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Discuss the continuity of f(x) at x = π4 where, f(x) =(sinx+cosx)3-22sin2x-1,for x≠π4=32,for x=π4

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प्रश्न

Discuss the continuity of f(x) at x = `pi/4` where, 

f(x) `{:(= ((sinx + cosx)^3 - 2sqrt(2))/(sin 2x - 1)",", "for"  x ≠ pi/4),(= 3/sqrt(2)",", "for"  x = pi/4):}`

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उत्तर

`"f"(pi/4) = 3/sqrt(2)`

= `lim_(x -> pi/4) "f"(x) =  lim_(x -> pi/4) ((sinx + cosx)^3 - 2sqrt(2))/(sin 2x - 1)`

= (sin x + cos x)3 = `[(sin x + cos x)^2]^(3/2)`

= `(1 + sin 2x)^(3/2)`

∴ `lim_(x -> pi/4) "f"(x) =  lim_(x -> pi/4) ((1 + sin 2x)^(3/2) - 2^(3/2))/(sin 2x - 1)`

Put 1 + sin 2x = t

∴ sin 2x = t – 1

As `x -> pi/4","  "t" -> 1 + sin 2(pi/4)`

i.e. `"t" -> 1 + sin  pi/(2)`

i.e. t → 1 + 1

i.e. t → 2

∴ `lim_(x -> pi/4) "f"(x) =  lim_("t" -> 2) ("t"^(3/2) - 2^(3/2))/("t" - 1 - 1)`

= `lim_("t" -> 2) ("t"^(3/2) - 2^(3/2))/("t" - 2)`

= `3/2(2)^(1/2)    ...[lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= `(3sqrt(2))/(2)`

= `3/sqrt(2)`

∴ `lim_(x -> pi/4) "f"(x) = "f"(pi/4)`

∴ f(x) is continuous at x = `pi/(4)`

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अध्याय 8: Continuity - EXERCISE 8.1 [पृष्ठ १७४]

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