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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find the value of k in this question, so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}3\text{x}-8,&\text{if x}\leq5\\2\text{k},&\text{if x}>5\end{cases}$ at x = 5.
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Answer

We have, $\text{f(x)}=\begin{cases}3\text{x}-8,&\text{if x}\leq5\\2\text{k},&\text{if x}>5\end{cases}$ at x = 5.
Since, f(x) is continuous at x = 5.
$\therefore$ L.H.L = R.H.L = f(5)
Now, $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow5^-}(3\text{x}-8)=\lim\limits_{\text{h}\rightarrow0}[(5-\text{h})-8]$
$=\lim\limits_{\text{h}\rightarrow0}\ [15-3\text{h}-8]=7$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow5^+}2\text{k}=\lim\limits_{\text{h}\rightarrow0}2\text{k}=2\text{x}=7$ $[\because\ \text{L.H.L}=\text{R.H.L}]$
And f(5) = 3 × 5 - 8 = 7
$2\text{k}=7\Rightarrow\ \text{k}=\frac{7}{2}$

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