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Find a and b if following function is continuous at the point or on the interval indicated against them: f(x) =4tanx+5sinxax-1, for x<0=9log2, for x=0=11x+7x⋅cosxbx-1, for x>0 - Mathematics and Statistics

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Question

Find a and b if following function is continuous at the point or on the interval indicated against them:

f(x) `{:(= (4tanx + 5sinx)/("a"^x - 1)",", "for"  x < 0),(= (9)/(log2)",", "for"  x = 0),(= (11x + 7x*cosx)/("b"^x - 1)",", "for"  x > 0):}`

Sum
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Solution

f(x) is continuous at x = 0

∴ `lim_(x -> 0^-) "f"(x)` = f(0)

∴ `lim_(x -> 0) [(4tanx + 5sinx)/("a"^x - 1)] = 9/log2`

∴ `lim_(x -> 0) [((4tanx + 5sinx)/x)/(("a"^x - 1)/x)]`  ...[∵ x → 0, x ≠ 0]

= `9/log2`

∴ `(lim_(x -> 0)((4tanx)/x + (5sinx)/x))/(lim_(x -> 0) ("a"^x - 1)/x) = 9/log2`

∴ `(4lim_(x -> 0) (tanx)/x + 5 lim_(x -> 0) (sinx)/x)/(lim_(x -> 0) ("a"^x - 1)/x) = 9/log2`

∴ `(4(1) + 5(1))/(log"a") = 9/log2   ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

∴ `9/log"a" = 9/log2`

∴ log a = log 2

∴ a = 2

Also `lim_(x -> 0^+) "f"(x)` = f(0)

∴ `lim_(x -> 0) (11x + 7x*cosx)/("b"^x - 1)  = 9/log2`

∴ `lim_(x -> 0) ((11x + 7x cosx)/x)/(("b"^x - 1)/x) = 9/log2`  ...[∵ x → 0, x ≠ 0]

∴ `(lim_(x -> 0)(11 + 7cosx))/(lim_(x -> 0)(("b"^x - 1)/x)) = 9/log2`

∴ `(11 + 7cos0)/log"b" = 9/log2  ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

∴ `(11 + 7(1))/log"b" = 9/log2`

∴ 9log b = 18log 2

∴ log b = 2log 2

= log(2)2

∴ log b = log 4

∴ b = 4

∴ a = 2 and b = 4

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Chapter 8: Continuity - MISCELLANEOUS EXERCISE-8 [Page 178]

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