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

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

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

Let f:[0,2] → R be a twice differentiable function such… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 8 Apr · Shift I

Let

f:[0,2] \to R

be a twice differentiable function such that

f''(x) > 0

, for all

x \in (0,2)

. If

\phi(x) = f(x) + f(2-x)

, then

\phi

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 $(0, 2)$

Show answer and solution

Correct option: C

Solution

φ'(x) = f'(x) - f'(2-x). Since f''(x) > 0, f' is strictly increasing on (0,2). For x ∈ (0,1), we have x < 2-x, so f'(x) < f'(2-x), giving φ'(x) < 0 → φ decreasing on (0,1). For x ∈ (1,2), x > 2-x, so f'(x) > f'(2-x), giving φ'(x) > 0 → φ increasing on (1,2).

[JEE Main 2019 · 8 Apr · Shift I]

Practise Application of Derivatives the way it is asked

A timed set of real JEE Main Application of Derivatives questions, graded as the exam grades. No sign-up.

Start a Application of Derivatives test

More: all JEE Main Application of Derivatives questions · JEE Main 2019 · Mathematics chapter list

JEE Main 2019 Mathematics question on Application of Derivatives