Let f : (0, ) R be a function given by f(x) = cases ( 87 ) 8x 7x, & 0 < x < 2 \ a - 8, & x = 2 \ (1 + | x|) ba| x|, & 2 < x < cases where a, b Z. If f is continuous at x = 2, then a2 + b2 is equal to .
Answer
81
Solution
Left-hand limit: put x = 2 - h (h 0+); then 8x = (4 - 8h) = - 8h and 7x = ( 7 2 - 7h ) = 7h
So 8x 7x = - 8h 7h 0
Hence LHL = ( 87 )0 = 1.
Continuity gives a - 8 = 1, so a = 9.
Right-hand limit: put x = 2 + h; then | x| = h and | x| = h
Since h = 1 h and (1 + t) 1t e as t 0:
RHL = h 0 [(1 + h) 1 h ] b9 = e b9
Since RHL = 1, e b9 = 1 gives b = 0. Therefore a2 + b2 = 81 + 0 = 81.
Answer: 81