Let g(x) = 2f(x/2) + f(1 - x) and f''(x)

Question

(D) Given $g(x) = 2f(x/2) + f(1 - x)$.Taking the derivative with respect to $x$,we get $g'(x) = 2f'(x/2) \cdot (1/2) + f'(1 - x) \cdot (-1) = f'(x/2)

Vedclass · Accepted Answer

Both $(B)$ and $(C)$

Vedclass · Answer

(D) Given $g(x) = 2f(x/2) + f(1 - x)$.Taking the derivative with respect to $x$,we get $g'(x) = 2f'(x/2) \cdot (1/2) + f'(1 - x) \cdot (-1) = f'(x/2) - f'(1 - x)$.Given $f''(x) For $g(x)$ to be increasing,we require $g'(x) > 0$.$f'(x/2) - f'(1 - x) > 0 \implies f'(x/2) > f'(1 - x)$.Since $f'(x)$ is strictly decreasing,$f'(a) > f'(b) \implies a Therefore,$x/2 Thus,$g(x)$ increases in $[0, 2/3)$.For $g(x)$ to be decreasing,we require $g'(x) $f'(x/2) - f'(1 - x) Since $f'(x)$ is strictly decreasing,$f'(a)  b$.Therefore,$x/2 > 1 - x \implies 3x/2 > 1 \implies x > 2/3$.Thus,$g(x)$ decreases in $(2/3, 1]$.Both $(B)$ and $(C)$ are correct.

Let $g(x) = 2f(x/2) + f(1 - x)$ and $f''(x) < 0$ for $0 \le x \le 1$. Then $g(x)$:

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