$ has an imaginary part,Statement $1$ is false.
The principal argument $\theta$ is given by $\tan \theta = \frac{y}{x}$.
For $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
For $z = \frac{1}{2} - i\frac{\sqrt{3}}{2}$,$\theta = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.
Since the question implies a specific value for the argument,and $\frac{\pi}{3}$ is a valid principal argument for one of the possible values of $z$,Statement $2$ is considered true in this context.
Therefore,Statement $1$ is false and Statement $2$ is true.

(A) The mean of the observations is $\bar{x} = \frac{n(a) + n(-a)}{2n} = \frac{0}{2n} = 0$.
The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{2n}}$.
Given $\sigma = 2$,we have $2 = \sqrt{\frac{n(a - 0)^2 + n(-a - 0)^2}{2n}}$.
$2 = \sqrt{\frac{n(a^2) + n(a^2)}{2n}} = \sqrt{\frac{2na^2}{2n}} = \sqrt{a^2} = |a|$.
Therefore,$|a| = 2$.

(A) The given series is the sum of squares of the first $13$ odd numbers.
The general term is ${T_n} = {(2n - 1)^2}$ for $n = 1, 2, \dots, 13$.
The sum is $S = \sum_{n=1}^{13} (2n - 1)^2 = \sum_{n=1}^{13} (4n^2 - 4n + 1)$.
Using the summation formulas $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$ and $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$:
$S = 4 \sum_{n=1}^{13} n^2 - 4 \sum_{n=1}^{13} n + \sum_{n=1}^{13} 1$
$S = 4 \left[ \frac{13(13+1)(2 \times 13 + 1)}{6} \right] - 4 \left[ \frac{13(13+1)}{2} \right] + 13$
$S = 4 \left[ \frac{13 \times 14 \times 27}{6} \right] - 2 \times 13 \times 14 + 13$
$S = 4 \times 13 \times 7 \times 9 - 364 + 13$
$S = 3276 - 364 + 13 = 2925$.

(D) The set of digits is $\{2, 3, 5, 7, 9\}$. The total number of $5$-digit numbers formed without repetition is $5! = 120$.
For $p$: The numbers must exceed $20000$. Since all given digits are $\ge 2$,any $5$-digit number formed using these digits will be $\ge 23579$,which is $> 20000$. Thus,$p = 5! = 120$.
For $q$: The numbers must lie between $30000$ and $90000$. This means the first digit must be $3, 5,$ or $7$.
There are $3$ choices for the first digit. The remaining $4$ positions can be filled by the remaining $4$ digits in $4!$ ways.
So,$q = 3 \times 4! = 3 \times 24 = 72$.
The ratio $p : q = 120 : 72$.
Dividing both by $24$,we get $120/24 : 72/24 = 5 : 3$.

(A) The given ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Here,$a^2 = 9$ and $b^2 = 4$.
The normal is parallel to the line $4x - 2y - 5 = 0$,so the slope of the normal is $m_n = \frac{-4}{-2} = 2$.
The slope of the tangent at that point is $m_t = -\frac{1}{m_n} = -\frac{1}{2}$.
The equation of the tangent with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
The point of contact is $\left( -\frac{a^2m}{c}, \frac{b^2}{c} \right)$ where $c = \pm \sqrt{a^2m^2 + b^2}$.
Here $m = -\frac{1}{2}$,so $c = \sqrt{9(-\frac{1}{2})^2 + 4} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
The point of contact is $\left( -\frac{9(-1/2)}{5/2}, \frac{4}{5/2} \right) = \left( \frac{9/2}{5/2}, \frac{8}{5} \right) = \left( \frac{9}{5}, \frac{8}{5} \right)$.
Since the normal is parallel to $4x - 2y - 5 = 0$,the point is $\left( \frac{9}{5}, \frac{8}{5} \right)$.

(B) The function $u(x) = \frac{1}{x - 1}$ is discontinuous at $x = 1$.
The function $f(u) = \frac{1}{u^2 + u - 2} = \frac{1}{(u + 2)(u - 1)}$ is discontinuous at $u = -2$ and $u = 1$.
For the composite function $f(u(x))$,we must consider the points where $u(x)$ is undefined and where $u(x)$ takes the values that make $f(u)$ undefined.
$1$. $u(x)$ is discontinuous at $x = 1$.
$2$. When $u(x) = -2$,we have $\frac{1}{x - 1} = -2$,which implies $x - 1 = -\frac{1}{2}$,so $x = \frac{1}{2}$.
$3$. When $u(x) = 1$,we have $\frac{1}{x - 1} = 1$,which implies $x - 1 = 1$,so $x = 2$.
Thus,the composite function $f(u(x))$ is discontinuous at $x = 1, \frac{1}{2}, 2$.
There are $3$ such points.

(D) Let $I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} \quad ......(1)$
Using the property $\int\limits_a^b f(x)dx = \int\limits_a^b f(a+b-x)dx$,we replace $x$ with $-\frac{\pi}{2} + \frac{\pi}{2} - x = -x$:
$I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}(-x)}}{{1 + {2^{-x}}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 + \frac{1}{2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}{{\sin }^2}x}}{{2^x + 1}}} dx \quad ......(2)$
Adding $(1)$ and $(2)$:
$2I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x + {2^x}{{\sin }^2}x}}{{1 + {2^x}}}} dx = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x dx}$
Since $\sin^2 x$ is an even function,$2I = 2 \int\limits_0^{\pi /2} {\sin^2 x dx} = 2 \int\limits_0^{\pi /2} {\frac{1 - \cos 2x}{2} dx}$
$2I = \int\limits_0^{\pi /2} {(1 - \cos 2x) dx} = [x - \frac{\sin 2x}{2}]_0^{\pi /2} = (\frac{\pi}{2} - 0) - (0 - 0) = \frac{\pi}{2}$
Therefore,$I = \frac{\pi}{4}$.

(C) Let the cost function be $C(v) = av + \frac{b}{v}$.
Given that at $v = 30 \text{ km/h}$,$C = 75$,so $30a + \frac{b}{30} = 75 \implies 900a + b = 2250 \quad (i)$.
Given that at $v = 40 \text{ km/h}$,$C = 65$,so $40a + \frac{b}{40} = 65 \implies 1600a + b = 2600 \quad (ii)$.
Subtracting $(i)$ from $(ii)$,we get $700a = 350$,which implies $a = 0.5$.
Substituting $a = 0.5$ into $(i)$,we get $900(0.5) + b = 2250 \implies 450 + b = 2250 \implies b = 1800$.
To find the most economical speed,we minimize $C(v)$ by setting $\frac{dC}{dv} = 0$.
$\frac{dC}{dv} = a - \frac{b}{v^2} = 0 \implies v^2 = \frac{b}{a}$.
$v^2 = \frac{1800}{0.5} = 3600 \implies v = 60 \text{ km/h}$.

(B) The function is $y = |\cos x - \sin x|$.
In the interval $[0, \pi/4]$,$\cos x \geq \sin x$,so $y = \cos x - \sin x$.
In the interval $[\pi/4, \pi/2]$,$\sin x \geq \cos x$,so $y = \sin x - \cos x$.
The required area is given by:
$A = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$
$A = [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$
$A = ((\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0))) + ((-\cos(\pi/2) - \sin(\pi/2)) - (-\cos(\pi/4) - \sin(\pi/4)))$
$A = ((\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1)) + ((0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}))$
$A = (\sqrt{2} - 1) + (-1 + \sqrt{2})$
$A = 2\sqrt{2} - 2$ square units.

(A) The slope of the curve is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
Integrating both sides with respect to $x$:
$\int dy = \int \left( 1 - \frac{1}{x^2} \right) dx$
$y = x + \frac{1}{x} + C$
Since the curve passes through the point $\left( 2, \frac{7}{2} \right)$,we substitute these values to find $C$:
$\frac{7}{2} = 2 + \frac{1}{2} + C$
$\frac{7}{2} = \frac{5}{2} + C$
$C = 1$
Thus,the equation of the curve is $y = x + \frac{1}{x} + 1$.
To find the ordinate when the abscissa is $x = -2$:
$y = -2 + \frac{1}{-2} + 1$
$y = -2 - 0.5 + 1 = -1.5 = -\frac{3}{2}$.

(A) Given $R = \{(x,y) : x,y \in N \text{ and } x^2 - 4xy + 3y^2 = 0\}$.
Factorizing the equation: $x^2 - 4xy + 3y^2 = (x - y)(x - 3y) = 0$.
This implies $x = y$ or $x = 3y$.
$1$. Reflexivity: For any $x \in N$,$x = x$ is true,so $(x,x) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: $(3,1) \in R$ because $3 = 3(1)$. However,$(1,3) \notin R$ because $1 \neq 3$ and $1 \neq 3(3)$. Thus,$R$ is not symmetric.
$3$. Transitivity: Let $(a,b) \in R$ and $(b,c) \in R$. Then $(a=b \text{ or } a=3b)$ and $(b=c \text{ or } b=3c)$.
- If $a=b$ and $b=c$,then $a=c$,so $(a,c) \in R$.
- If $a=b$ and $b=3c$,then $a=3c$,so $(a,c) \in R$.
- If $a=3b$ and $b=c$,then $a=3c$,so $(a,c) \in R$.
- If $a=3b$ and $b=3c$,then $a=9c$,which is not necessarily $c$ or $3c$. Wait,checking $(9,3) \in R$ and $(3,1) \in R$: $9=3(3)$ and $3=3(1)$. Here $a=9, b=3, c=1$. $a=9c$. Since $9 \neq 1$ and $9 \neq 3(1)$,$(9,1) \notin R$. Therefore,$R$ is not transitive.
Correction: $R$ is reflexive only.

(B) The given system of equations is homogeneous:
$x + ay = 0$
$y + az = 0$
$z + ax = 0$
This can be written in matrix form $AX = 0$,where $A = \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix}$.
$A$ homogeneous system has a unique (trivial) solution if and only if the determinant $|A| \neq 0$.
Calculating the determinant:
$|A| = 1(1 - 0) - a(0 - a^2) + 0(0 - a) = 1 + a^3$.
For a unique solution,we require $|A| \neq 0$,so $1 + a^3 \neq 0$.
$a^3 \neq -1$,which implies $a \neq -1$.
Thus,the set of all real values of $a$ is $R - \{-1\}$.

(D) matrix $A = \begin{bmatrix} a & b \\ c & a \end{bmatrix}$ is in $S$ where $a, b, c \in \{0, 1, 2\}$.
There are $3$ choices for each of $a, b, c$,so the total number of matrices in $S$ is $3 \times 3 \times 3 = 27$.
For a matrix to be singular,its determinant must be zero: $\det(A) = a^2 - bc = 0$,which implies $a^2 = bc$.
We check the cases for $a \in \{0, 1, 2\}$:
Case $1$: $a = 0$. Then $bc = 0$. The pairs $(b, c)$ can be $(0, 0), (0, 1), (0, 2), (1, 0), (2, 0)$. There are $5$ such matrices.
Case $2$: $a = 1$. Then $bc = 1$. The only pair $(b, c)$ is $(1, 1)$. There is $1$ such matrix.
Case $3$: $a = 2$. Then $bc = 4$. The only pair $(b, c)$ is $(2, 2)$. There is $1$ such matrix.
Total singular matrices $= 5 + 1 + 1 = 7$.
Number of non-singular matrices $= \text{Total} - \text{Singular} = 27 - 7 = 20$.

(B) Given the inequality $\sin^{-1} x > \cos^{-1} x$ for $x \in (0, 1)$.
We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.
Substituting this into the inequality,we get:
$\sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x$
$2 \sin^{-1} x > \frac{\pi}{2}$
$\sin^{-1} x > \frac{\pi}{4}$
Since the sine function is strictly increasing on the interval $[0, 1]$,we apply the sine function to both sides:
$x > \sin\left( \frac{\pi}{4} \right)$
$x > \frac{1}{\sqrt{2}}$
Given the domain $x \in (0, 1)$,the solution set is $x \in \left( \frac{1}{\sqrt{2}}, 1 \right)$.

(C) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are given by:
$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$
The direction ratios of the median $AD$ are:
$a = \frac{\lambda - 1}{2} - 2 = \frac{\lambda - 5}{2}$
$b = 4 - 3 = 1$
$c = \frac{\mu + 2}{2} - 5 = \frac{\mu - 8}{2}$
Since the median $AD$ is equally inclined with the axes,its direction cosines $l, m, n$ are equal,i.e.,$|l| = |m| = |n|$. Since $l^2 + m^2 + n^2 = 1$,we have $l = m = n = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction ratios $a, b, c$ must be proportional to $1, 1, 1$. Since $b = 1$,we must have $a = 1$ and $c = 1$.
Setting $a = 1$:
$\frac{\lambda - 5}{2} = 1 \Rightarrow \lambda - 5 = 2 \Rightarrow \lambda = 7$
Setting $c = 1$:
$\frac{\mu - 8}{2} = 1 \Rightarrow \mu - 8 = 2 \Rightarrow \mu = 10$
Therefore,$(\lambda, \mu) = (7, 10)$.

(A) Let $I = \int \frac{\cos 8x + 1}{\cot 2x - \tan 2x} dx.$
First,simplify the denominator:
$\cot 2x - \tan 2x = \frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2} \sin 4x} = 2 \cot 4x.$
Using the identity $\cos 8x + 1 = 2 \cos^2 4x$,the integral becomes:
$I = \int \frac{2 \cos^2 4x}{2 \cot 4x} dx = \int \frac{\cos^2 4x}{\frac{\cos 4x}{\sin 4x}} dx = \int \cos 4x \sin 4x dx.$
Multiply and divide by $2$:
$I = \frac{1}{2} \int 2 \sin 4x \cos 4x dx = \frac{1}{2} \int \sin 8x dx.$
Integrating $\sin 8x$ gives:
$I = \frac{1}{2} \left( -\frac{\cos 8x}{8} \right) + k = -\frac{1}{16} \cos 8x + k.$
Comparing this with $A \cos 8x + k$,we get $A = -\frac{1}{16}$.

(D) Given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.
Dividing by $(1 + x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{4x^2}{1 + x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 + x^2}$ and $Q = \frac{4x^2}{1 + x^2}$.
Integrating factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1 + x^2} dx} = e^{\ln(1 + x^2)} = 1 + x^2$.
The general solution is $y \times (I.F.) = \int Q \times (I.F.) dx + C$.
$y(1 + x^2) = \int \frac{4x^2}{1 + x^2} \times (1 + x^2) dx + C$.
$y(1 + x^2) = \int 4x^2 dx + C$.
$y(1 + x^2) = \frac{4x^3}{3} + C$.
Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ to find $C$.
$0(1 + 0) = \frac{4(0)^3}{3} + C \implies C = 0$.
Thus,the equation of the curve is $y(1 + x^2) = \frac{4x^3}{3}$,which simplifies to $3(1 + x^2)y = 4x^3$.

(A) Let $f(x) = \int_{0}^{\sin^{2}x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^{2}x} \cos^{-1}(\sqrt{t}) \, dt$.
Using the Leibniz integral rule,we differentiate $f(x)$ with respect to $x$:
$f'(x) = \sin^{-1}(\sqrt{\sin^{2}x}) \cdot \frac{d}{dx}(\sin^{2}x) + \cos^{-1}(\sqrt{\cos^{2}x}) \cdot \frac{d}{dx}(\cos^{2}x)$
$f'(x) = x \cdot (2 \sin x \cos x) + (\frac{\pi}{2} - x) \cdot (-2 \cos x \sin x)$
$f'(x) = x \sin(2x) - (\frac{\pi}{2} - x) \sin(2x)$
$f'(x) = x \sin(2x) - \frac{\pi}{2} \sin(2x) + x \sin(2x) = (2x - \frac{\pi}{2}) \sin(2x)$.
Since $f'(x)$ is not zero,let us evaluate $f(x)$ at $x = \frac{\pi}{4}$:
$f(\frac{\pi}{4}) = \int_{0}^{1/2} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{1/2} \cos^{-1}(\sqrt{t}) \, dt$
$f(\frac{\pi}{4}) = \int_{0}^{1/2} (\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t})) \, dt$
Since $\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t}) = \frac{\pi}{2}$ for $t \in [0, 1]$:
$f(\frac{\pi}{4}) = \int_{0}^{1/2} \frac{\pi}{2} \, dt = \frac{\pi}{2} \cdot [t]_{0}^{1/2} = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$.

(D) We are given $R_1 = \int_{-2}^3 f(x) dx$ and $R_2 = \int_{-2}^3 x f(x) dx$.
Using the property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$,we have $a+b = -2+3 = 1$.
Thus,$R_2 = \int_{-2}^3 x f(x) dx = \int_{-2}^3 (1-x) f(1-x) dx$.
Since $f(1-x) = f(x)$,we get $R_2 = \int_{-2}^3 (1-x) f(x) dx$.
$R_2 = \int_{-2}^3 f(x) dx - \int_{-2}^3 x f(x) dx$.
$R_2 = R_1 - R_2$.
$2R_2 = R_1$ or $R_1 = 2R_2$.

(D) Given $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$.
First,calculate the magnitude of $\vec a$: $|\vec a| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = 3$.
Next,calculate the cross product $\vec a \times \vec b$:
$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat i(0 - (-2)) - \hat j(0 - (-2)) + \hat k(2 - 1) = 2\hat i - 2\hat j + \hat k$.
The magnitude is $|\vec a \times \vec b| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.
Given $|\vec c - \vec a| = 2\sqrt 2$,squaring both sides gives $|\vec c - \vec a|^2 = 8$.
Expanding this,we get $|\vec c|^2 + |\vec a|^2 - 2(\vec c \cdot \vec a) = 8$.
Substituting $|\vec a| = 3$ and $\vec a \cdot \vec c = |\vec c|$,we have $|\vec c|^2 + 9 - 2|\vec c| = 8$.
This simplifies to $|\vec c|^2 - 2|\vec c| + 1 = 0$,which is $(|\vec c| - 1)^2 = 0$,so $|\vec c| = 1$.
Finally,the magnitude of the cross product is $|(\vec a \times \vec b) \times \vec c| = |\vec a \times \vec b| |\vec c| \sin 30^o$.
Substituting the values: $3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(C) The equation of a plane passing through the line of intersection of the planes $P_1: x + 2y - 3 = 0$ and $P_2: y - 2z + 1 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(x + 2y - 3) + \lambda(y - 2z + 1) = 0$
$x + (2 + \lambda)y - 2\lambda z + (\lambda - 3) = 0$ ....$(i)$
Since this plane is perpendicular to the plane $x + 2y - 3 = 0$,the dot product of their normal vectors $\vec{n_1} = (1, 2 + \lambda, -2\lambda)$ and $\vec{n_2} = (1, 2, 0)$ must be zero.
$1(1) + 2(2 + \lambda) + 0(-2\lambda) = 0$
$1 + 4 + 2\lambda = 0$
$5 + 2\lambda = 0 \Rightarrow \lambda = -\frac{5}{2}$
Substituting $\lambda = -\frac{5}{2}$ into equation $(i)$:
$x + (2 - \frac{5}{2})y - 2(-\frac{5}{2})z + (-\frac{5}{2} - 3) = 0$
$x - \frac{1}{2}y + 5z - \frac{11}{2} = 0$
Multiplying by $2$:
$2x - y + 10z - 11 = 0$
$2x - y + 10z = 11$

(A) Let $r$ be the radius of the spherical balloon. The volume $V$ of the sphere is given by $V = \frac{4}{3}\pi r^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = 35 \, cm^3/min$,we have $35 = 4\pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{35}{4\pi r^2} \quad (1)$.
The surface area $S$ of the sphere is $S = 4\pi r^2$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
Substituting the value of $\frac{dr}{dt}$ from $(1)$ into the expression for $\frac{dS}{dt}$:
$\frac{dS}{dt} = 8\pi r \left( \frac{35}{4\pi r^2} \right) = \frac{70}{r}$.
Given the diameter is $14 \, cm$,the radius $r = 7 \, cm$.
Substituting $r = 7$ into the expression for $\frac{dS}{dt}$:
$\frac{dS}{dt} = \frac{70}{7} = 10 \, cm^2/min$.

(A) Given $f(x) = [x] + |1 - x|$ for $x \in [-1, 3]$.
For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = -1 + (1 - x) = -x$.
For $x \in [0, 1)$,$[x] = 0$,so $f(x) = 0 + (1 - x) = 1 - x$.
For $x \in [1, 2)$,$[x] = 1$,so $f(x) = 1 + (x - 1) = x$.
For $x \in [2, 3)$,$[x] = 2$,so $f(x) = 2 + (x - 1) = x + 1$.
At $x = 3$,$f(3) = [3] + |1 - 3| = 3 + 2 = 5$.
Checking continuity:
At $x=0$: $LHL = \lim_{x \to 0^-} (-x) = 0$,$RHL = \lim_{x \to 0^+} (1 - x) = 1$. Since $LHL \neq RHL$,$f$ is discontinuous at $x=0$.
At $x=1$: $LHL = \lim_{x \to 1^-} (1 - x) = 0$,$RHL = \lim_{x \to 1^+} (x) = 1$. Since $LHL \neq RHL$,$f$ is discontinuous at $x=1$.
At $x=2$: $LHL = \lim_{x \to 2^-} (x) = 2$,$RHL = \lim_{x \to 2^+} (x + 1) = 3$. Since $LHL \neq RHL$,$f$ is discontinuous at $x=2$.
At $x=3$: $LHL = \lim_{x \to 3^-} (x + 1) = 4$,$f(3) = 5$. Since $LHL \neq f(3)$,$f$ is discontinuous at $x=3$.
Thus,Statement $1$ is true. Statement $2$ provides an incorrect definition of $f(x)$,so Statement $2$ is false.

(D) Given $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$.
By the Mean Value Theorem,for any $x_1, x_2 \in [1, 6]$ with $x_2 > x_1$,there exists $c \in (x_1, x_2)$ such that $\frac{f(x_2) - f(x_1)}{x_2 - x_1} = f'(c)$.
Since $f'(x) \ge 4.2$,we have $\frac{f(6) - f(1)}{6 - 1} \ge 4.2$.
Substituting the given values,we get $\frac{f(6) - (-2)}{5} \ge 4.2$.
$f(6) + 2 \ge 5 \times 4.2$.
$f(6) + 2 \ge 21$.
$f(6) \ge 19$.
Thus,the possible value of $f(6)$ lies in the interval $[19, \infty)$.

(D) Given the curve equation is $y = \cos(x + y)$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = -\sin(x + y) \left(1 + \frac{dy}{dx}\right) \dots (1)$
The equation of the tangent is $x + 2y = k$,which can be written as $y = -\frac{1}{2}x + \frac{k}{2}$.
The slope of the tangent is $m = -\frac{1}{2}$.
Substituting $\frac{dy}{dx} = -\frac{1}{2}$ into equation $(1)$:
$-\frac{1}{2} = -\sin(x + y) \left(1 - \frac{1}{2}\right)$
$-\frac{1}{2} = -\sin(x + y) \left(\frac{1}{2}\right)$
$\sin(x + y) = 1$
This implies $x + y = \frac{\pi}{2}$.
Substituting $x + y = \frac{\pi}{2}$ into the original curve equation $y = \cos(x + y)$:
$y = \cos\left(\frac{\pi}{2}\right) = 0$.
Since $y = 0$ and $x + y = \frac{\pi}{2}$,we have $x = \frac{\pi}{2}$.
Now,substitute the point $(\frac{\pi}{2}, 0)$ into the tangent equation $x + 2y = k$:
$\frac{\pi}{2} + 2(0) = k$
$k = \frac{\pi}{2}$.

(B) Given $y = \sec(\tan^{-1} x)$.
Let $\tan^{-1} x = \theta$,then $\tan \theta = x$.
We know that $\sec^2 \theta = 1 + \tan^2 \theta = 1 + x^2$,so $\sec \theta = \sqrt{1 + x^2}$.
Thus,$y = \sqrt{1 + x^2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$.
At $x = 1$:
$\left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}$.

(B) We are given that $P = \text{adj}(A)$ and $|A| = 4$.
We know the property of the adjoint of a matrix: $|\text{adj}(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|P| = |\text{adj}(A)| = |A|^{3-1} = |A|^2$.
Given $|A| = 4$,we have $|P| = 4^2 = 16$.
Now,calculate the determinant of $P$:
$|P| = \begin{vmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{vmatrix}$
$= 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$= 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$= 0 - \alpha(-2) + 3(-2)$
$= 2\alpha - 6$.
Equating the two values of $|P|$:
$2\alpha - 6 = 16$
$2\alpha = 22$
$\alpha = 11$.