If f(x) = begin{cases} 2x+3, & x leq 1 ax^{2 | vedclass

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(B) Given $f(x) = \begin{cases} 2x+3, & x \leq 1 \ ax^{2}+bx, & x > 1 \end{cases}$.Since $f(x)$ is differentiable for all $x \in R$,it must be contin

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$4$

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(B) Given $f(x) = \begin{cases} 2x+3, & x \leq 1 \ ax^{2}+bx, & x > 1 \end{cases}$.Since $f(x)$ is differentiable for all $x \in R$,it must be continuous at $x = 1$.$\lim_{x 	o 1^{-}} f(x) = \lim_{x 	o 1^{+}} f(x) = f(1) \Rightarrow 2(1) + 3 = a(1)^2 + b(1) \Rightarrow a + b = 5 \quad \dots(i)$Also,$f'(x)$ must be continuous at $x = 1$.$f'(x) = \begin{cases} 2, & x  1 \end{cases}$$\lim_{x 	o 1^{-}} f'(x) = \lim_{x 	o 1^{+}} f'(x) \Rightarrow 2 = 2a(1) + b \Rightarrow 2a + b = 2 \quad \dots(ii)$Subtracting $(i)$ from $(ii)$: $(2a + b) - (a + b) = 2 - 5 \Rightarrow a = -3$.Substituting $a = -3$ in $(i)$: $-3 + b = 5 \Rightarrow b = 8$.Thus,for $x > 1$,$f(x) = -3x^2 + 8x$.We need to find $f(2)$.$f(2) = -3(2)^2 + 8(2) = -3(4) + 16 = -12 + 16 = 4$.

If $f(x) = \begin{cases} 2x+3, & x \leq 1 \\ ax^{2}+bx, & x > 1 \end{cases}$ is differentiable $\forall x \in R$,then $f(2) = $ . . . . . . .

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