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Let $f(x)=a x^{2}+b x+c$ such that $f(1)=f(-1)$ and $a, b, c$ are in Arithmetic Progression.
$\mathrm{f}^{\prime}(\mathrm{a}), \mathrm{f}^{\prime}(\mathrm{b}), \mathrm{f}^{\prime}(\mathrm{c})$ are
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$\mathrm{f}^{\prime}(\mathrm{a}), \mathrm{f}^{\prime}(\mathrm{b}), \mathrm{f}^{\prime}(\mathrm{c})$ are
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The correct answer is:
A.P.
We have $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{ax}$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{a})=2 \mathrm{a}^{2}, \mathrm{f}^{\prime}(\mathrm{b})=2 \mathrm{ab}=0$
and $\mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{ac} \quad(\because \mathrm{b}=0)$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{a})=2 \mathrm{a}^{2}$
$\mathrm{f}^{\prime}(\mathrm{b})=0$
and $\mathrm{f}^{\prime}(\mathrm{c})=-2 \mathrm{a}^{2} \quad(\because 2 \mathrm{~b}=\mathrm{a}+\mathrm{c} \Rightarrow \mathrm{c}=-\mathrm{a})$
Hence $\mathrm{f}^{\prime}(\mathrm{a}), \mathrm{f}^{\prime}(\mathrm{b})$ and $\mathrm{f}^{\prime}(\mathrm{c})$ are in $\mathrm{AP}$.
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{a})=2 \mathrm{a}^{2}, \mathrm{f}^{\prime}(\mathrm{b})=2 \mathrm{ab}=0$
and $\mathrm{f}^{\prime}(\mathrm{c})=2 \mathrm{ac} \quad(\because \mathrm{b}=0)$
$\therefore \quad \mathrm{f}^{\prime}(\mathrm{a})=2 \mathrm{a}^{2}$
$\mathrm{f}^{\prime}(\mathrm{b})=0$
and $\mathrm{f}^{\prime}(\mathrm{c})=-2 \mathrm{a}^{2} \quad(\because 2 \mathrm{~b}=\mathrm{a}+\mathrm{c} \Rightarrow \mathrm{c}=-\mathrm{a})$
Hence $\mathrm{f}^{\prime}(\mathrm{a}), \mathrm{f}^{\prime}(\mathrm{b})$ and $\mathrm{f}^{\prime}(\mathrm{c})$ are in $\mathrm{AP}$.
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