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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability1 Mark
Question
Find the value of $\mathrm{k}$ so that the function $\mathrm{f}$ is continuous at the indicated point $: f(x)=\left\{\begin{array}{l} k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x>5 \end{array}\right.$ at $\mathrm{x}=5 $
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Answer

Given that, $f\left( x \right) = \left\{ \begin{gathered} kx + 1,\,\,if\,x \leq 5 \hfill \\ 3x - 5,\,\,if\,x > 5 \hfill \\ \end{gathered} \right.$
When $x < 5$ we have $f(x) = kx + 1 $ which being a polynomial is continuous at each point $x < 5.$
And, when $x > 5,$ we have $f(x) = 3x - 5$ which being a polynomial is continuous at each point $x > 5$.
Now $f\left( 5 \right) = 5k + 1 $
$=\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) $
$= \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right) = 3(5+h)-5 $
$= \mathop {\lim }\limits_{h \to {0 }} (3h+10)=10$
Since function is continuous at $x = 5,$
therefore,
$\mathop {\lim }\limits_{x \to {5^ + }} f(x) = f(5)$
$\Rightarrow 10 = 5k + 1$
$\Rightarrow 5k = 9$
$\Rightarrow k = \frac{9}{5}$

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