Let fx=x−4|x−4|+a x 4 Then f(x) is continuous at x=4 when

The given function is fx=x−4|x−4|+a x 4Given that the function is continuous at x=4Hence,limx→4-fx=f4=limx→4+fxSubstitute the functional valueslimx→4-x-4x-4+a=a+b=limx→4+x-4x-4+b-1+a=a+b=1+bHence, a=1, b=-1

Let fx=x−4|x−4|+a x 4 Then f(x) is continuous at x=4 when

The given function is fx=x−4|x−4|+a x 4Given that the function is continuous at x=4Hence,limx→4-fx=f4=limx→4+fxSubstitute the functional valueslimx→4-x-4x-4+a=a+b=limx→4+x-4x-4+b-1+a=a+b=1+bHence, a=1, b=-1

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Let $f (x) = \{\begin{cases} \frac{x - 4}{| x - 4 |} + a x < 4 \\ a + b x = 4 \\ \frac{x - 4}{| x - 4 |} + b x > 4 \end{cases}$
$Then f (x) is continuous at x = 4 when$

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$a = 1, b = 1$

$a = - 1, b = - 1$

$a = 0, b = 0$

$a = 1, b = - 1$

answer is D.

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Detailed Solution

The given function is $f (x) = \{\begin{cases} \frac{x - 4}{| x - 4 |} + a x < 4 \\ a + b x = 4 \\ \frac{x - 4}{| x - 4 |} + b x > 4 \end{cases}$

Given that the function is continuous at $x = 4$

Hence,

$\lim_{x \to 4^{-}} f (x) = f (4) = \lim_{x \to 4^{+}} f (x)$

Substitute the functional values

$\begin{array}{rcl} \lim_{x \to 4^{-}} \frac{x - 4}{|x - 4|} + a & = & a + b = \lim_{x \to 4^{+}} \frac{x - 4}{|x - 4|} + b \\ - 1 + a & = & a + b = 1 + b \end{array}$