VITEEE 2014 Paper

It is clear that f (x) has a definite and unique value for each x ∊ [ 1,5]. Thus, for every point in the interval [1,5], the value of f (x) exists. So, f(x) is continuous in the interval [1,5]. Also, f ' (x) = $\frac{-x}{\sqrt{25-x^2}}$ , , which clearly exists for all x in an open interval (1,5). So, f (x) is differentiable in (1,5). So, there must be a value c ∊ [1,5] such that f ' (c) = $\frac{f(5)-f(1)}{5-1}$ = $\frac{-\sqrt{24}}{4}$ = $\frac{0-2\sqrt{6}}{4}$ = $-\frac{\sqrt{6}}{2}$ But f ' (c) = $\frac{-c}{\sqrt{25-c^2}}$ ⇒ $\frac{-c}{\sqrt{25-c^2}}$ = - $\frac{\sqrt{6}}{2}$ ⇒ $4x^2$ = 6 (25 - $x^2$) ⇒ $4c^2$ = 150 - $6c^2$ ⇒ $10c^2$ = 150 ⇒ $c^2$ = 15 ⇒ x = ± $\sqrt{15}$ ∴ c = $\sqrt{15}$ ∊ [1 , 5]