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Question:
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The function defined by
$$
f(x)=\left\{\begin{array}{cc}
\frac{x-4}{|x-4|}+a & \text { if } x<4 \\
a+b & \text { if } x=4 \\
\frac{x-4}{|x-4|}+b & \text { if } x>4
\end{array}\right.
$$
is continuous at $x=4$, are
Options:
$$
f(x)=\left\{\begin{array}{cc}
\frac{x-4}{|x-4|}+a & \text { if } x<4 \\
a+b & \text { if } x=4 \\
\frac{x-4}{|x-4|}+b & \text { if } x>4
\end{array}\right.
$$
is continuous at $x=4$, are
Solution:
2651 Upvotes
Verified Answer
The correct answer is:
$\mathrm{a}=1, \mathrm{~b}=-1$
L.H.L at $x=4$
$\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{|-h|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a=-1+a$
R.H.L at $x=4$
$\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{|h|}+b=\lim _{h \rightarrow 0} \frac{h}{h}+b=1+b$
For continuity at $\mathrm{x}=4$
$-1+\mathrm{a}=\mathrm{a}+\mathrm{b}=1+\mathrm{b} \Rightarrow \mathrm{a}=1$ and $\mathrm{b}=-1$
$\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{|-h|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a=-1+a$
R.H.L at $x=4$
$\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{|h|}+b=\lim _{h \rightarrow 0} \frac{h}{h}+b=1+b$
For continuity at $\mathrm{x}=4$
$-1+\mathrm{a}=\mathrm{a}+\mathrm{b}=1+\mathrm{b} \Rightarrow \mathrm{a}=1$ and $\mathrm{b}=-1$
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