Question

Find the values of a and b such that the function f defined by
f(x) = ${\displaystyle \{\begin{array}{ll}\frac{x-4}{|x-4|}+\text{a}\text{,}& \text{if} x<4\\ \text{a}+\text{b}\text{,}& \text{if} x=4\\ \frac{x-4}{|x-4|}+\text{b}\text{,}& \text{if} x>4\end{array}}$
is a continuous function at x = 4.

Answer 1

We have, f(x) = ${\displaystyle \{\begin{array}{ll}\frac{x-4}{|x-4|}+\text{a}\text{,}& \text{if} x<4\\ \text{a}+\text{b}\text{,}& \text{if} x=4\\ \frac{x-4}{|x-4|}+\text{b}\text{,}& \text{if} x>4\end{array}}$

At x = 4

L.H.L. = ${\displaystyle \underset{x\to 4^{-}}{lim}\left(\frac{x-4}{|x-4|+\text{a}}\right)}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}(\frac{4-\text{h}-4}{|4-\text{h}-4|}+\text{a})}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}(\frac{-\text{h}}{\text{h}}+\text{a})}$

= ${\displaystyle -1+\text{a}}$

R.H.L. = ${\displaystyle \underset{x\to 4^{+}}{lim}(\frac{x-4}{|x-4|}+\text{b})}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}(\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b})}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}(\frac{\text{h}}{\text{h}}+\text{b})}$

= 1 + b

Also f(4) = a + b  ....(Given)

Since f(x) is continuous at x = 4

–1 + a = 1 + b = a + b

Solving we get, b = –1  and a = 1