Let f : 0 ,   2 → R be a twice differentiable function such that f ' ' x > 0 , for…

Let f:0, 2→R be a twice differentiable function such that f''x>0, for all  x∈0, 2. If ϕx= fx+ f2–x, then ϕ is
  1. decreasing on 0,2 
  2. increasing on 0,2 
  3. increasing on (0,1) and decreasing on 1,2 
  4. decreasing on 0,1  and increasing on (1,2)

Solution

fx:0,2→R and f''x> 0  for x∈ [0,2]
    ⇒f'(x) is increasing for x∈ [0,2]
Now, ϕ(x)=f(x)+f(2-x)
    ⇒ϕ'x=f'x-f'(2-x)
For  x∈ [0,1) , x < 2 – x  

    ⇒f'(x)<f'(2-x)⇒ϕ'(x)<0
For x∈ (1, 2] , x > 2 – x

    ⇒f'x>f'2-x⇒ϕ'x>0
Hence, ϕ is decreasing on 0,1 and increasing on 1,2.

Asked in: JEE Main 2019 (08 Apr Shift 1)

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