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Find k if following function is continuous at the point indicated against them: f(x) =45x-9x-5x+1(kx-1)(3x-1),for x≠0=23,for x=0} at x = 0 - Mathematics and Statistics

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Question

Find k if following function is continuous at the point indicated against them:

f(x) `{:(= (45^x - 9^x - 5^x + 1)/(("k"^x - 1)(3^x - 1))",", "for"  x ≠ 0),(= 2/3",", "for"  x = 0):}}` at 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) ((45)^x - 9^x - 5^x + 1)/(("k"^x - 1)(3^x - 1)) = 2/3`

∴ `lim_(x -> 0) (9^x * 5^x - 9^x - 5^x + 1)/(("k"^x - 1)(3^x - 1)) = 2/3`

∴ `lim_(x -> 0) (9^x (5^x - 1) - 1(5^x - 1))/(("k"^x - 1)(3^x - 1)) = 2/3`

∴ `lim_(x -> 0) ((5^x - 1)(9^x - 1))/(("k"^x - 1)(3^x - 1)) = 2/3`

∴ `lim_(x -> 0) (((5^x - 1)(9^x - 1))/(x^2))/((("k"^x - 1)(3^x - 1))/(x^2)) = 2/3    ...[("Divide Numerator Denominator"),("by"  x^2),(because x -> 0","  therefore x ≠ 0  therefore x^2 ≠ 0)]`

∴ `(lim_(x -> 0) ((5^x - 1)/x)((9^x - 1)/x))/(lim_(x -> 0)(("k"^x - 1)/x)((3^x - 1)/x)) = 2/3`

∴ `((lim_(x -> 0) (5^x - 1)/x)*(lim_(x -> 0) (9^x - 1)/x))/((lim_(x -> 0) ("k"^x - 1)/x)* (lim_(x -> 0) (3^x - 1)/x)) = 2/3`

∴ `(log5 * log9)/(log"k" * log3) = 2/3   ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

∴ `(log5 * log(3)^2)/(log"k" * log3) = 2/3`

∴ `(log5 xx log 3)/(log "k" xx log3) = 1/3`

∴ 3 log 5 = log k

∴ log(5)3 = log k

∴ (5)3 = k

∴ k = 125

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Continuous and Discontinuous Functions
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Chapter 8: Continuity - MISCELLANEOUS EXERCISE-8 [Page 178]

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