The correct option is
B [2,∞)Substitute x=3x,y=0 in f(x+y2)=2+f(x)+f(y)3, we get
f(3x+03)=2+f(3x)+f(0)3⇒f(x)=2+f(3x)+f(0)3 ...(1)
Substitute x=y=0
f(0)=2+f(0)+f(0)3⇒3f(0)=2+2f(0)⇒f(0)=2 ....(2)
From (1)
f(x)=2+f(3x)+23⇒3f(x)=f(3x)+4 ...(3)
Now f′(x)=limh→0f(x+h)−f(x)h=limh→0f(3x+3h3)−f(x)h
=limh→0⎧⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪⎩2+f(3x)+f(3h)3−f(x)h⎫⎪
⎪
⎪
⎪⎬⎪
⎪
⎪
⎪⎭=limh→0[2+f(3x)+f(3h)−3f(x)3h]
=limh→0[2+3f(x)−4+f(3h)−3f(x)3h]
f′(x)=limh→0[f(3h)−23h]
Since f(x) is differentiable ∀x;limh→0f(3h)=2
⇒f(0)=2⇒f′(x)=limh→03f′(3h)3=f′(0)=k (say)
⇒f′(x)=k⇒f(x)=kx+c
f′(x)=k⇒f′(2)=k
But f′(2)=2⇒k=2
∴f(x)=2x+c
Also f(0)=2⇒c=2
∴f(x)=2x+2
⇒f(|x|)=2|x|+2
⇒f(∣∣∣x2∣∣∣)=2∣∣∣x2∣∣∣+2=|x|+2;
⇒g(x)=∣∣∣f∣∣∣x2∣∣∣∣∣∣=||x|+2|=|x|+2{x+2x≥0−x+2x<0
Graph of ||x|+2| is shown below
Clearly the range of g(x) is [2,∞)