First slide

Functions (XII)

Question

$The function, f (x) = \{\begin{array}{cc} \frac{(x^{2 n})}{{(x^{2 n} sgn x)}^{2 n + 1}} (\frac{e^{\frac{1}{x}} - e^{- \frac{1}{x}}}{e^{\frac{1}{x}} + e^{- \frac{1}{x}}}) & x \neq 0 \\ 1 & x = 0 \end{array}$ $n \in N is:$

Moderate

A

Odd function
B

Even function
C

Neither odd nor even function
D

Constant function

Solution

$f (x) = \frac{x^{2 n}}{{(x^{2 n} sgn x)}^{2 n + 1}} [\frac{e^{\frac{1}{x}} - e^{- \frac{1}{x}}}{e^{\frac{1}{x}} + e^{- \frac{1}{x}}}] x \neq 0 and f (0) = 1$
$when f (x) = \frac{(x^{2 n})}{{(x^{2 n})}^{2 n + 1}} [\frac{e^{\frac{1}{x}} - e^{- \frac{1}{x}}}{e^{\frac{1}{x}} + e^{- \frac{1}{x}}}]; x > 0$
$f (x) = \frac{x^{2 n}}{- {(x^{2 n})}^{2 n + 1}} [\frac{e^{\frac{1}{x}} - e^{- \frac{1}{x}}}{e^{\frac{1}{x}} + e^{- \frac{1}{x}}}]; x < 0$
$Clearly, f (x) = f (- x) . Hence, f (x) is even function.$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App