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For what value of \( \lambda \) is the function defined by \(f\) defined by $ f(x) = \left\{ \begin{array} {1 1} \lambda(x^2 - 2x) ,& \quad\text{ if $ x $ \(\leq 0\)}\\ 4x + 1,& \quad \text{if $x$ > 0}\\ \end{array} \right. $ Continuous at \(x = 0\)?

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A)
Toolbox:
  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
At $x=0$
LHL=$\lim\limits_{\large x\to 0}\lambda(x^2-2x)$
$\quad\;\;\;=0$
RHL=$\lim\limits_{\large x\to 0}(4x+1)$
$\quad\;\;\;=1$
$f(0)$=LHL $\neq$ RHL
$\Rightarrow f$ is not continuous at $x=0$ whatever value of $\lambda\in R$
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