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(A) Since $f(x)$ is continuous everywhere, it must be continuous at $x = 0, 1, 2$.At $x = 0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$

Vedclass · Accepted Answer

$133$

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(A) Since $f(x)$ is continuous everywhere, it must be continuous at $x = 0, 1, 2$.At $x = 0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.$\lim_{x 	o 0^-} \cos(2x + \pi) = \cos(\pi) = -1$.$\lim_{x 	o 0^+} (ax^2 + b) = b$.Thus, $b = -1$.At $x = 1$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$\lim_{x 	o 1^-} (ax^2 + b) = a + b = a - 1$.$\lim_{x 	o 1^+} (cx + 4) = c + 4$.Thus, $a - 1 = c + 4 \implies a = c + 5$.At $x = 2$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2)$.$\lim_{x 	o 2^-} (cx + 4) = 2c + 4$.$\lim_{x 	o 2^+} (3a + 1) = 3a + 1$.Thus, $2c + 4 = 3a + 1$.Substituting $a = c + 5$: $2c + 4 = 3(c + 5) + 1 \implies 2c + 4 = 3c + 16 \implies c = -12$.Then $a = -12 + 5 = -7$.Now, $b^2 - bc + c^2 = (-1)^2 - (-1)(-12) + (-12)^2 = 1 - 12 + 144 = 133$.

Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere, then $b^2 - bc + c^2 =$

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