If the function f(x) = begin{cases} a|pi - x| | vedclass

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(A) For the function $f(x)$ to be continuous at $x = 5$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = 5$

Vedclass · Accepted Answer

$\frac{2}{5 - \pi}$

Vedclass · Answer

(A) For the function $f(x)$ to be continuous at $x = 5$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = 5$ must be equal. $LHL = \lim_{x 	o 5^-} f(x) = a|\pi - 5| + 1 = a(5 - \pi) + 1$ (since $5 > \pi$,$|\pi - 5| = 5 - \pi$). $RHL = \lim_{x 	o 5^+} f(x) = b|\pi - 5| + 3 = b(5 - \pi) + 3$. $f(5) = a|\pi - 5| + 1 = a(5 - \pi) + 1$. Equating $LHL$ and $RHL$: $a(5 - \pi) + 1 = b(5 - \pi) + 3$. Rearranging the terms: $a(5 - \pi) - b(5 - \pi) = 3 - 1$. $(a - b)(5 - \pi) = 2$. Therefore,$a - b = \frac{2}{5 - \pi}$.

If the function $f(x) = \begin{cases} a|\pi - x| + 1, & x \le 5 \\ b|\pi - x| + 3, & x > 5 \end{cases}$ is continuous at $x = 5$,then the value of $a - b$ is

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