Sarvesh K Verma Solutions for Chapter: Functions and Graphs, Exercise 2: Introductory Exercise

Let $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}, x\ne 0$ then $f\left(x\right)$ is equals to

If a function $f$ satisfies the conditions $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)\forall ,\mathrm{x},\mathrm{y}\in \mathrm{R}$, then $\mathrm{f}$ is :

Which of the following function is an even function?

Which of the following function is odd?

If $\mathrm{f}\left(\mathrm{x}\right)=\sqrt[3]{\left(1-{\mathrm{x}}^2\right)}+\sqrt[3]{\left(1+{\mathrm{x}}^2\right)}$, then $\mathrm{f}\left(\mathrm{x}\right)$ is:

Which of the following function is an odd function?

The graph of the function $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$ is symmetrical about the line $\mathrm{x}=2$, then :

Given $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{(1-\mathrm{x})},\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right)$, then the value of $\mathrm{f}\left(\mathrm{x}\right)·\mathrm{g}\left(\mathrm{x}\right)·\mathrm{h}\left(\mathrm{x}\right)$ is :