The function f(x) = _0^x , , 1 , , - , , t^4 dt is such that - Vidyadip

MCQ

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

A
it is defined on the interval $[- 1, 1]$
B
it is an increasing function
C
it is an odd function
✓
All of the Above

✓

Answer

Correct option: D.

All of the Above

d
$f '(x) = \sqrt {1 - {x^4}} > 0$ in $(-1, 1)$ ==> $f$ is $\uparrow$

Now $f (x) + f (- x) = \int\limits_0^x {\sqrt {1 - {t^4}} } \,dt\, + \,\,\int\limits_0^{ - x} {\sqrt {1 - {t^4}} } \,\,dt\,$

==> $\int\limits_0^x {\sqrt {1 - {t^4}} } \,dt\, + \,\,\left( { - \,\,\int\limits_0^y {\sqrt {1 - {y^4}} } \,\,dy} \right)\, \,( t = - y)$

$ = 0 $ ==> $f (x)$ is odd

again $f''' (x) =\frac{{ - 4{x^3}}}{{2\sqrt {1 - {x^4}} }}$ which vanished at $x = 0$

and changes sign ==> $(0, 0)$ is inflection since $f$ is well defined in $[-1, 1] $

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