\\\\(f(x) = \\begin{cases} \\lambda(x^2 - 2x), \ ext{if} x \\le 0 \\\\4x + 1, \ ext{if} x > 0 \\end{cases}\\\\)If \\\\(f\\\\) is continuous at \\\\(x = 0\\\\), then \\\\(\\displaystyle \\underset{x \\rightarrow 0^-}{\\lim} f(x) = \\underset{x \\rightarrow 0^+}{\\lim} f(x) = f(0)\\\\)\\\\(\\Rightarrow\\displaystyle \\underset{x \\rightarrow 0}{\\lim}\\lambda(x^2 - 2x) = \\underset{x \\rightarrow 0}{\\lim}(4x + 1) = \\lambda(0^2 - 2 x 0)\\\\)\\\\(\\Rightarrow \\lambda(0^2 - 2 \\cdot 0) = 4 \\cdot 0 + 1 = 0\\\\)\\\\(\\Rightarrow 0 = 1 = 0,\\\\) which is not possibleTherefore, there is no value of \\\\(\\lambda\\\\) for which \\\\(f\\\\) is continuous at \\\\(x = 0\\\\)At \\\\(x = 1,\\\\)\\\\(f(1) = 4x + 1 = 4 \\cdot 1 + 1 = 5\\\\)\\\\(\\displaystyle \\underset{x \\rightarrow 1}{\\lim} (4x + 1) = 4 \\cdot 1 + 1 = 5\\\\)\\\\(\ herefore\\displaystyle \\underset{x \\rightarrow 1}{\\lim} f(x) = f(1)\\\\)Therefore, for any values of \\\\(\\lambda\\\\), f is continuous at \\\\(x = 1\\\\)