Get Answers to all your Questions

header-bg qa

\mathrm{\text { Let } f(x)=\left\{\begin{array}{cc} \frac{x-4}{|x-4|}+a, & x<4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|}+b, & x>4 \end{array}\right.}      Then f(x) is continuous at x=4 when

Option: 1

a=0, b=0


Option: 2

a=-1, b=1


Option: 3

a=1, b=1


Option: 4

a=1, b=-1


Answers (1)

best_answer

We have 

\mathrm{ \lim _{x \rightarrow 4^{-}} f(x) =\lim _{h \rightarrow 0} f(4-h)=\lim _{h \rightarrow 0} \frac{4-h-4}{|4-| h-4 \mid}+a \\ }

\mathrm{ =\lim _{h \rightarrow 0}-\frac{h}{h}+a=a-1 .}

\mathrm{ \lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h)=\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=b+1 .}
and, f(4)=a+b.
Since f(x) is continuous at x=4, therefore

\mathrm{ \lim _{x \rightarrow 4^{-}} f(x)=f(4)=\lim _{x \rightarrow 4^{+}} f(x) \\ .}
\mathrm{ \Rightarrow \quad a-1=a+b=b+1 \Rightarrow b=-1 \text { and } a=1 . .}
 

Posted by

Rakesh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE