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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Functions

BITSAT 2012 · Mathematics · Functions

If f is an even function and g is an odd function, then the...

If f\mathrm{f} is an even function and g\mathrm{g} is an odd function, then the function fog is

Options

  1. Aan even function
  2. Ban odd function
  3. Cneither even nor odd
  4. Da periodic function

Correct answer

A. an even function

Step-by-step solution

We have, fog(−x)=f[g(−x)]=f[−g(x)](∵gf o g(-x)=f[g(-x)]=f[-g(x)](\because g is odd ))=f[g(x)](∵fis even )=fog⁡(x)∀x∈R\begin{array}{l} =\mathrm{f}[\mathrm{g}(\mathrm{x})] \quad(\because \mathrm{fis} \text{ even }) \\ =\operatorname{fog}(\mathrm{x}) \forall \mathrm{x} \in \mathrm{R} \end{array}$∴\therefore fog is an even function.$
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