Let f: R be defined as f (x) = \ cases 2 (- ( )/(2)), & ifx 1 cases...

$f$\left(x\right)$is continuous on$R$</p><p>$⇒ f\left({1}^{-}\right) = f\left(1\right) = f\left({1}^{+}\right)$</p><p>\left|a + 1 + b\right| = \lim_{x \to {1}^{+}}\sin \left(\pi x\right)$

\left|a + 1 + b\right| = 0 ⇒ a + b = - 1$</p><p>$⇒$Also</p><p>$f\left(- {1}^{-}\right) = f\left(- 1\right) = f\left(- {1}^{+}\right)$</p><p>$\lim_{x \to - 1}2\sin \left(\dfrac{- \pix}{2}\right) = \left|a - 1 + b\right|$</p><p>\left|a - 1 + b\right| = 2$

Either $a - 1 + b = 2$ or $a - 1 + b = - 2$

$a + b = 3 . . . \left(2\right)$ or $a + b = - 1 . . .$\left(3\right)$</p><p>from \left(1\right)$ and $\left(2\right)$⇒ a + b = 3 = - 1$(reject) </p><p>from \left(1\right)$ and \left(3\right)$⇒ a + b = - 1$