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

∵ Continuous at x = 2 ⇒ p 2 + q = 2 2 - 5 2 + 6 ⇒ q = - 2 p … i Continuous at x = 3 ⇒ a 2 9 + b 3 + 1 = 3 2 - 5 3 + 6 ⇒ 9 a 2 + 3 b + 1 = 0 … ii f ' x = p x ≤ 2 2 x − 5 2 < x < 3 2 a 2 x + b x ≥ 3 ∵ Differentiable at x = 2 ⇒ p = 2 2 - 5 ⇒ p = - 1 … . . iii ∵ Differentiable at x = 3 ⇒ 2 a 2 3 + b = 2 3 - 5 ⇒ 6 a 2 + b = 1 … … iv ∴ p = - 1 , q = 4 , a = 2 3 , b = - 5 3 from equations i , i i