Question

The function $f(x)=\int\limits_0^x \sqrt{1-t^4} d t$ is such that :

• A. it is defined on the interval $[-1,1]$
• B. it is an increasing function
• C. it is an odd function
• D. the point $(0,0)$ is the point of inflection

Answer 1

Answer: (A), (B), (C), (D)

$f(x) =\int\limits_0^x \sqrt{1-t^4} d t$
$f(-x) =\int\limits_0^{-x} \sqrt{1-t^4} d t$
$\left.=-\int\limits_0^x \sqrt{1-u^4} d u \,\,\, \text { (Put } t=-u\right)$
$f(-x)=-f(x) \Rightarrow $ ' $f$ ' is odd function.
Check other options.