Question

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}\right.$. Then f(x) is continuous at x = 4, then a + b = _____________.

Answer 1

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}\right.$ is continuous at x = 4.


$\therefore f\left(4\right)=\underset{x\to 4}{\lim }f\left(x\right)$

$\Rightarrow f\left(4\right)=\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)$ .....(1)

$\left|x-4\right|=\left\{\begin{array}{ll}-\left(x-4\right),& x<4\\ x-4,& x\ge 4\end{array}\right.$

Now,

f(4) = a + b .....(2)

$\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4}{\lim }\left[\frac{x-4}{-\left(x-4\right)}+a\right]=\underset{x\to 4}{\lim }\frac{x-4}{-\left(x-4\right)}+\underset{x\to 4}{\lim }a=-1+a$ .....(3)

$\underset{x\to 4^{+}}{\lim }f\left(x\right)=\underset{x\to 4}{\lim }\left[\frac{x-4}{\left(x-4\right)}+b\right]=\underset{x\to 4}{\lim }\frac{x-4}{\left(x-4\right)}+\underset{x\to 4}{\lim }b=1+b$ .....(4)

From (1), (2), (3) and (4), we get

a + b = −1 + a = 1 + b

So,

a + b = −1 + a

⇒ b = −1

Also,

a + b = 1 + b

⇒ a = 1

∴ a + b = 1 + (−1) = 0

Thus, the value of a + b is 0.


If $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}\right.$. Then f(x) is continuous at x = 4, then a + b = ___0___.