If log2x≥0, then log1/πsin−12x1+x2+2tan−1x is equals
log1/π4tan−1x
0
−1
none of these
We have,
log2x≥0⇒x≥2∘=1.
For x≥1.we have
sin−12x1+x2=π−2tan−1x.∴ log1/πsin−12x1+x2+2tan−1x= log1/ππ−2tan−1x+2tan−1x= log1/ππ=−1.