If $f(x) = \frac{x}{{x - 1}}$, then $\frac{{f(a)}}{{f(a + 1)}} = $

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 :-

The domain of the function $f(x){ = ^{16 - x}}{\kern 1pt} {C_{2x - 1}}{ + ^{20 - 3x}}{\kern 1pt} {P_{4x - 5}}$, where the symbols have their usual meanings, is the set

Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.

If $\phi (x) = {a^x}$, then ${\{ \phi (p)\} ^3} $ is equal to

Let $f, g: N -\{1\} \rightarrow N$ be functions defined by $f(a)=\alpha$, where $\alpha$ is the maximum of the powers of those primes $p$ such that $p^{\alpha}$ divides $a$, and $g(a)=a+1$, for all $a \in N -\{1\}$. Then, the function $f+ g$ is.