Question

# For what value of $$\lambda$$ is the function defined by$$f(x) = \begin{cases} \lambda(x^2 - 2x), \text{if} x \le 0 \\4x + 1, \text{if} x > 0 \end{cases}$$continuous at $$x = 0$$? What about continuity at $$x = 1$$ ?

Solution

## $$f(x) = \begin{cases} \lambda(x^2 - 2x), \text{if} x \le 0 \\4x + 1, \text{if} x > 0 \end{cases}$$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 $$x = 1,$$$$f(1) = 4x + 1 = 4 \cdot 1 + 1 = 5$$$$\displaystyle \underset{x \rightarrow 1}{\lim} (4x + 1) = 4 \cdot 1 + 1 = 5$$$$\therefore\displaystyle \underset{x \rightarrow 1}{\lim} f(x) = f(1)$$Therefore, for any values of $$\lambda$$, f is continuous at $$x = 1$$MathematicsRS AgarwalStandard XII

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