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Question
If the function f is continuous at x = 2, then find 'k' where
f(x) = `(x^2 + 5)/(x - 1),` for 1< x ≤ 2
= kx + 1 , for x > 2
Solution
Since f is continuous at x = 2
∴ `lim_(x → 2^-) f(x) = lim_(x → 2^+) f(x) = f(2)`
∴ `lim_(x → 2^-)[(x^2 + 5)/(x - 1)] = lim_(x → 2^+) (kx + 1) = (2^2 + 5)/(2 - 1)`
∴ `(4 + 5)/(2 - 1) = k(2) + 1 = 9`
∴ 9 = 2k + 1
⇒ ∴ k = 4
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