JEE MAIN - Mathematics (2003 - No. 53)
Let $$f\left( x \right)$$ be a polynomial function of second degree. If $$f\left( 1 \right) = f\left( { - 1} \right)$$ and $$a,b,c$$ are in $$A.P, $$ then $$f'\left( a \right),f'\left( b \right),f'\left( c \right)$$ are in
Arithmetic -Geometric Progression
$$A.P$$
$$G.P$$
$$H.P$$
Explanation
$$f\left( x \right) = a{x^2} + bx + c$$
$$f\left( 1 \right) = f\left( { - 1} \right)$$
$$ \Rightarrow a + b + c = a - b + c$$
or $$b = 0$$
$$\therefore$$ $$f\left( x \right) = a{x^2} + c$$
or $$f'\left( x \right) = 2ax$$
Now $$f'\left( a \right);f'\left( b \right);$$
and $$f'\left( c \right)$$ are $$2a\left( a \right);2a\left( b \right);2a\left( c \right)$$
i.e.$$\,2{a^2},\,2ab,\,2ac.$$
$$ \Rightarrow $$ If $$a,b,c$$ are in $$A.P.$$ then
$$f'\left( a \right);f'\left( b \right)$$ and
$$f'\left( c \right)$$ are also in $$A.P.$$
$$f\left( 1 \right) = f\left( { - 1} \right)$$
$$ \Rightarrow a + b + c = a - b + c$$
or $$b = 0$$
$$\therefore$$ $$f\left( x \right) = a{x^2} + c$$
or $$f'\left( x \right) = 2ax$$
Now $$f'\left( a \right);f'\left( b \right);$$
and $$f'\left( c \right)$$ are $$2a\left( a \right);2a\left( b \right);2a\left( c \right)$$
i.e.$$\,2{a^2},\,2ab,\,2ac.$$
$$ \Rightarrow $$ If $$a,b,c$$ are in $$A.P.$$ then
$$f'\left( a \right);f'\left( b \right)$$ and
$$f'\left( c \right)$$ are also in $$A.P.$$
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