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Question

Let g(x)=2f(x2)+f(2−x) and f′′(x)<0, ∀ x∈(0,2). Then which of the following is (are) TRUE?

A
g(x) is decreasing in (0,43)
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B
g(x) is decreasing in (43,2)
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C
g(110)<g(1110)
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D
g(x) has a local minimum at x=43
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Solution

The correct options are
B g(x) is decreasing in (43,2)
C g(110)<g(1110)
g(x)=2f(x2)+f(2x)
g(x)=f(x2)f(2x)
Given, f′′(x)<0, x(0,2)
f(x) is decreasing in (0,2).

Case 1: x2<2x i.e., x<43
f(x2)>f(2x)
g(x)>0
g(x) is increasing in (0,43)

Case 2: x2>2x i.e., x>43
f(x2)<f(2x)
g(x)<0
g(x) is decreasing in (43,2)

Since g(x) is increasing in (0,43) and 110<1110
g(110)<g(1110)

g(x)>0 for x(0,43) and g(x)<0 for x(43,2).
So, g(x) has a local maximum at x=43

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