Sol. fog exists if range of g is a subset of domain of f and in that case the domain of fog is same as that of g.

g(x) = 2 – |x|, where – 1 ≤ x ≤ 3

⇒ 1 ≤ |x| ≤ 3 ⇒ – 1 ≥ –|x| ≥ –3 ⇒ 2 – 1 ≥ 2 – |x| ≥ 2 – 3

⇒ 1 ≥ g(x) ≥ –1

⇒ – 1 ≤ g (x) ≤ 1 ∴ Range of g = [–1, 1]

Clearly range (g) is not a subset of domain of f. So, let us restrict the domain of g so that its range is contained in the domain of f. Thus,

If 0 ≤ g(x) ≤ 4, then 0 ≤ 2 – |x| ≤ 4 ⇒ – 2 ≤ –|x| ≤ 2

⇒ – 2 ≤ |x| ≤ 2 ⇒ – 2 ≤ x ≤ 2

But x should also be in the domain of g i.e. –1 ≤ x ≤ 3. Thus range of g is contained in the domain of f, if we take [–1, 2] as the domain of g.

for x ∈ [–1, 2], we have fog (x) = f (2 –|x|) = –1 + |2 –|x| –2| = –1 + |x|

∴ the graph of fog (x) is given below It is evident from the graph that fog is continuous ∀ ∈ [–1, 2]

gof exists if range of f is contained in the domain of g.

– 1 ≤ –1+ |x –2| ≤ 1 ∀ x ∈ [0, 4]

⇒ –1 ≤ f(x) ≤ 1 ∀ x ∈ [0, 4]

⇒ range of f = [–1, 1] [–1, 3]

⇒ range of f domain of g.

so, gof exists for all x ∈ [0, 4]

gof (x) = g(f(x)) = g(–1 + | x –2| ) = 2– |–1 +| x –2||

⇒ gof (x) =

=

The graph of gof (x) is

clearly gof is continuous in [0, 4]