Let $f(x)=1-\sqrt{\left(x^{2}\right)}$ where the square root is to be taken positive, then

• A

  $f$ has no extrema at $x=0$
• B

  $f$ has minima at $x=0$
• Correct answer
  C

  $f$ has maxima at $x=0$
• D

  $\mathrm{f}^{\prime}$ exists at 0

Hence, the correct answer is option: 'C'.