The correct option is B f is strictly decreasing in the interval (12k+2,12k+1), k∈Z
f(x)=cos(πx)
The function is defined for all x∈R except x≠0
f′(x)=−sin(πx)×(−πx2)⇒f′(x)=πx2sin(πx) ⋯(i)
For f(x) to be strictly increasing,
f′(x)>0⇒πx2sin(πx)>0⇒sin(πx)>0 (∵x2>0)⇒2kπ<πx<(2k+1)π, k∈Z⇒12k+1<x<12k
Hence, f(x) is strictly increasing in the interval (12k+1,12k), k∈Z
And, for strictly decreasing
f′(x)<0⇒πx2sin(πx)<0⇒sin(πx)<0⇒(2k+1)π<πx<(2k+2)π, k∈Z⇒12k+2<x<12k+1
Hence, f(x) is strictly decreasing in the interval (12k+2,12k+1), k∈Z