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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
Question
Determine the value of ‘k’ for which the following function is continuous at x = 3:
$ \text{f(x)} = \begin{cases} \frac{(\text{x + 3)}^{2} \text{ - } 36}{\text{x - 3}} & , & \text{x}\neq 3 \\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{k}& , & \text{x = 3}\\ \end{cases}$
✓

Answer

It is given that the function f(x) is continuous at x = 3.
$\therefore \lim\limits_{x \rightarrow 3^{-}} f (x) = \lim\limits_{x \rightarrow 3^{+}} f(x) = f(3)$
$\Rightarrow f(3) = \lim\limits_{x \rightarrow 3} f (x)$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} \frac{(x + 3)^{2} - 36}{x -3}$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} \frac{(x + 3 - 6) (x + 3 + 6)}{x - 3}$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} \frac{(x - 3) (x + 9)}{x - 3}$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} (x + 9) = 3 + 9 = 12$
Thus, the value of k is 12.

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