Question

For the function $f\left(x\right)=\sin 2^x,\forall x\in R$, which of the following is/are correct ?

• A. For $x<0,f\left(x\right)<\sin 1$
• B. $x=0$ is an asymptote
• C. $y=0$ is an asymptote
• D. For $f\left(x\right)=0,x={\log }_2\left(n\pi \right),n\in Z$

Answer 1

Answer: (A), (C)

$y=0$ is an asymptote

$f\left(x\right)=\sin 2^x$

$f\left(x\right)=0\Rightarrow \sin 2^x=0\Rightarrow 2^x=n\pi ,n\in N\Rightarrow x={\log }_2\left(n\pi \right),n\in N$

When
$x\to -\infty ,\sin 2^x\to 0$
So, $y=0$ is an asymptote.

For $x<0$,
$0<2^x<1$
$f\left(x\right)=\sin 2^x\Rightarrow f^{\prime }\left(x\right)=\cos 2^x\times 2^x\ln 2\ge 0$
So $f\left(x\right)$ is a increasing function.
$f\left(x\right)<f\left(0\right)\Rightarrow f\left(x\right)<\sin 1$