Permutation and CombinationHard
Question
Let g(x) = 2f(x/2) + f(1 - x) and f″(x) < 0 in 0 ≤ x ≤ 1 then g(x)
Options
A.decreases in 
B.decreases in 
C.increases in
D.increases in
Solution
g(x) = 2f 
+ f(1 - 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]
⇒ x/2 ≤ 1- x ⇒ x ≤ 2 - 2x
⇒ 3x ≤ 2 ⇒ x ≤ 2/3 ⇒ 0 ≤ x ≤ 2/3
⇒ g(x) increases in 0 ≤ x ≤ 2/3
and g′(x) ≤ 0 for decreasing
⇒ f′
≤ f′(1 - x) ⇒ x/2 ≥ 1- x
⇒ x ≥ 2/3
⇒ 2/3 ≤ x ≤ 1
+ f(1 - 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]
⇒ x/2 ≤ 1- x ⇒ x ≤ 2 - 2x
⇒ 3x ≤ 2 ⇒ x ≤ 2/3 ⇒ 0 ≤ x ≤ 2/3
⇒ g(x) increases in 0 ≤ x ≤ 2/3
and g′(x) ≤ 0 for decreasing
⇒ f′
≤ f′(1 - x) ⇒ x/2 ≥ 1- x⇒ x ≥ 2/3
⇒ 2/3 ≤ x ≤ 1
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