Question

Find a and b if the following function is continuous at the point indicated against them.

`f(x) = (32^x - 1)/(8^x - 1) + a` , for x > 0
         = 2                      , for x = 0
         = x + 5 − 2b       , for x < 0
is continuous at x = 0

Answer 1

f is continuous at x = 0

∴ `lim_(x→0^-) "f"(x)` = f(0)

∴ `lim_{x→0^-} (x + 5 - 2b) = 2`

∴ 0 + 5 – 2b = 2
∴ 5 – 2 = 2b
∴ 2b = 3

∴ b = `3/2`

Also `lim_(x→0^+) "f"(x)` = f(0)

∴ `lim_(x→0^+) ((32^x - 1)/(8^x - 1) + "a") = 2`

∴ `lim_(x→0^+) (((32^x - 1)/x)/((8^x - 1)/x)) + lim_(x→0^+) "a" = 2`

∴ `log 32/log 8 + a = 2` ....`[∵ lim_(x→0) ("a"^x - 1)/x = log "a"]`

∴ `log (2)^5/log(2)^3  + "a"` = 2

∴ `(5 log2)/(3log2) + "a"` = 2

∴ `5/3 + a = 2`

∴ a = `2 - 5/3`

∴ a = `1/3`

∴ a = `1/3`  and b = `3/2`