Let f:[-1,1] rightarrow R be defined as f(x)= | vedclass

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(C) Given $f(x) = ax^2 + bx + c$.$f(-1) = a - b + c = 2$  --- $(1)$$f^{\prime}(x) = 2ax + b \Rightarrow f^{\prime}(-1) = -2a + b = 1$ --- $(2)$$f^{\pr

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$5$

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(C) Given $f(x) = ax^2 + bx + c$.$f(-1) = a - b + c = 2$  --- $(1)$$f^{\prime}(x) = 2ax + b \Rightarrow f^{\prime}(-1) = -2a + b = 1$ --- $(2)$$f^{\prime\prime}(x) = 2a$.Given the maximum value of $f^{\prime\prime}(x)$ is $\frac{1}{2}$,so $2a = \frac{1}{2} \Rightarrow a = \frac{1}{4}$.Substituting $a = \frac{1}{4}$ in $(2)$: $-2(\frac{1}{4}) + b = 1 \Rightarrow -\frac{1}{2} + b = 1 \Rightarrow b = \frac{3}{2}$.Substituting $a$ and $b$ in $(1)$: $\frac{1}{4} - \frac{3}{2} + c = 2 \Rightarrow \frac{1-6}{4} + c = 2 \Rightarrow c = 2 + \frac{5}{4} = \frac{13}{4}$.Thus,$f(x) = \frac{1}{4}x^2 + \frac{3}{2}x + \frac{13}{4}$.To find the maximum value of $f(x)$ on $[-1, 1]$,we check the endpoints and critical points.$f^{\prime}(x) = \frac{1}{2}x + \frac{3}{2}$. Setting $f^{\prime}(x) = 0 \Rightarrow x = -3$,which is outside $[-1, 1]$.Since $f^{\prime}(x) > 0$ for all $x \in [-1, 1]$,$f(x)$ is strictly increasing.Therefore,the maximum value occurs at $x = 1$.$f(1) = \frac{1}{4}(1)^2 + \frac{3}{2}(1) + \frac{13}{4} = \frac{1}{4} + \frac{6}{4} + \frac{13}{4} = \frac{20}{4} = 5$.Since $f(x) \leq \alpha$,the least value of $\alpha$ is $5$.

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