Let f(x)=limh→0(sin(x+h))ln(x+h)−(sinx)lnxh then f(π2) is
A
Equal to 0
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B
Equal to 1
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C
lnπ2
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D
Non - existent
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Solution
The correct option is A Equal to 0 Assuming g(x)=(sinx)lnx
So, f(x)=g′(x)
Taking log of g(x), lng(x)=lnx×lnsinx⇒g′(x)g(x)=lnsinxx+lnx×cotx⇒g′(x)=(sinx)lnx[lnsinxx+lnx×cotx]⇒f(π2)=g′(π2)=0