Question

If $f\left(x\right)=x^2$; when $x<0,f\left(x\right)=x$; when $0<x<1,f\left(x\right)=\frac{1}{x}$; when $x\ge 1$, then

• A. $3f\left(\frac{1}{2}\right)-6=-\frac{3}{2}$
• B. $f\left(2\right)=\frac{1}{3}$
• C. $f(-6)+f\left(\frac{1}{2}\right)=f\left(2\right)$
• D. $3\sqrt{f(-4)}+4=16$

Answer 1

The correct option is D $3\sqrt{f(-4)}+4=16$

(A) $3f\left(\frac{1}{2}\right)-6=3\times \frac{1}{2}-6$
$=\frac{9}{2}$

So, A is not correct

(B) $f\left(2\right)=\frac{1}{2}$

So, B is not correct

(C) $f(-6)+f\left(\frac{1}{2}\right)=36+\frac{1}{2}$
$=\frac{73}{2}\ne \frac{1}{2}=f\left(2\right)$

So, C is not correct

(D) $3\sqrt{f(-4)}+4=3\times \sqrt{16}+4$
$=16$

Hence, D is correct.