Saved Bookmarks
| 1. |
Which of the following function is neither even nor odd. (A) \( f(x)=\left(\left[\frac{x}{\pi}\right]+\frac{1}{2}\right) \sin x \)(B) \( f(x)=\frac{\left(a^{x}+1\right)^{5}}{a^{x}}, a>0 \)(C) \( f(x)=\frac{x}{e^{x}-1}+\frac{x}{2}+1 \)(D) \( f(x)=\frac{g(x)-g(-x)}{5} \), where \( g(x) \) is a real valued function \( x \in R \) |
|
Answer» (a) f(-x) = \(([\frac{-x}{\pi}]+\frac{1}{2})sin(-x)\) \(=-([\frac{-4}{\pi}]+\frac{1}{2})sinx\) \(≠\) f(x) or \(≠\) -f(x) ∴ f(x) is neither even nor odd function (b) f(-x) = \(\frac{(a^{-x}+1)^5}{a^{-x}}=\frac{(1+a^x)^5}{a^{5x}a^{-x}}\) \(=\frac{(1+a^x)^5}{a^{4x}}≠ f(x)\) or \(≠-f(x)\) ∴ f(x) is nether even nor odd function (c) f(-x) = \(\frac{-x}{e^{-x}-1}-\frac{x}{2}+1\) \(=-\frac{xe^x}{1-e^x}-\frac{x}{2}+1\) \(≠ \) f(x) or \(≠\) -f (x) ∴ f(x) is neither even nor odd function (d) f(-x) \(=\frac{g(-x)-g(x)}{5}=-(\frac{g(x)-g(-x)}{5})\) = -f (x) ∴ f(x) B odd function |
|