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Examine whether the function is continuous at the points indicated against them:f(x) = x2+18x-19x-1 for x ≠ 1 = 20 for x = 1, at x = 1

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प्रश्न

Examine whether the function is continuous at the points indicated against them:
f(x) = `(x^2 + 18x - 19)/(x - 1)`        for x ≠ 1

      = 20                               for x = 1, at x = 1

योग
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उत्तर

`lim_(x→1) "f"(x) = lim_(x→1) (x^2 + 18x - 19)/(x - 1)`

= `lim_(x→1) (x^2 + 19x - x - 19)/(x - 1)`

= `lim_(x→1) (x(x + 19) - 1(x + 19))/(x - 1)`

= `lim_(x→1) ((x - 1)(x + 19))/((x - 1))`

= `lim_(x→1) (x + 19)`  ....[∵ x → 1, ∴ x ≠ 1, ∴ x - 1 ≠ 0]

= 1 + 19 = 20
Also, f(1) = 20

∴ `lim_(x→1) "f"(x) = "f"(1)`

∴ f(x) is continuous at x = 1

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Continuity - Exercise 8.1 [पृष्ठ ११२]

APPEARS IN

बालभारती Mathematics and Statistics (Commerce) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 8 Continuity
Exercise 8.1 | Q 2. (ii) | पृष्ठ ११२

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