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Question

A function f is defined by f(x)=exsinx in [0,π].

Which of the following is not correct?


A

fis continuous in [0,π]

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B

fis differentiable in [0,π]

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C

f(0)=f(π)

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D

Roll's theorem is not true in[0,π]

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Solution

The correct option is D

Roll's theorem is not true in[0,π]


The explanation for the correct answer:

Step 1: Finding continuity of the function:

Given, f(x)=exsinx

Since sinxis continuous all x0,π and exis continuous all xR

Thus, fx is continuous in 0,π

Step 2: Finding differentiability of the function:

We have f'x=excosx+sinxex

Since sinx and cosx is differentiable all x0,π and exis differentiable all xR

Thus fx is differentiable in 0,π

Step 3: Finding f0and fπ:

f0=e0sin0=10=0fπ=eπsinπ=eπ0=0

Thus f0=fπ.

Therefore, all three conditions are true.

So the function is satisfied all the conditions of rolle's theorem in [0,π].

Hence, option (D) is the correct option.


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