Question

Pls solve this for me....

Answer 1

Let's redefine f(x)

$f(x)=[\frac{(x-4)}{-(x-4)}+a\ ;\ x<4 ]$

$[a+b\ ; \ x=4]$

$f(x)=[\frac{(x-4)}{(x-4)}+b\ ;\ x>4 ]$

We know that $x<4$

∴ $\sf{ | x \: - \: 4 | } =-(x-4)$

And if $x>4$

∴ $\sf{ | x \: - \: 4 | }=(x-4)$

⇒ $f(x)=[-1+a\ ; \ x<4]$

$[a+b \ ; \ x=4]$

$[1+b\ ; \ x>4]$

LHL = $(-1+a)$

RHL = $(1+b)$

$f(4)=a+b$

Since f(x) is continuous at x = 4

LHL=RHL=f(4)

⇒ $-1+a=1+b=a-n$

Here,

$1+h=a+b$

$a=1$

And also,

$-1=a=a+b$

$b=-1$

∴ f(x) is continuous at x = 4 where a = 1 and b = (-1)