Question
The point in the interval [0,2π] where f(x)=ex sin x has maximum slope, is
Answer: Option B
:
B
We have, f′(x)=ex+cosx+sinxexAndf′(x)=−sinxex+cosxex+cosxex+sinxcosxex.Now,f′(x)=2cosxcosxex=0⇒cosx=0⇒x=π2.Also,f′(x)=−2sinxex+2cosxex=−ve
∴ Slope is maximum at x=π2.
Hence (b) is the correct answer.
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:
B
We have, f′(x)=ex+cosx+sinxexAndf′(x)=−sinxex+cosxex+cosxex+sinxcosxex.Now,f′(x)=2cosxcosxex=0⇒cosx=0⇒x=π2.Also,f′(x)=−2sinxex+2cosxex=−ve
∴ Slope is maximum at x=π2.
Hence (b) is the correct answer.
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