Question

The function y = a log x+bx² + x has extreme values at x=1 and x=2. Find a and b ?

Answer 1

\[\text { Given }: f\left( x \right) = y = a \log x + b x^2 + x\]

\[ \Rightarrow f'\left( x \right) = \frac{a}{x} + 2bx + 1\]

\[\text { Since }, f'\left( x \right) \text { has extreme values at x = 1 and x = 2,} f'\left( 1 \right) = 0 . \]

\[ \Rightarrow \frac{a}{1} + 2b\left( 1 \right) + 1 = 0\]

\[ \Rightarrow a = - 1 - 2b . . . \left( 1 \right)\]

\[f'\left( 2 \right) = 0\]

\[ \Rightarrow \frac{a}{2} + 2b\left( 2 \right) + 1 = 0\]

\[ \Rightarrow a + 8b = - 2 \]

\[ \Rightarrow a = - 2 - 8b . . . \left( 2 \right)\]

\[\text { From eqs } . \left( 1 \right) \text { and } \left( 2 \right), \text { we get }\]

\[ - 2 - 8b = - 1 - 2b\]

\[ \Rightarrow 6b = - 1\]

\[ \Rightarrow b = \frac{- 1}{6}\]

\[\text { Substituting b } = \frac{- 1}{6} \text { in eq } . \left( 1 \right), \text{we get }\]

\[a = - 1 + \frac{1}{3} = \frac{- 2}{3}\]