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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 possible
Therefore, 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 $$x = 1$$

Mathematics
RS Agarwal
Standard XII

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