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

1 Answer

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 $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$ f(x) = \left\{ \begin{array}{l l}\lambda (x^2-2x), & \quad if { x \leq 0 } \\ 4x+1, & \quad if { x > 0 } \end{array} \right. $ is continuous at \( x=0.\)
$\therefore$ LHL = RHL
LHL =$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0}\lambda(x^2-2x)=0$ for all value of $\lambda$
Step 2:
RHL =$\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0}4x+1$
$\Rightarrow RHL=1$
Clearly $\lim\limits_{x\to 0^-}f(x)\neq \lim\limits_{x\to 0^+}f(x)$ for all values of $\lambda$
So,$f(x)$ cannot be made continuous for any values of $\lambda$
answered Nov 8, 2013 by sreemathi.v
 

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