Question

# f is an odd function. It is also known that f(x) is continuous for all values of x and is periodic with period 2. If g(x)=x∫0f(t) dt, then

A
g(x) is even
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B
g(n)=0,nN
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C
g(2n)=0,nN
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D
g(x) is non-periodic
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Solution

## The correct option is C g(2n)=0,n∈Ng(x)=x∫0f(t) dt g(−x)=−x∫0f(t) dt=−x∫0f(−t) dt⇒g(−x)=x∫0f(t) dt[∵f(−t)=−f(t)] ⇒g(x)=g(−x) Thus, g(x) is even function. Also, g(x+2)=x+2∫0f(t) dt⇒g(x+2)=2∫0f(t) dt+x+2∫2f(t) dt⇒g(x+2)=g(2)+x∫0f(t+2) dt⇒g(x+2)=g(2)+x∫0f(t) dt⇒g(x+2)=g(2)+g(x) Now, g(2)=2∫0f(t) dt⇒g(2)=1∫0f(t) dt+2∫1f(t) dt⇒g(2)=1∫0f(t) dt+0∫−1f(t+2) dt⇒g(2)=1∫0f(t) dt+0∫−1f(t) dt⇒g(2)=1∫−1f(t) dt=0 As f(t) is odd function. So, g(2)=0⇒g(x+2)=g(x) i.e. g(x) is periodic with period 2. ∴g(4)=0 or f(6)=0, ⇒g(2n)=0,n∈N.

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