Let F(x) = (f(x)) 2 + (f ′ (x)) 2 , F(0) = 7, where f(x) is thrice differentiable function such that
|f(x)| ≤ 1 ∀ x ∈ [–1, 1], then prove the followings.
(i) there is atleast one point in each of the intervals (–1, 0) and (0, 1) where |f ′ (x)| ≤ 2
(ii) there is atleast one point in each of the intervals (–1, 0) and (0, 1) where F(x) ≤ 5
(iii) there exits atleast one maxima of F(x) in (–1, 1)
(iv) for some c ∈ (–1, 1), F(c) ≥ 7, F ′
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) 
= f ′ ( α ), where – 1 < α < 0
⇒ |f ′ ( α ))| = |f(0) – f(1)| ≤ |f(0)| + |f(–1)|
⇒ |f ′ ( α )| ≤ 1 + 1 = 2
similarly for 0 < β < 1
(ii) 
(iii) Obvious from (i) and (ii) that there exits atleast one max.
(iv) Also from (i) and (ii) option iv is quite obvious.
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems