If f(x) = begin{cases} ax^2 + bx - frac{13}{8 | vedclass

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(A) Since $f(x)$ is differentiable $\forall x \in R$,it must be continuous $\forall x \in R$. For continuity at $x = 2$: $\lim_{x \rightarrow 2^+} f(x

Vedclass · Accepted Answer

$\frac{9}{8}$

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(A) Since $f(x)$ is differentiable $\forall x \in R$,it must be continuous $\forall x \in R$. For continuity at $x = 2$: $\lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^-} f(x)$ $\Rightarrow \lim_{x \rightarrow 2} (bx^3 + 1) = \lim_{x \rightarrow 2} (3x - 3)$ $\Rightarrow 8b + 1 = 3(2) - 3 = 3$ $\Rightarrow 8b = 2 \Rightarrow b = \frac{1}{4}$. For continuity at $x = 1$: $\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x)$ $\Rightarrow \lim_{x \rightarrow 1} (ax^2 + bx - \frac{13}{8}) = \lim_{x \rightarrow 1} (3x - 3)$ $\Rightarrow a + b - \frac{13}{8} = 3(1) - 3 = 0$ $\Rightarrow a + \frac{1}{4} - \frac{13}{8} = 0$ $\Rightarrow a = \frac{13}{8} - \frac{2}{8} = \frac{11}{8}$. Therefore,$a - b = \frac{11}{8} - \frac{1}{4} = \frac{11}{8} - \frac{2}{8} = \frac{9}{8}$.

If $f(x) = \begin{cases} ax^2 + bx - \frac{13}{8}, & x \leq 1 \\ 3x - 3, & 1 < x \leq 2 \\ bx^3 + 1, & x > 2 \end{cases}$ is differentiable $\forall x \in R$,then $a - b =$

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