Let $f(x) = \frac{\sqrt{x - 2\sqrt{x - 1}}}{\sqrt{x - 1} - 1}$. Then:

If $R \subset A \times B$ and $S \subset B \times C$,then the relation $(SoR)^{-1} = $

Let $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32}$. Then the value of $8 \left( f \left( \frac{1}{15} \right) + f \left( \frac{2}{15} \right) + \dots + f \left( \frac{59}{15} \right) \right)$ is equal to

If a function $g(x)$ is defined in $[-1, 1]$ and two vertices of an equilateral triangle are $(0, 0)$ and $(x, g(x))$ and its area is $\frac{\sqrt{3}}{4}$,then $g(x)$ equals :-

Let $R$ denote the set of all real numbers and $R^{+}$ denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$, define $f: A \rightarrow B$ by $f(x) = x^2$ for $x \in A$. Match the following lists:
| Column $I$ | Column $II$ |
| :--- | :--- |
| $A$. $f$ is one-one and onto, if | $1$. $A = R^{+}, B = R$ |
| $B$. $f$ is one-one but not onto, if | $2$. $A = B = R$ |
| $C$. $f$ is onto but not one-one, if | $3$. $A = R, B = R^{+}$ |
| $D$. $f$ is neither one-one nor onto, if | $4$. $A = B = R^{+}$ |

If $f : R \rightarrow R$ is defined by $f(x) = \begin{cases} x + 4, & x < -4 \\ 3x + 2, & -4 \leq x < 4 \\ x - 4, & x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is :
List-$I$
$(A) f(-5) + f(-4)$
$(B) f(|f(-8)|)$
$(C) f(f(-7) + f(3))$
$(D) f(f(f(f(0)))) + 1$
List-$II$
$(i) 14$
$(ii) 4$
$(iii) -11$
$(iv) -1$
$(v) 1$
$(vi) 0$