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(D) Let the second-degree polynomial be $f(x) = px^2 + qx + r$.Given $f(1) = f(-1)$,we have $p(1)^2 + q(1) + r = p(-1)^2 + q(-1) + r$.This simplifies

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$A.P.$

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(D) Let the second-degree polynomial be $f(x) = px^2 + qx + r$.Given $f(1) = f(-1)$,we have $p(1)^2 + q(1) + r = p(-1)^2 + q(-1) + r$.This simplifies to $p + q + r = p - q + r$,which implies $2q = 0$,so $q = 0$.Thus,the function is $f(x) = px^2 + r$.The derivative is $f'(x) = 2px$.Since $a, b, c$ are in $A.P.$,there exists a common difference $d$ such that $b = a + d$ and $c = a + 2d$.Then $f'(a) = 2pa$,$f'(b) = 2p(a + d) = 2pa + 2pd$,and $f'(c) = 2p(a + 2d) = 2pa + 4pd$.Since $f'(b) - f'(a) = 2pd$ and $f'(c) - f'(b) = 2pd$,the terms $f'(a), f'(b), f'(c)$ have a common difference of $2pd$.Therefore,$f'(a), f'(b), f'(c)$ are in $A.P.$

Let $f(x)$ be a polynomial function of second degree. If $f(1) = f(-1)$ and $a, b, c$ are in $A.P.$,then $f'(a), f'(b)$ and $f'(c)$ are in

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