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Discuss the continuity of the following function at the point(s) or on the interval indicated against them: f(x) = cos4x-cos9x1-cosx, for x ≠ 0 f(0) = 6815, at x = 0 on -π2≤x≤π2

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प्रश्न

Discuss the continuity of the following function at the point(s) or on the interval indicated against them:

f(x) = `(cos4x - cos9x)/(1 - cosx)`, for x ≠ 0

f(0) = `68/15`, at x = 0 on `- pi/2 ≤ x ≤ pi/2`

योग
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उत्तर

The domain of f(x) is `[- pi/2, pi/2]`

i. For `[- pi/2, pi/2] - {0}`

f(x) = `(cos4x - cos9x)/(1 - cosx)`

It is a rational function and is continuous everwhere except at points where its denominator becomes zero.

Denominator becomes zero when cos x = 1

i.e. x = 0

But x = 0 does not lie in the interval

∴ f(x) is continuous at all points in `[- pi/2, pi/2] - {0}`

ii. For continuity at x = 0

f(0) = `68/15`

`lim_(x -> 0) "f"(x) =  lim_(x -> 0) (cos4x - cos9x)/(1 - cosx)`

= `lim_(x -> 0) (2sin ((4x + 9x)/2) * sin((9x - 4x)/2))/(2sin^2  x/2)`

= `lim_(x -> 0) (sin((13x)/2)* sin((5x)/(2)))/(sin  x/2)^2`

= `lim_(x -> 0) [((sin((13x)/2) * sin ((5x)/2))/(x^2))/((sin  x/2)^2/(x^2))]   ...[("Divide numerator and denominator by"  x^2),("As"  x -> 0","  x ≠ 0 therefore x^2 ≠0)]`

= `(lim_(x -> 0) sin((13x)/2)/x * (sin ((5x)/2))/x)/(lim_(x -> 0) ((sin  x/2)/x)^2`

= `(lim_(x -> 0) sin((13x)/2)/((13x)/2) xx 13/2 * lim_(x -> 0) sin ((5x)/2)/((5x)/2) xx 5/2)/(lim_(x -> 0) ((sin  x/2)/(x/2))^2 xx 1/4`

= `(1 xx 13/2 xx 1 xx 5/2)/((1)^2 xx 1/4)  ...[(because  x -> 0","  (13x)/2 -> 0),((5x)/2 -> 0  "and" lim_(theta -> 0)  sintheta/theta = 1)]`

= 65

∴ `lim_(x -> 0) "f"(x) ≠ "f"(0)`

∴ f(x) is discontinuous at x = 0

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अध्याय 8: Continuity - MISCELLANEOUS EXERCISE-8 [पृष्ठ १७७]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
MISCELLANEOUS EXERCISE-8 | Q (II) (3) | पृष्ठ १७७

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