JEE Advance - Mathematics (1996 - No. 31)
Determine the points of maxima and minima of the function
$$f\left( x \right) = {1 \over 8}\ell n\,x - bx + {x^2},x > 0,$$ where $$b \ge 0$$ is a constant.
$$f\left( x \right) = {1 \over 8}\ell n\,x - bx + {x^2},x > 0,$$ where $$b \ge 0$$ is a constant.
Minimum at $$x = {1 \over 4}\left( {b + \sqrt {{b^2} - 1} } \right)$$, Maximum at $$x = {1 \over 4}\left( {b - \sqrt {{b^2} - 1} } \right)$$
Minimum at $$x = {1 \over 4}\left( {b - \sqrt {{b^2} - 1} } \right)$$, Maximum at $$x = {1 \over 4}\left( {b + \sqrt {{b^2} + 1} } \right)$$
Maximum at $$x = {1 \over 4}\left( {b + \sqrt {{b^2} - 1} } \right)$$, Minimum at $$x = {1 \over 4}\left( {b - \sqrt {{b^2} - 1} } \right)$$
Minimum at $$x = {1 \over 4}\left( {b + \sqrt {{b^2} - 1} } \right)$$, Maximum at $$x = {1 \over 4}\left( {b - \sqrt {{b^2} + 1} } \right)$$
Minimum at $$x = {1 \over 4}\left( {b + \sqrt {{b^2} - 1} } \right)$$, Maximum at $$x = {1 \over 4}\left( {b - \sqrt {{b^2} - 1} } \right)$$