Let f(x) = begin{cases} x^2 - a & x

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(B) For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 3^-} (x^2 - a) = \lim_{x 	o 3^+} (b\sqrt

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$-2$

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(B) For $f(x)$ to be continuous at $x = 3$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 3^-} (x^2 - a) = \lim_{x 	o 3^+} (b\sqrt{x - 2} + a) \implies 9 - a = b(1) + a \implies 2a + b = 9$. For $f(x)$ to be continuous at $x = 6$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 6^-} (b\sqrt{x - 2} + a) = \lim_{x 	o 6^+} (2x + b) \implies b\sqrt{4} + a = 12 + b \implies 2b + a = 12 + b \implies a + b = 12$. Solving the system of equations: $2a + b = 9$ and $a + b = 12$. Subtracting the second from the first gives $a = -3$. Substituting $a = -3$ into $a + b = 12$ gives $b = 15$. Now,$f(1) = 1^2 - (-3) = 4$ and $f(3) = b\sqrt{3 - 2} + a = 15(1) - 3 = 12$. Finally,$\frac{f(1) - f(3)}{4} = \frac{4 - 12}{4} = \frac{-8}{4} = -2$.

Let $f(x) = \begin{cases} x^2 - a & x < 3 \\ b\sqrt{x - 2} + a & 3 \leqslant x < 6 \\ 2x + b & x \geqslant 6 \end{cases}$. If $f(x)$ is continuous $\forall x \in R$,then find the value of $\frac{f(1) - f(3)}{4}$.

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