Question 2
The value of c in Mean value theorem for the function f (x) = x (x – 2), x ∈ [1, 2] is
(A) 3/2 (B) 2/3 (C) 1/2 (D) 3/2
𝑓(𝑥)=𝑥" (" 𝑥" – 2)"
𝑓(𝑥) = 𝑥^2 – 2𝑥 in interval [1, 2].
Checking conditions for
Mean value Theorem
Conditions of Mean value theorem
𝑓(𝑥) is continuous at {𝑎, 𝑏}
𝑓(𝑥) is differentiable at (𝑎 , 𝑏)
If both conditions satisfied, then there exist some c in (𝑎 , 𝑏)
such that 𝑓′(𝑐) = (𝑓(𝑏) − 𝑓(𝑎))/(𝑏 − 𝑎)
Condition 1
We need to check if𝑓(𝑥) = 𝑥^2 – 2𝑥 is continuous in interval [1, 2]
Since 𝑓(𝑥) is polynomial .
it is continuous
∴ 𝒇(𝒙) is continuous at [1, 2]
Condition 2
We need to check if 𝑓(𝑥) = 𝑥^2 – 2𝑥 is differentiable in interval (1, 2)
Since 𝑓(𝑥) is a polynomial .
it is Differentiable
∴ 𝒇(𝒙) is differentiable in (1, 2)
Since both conditions are satisfied
From Mean Value Theorem,
There exists a c ∈ (1, 2) such that,
𝒇^′ (𝒄) = (𝒇(𝟐) − 𝒇(𝟏))/(𝟐 − 𝟏)
2𝑐−2= ((2^2 − 2(2)) − (1^2 − 2(1)))/1
𝟐𝒄−𝟐 = ((𝟒 − 𝟒) − (𝟏 − 𝟐))/𝟏
𝟐𝒄−𝟐 = (𝟎 − (−𝟏))/𝟏
2𝑐−2 = 1
2𝑐 = 1+2
2𝑐 = 3
𝒄=𝟑/𝟐
Since c = 𝟑/𝟐 ∈(𝟏, 𝟐)
So, the correct answer is (A)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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