MCQ
જો $f(x) = \left\{ \begin{array}{l}x + \lambda ,\;x\, < 3\\\,\,\,\,\,\,\,\,\,4,\,\,x = 3\\3x - 5,\,\,x > 3\end{array} \right.$ એ $x = 3$ આગળ સતત હોય , તો $\lambda = $
- A$4$
- B$3$
- C$2$
- ✓$1$
$\mathop {\lim }\limits_{x \to 3 + } f(x) = f(3) = \mathop {\lim }\limits_{x \to 3 - } f(x)$
$ \Rightarrow \,\mathop {\lim }\limits_{x \to 3 - } f(x) = 4$ or $\mathop {\lim }\limits_{h \to 0} 3 - h + \lambda = 4$
==> $3 + \lambda = 4 ==> \lambda = 1$
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