Question

If $A=\{0,\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{2}\}$ and $f:A\to R$ is defined by $f\left(x\right)=\sin x$ then the range of $f$

• A. $\{0,\frac{1}{2},\frac{1}{3},\frac{\sqrt{3}}{2}\}$
• B. $\{0,\frac{1}{2},\frac{1}{\sqrt{2}},1\}$
• C. $\{0,\frac{1}{2},\frac{1}{3},1\}$
• D. $\{0,\frac{1}{\sqrt{2}},-\frac{1}{2},1\}$

Answer 1

Answer: (B)

$\{0,\frac{1}{2},\frac{1}{\sqrt{2}},1\}$

$f\left(x\right)=\sin x$
$f\left(0\right)=\sin 0$
$=0$
$f\left(\frac{\pi }{6}\right)=\sin \frac{\pi }{6}$
$=\frac{1}{2}$
$f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}$
$f\left(\frac{\pi }{2}\right)=\sin \frac{\pi }{2}$
$=1$