If f x = e 2 x 3 + x x > 0 a x + b x ≤ 0 is differentiable at x = 0 , then

Step-by-step solution

f ' 0 − = lim h → 0 + f 0 − h − f 0 − h ⇒ lim h → 0 + − a h + b − b − h = a f ' 0 + = lim h → 0 + f 0 + h − f 0 h ⇒ lim h → 0 + e 2 h 3 + h − b h … … i For this limit to exist, it must be of 0 0 form. ∴ e 2 0 + 0 - b = 0 ⇒ b = 1 … … . . i i From i f ' 0 + = lim h → 0 + e 2 h 3 + h − 1 h ⇒ lim h → 0 + e 2 h 3 + h − 1 2 h 3 + h × 2 h 3 + h h ⇒ 1 × 1 = 1 ∵ f ' 0 − = f ' 0 + ⇒ a = 1