Published by:
CGP EDU Academic Team
Published on: August 14, 2026
Let g(x) = 2f(x/2) + f(1 – x) and f ′′ (x) < 0 in 0 ≤ x ≤ 1 then g(x)
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(b,c)
g(x) =
and
g ′ (x) = f ′ (x/2) – f ′ (1 – x)
Now g(x) is increasing if g ′ (x) ≥ 0
f ′
≥ f ′ (1 – x)
[ f ′′ (x) < 0 i.e. f ′ (x) is decreasing]
⇒
≤ 1 – x ⇒ x ≤ 2 – 2x
⇒ 3x ≤ 2 ⇒ x ≤ 2/3 ⇒ 0 ≤ x ≤ 
⇒ g(x) increases in 0 ≤ x ≤ 2/3
and g ′ (x) ≤ 0 for decreasing
⇒
≤ f ′ (1 – x) ⇒
≥ 1 – x
⇒ x ≥ 2/3
⇒ 2/3 ≤ x ≤ 1
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