Let ƒ : [0, 2] \to R be a twice differentiable function such that ƒ''(x) > 0, for all x \in (0, 2). If \phi (x) = ƒ(x) + ƒ(2 – x), then \phi is :

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

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

Application of Derivatives PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q45 with Solution

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Correct answer: B

We are given $f''(x) > 0 \,\text{for all } x \in (0,2).$ Hence $f$ is strictly convex on $(0,2)$ , so $f'(x)$ is strictly increasing on $(0,2)$ .
Define $\phi(x)=f(x)+f(2-x).$ We differentiate: $\phi'(x)=f'(x)+\frac{d}{dx}f(2-x)=f'(x)-f'(2-x).$
Now analyze the sign of $\phi'(x)$ .

Since $f'(x)$ is strictly increasing:
- If $x \in (0,1)$ , then $x < 2-x$ . Therefore, $f'(x) < f'(2-x).$ So, $\phi'(x)=f'(x)-f'(2-x)<0.$ Hence $\phi$ is decreasing on $(0,1)$ .
- If $x \in (1,2)$ , then $x > 2-x$ . Therefore, $f'(x) > f'(2-x).$ So, $\phi'(x)=f'(x)-f'(2-x)>0.$ Hence $\phi$ is increasing on $(1,2)$ .
Also at $x=1$ , $\phi'(1)=f'(1)-f'(1)=0.$ So $x=1$ is the turning point.
Therefore, $\phi$ is $\boxed{\text{decreasing on }(0,1)\text{ and increasing on }(1,2).}$
Checking options:
- A: false
- B: true
- C: false
- D: false

Thus the correct option is B.