Let f : R → R and g : R → R be defined as
f(x) = { x + a,, x < 0; |x - 1|,, x ≥ 0 and
g(x) = { x + 1,, x < 0; (x - 1)² + b,, x ≥ 0 ,
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x ∈ R, then a + b is equal to ____________.

Solution

Maths sir · 5 steps

Do not test g ° f only at 0: g also switches rule where f(x) crosses zero, at x=-a.

1. Track the sign. g changes formula depending on whether its input is negative, so look at the sign of f(x). For x ≥ 0, f(x) = |x-1| ≥ 0. For x < 0, f(x) = x + a is negative only when x < -a.

   (g° f)(x) = x+a+1, x<-a ; (x+a-1)²+b, -a≤ x<0 ; (|x-1|-1)²+b, x≥ 0

2. Check at minus a. Take a > 0. From the left the limit at x=-a is -a+a+1 = 1. The value at x=-a is (-1)² + b = 1 + b. Continuity needs 1 + b = 1, so b = 0.
3. Check at zero. From the left the limit at x=0 is (a-1)² + b. The value at 0 is (|0-1|-1)² + b = b. So (a-1)² = 0, which gives a = 1.
4. Case a equals zero. If a = 0, the middle piece disappears and g(f(x)) = x + 1 for all x < 0. At x=0 the left limit is 1 and the value is b, so b = 1.
5. So. Either way (a, b) is (1, 0) or (0, 1), so a + b = 1.

The idea to keep. For g ° f, list every break point: where f switches rule, and where f(x) reaches the point where g switches rule. Check continuity at each one.

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