The correct options are
A For x<0, f(x)<sin1
C y=0 is an asymptote
f(x)=sin2x
f(x)=0⇒sin2x=0⇒2x=nπ, n∈N⇒x=log2(nπ), n∈N
When
x→−∞, sin2x→0
So, y=0 is an asymptote.
For x<0,
0<2x<1
f(x)=sin2x⇒f′(x)=cos2x×2xln2≥0
So f(x) is a increasing function.
f(x)<f(0)⇒f(x)<sin1