The correct options are
B f(π2)−f(0)=4
C Graph of y=f(x) intersects x-axis at only one point in the interval [0,π]
f(x)=4sinx+cos2x=4sinx+1−sin2x⇒f(x)=5−(sin2x−4sinx+4)⇒f(x)=5−(sinx−2)2
We know that −1≤sinx≤1
⇒−3≤sinx−2≤−1⇒1≤(sinx−2)2≤9⇒−4≤5−(sinx−2)2≤4
Hence, range of f is [−4,4]
f(π2)−f(0)=4−1=3
If x∈[0,π], then f(x)>0
Hence, graph of y=f(x) can not intersect x-axis in the interval [0,π].
f(−x)=f(x)⇒−4sinx+cos2x=4sinx+cos2x⇒8sinx=0
⇒x=nπ, where n∈Z