Published by:
CGP EDU Academic Team
Published on: August 14, 2026
Find positive real numbers ‘a’ and ‘b’ such that f(x) = ax – bx 3 has four extrema on [–1 , 1] at each of which | f(x) | = 1
Text Solution
Verified by ExpertsThe correct answer is:
A
(a = 3 , b = 4 )
f’(x) = 0 ⇒ x = 
f
=

f
=

f(–1) = b – a
f(1) = a – b
Given that
=
= | b – a | = | a – b | = 1
⇒
= 1 ⇒ b =
⇒ a – b = 1 ⇒ a –
= 1 ⇒ 4a 3 – 27a + 27 = 0
a = –3, 
⇒ a = – 3 – 1 : b =
= – 
= – 4
Also , b – a = 1 ⇒
– a ⇒ 4a 3 – 27a – 27 = 0 ⇒ (a – 3) (2a + 3) 2 = 0 ⇒ a = 3 ⇒ b = 4
Rejecting –ve values , therefore a = 3 , b = 4
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