VITEEE 2015 Paper

Given: f(t) = $\frac{\sin t}{t}$ At t = 0, we will check continuity of the function. LHL = f(0 - h) = $\lim\limits_{h\rightarrow0}$ $\frac{\sin(0-h)}{0-h}$ = $\lim\limits_{h\rightarrow0}$ $\frac{-\sin h}{-h}$ = 1 RHL = f(0 + h) $\lim\limits_{h\rightarrow0}$ $\frac{\sin(0+h)}{0+h}$ = $\lim\limits_{h\rightarrow0}$ $\frac{\sin h}{h}$ = 1 and f(0) = 1 LHL = RHL = f(0) So, the function is continuous at t = 0 Now, we check the function is maximum or minimum. f ' (t) = $\frac{1}{t}$ cos t - $\frac{1}{t^2}$ sin t and f " (t) = $-\frac{1}{t}$ sin t - $\frac{1}{t^2}$ cos t - $\frac{1}{t^2} \cos t$ + $\frac{2}{t^3}$ sin t = $-\frac{\sin t}{t}$ - $\frac{2\cos t}{t^2}$ + $\frac{2\sin t}{t^3}$ For maximum or minimum value of f(x), put ⇒ $\frac{\cos t}{t}$f (x) = 0 - $\frac{\sin t}{t}$ = 0 ⇒ $\frac{\tan t}{t}$ = 1 Now , $\lim\limits_{t\rightarrow0}$ f " (t) = - $\lim\limits_{t\rightarrow0}$ $\left( \frac{\sin t}{t} \right)$ - 2 $\lim\limits_{t\rightarrow0}$ $\left( \frac{t\cos t - \sin t}{t^3} \right)$ [0/0 form] = - 1 - 2 $\lim\limits_{t\rightarrow0}$ $\left( \frac{\cos t - t\sin t - \cos t}{3t^2} \right)$ [using L’ Hospital rule] = - 1 + $\frac{2}{3}$ $\lim\limits_{t\rightarrow0}$ $\frac{\sin t}{t}$ = - 1 + $\frac{2}{3}$ × 1 = $-\frac{1}{3}$ < 0 So, function f(t) is maximum at t = 0