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Discuss the continuity of the following function at the point(s) or on the interval indicated against them: f(x) =|x+1|2x2+x-1,for x≠-1=0,for x=-1} at x = – 1 - Mathematics and Statistics

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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) `{:(= (|x + 1|)/(2x^2 + x - 1)",", "for"  x ≠ -1),(= 0",", "for"  x = -1):}}` at x = – 1

योग
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उत्तर

|x + 1| `{:(= x + 1, ";"  x  ≥ -1),(= - (x + 1), ";"  x < - 1):}`

∴ f(x) `{:(= (-(x + 1))/(2x^2 + x - 1), ";"  x < -1),(= 0, ";"  x = -1),(=(x + 1)/(2x^2 + x - 1), ";"  x > - 1):}`

f(–1) = 0

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

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

= `lim_(x -> -1^-) (-1)/(2x - 1)  ...[(because x -> -1","  therefore x ≠ -1),(therefore x + 1 ≠ 0)]`

= `(-1)/(2(-1) - 1)`

= `1/3`

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

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

= `lim_(x -> -1^+) (-1)/(2x - 1)   ...[("As"   x -> -1","  x ≠ -1),(therefore x + 1 ≠ 0)]`

= `1/(2(-1) - 1)`

= `(-1)/3`

∴ `lim_(x -> -1^-) "f"(x) ≠ lim_(x -> -1^+) "f"(x)`

∴ f(x) is discontinuous at x = – 1

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Continuous and Discontinuous Functions
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अध्याय 8: Continuity - MISCELLANEOUS EXERCISE-8 [पृष्ठ १७७]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
MISCELLANEOUS EXERCISE-8 | Q (II) (5) | पृष्ठ १७७

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