Let f : R → R be defined as
f(x) = { 2 sin ( - (π x)/2 ),if x < - 1; |ax² + x + b|, if - 1 ≤ x ≤ 1; sin (π x), if x > 1 If f(x) is continuous on R, then a + b equals :

Solution

Maths sir · 4 steps

Start at x = 1: the right piece tends to 0, so the modulus must vanish, and that pins down a + b.

1. Left joint. At x = -1, the left piece tends to 2 sin(π/2) = 2. The middle piece gives f(-1) = |a - 1 + b|. Continuity needs |a + b - 1| = 2, so a + b = 3 or a + b = -1.
2. Right joint. At x = 1, the right piece tends to sin π = 0. The middle piece gives f(1) = |a + 1 + b|. A modulus equals zero only when the expression inside it is zero.

   |a + b + 1| = 0 ⇒ a + b = -1

3. Check both joints. With a + b = -1, the condition at x = -1 reads |-1 - 1| = 2, which holds. The other root, a + b = 3, fails at x = 1, because |3 + 1| = 4 ≠ 0.
4. So. a + b = -1, which is option (c).

The idea to keep. Write one continuity equation at every joint of a piecewise function. A modulus set equal to zero gives one clean equation, so solve that joint first and use it to choose between the other roots.

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