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

Answer 1

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