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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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