Continuity and Differentiability · Question 17

Let f(x) = cases (1 + ax)1/x, & x < 0 \ 1 + b, & x = 0 \ (x+4)1/2 - 2(x+c)1/3 - 2, & x > 0 cases be continuous at x = 0. Then ea bc is equal to:

1. 64
2. 48
3. 36
4. 72

Answer

(2) 48

Solution

LHL: x 0- (1 + ax)1/x = ea

RHL: numerator 0 as x 0+, so for a finite non-zero limit the denominator must also 0

⇒ c1/3 = 2 ⇒ c = 8

RHL = x 0+ (x+4)1/2 - 2(x+8)1/3 - 2

= 12(4)-1/2 13(8)-2/3 (L'Hôpital's rule)

= 1/41/12 = 3

Continuity ⇒ ea = 1 + b = 3 ⇒ b = 2

∴ ea bc = 3 2 8 = 48

Answer: (2)