## The value of c in Rolleβs Theorem for the function f (x) = e^x sin x,

## xβ [0, Ο] is

## (A) Ο/6Β (B) Ο/4Β

## (C) Ο/2Β

## (D) 3Ο/4

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Ex 5.8, 1 - Chapter 5 Class 12 -Continuity and Differentiability

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Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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This question is similar to

Ex 5.8, 1 - Chapter 5 Class 12 -Continuity and Differentiability

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

Question 11 The value of c in Rolleβs Theorem for the function f (x) = π^π₯ sin π₯, π₯β [0, π] is (A) π/6 (B) π/4 (C) π/2 (D) 3π/4 π (π₯)= π^π₯ sinβ‘γπ₯, π₯ β [0,π] γ First, we will check if the conditions of Rolleβs theorem are satisfied Condition 1 We need to check if π(π₯)=π^π₯ sinβ‘γπ₯ γis continuous at [π,π ] Let π(π)=ππ and π(π)=π¬π’π§β‘π We know that, π(π₯)=π^π₯ is continuous as it is an exponential function β(π₯)=sinβ‘π₯ is continous as sin is continuous at [π,π ] Hence, π(π)=π(π)π(π) is also continuous at [0,π] β΄ π(π)=π^π₯ sinβ‘π₯ is continuous at π₯β[0,π] Condition 2 We need to check if π(π₯)=π^π₯ sinβ‘π₯ is differentiable at (π,π ) A function is said to be differentiable if the derivative of the function exists. Differentiating π(π₯) wrt π₯ π^β² (π)=π^π πππβ‘π+π^π ππ¨π¬β‘π Since, derivative of the given function exists Hence, π(π) is differentiable at (0,π) Finding π(π) π(π₯)=π^π₯ sinβ‘π₯ π(π) = π^0 sinβ‘0 = π^0 (0) = 0 Findingπ(π ) π(π₯) = π^π₯ sinβ‘π₯ π(π ) = π^π sinβ‘π = π^π (0) = 0 Condition 3 Hence π(π) = π(π ) Now, π(π₯) = π^π₯ sinβ‘π₯ π^β² (π₯) =π^π₯ sinβ‘π₯+π^π₯ cosβ‘π₯ π^β² (π₯) =π^π₯ γ(sinγβ‘π₯+cosβ‘π₯) π^β² (π) = π^π γ(πππγβ‘π+πππβ‘π) Since all three condition satisfied π^β² (π) = 0 π^π γ(πππγβ‘π+πππβ‘π) = π Either, π^π=π It is not possible, since it is an exponential function. Or, γ(πππγβ‘π+πππβ‘π)=π sinβ‘c=βcosβ‘c Dividing by πππβ‘π on both sides sinβ‘c/cosβ‘c =βcosβ‘c/cosβ‘c πππ§β‘γπ=βπγ c=3Ο/4, 7Ο/4, β¦ Since, c β(0,Ο) β΄ c = ππ /π So, the correct answer is (D)