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Question
The function y = a log x+bx2 + x has extreme values at x=1 and x=2. Find a and b ?
Solution
\[\text { Given }: f\left( x \right) = y = a \log x + b x^2 + x\]
\[ \Rightarrow f'\left( x \right) = \frac{a}{x} + 2bx + 1\]
\[\text { Since }, f'\left( x \right) \text { has extreme values at x = 1 and x = 2,} f'\left( 1 \right) = 0 . \]
\[ \Rightarrow \frac{a}{1} + 2b\left( 1 \right) + 1 = 0\]
\[ \Rightarrow a = - 1 - 2b . . . \left( 1 \right)\]
\[f'\left( 2 \right) = 0\]
\[ \Rightarrow \frac{a}{2} + 2b\left( 2 \right) + 1 = 0\]
\[ \Rightarrow a + 8b = - 2 \]
\[ \Rightarrow a = - 2 - 8b . . . \left( 2 \right)\]
\[\text { From eqs } . \left( 1 \right) \text { and } \left( 2 \right), \text { we get }\]
\[ - 2 - 8b = - 1 - 2b\]
\[ \Rightarrow 6b = - 1\]
\[ \Rightarrow b = \frac{- 1}{6}\]
\[\text { Substituting b } = \frac{- 1}{6} \text { in eq } . \left( 1 \right), \text{we get }\]
\[a = - 1 + \frac{1}{3} = \frac{- 2}{3}\]
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