Let the functions f : R → R and g : R → R be defined as :
f(x) = { x + 2,, x < 0; x²,, x ≥ 0 and
g(x) = { x³,, x < 1; 3x - 2,, x ≥ 1
Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :

Solution

Maths sir · 4 steps

Write fog piece by piece first, then test only the joining points x = 0 and x = 1.

1. Sign of g decides. f uses g + 2 when g(x) < 0 and g² when g(x) ≥ 0. So split the real line wherever g changes formula or changes sign: at x = 1 and at x = 0.

   Interval	g(x)	Sign of g	(f ° g)(x)
   x < 0	x³	negative	x³ + 2
   0 ≤ x < 1	x³	≥ 0	x⁶
   x ≥ 1	3x - 2	≥ 1	(3x - 2)²

2. Check x = 0. From the left (f ° g)(x) → 0 + 2 = 2, but from the right it goes to 0. The function jumps, so it is not continuous, and therefore not differentiable at x = 0.
3. Check x = 1. Both pieces give the value 1, so it is continuous. The left derivative is 6x⁵ = 6 and the right derivative is 6(3x - 2) = 6. They match, so it is differentiable at x = 1.
4. So. Each piece is a polynomial, so the only bad point is x = 0. The number of such points is 1, option (c).

The idea to keep. For a composite of piecewise functions, first find where the inner function crosses the outer function's breakpoint. Then check only those joints: a jump means not differentiable straight away.

Answer from the NTA final answer key. Solution by PrepWiser.