Let f(x)= x-4 |x-4| +a,&if x 4 Then, f(x) is continus at x = 4 when:

Let f(x)= x-4 |x-4| +a,&if x < 4 +b,&if x =4 x-4 |x-4| +b,&if x >…

MCQ

Let $\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$ Then, f(x) is continus at x = 4 when:

A
a = 0, b = 0
✓
a = 1, b = 1
C
a = -1, b = 1
D
a = 1, b = -1

✓

Answer

Correct option: B.

a = 1, b = 1

Given,
$\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$

We have

$(\text{LHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(4-\text{h})$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4- \text{h}-4}{|4-\text{h}-4|}+\text{a}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{-\text{h}}{|-\text{h}|}+\text{a}\Big)=\text{a}-1$

$(\text{RHL at x = 4})=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f(4+h)}$

$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b}\Big)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\text{h}}{|\text{h}|}+\text{b}\Big)=\text{b}+1$

Also,

$\text{f}(4)=\text{a+b}$

if(x) is continuous at x = 4, then

$\lim\limits_{\text{x}\rightarrow4^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\text{f}(4)$

$\Rightarrow\text{a}-1=\text{b}+1=\text{a + b}$

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

$\Rightarrow\text{b}=-1$ and $\text{a}=1$