If f(x) = begin{cases} x^2 + 3x + a, & x leq | vedclass

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(A) For $f(x)$ to be differentiable at $x = 1$, it must first be continuous at $x = 1$.$	ext{LHL} = \lim_{x 	o 1^-} (x^2 + 3x + a) = 1 + 3 + a = 4 +

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$a = 3, b = 5$

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(A) For $f(x)$ to be differentiable at $x = 1$, it must first be continuous at $x = 1$.$	ext{LHL} = \lim_{x 	o 1^-} (x^2 + 3x + a) = 1 + 3 + a = 4 + a$.$	ext{RHL} = \lim_{x 	o 1^+} (bx + 2) = b + 2$.Since $	ext{LHL} = 	ext{RHL}$, we have $4 + a = b + 2$, which implies $b - a = 2$ (Equation $1$).For differentiability, $	ext{LHD} = 	ext{RHD}$ at $x = 1$.$	ext{LHD} = \frac{d}{dx}(x^2 + 3x + a) = 2x + 3$. At $x = 1$, $	ext{LHD} = 2(1) + 3 = 5$.$	ext{RHD} = \frac{d}{dx}(bx + 2) = b$.Thus, $b = 5$.Substituting $b = 5$ into Equation $1$: $5 - a = 2 \Rightarrow a = 3$.

If $f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 \\ bx + 2, & x > 1 \end{cases}$ is everywhere differentiable, then:

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