For a, b > 0, let f(x) = cases ((a + 1)x) + b xx, & x < 0 \ 3, & x = 0 \ ax + b2x2 - axb a x x, & x > 0 cases be a continuous function at x = 0. Then ba is equal to :
- 4
- 5
- 8
- 6
Answer
(4) 6
Solution
continuity at 0: LHL = f(0) = RHL = 3
LHL = x 0- ( (a + 1)xx + b xx ) = a + 1 + b = 3
RHL: take x common in the numerator:
x 0+ a + b2x - ab a x = x 0+ b2xb a x ( a + b2x + a ) (rationalise)
= b2b a 2 a = b2a = 3
ba = 6
Answer: (4)