Correct option is (C) y = h(f(x)) is not differentiable at 3 points

f(x) = 2|x + 1| - |x - 1| + 3|x - 2| + 2x + 1

 = \(\begin{cases}-2(x+1)+x-1-3(x-2)+2x+1;&x\leq -1\\2(x+1)+x-1-3(x-2)+2x+1;&-1\leq x\leq1\\2(x+1)-(x-1)-3(x-2)+2x+1;&1\leq x\leq2\\2(x+1)-(x-1)+3(x-2)+2x+1;&x\geq2\end{cases}\)

=\(\begin{cases}-2x+4&;&x\leq -1\\2x+8&;&-1\leq x\leq1\\10&;&1\leq x\leq2\\6x-2&;&x\geq2\end{cases}\)

\(\because\) f(x) is continuous

\(\therefore\) h(f(x)) is discontinuous only where h is discontinuous and h(x) can be discontinuous only at x = 3

Hence, h(f(x)) is discontinuous only when f(x) = 3

when x \(\leq\) - 1, 2x + 4 = 3 ⇒ 2x = 4 - 3 = 1 

⇒ x = 1/2 (Not satisfied)

when -1 \(\leq\) x \(\leq\) 1; 2x + 8 = 3 ⇒ 2x = 3 - 8 = -5

⇒ x = -5/2 (not satisfied)

when 1 \(\leq\) x \(\leq\) 2 ; f(x) = 10 \(\neq3\)

when x \(\geq\) 2, 6x - 2 = 3

⇒ 6x = 2 + 3 = 5

⇒ x = 5/6 \(\ngeq\)2 (not satisfied)

Hence, h(f(x)) is always continuous.

Also f(x) \(\geq\) 6 \(\forall\) x \(\in\) R

\(\therefore\) h(f(x)) = 2f(x) which is always continuous but not differentiable at x = 1, -1, & 2.