Let f(x)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩0x=0x2sin(πx)−1<x<1(x≠0)x|x|−1>x>1
f(x) is an odd function
f(x) is an even function
f′(x) is an even function
f(x) is neither odd nor even
A
f(x) is an even function
B
f(x) is neither odd nor even
C
f(x) is an odd function
D
f′(x) is an even function
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Solution
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f(x)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩0x=0x2sin(πx)−1<x<1(x≠0)x|x|−1>x>1 f(−x)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩0−x=0(−x)2sin(−πx)−1<−x<1(x≠0)−x|−x|−1>−x>1 =⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩0x=0−x2sin(πx)−1<−x<1(x≠0)−x|x|−1>−x>1 =−f(x) Hence f is an odd function, and f′ will be an even function.
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