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Question

# If g(x)=f(x)+f(x−1) and f′′(x)<0,0≤x≤1 then find the intervals of monotonicity of g(x).

A
xϵ[12,1]
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B
xϵ[12,1]
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C
xϵ[13,1]
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D
xϵ[13,1]
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Solution

## The correct option is A xϵ[12,1]g′(x)=f′(x)−f′(1−x) Now f′′(x)<0 when 0≤x≤1 Hence f'(x) is a decreasing function when 0≤x≤1. Now g(x) is increasing or decreasing according g′(x)=f′(x)−f(1−x)>0 or <0 or f′(x)>f′(1−x) for increasing i.e.f′(1−x)<f′(x)...(1) Since f' is a decreasing function then by definition later value is less than the earlier value i.e.f′(a)<f′(b) when a>b. or f′(x)<f′(1−x) for decreasing ...(2) Hence (1)⇒(1−x)>x or 1>2x or x<12 ∴xϵ[0,12] for g(x) to be increasing and ⇒x>(1−x) or 2x>1 or x>12 ∴xϵ[12,1] for g(x) to be decreasing.

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