JEE Mains Model Paper 15

f'(x) = $\frac{1}{8x}$ - b + 2x ; f''(x) = $-\frac{1}{8x^2}$ + 2 For max or min , f'(x) = 0 1 - 8bx + 16$x^2$ = 0 ∴ x = $\frac{8b\pm\sqrt{64b^2-64}}{32}$ = $\frac{b\pm\sqrt{b^2-1}}{4}$ Case I. When b = 0 , f' (x) = $\frac{1}{8x}$ + 2x ≠ 0 Hence there is no extreme value for b = 0 Case II. When 0 < b < 1 , then $b^2$ - 1 < 0 ∴ x is not real ∴ there is no extreme value for 0 < b < 1 Case III. When b = 1 , then x = $\frac{1}{4}$ f''(x) = 2 - 2 = 0 f''' (x) = $\frac{1}{4x^3}$ = 16 ( ≠ 0 ) for x = $\frac{1}{4}$ ∴ x = $\frac{1}{4}$ is a point of inflexion Case IV. When b > 1 Clearly α < β We have f'(x) = $\frac{1}{8x}$ - b + 2x = $\frac{2}{x}\left(x^2-\frac{1}{2}bx+\frac{1}{16}\right)$ = $\frac{2}{x}\left[\left(x-\frac{b}{4}\right)^2-\frac{1}{16}(b^2-1)\right]$ = $\frac{2}{x}[\left(x-\frac{b}{4}-\frac{1}{4}\sqrt{b^2-1}\right)$ $\left(x-\frac{b}{4}+\frac{1}{4}\sqrt{b^2-1}\right)]$ = $\frac{2}{x}$ (x - β) (x - α) = $\frac{2}{x}$ (x - α) (x - β) ∴ f'(x) > 0 , for 0 < x < α f'(x) < 0 , for α < x < β and f'(x) > 0 for x > β ∴ f(x) has a local max at x = α = $\frac{1}{4}$ (b - $\sqrt{b^2-1}$) and local min at x = β = $\frac{1}{4}$ (b + $\sqrt{b^2-1}$)