Question

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

 

 

विकल्प

•  a = 0, b = 0

• a = 1, b = 1

• a = −1, b = 1

• a = 1, b = −1.

Answer 1

a = 1, b = −1. 

Given: 

\[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if }  x < 4 \\ a + b, \text{ if }  x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } x > 4\end{cases}\]

We have
(LHL at x = 4) = 

\[\lim_{x \to 4^-} f\left( x \right) = \lim_{h \to 0} f\left( 4 - h \right)\]

\[= \lim_{h \to 0} \left( \frac{4 - h - 4}{\left| 4 - h - 4 \right|} + a \right) = \lim_{h \to 0} \left( \frac{- h}{\left| - h \right|} + a \right) = a - 1\]

(RHL at x = 4) =  \[\lim_{x \to 4^+} f\left( x \right) = \lim_{h \to 0} f\left( 4 + h \right)\]

\[= \lim_{h \to 0} \left( \frac{4 + h - 4}{\left| 4 + h - 4 \right|} + b \right) = \lim_{h \to 0} \left( \frac{h}{\left| h \right|} + b \right) = b + 1\]

Also, 

\[f\left( 4 \right) = a + b\]

If f(x) is continuous at x = 4, then 

\[\lim_{x \to 4^-} f\left( x \right) = \lim_{x \to 4^+} f\left( x \right) = f\left( 4 \right)\]

\[\Rightarrow a - 1 = b + 1 = a + b\]

\[\Rightarrow a - 1 = a + b \text{ and } b + 1 = a + b\]

\[\Rightarrow b = - 1 \text{ and } a = 1\]