If f x = p x + q : x ≤ 2 x 2 - 5 x + 6 : 2 < x < 3 a x 2 + b x + 1 : x ≥ 3 is differentiable everywhere, then p + q + 1 a + 1 b is equal to

Step-by-step solution

Continuity at   x = 2 ⇒ p 2 + q = 2 2 - 5 2 + 6 ⇒ q = - 2 p Continuity at x = 3 ⇒ a 9 + b 3 + 1 = 0 Differentiable at x = 2 ⇒ p = 2 2 - 5 ⇒ p = - 1 Differentiable at x = 3 ⇒ 2 a 3 + b = 2 3 - 5 ⇒ 6 a + b = 1 ∴ p = - 1 ,   q = 2 ,   a = 4 9 ,   b = - 5 3 p + q + 1 a + 1 b = 1 + 2 + 9 4 + 3 5