Question 25 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 18, 2021 by Teachoo

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The value of c in Rolle’s theorem for the function f (x) = x
^{
3
}
– 3x in the interval [0,√3] is

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Question 25
The value of c in Rolle’s theorem for the function f (x) = x3 – 3x in the interval [0,√3] is
(A) 1 (B) −1
(C) 3/2 (D) 1/3
𝑓 (𝑥)= 𝑥3 −3𝑥 , 𝑥 ∈ [0,√3]
First, we will check if the
conditions of Rolle’s theorem
are satisfied
Conditions of Rolle’s theorem
𝑓(𝑥) is continuous at {𝑎 , 𝑏}
𝑓(𝑥) is derivable at (𝑎 , 𝑏)
𝑓(𝑎) = 𝑓(𝑏)
If all 3 conditions satisfied then there exist some c in (𝑎 , 𝑏)
such that 𝑓′(𝑐) = 0
Condition 1
We need to check if 𝑓(𝑥)=𝑥3 −3𝑥 is continuous at [𝟎,√𝟑]
Since 𝑓(𝑥)=𝑥3 −3𝑥 is a polynomial
& Every polynomial function is continuous for all 𝑥 ∈𝑅
∴ 𝑓(𝑥)=𝑥3 −3𝑥 is continuous at 𝑥∈[0,√3]
Condition 2
We need to check if 𝑓(𝑥)=𝑥3 −3𝑥 is differentiable at (0,√3)
𝑓(𝑥) =𝑥3 −3𝑥 is a polynomial
& Every polynomial function is differentiable for all 𝑥 ∈𝑅
∴ 𝒇(𝒙) is differentiable at (0,√3)
Condition 3
Finding 𝒇(𝟎)
𝑓(𝑥) = 𝑥3 −3𝑥
𝑓(𝟎) = 03 −3(0)
= 0
Finding 𝒇(𝝅)
𝑓(𝑥) = 𝑥3 −3𝑥
𝑓(√𝟑) = (√3)3 −3(√3)
= 3(√3)−3(√3)
= 0
Hence, 𝒇(𝟎) = 𝒇(√𝟑)
Now,
𝑓(𝑥) = 𝑥3 −3𝑥
𝑓^′ (𝑥) = 3𝑥^2−3
𝒇^′ (𝒄) = 𝟑𝒄^𝟐−𝟑
Since all three condition satisfied
𝒇^′ (𝒄) = 𝟎
3𝑐^2−3= 0
3𝑐^2 = 3
𝑐^2 = 3/3
𝑐^2 = 1
c = 1
Since value of c = 1 ∈(𝟎,√𝟑)
So, the correct answer is (A)

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