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Exercise 2.3 · Q45

Q.Check the validity of the Rolle's theorem for the function f(x)=e−xsin⁡x, x∈[0,π]f(x) = e^{-x}\sin x,\ x \in [0, \pi].

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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✓ Free question

e−xe^{-x} and sin⁡x\sin x are both continuous and differentiable everywhere, so their product f(x)=e−xsin⁡xf(x)=e^{-x}\sin x is continuous on [0,π][0,\pi] and differentiable on (0,π)(0,\pi).

f(0)=e0sin⁡0=1(0)=0f(0)=e^0\sin0=1(0)=0. f(π)=e−πsin⁡π=e−π(0)=0f(\pi)=e^{-\pi}\sin\pi=e^{-\pi}(0)=0. So f(0)=f(π)=0f(0)=f(\pi)=0.

f′(x)=−e−xsin⁡x+e−xcos⁡x=e−x(cos⁡x−sin⁡x)f'(x)=-e^{-x}\sin x+e^{-x}\cos x=e^{-x}(\cos x-\sin x). Setting f′(c)=0f'(c)=0: since e−c≠0e^{-c}\ne0, need cos⁡c=sin⁡c⇒tan⁡c=1⇒c=π4\cos c=\sin c\Rightarrow\tan c=1\Rightarrow c=\dfrac{\pi}{4} (within (0,π)(0,\pi); the other solution 5π4\tfrac{5\pi}4 is out of range).

✓Final answer

Rolle's theorem holds; c=π4∈(0,π)c = \dfrac{\pi}{4} \in (0,\pi)

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