JEE Main 2014 Mathematics Question Paper with Answer and Solution

(C) Given the expression $\sim (p \vee \sim q) \vee \sim (p \vee q)$.
Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$.
So,$\sim (p \vee \sim q) \equiv \sim p \wedge q$ and $\sim (p \vee q) \equiv \sim p \wedge \sim q$.
The expression becomes $(\sim p \wedge q) \vee (\sim p \wedge \sim q)$.
Using the Distributive Law,we factor out $\sim p$:
$\sim p \wedge (q \vee \sim q)$.
Since $(q \vee \sim q)$ is a tautology $(T)$,
$\sim p \wedge T \equiv \sim p$.

(B) Given the set $A = \{x : |x| < 3, x \in Z\}$.
Since $|x| < 3$ and $x$ is an integer,the elements of $A$ are $\{-2, -1, 0, 1, 2\}$.
The relation $R$ is defined as $R = \{(x, y) : y = |x|, x \neq -1\}$.
For each $x \in A$ such that $x \neq -1$,we find the corresponding $y = |x|$:
If $x = -2$,$y = |-2| = 2$. So,$(-2, 2) \in R$.
If $x = 0$,$y = |0| = 0$. So,$(0, 0) \in R$.
If $x = 1$,$y = |1| = 1$. So,$(1, 1) \in R$.
If $x = 2$,$y = |2| = 2$. So,$(2, 2) \in R$.
Thus,$R = \{(-2, 2), (0, 0), (1, 1), (2, 2)\}$.
The number of elements in $R$ is $n(R) = 4$.
The number of elements in the power set of $R$ is given by $2^{n(R)} = 2^4 = 16$.

(C) Let $z = x + iy$.
Given that $\frac{z - i}{z + i}$ is purely imaginary,its real part must be zero.
$\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x + i(y + 1)} \times \frac{x - i(y + 1)}{x - i(y + 1)} = \frac{x^2 + i(x(y - 1) - x(y + 1)) + (y - 1)(y + 1)}{x^2 + (y + 1)^2} = \frac{x^2 + y^2 - 1 - 2xi}{x^2 + (y + 1)^2}$.
For the expression to be purely imaginary,the real part must be zero:
$\frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2} = 0 \Rightarrow x^2 + y^2 = 1$.
Since $|z|^2 = x^2 + y^2 = 1$,we have $z \bar{z} = 1$,which implies $\bar{z} = \frac{1}{z}$.
Then $z + \frac{1}{z} = z + \bar{z} = (x + iy) + (x - iy) = 2x$.
Since $z \neq -i$,$x^2 + y^2 = 1$ implies $z$ lies on the unit circle excluding the point $(0, -1)$. Thus,$x$ can take any real value in the interval $(-1, 1)$,but $x$ cannot be $0$ because if $x=0$,then $y^2=1$,so $y=1$ (as $y=-1$ is excluded). Thus $2x$ can be any real number except $0$.

(C) Given the equation: $x^2 + |2x - 3| - 4 = 0$.
Case $1$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.
$x^2 + 2x - 3 - 4 = 0 \implies x^2 + 2x - 7 = 0$.
Using the quadratic formula: $x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$.
Since $x \ge \frac{3}{2}$,we check the values: $-1 + 2\sqrt{2} \approx -1 + 2(1.414) = 1.828 > 1.5$ (Valid).
$-1 - 2\sqrt{2} < 0 < 1.5$ (Invalid).
So,$x_1 = 2\sqrt{2} - 1$.
Case $2$: If $x < \frac{3}{2}$,then $|2x - 3| = -(2x - 3) = -2x + 3$.
$x^2 - 2x + 3 - 4 = 0 \implies x^2 - 2x - 1 = 0$.
Using the quadratic formula: $x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.
Since $x < \frac{3}{2}$,we check the values: $1 + \sqrt{2} \approx 2.414 > 1.5$ (Invalid).
$1 - \sqrt{2} \approx -0.414 < 1.5$ (Valid).
So,$x_2 = 1 - \sqrt{2}$.
Sum of the roots: $x_1 + x_2 = (2\sqrt{2} - 1) + (1 - \sqrt{2}) = \sqrt{2}$.

(B) The total number of digits is $8$. The odd places are $1, 3, 5, 7$ ($4$ places) and the even places are $2, 4, 6, 8$ ($4$ places).
The given digits are $1, 1, 2, 2, 2, 3, 4, 4$. The odd digits are $1, 1, 3$ (total $3$ digits) and the even digits are $2, 2, 2, 4, 4$ (total $5$ digits).
We are given that odd digits do not occupy odd places. This means the $3$ odd digits must be placed in the $4$ available even places.
The number of ways to arrange the $3$ odd digits $(1, 1, 3)$ in $4$ even places is given by $\frac{4!}{2!} = 12$.
After placing the $3$ odd digits,we have $5$ remaining places (the $1$ remaining even place and the $4$ odd places) to fill with the remaining $5$ even digits $(2, 2, 2, 4, 4)$.
The number of ways to arrange these $5$ digits is $\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10$.
Therefore,the total number of such $8$-digit numbers is $12 \times 10 = 120$.

(A) Let the $(r+1)^{\text{th}}$ and $(r+2)^{\text{th}}$ terms have equal coefficients.
${\left(2+\frac{x}{3}\right)^{55} = 2^{55}\left(1+\frac{x}{6}\right)^{55}}$
The $(r+1)^{\text{th}}$ term is $2^{55} \cdot {}^{55}C_r \left(\frac{x}{6}\right)^r$. The coefficient of $x^r$ is $2^{55} \cdot {}^{55}C_r \cdot \frac{1}{6^r}$.
The $(r+2)^{\text{th}}$ term is $2^{55} \cdot {}^{55}C_{r+1} \left(\frac{x}{6}\right)^{r+1}$. The coefficient of $x^{r+1}$ is $2^{55} \cdot {}^{55}C_{r+1} \cdot \frac{1}{6^{r+1}}$.
Equating the coefficients:
${}^{55}C_r \cdot \frac{1}{6^r} = {}^{55}C_{r+1} \cdot \frac{1}{6^{r+1}}$
${}^{55}C_r = {}^{55}C_{r+1} \cdot \frac{1}{6}$
$6 \cdot {}^{55}C_r = {}^{55}C_{r+1}$
$6 \cdot \frac{55!}{r!(55-r)!} = \frac{55!}{(r+1)!(54-r)!}$
$6 \cdot \frac{1}{r!(55-r)(54-r)!} = \frac{1}{(r+1)r!(54-r)!}$
$\frac{6}{55-r} = \frac{1}{r+1}$
$6(r+1) = 55-r$
$6r + 6 = 55 - r$
$7r = 49$
$r = 7$
Thus,the terms are $(r+1)^{\text{th}} = 8^{\text{th}}$ and $(r+2)^{\text{th}} = 9^{\text{th}}$.

(A) $G = \sqrt{ab}$
$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a + b}{2ab}$
Given that $\frac{1}{M} : G = 4 : 5$,we have $\frac{2ab}{(a + b)\sqrt{ab}} = \frac{4}{5}$
$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$
Using Componendo and Dividendo:
$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$
$\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 9$
$\Rightarrow \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = 3$ (assuming $a > b$ for positive ratio)
$\Rightarrow \sqrt{a} + \sqrt{b} = 3\sqrt{a} - 3\sqrt{b}$
$\Rightarrow 4\sqrt{b} = 2\sqrt{a}$ $\Rightarrow \sqrt{\frac{a}{b}} = 2$ $\Rightarrow \frac{a}{b} = 4$
Since the ratio $a:b$ can be $1:4$ or $4:1$,and $1:4$ is given in the options,the correct answer is $1:4$.

(C) The given expression is $1 - \sum_{k=1}^{n-1} \frac{2}{3^k} < \frac{1}{100}$.
This can be written as $1 - 2 \left( \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^{n-1}} \right) < \frac{1}{100}$.
The sum inside the bracket is a geometric progression with $a = \frac{1}{3}$,$r = \frac{1}{3}$,and $n-1$ terms.
The sum $S_{n-1} = \frac{a(1-r^{n-1})}{1-r} = \frac{\frac{1}{3}(1 - (\frac{1}{3})^{n-1})}{1 - \frac{1}{3}} = \frac{\frac{1}{3}(1 - \frac{1}{3^{n-1}})}{\frac{2}{3}} = \frac{1}{2} (1 - \frac{1}{3^{n-1}}) = \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}}$.
Substituting this back: $1 - 2 \left( \frac{1}{2} - \frac{1}{2 \cdot 3^{n-1}} \right) < \frac{1}{100}$.
$1 - 1 + \frac{1}{3^{n-1}} < \frac{1}{100}$.
$\frac{1}{3^{n-1}} < \frac{1}{100} \Rightarrow 3^{n-1} > 100$.
For $n=5$,$3^{5-1} = 3^4 = 81$ (which is not $> 100$).
For $n=6$,$3^{6-1} = 3^5 = 243$ (which is $> 100$).
Thus,the least positive integer $n$ is $6$.

(A) Given $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i.$
Let $y = 1 + x$,so $x = y - 1$.
Substituting this into the expression:
$1 + (y - 1)^4 + (y - 1)^5 = \sum\limits_{i = 0}^5 a_i y^i.$
Expanding the terms using the Binomial Theorem:
$(y - 1)^4 = y^4 - 4y^3 + 6y^2 - 4y + 1$
$(y - 1)^5 = y^5 - 5y^4 + 10y^3 - 10y^2 + 5y - 1$
Adding these to $1$:
$1 + (y^4 - 4y^3 + 6y^2 - 4y + 1) + (y^5 - 5y^4 + 10y^3 - 10y^2 + 5y - 1) = y^5 - 4y^4 + 6y^3 - 4y^2 + y + 1.$
Comparing this with $\sum\limits_{i = 0}^5 a_i y^i = a_5 y^5 + a_4 y^4 + a_3 y^3 + a_2 y^2 + a_1 y + a_0$,we get:
$a_5 = 1, a_4 = -4, a_3 = 6, a_2 = -4, a_1 = 1, a_0 = 1.$
Thus,$a_2 = -4$.

(B) Let the line intersect the $x$-axis at $B(a, 0)$ and the $y$-axis at $C(0, b)$.
Since the point $A(4, 3)$ is nearer to the $x$-axis and trisects the line segment $BC$,it divides $BC$ in the ratio $2:1$ from $C$ to $B$.
Using the section formula,the coordinates of $A$ are given by:
$A = \left( \frac{1 \times 0 + 2 \times a}{1 + 2}, \frac{1 \times b + 2 \times 0}{1 + 2} \right) = \left( \frac{2a}{3}, \frac{b}{3} \right)$
Given $A(4, 3)$,we have:
$\frac{2a}{3} = 4$ $\Rightarrow 2a = 12$ $\Rightarrow a = 6$
$\frac{b}{3} = 3 \Rightarrow b = 9$
Thus,the intercepts are $a = 6$ and $b = 9$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values,we get $\frac{x}{6} + \frac{y}{9} = 1$.
Multiplying by $18$,we get $3x + 2y = 18$.

(C) The given lines are:
$L_1: x + 2ay + a = 0$ $(1)$
$L_2: x + 3by + b = 0$ $(2)$
$L_3: x + 4ay + a = 0$ $(3)$
Since the lines are concurrent,they intersect at a common point.
Subtracting equation $(1)$ from $(3)$:
$(x + 4ay + a) - (x + 2ay + a) = 0$
$2ay = 0$
Since the lines are distinct,$a \neq 0$,therefore $y = 0$.
Substituting $y = 0$ into equation $(1)$:
$x + 2a(0) + a = 0 \Rightarrow x = -a$.
The point of concurrency is $(-a, 0)$.
Since this point must lie on line $(2)$:
$-a + 3b(0) + b = 0$
$-a + b = 0 \Rightarrow b = a$.
The point $(a, b)$ satisfies the equation $y = x$,which represents a straight line.

(D) For the circle $x^2 + y^2 = 16$,the center $C_1 = (0, 0)$ and radius $r_1 = 4$.
For the circle $x^2 + y^2 - 2y = 0$,we rewrite it as $x^2 + (y - 1)^2 = 1$,so the center $C_2 = (0, 1)$ and radius $r_2 = 1$.
The distance between the centers is $d = \sqrt{(0 - 0)^2 + (1 - 0)^2} = 1$.
The sum of the radii is $r_1 + r_2 = 4 + 1 = 5$.
The difference of the radii is $|r_1 - r_2| = |4 - 1| = 3$.
Since $d < |r_1 - r_2|$ (because $1 < 3$),the smaller circle lies entirely inside the larger circle.
Therefore,there are no common tangents to these two circles.

(D) The locus of the point of intersection of tangents to the parabola $y^2 = 4ax$ inclined at an angle $\alpha$ to each other is given by the equation:
$\tan^2 \alpha (x + a)^2 = y^2 - 4ax$
Given the equation of the parabola $y^2 = 4x$,we have $a = 1$.
The point of intersection is $(x, y) = (-2, -1)$.
Substituting these values into the locus equation:
$\tan^2 \alpha (-2 + 1)^2 = (-1)^2 - 4(1)(-2)$
$\tan^2 \alpha (-1)^2 = 1 + 8$
$\tan^2 \alpha (1) = 9$
$\tan^2 \alpha = 9$
Taking the square root on both sides:
$|\tan \alpha| = 3$

(D) Let the point of tangency be $(h, k)$ on the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$.
The equation of the tangent at $(h, k)$ is given by $\frac{xh}{16} + \frac{yk}{81} = 1$.
The $x$-intercept is found by setting $y=0$,which gives $x = \frac{16}{h}$. So,point $B = (\frac{16}{h}, 0)$.
The $y$-intercept is found by setting $x=0$,which gives $y = \frac{81}{k}$. So,point $A = (0, \frac{81}{k})$.
The area of the triangle $OAB$ is $A = \frac{1}{2} \times |\frac{16}{h}| \times |\frac{81}{k}| = \frac{648}{|hk|}$.
Since $(h, k)$ lies on the ellipse,$\frac{h^2}{16} + \frac{k^2}{81} = 1$. By the $AM$-$GM$ inequality,$\frac{\frac{h^2}{16} + \frac{k^2}{81}}{2} \ge \sqrt{\frac{h^2 k^2}{16 \times 81}}$.
$\frac{1}{2} \ge \frac{|hk|}{4 \times 9} \Rightarrow |hk| \le 18$.
Therefore,the minimum area $A = \frac{648}{|hk|} \ge \frac{648}{18} = 36$.
The minimum area of the triangle is $36$ square units.

(A) We are given the inequality $\frac{(x-10)(x-50)}{(x-30)} \ge 0$.
Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-10)(x-50)}{(x-30)}$,the critical points are $x = 10, 30, 50$.
The sign of $f(x)$ in different intervals is:
- For $x < 10$: $f(x) < 0$
- For $10 \le x < 30$: $f(x) \ge 0$
- For $30 < x < 50$: $f(x) < 0$
- For $x \ge 50$: $f(x) \ge 0$
Since $x \in \{1, 2, \dots, 100\}$,we look for integers $x$ in the intervals $[10, 30)$ and $[50, 100]$.
In the interval $[10, 30)$,the integers are $\{10, 11, \dots, 29\}$. The number of such integers is $29 - 10 + 1 = 20$.
In the interval $[50, 100]$,the integers are $\{50, 51, \dots, 100\}$. The number of such integers is $100 - 50 + 1 = 51$.
The total number of favorable outcomes is $20 + 51 = 71$.
The total number of possible outcomes is $100$.
Therefore,$P(A) = \frac{71}{100} = 0.71$.

(B) The given statement is $p \Rightarrow (q \vee r)$.
Using the logical equivalence $A$ $\Rightarrow (B \vee C) \equiv (A$ $\Rightarrow B) \vee (A$ $\Rightarrow C)$,we can rewrite the expression.
Thus,$p \Rightarrow (q \vee r)$ is equivalent to $(p$ $\Rightarrow q) \vee (p$ $\Rightarrow r)$.

(B) Given $z = 1 + i\alpha$,where $\alpha \in R$.
Squaring both sides,we get $z^2 = (1 + i\alpha)^2 = 1^2 + (i\alpha)^2 + 2(1)(i\alpha)$.
$z^2 = 1 - \alpha^2 + 2i\alpha$.
Given $z^2 = x + iy$,we equate the real and imaginary parts:
$x = 1 - \alpha^2$ and $y = 2\alpha$.
From $y = 2\alpha$,we have $\alpha = \frac{y}{2}$.
Substituting this into the expression for $x$:
$x = 1 - (\frac{y}{2})^2$
$x = 1 - \frac{y^2}{4}$
Multiplying by $4$:
$4x = 4 - y^2$
$y^2 + 4x - 4 = 0$.

(A) Given the equation $\sqrt{3x^2 + x + 5} = x - 3$.
For the square root to be defined and equal to a real value,we must have $x - 3 \geq 0$,which implies $x \geq 3$.
Squaring both sides:
$3x^2 + x + 5 = (x - 3)^2$
$3x^2 + x + 5 = x^2 - 6x + 9$
$2x^2 + 7x - 4 = 0$
Solving the quadratic equation $2x^2 + 7x - 4 = 0$ using the quadratic formula or factorization:
$2x^2 + 8x - x - 4 = 0$
$2x(x + 4) - 1(x + 4) = 0$
$(2x - 1)(x + 4) = 0$
So,$x = \frac{1}{2}$ or $x = -4$.
Checking these values against the condition $x \geq 3$:
For $x = \frac{1}{2}$,$\frac{1}{2} < 3$ (Invalid).
For $x = -4$,$-4 < 3$ (Invalid).
Since neither value satisfies the condition $x \geq 3$,the equation has no real solution.

(C) Let the number of men be $n$.
Total participants $= n + 2$.
Each participant plays $2$ games with every other participant.
The number of games played between $n$ men is $2 \times \binom{n}{2} = 2 \times \frac{n(n-1)}{2} = n(n-1)$.
The number of games played between $n$ men and $2$ women is $n \times 2 \times 2 = 4n$ (since each man plays $2$ games with each of the $2$ women).
Given that the number of games between men exceeds the number of games between men and women by $66$:
$n(n-1) - 4n = 66$
$n^2 - n - 4n = 66$
$n^2 - 5n - 66 = 0$
$(n - 11)(n + 6) = 0$
Since $n > 0$,we have $n = 11$.
The value $n = 11$ lies in the interval $(11, 13]$.

(B) The given expression is $(1 + x^n + x^{253})^{10}$.
Using the multinomial theorem,the general term is given by $\frac{10!}{a!b!c!} (1)^a (x^n)^b (x^{253})^c$,where $a + b + c = 10$.
We need the coefficient of $x^{1012}$,so we set the exponent of $x$ to $1012$:
$nb + 253c = 1012$.
Since $c \leq 10$ and $253 \times 4 = 1012$,we test values for $c$:
If $c = 4$,then $nb = 1012 - 253(4) = 0$. Since $n$ is a positive integer,$b$ must be $0$.
Then $a = 10 - b - c = 10 - 0 - 4 = 6$.
The coefficient is $\frac{10!}{6!0!4!} = ^{10}C_4$.
If $c < 4$,then $nb = 253(4-c)$. For $c=3$,$nb = 253$,which is impossible for $n \leq 22$ and $b \leq 10$ (as $22 \times 10 = 220 < 253$).
Thus,the only solution is $a=6, b=0, c=4$.
The coefficient is $^{10}C_4$.

(B) Let the total number of terms be $2n$,the first term be $a$,and the common difference be $d$.
The terms are $a, a+d, a+2d, ..., a+(2n-1)d$.
The odd-positioned terms are $a, a+2d, ..., a+(2n-2)d$. There are $n$ such terms.
Sum of odd terms $S_o = \frac{n}{2}[2a + (n-1)(2d)] = n[a + (n-1)d] = 24$ --- $(i)$
The even-positioned terms are $a+d, a+3d, ..., a+(2n-1)d$. There are $n$ such terms.
Sum of even terms $S_e = \frac{n}{2}[2(a+d) + (n-1)(2d)] = n[a+d+(n-1)d] = 30$ --- $(ii)$
Subtracting $(i)$ from $(ii)$:
$n(a+d+(n-1)d) - n(a+(n-1)d) = 30 - 24$
$nd = 6$ --- $(iii)$
Given the last term exceeds the first term by $10\frac{1}{2} = \frac{21}{2}$:
$(a+(2n-1)d) - a = \frac{21}{2}$
$(2n-1)d = \frac{21}{2}$
$2nd - d = \frac{21}{2}$
Substitute $nd = 6$ into the equation:
$2(6) - d = \frac{21}{2}$
$12 - d = 10.5$
$d = 1.5 = \frac{3}{2}$
Using $nd = 6$:
$n(\frac{3}{2}) = 6 \Rightarrow n = 4$
Total number of terms $= 2n = 2 \times 4 = 8$.

(D) Given $f(n) = \left[ \frac{1}{3} + \frac{3n}{100} \right]n$.
For $1 \le n \le 22$,$\frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{66}{100} = 0.333 + 0.66 = 0.993 < 1$. Thus,$\left[ \frac{1}{3} + \frac{3n}{100} \right] = 0$,so $f(n) = 0$.
For $23 \le n \le 55$,$1 \le \frac{1}{3} + \frac{3n}{100} < \frac{1}{3} + \frac{165}{100} = 0.333 + 1.65 = 1.983 < 2$. Thus,$\left[ \frac{1}{3} + \frac{3n}{100} \right] = 1$,so $f(n) = n$.
For $n = 56$,$\frac{1}{3} + \frac{3(56)}{100} = 0.333 + 1.68 = 2.013$. Thus,$\left[ \frac{1}{3} + \frac{3(56)}{100} \right] = 2$,so $f(56) = 2 \times 56 = 112$.
The sum is $\sum_{n=1}^{56} f(n) = \sum_{n=1}^{22} 0 + \sum_{n=23}^{55} n + f(56)$.
$= 0 + \frac{(55-23+1)}{2}(23+55) + 112 = \frac{33}{2}(78) + 112 = 33 \times 39 + 112 = 1287 + 112 = 1399$.

(D) We have $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ for $x \in R$.
Completing the square for $f(x)$:
$f(x) = (x + b)^2 + 2c^2 - b^2$.
Thus,the minimum value of $f(x)$ is $f_{\min} = 2c^2 - b^2$.
Completing the square for $g(x)$:
$g(x) = -(x^2 + 2cx) + b^2 = -(x + c)^2 + c^2 + b^2$.
Thus,the maximum value of $g(x)$ is $g_{\max} = c^2 + b^2$.
Given the condition $\min f(x) > \max g(x)$:
$2c^2 - b^2 > c^2 + b^2$.
Subtracting $c^2$ and adding $b^2$ to both sides:
$c^2 > 2b^2$.
Since $b$ and $c$ are non-zero,we divide by $b^2$:
$\frac{c^2}{b^2} > 2$.
Taking the square root on both sides:
$\left| \frac{c}{b} \right| > \sqrt{2}$.
Therefore,$\left| \frac{c}{b} \right| \in (\sqrt{2}, \infty)$.

(D) Let the circumcentre be $O = (0, 0)$.
The centroid $G$ is the midpoint of the segment joining $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$.
$G = \left( \frac{a^2 + 1 + 2a}{2}, \frac{a^2 + 1 - 2a}{2} \right) = \left( \frac{(a + 1)^2}{2}, \frac{(a - 1)^2}{2} \right)$.
We know that the orthocentre $H$,centroid $G$,and circumcentre $O$ are collinear,and $G$ divides $OH$ in the ratio $1:2$ internally.
Using the section formula,$G = \frac{1 \cdot H + 2 \cdot O}{1 + 2} = \frac{H}{3}$.
Thus,$H = 3G = \left( \frac{3(a + 1)^2}{2}, \frac{3(a - 1)^2}{2} \right)$.
Let the line be $Ax + By = 0$. Substituting the coordinates of $H$ into the equation in option $(d)$:
$(a - 1)^2 \left( \frac{3(a + 1)^2}{2} \right) - (a + 1)^2 \left( \frac{3(a - 1)^2}{2} \right) = \frac{3}{2} (a - 1)^2 (a + 1)^2 - \frac{3}{2} (a + 1)^2 (a - 1)^2 = 0$.
Since the coordinates of $H$ satisfy the equation in option $(d)$,the orthocentre lies on this line.

(B) The equation of a line perpendicular to $5x - y = 1$ is of the form $x + 5y = c$.
The intercepts of this line on the coordinate axes are found by setting $y=0$ and $x=0$:
For $y=0$,$x=c$. So,the $x$-intercept is $(c, 0)$.
For $x=0$,$5y=c$,so $y=c/5$. The $y$-intercept is $(0, c/5)$.
The area of the triangle formed by this line and the coordinate axes is given by $\frac{1}{2} \times |\text{base}| \times |\text{height}| = 5$.
$\frac{1}{2} \times |c| \times |\frac{c}{5}| = 5$
$\frac{c^2}{10} = 5$ $\Rightarrow c^2 = 50$ $\Rightarrow c = \pm 5\sqrt{2}$.
Thus,the equation of line $L$ is $x + 5y = \pm 5\sqrt{2}$.
The distance $d$ between the parallel lines $x + 5y = c$ and $x + 5y = 0$ is given by $d = \frac{|c - 0|}{\sqrt{1^2 + 5^2}} = \frac{|c|}{\sqrt{26}}$.
Substituting $c = \pm 5\sqrt{2}$:
$d = \frac{5\sqrt{2}}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{2} \times \sqrt{13}} = \frac{5}{\sqrt{13}}$.

(A) The given circle is $S: x^2 + y^2 - 16 = 0$.
The equation of the chord is $L: 3x + y + 5 = 0$.
The equation of the family of circles passing through the intersection of $S$ and $L$ is $S + \lambda L = 0$.
Thus,$x^2 + y^2 - 16 + \lambda(3x + y + 5) = 0$.
$x^2 + y^2 + 3\lambda x + \lambda y + (5\lambda - 16) = 0$.
The center of this circle is $C = (-\frac{3\lambda}{2}, -\frac{\lambda}{2})$.
Since the chord $3x + y + 5 = 0$ is the diameter of this circle,the center $C$ must lie on the line $3x + y + 5 = 0$.
Substituting the center into the line equation: $3(-\frac{3\lambda}{2}) + (-\frac{\lambda}{2}) + 5 = 0$.
$-\frac{9\lambda}{2} - \frac{\lambda}{2} + 5 = 0$.
$-5\lambda + 5 = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$ into the circle equation: $x^2 + y^2 + 3(1)x + (1)y + 5(1) - 16 = 0$.
$x^2 + y^2 + 3x + y - 11 = 0$.

(A) The equation of the parabola is $y^2 = 6x$,which is of the form $y^2 = 4ax$ with $4a = 6$,so $a = \frac{3}{2}$.
The focus is $S = (\frac{3}{2}, 0)$ and the vertex is $V = (0, 0)$.
Let the equation of the chord passing through the focus be $y - 0 = m(x - \frac{3}{2})$,which simplifies to $mx - y - \frac{3m}{2} = 0$.
The distance $d$ of this line from the vertex $(0, 0)$ is given by $d = \frac{|m(0) - 0 - \frac{3m}{2}|}{\sqrt{m^2 + (-1)^2}} = \frac{|-\frac{3m}{2}|}{\sqrt{m^2 + 1}}$.
Given $d = \frac{\sqrt{5}}{2}$,we have $\frac{|\frac{3m}{2}|}{\sqrt{m^2 + 1}} = \frac{\sqrt{5}}{2}$.
Squaring both sides,we get $\frac{9m^2}{4(m^2 + 1)} = \frac{5}{4}$.
$9m^2 = 5(m^2 + 1)$ $\Rightarrow 9m^2 = 5m^2 + 5$ $\Rightarrow 4m^2 = 5$.
$m^2 = \frac{5}{4} \Rightarrow m = \pm \frac{\sqrt{5}}{2}$.
Thus,the slope can be $\frac{\sqrt{5}}{2}$.

(A) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$.
Here,$a^2 = 4$ and $b^2 = 5$.
The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
The extremity of the latus rectum in the first quadrant is $L = (ae, \frac{b^2}{a}) = (2 \times \frac{3}{2}, \frac{5}{2}) = (3, \frac{5}{2})$.
The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.
Substituting $(x_1, y_1) = (3, \frac{5}{2})$,we get $\frac{3x}{4} - \frac{y(5/2)}{5} = 1$,which simplifies to $\frac{3x}{4} - \frac{y}{2} = 1$.
This can be written as $\frac{x}{4/3} + \frac{y}{-2} = 1$.
The $x$-intercept is $OA = \frac{4}{3}$ and the $y$-intercept is $OB = -2$.
Therefore,$(OA)^2 - (OB)^2 = (\frac{4}{3})^2 - (-2)^2 = \frac{16}{9} - 4 = \frac{16 - 36}{9} = -\frac{20}{9}$.

(B) Given $d_i = -x_i - a$.
Statement $I$: The variance of a set of observations is invariant under change of origin and scale factor $-1$. Specifically,if $y_i = c x_i + k$,then $\text{Var}(y) = c^2 \text{Var}(x)$. Here,$c = -1$ and $k = -a$. Thus,$\text{Var}(d) = (-1)^2 \sigma^2 = \sigma^2$. So,Statement $I$ is true.
Statement $II$: The mean of $d_i$ is $\bar{d} = \frac{1}{n} \sum (-x_i - a) = -\bar{x} - a$. This is correct.
For the mode,if $M$ is the mode of $x_i$,then the mode of $d_i = -x_i - a$ is $-M - a$. This is also correct.
Therefore,both statements are true.

(B) Let $f(x) = |\sin 4x| + |\cos 2x|$.
We know that the period of $|\sin ax|$ is $\frac{\pi}{|a|}$ and the period of $|\cos ax|$ is $\frac{\pi}{|a|}$.
For $|\sin 4x|$,the period is $T_1 = \frac{\pi}{4}$.
For $|\cos 2x|$,the period is $T_2 = \frac{\pi}{2}$.
The period of the sum of two periodic functions is the least common multiple $(LCM)$ of their individual periods.
$T = \text{LCM}\left(\frac{\pi}{4}, \frac{\pi}{2}\right) = \frac{\text{LCM}(\pi, \pi)}{\text{HCF}(4, 2)} = \frac{\pi}{2}$.
Thus,the period of $f(x)$ is $\frac{\pi}{2}$.

(C) The given statement is in the form $p \Rightarrow q$,where $p$ is "$I$ am not feeling well" and $q$ is "$I$ will go to the doctor".
The contrapositive of $p \Rightarrow q$ is defined as $\neg q \Rightarrow \neg p$.
Here,$\neg q$ is "$I$ will not go to the doctor" and $\neg p$ is "$I$ am feeling well".
Therefore,the contrapositive is "If $I$ will not go to the doctor,then $I$ am feeling well".

(D) Given that $\alpha$ and $\beta$ are roots of $p x^2 + q x + r = 0$. Since $p, q, r$ are in $AP$,we have $2q = p + r$.
From $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,we get $\frac{\alpha + \beta}{\alpha \beta} = 4$.
Using Vieta's formulas,$\alpha + \beta = -\frac{q}{p}$ and $\alpha \beta = \frac{r}{p}$.
Substituting these,$\frac{-q/p}{r/p} = -\frac{q}{r} = 4$,so $q = -4r$.
Since $p + r = 2q$,we have $p + r = 2(-4r) = -8r$,which implies $p = -9r$.
Now,$|\alpha - \beta| = \frac{\sqrt{D}}{|p|} = \frac{\sqrt{q^2 - 4pr}}{|p|}$.
Substituting $q = -4r$ and $p = -9r$:
$|\alpha - \beta| = \frac{\sqrt{(-4r)^2 - 4(-9r)(r)}}{|-9r|} = \frac{\sqrt{16r^2 + 36r^2}}{9|r|} = \frac{\sqrt{52r^2}}{9|r|} = \frac{2|r|\sqrt{13}}{9|r|} = \frac{2\sqrt{13}}{9}$.
Thus,the correct option is $D$.

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$
Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \sin ^2 x$.
Using the identity $\sin (\pi - \theta) = \sin \theta$,we get $\sin (\pi - \pi \sin ^2 x) = \sin (\pi \sin ^2 x)$.
Now,the limit becomes: $\lim _{x \rightarrow 0} \frac{\sin (\pi \sin ^2 x)}{x^2}$.
Multiplying and dividing by $\pi \sin ^2 x$:
$\lim _{x}$ ${\rightarrow 0} \left( \frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} \right) \times \left( \frac{\pi \sin ^2 x}{x^2} \right)$.
As $x \rightarrow 0$,$\sin ^2 x \rightarrow 0$,so $\frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} \rightarrow 1$.
Also,$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} = 1$.
Therefore,the limit is $1 \times \pi \times 1 = \pi$.

(B) The first $50$ even natural numbers are $2, 4, 6, \ldots, 100$.
Mean,$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{50} = \frac{2(1+2+3+\ldots+50)}{50} = \frac{2 \times \frac{50 \times 51}{2}}{50} = 51$.
Variance,$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 4^2 + \ldots + 100^2 = 4(1^2 + 2^2 + \ldots + 50^2)$.
Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$\sum x_i^2 = 4 \times \frac{50(51)(101)}{6} = \frac{2 \times 50 \times 51 \times 101}{3} = 2 \times 50 \times 17 \times 101 = 171700$.
$\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

(C) Given $f(\theta ) = \left| \begin{array}{ccc} 1 & \cos \theta & 1 \\ - \sin \theta & 1 & - \cos \theta \\ - 1 & \sin \theta & 1 \end{array} \right|$.
Expanding the determinant along the first row:
$f(\theta ) = 1(1 - (-\sin \theta \cos \theta )) - \cos \theta (-\sin \theta - \cos \theta ) + 1(-\sin^2 \theta + 1)$
$f(\theta ) = 1 + \sin \theta \cos \theta + \sin \theta \cos \theta + \cos^2 \theta - \sin^2 \theta + 1$
$f(\theta ) = 2 + 2 \sin \theta \cos \theta + \cos 2\theta$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$f(\theta ) = 2 + \sin 2\theta + \cos 2\theta$.
The expression $a \sin x + b \cos x$ has a maximum value of $\sqrt{a^2 + b^2}$ and a minimum value of $-\sqrt{a^2 + b^2}$.
Here,$a = 1$ and $b = 1$,so the range of $\sin 2\theta + \cos 2\theta$ is $[-\sqrt{2}, \sqrt{2}]$.
Therefore,the maximum value $A = 2 + \sqrt{2}$ and the minimum value $B = 2 - \sqrt{2}$.
Thus,$(A, B) = (2 + \sqrt{2}, 2 - \sqrt{2})$.

(D) Given $f(x) = \frac{|x| - 1}{|x| + 1}$.
For a function to be one-one,if $f(x_1) = f(x_2)$,then $x_1$ must be equal to $x_2$.
Let $f(x_1) = f(x_2)$:
$\frac{|x_1| - 1}{|x_1| + 1} = \frac{|x_2| - 1}{|x_2| + 1}$
$(|x_1| - 1)(|x_2| + 1) = (|x_2| - 1)(|x_1| + 1)$
$|x_1||x_2| + |x_1| - |x_2| - 1 = |x_1||x_2| - |x_1| + |x_2| - 1$
$2|x_1| = 2|x_2| \implies |x_1| = |x_2|$
This implies $x_1 = x_2$ or $x_1 = -x_2$. Since $f(1) = f(-1) = 0$,the function is not one-one (it is many-one).
For onto: Let $y = \frac{|x| - 1}{|x| + 1}$.
$y(|x| + 1) = |x| - 1 \implies y|x| + y = |x| - 1 \implies |x|(y - 1) = -1 - y \implies |x| = \frac{1 + y}{1 - y}$.
Since $|x| \ge 0$,we must have $\frac{1 + y}{1 - y} \ge 0$. Solving this inequality,we get $y \in [-1, 1)$.
The range of $f$ is $[-1, 1)$,which is not equal to the codomain $R$. Therefore,the function is not onto.

(B) Given that $A$ is a symmetric matrix,so $A^T = A$.
Given that $B$ is a skew-symmetric matrix,so $B^T = -B$.
We need to check the nature of the matrix $M = AB - BA$.
Taking the transpose of $M$:
$M^T = (AB - BA)^T = (AB)^T - (BA)^T$
Using the property $(XY)^T = Y^T X^T$:
$M^T = B^T A^T - A^T B^T$
Substituting $A^T = A$ and $B^T = -B$:
$M^T = (-B)(A) - (A)(-B)$
$M^T = -BA + AB = AB - BA = M$
Since $M^T = M$,the matrix $AB - BA$ is a symmetric matrix.

(D) We are given the determinant ${\Delta _r} = \left| {\begin{array}{*{20}{c}} r&{2r - 1}&{3r - 2} \\ {\frac{n}{2}}&{n - 1}&a \\ {\frac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\frac{1}{2}\left( {n - 1} \right)\left( {3n - 4} \right)} \end{array}} \right|$.
We need to calculate $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $. Since the summation operator is linear with respect to the rows of a determinant,we can write:
$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} = \left| {\begin{array}{*{20}{c}} {\sum\limits_{r = 1}^{n - 1} r }&{\sum\limits_{r = 1}^{n - 1} {\left( {2r - 1} \right)} }&{\sum\limits_{r = 1}^{n - 1} {\left( {3r - 2} \right)} } \\ {\frac{n}{2}}&{n - 1}&a \\ {\frac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\frac{1}{2}\left( {n - 1} \right)\left( {3n - 4} \right)} \end{array}} \right|$.
Calculating the sums:
$1. \sum\limits_{r = 1}^{n - 1} r = \frac{{\left( {n - 1} \right)n}}{2}$
$2. \sum\limits_{r = 1}^{n - 1} {\left( {2r - 1} \right)} = 2\frac{{\left( {n - 1} \right)n}}{2} - \left( {n - 1} \right) = n(n-1) - (n-1) = {(n-1)^2}$
$3. \sum\limits_{r = 1}^{n - 1} {\left( {3r - 2} \right)} = 3\frac{{\left( {n - 1} \right)n}}{2} - 2(n-1) = \frac{{3n(n-1) - 4(n-1)}}{2} = \frac{{\left( {n - 1} \right)\left( {3n - 4} \right)}}{2}$
Substituting these back into the determinant:
$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} = \left| {\begin{array}{*{20}{c}} {\frac{{n\left( {n - 1} \right)}}{2}}&{{{\left( {n - 1} \right)}^2}}&{\frac{{\left( {n - 1} \right)\left( {3n - 4} \right)}}{2}} \\ {\frac{n}{2}}&{n - 1}&a \\ {\frac{{n\left( {n - 1} \right)}}{2}}&{{{\left( {n - 1} \right)}^2}}&{\frac{{\left( {n - 1} \right)\left( {3n - 4} \right)}}{2}} \end{array}} \right|$.
Since the first row $(R_1)$ and the third row $(R_3)$ are identical,the value of the determinant is $0$.
Therefore,the value is independent of both $a$ and $n$.

(D) For the function $f(x)$ to be continuous at $x = \pi$,we must have $\lim_{x \to \pi} f(x) = f(\pi) = k$.
Let $x = \pi + h$,where $h \to 0$ as $x \to \pi$. Then $(\pi - x)^2 = (-h)^2 = h^2$.
$\lim_{h \to 0} \frac{\sqrt{2 + \cos(\pi + h)} - 1}{h^2} = k$.
Since $\cos(\pi + h) = -\cos h$,we have:
$k = \lim_{h \to 0} \frac{\sqrt{2 - \cos h} - 1}{h^2}$.
Rationalizing the numerator:
$k = \lim_{h \to 0} \frac{(\sqrt{2 - \cos h} - 1)(\sqrt{2 - \cos h} + 1)}{h^2(\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^2(\sqrt{2 - \cos h} + 1)}$.
$k = \lim_{h \to 0} \frac{1 - \cos h}{h^2(\sqrt{2 - \cos h} + 1)}$.
Using the identity $1 - \cos h = 2 \sin^2(h/2)$:
$k = \lim_{h \to 0} \frac{2 \sin^2(h/2)}{h^2(\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{2 \sin^2(h/2)}{4(h/2)^2(\sqrt{2 - \cos h} + 1)}$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$k = \frac{2}{4} \times \frac{1}{\sqrt{2 - 1} + 1} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$.

(B) Given $|f(x)| \leq x^2$ for all $x \in R$.
At $x = 0$,$|f(0)| \leq 0^2 = 0$,which implies $f(0) = 0$.
To check differentiability at $x = 0$,we evaluate the limit of the derivative:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h - 0} = \lim_{h \to 0} \frac{f(h)}{h}$.
From the given inequality,$|f(h)| \leq h^2$,so $|\frac{f(h)}{h}| \leq |h|$.
By the Sandwich Theorem,as $h \to 0$,$|h| \to 0$,therefore $\lim_{h \to 0} \frac{f(h)}{h} = 0$.
Thus,$f'(0) = 0$,which means $f$ is differentiable at $x = 0$.
Since differentiability implies continuity,$f$ is also continuous at $x = 0$.
Therefore,$f$ is continuous as well as differentiable at $x = 0$.

(C) The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given that $\frac{dV}{dt} = 4\pi \, cc/sec$,we substitute this into the equation:
$4\pi = 4\pi r^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{r^2}$.
We are given that the volume $V = 288\pi \, cc$.
Using the volume formula:
$288\pi = \frac{4}{3}\pi r^3
\Rightarrow r^3 = \frac{288 \times 3}{4} = 216
\Rightarrow r = 6 \, cm$.
Substituting $r = 6$ into the expression for $\frac{dr}{dt}$:
$\frac{dr}{dt} = \frac{1}{6^2} = \frac{1}{36} \, cm/sec$.

(B) Given integral: $I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{\left(x^{2 m}+x^{m}+1\right)^{3}} d x$
Divide numerator and denominator by $x^{6m}$:
$I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{x^{6m}\left(x^{-2m}+x^{-m}+1\right)^{3}} d x$
$I = \int \frac{x^{-m-1}+2 x^{-2 m-1}}{\left(1+x^{-m}+x^{-2 m}\right)^{3}} d x$
Let $t = 1+x^{-m}+x^{-2 m}$.
Then $dt = (-m x^{-m-1} - 2m x^{-2m-1}) dx = -m(x^{-m-1} + 2x^{-2m-1}) dx$.
So,$(x^{-m-1} + 2x^{-2m-1}) dx = -\frac{dt}{m}$.
Substituting into the integral:
$I = \int \frac{-dt/m}{t^3} = -\frac{1}{m} \int t^{-3} dt = -\frac{1}{m} \left( \frac{t^{-2}}{-2} \right) + C = \frac{1}{2mt^2} + C$.
Substituting $t$ back:
$I = \frac{1}{2m(1+x^{-m}+x^{-2m})^2} + C = \frac{1}{2m(\frac{x^{2m}+x^m+1}{x^{2m}})^2} + C = \frac{x^{4m}}{2m(x^{2m}+x^m+1)^2} + C$.
Thus,$f(x) = \frac{x^{4m}}{2m(x^{2m}+x^m+1)^2}$.

(D) Given $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt$ for $x > 0$.
Let $I = \int_{1}^{x} \frac{e^{t}}{t+a} dt$.
Substitute $t+a = z$,so $t = z-a$ and $dt = dz$.
When $t = 1$,$z = 1+a$. When $t = x$,$z = x+a$.
Substituting these into the integral:
$I = \int_{1+a}^{x+a} \frac{e^{z-a}}{z} dz = e^{-a} \int_{1+a}^{x+a} \frac{e^{z}}{z} dz$.
Using the property of definite integrals $\int_{b}^{c} f(z) dz = \int_{1}^{c} f(z) dz - \int_{1}^{b} f(z) dz$:
$I = e^{-a} \left[ \int_{1}^{x+a} \frac{e^{z}}{z} dz - \int_{1}^{1+a} \frac{e^{z}}{z} dz \right]$.
By the definition of $F(x)$,this becomes:
$I = e^{-a} [F(x+a) - F(1+a)]$.
Thus,the correct option is $D$.

(A) The given curve is $y = \tan x$ ... $(1)$
When $x = \frac{\pi}{4}$,$y = \tan(\frac{\pi}{4}) = 1$. So,the point of tangency is $P(\frac{\pi}{4}, 1)$.
The derivative is $\frac{dy}{dx} = \sec^2 x$. At $x = \frac{\pi}{4}$,the slope $m = \sec^2(\frac{\pi}{4}) = 2$.
The equation of the tangent at $P$ is $y - 1 = 2(x - \frac{\pi}{4})$,which simplifies to $y = 2x + 1 - \frac{\pi}{2}$ ... $(2)$.
The $x-$intercept of the tangent is found by setting $y = 0$: $0 = 2x + 1 - \frac{\pi}{2} \implies x = \frac{\pi - 2}{4}$. Let this point be $L(\frac{\pi - 2}{4}, 0)$.
The area of the region is the area under the curve $y = \tan x$ from $x = 0$ to $x = \frac{\pi}{4}$ minus the area of the triangle formed by the tangent line,the $x-$axis,and the vertical line $x = \frac{\pi}{4}$.
Area $= \int_{0}^{\frac{\pi}{4}} \tan x \, dx - \text{Area of } \Delta PLM$
$= [\log |\sec x|]_{0}^{\frac{\pi}{4}} - \frac{1}{2} \times \text{base} \times \text{height}$
$= \log(\sec \frac{\pi}{4}) - \log(\sec 0) - \frac{1}{2} \times (\frac{\pi}{4} - \frac{\pi - 2}{4}) \times 1$
$= \log(\sqrt{2}) - 0 - \frac{1}{2} \times (\frac{2}{4}) = \frac{1}{2} \log 2 - \frac{1}{4} = \frac{1}{2}(\log 2 - \frac{1}{2})$ sq units.

(C) The given differential equation is $\frac{dy}{dx} + y \tan x = \sin 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sin 2x$.
The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int \tan x dx} = e^{\ln(\sec x)} = \sec x$.
The general solution is $y(IF) = \int Q(IF) dx + c$.
$y \sec x = \int \sin 2x \sec x dx + c$.
$y \sec x = \int (2 \sin x \cos x) \sec x dx + c$.
$y \sec x = 2 \int \sin x dx + c$.
$y \sec x = -2 \cos x + c$ .....$(1)$.
Given $y(0) = 1$,substitute $x = 0$ and $y = 1$ into $(1)$:
$1 \cdot \sec(0) = -2 \cos(0) + c \Rightarrow 1(1) = -2(1) + c \Rightarrow c = 3$.
Substituting $c = 3$ into $(1)$,we get $y \sec x = -2 \cos x + 3$.
To find $y(\pi)$,substitute $x = \pi$:
$y \sec(\pi) = -2 \cos(\pi) + 3$.
$y(-1) = -2(-1) + 3$.
$-y = 2 + 3 = 5$.
$y = -5$.

(B) Let the two lines be $L_1: \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} = r_1$ and $L_2: \frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1} = r_2$.
Any point on $L_1$ is $P(r_1, -r_1, r_1)$ and any point on $L_2$ is $Q(1, -2r_2 - 1, r_2)$.
The direction ratios of the line $PQ$ are $(1 - r_1, -2r_2 - 1 + r_1, r_2 - r_1)$.
Since $PQ$ is the shortest distance line,it is perpendicular to both $L_1$ (direction $\vec{v_1} = \langle 1, -1, 1 \rangle$) and $L_2$ (direction $\vec{v_2} = \langle 0, -2, 1 \rangle$).
$1$) $(1 - r_1)(1) + (-2r_2 - 1 + r_1)(-1) + (r_2 - r_1)(1) = 0 \Rightarrow 1 - r_1 + 2r_2 + 1 - r_1 + r_2 - r_1 = 0 \Rightarrow -3r_1 + 3r_2 + 2 = 0$.
$2$) $(1 - r_1)(0) + (-2r_2 - 1 + r_1)(-2) + (r_2 - r_1)(1) = 0 \Rightarrow 4r_2 + 2 - 2r_1 + r_2 - r_1 = 0 \Rightarrow -3r_1 + 5r_2 + 2 = 0$.
Subtracting the equations: $2r_2 = 0 \Rightarrow r_2 = 0$.
Then $-3r_1 + 2 = 0 \Rightarrow r_1 = 2/3$.
The points are $P(2/3, -2/3, 2/3)$ and $Q(1, -1, 0)$.
The direction ratios of $PQ$ are $(1 - 2/3, -1 + 2/3, 0 - 2/3) = (1/3, -1/3, -2/3)$,which is proportional to $(1, -1, -2)$.
The line passes through $Q(1, -1, 0)$,so the equation is $\frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z}{-2}$.

(C) The given equation of the line is $2(x + 1) = y = z + 4$. Dividing by $2$,we get the symmetric form: $\frac{x + 1}{1} = \frac{y}{2} = \frac{z + 4}{2}$.
Thus,the direction vector of the line is $\vec{b} = (1, 2, 2)$.
The equation of the plane is $2x + 0y - \sqrt{\lambda} z + 4 = 0$. The normal vector to the plane is $\vec{n} = (2, 0, -\sqrt{\lambda})$.
Let $\theta$ be the angle between the line and the plane. Then $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\theta = \frac{\pi}{6}$,so $\sin \frac{\pi}{6} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \frac{|(1)(2) + (2)(0) + (2)(-\sqrt{\lambda})|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + 0^2 + (-\sqrt{\lambda})^2}}$.
$\frac{1}{2} = \frac{|2 - 2\sqrt{\lambda}|}{\sqrt{9} \sqrt{4 + \lambda}} = \frac{2|1 - \sqrt{\lambda}|}{3 \sqrt{4 + \lambda}}$.
$3 \sqrt{4 + \lambda} = 4 |1 - \sqrt{\lambda}|$.
Squaring both sides: $9(4 + \lambda) = 16(1 - 2\sqrt{\lambda} + \lambda)$.
$36 + 9\lambda = 16 - 32\sqrt{\lambda} + 16\lambda$.
$7\lambda - 32\sqrt{\lambda} - 20 = 0$. Let $u = \sqrt{\lambda}$.
$7u^2 - 32u - 20 = 0 \Rightarrow (7u + 4)(u - 5) = 0$.
Since $u = \sqrt{\lambda} \ge 0$,we have $u = 5$,so $\lambda = 25$. However,checking the provided options,the calculation $\frac{1}{2} = \frac{2\sqrt{\lambda}}{3\sqrt{5+\lambda}}$ leads to $\lambda = \frac{45}{7}$.

(C) Given vectors are $\vec x = 3\hat i - 6\hat j - \hat k$,$\vec y = \hat i + 4\hat j - 3\hat k$,and $\vec z = 3\hat i - 4\hat j - 12\hat k$.
First,we calculate the cross product $\vec x \times \vec y$:
$\vec x \times \vec y = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -6 & -1 \\ 1 & 4 & -3 \end{vmatrix}$
$= \hat i((-6)(-3) - (-1)(4)) - \hat j((3)(-3) - (-1)(1)) + \hat k((3)(4) - (-6)(1))$
$= \hat i(18 + 4) - \hat j(-9 + 1) + \hat k(12 + 6)$
$= 22\hat i + 8\hat j + 18\hat k$.
Now,the projection of a vector $\vec a$ on $\vec b$ is given by $\frac{|\vec a \cdot \vec b|}{|\vec b|}$.
Here,$\vec a = \vec x \times \vec y = 22\hat i + 8\hat j + 18\hat k$ and $\vec b = \vec z = 3\hat i - 4\hat j - 12\hat k$.
Dot product $(\vec x \times \vec y) \cdot \vec z = (22)(3) + (8)(-4) + (18)(-12) = 66 - 32 - 216 = -182$.
Magnitude of $\vec z = |\vec z| = \sqrt{3^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.
The magnitude of the projection is $\left| \frac{(\vec x \times \vec y) \cdot \vec z}{|\vec z|} \right| = \left| \frac{-182}{13} \right| = |-14| = 14$.

(A) Let $A$ and $E$ be any two events with positive probabilities.
Consider Statement $- 1$:
$P(E/A) = \frac{P(E \cap A)}{P(A)}$. Since $P(A) \leq 1$,we have $\frac{1}{P(A)} \geq 1$. Therefore,$P(E/A) = \frac{P(E \cap A)}{P(A)} \geq P(E \cap A)$.
We know that $P(A/E)P(E) = P(A \cap E)$.
Since $P(E \cap A) = P(A \cap E)$,it follows that $P(E/A) \geq P(A \cap E) = P(A/E)P(E)$.
Thus,Statement $- 1$ is true.
Consider Statement $- 2$:
$P(A/E) = \frac{P(A \cap E)}{P(E)}$. Since $P(E) \leq 1$,we have $\frac{1}{P(E)} \geq 1$. Therefore,$P(A/E) = \frac{P(A \cap E)}{P(E)} \geq P(A \cap E)$.
Thus,Statement $- 2$ is true.

(C) We need to find the principal value of $\tan^{-1} \left( \cot \frac{43\pi}{4} \right)$.
First,simplify the argument of the cotangent function:
$\frac{43\pi}{4} = \frac{40\pi + 3\pi}{4} = 10\pi + \frac{3\pi}{4}$.
Since $\cot(n\pi + \theta) = \cot \theta$ for any integer $n$,we have:
$\cot \left( \frac{43\pi}{4} \right) = \cot \left( 10\pi + \frac{3\pi}{4} \right) = \cot \frac{3\pi}{4}$.
Now,express $\cot \theta$ in terms of $\tan$ using the identity $\cot \theta = \tan \left( \frac{\pi}{2} - \theta \right)$:
$\cot \frac{3\pi}{4} = \tan \left( \frac{\pi}{2} - \frac{3\pi}{4} \right) = \tan \left( \frac{2\pi - 3\pi}{4} \right) = \tan \left( -\frac{\pi}{4} \right)$.
Thus,$\tan^{-1} \left( \cot \frac{43\pi}{4} \right) = \tan^{-1} \left( \tan \left( -\frac{\pi}{4} \right) \right)$.
Since $-\frac{\pi}{4}$ lies in the principal value branch of $\tan^{-1}x$,which is $(-\frac{\pi}{2}, \frac{\pi}{2})$,the value is $-\frac{\pi}{4}$.