Question

The function $f\left(x\right)$ is defined in $[0,1]$, then the domain of definition of the function $f\left[\log \left(1-x^2\right)\right]$

• A. $x\in \left\{0\right\}$
• B. $x\in \left[-\sqrt{1+e,}-1\right]\cup \left[1+\sqrt{1+e}\right]$
• C. $x\in \left(-\infty ,\infty \right)$
• D. None of the above

Answer 1

Answer: (A)

$x\in \left\{0\right\}$

$f\left(x\right)$ in $[0,1]$ defined,

$f\left[log\right(1-x^2\left)\right]$

For $x=0,f\left[log\right(1\left)\right]$

$\Rightarrow f\left(10\right)$

$=$ Defined.

if $x=1$

$f\left[log\right(1-1^2\left)\right]=f\left[log0\right]$

$=$ undefined.

If $0<x<1,$ the $x=0.5$

$f\left[log\right(1-\left(0.5{)}^2\right)=f\left[log0.75\right)]$

As we know $log\left(x\right)$ is negative.

If $x$ is in $(0,1)$ so, this is outside the domain.