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If f(x) = 4x-π+4π-x-2(x-π)2 for x ≠ π, is continuous at x = π, then find f(π) - Mathematics and Statistics

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प्रश्न

If f(x) = `(4^(x - π) + 4^(π - x) - 2)/(x - π)^2` for x ≠ π, is continuous at x = π, then find f(π).

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उत्तर

f is given to be continuous at x = π

∴ by definition,

f(π) = `lim_(x -> pi) "f"(x)`

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

Put x = π + h. Then as x → π, h → 0 and x – π = h

∴ f(π) = `lim_("h" -> 0) (4^"h" + 4^(-"h") - 2)/"h"^2`

= `lim_("h" -> 0) (4^"h"[4^"h" + 4^(-"h") - 2])/(4^"h" * "h"^2)`

= `lim_("h" -> 0) ((4^"h")^2 + 1 - 2(4^"h"))/(4^"h" * "h"^2)`

= `lim_("h" -> 0) (4^"h" - 1)^2/(4^"h" * "h"^2)`

= `[lim_("h" -> 0) (4^"h" - 1)/"h"]^2/(lim_("h" -> 0) 4^"h")`

= `(log 4)^2/4^0    ...[lim_(x -> 0) ("a"^x - 1)/x = log "a"]`

= (2 log 2)2

∴ f(π) = 4(log 2)2

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Continuous and Discontinuous Functions
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
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 10) (iii) | पृष्ठ १७४

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