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Question

# If f(x)=cos(πx) where x≠0, then which of the following is/are CORRECT?

A
f is strictly increasing in the interval (12k+1,12k), kZ
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B
f is strictly decreasing in the interval (12k+2,12k+1), kZ
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C
f is strictly increasing in the interval (12k+2,12k+1), kZ
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D
f is strictly decreasing in the interval (12k+3,12k+2), kZ
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Solution

## The correct option is B f is strictly decreasing in the interval (12k+2,12k+1), k∈Zf(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

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