The value of c in Rolle’s Theorem for the function f (x) = e^x sin x,

x∈ [0, π] is

(A) π/6  (B) π/4 

(C) π/2 

(D) 3π/4

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



  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise


Question 11 The value of c in Rolle’s Theorem for the function f (x) = 𝑒^π‘₯ sin π‘₯, π‘₯∈ [0, πœ‹] is (A) πœ‹/6 (B) πœ‹/4 (C) πœ‹/2 (D) 3πœ‹/4 𝑓 (π‘₯)= 𝑒^π‘₯ sin⁑〖π‘₯, π‘₯ ∈ [0,πœ‹] γ€— First, we will check if the conditions of Rolle’s theorem are satisfied Condition 1 We need to check if 𝑓(π‘₯)=𝑒^π‘₯ sin⁑〖π‘₯ γ€—is continuous at [𝟎,𝝅] Let π’ˆ(𝒙)=𝒆𝒙 and 𝒉(𝒙)=𝐬𝐒𝐧⁑𝒙 We know that, 𝑔(π‘₯)=𝑒^π‘₯ is continuous as it is an exponential function β„Ž(π‘₯)=sin⁑π‘₯ is continous as sin is continuous at [𝟎,𝝅] Hence, 𝒇(𝒙)=π’ˆ(𝒙)𝒉(𝒙) is also continuous at [0,πœ‹] ∴ 𝒇(𝒙)=𝑒^π‘₯ sin⁑π‘₯ is continuous at π‘₯∈[0,πœ‹] Condition 2 We need to check if 𝑓(π‘₯)=𝑒^π‘₯ sin⁑π‘₯ is differentiable at (𝟎,𝝅) A function is said to be differentiable if the derivative of the function exists. Differentiating 𝑓(π‘₯) wrt π‘₯ 𝒇^β€² (𝒙)=𝒆^𝒙 π’”π’Šπ’β‘π’™+𝒆^𝒙 πœπ¨π¬β‘π’™ Since, derivative of the given function exists Hence, 𝒇(𝒙) is differentiable at (0,πœ‹) Finding 𝒇(𝟎) 𝑓(π‘₯)=𝑒^π‘₯ sin⁑π‘₯ 𝑓(𝟎) = 𝑒^0 sin⁑0 = 𝑒^0 (0) = 0 Finding𝒇(𝝅) 𝑓(π‘₯) = 𝑒^π‘₯ sin⁑π‘₯ 𝑓(𝝅) = 𝑒^πœ‹ sinβ‘πœ‹ = 𝑒^πœ‹ (0) = 0 Condition 3 Hence 𝒇(𝟎) = 𝒇(𝝅) Now, 𝑓(π‘₯) = 𝑒^π‘₯ sin⁑π‘₯ 𝑓^β€² (π‘₯) =𝑒^π‘₯ sin⁑π‘₯+𝑒^π‘₯ cos⁑π‘₯ 𝑓^β€² (π‘₯) =𝑒^π‘₯ γ€–(sin〗⁑π‘₯+cos⁑π‘₯) 𝒇^β€² (𝒄) = 𝒆^𝒄 γ€–(π’”π’Šπ’γ€—β‘π’„+𝒄𝒐𝒔⁑𝒄) Since all three condition satisfied 𝑓^β€² (𝑐) = 0 𝒆^𝒄 γ€–(π’”π’Šπ’γ€—β‘π’„+𝒄𝒐𝒔⁑𝒄) = 𝟎 Either, 𝒆^𝒄=𝟎 It is not possible, since it is an exponential function. Or, γ€–(π’”π’Šπ’γ€—β‘π’„+𝒄𝒐𝒔⁑𝒄)=𝟎 sin⁑c=βˆ’cos⁑c Dividing by 𝒄𝒐𝒔⁑𝒄 on both sides sin⁑c/cos⁑c =βˆ’cos⁑c/cos⁑c π­πšπ§β‘γ€–πœ=βˆ’πŸγ€— c=3Ο€/4, 7Ο€/4, … Since, c ∈(0,Ο€) ∴ c = πŸ‘π…/πŸ’ So, the correct answer is (D)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.