Question

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

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

Answer 1

Answer: (A), (B)

$g\left(f\right(1\left)\right)=1$
$g\left(f\right(0\left)\right)=1$

$f\left(x\right)=\left(max\right[1+sinx,1,1-cos\left(x\right)]$ for $xϵ[0,2\pi ]$
$f\left(0\right)$
$=\left(max\right(1+0,1,0\left)\right)$
$=1$
$f\left(1\right)$
$=\left(max\right(1+sin\left(1\right)),1,1-cos1)$
$=1+sin\left(1\right)$
$g\left(x\right)=max(1,|x-1|)$
Hence,
$g\left(f\right(0\left)\right)$
$=g\left(1\right)$
$=max(1,0)$
$=1$
And, $g\left(f\right(1\left)\right)$
$=max(1,|1+sin(1)-1|)$
$=max(1,|sin(1\left)|\right)$
$=1$
$g\left(0\right)=1$
$f\left(g\right(0\left)\right)=1+sin1$