Let $\mathrm{f}:[0,2] \rightarrow \mathrm{R}$ be a twice differentiable function such that $\mathrm{f}^{\prime \prime}(\mathrm{x})>0$, for all $\mathrm{x} \in(0,2)$. If $\phi(x)=f(x)+f(2-x)$, then $\phi$ is:

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

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

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Application of Derivatives

Let

f : [0, 2] \to R

be a twice differentiable function such that

f^{''} (x) > 0

, for all

x \in (0, 2)

. If

ϕ (x) = f (x) + f (2 - x)

, then

ϕ

is:

Answer: (C)

Differentiate

ϕ (x) = f (x) + f (2 - x)

to get

ϕ^{'} (x) = f^{'} (x) - f^{'} (2 - x)

. Because

f^{''} > 0

throughout

(0, 2)

, the derivative

f^{'}

is strictly increasing. For

x < 1

we have

x < 2 - x

, so

f^{'} (x) < f^{'} (2 - x)

and

ϕ^{'} (x) < 0

, making

ϕ

decreasing on

(0, 1)

; for

x > 1

we have

x > 2 - x

, so

f^{'} (x) > f^{'} (2 - x)

and

ϕ^{'} (x) > 0

, making

ϕ

increasing on

(1, 2)

. Thus

ϕ

decreases on

(0, 1)

and increases on

(1, 2)

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option