Show that the function $f: N \rightarrow N$ given by $f(x)=2 x,$ is one-one but not onto.
The function $f$ is one-one, for $f\left(x_{1}\right)$ $=f\left(x_{2}\right) \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}$ $=x_{2},$ Further, $f$ is not onto, as for $1 \in N ,$ there does not exist any $x$ in $N$ such that $f(x)=2 x=1$
If $f(x) = \frac{x}{{x - 1}} = \frac{1}{y}$, then $f(y) = $
The domain of $f(x) = \frac{1}{{\sqrt {{{\log }_{\frac{\pi }{4}}}({{\sin }^{ - 1}}x) - 1} }}$,is
Let $f(x) = cos(\sqrt P \,x),$ where $P = [\lambda], ([.]$ is $G.I.F.)$ If the period of $f(x)$ is $\pi$. then
Let $f : R \to R$ be a function defined by $f(x) = - \frac{{|x{|^3} + |x|}}{{1 + {x^2}}}$; then the graph of $f(x)$ is lies in the :-
If $f(x)$ be a polynomial function satisfying $f(x).f (\frac{1}{x}) = f(x) + f (\frac{1}{x})$ and $f(4) = 65$ then value of $f(6)$ is