If a function f(x) = begin{cases} ax+b, & x l | vedclass

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(A) Given that $f(x)$ is continuous on $\mathbb{R}$, it must be continuous at $x = -1$ and $x = 1$.At $x = -1$:$\lim_{x 	o -1^-} f(x) = \lim_{x 	o -

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$(-22, -3)$

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(A) Given that $f(x)$ is continuous on $\mathbb{R}$, it must be continuous at $x = -1$ and $x = 1$.At $x = -1$:$\lim_{x 	o -1^-} f(x) = \lim_{x 	o -1^+} f(x)$$a(-1) + b = 2(-1)^2 + 2b(-1) - \frac{a}{2}$$-a + b = 2 - 2b - \frac{a}{2}$$-\frac{a}{2} + 3b = 2 \quad \dots (i)$At $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x)$$2(1)^2 + 2b(1) - \frac{a}{2} = 7$$2 + 2b - \frac{a}{2} = 7$$-\frac{a}{2} + 2b = 5 \quad \dots (ii)$Subtracting equation $(ii)$ from equation $(i)$:$(-\frac{a}{2} + 3b) - (-\frac{a}{2} + 2b) = 2 - 5$$b = -3$Substituting $b = -3$ into equation $(ii)$:$-\frac{a}{2} + 2(-3) = 5$$-\frac{a}{2} - 6 = 5$$-\frac{a}{2} = 11$$a = -22$Therefore, $(a, b) = (-22, -3)$.

If a function $f(x) = \begin{cases} ax+b, & x \leq -1 \\ 2x^2+2bx-\frac{a}{2}, & -1 < x < 1 \\ 7, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$, then $(a, b) =$

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