KCET 2020 Mathematics Question Paper with Answer and Solution

(A) We know that $\sin(90^{\circ} - \theta) = \cos \theta$.
Given expression: $\sin^{2} 51^{\circ} + \sin^{2} 39^{\circ}$.
We can write $39^{\circ}$ as $(90^{\circ} - 51^{\circ})$.
So,$\sin^{2} 39^{\circ} = \sin^{2}(90^{\circ} - 51^{\circ}) = \cos^{2} 51^{\circ}$.
Substituting this into the expression: $\sin^{2} 51^{\circ} + \cos^{2} 51^{\circ}$.
Using the identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we get $1$.

(C) We are given the statement $P(n): 2^{n} < n!$.
We test for positive integer values of $n$:
For $n=1$: $2^{1} = 2$ and $1! = 1$. Since $2 < 1$ is false,$P(1)$ is false.
For $n=2$: $2^{2} = 4$ and $2! = 2$. Since $4 < 2$ is false,$P(2)$ is false.
For $n=3$: $2^{3} = 8$ and $3! = 6$. Since $8 < 6$ is false,$P(3)$ is false.
For $n=4$: $2^{4} = 16$ and $4! = 24$. Since $16 < 24$ is true,$P(4)$ is true.
Therefore,the smallest positive integer for which $P(n)$ is true is $n=4$.

(D) Given,$z=x+iy$.
The equation is $|z+1|=|z-1|$.
Squaring both sides,we get $|z+1|^2 = |z-1|^2$.
Substituting $z=x+iy$,we have $|(x+1)+iy|^2 = |(x-1)+iy|^2$.
This implies $(x+1)^2 + y^2 = (x-1)^2 + y^2$.
Simplifying,$(x+1)^2 - (x-1)^2 = 0$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get $(x+1-x+1)(x+1+x-1) = 0$.
$(2)(2x) = 0$,which gives $4x = 0$,so $x = 0$.
The equation $x=0$ represents the $Y$-axis in the complex plane.

(C) Given set $A = \{1, 2, 3, 4, 5, 6\}$.
The total number of subsets of $A$ is $2^n$,where $n = 6$,so $2^6 = 64$.
The subsets containing at least two elements are the total subsets minus the subsets with zero elements (empty set) and the subsets with exactly one element.
Number of subsets with zero elements = ${}^6C_0 = 1$.
Number of subsets with one element = ${}^6C_1 = 6$.
Therefore,the number of subsets with at least two elements = $64 - (1 + 6) = 64 - 7 = 57$.

(A) Given,$\tan A + \cot A = 2$.
Squaring both sides,we get:
$(\tan A + \cot A)^{2} = 2^{2}$
$\tan^{2} A + \cot^{2} A + 2 \tan A \cot A = 4$
Since $\tan A \cot A = 1$,we have:
$\tan^{2} A + \cot^{2} A + 2 = 4$
$\tan^{2} A + \cot^{2} A = 2$
Now,squaring both sides again:
$(\tan^{2} A + \cot^{2} A)^{2} = 2^{2}$
$\tan^{4} A + \cot^{4} A + 2 \tan^{2} A \cot^{2} A = 4$
$\tan^{4} A + \cot^{4} A + 2(1)^{2} = 4$
$\tan^{4} A + \cot^{4} A = 4 - 2 = 2$.

(B) Given,$P(A) = \frac{1}{2}, P(B) = \frac{1}{4}, P(C) = \frac{1}{3}$.
Therefore,the probabilities of not solving the problem are $P(\bar{A}) = 1 - \frac{1}{2} = \frac{1}{2}$,$P(\bar{B}) = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(\bar{C}) = 1 - \frac{1}{3} = \frac{2}{3}$.
The problem is solved by exactly two persons if ($A$ and $B$ solve,$C$ fails) or ($A$ and $C$ solve,$B$ fails) or ($B$ and $C$ solve,$A$ fails).
Required Probability $= P(A)P(B)P(\bar{C}) + P(A)P(\bar{B})P(C) + P(\bar{A})P(B)P(C)$.
$= (\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{1}{3}) + (\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3})$.
$= \frac{2}{24} + \frac{3}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$.

(A) The number of terms in the expansion of $(x_1 + x_2 + \dots + x_r)^n$ is given by the formula $\binom{n+r-1}{r-1}$.
Here,$n = 10$ and $r = 3$.
Substituting these values,we get:
$\binom{10+3-1}{3-1} = \binom{12}{2}$.
Calculating the value: $\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66$.

(C) Given the sum of $n$ terms of an $AP$ is $S_{n} = n^{2} + n$.
We know that the first term $a_{1} = S_{1} = 1^{2} + 1 = 2$.
The sum of the first two terms is $S_{2} = 2^{2} + 2 = 4 + 2 = 6$.
The second term $a_{2} = S_{2} - S_{1} = 6 - 2 = 4$.
The common difference $d = a_{2} - a_{1} = 4 - 2 = 2$.

(A) The equation of the first line is $lx + my = n$,which can be written as $y = -\frac{l}{m}x + \frac{n}{m}$. The slope $m_1 = -\frac{l}{m}$.
The equation of the second line is $l'x + m'y = n'$,which can be written as $y = -\frac{l'}{m'}x + \frac{n'}{m'}$. The slope $m_2 = -\frac{l'}{m'}$.
Two lines are perpendicular if the product of their slopes is $-1$,i.e.,$m_1 \times m_2 = -1$.
$\left(-\frac{l}{m}\right) \times \left(-\frac{l'}{m'}\right) = -1$
$\frac{ll'}{mm'} = -1$
$ll' = -mm'$
$ll' + mm' = 0$.

(B) The equation of the parabola is $x^{2} = 4ay$.
Since the parabola passes through the point $(2, 1)$,we substitute $x = 2$ and $y = 1$ into the equation:
$(2)^{2} = 4a(1)$
$4 = 4a$
The length of the latus rectum of the parabola $x^{2} = 4ay$ is defined as $4a$.
From the equation $4 = 4a$,we can see that the length of the latus rectum is $4$.

(C) The given statement is a universal quantification: "For all $x, y \in \mathbb{R}, x+y=y+x$."
To find the negation of a statement involving the universal quantifier "For all" $(\forall)$,we replace it with the existential quantifier "There exists" (or "For some") and negate the predicate.
The negation of "For all $x$ and $y, P(x, y)$" is "There exist $x$ and $y$ such that $\neg P(x, y)$."
Here,the predicate $P(x, y)$ is $x+y=y+x$.
The negation of $x+y=y+x$ is $x+y \neq y+x$.
Therefore,the negation of the statement is "For some real numbers $x$ and $y, x+y \neq y+x$."

(D) Given that $n(A) = 2$. Let $n(B) = m$.
The total number of relations from set $A$ to set $B$ is given by the formula $2^{n(A) \times n(B)}$.
According to the problem,$2^{2 \times m} = 1024$.
Since $1024 = 2^{10}$,we have $2^{2m} = 2^{10}$.
Equating the exponents,$2m = 10$,which gives $m = 5$.
Therefore,$n(B) = 5$.

(C) Since $A, B, C$ are mutually exclusive and exhaustive events,their sum of probabilities must be $1$:
$P(A) + P(B) + P(C) = 1$
Given that $P(A) = 2P(B) = 3P(C)$,we can express $P(A)$ and $P(C)$ in terms of $P(B)$:
$P(A) = 2P(B)$
$P(C) = \frac{2P(B)}{3}$
Substituting these into the sum equation:
$2P(B) + P(B) + \frac{2P(B)}{3} = 1$
Multiply the entire equation by $3$ to clear the denominator:
$6P(B) + 3P(B) + 2P(B) = 3$
$11P(B) = 3$
$P(B) = \frac{3}{11}$

(A) Given data: $6, 7, 8, 9, 10$
Mean $(\bar{x})$ = $\frac{6+7+8+9+10}{5} = \frac{40}{5} = 8$
Standard Deviation $(SD)$ = $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$
$SD$ = $\sqrt{\frac{(6-8)^2 + (7-8)^2 + (8-8)^2 + (9-8)^2 + (10-8)^2}{5}}$
$SD$ = $\sqrt{\frac{(-2)^2 + (-1)^2 + (0)^2 + (1)^2 + (2)^2}{5}}$
$SD$ = $\sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$

(B) Given $f(x) = \begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ 0, & x=0 \end{cases}$
Right hand limit $(RHL)$:
$\lim_{x \rightarrow 0^{+}} f(x) = \lim_{h \rightarrow 0} f(0+h) = \lim_{h \rightarrow 0} \frac{e^{1/h}-1}{e^{1/h}+1}$
Dividing numerator and denominator by $e^{1/h}$:
$\lim_{h \rightarrow 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$
Left hand limit $(LHL)$:
$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{h \rightarrow 0} f(0-h) = \lim_{h \rightarrow 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$
As $h \rightarrow 0^{+}$,$e^{-1/h} \rightarrow 0$:
$\frac{0 - 1}{0 + 1} = -1$
Thus,the $RHL$ is $1$ and the $LHL$ is $-1$.

(A) We use the property of combinations: ${ }^{n} C_{r} = { }^{n} C_{n-r}$.
Applying this to the terms ${ }^{16} C_{6}$ and ${ }^{16} C_{7}$:
${ }^{16} C_{6} = { }^{16} C_{16-6} = { }^{16} C_{10}$
${ }^{16} C_{7} = { }^{16} C_{16-7} = { }^{16} C_{9}$
Substituting these into the original expression:
${ }^{16} C_{9} + { }^{16} C_{10} - { }^{16} C_{6} - { }^{16} C_{7} = { }^{16} C_{9} + { }^{16} C_{10} - { }^{16} C_{10} - { }^{16} C_{9}$
$= ({ }^{16} C_{9} - { }^{16} C_{9}) + ({ }^{16} C_{10} - { }^{16} C_{10}) = 0 + 0 = 0$.

(C) In a three-dimensional Cartesian coordinate system,the octants are determined by the signs of the coordinates $(x, y, z)$.
For the point $(1, -3, 4)$,we have $x > 0$,$y < 0$,and $z > 0$.
The octants are defined as follows:
$I: (+, +, +)$
$II: (-, +, +)$
$III: (-, -, +)$
$IV: (+, -, +)$
$V: (+, +, -)$
$VI: (-, +, -)$
$VII: (-, -, -)$
$VIII: (+, -, -)$
Since the signs are $(+, -, +)$,the point $(1, -3, 4)$ lies in the $IV$ octant.

(A) We have,$\lim _{x \rightarrow 0} \left( \frac{\tan x}{\sqrt{2x+4}-2} \right)$.
To evaluate this limit,we rationalize the denominator:
$= \lim _{x \rightarrow 0} \frac{\tan x (\sqrt{2x+4}+2)}{(\sqrt{2x+4}-2)(\sqrt{2x+4}+2)}$
$= \lim _{x \rightarrow 0} \frac{\tan x (\sqrt{2x+4}+2)}{(2x+4)-4}$
$= \lim _{x \rightarrow 0} \frac{\tan x (\sqrt{2x+4}+2)}{2x}$
$= \lim _{x \rightarrow 0} \left( \frac{\tan x}{x} \right) \times \frac{\sqrt{2x+4}+2}{2}$
Using the standard limit $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$:
$= 1 \times \frac{\sqrt{2(0)+4}+2}{2}$
$= \frac{\sqrt{4}+2}{2} = \frac{2+2}{2} = \frac{4}{2} = 2$.

(A) Since $E_{1}$ and $E_{2}$ form a partition of the sample space $S$,we have $P(E_{1}) + P(E_{2}) = 1$.
Given $P(E_{1}) = P(E_{2}) = \frac{1}{2}$.
By the definition of conditional probability,$P(E_{1} | A) + P(E_{2} | A) = 1$ because $E_{1}$ and $E_{2}$ are exhaustive and mutually exclusive events.
Given $P(E_{2} | A) = \frac{1}{2}$.
Therefore,$P(E_{1} | A) = 1 - P(E_{2} | A) = 1 - \frac{1}{2} = \frac{1}{2}$.

(D) Let the common difference of the $AP$ be $d$. Then $a_{n} = a_{1} + (n-1)d$.
Applying the row operation $R_{1} \rightarrow R_{1} + R_{3} - 2R_{2}$:
The elements of the first row become:
$a_{1} + a_{7} - 2a_{4} = (a_{1}) + (a_{1} + 6d) - 2(a_{1} + 3d) = 2a_{1} + 6d - 2a_{1} - 6d = 0$.
$a_{2} + a_{8} - 2a_{5} = (a_{1} + d) + (a_{1} + 7d) - 2(a_{1} + 4d) = 2a_{1} + 8d - 2a_{1} - 8d = 0$.
$a_{3} + a_{9} - 2a_{6} = (a_{1} + 2d) + (a_{1} + 8d) - 2(a_{1} + 5d) = 2a_{1} + 10d - 2a_{1} - 10d = 0$.
Since all elements of the first row are $0$,the value of the determinant is $0$.

(A) Given $P(A)=\frac{1}{3}$,$P(B)=\frac{1}{2}$ and $P(A \cap B)=\frac{1}{6}$.
We need to find $P(A^{\prime} | B)$.
Using the formula for conditional probability,$P(A^{\prime} | B) = \frac{P(A^{\prime} \cap B)}{P(B)}$.
Since $P(A^{\prime} \cap B) = P(B) - P(A \cap B)$,
$P(A^{\prime} | B) = \frac{P(B) - P(A \cap B)}{P(B)}$.
Substituting the values:
$P(A^{\prime} | B) = \frac{\frac{1}{2} - \frac{1}{6}}{\frac{1}{2}} = \frac{\frac{3-1}{6}}{\frac{1}{2}} = \frac{\frac{2}{6}}{\frac{1}{2}} = \frac{1}{3} \times 2 = \frac{2}{3}$.

(D) binary operation on a set $A$ is a function from $A \times A$ to $A$.
If the number of elements in set $A$ is $n$,then the number of elements in $A \times A$ is $n^{2}$.
The number of binary operations on set $A$ is given by the formula $n^{(n^{2})}$.
Here,the set $A = \{a, b, c\}$,so the number of elements $n = 3$.
Substituting $n = 3$ into the formula,we get the number of binary operations as $3^{(3^{2})} = 3^{9}$.

(B) Given the equation $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A = I$,where $I$ is the identity matrix.
Let $B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$. Then $BA = I$,which implies $A = B^{-1}$.
The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Here,$ad-bc = (2)(2) - (1)(3) = 4 - 3 = 1$.
Thus,$B^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.
Therefore,$A = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.

(D) We know that the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$.
Here,the argument is $\frac{\pi}{3}$.
Since $\pi \approx 3.14159$,we have $\frac{\pi}{3} \approx 1.047$.
Because $1.047 > 1$,the value $\frac{\pi}{3}$ lies outside the domain $[-1, 1]$.
Therefore,$\sin ^{-1} \left(\frac{\pi}{3}\right)$ and $\cos ^{-1} \left(\frac{\pi}{3}\right)$ are not defined in the set of real numbers.
Thus,the expression $\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)$ does not exist.

(C) Given $A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$.
First,we calculate $A^{2} = A \cdot A$:
$A^{2} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Now,we calculate $A^{4} = A^{2} \cdot A^{2}$:
$A^{4} = I \cdot I = I$.
Therefore,$A^{4} = I$.

(D) Given that $B$ is a skew-symmetric matrix,we have $B^{\prime} = -B$.
To check if $A^{\prime} B A$ is symmetric or skew-symmetric,we take its transpose:
$(A^{\prime} B A)^{\prime} = A^{\prime} B^{\prime} (A^{\prime})^{\prime}$
Using the property $(XYZ)^{\prime} = Z^{\prime} Y^{\prime} X^{\prime}$,we get:
$(A^{\prime} B A)^{\prime} = A^{\prime} B^{\prime} A$
Since $B^{\prime} = -B$,we substitute this into the expression:
$(A^{\prime} B A)^{\prime} = A^{\prime} (-B) A = -(A^{\prime} B A)$
Since the transpose of the matrix $A^{\prime} B A$ is equal to its negative,$A^{\prime} B A$ is a skew-symmetric matrix.

(B) Given the function $f(x) = x^{2} - 4x + 5$ defined on the domain $[2, \infty)$.
We can rewrite the function by completing the square:
$f(x) = (x^{2} - 4x + 4) + 1$
$f(x) = (x - 2)^{2} + 1$
Since the domain is $x \in [2, \infty)$,we have $x - 2 \geq 0$.
Therefore,$(x - 2)^{2} \geq 0$.
Adding $1$ to both sides,we get $(x - 2)^{2} + 1 \geq 1$.
Thus,$f(x) \geq 1$.
The range of the function $f$ is $[1, \infty)$.

(C) Let $A = \{1, 2, 3\}$.
For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Since $(2, 2) \notin R$ and $(3, 3) \notin R$,$R$ is not reflexive.
For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. Here,$(1, 1) \in R$ implies $(1, 1) \in R$,which holds true. Thus,$R$ is symmetric.
For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. Since there is only one element $(1, 1)$,the condition holds vacuously. Thus,$R$ is transitive.
Therefore,$R$ is symmetric and transitive.

(A) We have the function $f(x) = \cos^{-1} \sqrt{x-1}$.
For the function $\cos^{-1}(u)$ to be defined,the argument $u$ must satisfy $u \in [-1, 1]$.
Since $\sqrt{x-1}$ is always non-negative,the condition becomes $0 \leq \sqrt{x-1} \leq 1$.
Squaring the inequality,we get $0 \leq x-1 \leq 1$.
Adding $1$ to all parts,we obtain $1 \leq x \leq 2$.
Therefore,the domain of $f(x)$ is $[1, 2]$.

(C) We have,$f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \\ x - a & x^2 - x & c + x \\ x - b & x - c & 0 \end{array} \right|$.
To find $f(0)$,we substitute $x = 0$ into the determinant:
$f(0) = \left| \begin{array}{ccc} 0^3 - 0 & a + 0 & b + 0 \\ 0 - a & 0^2 - 0 & c + 0 \\ 0 - b & 0 - c & 0 \end{array} \right| = \left| \begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array} \right|$.
Let $A = \left[ \begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array} \right]$.
Since $A^T = \left[ \begin{array}{ccc} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{array} \right] = -A$,the matrix $A$ is a skew-symmetric matrix of order $3$.
The determinant of a skew-symmetric matrix of odd order is always $0$.
Therefore,$f(0) = 0$.

(B) We know that for a square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A \text{ adj. } A = |A| I$,where $I$ is the identity matrix of order $n$.
Taking the determinant on both sides,we get $|A \text{ adj. } A| = | |A| I |$.
Since $|kI| = k^n |I|$ for a matrix of order $n$,we have $|A \text{ adj. } A| = |A|^n |I|$.
Since $|I| = 1$,the formula becomes $|A \text{ adj. } A| = |A|^n$.
Given that $n = 3$ and $|A| = 5$,we substitute these values into the formula:
$|A \text{ adj. } A| = (5)^3 = 125$.

(B) Given equation is $2^{x}+2^{y}=2^{x+y} \quad ...(i)$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2^{x}) + \frac{d}{dx}(2^{y}) = \frac{d}{dx}(2^{x+y})$
$2^{x} \ln 2 + 2^{y} \ln 2 \cdot \frac{dy}{dx} = 2^{x+y} \ln 2 \cdot (1 + \frac{dy}{dx})$
Dividing by $\ln 2$:
$2^{x} + 2^{y} \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$2^{y} \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^{x}$
$\frac{dy}{dx} (2^{y} - 2^{x+y}) = 2^{x+y} - 2^{x}$
From equation $(i)$,we know $2^{x+y} = 2^{x} + 2^{y}$. Substituting this:
$\frac{dy}{dx} (2^{y} - (2^{x} + 2^{y})) = (2^{x} + 2^{y}) - 2^{x}$
$\frac{dy}{dx} (-2^{x}) = 2^{y}$
$\frac{dy}{dx} = -\frac{2^{y}}{2^{x}} = -2^{y-x}$

(D) Given that $f(x)$ is continuous at $x=0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Therefore,$\lim_{x \to 0} \frac{1-\cos Kx}{x \sin x} = \frac{1}{2}$.
Using the trigonometric identity $1-\cos \theta = 2 \sin^2(\frac{\theta}{2})$,we get:
$\lim_{x \to 0} \frac{2 \sin^2(\frac{Kx}{2})}{x \sin x} = \frac{1}{2}$.
Dividing numerator and denominator by $x^2$,we get:
$\lim_{x \to 0} \frac{2 \left(\frac{\sin(Kx/2)}{x}\right)^2}{\frac{\sin x}{x}} = \frac{1}{2}$.
Multiplying and dividing by $(\frac{K}{2})^2$ in the numerator:
$\lim_{x \to 0} \frac{2 \cdot (\frac{K}{2})^2 \cdot (\frac{\sin(Kx/2)}{Kx/2})^2}{\frac{\sin x}{x}} = \frac{1}{2}$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we have:
$2 \cdot \frac{K^2}{4} \cdot \frac{1^2}{1} = \frac{1}{2}$.
$\frac{K^2}{2} = \frac{1}{2} \implies K^2 = 1$.
Thus,$K = \pm 1$.

(B) We have,$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$.
Let $x=\tan \theta$,then $\theta=\tan ^{-1} x$.
Substituting this into the function,we get $f(x)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$.
Since $\sin 2\theta = \frac{2 \tan \theta}{1+\tan ^{2} \theta}$,we have $f(x)=\sin ^{-1}(\sin 2 \theta) = 2 \theta = 2 \tan ^{-1} x$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = \frac{2}{1+x^{2}}$.
Substituting $x=\sqrt{3}$,we get $f^{\prime}(\sqrt{3}) = \frac{2}{1+(\sqrt{3})^{2}} = \frac{2}{1+3} = \frac{2}{4} = \frac{1}{2}$.

(C) The line $y=2x+1$ intersects the $X$-axis at $x=-\frac{1}{2}$.
For $x \in [-1, -\frac{1}{2}]$,$y \le 0$,and for $x \in [-\frac{1}{2}, 1]$,$y \ge 0$.
The required area is given by:
$\text{Area} = \int_{-1}^{1} |y| dx = \int_{-1}^{-1/2} -(2x+1) dx + \int_{-1/2}^{1} (2x+1) dx$
$= -[x^2+x]_{-1}^{-1/2} + [x^2+x]_{-1/2}^{1}$
$= -[(\frac{1}{4} - \frac{1}{2}) - (1 - 1)] + [(1+1) - (\frac{1}{4} - \frac{1}{2})]$
$= -[-\frac{1}{4}] + [2 - (-\frac{1}{4})]$
$= \frac{1}{4} + 2 + \frac{1}{4} = \frac{1}{2} + 2 = \frac{5}{2} \text{ sq units}$.

(B) Given,$y = 2 x^{n+1} + 3 x^{-n} \dots (i)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2}$
$\frac{d^2y}{dx^2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}$
$\frac{d^2y}{dx^2} = n(n+1) [2x^{n-1} + 3x^{-n-2}]$
Multiplying both sides by $x^2$:
$x^2 \frac{d^2y}{dx^2} = n(n+1) [2x^{n+1} + 3x^{-n}]$
Substituting $y$ from equation $(i)$:
$x^2 \frac{d^2y}{dx^2} = n(n+1)y$

(A) Given equation is $(x e)^{y}=e^{x}$.
Taking natural logarithm on both sides,we get:
$y \log(x e) = x \log e$
Since $\log(x e) = \log x + \log e$ and $\log e = 1$,we have:
$y(\log x + 1) = x$
$y = \frac{x}{\log x + 1}$
Differentiating with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
$\frac{dy}{dx} = \frac{(\log x + 1)(1) - x(\frac{1}{x} + 0)}{(\log x + 1)^2}$
$\frac{dy}{dx} = \frac{\log x + 1 - 1}{(\log x + 1)^2}$
$\frac{dy}{dx} = \frac{\log x}{(\log x + 1)^2}$

(D) Given curves are $2x = y^2$ $(i)$ and $2xy = K$ $(ii)$.
Solving $(i)$ and $(ii)$,substitute $2x = y^2$ into $(ii)$:
$y^2 \cdot y = K \Rightarrow y^3 = K \Rightarrow y = K^{1/3}$.
Then $2x = (K^{1/3})^2 = K^{2/3} \Rightarrow x = \frac{K^{2/3}}{2}$.
The intersection point is $(\frac{K^{2/3}}{2}, K^{1/3})$.
Differentiating $(i)$ w.r.t. $x$: $2 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{y}$.
Slope $m_1$ at the intersection point is $\frac{1}{K^{1/3}}$.
Differentiating $(ii)$ w.r.t. $x$: $2(y + x \frac{dy}{dx}) = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
Slope $m_2$ at the intersection point is $-\frac{K^{1/3}}{K^{2/3}/2} = -\frac{2K^{1/3}}{K^{2/3}} = -2K^{-1/3}$.
Since the curves intersect perpendicularly,$m_1 \cdot m_2 = -1$.
$(\frac{1}{K^{1/3}}) \cdot (-2K^{-1/3}) = -1$.
$-2K^{-2/3} = -1 \Rightarrow K^{-2/3} = \frac{1}{2}$.
$K^{2/3} = 2 \Rightarrow (K^{2/3})^3 = 2^3 \Rightarrow K^2 = 8$.

(D) Let the side of the cube be $x$. The surface area $S$ of the cube is given by $S = 6x^2$.
When the side $x$ is increased by $5 \%$,the new side becomes $x' = x + 0.05x = 1.05x$.
The new surface area $S'$ is given by $S' = 6(1.05x)^2$.
$S' = 6(1.1025x^2) = 1.1025(6x^2) = 1.1025S$.
The percentage increase in surface area is $\frac{S' - S}{S} \times 100 \%$.
$= \frac{1.1025S - S}{S} \times 100 \% = 0.1025 \times 100 \% = 10.25 \%$.

(A) We have,$I = \int e^{\sin x} \sin 2x \, dx$
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int e^{\sin x} (2 \sin x \cos x) \, dx = 2 \int e^{\sin x} \sin x \cos x \, dx$
Let $\sin x = t$,then $\cos x \, dx = dt$.
Substituting these into the integral:
$I = 2 \int e^t \cdot t \, dt$
Using integration by parts $\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$:
$I = 2 [t e^t - \int e^t \, dt] = 2 [t e^t - e^t] + C$
$I = 2 e^t (t - 1) + C$
Substituting $t = \sin x$ back:
$I = 2 e^{\sin x} (\sin x - 1) + C$

(B) We have,$I = \int \frac{1+x^{4}}{1+x^{6}} dx$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{\frac{1}{x^2} + x^2}{\frac{1}{x^2} + x^4} dx$.
Alternatively,we can write:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}}{1+x^{6}} dx - \int \frac{x^{2}}{1+x^{6}} dx$.
However,a simpler approach is:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx$ is not direct.
Let's use the substitution method:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{1+x^{6}} dx$.
Actually,the standard way is:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx$.
Using the identity $1+x^6 = (1+x^2)(1-x^2+x^4)$:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}+x^{4}-x^{2}}{(1+x^{2})(1-x^{2}+x^{4})} dx = \int \frac{1}{1+x^2} dx + \int \frac{x^2(x^2-1)}{1+x^6} dx$.
Following the provided steps:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx$ is complex.
Correct approach:
$\int \frac{1+x^{4}}{1+x^{6}} dx = \int \frac{1+x^{2}}{1+x^{6}} dx + \int \frac{x^{4}-x^{2}}{1+x^{6}} dx = \tan^{-1} x + \int \frac{x^2}{1+(x^3)^2} dx$.
Let $t = x^3$,then $dt = 3x^2 dx$,so $x^2 dx = \frac{dt}{3}$.
Thus,the integral becomes $\tan^{-1} x + \frac{1}{3} \int \frac{dt}{1+t^2} = \tan^{-1} x + \frac{1}{3} \tan^{-1} t + C = \tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + C$.

(C) Let $f(x) = \frac{\log _{e} x}{x}$.
To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log _{e} x) - \log _{e} x \cdot \frac{d}{dx}(x)}{x^{2}} = \frac{x \cdot \frac{1}{x} - \log _{e} x \cdot 1}{x^{2}} = \frac{1 - \log _{e} x}{x^{2}}$.
For critical points,set $f'(x) = 0$:
$1 - \log _{e} x = 0 \Rightarrow \log _{e} x = 1 \Rightarrow x = e$.
Now,we check the second derivative $f''(x)$ to confirm the maximum:
$f''(x) = \frac{x^{2}(-\frac{1}{x}) - (1 - \log _{e} x)(2x)}{x^{4}} = \frac{-x - 2x(1 - \log _{e} x)}{x^{4}}$.
At $x = e$,$f''(e) = \frac{-e - 2e(1 - 1)}{e^{4}} = \frac{-e}{e^{4}} = -\frac{1}{e^{3}} < 0$.
Since $f''(e) < 0$,the function has a local maximum at $x = e$.
The maximum value is $f(e) = \frac{\log _{e} e}{e} = \frac{1}{e}$.

(D) We have the integral $\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx$.
Using partial fractions,let $\frac{3x+1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$.
Multiplying both sides by $(x-1)(x-2)(x-3)$,we get:
$3x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$.
Setting $x=1$: $3(1)+1 = A(1-2)(1-3) \Rightarrow 4 = A(-1)(-2) \Rightarrow 4 = 2A \Rightarrow A = 2$.
Setting $x=2$: $3(2)+1 = B(2-1)(2-3) \Rightarrow 7 = B(1)(-1) \Rightarrow 7 = -B \Rightarrow B = -7$.
Setting $x=3$: $3(3)+1 = C(3-1)(3-2) \Rightarrow 10 = C(2)(1) \Rightarrow 10 = 2C \Rightarrow C = 5$.
Thus,the integral becomes $\int (\frac{2}{x-1} - \frac{7}{x-2} + \frac{5}{x-3}) dx$.
Integrating term by term,we get $2 \log |x-1| - 7 \log |x-2| + 5 \log |x-3| + K$.
Comparing this with $A \log |x-1| + B \log |x-2| + C \log |x-3| + K$,we find $A=2, B=-7, C=5$.

(B) We have,$I = \int_{-1 / 2}^{1 / 2} \cos ^{-1} x \, dx$.
Using the property $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$,we get:
$I = \int_{-1 / 2}^{1 / 2} (\frac{\pi}{2} - \sin^{-1} x) \, dx$.
$I = \int_{-1 / 2}^{1 / 2} \frac{\pi}{2} \, dx - \int_{-1 / 2}^{1 / 2} \sin^{-1} x \, dx$.
Since $\sin^{-1} x$ is an odd function,the integral $\int_{-a}^{a} \sin^{-1} x \, dx = 0$.
Therefore,$I = \frac{\pi}{2} [x]_{-1 / 2}^{1 / 2} - 0$.
$I = \frac{\pi}{2} (\frac{1}{2} - (- \frac{1}{2})) = \frac{\pi}{2} (1) = \frac{\pi}{2}$.

(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$.
Using the property $\int_{-a}^{a} f(x) d x = \int_{0}^{a} (f(x) + f(-x)) d x$,we get:
$I = \int_{0}^{\pi / 2} \left( \frac{\cos x}{1+e^{x}} + \frac{\cos(-x)}{1+e^{-x}} \right) d x$.
Since $\cos(-x) = \cos x$ and $\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1}$,the expression becomes:
$I = \int_{0}^{\pi / 2} \left( \frac{\cos x}{1+e^{x}} + \frac{e^x \cos x}{1+e^{x}} \right) d x$.
$I = \int_{0}^{\pi / 2} \frac{\cos x (1+e^x)}{1+e^x} d x$.
$I = \int_{0}^{\pi / 2} \cos x d x$.
$I = [\sin x]_{0}^{\pi / 2} = \sin(\pi / 2) - \sin(0) = 1 - 0 = 1$.

(D) Let $I = \int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.
Substitute $x = \tan \theta$,then $d x = \sec^{2} \theta d \theta$.
When $x = 0$,$\theta = 0$,and when $x = 1$,$\theta = \frac{\pi}{4}$.
Thus,$I = \int_{0}^{\pi/4} \frac{\log (1+\tan \theta)}{1+\tan^{2} \theta} (\sec^{2} \theta) d \theta = \int_{0}^{\pi/4} \log (1+\tan \theta) d \theta \quad ...(i)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have:
$I = \int_{0}^{\pi/4} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] d \theta$
$I = \int_{0}^{\pi/4} \log \left[1+\frac{1-\tan \theta}{1+\tan \theta}\right] d \theta = \int_{0}^{\pi/4} \log \left(\frac{2}{1+\tan \theta}\right) d \theta$
$I = \int_{0}^{\pi/4} [\log 2 - \log (1+\tan \theta)] d \theta \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_{0}^{\pi/4} \log 2 d \theta = \log 2 [\theta]_{0}^{\pi/4} = \frac{\pi}{4} \log 2$.
Therefore,$I = \frac{\pi}{8} \log 2$.

(B) Given equations of the curve and the line are $y^{2}=8x$ and $y=2x$.
To find the intersection points,substitute $y=2x$ into $y^{2}=8x$:
$(2x)^{2}=8x$
$4x^{2}=8x$
$4x^{2}-8x=0$
$4x(x-2)=0$
So,$x=0$ and $x=2$.
When $x=0$,$y=0$. When $x=2$,$y=4$.
The intersection points are $(0,0)$ and $(2,4)$.
The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=2$:
$A = \int_{0}^{2} (\sqrt{8x} - 2x) dx$
$A = \int_{0}^{2} (2\sqrt{2}x^{1/2} - 2x) dx$
$A = 2\sqrt{2} \int_{0}^{2} x^{1/2} dx - 2 \int_{0}^{2} x dx$
$A = 2\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} - 2 \left[ \frac{x^{2}}{2} \right]_{0}^{2}$
$A = 2\sqrt{2} \cdot \frac{2}{3} [x^{3/2}]_{0}^{2} - [x^{2}]_{0}^{2}$
$A = \frac{4\sqrt{2}}{3} (2^{3/2}) - (2^{2})$
$A = \frac{4\sqrt{2}}{3} (2\sqrt{2}) - 4$
$A = \frac{4 \cdot 2 \cdot 2}{3} - 4 = \frac{16}{3} - 4 = \frac{16-12}{3} = \frac{4}{3}$ sq. units.

(A) Given the equation: $c_{1} y = (c_{2} + c_{3}) e^{x + c_{4}}$
We can simplify this by grouping the constants: $y = \left( \frac{c_{2} + c_{3}}{c_{1}} \right) e^{c_{4}} \cdot e^{x}$
Let $C = \left( \frac{c_{2} + c_{3}}{c_{1}} \right) e^{c_{4}}$,where $C$ is a single arbitrary constant.
Thus,the equation becomes $y = C e^{x}$.
Since there is only one independent arbitrary constant $C$,the order of the differential equation obtained by eliminating it is $1$.
Differentiating $y = C e^{x}$ with respect to $x$,we get $\frac{dy}{dx} = C e^{x}$.
Substituting $y = C e^{x}$ into the derivative,we get $\frac{dy}{dx} = y$.
This is a first-order differential equation,so the order is $1$.

(D) Given the slope of the tangent $\frac{dy}{dx} = \frac{3x}{y}$.
By separating the variables,we get $y \, dy = 3x \, dx$.
Integrating both sides,we have $\int y \, dy = \int 3x \, dx$,which gives $\frac{y^2}{2} = \frac{3x^2}{2} + C$.
Multiplying by $2$,we get $y^2 = 3x^2 + 2C$,or $y^2 - 3x^2 = K$ where $K = 2C$.
Since the curve passes through $(1, 2)$,we substitute $x=1$ and $y=2$ into the equation: $2^2 - 3(1)^2 = K$,so $4 - 3 = K$,which means $K = 1$.
The equation of the curve is $y^2 - 3x^2 = 1$.
This equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,which represents a Hyperbola.

(A) Given the differential equation: $x^{2} dy - 2xy dx = x^{4} \cos x dx$.
Dividing both sides by $x^{2} dx$ (assuming $x \neq 0$),we get:
$\frac{dy}{dx} - \frac{2}{x}y = x^{2} \cos x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{2}{x}$ and $Q = x^{2} \cos x$.
The integrating factor $(IF)$ is given by:
$IF = e^{\int P dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln |x|} = e^{\ln |x^{-2}|} = \frac{1}{x^{2}}$.
The general solution is $y \cdot IF = \int (Q \cdot IF) dx + c$.
Substituting the values:
$y \cdot \frac{1}{x^{2}} = \int (x^{2} \cos x) \cdot \frac{1}{x^{2}} dx + c$.
$\frac{y}{x^{2}} = \int \cos x dx + c$.
$\frac{y}{x^{2}} = \sin x + c$.
Multiplying by $x^{2}$,we get the general solution:
$y = x^{2} \sin x + cx^{2}$.

(D) Let $\vec{AB} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{AC} = \hat{i}+3\hat{j}+5\hat{k}$.
Let $M$ be the midpoint of side $BC$. The median through $A$ is represented by the vector $\vec{AM}$.
In $\triangle ABC$,by the triangle law of vector addition,the position vector of the midpoint $M$ of $BC$ relative to $A$ is given by the average of the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AM} = \frac{1}{2}(\vec{AB} + \vec{AC})$
Substituting the given vectors:
$\vec{AM} = \frac{1}{2}((\hat{i}+\hat{j}+\hat{k}) + (\hat{i}+3\hat{j}+5\hat{k}))$
$\vec{AM} = \frac{1}{2}(2\hat{i} + 4\hat{j} + 6\hat{k})$
$\vec{AM} = \hat{i} + 2\hat{j} + 3\hat{k}$
The length of the median is the magnitude of vector $\vec{AM}$:
$|\vec{AM}| = \sqrt{(1)^2 + (2)^2 + (3)^2}$
$|\vec{AM}| = \sqrt{1 + 4 + 9}$
$|\vec{AM}| = \sqrt{14}$

(C) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$.
We know that $|a-b|^2 = (a-b) \cdot (a-b) = |a|^2 - 2(a \cdot b) + |b|^2$.
Since $a \cdot b = |a||b| \cos \theta$,we have $|a-b|^2 = 1^2 - 2(1)(1) \cos \theta + 1^2 = 2 - 2 \cos \theta$.
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get $|a-b|^2 = 2(2 \sin^2 \frac{\theta}{2}) = 4 \sin^2 \frac{\theta}{2}$.
Taking the square root on both sides,we get $|a-b| = 2 \sin \frac{\theta}{2}$.
Therefore,$\sin \frac{\theta}{2} = \frac{|a-b|}{2}$.

(NONE) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
The given vectors are $\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$,$\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$,and $\vec{c} = \lambda \hat{i} - \hat{j} + 2 \hat{k}$.
The condition for coplanarity is the determinant of the components being zero:
$\begin{vmatrix} 2 & 3 & 4 \\ 2 & 1 & -1 \\ \lambda & -1 & 2 \end{vmatrix} = 0$
Expanding along the first row:
$2(1(2) - (-1)(-1)) - 3(2(2) - (-1)(\lambda)) + 4(2(-1) - 1(\lambda)) = 0$
$2(2 - 1) - 3(4 + \lambda) + 4(-2 - \lambda) = 0$
$2(1) - 12 - 3\lambda - 8 - 4\lambda = 0$
$2 - 12 - 8 - 7\lambda = 0$
$-18 - 7\lambda = 0$
$-7\lambda = 18$
$\lambda = -\frac{18}{7}$

(C) Given that $|a+b|^{2}+|a \cdot b|^{2}=144$ and $|a|=6$.
We know the identity $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$,but here we use the property related to vectors.
Actually,the expression $|a+b|^{2} + |a \cdot b|^{2}$ is not a standard identity.
Assuming the intended expression is $|a+b|^{2} + |a-b|^{2} = 2(|a|^{2} + |b|^{2})$ or similar,let us re-evaluate the given equation $|a+b|^{2} + |a \cdot b|^{2} = 144$.
If we assume the question implies $|a|^{2}|b|^{2} = 144$,then $|6|^{2}|b|^{2} = 144$.
$36|b|^{2} = 144$.
$|b|^{2} = \frac{144}{36} = 4$.
Therefore,$|b| = 2$.

(B) Let the given point be $A(1, 2, -4)$ and the line be $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6} = \lambda$.
Any point $P$ on the line is given by $(2\lambda+3, 3\lambda+3, 6\lambda-5)$.
The direction ratios of the line $AP$ are $(2\lambda+3-1, 3\lambda+3-2, 6\lambda-5-(-4))$,which simplifies to $(2\lambda+2, 3\lambda+1, 6\lambda-1)$.
Since $AP$ is perpendicular to the given line with direction ratios $(2, 3, 6)$,their dot product must be zero:
$2(2\lambda+2) + 3(3\lambda+1) + 6(6\lambda-1) = 0$
$4\lambda + 4 + 9\lambda + 3 + 36\lambda - 6 = 0$
$49\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{49}$.
The square of the distance $AP$ is $AP^2 = (2\lambda+2)^2 + (3\lambda+1)^2 + (6\lambda-1)^2$.
Substituting $\lambda = -\frac{1}{49}$:
$AP^2 = 49\lambda^2 + 2\lambda + 6 = 49(-\frac{1}{49})^2 + 2(-\frac{1}{49}) + 6$
$AP^2 = \frac{1}{49} - \frac{2}{49} + 6 = 6 - \frac{1}{49} = \frac{294-1}{49} = \frac{293}{49}$.
Therefore,$AP = \sqrt{\frac{293}{49}} = \frac{\sqrt{293}}{7}$.

(D) The given line is $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$. The direction vector of the line is $\vec{b} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
The given plane is $2x - 2y + z = 5$. The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + 1\hat{k}$.
Let $\theta$ be the angle between the line and the plane. The sine of the angle is given by $\sin \theta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} \right|$.
Calculating the dot product: $\vec{b} \cdot \vec{n} = (3)(2) + (4)(-2) + (5)(1) = 6 - 8 + 5 = 3$.
Calculating the magnitudes: $|\vec{b}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.
$|\vec{n}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,$\sin \theta = \left| \frac{3}{(5\sqrt{2})(3)} \right| = \frac{3}{15\sqrt{2}} = \frac{1}{5\sqrt{2}}$.
Rationalizing the denominator: $\frac{1}{5\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{10}$.

(A) Let the direction cosines of the line be $l, m, n$. Given that the line makes an angle $\alpha = \frac{\pi}{3}$ with both $X$ and $Y$-axes,we have $l = \cos(\frac{\pi}{3}) = \frac{1}{2}$ and $m = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
We know that $l^2 + m^2 + n^2 = 1$.
Substituting the values: $(\frac{1}{2})^2 + (\frac{1}{2})^2 + n^2 = 1$.
$\frac{1}{4} + \frac{1}{4} + n^2 = 1$ $\Rightarrow \frac{1}{2} + n^2 = 1$ $\Rightarrow n^2 = \frac{1}{2}$.
Thus,$n = \pm \frac{1}{\sqrt{2}}$.
For the acute angle $\gamma$ with the $Z$-axis,$\cos \gamma = |n| = \frac{1}{\sqrt{2}}$.
Therefore,$\gamma = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.

(B) The objective function is $z = px + qy$.
If the minimum value of $z$ occurs at two distinct points $(x_1, y_1)$ and $(x_2, y_2)$,then the value of $z$ at these points must be equal.
Given points are $(3, 0)$ and $(1, 1)$.
Equating the values of $z$ at these points:
$p(3) + q(0) = p(1) + q(1)$
$3p = p + q$
$2p = q$
$p = \frac{q}{2}$

(D) Given,maximize $z=11x+7y$.
The corner points of the feasible region are determined by the intersection of the lines and the axes.
$1$. The intersection of $x+y=5$ and $x+3y=9$ is found by subtracting the equations: $(x+3y)-(x+y) = 9-5 \Rightarrow 2y=4 \Rightarrow y=2$. Substituting $y=2$ into $x+y=5$ gives $x=3$. So,point $B$ is $(3,2)$.
$2$. The intersection of $x+3y=9$ with the $y$-axis $(x=0)$ is $(0,3)$. So,point $A$ is $(0,3)$.
$3$. The intersection of $x+y=5$ with the $y$-axis $(x=0)$ is $(0,5)$. So,point $C$ is $(0,5)$.
Now,we evaluate $z=11x+7y$ at these corner points:
At $A(0,3): z = 11(0) + 7(3) = 21$.
At $B(3,2): z = 11(3) + 7(2) = 33 + 14 = 47$.
At $C(0,5): z = 11(0) + 7(5) = 35$.
Comparing these values,the maximum value of $z$ is $47$,which occurs at $(3,2)$.

(C) Given,$n=10$.
Probability of getting an odd number in a single throw,$p = \frac{3}{6} = \frac{1}{2}$.
Probability of not getting an odd number,$q = 1 - p = \frac{1}{2}$.
We need to find the probability that an odd number appears at least once,which is $P(X \geq 1)$.
Using the complement rule,$P(X \geq 1) = 1 - P(X=0)$.
Using the binomial distribution formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:
$P(X=0) = {}^{10}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{10} = 1 \times 1 \times \frac{1}{1024} = \frac{1}{1024}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{1024} = \frac{1023}{1024}$.