If the function f(x) = \ array ll a|π - x| + 1, x ≤ 5 b|x - π| + 3,...

For continuity at $x = 5$

$\lim_{x \to {5}^{+}}f\left(x\right) = \lim_{x \to {5}^{-}}f\left(x\right) = f\left(5\right)$

$⇒ \lim_{x \to {5}^{+}}\left(b\left|x - \pi\right| + 3\right) = \lim_{x \to {5}^{-}}a\left(\left|\pi - x\right| + 1\right) = a\left(5 - \pi\right) + 1$

We know that \left|x\right| = \left\{$\begin{cases}x, & x \ge 0 \\ - x, & x < 0\end{cases}$\right)

$⇒ b\left(5 - \pi\right) + 3 = a\left(5 - \pi\right) + 1 = a\left(5 - \pi\right) + 1$

$⇒ b\left(5 - \pi\right) + 3 = a\left(5 - \pi\right) + 1$

$⇒ 3 - 1 = a\left(5 - \pi\right) - b\left(5 - \pi\right)$

$⇒ \left(a - b\right)\left(5 - \pi\right) = 2$

$⇒ a - b = \dfrac{2}{5 - \pi} .$