If $f(x) = sgn(x)=\left\{\begin{array}{cc}\frac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$ and g(x) = f(f(x)), then at x = 0 g(x), is
Answer & explanation
Correct answer: option 4
We have,
$f(x)=\left\{\begin{array}{rr} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{array}\right.$
Clearly, it is neither continuous nor differentiable at x = 0.
Now,
$g(x)=f(f(x))=\left\{\begin{array}{rr} f(1), & x>0 \\ f(0), & x=0 \\ f(-1), & x<0 \end{array}\right.$
$\Rightarrow g(x)=\left\{\begin{array}{rr} 1, & x>0 \\ 0, & x=0 \\ -1, & x<0 \end{array}\right.$
⇒ g(x) = f(x)
Hence, g(x) is neither continuous nor differentiable at x = 0.