The function f(x)=1+x(sinx)cosx, 0<x≤π2
is continuous on 0,π2
is strictly increasing on 0,π2
is strictly decreasing on 0,π2
None of these
Explanation for the correct option
Given function, f(x)=1+x(sinx)cosx
Since, 0<x≤π2
⇒0≤cosx<1
So, [cosx]=0
∴f(x)=1
So, f(x) is a constant function and is neither an increasing nor decreasing function.
Thus, f(x) is a continuous function.
Hence, correct option is A