KCET 2016 Mathematics Question Paper with Answer and Solution

(A) Given the expression: $i^{n} + i^{n+1} + i^{n+2} + i^{n+3}$
Factor out $i^{n}$ from the expression: $i^{n}(1 + i + i^{2} + i^{3})$
We know that $i^{2} = -1$ and $i^{3} = -i$.
Substituting these values: $i^{n}(1 + i - 1 - i)$
Simplifying the terms inside the parentheses: $i^{n}(0) = 0$
Therefore,the simplified form is $0$.

(A) Given the expression $(1-\cos \theta+i \sin \theta)^{-1} = \frac{1}{1-\cos \theta+i \sin \theta}$.
Using trigonometric identities $1-\cos \theta = 2 \sin^2(\theta/2)$ and $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$:
$\frac{1}{2 \sin^2(\theta/2) + i(2 \sin(\theta/2) \cos(\theta/2))} = \frac{1}{2 \sin(\theta/2) [\sin(\theta/2) + i \cos(\theta/2)]}$.
Multiplying the numerator and denominator by the conjugate $[\sin(\theta/2) - i \cos(\theta/2)]$:
$= \frac{\sin(\theta/2) - i \cos(\theta/2)}{2 \sin(\theta/2) [\sin^2(\theta/2) + \cos^2(\theta/2)]} = \frac{\sin(\theta/2) - i \cos(\theta/2)}{2 \sin(\theta/2)}$.
$= \frac{\sin(\theta/2)}{2 \sin(\theta/2)} - i \frac{\cos(\theta/2)}{2 \sin(\theta/2)} = \frac{1}{2} - i \frac{1}{2} \cot(\theta/2)$.
The real part is $1/2$.

(C) Given the infinite geometric series: $1+\sin \theta+\sin ^{2} \theta+\ldots \infty = 2 \sqrt{3}+4$.
The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$,where $a = 1$ and $r = \sin \theta$.
Thus,$\frac{1}{1-\sin \theta} = 2 \sqrt{3}+4$.
Taking the reciprocal: $1-\sin \theta = \frac{1}{2 \sqrt{3}+4}$.
Rationalizing the denominator: $1-\sin \theta = \frac{2 \sqrt{3}-4}{(2 \sqrt{3}+4)(2 \sqrt{3}-4)} = \frac{2 \sqrt{3}-4}{12-16} = \frac{2 \sqrt{3}-4}{-4}$.
Simplifying: $1-\sin \theta = -\frac{\sqrt{3}}{2} + 1$.
Therefore,$\sin \theta = \frac{\sqrt{3}}{2}$.
Since $\sin \theta = \frac{\sqrt{3}}{2}$,the value of $\theta$ is $\frac{\pi}{3}$.

(B) The $n^{th}$ term of the series is given by $t_n = \frac{1^2 + 2^2 + \ldots + n^2}{1 + 2 + \ldots + n}$.
Using the formulas for the sum of the first $n$ natural numbers and their squares:
$t_n = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{2n+1}{3}$.
Now,the sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} \frac{2k+1}{3}$.
$S_n = \frac{1}{3} [2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1]$.
$S_n = \frac{1}{3} [2 \cdot \frac{n(n+1)}{2} + n] = \frac{1}{3} [n(n+1) + n] = \frac{1}{3} [n^2 + n + n] = \frac{n(n+2)}{3}$.

(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
Here,$n=14$,$a=x$,$b=\frac{1}{\sqrt{x}}$,and we need the $11^{th}$ term,so $r+1=11$,which implies $r=10$.
Substituting these values:
$T_{11} = {}^{14}C_{10} (x)^{14-10} \left(\frac{1}{\sqrt{x}}\right)^{10}$
$T_{11} = {}^{14}C_{10} (x)^4 \left(\frac{1}{x^{1/2}}\right)^{10}$
$T_{11} = {}^{14}C_{10} (x)^4 \left(\frac{1}{x^5}\right)$
$T_{11} = {}^{14}C_{10} \cdot \frac{1}{x}$
Calculating ${}^{14}C_{10} = {}^{14}C_{4} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$.
Thus,$T_{11} = \frac{1001}{x}$.

(C) We know the formula $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$.
Setting $\theta = \frac{\pi}{4}$,we get $\tan \frac{\pi}{8} = \frac{\sin(\pi/4)}{1 + \cos(\pi/4)}$.
Substituting the values $\sin(\pi/4) = \frac{1}{\sqrt{2}}$ and $\cos(\pi/4) = \frac{1}{\sqrt{2}}$,we have:
$\tan \frac{\pi}{8} = \frac{1/\sqrt{2}}{1 + 1/\sqrt{2}} = \frac{1/\sqrt{2}}{(\sqrt{2}+1)/\sqrt{2}} = \frac{1}{\sqrt{2}+1}$.

(A) The given expression is $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$.
We know that $\sin(360^{\circ} - \theta) = -\sin \theta$.
Thus,$\sin 359^{\circ} = \sin(360^{\circ} - 1^{\circ}) = -\sin 1^{\circ}$.
Similarly,$\sin 358^{\circ} = -\sin 2^{\circ}$,and so on.
We can pair the terms as $(\sin 1^{\circ} + \sin 359^{\circ}) + (\sin 2^{\circ} + \sin 358^{\circ}) + \ldots + (\sin 179^{\circ} + \sin 181^{\circ}) + \sin 180^{\circ}$.
Since $\sin(180^{\circ} + \theta) = -\sin \theta$,we have $\sin 181^{\circ} = -\sin 1^{\circ}$,$\sin 182^{\circ} = -\sin 2^{\circ}$,etc.
Each pair sums to $0$,and $\sin 180^{\circ} = 0$.
Therefore,the total sum is $0$.

(B) Given that,$\cot \theta + \tan \theta = 2$
$\Rightarrow \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = 2$
$\Rightarrow \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = 2$
$\Rightarrow \frac{1}{\sin \theta \cos \theta} = 2$
$\Rightarrow 2 \sin \theta \cos \theta = 1$
$\Rightarrow \sin 2 \theta = 1 = \sin \left(\frac{\pi}{2}\right)$
As we know,if $\sin x = \sin \alpha$,then $x = n \pi + (-1)^n \alpha$
Therefore,$2 \theta = n \pi + (-1)^n \frac{\pi}{2}$
$\Rightarrow \theta = \frac{n \pi}{2} + (-1)^n \frac{\pi}{4}$

(C) The given equations of the lines are:
$2x + 3y - 3 = 0$ --- $(1)$
$x + ky + 7 = 0$ --- $(2)$
The slope of line $(1)$ is $m_1 = -\frac{2}{3}$.
The slope of line $(2)$ is $m_2 = -\frac{1}{k}$.
Since the lines are perpendicular,the product of their slopes must be $-1$:
$m_1 \times m_2 = -1$
$(-\frac{2}{3}) \times (-\frac{1}{k}) = -1$
$\frac{2}{3k} = -1$
$2 = -3k$
$k = -\frac{2}{3}$

(B) Given that,
$x = 2 + 3 \cos \theta \implies x - 2 = 3 \cos \theta \implies \cos \theta = \frac{x - 2}{3}$
$y = 1 - 3 \sin \theta \implies y - 1 = -3 \sin \theta \implies \sin \theta = \frac{y - 1}{-3}$
We know the identity $\sin^2 \theta + \cos^2 \theta = 1$.
Substituting the values,we get:
$\left(\frac{y - 1}{-3}\right)^2 + \left(\frac{x - 2}{3}\right)^2 = 1$
$\frac{(y - 1)^2}{9} + \frac{(x - 2)^2}{9} = 1$
$(x - 2)^2 + (y - 1)^2 = 9$
Comparing this with the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$,we get the centre $(h, k) = (2, 1)$ and radius $r = \sqrt{9} = 3$.

(D) Given the equation of the parabola: $4y^{2} + 3x + 3y + 1 = 0$.
Rearranging the terms to isolate $y$ terms: $4y^{2} + 3y = -3x - 1$.
Dividing by $4$: $y^{2} + \frac{3}{4}y = -\frac{3}{4}x - \frac{1}{4}$.
Completing the square on the left side: $(y + \frac{3}{8})^{2} - \frac{9}{64} = -\frac{3}{4}x - \frac{1}{4}$.
$(y + \frac{3}{8})^{2} = -\frac{3}{4}x - \frac{1}{4} + \frac{9}{64}$.
$(y + \frac{3}{8})^{2} = -\frac{3}{4}x - \frac{16}{64} + \frac{9}{64}$.
$(y + \frac{3}{8})^{2} = -\frac{3}{4}x - \frac{7}{64}$.
$(y + \frac{3}{8})^{2} = -\frac{3}{4}(x + \frac{7}{48})$.
Comparing this with the standard form $(y - k)^{2} = -4a(x - h)$,we get $4a = \frac{3}{4}$.
Thus,the length of the latus rectum is $4a = \frac{3}{4}$.

(C) We are given the limit: $\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}$.
By splitting the fraction,we get:
$\lim _{x \rightarrow 0} \left( \frac{x e^{x}}{x} - \frac{\sin x}{x} \right)$
$= \lim _{x \rightarrow 0} e^{x} - \lim _{x \rightarrow 0} \frac{\sin x}{x}$
Using the standard limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and evaluating $e^{0} = 1$:
$= 1 - 1 = 0$.

(D) Let $p: x$ is a prime number and $q: x$ is an odd number.
The given statement is $p \rightarrow q$.
The converse of the statement is $q \rightarrow p$.
The contrapositive of the converse $(q \rightarrow p)$ is $\sim p \rightarrow \sim q$.
Thus,the contrapositive of the converse is: "If $x$ is not a prime number,then $x$ is not odd."

(A) The coefficient of variation $(CV)$ is given by the formula: $CV = \frac{\sigma}{\bar{x}} \times 100$,where $\sigma$ is the standard deviation and $\bar{x}$ is the mean.
For the first distribution: $60 = \frac{21}{\bar{x}_1} \times 100 \Rightarrow \bar{x}_1 = \frac{2100}{60} = 35$.
For the second distribution: $70 = \frac{16}{\bar{x}_2} \times 100 \Rightarrow \bar{x}_2 = \frac{1600}{70} \approx 22.86$.
Thus,the means are $35$ and $22.86$ respectively.

(C) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcomes for obtaining a total score of $5$ are the pairs $(1, 4), (4, 1), (2, 3), \text{ and } (3, 2)$.
The number of favorable outcomes is $4$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes:
$P(E) = \frac{4}{36} = \frac{1}{9}$.

(B) The total number of cards in a pack is $52$. The number of Aces in a pack is $4$.
We need to select $2$ cards out of $52$,which can be done in ${}^{52}C_{2}$ ways.
The number of ways to select $2$ Aces out of $4$ is ${}^{4}C_{2}$.
The probability $P$ is given by:
$P = \frac{{}^{4}C_{2}}{{}^{52}C_{2}} = \frac{\frac{4 \times 3}{2 \times 1}}{\frac{52 \times 51}{2 \times 1}} = \frac{4 \times 3}{52 \times 51} = \frac{12}{2652} = \frac{1}{221}$.

(A) Let $f(x) = (\frac{1}{x})^x = x^{-x}$.
Taking the natural logarithm on both sides,we get $\ln(f(x)) = -x \ln(x)$.
To find the critical points,we differentiate with respect to $x$:
$\frac{d}{dx}(\ln(f(x))) = \frac{d}{dx}(-x \ln(x))$
$\frac{f'(x)}{f(x)} = -[\ln(x) + x \cdot \frac{1}{x}] = -(\ln(x) + 1)$.
Setting $f'(x) = 0$,we get $\ln(x) + 1 = 0$,which implies $\ln(x) = -1$,so $x = e^{-1} = \frac{1}{e}$.
Now,we check the second derivative or the sign of $f'(x)$ around $x = \frac{1}{e}$.
For $x < \frac{1}{e}$,$f'(x) > 0$ and for $x > \frac{1}{e}$,$f'(x) < 0$,so $x = \frac{1}{e}$ is a point of local maximum.
The maximum value is $f(\frac{1}{e}) = (\frac{1}{1/e})^{1/e} = e^{1/e}$.

(B) Given that,$a * b = \frac{a+b}{4}$.
For commutativity,we check $b * a = \frac{b+a}{4} = \frac{a+b}{4} = a * b$. Thus,the operation is commutative.
For associativity,we check $a * (b * c)$ and $(a * b) * c$.
$a * (b * c) = a * \left(\frac{b+c}{4}\right) = \frac{a + \frac{b+c}{4}}{4} = \frac{4a + b + c}{16}$.
$(a * b) * c = \left(\frac{a+b}{4}\right) * c = \frac{\frac{a+b}{4} + c}{4} = \frac{a + b + 4c}{16}$.
Since $\frac{4a + b + c}{16} \neq \frac{a + b + 4c}{16}$,the operation is not associative.
Therefore,the operation is commutative but not associative.

(D) Given matrices are $A = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(x\pi) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & \frac{1}{\pi} \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1}(x\pi) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & -\frac{1}{\pi} \tan^{-1}(\pi x) \end{bmatrix}$.
Subtracting $B$ from $A$:
$A-B = \begin{bmatrix} \frac{1}{\pi}(\sin^{-1}(x\pi) + \cos^{-1}(x\pi)) & \frac{1}{\pi}(\tan^{-1}(\frac{x}{\pi}) - \tan^{-1}(\frac{x}{\pi})) \\ \frac{1}{\pi}(\sin^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi})) & \frac{1}{\pi}(\cot^{-1}(\pi x) + \tan^{-1}(\pi x)) \end{bmatrix}$.
Using the identities $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$ and $\tan^{-1}(\theta) + \cot^{-1}(\theta) = \frac{\pi}{2}$,we get:
$A-B = \begin{bmatrix} \frac{1}{\pi} \cdot \frac{\pi}{2} & 0 \\ 0 & \frac{1}{\pi} \cdot \frac{\pi}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} I$.

(A) Given that $ A = \begin{bmatrix} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{bmatrix} $.
The transpose of $ A $ is $ A^{T} = \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{bmatrix} $.
Adding $ A $ and $ A^{T} $,we get:
$ A+A^{T} = \begin{bmatrix} \cos 2 \theta + \cos 2 \theta & -\sin 2 \theta + \sin 2 \theta \\ \sin 2 \theta - \sin 2 \theta & \cos 2 \theta + \cos 2 \theta \end{bmatrix} = \begin{bmatrix} 2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta \end{bmatrix} $.
This can be written as $ (2 \cos 2 \theta) I $,where $ I $ is the identity matrix.
Given $ A+A^{T} = I $,we have $ (2 \cos 2 \theta) I = I $.
Comparing the elements,we get $ 2 \cos 2 \theta = 1 $,which implies $ \cos 2 \theta = \frac{1}{2} $.
Since $ \cos \frac{\pi}{3} = \frac{1}{2} $,we have $ 2 \theta = \frac{\pi}{3} $,which gives $ \theta = \frac{\pi}{6} $.

(D) Let the order of matrix $B$ be $x \times y$.
Then the order of $B^{\prime}$ is $y \times x$.
For the product $AB^{\prime}$ to be defined,the number of columns in $A$ must equal the number of rows in $B^{\prime}$.
Since $A$ is $m \times n$,we have $n = y$.
For the product $B^{\prime}A$ to be defined,the number of columns in $B^{\prime}$ must equal the number of rows in $A$.
Thus,$x = m$.
Therefore,the order of matrix $B$ is $m \times n$.

(D) Given that,$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
First,we calculate $A^{2} = A \times A$:
$A^{2} = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
Next,we calculate $5A$:
$5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$.
Finally,we compute $A^{2} - 5A$:
$A^{2} - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$.
This can be written as $-7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -7I$.

(D) Given the determinant equation:
$\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$
Divide the first row by $x$,the second row by $y$,and the third row by $z$:
$xyz \left|\begin{array}{ccc}\frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}\right|=0$
Since $x, y, z \neq 0$,we can divide by $xyz$:
$\left|\begin{array}{ccc}\frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}\right|=0$
Apply the operation $C_1 \rightarrow C_1 + C_2 + C_3$:
$\left|\begin{array}{ccc}1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{x} & \frac{1}{x} \\ 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{y}+1 & \frac{1}{y} \\ 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}\right|=0$
Factor out $(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$:
$(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \left|\begin{array}{ccc}1 & \frac{1}{x} & \frac{1}{x} \\ 1 & \frac{1}{y}+1 & \frac{1}{y} \\ 1 & \frac{1}{z} & \frac{1}{z}+1\end{array}\right|=0$
Since $x, y, z$ are distinct,the determinant part is non-zero.
Therefore,$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 0$
$x^{-1}+y^{-1}+z^{-1} = -1$

(C) We know that for any square matrix $A$ of order $n \times n$,the property of the determinant is given by $|kA| = k^n |A|$.
Here,the order of the matrix $A$ is $3 \times 3$,so $n = 3$.
Substituting the values,we get $|3A| = 3^3 |A|$.
Calculating the power,$3^3 = 3 \times 3 \times 3 = 27$.
Therefore,$|3A| = 27|A|$.

(C) Given the determinant $D = \begin{vmatrix} \log x & \log y & \log z \\ \log 2x & \log 2y & \log 2z \\ \log 3x & \log 3y & \log 3z \end{vmatrix}$.
Using the property $\log(ab) = \log a + \log b$,we can rewrite the rows:
$R_2 \rightarrow R_2 - R_1$ gives $\log 2x - \log x = \log 2$ for each element in the row.
$R_3 \rightarrow R_3 - R_1$ gives $\log 3x - \log x = \log 3$ for each element in the row.
Thus,the determinant becomes:
$D = \begin{vmatrix} \log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 \end{vmatrix}$.
Factoring out $\log 2$ from $R_2$ and $\log 3$ from $R_3$:
$D = (\log 2)(\log 3) \begin{vmatrix} \log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}$.
Since $R_2$ and $R_3$ are identical,the value of the determinant is $0$.

(D) Given expression: $\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$
We can write $\frac{53\pi}{5}$ as $10\pi + \frac{3\pi}{5}$.
So,$\cos\left(10\pi + \frac{3\pi}{5}\right) = \cos\left(\frac{3\pi}{5}\right)$.
Now,$\sin^{-1}\left(\cos \frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right)\right)$ is incorrect,let us use $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$.
$\sin^{-1}\left(\cos \frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right)\right)$
$= \sin^{-1}\left(\sin\left(\frac{5\pi - 6\pi}{10}\right)\right)$
$= \sin^{-1}\left(\sin\left(-\frac{\pi}{10}\right)\right)$
$= -\frac{\pi}{10}$.

(B) Given the equation: $3 \tan^{-1} x + \cot^{-1} x = \pi$.
We know that $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
We can rewrite the given equation as: $2 \tan^{-1} x + (\tan^{-1} x + \cot^{-1} x) = \pi$.
Substituting the identity $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ into the equation:
$2 \tan^{-1} x + \frac{\pi}{2} = \pi$.
Subtracting $\frac{\pi}{2}$ from both sides:
$2 \tan^{-1} x = \frac{\pi}{2}$.
Dividing by $2$:
$\tan^{-1} x = \frac{\pi}{4}$.
Taking the tangent of both sides:
$x = \tan(\frac{\pi}{4}) = 1$.

(A) Given that,$\sin ^{-1} x + \sin ^{-1} y = \frac{\pi}{2} \dots (1)$
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2} \dots (2)$
Comparing equations $(1)$ and $(2)$,we get $\sin ^{-1} y = \cos ^{-1} x$
Taking $\cos$ on both sides,we have $\cos(\sin ^{-1} y) = \cos(\cos ^{-1} x)$
Since $\cos(\sin ^{-1} y) = \sqrt{1 - y^{2}}$,we get $x = \sqrt{1 - y^{2}}$
Squaring both sides,we obtain $x^{2} = 1 - y^{2}$

(B) Given expression: $\tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{x-y}{x+y}\right)$.
Divide the numerator and denominator of the second term by $x$:
$\tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{1 - \frac{y}{x}}{1 + \frac{y}{x}}\right)$.
Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we have:
$\tan^{-1}\left(\frac{1 - \frac{y}{x}}{1 + 1 \cdot \frac{y}{x}}\right) = \tan^{-1}(1) - \tan^{-1}\left(\frac{y}{x}\right)$.
Substituting this back into the expression:
$= \tan^{-1}\left(\frac{x}{y}\right) - \left(\tan^{-1}(1) - \tan^{-1}\left(\frac{y}{x}\right)\right)$
$= \tan^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) - \tan^{-1}(1)$.
Since $\tan^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$ and $\cot^{-1}(z) = \tan^{-1}\left(\frac{1}{z}\right)$:
$= \tan^{-1}\left(\frac{x}{y}\right) + \cot^{-1}\left(\frac{x}{y}\right) - \tan^{-1}(1)$
$= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(D) Given that,$n(A) = 4$ and $n(B) = 5$.
An injective mapping (one-to-one function) from set $A$ to set $B$ exists if each element of $A$ is mapped to a unique element of $B$.
The number of injective mappings from a set with $m$ elements to a set with $n$ elements (where $n \ge m$) is given by the formula $P(n, m) = \frac{n!}{(n-m)!}$.
Here,$n = 5$ and $m = 4$.
Therefore,the number of injective mappings is $P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

(A) Given the function $f(x) = 2x + 6$,where $f: R \rightarrow R$.
To find the inverse $f^{-1}(x)$,let $y = f(x)$.
So,$y = 2x + 6$.
Solving for $x$ in terms of $y$:
$2x = y - 6$
$x = \frac{y - 6}{2}$
$x = \frac{y}{2} - 3$.
By definition,$f^{-1}(y) = x$,so $f^{-1}(y) = \frac{y}{2} - 3$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x}{2} - 3$.

(A) The greatest integer function $f(x) = [x]$ is defined as the greatest integer less than or equal to $x$.
It is a well-known property that the greatest integer function is discontinuous at every integer value $n \in \mathbb{Z}$.
Conversely,the function is continuous at all non-integer values.
Comparing the given options:
$A) 1.5$ (Non-integer)
$B) 4$ (Integer)
$C) 1$ (Integer)
$D) -2$ (Integer)
Since $1.5$ is not an integer,the function $f(x) = [x]$ is continuous at $x = 1.5$.

(A) Given that,$ x^{y}=e^{x-y} $.
Taking the natural logarithm on both sides,we get:
$ y \log x = x - y $
Rearranging the terms to isolate $ y $:
$ y + y \log x = x $
$ y(1 + \log x) = x $
$ y = \frac{x}{1 + \log x} $
Now,differentiating with respect to $ x $ using the quotient rule $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2} $:
$ \frac{dy}{dx} = \frac{(1 + \log x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1 + \log x)}{(1 + \log x)^2} $
$ \frac{dy}{dx} = \frac{(1 + \log x)(1) - x \cdot (\frac{1}{x})}{(1 + \log x)^2} $
$ \frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2} $
$ \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} $

(D) Given that,$x^{m} y^{n}=(x+y)^{m+n}$.
Taking the natural logarithm on both sides,we get:
$m \ln x + n \ln y = (m+n) \ln (x+y)$
Differentiating both sides with respect to $x$,we have:
$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = (m+n) \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx}$
$\left(\frac{n}{y} - \frac{m+n}{x+y}\right) \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x}$
$\left(\frac{nx + ny - my - ny}{y(x+y)}\right) \frac{dy}{dx} = \frac{mx + nx - mx - my}{x(x+y)}$
$\left(\frac{nx - my}{y(x+y)}\right) \frac{dy}{dx} = \frac{nx - my}{x(x+y)}$
Canceling the common term $(nx - my)$ from both sides:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x}$
Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

(A) Given that,$\tan^{-1}(x^2 + y^2) = \alpha$.
Taking $\tan$ on both sides,we get $x^2 + y^2 = \tan \alpha$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(\tan \alpha)$.
Since $\alpha$ is a constant,$\frac{d}{dx}(\tan \alpha) = 0$.
So,$2x + 2y \frac{dy}{dx} = 0$.
$2y \frac{dy}{dx} = -2x$.
$\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$.

(B) Given that,$y=e^{\sin ^{-1}(t^{2}-1)}$ and $x=e^{\sec ^{-1}(\frac{1}{t^{2}-1})}$.
Taking natural logarithm on both sides:
$\log y = \sin ^{-1}(t^{2}-1)$
$\log x = \sec ^{-1}(\frac{1}{t^{2}-1}) = \cos ^{-1}(t^{2}-1)$.
Adding the two equations:
$\log y + \log x = \sin ^{-1}(t^{2}-1) + \cos ^{-1}(t^{2}-1)$.
Using the identity $\sin ^{-1}(\theta) + \cos ^{-1}(\theta) = \frac{\pi}{2}$,we get:
$\log(xy) = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\log y + \log x) = \frac{d}{dx}(\frac{\pi}{2})$
$\frac{1}{y} \frac{dy}{dx} + \frac{1}{x} = 0$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(B) Given the parametric equations of the curve:
$x = t^{2} + 3t - 8 \quad (1)$
$y = 2t^{2} - 2t - 5 \quad (2)$
At the point $(2, -1)$,we first find the value of $t$.
From equation $(2)$:
$-1 = 2t^{2} - 2t - 5 \Rightarrow 2t^{2} - 2t - 4 = 0 \Rightarrow t^{2} - t - 2 = 0$
$(t - 2)(t + 1) = 0 \Rightarrow t = 2$ or $t = -1$.
From equation $(1)$:
$2 = t^{2} + 3t - 8 \Rightarrow t^{2} + 3t - 10 = 0$
$(t + 5)(t - 2) = 0 \Rightarrow t = 2$ or $t = -5$.
The common value for $t$ is $t = 2$.
Now,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2t + 3$
$\frac{dy}{dt} = 4t - 2$
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$.
At $t = 2$:
$\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

(B) Let $y = \log _{10} x$ and $z = \log _{x} 10$.
We know that $\log _{a} b = \frac{\ln b}{\ln a}$.
So,$y = \frac{\ln x}{\ln 10}$ and $z = \frac{\ln 10}{\ln x}$.
Thus,$y = \frac{1}{z}$.
We need to find the derivative of $y$ with respect to $z$,which is $\frac{dy}{dz}$.
Since $y = z^{-1}$,differentiating with respect to $z$ gives $\frac{dy}{dz} = -z^{-2} = -\frac{1}{z^{2}}$.
Substituting $z = \frac{1}{y}$ back into the expression,we get $\frac{dy}{dz} = -y^{2}$.
Since $y = \log _{10} x$,the result is $-\left(\log _{10} x\right)^{2}$.

(B) Given curves are:
$x^{3}-3xy^{2}+2=0 \quad (1)$
$3x^{2}y-y^{3}=2 \quad (2)$
Differentiating equation $(1)$ with respect to $x$:
$3x^{2} - 3(y^{2} + x(2yy')) = 0$
$x^{2} - y^{2} - 2xyy' = 0$
$y' = \frac{x^{2}-y^{2}}{2xy} = m_{1}$
Differentiating equation $(2)$ with respect to $x$:
$3(2xy + x^{2}y') - 3y^{2}y' = 0$
$2xy + x^{2}y' - y^{2}y' = 0$
$y' = \frac{-2xy}{x^{2}-y^{2}} = m_{2}$
Now,calculating the product of slopes:
$m_{1} \times m_{2} = \left(\frac{x^{2}-y^{2}}{2xy}\right) \times \left(\frac{-2xy}{x^{2}-y^{2}}\right) = -1$
Since the product of the slopes is $-1$,the two curves intersect at a right angle.

(A) Given curve: $y(1+x^{2})=2-x$ $(1)$
At the $x$-axis,$y=0$. Substituting $y=0$ in Eq. $(1)$:
$0 = 2-x \Rightarrow x=2$.
So,the point of intersection is $(2, 0)$.
Differentiating Eq. $(1)$ with respect to $x$:
$y'(1+x^{2}) + y(2x) = -1$.
At point $(2, 0)$,we have:
$y'(1+2^{2}) + 0(2 \times 2) = -1$
$y'(5) = -1 \Rightarrow y' = -\frac{1}{5}$.
This is the slope of the tangent at $(2, 0)$.
The slope of the normal is $-\frac{1}{y'} = -\frac{1}{-1/5} = 5$.
The equation of the normal at $(2, 0)$ with slope $m=5$ is:
$y - 0 = 5(x - 2)$
$y = 5x - 10$
$5x - y - 10 = 0$.

(D) The area of a circle $A$ is given by the formula:
$A = \pi r^2$
To find the rate of change of the area with respect to the radius $r$,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$
Now,we evaluate this derivative at $r = 2 \text{ cm}$:
$\left(\frac{dA}{dr}\right)_{r=2} = 2\pi(2) = 4\pi \text{ cm}^2/\text{cm}$
Thus,the rate of change is $4\pi$.

(B) Let $I = \int \frac{e^{x}(1+x) dx}{\cos^{2}(x e^{x})}$.
Substitute $t = x e^{x}$.
Then,$dt = (1 \cdot e^{x} + x \cdot e^{x}) dx = e^{x}(1+x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{\cos^{2} t}$.
$I = \int \sec^{2} t dt$.
Integrating $\sec^{2} t$,we get $I = \tan t + c$.
Substituting back $t = x e^{x}$,we get $I = \tan(x e^{x}) + c$.

(A) Given that,$ I = \int \frac{e^{x}(x^{2} \tan^{-1} x + \tan^{-1} x + 1)}{x^{2} + 1} dx $.
We can rewrite the numerator by grouping terms:
$ I = \int e^{x} \left( \frac{(x^{2} + 1) \tan^{-1} x + 1}{x^{2} + 1} \right) dx $.
Dividing each term by $ x^{2} + 1 $:
$ I = \int e^{x} \left( \tan^{-1} x + \frac{1}{x^{2} + 1} \right) dx $.
Let $ f(x) = \tan^{-1} x $. Then $ f'(x) = \frac{1}{x^{2} + 1} $.
Using the standard integral formula $ \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c $:
$ I = e^{x} \tan^{-1} x + c $.

(B) Given the integral: $ I = \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} dx $
Using the logarithmic property $ e^{k \log x} = x^k $,we can simplify the expression:
$ I = \int \frac{x^6 - x^5}{x^4 - x^3} dx $
Factor out the common terms in the numerator and denominator:
$ I = \int \frac{x^5(x - 1)}{x^3(x - 1)} dx $
Assuming $ x \neq 1 $,we can cancel the term $ (x - 1) $:
$ I = \int \frac{x^5}{x^3} dx = \int x^2 dx $
Integrating $ x^2 $ with respect to $ x $ gives:
$ I = \frac{x^3}{3} + c $

(D) Let $f(x) = \sin^{103} x \cdot \cos^{101} x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \sin^{103}(-x) \cdot \cos^{101}(-x)$
Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\sin x)^{103} \cdot (\cos x)^{101} = -\sin^{103} x \cdot \cos^{101} x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-\pi/4}^{\pi/4} \sin^{103} x \cdot \cos^{101} x \, dx = 0$.

(D) Let $ I = \int_{2}^{8} \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x \quad (1) $
Using the property $ \int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x $,where $ a=2 $ and $ b=8 $,we have $ a+b-x = 2+8-x = 10-x $.
Substituting this into the integral,we get:
$ I = \int_{2}^{8} \frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}} d x $
$ I = \int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}} d x \quad (2) $
Adding equations $(1)$ and $(2)$:
$ 2I = \int_{2}^{8} \frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}} d x $
$ 2I = \int_{2}^{8} 1 d x $
$ 2I = [x]_{2}^{8} = 8 - 2 = 6 $
$ I = \frac{6}{2} = 3 $

(D) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $.
Using the property $ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx $,we get:
$ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} (\frac{\pi}{2}-x)}{\sin ^{1000} (\frac{\pi}{2}-x)+\cos ^{1000} (\frac{\pi}{2}-x)} \, dx $
Since $ \sin(\frac{\pi}{2}-x) = \cos x $ and $ \cos(\frac{\pi}{2}-x) = \sin x $,we have:
$ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{1000} x}{\cos ^{1000} x+\sin ^{1000} x} \, dx $
Adding the two expressions for $ I $:
$ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x + \cos ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $
$ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $
Therefore,$ I = \frac{\pi}{4} $.

(A) Given curves are $y^{2}=2x$ and $y=x$.
To find the points of intersection,substitute $y=x$ into $y^{2}=2x$:
$x^{2}=2x \implies x^{2}-2x=0 \implies x(x-2)=0$.
So,$x=0$ and $x=2$.
When $x=0, y=0$ and when $x=2, y=2$. The points of intersection are $(0,0)$ and $(2,2)$.
The required area is the integral of the upper curve minus the lower curve from $x=0$ to $x=2$:
Area $= \int_{0}^{2} (\sqrt{2x} - x) dx$
$= \int_{0}^{2} (\sqrt{2}x^{1/2} - x) dx$
$= \left[ \sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^{2}}{2} \right]_{0}^{2}$
$= \left[ \frac{2\sqrt{2}}{3} x^{3/2} - \frac{x^{2}}{2} \right]_{0}^{2}$
$= \left( \frac{2\sqrt{2}}{3} (2)^{3/2} - \frac{2^{2}}{2} \right) - (0 - 0)$
$= \left( \frac{2\sqrt{2}}{3} \cdot 2\sqrt{2} - \frac{4}{2} \right)$
$= \left( \frac{4 \cdot 2}{3} - 2 \right)$
$= \frac{8}{3} - 2 = \frac{8-6}{3} = \frac{2}{3} \text{ sq. units.}$

(D) The given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^{2}+\sin \left(\frac{dy}{dx}\right)\right]^{\frac{3}{4}}=\frac{d^{2}y}{dx^{2}}$.
To find the degree,the equation must be a polynomial in terms of its derivatives.
The term $\sin \left(\frac{dy}{dx}\right)$ makes the equation non-polynomial in terms of the derivative $\frac{dy}{dx}$.
Therefore,the degree of this differential equation is not defined.
The order of the differential equation is the highest derivative present,which is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
Thus,the correct option is $D$.

(C) Given differential equation is,$\frac{dy}{y} + \frac{dx}{x} = 0$.
Integrating both sides with respect to their variables:
$\int \frac{dy}{y} + \int \frac{dx}{x} = \int 0 \, dx$
Using the standard integral formula $\int \frac{1}{u} du = \log |u| + C$,we get:
$\log |y| + \log |x| = \log |c|$
Using the logarithmic property $\log a + \log b = \log(ab)$:
$\log |xy| = \log |c|$
Taking the exponential on both sides:
$xy = c$

(C) The given differential equation is $x \frac{dy}{dx} - y = x^4 - 3x$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} - \frac{1}{x}y = x^3 - 3$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = x^3 - 3$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx}$.
$I$.$F$. $= e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(x^{-1})} = x^{-1} = \frac{1}{x}$.

(A) Given that,$|\vec{a}| = 1$ and $|\vec{b}| = 1$.
Since $\sqrt{3}\vec{a} - \vec{b}$ is a unit vector,its magnitude is $1$,so $|\sqrt{3}\vec{a} - \vec{b}| = 1$.
Squaring both sides,we get $|\sqrt{3}\vec{a} - \vec{b}|^2 = 1^2$.
Using the property $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we have:
$3|\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3}(\vec{a} \cdot \vec{b}) = 1$.
Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$:
$3(1)^2 + (1)^2 - 2\sqrt{3}|\vec{a}||\vec{b}|\cos \theta = 1$.
$3 + 1 - 2\sqrt{3}(1)(1)\cos \theta = 1$.
$4 - 2\sqrt{3}\cos \theta = 1$.
$2\sqrt{3}\cos \theta = 3$.
$\cos \theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Therefore,$\theta = 30^{\circ}$.

(C) Given that,$\vec{a}+\vec{b}+\vec{c}=0$.
We can write this as $\vec{a}+\vec{b}=-\vec{c}$.
Taking the dot product of both sides with themselves:
$(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = (-\vec{c}) \cdot (-\vec{c})$.
$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.
Substituting the given values $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$:
$3^2 + 5^2 + 2|\vec{a}||\vec{b}| \cos \theta = 7^2$.
$9 + 25 + 2(3)(5) \cos \theta = 49$.
$34 + 30 \cos \theta = 49$.
$30 \cos \theta = 49 - 34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(A) Given that each vector is perpendicular to the sum of the other two:
$\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$ $(1)$
$\vec{b} \cdot (\vec{a} + \vec{c}) = 0 \implies \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} = 0$ $(2)$
$\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$ $(3)$
Adding equations $(1), (2),$ and $(3)$,we get:
$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Now,consider the magnitude squared:
$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$
Substituting the given values and the sum of dot products:
$|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 0$
$|\vec{a} + \vec{b} + \vec{c}|^2 = 9 + 16 + 25 = 50$
$|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$

(A) Given that the vector $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector,its magnitude must be equal to $1$.
So,$|x(\hat{i}+\hat{j}+\hat{k})| = 1$.
Using the property $|k\vec{a}| = |k| |\vec{a}|$,we get:
$|x| |\hat{i}+\hat{j}+\hat{k}| = 1$.
The magnitude of the vector $(\hat{i}+\hat{j}+\hat{k})$ is $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.
Substituting this value,we have $|x| \sqrt{3} = 1$.
Therefore,$|x| = \frac{1}{\sqrt{3}}$,which implies $x = \pm \frac{1}{\sqrt{3}}$.

(D) Given that,$ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $.
We know that the dot product of two vectors is defined as $ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta $,where $ \theta $ is the angle between the vectors.
Substituting this into the given equation:
$ 2 (|\vec{a}| |\vec{b}| \cos \theta) = |\vec{a}| |\vec{b}| $.
Dividing both sides by $ |\vec{a}| |\vec{b}| $ (assuming non-zero vectors):
$ 2 \cos \theta = 1 $.
$ \cos \theta = \frac{1}{2} $.
Since $ \cos 60^{\circ} = \frac{1}{2} $,the angle $ \theta = 60^{\circ} $.

(C) We know that the identity for $\cos 2\theta$ is $\cos 2\theta = 2\cos^2 \theta - 1$.
Substituting this for each term,we get:
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma = (2\cos^2 \alpha - 1) + (2\cos^2 \beta - 1) + (2\cos^2 \gamma - 1)$.
This simplifies to:
$2(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 3$.
Since $\cos \alpha, \cos \beta, \cos \gamma$ are the direction cosines of a vector $\vec{a}$,we know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting this value into the expression:
$2(1) - 3 = 2 - 3 = -1$.

(A) The vector equation of a plane at a distance $ d $ from the origin with a unit normal vector $ \hat{n} $ is given by $ \vec{r} \cdot \hat{n} = d $.
Given the normal vector $ \vec{N} = 2 \hat{i} - 3 \hat{j} + \hat{k} $.
First,find the magnitude of $ \vec{N} $: $ |\vec{N}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} $.
The unit normal vector is $ \hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2 \hat{i} - 3 \hat{j} + \hat{k}}{\sqrt{14}} $.
The distance from the origin is $ d = \frac{3}{\sqrt{14}} $.
Substituting these into the equation $ \vec{r} \cdot \hat{n} = d $:
$ \vec{r} \cdot \left( \frac{2 \hat{i} - 3 \hat{j} + \hat{k}}{\sqrt{14}} \right) = \frac{3}{\sqrt{14}} $.
Multiplying both sides by $ \sqrt{14} $,we get:
$ \vec{r} \cdot (2 \hat{i} - 3 \hat{j} + \hat{k}) = 3 $.

(D) The given equation of the plane is $5y + 8 = 0$.
The normal vector to the plane is $\vec{n} = (0, 5, 0)$.
The line passing through the origin $(0, 0, 0)$ and perpendicular to the plane has the direction ratios of the normal vector.
Thus,the equation of the line is $\frac{x-0}{0} = \frac{y-0}{5} = \frac{z-0}{0} = \lambda$.
This gives $x = 0$,$y = 5\lambda$,and $z = 0$.
Since the foot of the perpendicular lies on the plane,we substitute these coordinates into the plane equation:
$5(5\lambda) + 8 = 0$
$25\lambda = -8$
$\lambda = -\frac{8}{25}$.
Substituting $\lambda$ back into the coordinates:
$x = 0$,$y = 5(-\frac{8}{25}) = -\frac{8}{5}$,$z = 0$.
Therefore,the coordinates of the foot of the perpendicular are $(0, -\frac{8}{5}, 0)$.

(C) Given that,$P(A \cap B) = \frac{7}{10}$ and $P(B) = \frac{17}{20}$.
We know the formula for conditional probability is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Substituting the given values into the formula:
$P(A \mid B) = \frac{\frac{7}{10}}{\frac{17}{20}}$
$P(A \mid B) = \frac{7}{10} \times \frac{20}{17}$
$P(A \mid B) = \frac{7 \times 2}{17} = \frac{14}{17}$.