Let f: R rightarrow R be defined as f(x) = be | vedclass

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(B) $f(x)$ is continuous on $R$,so it must be continuous at $x = 1$ and $x = -1$.For continuity at $x = 1$:$\lim_{x \rightarrow 1^-} f(x) = f(1) = \li

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$-1$

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(B) $f(x)$ is continuous on $R$,so it must be continuous at $x = 1$ and $x = -1$.For continuity at $x = 1$:$\lim_{x \rightarrow 1^-} f(x) = f(1) = \lim_{x \rightarrow 1^+} f(x)$$|a(1)^2 + 1 + b| = \lim_{x \rightarrow 1^+} \sin(\pi x)$$|a + 1 + b| = \sin(\pi) = 0$$a + b + 1 = 0 \Rightarrow a + b = -1$ ... $(1)$For continuity at $x = -1$:$\lim_{x \rightarrow -1^-} f(x) = f(-1) = \lim_{x \rightarrow -1^+} f(x)$$\lim_{x \rightarrow -1^-} 2 \sin \left(-\frac{\pi x}{2}\right) = |a(-1)^2 + (-1) + b|$$2 \sin \left(\frac{\pi}{2}\right) = |a - 1 + b|$$2(1) = |a + b - 1|$$|a + b - 1| = 2$This implies $a + b - 1 = 2$ or $a + b - 1 = -2$.$a + b = 3$ or $a + b = -1$.Since we already established $a + b = -1$ from the continuity at $x = 1$,the value of $a + b$ is $-1$.

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} 2 \sin \left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^2 + x + b|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{cases}$. If $f(x)$ is continuous on $R$,then $a + b$ equals ..... .

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