In this page mean value theorem questions we are going to see some practice questions of the topic Rolle's theorem.
(1) Verify Rolle's theorem for the following functions:
(i) f (x) = sin x 0 ≤ x ≤ π
(ii) f (x) = x², 0 ≤ x ≤ 1
(iii) f (x) = |x - 1|, 0 ≤ x ≤ 2
(iv) f (x) = 4 x³ - 9 x, -3/2 ≤ x ≤ 3/2
(2) Using Rolle's theorem find the points on the curve y = x² + 1,
- 2 ≤ x
≤ 2 where the tangent is parallel to x - axis.
Solution
(1) Verify Rolle's theorem for the following functions:
(i) f (x) = sin x 0 ≤ x ≤ π
If f(x) be a real valued function that satisfies the following three conditions.
1. |
f(x) is defined and continuous on the closed interval [0,π] |
2. |
f(x) is differentiable on the open interval (0,π). |
f (X) = sin x f (0) = sin 0 = 0 f (π) = sin π = 0 f (0) = f (π) from this we come to know that the given function satisfies all the conditions of Rolle's theorem. c ∈ (0,π) we can find the value of c by using the condition f '(c) = 0. f (x) = sin x f '(x) = cos x f '(c) = cos c f '(c) = 0 Cos c = 0 c = (2 n + 1) (π/2) c = π/2,3π/2,5π/2,............. Here π/2 is the only value that is in the given interval.So the value of c = π/2 mean value theorem questions |
(ii) f (x) = x², 0 ≤ x ≤ 1
If f(x) be a real valued function that satisfies the following three conditions.
1. |
f(x) is defined and continuous on the closed interval [0,1] |
2. |
f(x) is differentiable on the open interval (0,1). |
f (X) = x² f (0) = 0² = 0 f (1) = 1² = 1 f (0) ≠ f (1) The given function is not satisfying all the conditions of mean value theorem. So we cannot find the value of c. |