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If f(x) =24x-8x-3x+112x-4x-3x+1, for x≠0=k, for x=0} is continuous at x = 0, find k - Mathematics and Statistics

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प्रश्न

If f(x) `{:(= (24^x - 8^x - 3^x + 1)/(12^x - 4^x - 3^x + 1)",",  "for"  x ≠ 0), (= "k"",",  "for"  x = 0):}}` is continuous at x = 0, find k

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उत्तर

f(x) is continuous at x = 0

∴ f(0) = `lim_(x -> 0) "f"(x)`

∴ k = `lim_(x -> 0) (24^x - 8^x - 3^x + 1)/(12^x - 4^x - 3^x + 1)`

= `lim_(x -> 0) (8^x * 3^x - 8^x - 3^x + 1)/(4^x * 3^x - 4^x - 3^x + 1)`

= `lim_(x -> 0) (8^x(3^x - 1) - 1(3^x - 1))/(4^x (3^x - 1) - 1(3^x - 1)`

= `lim_(x -> 0) ((3^x - 1)(8^x - 1))/((3^x - 1)(4^x - 1)) ...[(because x -> 0","  3x -> 3^0),(therefore 3^x -> 1 therefore 3^x ≠ 1),(therefore 3^x - 1 ≠ 0)]` 

= `lim_(x -> 0) (8^x - 1)/(4^x - 1)`

= `lim_(x -> 0) (((8^x - 1)/x)/((4^x - 1)/x))` ...[∵ x → 0, ∴ x ≠ 0]

= `(lim_(x -> 0) (8^x - 1)/x)/(lim_(x -> 0) (4^x - 1)/x)`

= `log8/log4    ...[because lim_(x -> 0) (("a"^x - 1)/x) = log"a"]`

= `log(2)^3/log(2)^2`

= `(3log2)/(2log2)`

∴ f(0) = `3/2`

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Continuous and Discontinuous Functions
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अध्याय 8: Continuity - EXERCISE 8.1 [पृष्ठ १७४]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
EXERCISE 8.1 | Q 11) (i) | पृष्ठ १७४

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