The range of values of the function $f\left( x \right) = \frac{1}{{2 - 3\sin x}}$ is
$\left[ { - 1,\frac{1}{5}} \right]$
$\left[ { - 1,5} \right]$
$\left( { - \infty , - 1} \right] \cup \left[ {\frac{1}{5},\infty } \right)$
$\left( { - \infty ,\frac{1}{5}} \right] \cup \left[ {1,\infty } \right)$
If $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, for what value of $\alpha $ is $f(f(x)) = x$
Show that the Modulus Function $f : R \rightarrow R$ given by $(x)=|x|$, is neither one - one nor onto, where $|x|$ is $x$, if $x$ is positive or $0$ and $| X |$ is $- x$, if $x$ is negative.
Period of $f(x) = nx + n - [nx + n]$, $n \in N$
where [ ] denotes the greatest integer function is :
If $f(x)$ satisfies the relation $f\left( {\frac{{5x - 3y}}{2}} \right)\, = \,\frac{{5f(x) - 3f(y)}}{2}\,\forall x,y\in R$ $f(0) = 1, f '(0) = 2$ then period of $sin \ (f(x))$ is
The graph of $y = f(x)$ is shown then number of solutions of the equation $f(f(x)) =2$ is