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Let f(x): [0, 2] to R be a twice differ...

Let `f(x): [0, 2] to R` be a twice differenctiable function such that `f''(x) gt 0`, for all `x in (0, 2)`. If `phi (x) = f(x) + f(2-x)`, then `phi` is (A) increasing on (0, 1) and decreasing on (1, 2) (B) decreasing on (0, 2) (C) decreasing on (0, 1) and increasing on (1, 2) (D) increasing on (0, 2)

A

increasing on (0, 1) and decreasing on (1, 2)

B

decreasing on (0, 2)

C

decreasing on (0, 1) and increasing on (1, 2)

D

increasing on (0, 2)

Text Solution

Verified by Experts

The correct Answer is:
C
Given `phi( x) = f(x) + f(2- x ), AA x in ( 0, 2)`
`rArr " " phi' (x) = f ' (x) - f'(2 - x)`
Also, we have `f''(x) gt 0 AA x in (0, 2)`
`rArr f'(x)` is strictly increasing function
Now, for `phi( x)` to be increasing,
`phi' (x) ge 0`
`rArr f ' (x) - f'(2-x ) ge 0 " "` [using Eq. (i)]
`rArr " " f'(x) ge f' (2 - x) rArr x gt 2- x `
`" " [because f'` is a strictly increasing function]
`rArr " " 2x gt 2 rArr " " x gt 1`
Thus ` phi(x)` is increasing on `(1, 2)`
Similarly, for `phi (x)` to the decreasing,
`" " phi'(x) le 0`
`rArr " " f'(x) - f ' (2 - x) le 0`
[ using Eq. (i)]
`rArr " " f ' (x) le f ' ( 2- x ) `
`rArr x lt 2 - x " " [ because f'` is a strictly increasing function ]
`rArr " " 2x lt 2`
`rArr " " x lt 1 `
Thus, `phi (x)` is decreasing on `(0, 1)`.
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