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Question
f(x) = `{{:(x^2/2",", "if" 0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",", "if" 1 < x ≤ 2):}` at x = 1
Solution
We have, f(x) = `{{:(x^2/2",", "if" 0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",", "if" 1 < x ≤ 2):}` at x = 1
At x = 1
L.H.L. = `lim_(x -> 1^-) x^2/2`
= `lim_("h" -> 0) (1 - "h")^2/2`
= `lim_("h" -> 0) (1 + "h"^2 - 2"h")/2`
= `1/2`
R.H.L. = `lim_(x -> 1^+) (2x^2 - 3x + 3/2)`
= `lim_("h" -> 0) [2(1 + "h")^2 - 3(1 + "h") + 3/2]`
= `2 - 3 + 3/2`
= `1/2`
Also f(1) = `1^2/2 = 1/2`
Thus L.H.L. = R.H.L. = f(1)
Hence, f(x) is continuous at x = 1.
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