Question

If $f\left(x\right)=2x+|x|,g\left(x\right)=\frac{1}{3}(2x-|x|)$ and $h\left(x\right)=f\left(g\left(x\right)\right)$ then $h\left(1\right)$

• A. $2$
• B. $1$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $1$

$f\left(x\right)=2x+|x|$

$g\left(x\right)=\frac{1}{3}(2x-|x|)$

$h\left(x\right)=f\left(g\right(x\left)\right)$

$h\left(1\right)=f\left(g\right(1\left)\right)$

$g\left(1\right)=\frac{1}{3}(2-1)=\frac{1}{3}$

$f\left(\frac{1}{3}\right)=\frac{2}{3}+\frac{1}{3}=\frac{3}{3}=1$