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Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension : f(x) = 1-cos2xsinx, for x ≠ 0

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प्रश्न

Show that following function have continuous extension to the point where f(x) is not defined. Also find the extension :

f(x) = `(1 - cos2x)/sinx`, for x ≠ 0

योग
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उत्तर

f(x) = `(1 - cos2x)/sinx`, for x ≠ 0

Here, f(0) is not defined.

Consider,

`lim_(x -> 0) "f"(x) =  lim_(x -> 0) (1 - cos 2x)/sinx`

= `lim_(x -> 0) (2sin^2x)/sinx`

= `2 lim_(x -> 0) (sin x)    ...[(because x -> 0","  therefore x ≠ 0),(therefore sin x ≠ 0)]`

= 2(sin 0)

= 2 × 0

= 0.

∴ `lim_(x -> 0) "f"(x)` exists.

But f(0) is not been defined.

∴ f(x) has a removable discontinuity at x = 0.

∴ The extension of the original function is

f(x) = `{:((1 - cos 2x)/sinx  ,";"   "for"  x ≠ 0),(= 0, ";"   "for"  x = 0):}`

∴ f(x) is continuous at x = 0

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अध्याय 8: Continuity - EXERCISE 8.1 [पृष्ठ १७३]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
EXERCISE 8.1 | Q 7) (i) | पृष्ठ १७३

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