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प्रश्न
For what value of `lambda` is the function defined by `f(x) = {(lambda(x^2 - 2x), "," if x <= 0),(4x+ 1, "," if x > 0):}` continuous at x = 0? What about continuity at x = 1?
उत्तर
`f(x) = {(lambda(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)`
`=> (0^2 - 2(0)) = (4 (0) + 1) = (0^2 - 2 (0))`
`=> 0 = 1 = 0`
Which cannot be true, i.e. for any value of `lambda` 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 `lambda` this function is continuous at x = 1.
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