MHT CET · Maths · Continuity and Differentiability
\(f(x)=\left\{\begin{array}{cc}
\frac{x-4}{|x-4|}+a, & \text { for } x < 4 \
a+b, & \text { for } x=4 \
\end{array}\right.\) \(\frac{x-4}{|x-4|}+b, \text { for } x>4\) Is continuous at \(x=4\), then
- A a=0,b=0
- B a=1,b=1
- C a=-1,b=1
- D a=1,b=-1
Answer & Solution
Correct Answer
(D) a=1,b=-1
Step-by-step Solution
Detailed explanation
For continuity at \(x=4\)
\(\begin{aligned}
& \lim _{x \rightarrow 4^{-}} f(x)=f(4)=\lim _{x \rightarrow 4^{+}} f(x) \\
& \Rightarrow \lim _{x \rightarrow 4^{-}} \frac{x-4}{|x-4|}+a=a+b=\lim _{x \rightarrow 4^{+}} \frac{x-4}{|x-4|}+b \\
& \Rightarrow-1+a=a+b=1+b \\
& \Rightarrow a=1 \text { and } b=-1
\end{aligned}\)
\(\begin{aligned}
& \lim _{x \rightarrow 4^{-}} f(x)=f(4)=\lim _{x \rightarrow 4^{+}} f(x) \\
& \Rightarrow \lim _{x \rightarrow 4^{-}} \frac{x-4}{|x-4|}+a=a+b=\lim _{x \rightarrow 4^{+}} \frac{x-4}{|x-4|}+b \\
& \Rightarrow-1+a=a+b=1+b \\
& \Rightarrow a=1 \text { and } b=-1
\end{aligned}\)
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