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Question
If f (x) = Ax2 + Bx + C is such that f (a) = f (b), then write the value of c in Rolle's theorem ?
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Solution
We have
\[f\left( x \right) = A x^2 + Bx + C\]
Differentiating the given function with respect to x, we get
\[f'\left( x \right) = 2Ax + B\]
\[\because f\left( a \right) = f\left( b \right)\]
\[ \therefore A a^2 + Ba + C = A b^2 + bB + C\]
\[ \Rightarrow A a^2 + Ba = A b^2 + bB\]
\[ \Rightarrow A\left( a^2 - b^2 \right) + B\left( a - b \right) = 0\]
\[ \Rightarrow A\left( a - b \right)\left( a + b \right) + B\left( a - b \right) = 0\]
\[ \Rightarrow \left( a - b \right)\left[ A\left( a + b \right) + B \right] = 0\]
\[ \Rightarrow a = b, A = \frac{- B}{\left( a + b \right)}\]
\[ \Rightarrow \left( a + b \right) = \frac{- B}{A} \left( \because a \neq b \right)\]
From (1), we have
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