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Select the correct answer from the given alternatives: f(x) = =2cotx-1π-2x,for x≠π2=log2,for x=π2

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प्रश्न

Select the correct answer from the given alternatives:

f(x) = `{:(= (2^(cotx) - 1)/(pi - 2x)",", "for"  x ≠ pi/2),(= log sqrt(2)",", "for"  x = pi/2):}`

विकल्प

  • f is continuous at x = `pi/2`

  • f has a jump discontinuity at x = `pi/2`

  • f has a removable discontinuity

  • `lim_(x -> pi/2) "f"(x)` = 2 log 3

MCQ
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उत्तर

f is continuous at x = `pi/2`

Explanation;

`"f"(pi/2) = log sqrt(2)`

`lim_(x -> pi/2) "f"(x) =  lim_(x -> pi/2) (2^(cotx) - 1)/(pi - 2x)`

= `lim_(x -> pi/2) (2^(tan(pi/2 - x)) - 1)/(2(pi/2 - x))`

Put `pi/2 - x` = h

As `x -> pi/2, "h" -> 0`

∴ `lim_(x -> pi/2) "f"(x) =  lim_("h" -> 0) (2^(tan"h") - 1)/(2"h")`

= `1/2 lim_("h" -> 0) ((2^(tan"h") - 1)/(tan"h") xx tan"h"/"h")`   ...[∵ h → 0, ∴ tan h → 0 ∴ tan h ≠ 0)

= `1/2 lim_("h" -> 0) (2^(tan"h") - 1)/tan"h" xx lim_("h" -> 0) tan"h"/"h"`

= `1/2*log 2*(1)`

= `log sqrt(2)`

= `"f"(pi/2)`

∴ f is continuous at x = `pi/2`

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अध्याय 8: Continuity - MISCELLANEOUS EXERCISE-8 [पृष्ठ १७६]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
MISCELLANEOUS EXERCISE-8 | Q (I) (1) | पृष्ठ १७६

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