Question

If $f\left(x\right)=max\{1+\sin x,1,1-\cos x\},\forall x\in [0,2\pi ]$ and $g\left(x\right)=max\{1,x-1\}\forall x\in R$, then

• A. $g\left(f\right(0\left)\right)=1$
• B. $g\left(f\right(0\left)\right)=1$
• C. $f\left(f\right(1\left)\right)=1$
• D. $f\left(g\right(0\left)\right)=1+\sin 1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $g\left(f\right(0\left)\right)=1$
B $g\left(f\right(0\left)\right)=1$
D $f\left(g\right(0\left)\right)=1+\sin 1$

$f\left(0\right)=max\{1+\sin 0,1,1-\cos 0\}=1$
$g\left(0\right)=max\{1,(0-1\left)\right\}=1$
$f\left(1\right)=max\{1+\sin 1,1,1-\cos 1\}=1+\sin 1$
$g\left(f\right(0\left)\right)=g\left(1\right)=max\{1,(1-1\left)\right\}=1$
$f\left(g\right(0\left)\right)=f\left(1\right)=1+\sin 1$
$g\left(f\right(1\left)\right)=g(1+\sin 1)=max\{1,(1+\sin 1-1\left)\right\}=1$
$f\left(f\right(1\left)\right)=f(1+\sin 1)\ne 1$