If f(x) = cases a|x| + x2 - 2( |x|)( |x|)x & , x 0 \ b & , x = 0 cases is continuous at x = 0, then a + b is equal to :
- 1
- 2
- 0
- 4
Answer
(2) 2
Solution
2 |x| |x| = 2|x|
Right-hand limit (x > 0, |x| = x):
x 0+ ax + x2 - 2xx = a + 0 - 2 = a - 2
Left-hand limit (x < 0, |x| = -x, 2|x| = - 2x):
x 0- -ax + x2 + 2xx = -a + 0 + 2 = 2 - a
Continuity at x = 0 RHL = LHL = f(0)
a - 2 = 2 - a a = 2
b = a - 2 = 0
a + b = 2 + 0 = 2
Answer: (2)