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Continuity — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity3 Marks
Question
Find the value of k for which the function $\text{f(x)}=\begin{cases}\frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2},&\text{x}\neq2\\\text{k},&\text{x} = {2}\end{cases}$ is continues at x = 2.
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Answer

Given
$\text{f(x)} = \frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} - 2}$
Continuity
$\text{x} = 2$
$\lim\limits_{\text{x} \rightarrow 2} \frac{\text{x}^{2} + 3\text{x} - 10}{\text{x} -2} = \text{k}$
$\lim\limits_{\text{x}\rightarrow 2} \frac{\text{x}^{2} + 5\text{x} - 2\text{x} - 10}{\text{x} - 2 } = \text{k}$
$\lim\limits_{\text{x} \rightarrow 2}\frac{\text{x} (\text{x} + 5) - 2 (\text{x} + 5)}{\text{x} - 2} = \text{k}$
$\lim\limits_{\text{x} \rightarrow 2} \frac{(\text{x} - 2) (\text{x} + 5)}{\text{(x} - 2)} = \text{k}$
When x = 2
x + 5 = k
k = 5 + 2 = 7
k = 7

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