(A) Given the function:$f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x 4 \end{cases}$For $x < 4$,$|

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(A) Given the function:$f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x 4 \end{cases}$For $x For $x > 4$,$|x-4| = (x-4)$,so $\frac{x-4}{|x-4|} = 1$. Thus,$f(x) = 1 + b$.Since the function is continuous at $x=4$,the Left Hand Limit $(LHL)$,Right Hand Limit $(RHL)$,and the value of the function $f(4)$ must be equal:$\text{LHL} = \lim_{x \to 4^-} f(x) = -1 + a$$\text{RHL} = \lim_{x \to 4^+} f(x) = 1 + b$$f(4) = a + b$Equating these: $-1 + a = a + b = 1 + b$.From $-1 + a = a + b$,we get $b = -1$.From $a + b = 1 + b$,we get $a = 1$.Therefore,$a=1$ and $b=-1$.

Class 12 Mathematics Continuity and Differentiability Continuity

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$f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$
If $f(x)$ given above is continuous at $x=4$,then find the values of '$a$' and '$b$'.

AP EAMCET 2020 All AP EAMCET

A
$a=1, b=-1$
B
$a=-1, b=1$
C
$a=1, b=1$
D
$a=-1, b=-1$

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$f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & x < 4 \\ a+b, & x=4 \\ \frac{x-4}{|x-4|} + b, & x > 4 \end{cases}$
If $f(x)$ given above is continuous at $x=4$,then find the values of '$a$' and '$b$'.