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Let f:[0,2]→ R be a twice differentiable function such that f''(x)>0, for all x∈(0,2). If φ(x)=f(x)+f(2-x), then φ is -
- increasing on (0,2)
- increasing on (0,1) and decreasing on (1,2)
- decreasing on (0,2)
- decreasing on (0,1) and increasing on (1,2)
Correct answer: decreasing on (0,1) and increasing on (1,2)
Solution
phi(x)=f(x)+f(2-x) so phi'(x)=f'(x)-f'(2-x). Since f''>0, f' is increasing: for x in (0,1), x<2-x gives phi'<0 (decreasing); for x in (1,2), phi'>0 (increasing). So phi decreases on (0,1) and increases on (1,2).
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