Let g(x)=2f(x/2)+f(2-x) and f''(x)<0 for every x belongs to (0,2). Then g(x) increases in
a. (1/2,2)
b. (4/3,2)
d. (0,4/3)

Dear Student,
Please find below the solution to the asked query:

Given thatf''x<0 for every x0,2f'x is a decreasing function.We have:gx=2.fx2+f2-xDifferentiating both sides with respect to x, we getg'x=2.f'x2.ddxx2+f'2-x.ddx2-xg'x=2.f'x2.12+f'2-x.-1g'x=f'x2-f'2-xTo get interval of increase put:g'x>0f'x2-f'2-x>0f'x2>f'2-xx2<2-x As f'x is decreasing function, hence inequality reversesx<4-2xx+2x<4x<43Also x0,2Hence gx increases in 0,43Hence optiond is correct.

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