Question 12
The value of c in Mean value theorem for the function f (x) = x (x โ 2), x โ [1, 2] is
(A) 3/2 (B) 2/3 (C) 1/2 (D) 3/2
๐(๐ฅ)=๐ฅ" (" ๐ฅ" โ 2)"
๐(๐ฅ) = ๐ฅ^2 โ 2๐ฅ in interval [1, 2].
Checking conditions for
Mean value Theorem
Conditions of Mean value theorem
๐(๐ฅ) is continuous at {๐, ๐}
๐(๐ฅ) is differentiable at (๐ , ๐)
If both conditions satisfied, then there exist some c in (๐ , ๐)
such that ๐โฒ(๐) = (๐(๐) โ ๐(๐))/(๐ โ ๐)
Condition 1
We need to check if๐(๐ฅ) = ๐ฅ^2 โ 2๐ฅ is continuous in interval [1, 2]
Since ๐(๐ฅ) is polynomial .
it is continuous
โด ๐(๐) is continuous at [1, 2]
Condition 2
We need to check if ๐(๐ฅ) = ๐ฅ^2 โ 2๐ฅ is differentiable in interval (1, 2)
Since ๐(๐ฅ) is a polynomial .
it is Differentiable
โด ๐(๐) is differentiable in (1, 2)
Since both conditions are satisfied
From Mean Value Theorem,
There exists a c โ (1, 2) such that,
๐^โฒ (๐) = (๐(๐) โ ๐(๐))/(๐ โ ๐)
2๐โ2= ((2^2 โ 2(2)) โ (1^2 โ 2(1)))/1
๐๐โ๐ = ((๐ โ ๐) โ (๐ โ ๐))/๐
๐๐โ๐ = (๐ โ (โ๐))/๐
2๐โ2 = 1
2๐ = 1+2
2๐ = 3
๐=๐/๐
Since c = ๐/๐ โ(๐, ๐)
So, the correct answer is (A)

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.