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F(x) = ,ifk,if{2x+2-164x-16, if x≠2k, if x=2 at x = 2 - Mathematics

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Question

f(x) = `{{:((2^(x + 2) - 16)/(4^x - 16)",",  "if"  x ≠ 2),("k"",",  "if"  x = 2):}` at x = 2

Sum
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Solution

We have, f(x) = `{{:((2^(x + 2) - 16)/(4^x - 16)",",  "if"  x ≠ 2),("k"",",  "if"  x = 2):}`

Since, f(x) is continuous at x = 2

∴ f(2) = `lim_(x -> 2) "f"(x)`

∴ k = `lim_(x -> 2) (2^(x + 2) - 16)/(4^x - 16)`

= `lim_(x -> 2) (4(2^x - 4))/((2^x - 4)(2^x + 4))`

= `lim_(x -> 2) 4/(2^x + 4)`

= `4/(4 + 4)`

= `1/2`

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Chapter 5: Continuity And Differentiability - Exercise [Page 108]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 12 | Page 108

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