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Chapter 5 Class 12 Continuity and Differentiability
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## (D) 3Ο/4

This question is similar to Ex 5.8, 1 - Chapter 5 Class 12 - Continuity and Differentiability

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Question 11 The value of c in Rolleβs 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‘γπ₯, π₯ β [0,π] γ First, we will check if the conditions of Rolleβs theorem are satisfied Condition 1 We need to check if π(π₯)=π^π₯ sinβ‘γπ₯ γis continuous at [π,π] Let π(π)=ππ 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^β² (π)=π^π πππβ‘π+π^π ππ¨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^β² (π₯) =π^π₯ sinβ‘π₯+π^π₯ cosβ‘π₯ π^β² (π₯) =π^π₯ γ(sinγβ‘π₯+cosβ‘π₯) π^β² (π) = π^π γ(πππγβ‘π+πππβ‘π) Since all three condition satisfied π^β² (π) = 0 π^π γ(πππγβ‘π+πππβ‘π) = π Either, π^π=π It is not possible, since it is an exponential function. Or, γ(πππγβ‘π+πππβ‘π)=π sinβ‘c=βcosβ‘c Dividing by πππβ‘π on both sides sinβ‘c/cosβ‘c =βcosβ‘c/cosβ‘c π­ππ§β‘γπ=βπγ c=3Ο/4, 7Ο/4, β¦ Since, c β(0,Ο) β΄ c = ππ/π So, the correct answer is (D)