JEE Main 2017 Mathematics Question Paper with Answer and Solution

(D) Let $\angle APC = \alpha$. In $\triangle APC$,$\tan \alpha = \frac{AC}{AP}$.
Since $C$ is the mid-point of $AB$,$AC = \frac{1}{2} AB$. Given $AP = 2AB$,we have $\tan \alpha = \frac{\frac{1}{2} AB}{2 AB} = \frac{1}{4}$.
Now,consider $\triangle ABP$. $\angle BAP = 90^{\circ}$. $\angle BPC = \beta$ and $\angle APC = \alpha$,so $\angle BAP = \alpha + \beta$.
$\tan(\alpha + \beta) = \frac{AB}{AP} = \frac{AB}{2 AB} = \frac{1}{2}$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$,we have:
$\frac{1}{2} = \frac{\frac{1}{4} + \tan \beta}{1 - \frac{1}{4} \tan \beta}$.
Multiplying both sides by $4(1 - \frac{1}{4} \tan \beta)$:
$2(1 - \frac{1}{4} \tan \beta) = 1 + 4 \tan \beta$
$2 - \frac{1}{2} \tan \beta = 1 + 4 \tan \beta$
$1 = 4.5 \tan \beta$
$1 = \frac{9}{2} \tan \beta$
$\tan \beta = \frac{2}{9}$.

(A) Given equation: $5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9$
Using $\cos 2x = 2\cos^2 x - 1$,we get:
$5\tan^2 x - 5\cos^2 x = 2(2\cos^2 x - 1) + 9$
$5\tan^2 x - 5\cos^2 x = 4\cos^2 x - 2 + 9$
$5\tan^2 x = 9\cos^2 x + 7$
Since $\tan^2 x = \sec^2 x - 1 = \frac{1}{\cos^2 x} - 1$,let $t = \cos^2 x$:
$5(\frac{1}{t} - 1) = 9t + 7$
$5 - 5t = 9t^2 + 7t$
$9t^2 + 12t - 5 = 0$
$(3t - 1)(3t + 5) = 0$
Since $t = \cos^2 x$ must be positive,$t = \frac{1}{3}$.
Now,$\cos 2x = 2\cos^2 x - 1 = 2(\frac{1}{3}) - 1 = -\frac{1}{3}$.
Finally,$\cos 4x = 2\cos^2 2x - 1 = 2(-\frac{1}{3})^2 - 1 = 2(\frac{1}{9}) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.

(B) $X$ has $4$ ladies and $3$ men. $Y$ has $3$ ladies and $4$ men.
We need to select $3$ ladies and $3$ men in total,such that $3$ friends are chosen from $X$'s group and $3$ friends from $Y$'s group.
Let $X$ choose $l_1$ ladies and $m_1$ men,and $Y$ choose $l_2$ ladies and $m_2$ men.
Constraints: $l_1 + m_1 = 3$,$l_2 + m_2 = 3$,$l_1 + l_2 = 3$,$m_1 + m_2 = 3$.
Possible cases $(l_1, m_1)$ for $X$ and $(l_2, m_2)$ for $Y$:
$1. (l_1, m_1) = (3, 0) \implies (l_2, m_2) = (0, 3)$. Ways: $\binom{4}{3}\binom{3}{0} \times \binom{3}{0}\binom{4}{3} = 4 \times 4 = 16$.
$2. (l_1, m_1) = (2, 1) \implies (l_2, m_2) = (1, 2)$. Ways: $\binom{4}{2}\binom{3}{1} \times \binom{3}{1}\binom{4}{2} = (6 \times 3) \times (3 \times 6) = 18 \times 18 = 324$.
$3. (l_1, m_1) = (1, 2) \implies (l_2, m_2) = (2, 1)$. Ways: $\binom{4}{1}\binom{3}{2} \times \binom{3}{2}\binom{4}{1} = (4 \times 3) \times (3 \times 4) = 12 \times 12 = 144$.
$4. (l_1, m_1) = (0, 3) \implies (l_2, m_2) = (3, 0)$. Ways: $\binom{4}{0}\binom{3}{3} \times \binom{3}{3}\binom{4}{0} = 1 \times 1 = 1$.
Total ways = $16 + 324 + 144 + 1 = 485$.

(A) The given expression is $S = \sum_{r=1}^{10} \binom{21}{r} - \sum_{r=1}^{10} \binom{10}{r}$.
We know that $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$.
Since $\binom{21}{r} = \binom{21}{21-r}$,we have $\sum_{r=0}^{10} \binom{21}{r} = \sum_{r=11}^{21} \binom{21}{r}$.
Thus,$2 \sum_{r=1}^{10} \binom{21}{r} + \binom{21}{0} + \binom{21}{11} = 2^{21}$ is not quite right; rather,$2 \sum_{r=1}^{10} \binom{21}{r} + \binom{21}{0} + \binom{21}{21} = 2^{21}$ is incorrect.
Correctly,$\sum_{r=0}^{21} \binom{21}{r} = 2^{21} \implies 2 \sum_{r=0}^{10} \binom{21}{r} = 2^{21} \implies \sum_{r=0}^{10} \binom{21}{r} = 2^{20}$.
Therefore,$\sum_{r=1}^{10} \binom{21}{r} = 2^{20} - \binom{21}{0} = 2^{20} - 1$.
Next,$\sum_{r=1}^{10} \binom{10}{r} = (1+1)^{10} - \binom{10}{0} = 2^{10} - 1$.
Substituting these into the original expression:
$S = (2^{20} - 1) - (2^{10} - 1) = 2^{20} - 2^{10}$.

(C) Given equation: $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$
Expanding the terms: $225a^2 + 9b^2 + 25c^2 - 75ac = 45ab + 15bc$
Rearranging: $225a^2 + 9b^2 + 25c^2 - 45ab - 15bc - 75ac = 0$
Multiply by $2$: $450a^2 + 18b^2 + 50c^2 - 90ab - 30bc - 150ac = 0$
This can be rewritten as: $(15a - 3b)^2 + (3b - 5c)^2 + (5c - 15a)^2 = 0$
For the sum of squares to be zero,each term must be zero:
$15a - 3b = 0$ $\Rightarrow 3b = 15a$ $\Rightarrow b = 5a$
$3b - 5c = 0 \Rightarrow 3b = 5c$
$5c - 15a = 0$ $\Rightarrow 5c = 15a$ $\Rightarrow c = 3a$
Now check the sequence $b, c, a$:
$b = 5a, c = 3a, a = a$
Common difference $d_1 = c - b = 3a - 5a = -2a$
Common difference $d_2 = a - c = a - 3a = -2a$
Since $d_1 = d_2$,the terms $b, c, a$ are in $A.P.$

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 28$.
Substituting the vertices $(k, -3k), (5, k), (-k, 2)$:
$\frac{1}{2} |k(k - 2) + 5(2 - (-3k)) + (-k)(-3k - k)| = 28$
$\frac{1}{2} |k^2 - 2k + 10 + 15k + 4k^2| = 28$
$|5k^2 + 13k + 10| = 56$
Case $1$: $5k^2 + 13k + 10 = 56$ $\Rightarrow 5k^2 + 13k - 46 = 0$ $\Rightarrow (k - 2)(5k + 23) = 0$. Since $k$ is an integer,$k = 2$.
Case $2$: $5k^2 + 13k + 10 = -56 \Rightarrow 5k^2 + 13k + 66 = 0$. The discriminant $D = 13^2 - 4(5)(66) = 169 - 1320 < 0$,so no real solutions.
For $k = 2$,the vertices are $A(2, -6), B(5, 2), C(-2, 2)$.
The side $BC$ is horizontal $(y = 2)$,so the altitude from $A$ is the vertical line $x = 2$.
The slope of $AC$ is $m_{AC} = \frac{2 - (-6)}{-2 - 2} = \frac{8}{-4} = -2$. The altitude from $B$ to $AC$ has slope $\frac{1}{2}$ and passes through $B(5, 2)$.
Equation of altitude from $B$: $y - 2 = \frac{1}{2}(x - 5)$ $\Rightarrow 2y - 4 = x - 5$ $\Rightarrow x - 2y = 1$.
Since the orthocentre $H$ lies on $x = 2$,substituting $x = 2$ into $x - 2y = 1$ gives $2 - 2y = 1$ $\Rightarrow 2y = 1$ $\Rightarrow y = \frac{1}{2}$.
Thus,the orthocentre is $\left( 2, \frac{1}{2} \right)$.

(C) The equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Since the foci are $(\pm 2, 0)$,we have $ae = 2$,so $a^{2}e^{2} = 4$.
Using the relation $b^{2} = a^{2}(e^{2} - 1) = a^{2}e^{2} - a^{2}$,we get $b^{2} = 4 - a^{2}$,which implies $a^{2} + b^{2} = 4$.
Since the hyperbola passes through $P(\sqrt{2}, \sqrt{3})$,we have $\frac{2}{a^{2}} - \frac{3}{b^{2}} = 1$.
Substituting $a^{2} = 4 - b^{2}$,we get $\frac{2}{4 - b^{2}} - \frac{3}{b^{2}} = 1$.
$2b^{2} - 3(4 - b^{2}) = b^{2}(4 - b^{2}) \Rightarrow 2b^{2} - 12 + 3b^{2} = 4b^{2} - b^{4}$.
$b^{4} + b^{2} - 12 = 0 \Rightarrow (b^{2} + 4)(b^{2} - 3) = 0$.
Since $b^{2} > 0$,we have $b^{2} = 3$,which gives $a^{2} = 4 - 3 = 1$.
The equation of the hyperbola is $x^{2} - \frac{y^{2}}{3} = 1$.
The tangent at $P(\sqrt{2}, \sqrt{3})$ is $\frac{\sqrt{2}x}{1} - \frac{\sqrt{3}y}{3} = 1$,which simplifies to $\sqrt{2}x - \frac{y}{\sqrt{3}} = 1$.
Checking option $C$: $\sqrt{2}(2\sqrt{2}) - \frac{3\sqrt{3}}{\sqrt{3}} = 4 - 3 = 1$. Thus,it passes through $(2\sqrt{2}, 3\sqrt{3})$.

(D) Let the circle be $x^2 + (y - k)^2 = r^2$. Since it touches $y = |x|$,the distance from $(0, k)$ to $x - y = 0$ is $r$,so $\frac{|-k|}{\sqrt{2}} = r$,which gives $k = r\sqrt{2}$.
Substituting $x^2 = 4 - y$ into the circle equation: $(4 - y) + (y - k)^2 = r^2$.
$4 - y + y^2 - 2ky + k^2 = \frac{k^2}{2}$.
$y^2 - (2k + 1)y + (4 + \frac{k^2}{2}) = 0$.
For tangency,the discriminant $D = 0$:
$(2k + 1)^2 - 4(4 + \frac{k^2}{2}) = 0$.
$4k^2 + 4k + 1 - 16 - 2k^2 = 0$.
$2k^2 + 4k - 15 = 0$.
Solving for $k$ (taking $k > 0$): $k = \frac{-4 + \sqrt{16 - 4(2)(-15)}}{4} = \frac{-4 + \sqrt{136}}{4} = \frac{-4 + 2\sqrt{34}}{4} = \frac{-2 + \sqrt{34}}{2}$.
Since $r = \frac{k}{\sqrt{2}}$,$r = \frac{-2 + \sqrt{34}}{2\sqrt{2}}$.

(C) Given eccentricity $e = \frac{1}{2}$.
The equation of the directrix is $x = -\frac{a}{e} = -4$,so $\frac{a}{1/2} = 4$,which gives $a = 2$.
We know $b^2 = a^2(1 - e^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.
The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Differentiating with respect to $x$,we get $\frac{2x}{4} + \frac{2y}{3} \frac{dy}{dx} = 0$,which simplifies to $\frac{dy}{dx} = -\frac{3x}{4y}$.
At the point $\left(1, \frac{3}{2}\right)$,the slope of the tangent is $m_t = -\frac{3(1)}{4(3/2)} = -\frac{3}{6} = -\frac{1}{2}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = 2$.
The equation of the normal at $\left(1, \frac{3}{2}\right)$ is $y - \frac{3}{2} = 2(x - 1)$.
Multiplying by $2$,we get $2y - 3 = 4x - 4$,which simplifies to $4x - 2y = 1$.

(C) Let $L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}}$.
We can rewrite the expression as:
$L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cos x \left( {\frac{1}{{\sin x}} - 1} \right)}}{{ - 8{{\left( {x - \frac{\pi }{2}} \right)}^3}}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cos x(1 - \sin x)}}{{\sin x \cdot 8{{\left( {\frac{\pi }{2} - x} \right)}^3}}}$.
Substitute $t = \frac{\pi }{2} - x$. As $x \to \frac{\pi }{2}$,$t \to 0$. Then $x = \frac{\pi }{2} - t$ and $\cos x = \sin t$,$\sin x = \cos t$.
$L = \mathop {\lim }\limits_{t \to 0} \frac{{\sin t(1 - \cos t)}}{{8{t^3}\cos t}}$.
$L = \frac{1}{8} \cdot \mathop {\lim }\limits_{t \to 0} \left( \frac{{\sin t}}{t} \right) \cdot \left( \frac{{1 - \cos t}}{{{t^2}}} \right) \cdot \frac{1}{{\cos t}}$.
Using standard limits $\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} = 1$ and $\mathop {\lim }\limits_{t \to 0} \frac{{1 - \cos t}}{{{t^2}}} = \frac{1}{2}$:
$L = \frac{1}{8} \cdot 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{{16}}$.

(A) The given equation is $\sum_{r=1}^{n} (x + r - 1)(x + r) = 10n$.
Expanding the terms: $\sum_{r=1}^{n} (x^2 + (2r - 1)x + r^2 - r) = 10n$.
This simplifies to $nx^2 + x \sum_{r=1}^{n} (2r - 1) + \sum_{r=1}^{n} (r^2 - r) = 10n$.
Using summation formulas: $nx^2 + n^2x + \frac{(n-1)n(n+1)}{3} = 10n$.
Dividing by $n$ (since $n$ is a positive integer): $x^2 + nx + \frac{n^2 - 1}{3} = 10$.
$x^2 + nx + \frac{n^2 - 31}{3} = 0$.
Let the two consecutive integral solutions be $\alpha$ and $\alpha + 1$.
Sum of roots: $2\alpha + 1 = -n \Rightarrow \alpha = \frac{-(n+1)}{2}$.
Product of roots: $\alpha(\alpha + 1) = \frac{n^2 - 31}{3}$.
Substituting $\alpha$: $(\frac{-(n+1)}{2})(\frac{-(n+1)}{2} + 1) = \frac{n^2 - 31}{3}$.
$(\frac{-(n+1)}{2})(\frac{1-n}{2}) = \frac{n^2 - 31}{3}$.
$\frac{n^2 - 1}{4} = \frac{n^2 - 31}{3}$.
$3n^2 - 3 = 4n^2 - 124$.
$n^2 = 121 \Rightarrow n = 11$.

(B) Let the set be $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Total number of ways to choose two distinct numbers is $n(S) = \binom{11}{2} = \frac{11 \times 10}{2} = 55$.
Let the two numbers be $x$ and $y$ where $x > y$.
We require $(x+y)$ to be a multiple of $4$ and $(x-y)$ to be a multiple of $4$.
If $(x+y)$ and $(x-y)$ are both multiples of $4$,then their sum $(x+y) + (x-y) = 2x$ must be a multiple of $4$,implying $x$ is an even number.
Similarly,their difference $(x+y) - (x-y) = 2y$ must be a multiple of $4$,implying $y$ is an even number.
Thus,both $x$ and $y$ must be even numbers from the set $\{0, 2, 4, 6, 8, 10\}$.
Let the chosen numbers be $x, y \in \{0, 2, 4, 6, 8, 10\}$.
For $(x+y)$ and $(x-y)$ to be multiples of $4$,$x+y \equiv 0 \pmod{4}$ and $x-y \equiv 0 \pmod{4}$.
This implies $x \equiv y \pmod{4}$.
Possible pairs $(x, y)$ with $x > y$ such that $x \equiv y \pmod{4}$:
If $x, y \equiv 0 \pmod{4}$: $(4, 0), (8, 0), (8, 4)$.
If $x, y \equiv 2 \pmod{4}$: $(6, 2), (10, 2), (10, 6)$.
Total favorable outcomes $n(E) = 3 + 3 = 6$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{55}$.

(B) To determine the nature of the statement $(p \to q) \to [(\sim p \to q) \to q]$,we construct a truth table:

$p$	$q$	$\sim p$	$p \to q$	$\sim p \to q$	$(\sim p \to q) \to q$	$(p \to q) \to [(\sim p \to q) \to q]$
$T$	$T$	$F$	$T$	$T$	$T$	$T$
$T$	$F$	$F$	$F$	$T$	$F$	$T$
$F$	$T$	$T$	$T$	$T$	$T$	$T$
$F$	$F$	$T$	$T$	$F$	$T$	$T$

Since the final column contains only $T$ (True) values for all possible truth values of $p$ and $q$,the statement is a tautology.

(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.
Given $p(0) = 1$,we have $c = 1$.
By the Remainder Theorem,$p(1) = 4$ and $p(-1) = 6$.
Substituting these into $p(x) = ax^2 + bx + 1$:
$p(1) = a(1)^2 + b(1) + 1 = 4 \Rightarrow a + b = 3$.
$p(-1) = a(-1)^2 + b(-1) + 1 = 6 \Rightarrow a - b = 5$.
Adding the two equations: $2a = 8 \Rightarrow a = 4$.
Subtracting the two equations: $2b = -2 \Rightarrow b = -1$.
Thus,$p(x) = 4x^2 - x + 1$.
Now,evaluating $p(-2)$:
$p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 3 = 19$.
Therefore,$p(-2) = 19$ is the correct statement.

(A) Let $z = x + iy$. The equation is $2|x + i(y + 3)| = |x + i(y - 1)|$.
Squaring both sides,we get $4(x^2 + (y + 3)^2) = x^2 + (y - 1)^2$.
$4x^2 + 4(y^2 + 6y + 9) = x^2 + y^2 - 2y + 1$.
$3x^2 + 3y^2 + 26y + 35 = 0$.
Dividing by $3$,we get $x^2 + y^2 + \frac{26}{3}y + \frac{35}{3} = 0$.
This is the equation of a circle $x^2 + y^2 + 2gy + c = 0$ where $g = \frac{13}{3}$.
The radius $r = \sqrt{g^2 - c} = \sqrt{(\frac{13}{3})^2 - \frac{35}{3}} = \sqrt{\frac{169}{9} - \frac{105}{9}} = \sqrt{\frac{64}{9}} = \frac{8}{3}$.

(C) The letters in the word $QUEEN$ are $E, E, N, Q, U$. Arranging them in alphabetical order: $E, E, N, Q, U$.
$(i)$ Words starting with $E$: The remaining letters are $E, N, Q, U$ (all distinct). Number of words $= 4! = 24$.
$(ii)$ Words starting with $N$: The remaining letters are $E, E, Q, U$. Number of words $= \frac{4!}{2!} = 12$.
$(iii)$ Words starting with $QE$: The remaining letters are $E, N, U$. Number of words $= 3! = 6$.
$(iv)$ Words starting with $QN$: The remaining letters are $E, E, U$. Number of words $= \frac{3!}{2!} = 3$.
$(v)$ The next word is $QUEEN$ itself,which is the $1^{st}$ word after the above arrangements.
Therefore,the rank of the word $QUEEN = 24 + 12 + 6 + 3 + 1 = 46^{th}$.

(D) We need to find the remainder when $(27)^{999}$ is divided by $7$.
We can write $27$ as $(28 - 1)$.
So,$(27)^{999} = (28 - 1)^{999}$.
Using the Binomial Theorem,$(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.
$(28 - 1)^{999} = \binom{999}{0} (28)^{999} (-1)^0 + \binom{999}{1} (28)^{998} (-1)^1 + \dots + \binom{999}{999} (28)^0 (-1)^{999}$.
Every term except the last one contains a factor of $28$,which is divisible by $7$.
$(27)^{999} = 7k + (-1)^{999}$,where $k$ is an integer.
$(27)^{999} = 7k - 1$.
To find the positive remainder,we write $7k - 1 = 7(k - 1) + 7 - 1 = 7(k - 1) + 6$.
Thus,the remainder is $6$.

(D) Given that the arithmetic mean $(A.M.)$ is five times the geometric mean $(G.M.)$:
$\frac{a + b}{2} = 5\sqrt{ab}$
Divide both sides by $\sqrt{ab}$:
$\frac{a + b}{\sqrt{ab}} = 10$
Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a}{b} = x^2$. The equation becomes:
$\frac{x^2 + 1}{x} = 10 \implies x^2 - 10x + 1 = 0$
Solving for $x$ using the quadratic formula:
$x = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm \sqrt{24} = 5 \pm 2\sqrt{6}$
Since $a > b$,$x > 1$,so $x = 5 + 2\sqrt{6}$.
Then $\frac{a}{b} = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6}$.
Using Componendo and Dividendo on $\frac{a}{b} = \frac{49 + 20\sqrt{6}}{1}$:
$\frac{a + b}{a - b} = \frac{49 + 20\sqrt{6} + 1}{49 + 20\sqrt{6} - 1} = \frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}} = \frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}}$
Rationalizing the denominator:
$\frac{(25 + 10\sqrt{6})(24 - 10\sqrt{6})}{24^2 - (10\sqrt{6})^2} = \frac{600 - 250\sqrt{6} + 240\sqrt{6} - 600}{576 - 600} = \frac{-10\sqrt{6}}{-24} = \frac{5\sqrt{6}}{12}$

(B) The given series is $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$
This can be written as $\sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + 13\sqrt{3} + \dots$
This is an Arithmetic Progression with first term $a = \sqrt{3}$ and common difference $d = 4\sqrt{3}$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given $S_n = 435\sqrt{3}$,we have:
$\frac{n}{2}[2\sqrt{3} + (n-1)4\sqrt{3}] = 435\sqrt{3}$
Divide both sides by $\sqrt{3}$:
$\frac{n}{2}[2 + 4n - 4] = 435$
$n(2n - 2) = 870$
$2n^2 - 2n - 870 = 0$
$n^2 - n - 435 = 0$
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{1 \pm \sqrt{1 - 4(1)(-435)}}{2} = \frac{1 \pm \sqrt{1 + 1740}}{2} = \frac{1 \pm \sqrt{1741}}{2}$
Wait,checking the series calculation: $1, 5, 9, 13$ are terms of an $AP$ with $a=1, d=4$. Sum is $\frac{n}{2}[2(1) + (n-1)4] = \frac{n}{2}[4n - 2] = n(2n-1) = 435$.
$2n^2 - n - 435 = 0$.
$n = \frac{1 \pm \sqrt{1 + 4(2)(435)}}{4} = \frac{1 \pm \sqrt{1 + 3480}}{4} = \frac{1 \pm \sqrt{3481}}{4} = \frac{1 \pm 59}{4}$.
Since $n$ must be positive,$n = \frac{60}{4} = 15$.

(B) Let $A = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x} - 3}}{{\sqrt {2x - 4} - \sqrt 2 }}$.
Rationalizing the numerator and the denominator:
$A = \mathop {\lim }\limits_{x \to 3} \frac{(\sqrt{3x} - 3)(\sqrt{3x} + 3)(\sqrt{2x - 4} + \sqrt{2})}{(\sqrt{2x - 4} - \sqrt{2})(\sqrt{2x - 4} + \sqrt{2})(\sqrt{3x} + 3)}$
$A = \mathop {\lim }\limits_{x \to 3} \frac{(3x - 9)(\sqrt{2x - 4} + \sqrt{2})}{(2x - 4 - 2)(\sqrt{3x} + 3)}$
$A = \mathop {\lim }\limits_{x \to 3} \frac{3(x - 3)(\sqrt{2x - 4} + \sqrt{2})}{2(x - 3)(\sqrt{3x} + 3)}$
Canceling $(x - 3)$:
$A = \frac{3}{2} \times \frac{\sqrt{2(3) - 4} + \sqrt{2}}{\sqrt{3(3)} + 3}$
$A = \frac{3}{2} \times \frac{\sqrt{2} + \sqrt{2}}{3 + 3} = \frac{3}{2} \times \frac{2\sqrt{2}}{6} = \frac{3}{2} \times \frac{\sqrt{2}}{3} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

(B) The equation of the circle is $x^2 + y^2 - 5x - y + 5 = 0$.
Completing the square,we get $(x - 5/2)^2 + (y - 1/2)^2 = 25/4 + 1/4 - 5 = 6/4 = 3/2$.
The center of the circle is $C = (5/2, 1/2)$ and the radius is $r = \sqrt{3/2}$.
The distance $PQ$ is maximized when $Q$ lies on the line passing through $P$ and $C$,specifically at the point on the circle furthest from $P$.
The distance $PC$ is $\sqrt{(5/2 - 0)^2 + (1/2 - (-2))^2} = \sqrt{25/4 + 25/4} = \sqrt{50/4} = \frac{5\sqrt{2}}{2}$.
The maximum distance $PQ$ is $PC + r = \frac{5\sqrt{2}}{2} + \sqrt{\frac{3}{2}} = \frac{5\sqrt{2} + \sqrt{6}}{2}$.
Therefore,the maximum value of $(PQ)^2$ is $\left( \frac{5\sqrt{2} + \sqrt{6}}{2} \right)^2 = \frac{50 + 6 + 10\sqrt{12}}{4} = \frac{56 + 20\sqrt{3}}{4} = 14 + 5\sqrt{3}$.

(B) The diameter of the circle is $4 \, \text{units}$,so the radius $r = 2 \, \text{units}$.
Let the angles subtended by the chords at the centre be $2\theta$ and $2\phi$.
Given $2\theta = \cos^{-1}(1/7) \Rightarrow \cos(2\theta) = 1/7$.
Using the formula $\cos(2\theta) = 2\cos^2\theta - 1$,we have $2\cos^2\theta - 1 = 1/7$ $\Rightarrow 2\cos^2\theta = 8/7$ $\Rightarrow \cos^2\theta = 4/7$ $\Rightarrow \cos\theta = 2/\sqrt{7}$.
The distance of the first chord from the centre is $d_1 = r \cos\theta = 2 \times (2/\sqrt{7}) = 4/\sqrt{7}$.
Given $2\phi = \sec^{-1}(7)$ $\Rightarrow \sec(2\phi) = 7$ $\Rightarrow \cos(2\phi) = 1/7$.
Using the formula $\cos(2\phi) = 2\cos^2\phi - 1$,we have $2\cos^2\phi - 1 = 1/7$ $\Rightarrow 2\cos^2\phi = 8/7$ $\Rightarrow \cos^2\phi = 4/7$ $\Rightarrow \cos\phi = 2/\sqrt{7}$.
The distance of the second chord from the centre is $d_2 = r \cos\phi = 2 \times (2/\sqrt{7}) = 4/\sqrt{7}$.
Since the chords lie on opposite sides of the centre,the total distance between them is $d_1 + d_2 = 4/\sqrt{7} + 4/\sqrt{7} = 8/\sqrt{7}$.

(D) The equation of a tangent to the circle $x^2 + y^2 = 4$ with slope $m$ is $y = mx \pm 2\sqrt{1 + m^2}$.
Since this line is also a tangent to the parabola $x^2 = 4y$,we substitute $y = mx + c$ into the parabola equation: $x^2 = 4(mx + c) \Rightarrow x^2 - 4mx - 4c = 0$.
For tangency,the discriminant $D = 0$,so $(-4m)^2 - 4(1)(-4c) = 0$ $\Rightarrow 16m^2 + 16c = 0$ $\Rightarrow c = -m^2$.
Comparing this with the circle's tangent form $c = \pm 2\sqrt{1 + m^2}$,we get $-m^2 = \pm 2\sqrt{1 + m^2}$.
Squaring both sides: $m^4 = 4(1 + m^2) \Rightarrow m^4 - 4m^2 - 4 = 0$.
Using the quadratic formula for $m^2$: $m^2 = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.
Since $m^2$ must be positive,$m^2 = 2 + 2\sqrt{2}$.

(D) Given eccentricity $e = \frac{3}{5}$ and distance between foci $2ae = 6$.
$2a(\frac{3}{5}) = 6 \Rightarrow a = 5$.
Using the relation $b^2 = a^2(1 - e^2)$:
$b^2 = 25(1 - \frac{9}{25}) = 25(\frac{16}{25}) = 16 \Rightarrow b = 4$.
The vertices of the ellipse are $(\pm a, 0)$ and $(0, \pm b)$,which are $(\pm 5, 0)$ and $(0, \pm 4)$.
The quadrilateral formed by these vertices is a rhombus with diagonals of length $2a = 10$ and $2b = 8$.
Area of the quadrilateral $= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 8 = 40$ sq. units.

(C) Let the sum of the ages of the $25$ teachers be $S$.
Given that the mean age is $40 \text{ years}$,we have:
$\frac{S}{25} = 40 \Rightarrow S = 1000$.
Let the age of the new teacher be $A$.
After the retirement of a $60 \text{ year}$ old teacher and the appointment of the new teacher,the new sum of ages is $S - 60 + A$.
The new mean age is $39 \text{ years}$,so:
$\frac{S - 60 + A}{25} = 39$
$1000 - 60 + A = 39 \times 25$
$940 + A = 975$
$A = 975 - 940 = 35$.
Thus,the age of the newly appointed teacher is $35 \text{ years}$.

(A) Let $P(P), P(Q), P(R)$ be the probabilities of hitting the target by $P, Q, R$ respectively.
$P(P) = \frac{3}{4}, P(Q) = \frac{1}{2}, P(R) = \frac{5}{8}$.
The probabilities of missing the target are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$,$P(Q') = 1 - \frac{1}{2} = \frac{1}{2}$,and $P(R') = 1 - \frac{5}{8} = \frac{3}{8}$.
The event that the target is hit by $P$ or $Q$ but not by $R$ is $(P \cap Q' \cap R') \cup (P' \cap Q \cap R') \cup (P \cap Q \cap R')$.
Required probability $= P(P)P(Q')P(R') + P(P')P(Q)P(R') + P(P)P(Q)P(R')$.
$= (\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}) + (\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8}) + (\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8})$.
$= \frac{9}{64} + \frac{3}{64} + \frac{9}{64} = \frac{21}{64}$.

(C) Given the equation $2^{(x - 1)(x^2 + 5x - 50)} = 1$.
Since $2^0 = 1$,we equate the exponent to $0$:
$(x - 1)(x^2 + 5x - 50) = 0$.
Factoring the quadratic expression $x^2 + 5x - 50$:
$(x^2 + 10x - 5x - 50) = x(x + 10) - 5(x + 10) = (x - 5)(x + 10)$.
So,the equation becomes $(x - 1)(x - 5)(x + 10) = 0$.
The real values of $x$ are $x = 1, 5, -10$.
The sum of these values is $1 + 5 + (-10) = 6 - 10 = -4$.

(C) Let $z = x + iy$.
Given $\text{Im}\left( \frac{i(x+iy) - 2}{x+iy - i} \right) + 1 = 0$.
$\text{Im}\left( \frac{ix - y - 2}{x + i(y-1)} \right) + 1 = 0$.
Multiply numerator and denominator by the conjugate $x - i(y-1)$:
$\text{Im}\left( \frac{(ix - y - 2)(x - i(y-1))}{x^2 + (y-1)^2} \right) + 1 = 0$.
The imaginary part is $\frac{x^2 - (y+2)(y-1)}{x^2 + (y-1)^2} = -1$.
$x^2 - (y^2 + y - 2) = -(x^2 + y^2 - 2y + 1)$.
$x^2 - y^2 - y + 2 = -x^2 - y^2 + 2y - 1$.
$2x^2 - 3y + 3 = 0$.
Wait,re-evaluating the expression: $\text{Im}\left( \frac{ix - y - 2}{x + i(y-1)} \right) = \frac{x^2 - (y+2)(y-1)}{x^2 + (y-1)^2} = \frac{x^2 - y^2 - y + 2}{x^2 + (y-1)^2}$.
Given $\frac{x^2 - y^2 - y + 2}{x^2 + (y-1)^2} = -1$.
$x^2 - y^2 - y + 2 = -x^2 - y^2 + 2y - 1$.
$2x^2 - 3y + 3 = 0$. This suggests a parabola. Let's re-check the original equation: $\text{Im}\left( \frac{iz - 2}{z - i} \right) = -1$.
$\frac{iz - 2}{z - i} = \frac{i(z - 2/i)}{z - i} = \frac{i(z + 2i)}{z - i}$.
Let $w = \frac{z + 2i}{z - i}$. Then $\text{Im}(iw) = -1 \implies \text{Re}(w) = 1$.
$\text{Re}\left( \frac{x + i(y+2)}{x + i(y-1)} \right) = 1$.
$\frac{x^2 + (y+2)(y-1)}{x^2 + (y-1)^2} = 1$.
$x^2 + y^2 + y - 2 = x^2 + y^2 - 2y + 1$.
$3y = 3 \implies y = 1$. This is a line.
Re-reading the problem: If the equation is $\text{Im}\left( \frac{iz - 2}{z - i} \right) = -1$,it represents a line. If the question implies a circle,there might be a typo in the provided equation. Based on the provided solution steps,the radius is $\frac{3}{4}$.

(A) Total number of people = $5 + 3 = 8$.
Total ways to seat $8$ people on a round table = $(8-1)! = 7!$.
Now,consider the case where $B_1$ and $G_1$ sit together. Treat $(B_1, G_1)$ as one unit.
Now we have $7$ units to arrange in a circle,which can be done in $(7-1)! = 6!$ ways.
Within the unit,$B_1$ and $G_1$ can be arranged in $2! = 2$ ways.
So,the number of ways they sit together = $2 \times 6!$.
The number of ways they never sit adjacent = Total ways - Ways they sit together.
$= 7! - 2 \times 6! = 7 \times 6! - 2 \times 6! = (7-2) \times 6! = 5 \times 6!$.

(A) Simplify the expression inside the bracket:
Let $u = x^{1/3}$ and $v = x^{1/2}$.
The first term is $\frac{u^3 + 1}{u^2 - u + 1} = \frac{(u+1)(u^2 - u + 1)}{u^2 - u + 1} = u + 1 = x^{1/3} + 1$.
The second term is $\frac{v^2 - 1}{v(v - 1)} = \frac{(v-1)(v+1)}{v(v-1)} = \frac{v+1}{v} = 1 + \frac{1}{v} = 1 + x^{-1/2}$.
Subtracting the two terms: $(x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$.
Now,we need the coefficient of $x^{-5}$ in $(x^{1/3} - x^{-1/2})^{10}$.
The general term $T_{r+1} = {^{10}C_r} (x^{1/3})^{10-r} (-x^{-1/2})^r = {^{10}C_r} (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.
Set the exponent equal to $-5$: $\frac{10-r}{3} - \frac{r}{2} = -5$.
Multiply by $6$: $2(10-r) - 3r = -30 \implies 20 - 2r - 3r = -30 \implies 50 = 5r \implies r = 10$.
The coefficient is ${^{10}C_{10}} (-1)^{10} = 1 \times 1 = 1$.

(A) Since $a, b, c$ are in $A.P.$,we can write $a = b - d$ and $c = b + d$ for some common difference $d$.
Given $abc = 8$,we have $(b - d)b(b + d) = 8$.
$b(b^2 - d^2) = 8$,which implies $b^2 - d^2 = \frac{8}{b}$.
Since $d^2 \ge 0$,we have $b^2 - \frac{8}{b} \ge 0$.
$b^3 - 8 \ge 0$,so $b^3 \ge 8$,which means $b \ge 2$.
The minimum value of $b$ is $2$ when $d = 0$ (i.e.,$a = b = c = 2$).

(A) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^n k}{\sum_{k=1}^n k^3}$.
Using the standard formulas $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$,we get:
$T_n = \frac{\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}\right)^2} = \frac{2}{n(n+1)}$.
We can write $T_n$ using partial fractions as $T_n = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$.
Now,$S_n = \sum_{k=1}^n T_k = 2 \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}\right)$.
This is a telescoping sum: $S_n = 2 \left( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) \right) = 2 \left( 1 - \frac{1}{n+1} \right) = \frac{2n}{n+1}$.
Given $100 S_n = n$,we substitute $S_n$:
$100 \left( \frac{2n}{n+1} \right) = n$.
Since $n \neq 0$,we can divide by $n$:
$\frac{200}{n+1} = 1 \implies n+1 = 200 \implies n = 199$.

(B) Let the vertices of the square be $O(0, 0)$,$A$,$B$,and $C$. The side $OA$ makes an angle of $30^o$ with the positive $x-$axis. Since the side length is $2$,the coordinates of $A$ are $(2 \cos 30^o, 2 \sin 30^o) = (2 \cdot \frac{\sqrt{3}}{2}, 2 \cdot \frac{1}{2}) = (\sqrt{3}, 1)$.
The side $OC$ is perpendicular to $OA$ and lies at an angle of $30^o + 90^o = 120^o$ with the positive $x-$axis. The coordinates of $C$ are $(2 \cos 120^o, 2 \sin 120^o) = (2 \cdot -\frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2}) = (-1, \sqrt{3})$.
The vertex $B$ is the sum of vectors $\vec{OA}$ and $\vec{OC}$,so $B = (\sqrt{3} - 1, 1 + \sqrt{3})$.
The $x-$coordinates of the vertices are $0$,$\sqrt{3}$,$-1$,and $\sqrt{3} - 1$.
The sum of the $x-$coordinates is $0 + \sqrt{3} - 1 + \sqrt{3} - 1 = 2\sqrt{3} - 2$.

(B) The power of a point $P(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $PA \cdot PB = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
For the given circle $x^2 + y^2 - 9 = 0$ and point $P(4, 7)$:
$PA \cdot PB = (4)^2 + (7)^2 - 9$
$PA \cdot PB = 16 + 49 - 9$
$PA \cdot PB = 65 - 9 = 56$.

(C) The equation of an ellipse with centre at $(0, 0)$ and axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through $(4, -1)$,we have $\frac{16}{a^2} + \frac{1}{b^2} = 1$,which implies $16b^2 + a^2 = a^2b^2$ $(i)$.
Since the ellipse passes through $(-2, 2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which implies $4b^2 + 4a^2 = a^2b^2$ $(ii)$.
Equating $(i)$ and $(ii)$,we get $16b^2 + a^2 = 4b^2 + 4a^2$.
$12b^2 = 3a^2$,so $a^2 = 4b^2$.
Substituting $a^2 = 4b^2$ into $(ii)$,we get $4b^2 + 4(4b^2) = (4b^2)b^2$,which simplifies to $20b^2 = 4b^4$,so $b^2 = 5$.
Then $a^2 = 4(5) = 20$.
Since $b^2 = a^2(1 - e^2)$,we have $5 = 20(1 - e^2)$.
$1 - e^2 = \frac{5}{20} = \frac{1}{4}$.
$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,$e = \frac{\sqrt{3}}{2}$.

(C) For the parabola $y^2 = 4ax$, we have $4a = 8$, so $a = 2$.
The focal distance of a point $(at^2, 2at)$ on the parabola is given by $a(1 + t^2)$.
Given the focal distance is $8$, we have $2(1 + t^2) = 8$, which implies $1 + t^2 = 4$, so $t^2 = 3$ and $t = \pm\sqrt{3}$.
The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t$ is $y = -tx + 2at + at^3$.
Comparing this with $y = mx + c$, we get $m = -t$ and $c = 2at + at^3 = at(2 + t^2)$.
Substituting $a = 2$ and $t = \sqrt{3}$ (or $t = -\sqrt{3}$ for the magnitude):
$|c| = |2(\sqrt{3})(2 + 3)| = |2\sqrt{3}(5)| = 10\sqrt{3}$.

(C) Total ways to choose a $4$-member committee from $15$ people is $^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
Let $S$ be the set of committees with at least one woman.
The number of committees with no women is $^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
So,the number of committees with at least one woman is $n(S) = 1365 - 210 = 1155$.
We want the probability that the committee has more women than men. This happens if the committee has $3$ women and $1$ man,or $4$ women and $0$ men.
Number of ways for $3$ women and $1$ man: $^{5}C_3 \times ^{10}C_1 = 10 \times 10 = 100$.
Number of ways for $4$ women and $0$ men: $^{5}C_4 \times ^{10}C_0 = 5 \times 1 = 5$.
Total favorable ways = $100 + 5 = 105$.
The required probability is $\frac{105}{1155} = \frac{1}{11}$.

(D) Given: $N = 100$,$\sum x_i = 400$,$\sum x_i^2 = 2475$.
Removing incorrect observations $3, 4, 5$:
New sum $\sum x_i' = 400 - (3 + 4 + 5) = 400 - 12 = 388$.
New sum of squares $\sum (x_i')^2 = 2475 - (3^2 + 4^2 + 5^2) = 2475 - (9 + 16 + 25) = 2475 - 50 = 2425$.
New number of observations $N' = 100 - 3 = 97$.
Variance $\sigma^2 = \frac{\sum (x_i')^2}{N'} - \left( \frac{\sum x_i'}{N'} \right)^2$.
$\sigma^2 = \frac{2425}{97} - \left( \frac{388}{97} \right)^2$.
Since $388 = 4 \times 97$,$\frac{388}{97} = 4$.
$\sigma^2 = 25 - (4)^2 = 25 - 16 = 9$.

(C) Let the sides be $2, 5, a, b$ in order. The angle between sides $2$ and $5$ is $60^{\circ}$. Let $c$ be the diagonal opposite to the $60^{\circ}$ angle.
Using the Law of Cosines in the first triangle:
$c^2 = 2^2 + 5^2 - 2(2)(5)\cos(60^{\circ}) = 4 + 25 - 20(0.5) = 29 - 10 = 19$.
So,$c = \sqrt{19}$.
Since the quadrilateral is cyclic,the opposite angle to $60^{\circ}$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$.
Using the Law of Cosines in the second triangle with sides $a, b$ and diagonal $c$:
$c^2 = a^2 + b^2 - 2ab\cos(120^{\circ}) \implies 19 = a^2 + b^2 - 2ab(-0.5) \implies a^2 + b^2 + ab = 19$.
The area of the quadrilateral is the sum of the areas of the two triangles:
$\text{Area} = \frac{1}{2}(2)(5)\sin(60^{\circ}) + \frac{1}{2}ab\sin(120^{\circ}) = 4\sqrt{3}$.
$5\left(\frac{\sqrt{3}}{2}\right) + \frac{ab}{2}\left(\frac{\sqrt{3}}{2}\right) = 4\sqrt{3}$.
Dividing by $\frac{\sqrt{3}}{2}$: $5 + \frac{ab}{2} = 8 \implies \frac{ab}{2} = 3 \implies ab = 6$.
Now,$a^2 + b^2 = 19 - ab = 19 - 6 = 13$.
We have $a^2 + b^2 = 13$ and $ab = 6$. Solving these,we find $a=2, b=3$ (or vice versa).
The perimeter is $2 + 5 + a + b = 7 + 2 + 3 = 12$.

(C) Let $p$ be the statement: 'Two numbers are not equal'.
Let $q$ be the statement: 'Their squares are not equal'.
The given statement is $p \rightarrow q$.
The contrapositive of $p \rightarrow q$ is defined as $\sim q \rightarrow \sim p$.
Here,$\sim q$ is: 'The squares of two numbers are equal'.
And $\sim p$ is: 'The numbers are equal'.
Therefore,the contrapositive is: 'If the squares of two numbers are equal,then the numbers are equal'.

(B) Using the distributive law: $(\sim p) \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge (\sim p \vee \sim q)$.
Since $(\sim p \vee p) \equiv T$ (Tautology),the expression becomes $T \wedge (\sim p \vee \sim q)$.
By the identity law,$T \wedge (\sim p \vee \sim q) \equiv \sim p \vee \sim q$.
Using the logical equivalence $p \rightarrow q \equiv \sim p \vee q$,we can rewrite $\sim p \vee \sim q$ as $p \rightarrow (\sim q)$.

(C) The equation of the line passing through $P(1, -2, 3)$ and parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is given by $\frac{x-1}{1} = \frac{y+2}{4} = \frac{z-3}{5} = \lambda$.
Any point $F$ on this line can be represented as $(\lambda+1, 4\lambda-2, 5\lambda+3)$.
Since $F$ lies on the plane $2x + 3y - 4z + 22 = 0$,we substitute the coordinates of $F$ into the plane equation:
$2(\lambda+1) + 3(4\lambda-2) - 4(5\lambda+3) + 22 = 0$
$2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0$
$-6\lambda + 6 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $F$,we get $F(2, 2, 8)$.
Since $F$ is the midpoint of $PQ$,the distance $PQ = 2PF$.
The distance $PF = \sqrt{(2-1)^2 + (2-(-2))^2 + (8-3)^2} = \sqrt{1^2 + 4^2 + 5^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$.
Therefore,$PQ = 2PF = 2\sqrt{42}$.

(C) Let the equation of the plane passing through $(1, -1, -1)$ be $a(x - 1) + b(y + 1) + c(z + 1) = 0$.
The normal vector $\vec{n}$ is perpendicular to the direction vectors of the two lines,$\vec{v_1} = (1, -2, 3)$ and $\vec{v_2} = (2, -1, -1)$.
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix} = \hat{i}(2 + 3) - \hat{j}(-1 - 6) + \hat{k}(-1 + 4) = 5\hat{i} + 7\hat{j} + 3\hat{k}$.
Thus,the equation of the plane is $5(x - 1) + 7(y + 1) + 3(z + 1) = 0$,which simplifies to $5x + 7y + 3z + 5 = 0$.
The distance of the point $(1, 3, -7)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Substituting the values,$d = \frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}} = \frac{|5 + 21 - 21 + 5|}{\sqrt{25 + 49 + 9}} = \frac{10}{\sqrt{83}}$.

(A) The region is bounded by $x=0$,$y=1+\sqrt{x}$,$x+y=3$,and $y=\frac{x^2}{4}$.
From the graph,the intersection points are $(1,2)$ and $(2,1)$.
The area is given by the integral:
$A = \int_{0}^{1} (1+\sqrt{x} - \frac{x^2}{4}) dx + \int_{1}^{2} (3-x - \frac{x^2}{4}) dx$
$A = \left[ x + \frac{2}{3}x^{3/2} - \frac{x^3}{12} \right]_{0}^{1} + \left[ 3x - \frac{x^2}{2} - \frac{x^3}{12} \right]_{1}^{2}$
$A = (1 + \frac{2}{3} - \frac{1}{12}) + [(6 - 2 - \frac{8}{12}) - (3 - \frac{1}{2} - \frac{1}{12})]$
$A = (\frac{12+8-1}{12}) + [(4 - \frac{2}{3}) - (2.5 - \frac{1}{12})]$
$A = \frac{19}{12} + (\frac{10}{3} - \frac{29}{12}) = \frac{19}{12} + \frac{40-29}{12} = \frac{19+11}{12} = \frac{30}{12} = \frac{5}{2}$ sq. units.

(C) Let $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} \quad (i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$:
$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos(\pi - x)} = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 - \cos x} \quad (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left( \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} \right) dx$
$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1 - \cos x + 1 + \cos x}{1 - \cos^2 x} dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{2}{\sin^2 x} dx$
$2I = 2 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \csc^2 x dx$
$I = [-\cot x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$
$I = -(\cot \frac{3\pi}{4} - \cot \frac{\pi}{4}) = -(-1 - 1) = 2$

(A) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
The determinant $D$ is given by:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & a & 1 \\ a & b & 1 \end{vmatrix} = 1(a - b) - 1(1 - a) + 1(b - a^2) = a - b - 1 + a + b - a^2 = -a^2 + 2a - 1 = -(a - 1)^2$.
For the system to have no solution or infinitely many solutions,we must have $D = 0$,which implies $-(a - 1)^2 = 0$,so $a = 1$.
Substituting $a = 1$ into the system:
$x + y + z = 1$
$x + y + z = 1$
$x + by + z = 0$
For the system to have no solution,the third equation must be inconsistent with the first two. Since the first two equations represent the same plane $x + y + z = 1$,the third equation $x + by + z = 0$ must represent a plane parallel to the first one but not identical to it.
Comparing $x + y + z = 1$ and $x + by + z = 0$,we see that for the planes to be parallel,the coefficients of $x, y, z$ must be proportional. Thus,$1/1 = b/1 = 1/1$,which implies $b = 1$.
However,if $b = 1$,the third equation becomes $x + y + z = 0$,which is parallel to $x + y + z = 1$ but distinct (since $0 \neq 1$). Thus,the system has no solution when $b = 1$.
Therefore,$S = \{1\}$,which is a singleton set.

(B) Given $2\omega + 1 = z$ and $z = \sqrt{-3} = i\sqrt{3}$.
Thus,$\omega = \frac{i\sqrt{3} - 1}{2}$,which is a complex cube root of unity.
We know that $1 + \omega + \omega^2 = 0$,so $-\omega^2 - 1 = \omega$.
Also,$\omega^3 = 1$,so $\omega^7 = \omega^{3 \times 2 + 1} = \omega$.
The determinant is $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{array} \right|$.
Applying $R_1 \to R_1 + R_2 + R_3$:
$\Delta = \left| \begin{array}{ccc} 3 & 1+\omega+\omega^2 & 1+\omega^2+\omega \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{array} \right| = \left| \begin{array}{ccc} 3 & 0 & 0 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{array} \right|$.
Expanding along $R_1$: $\Delta = 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$.
Since $\omega = \frac{-1 + i\sqrt{3}}{2}$,$\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.
$\Delta = 3\left( \frac{-1 - i\sqrt{3}}{2} - \frac{-1 + i\sqrt{3}}{2} \right) = 3\left( \frac{-2i\sqrt{3}}{2} \right) = -3i\sqrt{3} = -3z$.
Given $\Delta = 3k$,we have $3k = -3z$,so $k = -z$.

(C) Given $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} 4+12 & -6-3 \\ -8-4 & 12+1 \end{bmatrix} = \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix}$.
Now,calculate $3A^2 = 3 \begin{bmatrix} 16 & -9 \\ -12 & 13 \end{bmatrix} = \begin{bmatrix} 48 & -27 \\ -36 & 39 \end{bmatrix}$.
Calculate $12A = 12 \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix}$.
Let $M = 3A^2 + 12A = \begin{bmatrix} 48+24 & -27-36 \\ -36-48 & 39+12 \end{bmatrix} = \begin{bmatrix} 72 & -63 \\ -84 & 51 \end{bmatrix}$.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$\text{adj}(M) = \begin{bmatrix} 51 & 63 \\ 84 & 72 \end{bmatrix}$.

(C) The equation of the curve is $y(x - 2)(x - 3) = x + 6$,which can be written as $y = \frac{x + 6}{x^2 - 5x + 6}$.
At the $y$-axis,$x = 0$. Substituting $x = 0$ into the equation,we get $y(0 - 2)(0 - 3) = 0 + 6$,which simplifies to $y(6) = 6$,so $y = 1$. Thus,the point of intersection is $(0, 1)$.
Now,differentiate $y = \frac{x + 6}{x^2 - 5x + 6}$ with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(x^2 - 5x + 6)(1) - (x + 6)(2x - 5)}{(x^2 - 5x + 6)^2}$.
At the point $(0, 1)$,we have $x = 0$ and $x^2 - 5x + 6 = 6$:
$\frac{dy}{dx} = \frac{(6)(1) - (6)(-5)}{(6)^2} = \frac{6 + 30}{36} = \frac{36}{36} = 1$.
The slope of the tangent at $(0, 1)$ is $1$. Therefore,the slope of the normal is $-\frac{1}{1} = -1$.
The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y - 1 = -x$,or $x + y = 1$.
Checking the options,the point $(\frac{1}{2}, \frac{1}{2})$ satisfies the equation $x + y = 1$ because $\frac{1}{2} + \frac{1}{2} = 1$.

(D) Let $r$ be the radius and $\theta$ be the central angle of the circular sector in radians.
The perimeter of the circular sector is given by $P = r + r + r\theta = 2r + r\theta$.
Given that the total length of the wire is $20 \ m$,we have:
$2r + r\theta = 20$
$\Rightarrow r\theta = 20 - 2r$
$\Rightarrow \theta = \frac{20 - 2r}{r}$
The area $A$ of the circular sector is given by:
$A = \frac{1}{2} r^2 \theta$
Substituting the value of $\theta$:
$A = \frac{1}{2} r^2 \left( \frac{20 - 2r}{r} \right) = \frac{1}{2} r (20 - 2r) = 10r - r^2$
To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 10 - 2r = 0$
$\Rightarrow r = 5 \ m$
To verify the maximum,we check the second derivative:
$\frac{d^2A}{dr^2} = -2 < 0$
Since the second derivative is negative,the area is maximum at $r = 5$.
Substituting $r = 5$ into the area formula:
$A_{max} = 10(5) - (5)^2 = 50 - 25 = 25 \ m^2$.

(B) Given the differential equation: $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.
This can be rewritten as: $\frac{d}{dx} [(2 + \sin x)(y + 1)] = 0$.
Integrating both sides with respect to $x$,we get: $(2 + \sin x)(y + 1) = C$,where $C$ is a constant.
Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$:
$(2 + \sin 0)(1 + 1) = C \Rightarrow (2 + 0)(2) = C \Rightarrow C = 4$.
Thus,the equation becomes: $(2 + \sin x)(y + 1) = 4$.
Solving for $y$: $y + 1 = \frac{4}{2 + \sin x} \Rightarrow y = \frac{4}{2 + \sin x} - 1$.
Now,we find $y(\frac{\pi}{2})$:
$y(\frac{\pi}{2}) = \frac{4}{2 + \sin(\frac{\pi}{2})} - 1 = \frac{4}{2 + 1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

(C) Given $I_n = \int \tan^n x dx$.
We need to evaluate $I_4 + I_6 = \int \tan^4 x dx + \int \tan^6 x dx$.
$I_4 + I_6 = \int (\tan^4 x + \tan^6 x) dx$.
Factor out $\tan^4 x$:
$I_4 + I_6 = \int \tan^4 x (1 + \tan^2 x) dx$.
Since $1 + \tan^2 x = \sec^2 x$,we have:
$I_4 + I_6 = \int \tan^4 x \sec^2 x dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
The integral becomes $\int u^4 du = \frac{u^5}{5} + C$.
Substituting back $u = \tan x$,we get:
$I_4 + I_6 = \frac{1}{5} \tan^5 x + C$.
Comparing this with $a \tan^5 x + b x^5 + C$,we get $a = \frac{1}{5}$ and $b = 0$.
Thus,the ordered pair $(a, b) = (\frac{1}{5}, 0)$.

(B) The problem follows a binomial distribution because the balls are drawn with replacement,making each trial independent.
Here,the total number of trials is $n = 10$.
The probability of drawing a green ball in a single trial is $p = \frac{15}{15 + 10} = \frac{15}{25} = \frac{3}{5}$.
The probability of not drawing a green ball (drawing a yellow ball) is $q = 1 - p = 1 - \frac{3}{5} = \frac{2}{5}$.
The variance of a binomial distribution is given by the formula $\text{Variance} = npq$.
Substituting the values: $\text{Variance} = 10 \times \frac{3}{5} \times \frac{2}{5} = 10 \times \frac{6}{25} = \frac{60}{25} = \frac{12}{5}$.

(C) Let $P(A), P(B), P(C)$ be the probabilities of events $A, B, C$.
Given:
$P(A) + P(B) - 2P(A \cap B) = \frac{1}{4}$ ... $(1)$
$P(B) + P(C) - 2P(B \cap C) = \frac{1}{4}$ ... $(2)$
$P(C) + P(A) - 2P(C \cap A) = \frac{1}{4}$ ... $(3)$
Adding $(1), (2),$ and $(3)$,we get:
$2[P(A) + P(B) + P(C)] - 2[P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3}{4}$
Dividing by $2$:
$[P(A) + P(B) + P(C)] - [P(A \cap B) + P(B \cap C) + P(C \cap A)] = \frac{3}{8}$
We know that $P(A \cup B \cup C) = [P(A) + P(B) + P(C)] - [P(A \cap B) + P(B \cap C) + P(C \cap A)] + P(A \cap B \cap C)$.
Substituting the values:
$P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16} = \frac{6+1}{16} = \frac{7}{16}$.

(D) Given $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right]$ defined by $f(x) = \frac{x}{1 + x^2}$.
First,check for injectivity:
$f'(x) = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} = \frac{(1 - x)(1 + x)}{(1 + x^2)^2}$.
Since $f'(x)$ changes sign at $x = 1$ and $x = -1$,the function is not monotonic,hence it is not injective.
Next,check for surjectivity:
Let $y = \frac{x}{1 + x^2}$. Then $yx^2 - x + y = 0$.
For $x$ to be a real number,the discriminant $D \ge 0$.
$D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \ge 0$.
$4y^2 \le 1 \Rightarrow y^2 \le \frac{1}{4} \Rightarrow y \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.
Since the range is equal to the codomain,the function is surjective.
Therefore,the function is surjective but not injective.

(C) 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 $|(\vec a \times \vec b) \times \vec c| = 3$ and the angle between $\vec c$ and $\vec a \times \vec b$ is $30^\circ$,we use the formula $|\vec u \times \vec v| = |\vec u||\vec v| \sin \theta$:
$|\vec a \times \vec b||\vec c| \sin 30^\circ = 3 \implies 3 \cdot |\vec c| \cdot \frac{1}{2} = 3 \implies |\vec c| = 2$.
Now,use the condition $|\vec c - \vec a| = 3$. Squaring both sides:
$|\vec c - \vec a|^2 = 3^2 \implies |\vec c|^2 + |\vec a|^2 - 2(\vec a \cdot \vec c) = 9$.
Substitute the known values $|\vec c| = 2$ and $|\vec a| = 3$:
$2^2 + 3^2 - 2(\vec a \cdot \vec c) = 9 \implies 4 + 9 - 2(\vec a \cdot \vec c) = 9 \implies 13 - 2(\vec a \cdot \vec c) = 9$.
$2(\vec a \cdot \vec c) = 4 \implies \vec a \cdot \vec c = 2$.

(D) Given $f(x) = 2^{10}x + 1$ and $g(x) = 3^{10}x - 1$.
We are given $(f \circ g)(x) = x$.
Substituting $g(x)$ into $f(x)$,we get $f(g(x)) = 2^{10}(3^{10}x - 1) + 1 = x$.
Expanding the expression: $2^{10} \cdot 3^{10}x - 2^{10} + 1 = x$.
Since $2^{10} \cdot 3^{10} = (2 \cdot 3)^{10} = 6^{10}$,we have $6^{10}x - x = 2^{10} - 1$.
Factoring out $x$: $x(6^{10} - 1) = 2^{10} - 1$.
Thus,$x = \frac{2^{10} - 1}{6^{10} - 1}$.
To match the options,divide the numerator and denominator by $6^{10}$:
$x = \frac{\frac{2^{10}}{6^{10}} - \frac{1}{6^{10}}}{1 - \frac{1}{6^{10}}} = \frac{3^{-10} - 6^{-10}}{1 - 6^{-10}}$.
Alternatively,divide the numerator and denominator by $2^{10} \cdot 3^{10} = 6^{10}$ or manipulate the expression:
$x = \frac{2^{10} - 1}{6^{10} - 1} = \frac{2^{10}(1 - 2^{-10})}{6^{10}(1 - 6^{-10})} = \frac{3^{-10}(1 - 2^{-10})}{1 - 6^{-10}}$.
Re-evaluating the expression $\frac{2^{10} - 1}{6^{10} - 1}$:
Divide numerator and denominator by $6^{10}$: $\frac{2^{10}/6^{10} - 1/6^{10}}{1 - 1/6^{10}} = \frac{3^{-10} - 6^{-10}}{1 - 6^{-10}}$.
Checking option $D$: $\frac{1 - 2^{-10}}{3^{10} - 2^{-10}} = \frac{(2^{10}-1)/2^{10}}{(3^{10} \cdot 2^{10} - 1)/2^{10}} = \frac{2^{10}-1}{6^{10}-1}$.
Thus,option $D$ is correct.

(B) For a homogeneous system of linear equations $AX = 0$ to have infinitely many solutions,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.
The coefficient matrix is:
$A = \begin{bmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{bmatrix}$
Setting the determinant to zero:
$|A| = 2(\lambda \cdot 2 - 2 \cdot 2) - 4(4 \cdot 2 - 2 \cdot \lambda) + (-\lambda)(4 \cdot 2 - \lambda \cdot \lambda) = 0$
$|A| = 2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2) = 0$
$|A| = 4\lambda - 8 - 32 + 8\lambda - 8\lambda + \lambda^3 = 0$
$|A| = \lambda^3 + 4\lambda - 40 = 0$
Let $f(\lambda) = \lambda^3 + 4\lambda - 40$.
Since $f'(\lambda) = 3\lambda^2 + 4 > 0$ for all real $\lambda$,the function $f(\lambda)$ is strictly increasing.
Therefore,it can have only one real root.
Thus,the number of real values of $\lambda$ is $1$.

(D) For a $3 \times 3$ matrix $A$,we have the following properties:
$1$. $adj(adj(A)) = |A|^{n-2} A$. Since $n=3$,$adj(adj(A)) = |A|^{3-2} A = |A| A$. Thus,option $B$ is true.
$2$. We know that $adj(A) = |A| A^{-1}$. Replacing $A$ with $adj(A)$,we get $adj(adj(A)) = |adj(A)| (adj(A))^{-1}$.
$3$. Since $|adj(A)| = |A|^{n-1} = |A|^{3-1} = |A|^2$,we have $adj(adj(A)) = |A|^2 (adj(A))^{-1}$. Thus,option $C$ is true.
$4$. Comparing the results,$adj(adj(A)) = |A| A$ and $adj(adj(A)) = |A|^2 (adj(A))^{-1}$.
$5$. Option $D$ states $adj(adj(A)) = |A| (adj(A))^{-1}$,which is generally false.
$6$. Option $A$ states $adj(A) = |A| (adj(A))^{-1}$. Since $adj(A) = |A| A^{-1}$,this implies $A^{-1} = (adj(A))^{-1}$,which is true. Therefore,option $D$ is the one that is not always true.

(D) Given curve: $x^2y^2 - 2x = 4 - 4y$.
Differentiating with respect to $x$:
$2xy^2 + 2x^2y \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$.
At point $(2, -2)$:
$2(2)(-2)^2 + 2(2)^2(-2) \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$.
$16 - 16 \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$.
$14 = 12 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{14}{12} = \frac{7}{6}$.
Equation of the tangent at $(2, -2)$:
$y - (-2) = \frac{7}{6}(x - 2)$.
$6(y + 2) = 7(x - 2) \Rightarrow 6y + 12 = 7x - 14 \Rightarrow 7x - 6y = 26$.
Checking the options:
For $(-2, -7)$: $7(-2) - 6(-7) = -14 + 42 = 28 \neq 26$.
Thus,the tangent does not pass through $(-2, -7)$.

(A) Given $y = {\left( {x + \sqrt {{x^2} - 1} } \right)^{15}} + {\left( {x - \sqrt {{x^2} - 1} } \right)^{15}}$.
Differentiating with respect to $x$:
$\frac{{dy}}{{dx}} = 15{\left( {x + \sqrt {{x^2} - 1} } \right)^{14}}\left( {1 + \frac{x}{{\sqrt {{x^2} - 1} }}} \right) + 15{\left( {x - \sqrt {{x^2} - 1} } \right)^{14}}\left( {1 - \frac{x}{{\sqrt {{x^2} - 1} }}} \right)$
$\frac{{dy}}{{dx}} = 15{\left( {x + \sqrt {{x^2} - 1} } \right)^{14}}\left( {\frac{{\sqrt {{x^2} - 1} + x}}{{\sqrt {{x^2} - 1} }}} \right) - 15{\left( {x - \sqrt {{x^2} - 1} } \right)^{14}}\left( {\frac{{x - \sqrt {{x^2} - 1} }}{{\sqrt {{x^2} - 1} }}} \right)$
$\frac{{dy}}{{dx}} = \frac{{15}}{{\sqrt {{x^2} - 1} }}\left[ {{{\left( {x + \sqrt {{x^2} - 1} } \right)}^{15}} - {{\left( {x - \sqrt {{x^2} - 1} } \right)}^{15}}} \right]$.
Note that this approach is slightly complex,so let $u = x + \sqrt{x^2-1}$ and $v = x - \sqrt{x^2-1}$. Then $uv = 1$ and $y = u^{15} + v^{15}$.
$\frac{dy}{dx} = 15u^{14} \frac{du}{dx} + 15v^{14} \frac{dv}{dx}$. Since $\frac{du}{dx} = \frac{u}{\sqrt{x^2-1}}$ and $\frac{dv}{dx} = \frac{-v}{\sqrt{x^2-1}}$,we get $\frac{dy}{dx} = \frac{15}{\sqrt{x^2-1}}(u^{15} - v^{15})$.
Let $y_1 = u^{15} + v^{15}$ and $y_2 = u^{15} - v^{15}$. Then $\sqrt{x^2-1} \frac{dy}{dx} = 15 y_2$.
Differentiating again: $\frac{x}{\sqrt{x^2-1}} \frac{dy}{dx} + \sqrt{x^2-1} \frac{d^2y}{dx^2} = 15 \frac{dy_2}{dx}$.
Since $\frac{dy_2}{dx} = 15(u^{14} \frac{du}{dx} - v^{14} \frac{dv}{dx}) = \frac{15}{\sqrt{x^2-1}}(u^{15} + v^{15}) = \frac{15y}{\sqrt{x^2-1}}$.
Multiplying by $\sqrt{x^2-1}$: $x \frac{dy}{dx} + (x^2-1) \frac{d^2y}{dx^2} = 15(15y) = 225y$.

(A) Let $I = \int \sqrt{1 + 2\cot x \csc x + 2\cot^2 x} \,dx$.
Since $1 + \cot^2 x = \csc^2 x$,we have $1 + 2\cot^2 x = \csc^2 x + \cot^2 x$.
Thus,the expression inside the square root is $\csc^2 x + \cot^2 x + 2\cot x \csc x = (\csc x + \cot x)^2$.
Since $0 < x < \frac{\pi}{2}$,$\csc x + \cot x > 0$,so $\sqrt{(\csc x + \cot x)^2} = \csc x + \cot x$.
$I = \int (\csc x + \cot x) \,dx$.
Using the standard integrals $\int \csc x \,dx = \log |\csc x - \cot x| + C$ and $\int \cot x \,dx = \log |\sin x| + C$,we get:
$I = \log |\csc x - \cot x| + \log |\sin x| + C$.
$I = \log \left| \frac{1 - \cos x}{\sin x} \cdot \sin x \right| + C = \log |1 - \cos x| + C$.
Using $1 - \cos x = 2\sin^2 \frac{x}{2}$,we get:
$I = \log |2\sin^2 \frac{x}{2}| + C = \log 2 + 2\log |\sin \frac{x}{2}| + C$.
Absorbing $\log 2$ into the constant $C$,we get $I = 2\log |\sin \frac{x}{2}| + C$.

(A) We know that $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$.
Substituting this into the integral:
$I = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{8 \cos 2x}{(\frac{2}{\sin 2x})^3} dx = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{8 \cos 2x \sin^3 2x}{8} dx = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \cos 2x \sin^3 2x dx$.
Let $u = \sin 2x$,then $du = 2 \cos 2x dx$,so $\cos 2x dx = \frac{du}{2}$.
When $x = \frac{\pi}{12}$,$u = \sin(\frac{\pi}{6}) = \frac{1}{2}$.
When $x = \frac{\pi}{4}$,$u = \sin(\frac{\pi}{2}) = 1$.
$I = \int_{1/2}^{1} u^3 \frac{du}{2} = \frac{1}{2} [\frac{u^4}{4}]_{1/2}^{1} = \frac{1}{8} [1^4 - (\frac{1}{2})^4] = \frac{1}{8} [1 - \frac{1}{16}] = \frac{1}{8} \times \frac{15}{16} = \frac{15}{128}$.

(D) Given curves are $x^2 + y^2 = 4$ (a circle with center $(0,0)$ and radius $2$) and $y^2 = 3x$ (a parabola).
To find the intersection points,substitute $y^2 = 3x$ into $x^2 + y^2 = 4$:
$x^2 + 3x - 4 = 0$
$(x+4)(x-1) = 0$
Since $x \ge 0$ for the parabola,we have $x = 1$.
At $x = 1$,$y^2 = 3(1) = 3$,so $y = \pm\sqrt{3}$.
The area of the smaller portion is symmetric about the $x$-axis.
Area $= 2 \times \left[ \int_{0}^{1} \sqrt{3x} \, dx + \int_{1}^{2} \sqrt{4-x^2} \, dx \right]$
$= 2 \times \left[ \sqrt{3} \left( \frac{x^{3/2}}{3/2} \right)_0^1 + \left( \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right)_1^2 \right]$
$= 2 \times \left[ \sqrt{3} \left( \frac{2}{3} \right) + \left( (0 + 2\sin^{-1}(1)) - (\frac{1}{2}\sqrt{3} + 2\sin^{-1}(1/2)) \right) \right]$
$= 2 \times \left[ \frac{2\sqrt{3}}{3} + \left( 2 \cdot \frac{\pi}{2} - \frac{\sqrt{3}}{2} - 2 \cdot \frac{\pi}{6} \right) \right]$
$= 2 \times \left[ \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} + \pi - \frac{\pi}{3} \right] = 2 \times \left[ \frac{4-3}{2\sqrt{3}} + \frac{2\pi}{3} \right]$
$= 2 \times \left[ \frac{1}{2\sqrt{3}} + \frac{2\pi}{3} \right] = \frac{1}{\sqrt{3}} + \frac{4\pi}{3}$

(C) First,find the normal vector $\vec{n}$ to the plane containing the two lines. The direction vectors of the lines are $\vec{v_1} = (6, 7, 8)$ and $\vec{v_2} = (3, 5, 7)$.
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{vmatrix} = \hat{i}(49-40) - \hat{j}(42-24) + \hat{k}(30-21) = (9, -18, 9)$.
We can simplify the normal vector to $\vec{n} = (1, -2, 1)$.
The plane passes through the point $(-1, 1, 3)$ (from the first line). The equation of the plane is $1(x+1) - 2(y-1) + 1(z-3) = 0$,which simplifies to $x - 2y + z = 0$.
Let the foot of the perpendicular from $P(1, -2, 1)$ to the plane be $F(x, y, z)$. The line passing through $P$ and perpendicular to the plane has direction ratios $(1, -2, 1)$.
Thus,$\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-1}{1} = k$.
$x = k+1, y = -2k-2, z = k+1$.
Since $F$ lies on the plane $x - 2y + z = 0$,we have $(k+1) - 2(-2k-2) + (k+1) = 0$.
$k+1 + 4k + 4 + k+1 = 0 \Rightarrow 6k + 6 = 0 \Rightarrow k = -1$.
Substituting $k = -1$,we get $x = 0, y = 0, z = 0$.
The coordinates of the foot of the perpendicular are $(0, 0, 0)$.

(C) The direction vector $\vec v$ of the line of intersection is given by the cross product of the normals $\vec n_1 = 3\hat i - \hat j + \hat k$ and $\vec n_2 = \hat i + 4\hat j - 2\hat k$.
$\vec v = \vec n_1 \times \vec n_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} = \hat i(2 - 4) - \hat j(-6 - 1) + \hat k(12 + 1) = -2\hat i + 7\hat j + 13\hat k$.
To find a point on the line,set $z = 0$ in the plane equations:
$3x - y = 1$ and $x + 4y = 2$.
Multiplying the first by $4$: $12x - 4y = 4$. Adding to the second: $13x = 6 \Rightarrow x = 6/13$.
Substituting $x$: $6/13 + 4y = 2 \Rightarrow 4y = 2 - 6/13 = 20/13 \Rightarrow y = 5/13$.
The point is $(6/13, 5/13, 0)$.
The line equation is $\frac{x - 6/13}{-2} = \frac{y - 5/13}{7} = \frac{z}{13}$,which is equivalent to $\frac{x - 6/13}{2} = \frac{y - 5/13}{-7} = \frac{z}{-13}$.

(B) Let the diagonals be $\vec{d_1} = 8\hat{i} - 6\hat{j} + 0\hat{k}$ and $\vec{d_2} = 3\hat{i} + 4\hat{j} - 12\hat{k}$.
The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula $\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -6 & 0 \\ 3 & 4 & -12 \end{vmatrix}$
$= \hat{i}((-6)(-12) - (0)(4)) - \hat{j}((8)(-12) - (0)(3)) + \hat{k}((8)(4) - (-6)(3))$
$= \hat{i}(72 - 0) - \hat{j}(-96 - 0) + \hat{k}(32 + 18)$
$= 72\hat{i} + 96\hat{j} + 50\hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{72^2 + 96^2 + 50^2}$
$= \sqrt{5184 + 9216 + 2500}$
$= \sqrt{16900} = 130$.
Therefore,the area of the parallelogram is $\frac{1}{2} \times 130 = 65$ sq. units.

(B) The total number of outcomes when a coin is tossed $8$ times is $2^8 = 256$.
The event of obtaining at least one head and at least one tail is the complement of the event of obtaining all heads or all tails.
$P(\text{All Heads}) = \frac{1}{2^8} = \frac{1}{256}$.
$P(\text{All Tails}) = \frac{1}{2^8} = \frac{1}{256}$.
The probability of obtaining either all heads or all tails is $P(\text{All Heads}) + P(\text{All Tails}) = \frac{1}{256} + \frac{1}{256} = \frac{2}{256} = \frac{1}{128}$.
Therefore,the required probability is $1 - \frac{1}{128} = \frac{127}{128}$.

(D) The given determinant is:
$D = 0(0 - \sin^2 x) - \cos x(0 - \cos^2 x) - \sin x(\sin^2 x - 0) = 0$
$\Rightarrow \cos^3 x - \sin^3 x = 0$
$\Rightarrow \cos^3 x = \sin^3 x$
$\Rightarrow \tan^3 x = 1$
Since $x \in [0, 2\pi]$,the solutions for $\tan x = 1$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
Thus,$S = \{ \frac{\pi}{4}, \frac{5\pi}{4} \}$.
We need to calculate $\sum_{x \in S} \tan \left( \frac{\pi}{3} + x \right) = \tan \left( \frac{\pi}{3} + \frac{\pi}{4} \right) + \tan \left( \frac{\pi}{3} + \frac{5\pi}{4} \right)$.
Using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
For $x = \frac{\pi}{4}$,$\tan \left( \frac{\pi}{3} + \frac{\pi}{4} \right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{3 + 1 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$.
For $x = \frac{5\pi}{4}$,$\tan \left( \frac{\pi}{3} + \frac{5\pi}{4} \right) = \tan \left( \frac{\pi}{3} + \frac{\pi}{4} + \pi \right) = \tan \left( \frac{\pi}{3} + \frac{\pi}{4} \right) = -2 - \sqrt{3}$.
Sum $= (-2 - \sqrt{3}) + (-2 - \sqrt{3}) = -4 - 2\sqrt{3}$.

(A) Let ${x^2} = \cos 2\theta$,which implies $2\theta = {\cos ^{ - 1}}{x^2}$ or $\theta = \frac{1}{2}{\cos ^{ - 1}}{x^2}$.
Substituting ${x^2} = \cos 2\theta$ into the expression:
${\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right]$
Using the identities $1 + \cos 2\theta = 2{\cos ^2}\theta$ and $1 - \cos 2\theta = 2{\sin ^2}\theta$:
${\tan ^{ - 1}}\left[ {\frac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}} \right] = {\tan ^{ - 1}}\left[ {\frac{{\sqrt 2 \cos \theta + \sqrt 2 \sin \theta }}{{\sqrt 2 \cos \theta - \sqrt 2 \sin \theta }}} \right]$
Dividing numerator and denominator by $\cos \theta$:
${\tan ^{ - 1}}\left[ {\frac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right] = {\tan ^{ - 1}}\left[ {\tan \left( {\frac{\pi }{4} + \theta } \right)} \right]$
$= \frac{\pi }{4} + \theta = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$.

(D) Given the function $f(x) = x - 5[\frac{x}{5}]$.
To check for one-one: Calculate $f(1) = 1 - 5[1/5] = 1 - 5(0) = 1$ and $f(6) = 6 - 5[6/5] = 6 - 5(1) = 1$.
Since $f(1) = f(6) = 1$ but $1 \neq 6$,the function is not one-one.
To check for onto: The codomain is $N = \{1, 2, 3, ...\}$.
Calculate $f(5) = 5 - 5[5/5] = 5 - 5(1) = 0$.
Since $0 \notin N$,the function is not onto.
Therefore,the function $f$ is neither one-one nor onto.

(B) Given equations are:
$(1) \ A + B = 2B^T$
$(2) \ 3A + 2B = I_3$
Taking the transpose of equation $(1)$:
$(A + B)^T = (2B^T)^T \Rightarrow A^T + B^T = 2B$
From $(1)$,$A = 2B^T - B$. Substitute this into $(2)$:
$3(2B^T - B) + 2B = I_3 \Rightarrow 6B^T - 3B + 2B = I_3 \Rightarrow 6B^T - B = I_3 \Rightarrow B = 6B^T - I_3$
Substitute $B$ into $(1)$:
$A + (6B^T - I_3) = 2B^T \Rightarrow A = I_3 - 4B^T$
Taking transpose of $A = I_3 - 4B^T$:
$A^T = I_3 - 4B$
From $A^T + B^T = 2B$,we have $B^T = 2B - A^T$. Substitute $A^T = I_3 - 4B$:
$B^T = 2B - (I_3 - 4B) = 6B - I_3$
Substitute $B^T$ back into $B = 6B^T - I_3$:
$B = 6(6B - I_3) - I_3 = 36B - 6I_3 - I_3 = 36B - 7I_3$
$35B = 7I_3 \Rightarrow B = \frac{1}{5}I_3$
Now find $A$:
$A = I_3 - 4B^T = I_3 - 4(6B - I_3) = I_3 - 24B + 4I_3 = 5I_3 - 24(\frac{1}{5}I_3) = \frac{25I_3 - 24I_3}{5} = \frac{1}{5}I_3$
Check the options:
$10A + 5B = 10(\frac{1}{5}I_3) + 5(\frac{1}{5}I_3) = 2I_3 + I_3 = 3I_3$
Thus,the correct option is $B$.

(C) The given system of linear equations is:
$x + 8y + 7z = 0$ $(1)$
$9x + 2y + 3z = 0$ $(2)$
$x + y + z = 0$ $(3)$
From equation $(3)$,$x = -y - z$. Substituting this into $(1)$ and $(2)$:
$(-y - z) + 8y + 7z = 0 \implies 7y + 6z = 0 \implies z = -\frac{7}{6}y$
$9(-y - z) + 2y + 3z = 0 \implies -7y - 6z = 0 \implies z = -\frac{7}{6}y$
Since the equations are homogeneous and the determinant of the coefficient matrix is $0$,there are infinitely many solutions. Let $y = 6\lambda$. Then $z = -7\lambda$ and $x = -6\lambda - (-7\lambda) = \lambda$.
So,$(a, b, c) = (\lambda, 6\lambda, -7\lambda)$.
Since this point lies on the plane $x + 2y + z = 6$:
$\lambda + 2(6\lambda) + (-7\lambda) = 6$
$\lambda + 12\lambda - 7\lambda = 6$
$6\lambda = 6 \implies \lambda = 1$.
Thus,$(a, b, c) = (1, 6, -7)$.
We need to find $2a + b + c = 2(1) + 6 + (-7) = 2 + 6 - 7 = 1$.

(B) Given: $y^{1/5} + y^{-1/5} = 2x$.
Differentiating with respect to $x$:
$\frac{1}{5} y^{-4/5} \frac{dy}{dx} - \frac{1}{5} y^{-6/5} \frac{dy}{dx} = 2$.
$\frac{1}{5} y^{-1} (y^{1/5} - y^{-1/5}) \frac{dy}{dx} = 2$.
$(y^{1/5} - y^{-1/5}) \frac{dy}{dx} = 10y$.
We know that $(y^{1/5} - y^{-1/5})^2 = (y^{1/5} + y^{-1/5})^2 - 4 = (2x)^2 - 4 = 4(x^2 - 1)$.
Thus,$y^{1/5} - y^{-1/5} = 2\sqrt{x^2 - 1}$.
Substituting this value: $2\sqrt{x^2 - 1} \frac{dy}{dx} = 10y$,which simplifies to $\sqrt{x^2 - 1} \frac{dy}{dx} = 5y$.
Differentiating again with respect to $x$:
$\frac{x}{\sqrt{x^2 - 1}} \frac{dy}{dx} + \sqrt{x^2 - 1} \frac{d^2y}{dx^2} = 5 \frac{dy}{dx}$.
Multiplying by $\sqrt{x^2 - 1}$:
$x \frac{dy}{dx} + (x^2 - 1) \frac{d^2y}{dx^2} = 5 \sqrt{x^2 - 1} \frac{dy}{dx}$.
Since $\sqrt{x^2 - 1} \frac{dy}{dx} = 5y$,we get $x \frac{dy}{dx} + (x^2 - 1) \frac{d^2y}{dx^2} = 5(5y) = 25y$.
Rearranging: $(x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} - 25y = 0$.
Comparing with $(x^2 - 1) \frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$,we get $\lambda = 1$ and $k = -25$.
Therefore,$\lambda + k = 1 - 25 = -24$.

(A) Given the function $f(x) = x^3 - 3x^2 + 5x + 7$.
To determine the intervals of increase or decrease,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 5x + 7) = 3x^2 - 6x + 5$.
Now,we analyze the sign of $f'(x) = 3x^2 - 6x + 5$.
We can rewrite this as $f'(x) = 3(x^2 - 2x) + 5 = 3(x^2 - 2x + 1 - 1) + 5 = 3(x-1)^2 - 3 + 5 = 3(x-1)^2 + 2$.
Since $(x-1)^2 \ge 0$ for all $x \in R$,it follows that $3(x-1)^2 + 2 \ge 2 > 0$ for all $x \in R$.
Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing in $R$.

(B) Let $f(x) = ax^n + \dots$ be a polynomial of degree $n$.
Comparing the degrees on both sides of $f(3x) = f'(x) \cdot f''(x)$:
$n = (n-1) + (n-2) = 2n - 3$,which gives $n = 3$.
Let $f(x) = ax^3 + bx^2 + cx + d$.
Then $f(3x) = a(3x)^3 + b(3x)^2 + c(3x) + d = 27ax^3 + 9bx^2 + 3cx + d$.
$f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.
$f'(x) \cdot f''(x) = (3ax^2 + 2bx + c)(6ax + 2b) = 18a^2x^3 + (6ab + 12ab)x^2 + (4b^2 + 6ac)x + 2bc$.
Equating coefficients of $x^3$: $27a = 18a^2 \Rightarrow a = \frac{3}{2}$ (since $a \neq 0$).
Equating coefficients of $x^2$: $9b = 18ab = 18(\frac{3}{2})b = 27b \Rightarrow b = 0$.
Equating coefficients of $x$: $3c = 4b^2 + 6ac = 0 + 6(\frac{3}{2})c = 9c \Rightarrow c = 0$.
Equating constant terms: $d = 2bc = 0$.
Thus,$f(x) = \frac{3}{2}x^3$.
Then $f'(x) = \frac{9}{2}x^2$ and $f''(x) = 9x$.
At $x = 2$,$f'(2) = \frac{9}{2}(4) = 18$ and $f''(2) = 9(2) = 18$.
Therefore,$f''(2) - f'(2) = 18 - 18 = 0$.

(B) Let $t = \frac{3x - 4}{3x + 4}$.
Then $3xt + 4t = 3x - 4$,which implies $x(3t - 3) = -4t - 4$,so $x = \frac{4t + 4}{3 - 3t}$.
Substituting this into the function: $f(t) = \frac{4t + 4}{3 - 3t} + 2 = \frac{4t + 4 + 6 - 6t}{3 - 3t} = \frac{10 - 2t}{3 - 3t}$.
Thus,$f(x) = \frac{10 - 2x}{3 - 3x} = \frac{2x - 10}{3x - 3}$.
Now,$\int f(x) dx = \int \frac{2x - 10}{3(x - 1)} dx = \frac{2}{3} \int \frac{x - 1 - 4}{x - 1} dx = \frac{2}{3} \int (1 - \frac{4}{x - 1}) dx$.
$= \frac{2}{3} x - \frac{8}{3} \ln |x - 1| + C = -\frac{8}{3} \ln |1 - x| + \frac{2}{3} x + C$.
Comparing with $A \log |1 - x| + Bx + C$,we get $A = -\frac{8}{3}$ and $B = \frac{2}{3}$.
Therefore,the ordered pair $(A, B) = \left( -\frac{8}{3}, \frac{2}{3} \right)$.

(A) Let $I = \int_{1}^{2} \frac{dx}{((x-1)^2 + 3)^{3/2}}$.
Substitute $x-1 = \sqrt{3} \tan \theta$,so $dx = \sqrt{3} \sec^2 \theta \, d\theta$.
When $x=1$,$\tan \theta = 0 \implies \theta = 0$.
When $x=2$,$\tan \theta = \frac{1}{\sqrt{3}} \implies \theta = \frac{\pi}{6}$.
$I = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 \theta \, d\theta}{(\sqrt{3} \sec \theta)^3} = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 \theta}{3\sqrt{3} \sec^3 \theta} \, d\theta$.
$I = \frac{1}{3} \int_{0}^{\pi/6} \cos \theta \, d\theta = \frac{1}{3} [\sin \theta]_{0}^{\pi/6} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.
Given $\frac{k}{k+5} = \frac{1}{6}$,we have $6k = k+5$,which implies $5k = 5$,so $k = 1$.

(A) The given limit is $\mathop {\lim }\limits_{n \to \infty } \frac{{\sum_{r=1}^n r^a}}{{(n+1)^{a-1} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$.
Divide the numerator and denominator by $n^{a+1}$:
$\mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^a}}{{(\frac{n+1}{n})^{a-1} \cdot \frac{1}{n^2} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$.
Using the definition of definite integral $\int_0^1 x^a dx = \frac{1}{a+1}$ and the sum of arithmetic progression $\sum_{r=1}^n (na + r) = n^2 a + \frac{n(n+1)}{2}$,we get:
$\frac{\int_0^1 x^a dx}{\lim_{n \to \infty} (1 + \frac{1}{n})^{a-1} (a + \frac{1}{2}(1 + \frac{1}{n}))} = \frac{1}{60}$.
$\frac{\frac{1}{a+1}}{a + \frac{1}{2}} = \frac{1}{60}$.
$\frac{1}{(a+1)(\frac{2a+1}{2})} = \frac{1}{60} \Rightarrow (a+1)(2a+1) = 30$.
$2a^2 + 3a + 1 = 30 \Rightarrow 2a^2 + 3a - 29 = 0$.
Wait,re-evaluating the denominator sum: $\sum_{r=1}^n (na+r) = n^2a + \frac{n(n+1)}{2}$. Dividing by $n^2$ gives $a + \frac{1}{2}(1 + \frac{1}{n}) \to a + \frac{1}{2}$.
Correcting the equation: $\frac{1/(a+1)}{a + 1/2} = 1/60$ $\Rightarrow \frac{2}{(a+1)(2a+1)} = 1/60$ $\Rightarrow (a+1)(2a+1) = 120$.
$2a^2 + 3a + 1 = 120 \Rightarrow 2a^2 + 3a - 119 = 0$.
$(2a + 17)(a - 7) = 0$.
Since $a > 0$,we have $a = 7$.

(C) Let the tangent at $P(x, y)$ be $Y - y = f'(x)(X - x)$.
For $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$. So $A = \left( x - \frac{y}{f'(x)}, 0 \right)$.
For $B$ (where $X=0$),$Y = y - x f'(x)$. So $B = (0, y - x f'(x))$.
Given $AP : BP = 1 : 3$,by section formula for $P(x, y)$ dividing $AB$ in ratio $1:3$:
$x = \frac{1 \cdot 0 + 3 \cdot (x - y/f'(x))}{1 + 3} \implies 4x = 3x - \frac{3y}{f'(x)} \implies x = -\frac{3y}{f'(x)}$.
Thus,$\frac{dy}{dx} = -\frac{3y}{x}$.
Separating variables: $\int \frac{dy}{y} = -3 \int \frac{dx}{x} \implies \ln|y| = -3 \ln|x| + C \implies y = \frac{k}{x^3}$.
Given $f(1) = 1$,we get $1 = \frac{k}{1^3} \implies k = 1$. So $y = \frac{1}{x^3}$.
Checking the options,for $x=2$,$y = \frac{1}{2^3} = \frac{1}{8}$.
Thus,the curve passes through $\left( 2, \frac{1}{8} \right)$.

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.
The distance of this plane from the origin $(0, 0, 0)$ is given as $3 \ units$. Thus,$\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 3$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{9}$.
The centroid $(x, y, z)$ of $\Delta ABC$ is given by $x = \frac{a+0+0}{3}$,$y = \frac{0+b+0}{3}$,and $z = \frac{0+0+c}{3}$.
Therefore,$a = 3x$,$b = 3y$,and $c = 3z$.
Substituting these values into the distance equation: $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{9}$.
$\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{9}$.
Multiplying by $9$,we get $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 1$.

(C) Since the line lies in the plane,the point $(3, -2, -\lambda)$ must satisfy the plane equation $2x - 4y + 3z = 2$.
Substituting the point: $2(3) - 4(-2) + 3(-\lambda) = 2$.
$6 + 8 - 3\lambda = 2 \implies 14 - 3\lambda = 2 \implies 3\lambda = 12 \implies \lambda = 4$.
Also,the direction vector of the line $(1, -1, -2)$ must be perpendicular to the normal of the plane $(2, -4, 3)$.
Check: $(1)(2) + (-1)(-4) + (-2)(3) = 2 + 4 - 6 = 0$. This is satisfied.
Now,consider the two lines:
$L_1: \frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + 4}{-2}$
$L_2: \frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$
Since $L_1$ lies in the plane $2x - 4y + 3z = 2$,we check if $L_2$ intersects the plane. For $L_2$,$x = 12k+1, y = 9k, z = 4k$.
Substituting into the plane: $2(12k+1) - 4(9k) + 3(4k) = 24k + 2 - 36k + 12k = 2$.
$2 = 2$. This is true for all $k$. Thus,the line $L_2$ also lies in the same plane.
Since both lines lie in the same plane,they are either parallel or intersecting. The direction vectors are $(1, -1, -2)$ and $(12, 9, 4)$,which are not proportional,so they are not parallel.
Therefore,the lines must intersect. The shortest distance between two intersecting lines is $0$.

(B) Given $\vec{b} = 3\hat{j} + 4\hat{k}$ and $\vec{a} = \hat{i} + \hat{j}$.
$\vec{b_1}$ is parallel to $\vec{a}$,so $\vec{b_1} = \text{proj}_{\vec{a}} \vec{b} = \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a}$.
$\vec{b} \cdot \vec{a} = (3\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j}) = 3$.
$|\vec{a}|^2 = |\hat{i} + \hat{j}|^2 = 1^2 + 1^2 = 2$.
Thus,$\vec{b_1} = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}$.
Since $\vec{b} = \vec{b_1} + \vec{b_2}$,we have $\vec{b_2} = \vec{b} - \vec{b_1} = (3\hat{j} + 4\hat{k}) - (\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j}) = -\frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} + 4\hat{k}$.
Now,calculate the cross product $\vec{b_1} \times \vec{b_2}$:
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3/2 & 3/2 & 0 \\ -3/2 & 3/2 & 4 \end{vmatrix}$.
$= \hat{i} \left( (3/2)(4) - (0)(3/2) \right) - \hat{j} \left( (3/2)(4) - (0)(-3/2) \right) + \hat{k} \left( (3/2)(3/2) - (3/2)(-3/2) \right)$.
$= \hat{i}(6) - \hat{j}(6) + \hat{k}(9/4 + 9/4) = 6\hat{i} - 6\hat{j} + \frac{9}{2}\hat{k}$.

(A) Given that $E$ and $F$ are independent events,we have $P(E \cap F) = P(E) \cdot P(F) = \frac{1}{12}$.
Also,$P(\bar{E} \cap \bar{F}) = P(\bar{E}) \cdot P(\bar{F}) = \frac{1}{2}$.
Let $P(E) = x$ and $P(F) = y$. Then $xy = \frac{1}{12}$.
The second equation becomes $(1-x)(1-y) = \frac{1}{2}$,which simplifies to $1 - (x+y) + xy = \frac{1}{2}$.
Substituting $xy = \frac{1}{12}$,we get $1 - (x+y) + \frac{1}{12} = \frac{1}{2}$.
Thus,$x+y = 1 + \frac{1}{12} - \frac{1}{2} = \frac{12+1-6}{12} = \frac{7}{12}$.
Now,$x$ and $y$ are roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{7}{12}t + \frac{1}{12} = 0$.
Multiplying by $12$,we get $12t^2 - 7t + 1 = 0$.
Factoring,$(4t-1)(3t-1) = 0$,so $t = \frac{1}{4}$ or $t = \frac{1}{3}$.
If $P(E) = \frac{1}{3}$,then $P(F) = \frac{1}{4}$,so $\frac{P(E)}{P(F)} = \frac{1/3}{1/4} = \frac{4}{3}$.
If $P(E) = \frac{1}{4}$,then $P(F) = \frac{1}{3}$,so $\frac{P(E)}{P(F)} = \frac{1/4}{1/3} = \frac{3}{4}$.
Since $\frac{4}{3}$ is an option,the correct answer is $A$.

(A) Let $\lambda = \cot^{-1}(1+x)$,then $\cot \lambda = 1+x$. From the right-angled triangle with base $(1+x)$ and perpendicular $1$,the hypotenuse is $\sqrt{(1+x)^2 + 1^2} = \sqrt{x^2 + 2x + 2}$. Thus,$\sin \lambda = \frac{1}{\sqrt{x^2 + 2x + 2}}$.
Let $\beta = \tan^{-1}x$,then $\tan \beta = x$. From the right-angled triangle with perpendicular $x$ and base $1$,the hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$. Thus,$\cos \beta = \frac{1}{\sqrt{x^2 + 1}}$.
Given the equation $\sin \lambda = \cos \beta$,we have:
$\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}}$
Squaring both sides:
$x^2 + 2x + 2 = x^2 + 1$
Subtracting $x^2$ from both sides:
$2x + 2 = 1$
$2x = -1$
$x = -\frac{1}{2}$

(B) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $f(\frac{\pi}{2}) = \lim_{x \to \frac{\pi}{2}} f(x)$.
First,evaluate the limit: $\lim_{x \to \frac{\pi}{2}} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}$.
Let $x = \frac{\pi}{2} + h$,where $h \to 0$. Then $\tan 4x = \tan(4(\frac{\pi}{2} + h)) = \tan(2\pi + 4h) = \tan 4h \approx 4h$.
And $\tan 5x = \tan(5(\frac{\pi}{2} + h)) = \tan(\frac{5\pi}{2} + 5h) = \cot 5h \approx \frac{1}{5h}$.
Thus,the exponent is $\frac{\tan 4x}{\tan 5x} = \frac{\tan 4h}{\cot 5h} = \tan 4h \cdot \tan 5h \to 0 \cdot \infty$ form.
Actually,$\lim_{x \to \frac{\pi}{2}} \frac{\tan 4x}{\tan 5x} = \lim_{h \to 0} \frac{\tan 4h}{\cot 5h} = \lim_{h \to 0} \tan 4h \tan 5h = 0 \cdot 0 = 0$.
Therefore,$\lim_{x \to \frac{\pi}{2}} f(x) = (\frac{4}{5})^0 = 1$.
Equating to $f(\frac{\pi}{2})$,we get $k + \frac{2}{5} = 1$.
$k = 1 - \frac{2}{5} = \frac{3}{5}$.

(D) Given the differential equation: $y \, dx - (x + 3y^2) \, dy = 0$
Rearranging the terms: $y \, dx = (x + 3y^2) \, dy$
Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x + 3y^2}{y} = \frac{x}{y} + 3y$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.
The Integrating Factor $(IF)$ is given by: $IF = e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF \, dy + c$.
Substituting the values: $x \cdot \frac{1}{y} = \int 3y \cdot \frac{1}{y} \, dy + c$
$\frac{x}{y} = \int 3 \, dy + c = 3y + c$
So,$x = 3y^2 + cy$.
Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$: $1 = 3(1)^2 + c(1) \Rightarrow 1 = 3 + c \Rightarrow c = -2$.
The equation of the curve is $x = 3y^2 - 2y$.
Checking option $(D)$ $(-\frac{1}{3}, \frac{1}{3})$: $x = 3(\frac{1}{3})^2 - 2(\frac{1}{3}) = 3(\frac{1}{9}) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.
Since the point satisfies the equation,the correct option is $(D)$.

(D) Let $y = \tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$.
We can rewrite the argument of $\tan^{-1}$ as follows:
$y = \tan^{-1}\left(\frac{2(3x\sqrt{x})}{1-(3x\sqrt{x})^2}\right)$.
Using the formula $2\tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,where $\theta = 3x\sqrt{x} = 3x^{3/2}$,we get:
$y = 2\tan^{-1}(3x^{3/2})$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3x^{3/2})^2} \cdot \frac{d}{dx}(3x^{3/2})$.
$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.
$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot \frac{9}{2} \sqrt{x} = \frac{9\sqrt{x}}{1+9x^3}$.
Given $\frac{dy}{dx} = \sqrt{x} \cdot g(x)$,we have $\sqrt{x} \cdot g(x) = \frac{9\sqrt{x}}{1+9x^3}$.
Therefore,$g(x) = \frac{9}{1+9x^3}$.

(A) Given $f(x) = ax^2 + bx + c$ and $a + b + c = 3$,we have $f(1) = a(1)^2 + b(1) + c = a + b + c = 3$.
Given the functional equation $f(x + y) = f(x) + f(y) + xy$.
Putting $y = 1$,we get $f(x + 1) = f(x) + f(1) + x$.
Substituting $f(1) = 3$,we have $f(x + 1) - f(x) = x + 3$.
Summing from $x = 1$ to $n - 1$:
$\sum_{x=1}^{n-1} (f(x+1) - f(x)) = \sum_{x=1}^{n-1} (x + 3)$.
This is a telescoping sum: $f(n) - f(1) = \frac{(n-1)n}{2} + 3(n-1)$.
Since $f(1) = 3$,we have $f(n) = 3 + \frac{n^2 - n}{2} + 3n - 3 = \frac{n^2 + 5n}{2}$.
Now,calculate $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (\frac{n^2}{2} + \frac{5n}{2})$.
Using summation formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:
$\sum_{n=1}^{10} f(n) = \frac{1}{2} \left( \frac{10 \cdot 11 \cdot 21}{6} \right) + \frac{5}{2} \left( \frac{10 \cdot 11}{2} \right)$.
$= \frac{385}{1} + \frac{275}{2} = 192.5 + 137.5 = 330$.