True

Let f be a function defined by f(x) = |1-x + |x|| .

Let g(x) = 1-x + |x| and h(x) = |x| be two functions defined on R.

Then,

hog(x) = h(g(x))

⇒ hog(x) = h(1-x + |x|)

⇒ hog(x) = |(1-x + |x|)|

⇒ hog(x) = f(x) ∀ x ∈ R.

Since (1-x) is a polynomial function and |X| is modulus function are continuous in R.

⇒ g(x) = 1-x + |x| is everywhere continuous.

⇒ h(x) = |x| is everywhere continuous.

Hence, f = hog is everywhere continuous.