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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`
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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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