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For what value of λ is the function defined by f(x) = {λ⁢(𝑥2 − 2⁢𝑥), if 𝑥 ≤ 0, 4⁢𝑥 + 1, if 𝑥 > 0 continuous at x = 0? What about continuity at x = 1? - Mathematics

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प्रश्न

For what value of λ is the function defined by f(x) = `{(λ(x^2 - 2x)", if"  x <= 0),(4x+ 1", if"  x > 0):}` continuous at x = 0? What about continuity at x = 1?

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उत्तर

f(x) = `{(λ(x^2 - 2x)", if"  x <= 0),(4x+ 1", if"  x > 0):}`

If f(x) is continuous at x = 0, it implies:

f(0) = `lim_(x -> 0^+)` f(x) = `lim_(x -> 0^-)` f(x)

⇒ [02 − 2(0)] = [4(0) + 1] = [02 − 2(0)]

⇒ 0 = 1 = 0

Which cannot be true, i.e. for any value of  λ this function is not continuous at x = 0.

If f(x) is continuous at x = 1, this implies:

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

⇒ 4(1) + 1 = 4(1) + 1 = 4(1) + 1

⇒ 5 = 5 = 5

Which is always true, i.e. for any value of  λ this function is continuous at x = 1.

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अध्याय 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १६०]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.1 | Q 18 | पृष्ठ १६०

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