Saved Bookmarks
| 1. |
\( f(x)=\left\{\begin{array}{cc}x^{2}\left|\cos \frac{\pi}{x}\right|, & x \neq 0 \\ 0, & x=0\end{array}, x \in R\right. \) then \( f \) isDifferentiable both at \( x=0 \) and at \( x=2 \)8) Differentiable at \( x=0 \) but not differentiable at \( x=2 \)C) Not differentiable at \( x=0 \) but differentiable at \( x=2 \)Differentiable neither at \( x=0 \) nor at \( x=2 \) |
|
Answer» Option(B) is correct. f(x) = \(\begin{cases}x^2|cos\frac{\pi}x|&;x\neq 0\\0&;x=0\end{cases}\) Left hand derivative of f(x) at x = 0 is Df(0-) = \(\lim\limits_{h\to 0}\frac{f(0-h)-f(0)}{-h}\) = \(\lim\limits_{h\to 0}\frac{(-h)^2|cos(\frac{\pi}{-h})|-0}{-h}\) = \(\lim\limits_{h\to 0}-h|cos(\frac{\pi}h)|\) = 0 (\(\because \lim\limits_{h\to 0}cos(\frac{\pi}h)\)) gives a bounded value) Right hand derivative of f(x) at x = 0 is Df(0+) = \(\lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h}\) = \(\lim\limits_{h\to 0}\frac{(+h)^2|cos(\frac{\pi}{-h})|-0}{h}\) = \(\lim\limits_{h\to 0}h|cos(\frac{\pi}h)|\) = 0 (\(\because\) \(\lim\limits_{h\to 0}cos\frac{\pi}4\) gives a bounded value) \(\because\) Df (0-) = Df(0+) \(\therefore\) f is differentiable at x = 0 Left hand derivative of f(x) at x = 2 is Df(2-) = \(\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{-h}\) = \(\lim\limits_{h\to 0}\frac{(2-h)^2|cos(\frac{\pi}{2-h})|-0}{-h}\) (\(\because f(2)=4\times|cos\frac{\pi}2|= 4\times0 = 0\)) = \(4\lim\limits_{h\to 0}\cfrac{-|cos(\frac{\pi}{2-h})|}{-h}\) (\(\because\frac{\pi}{2-h}>\frac{\pi}2\)) (⇒ cos (\(\frac{\pi}{2-h}\)) lies in 2nd quadrant so negative.) = \(\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2-h})\times\frac{-\pi}{(2-h)^2}\times{-1}}{1}\) = -4π\(\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2-h})}{(2-h)^2}\) = -4π\(\cfrac{sin\frac{\pi}2}4\) = \(\frac{-4\pi}4=-\pi\) Right hand derivative of f(x) at x = 2 is Df(2+) = \(\lim\limits_{h\to 0}\frac{f(2+h)-f(2)}{h}\) = \(\lim\limits_{h\to 0}\frac{(2+h)^2|cos(\frac{\pi}{2+h})|-0}{h}\) (\(\because\) f(2) = 0) = \(4\lim\limits_{h\to 0}\cfrac{cos(\frac{\pi}{2+h})}{h}\) (\(\because \frac{\pi}{2+h}<\frac{\pi}2\)) (⇒|cos(\(\frac{\pi}{2+h}\))|) = cos(\(\frac{\pi}{2+h}\)) = \(4\lim\limits_{h\to 0}\cfrac{-sin(\frac{\pi}{2+h})\times\frac{-\pi}{(2+h)^2}}{1}\) = \(4\pi\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2+h})}{(2+h)^2}\) = \(4\pi\times\cfrac{sin\frac{\pi}2}4\) = 4π x \(\cfrac{sin\frac{\pi}2}4\) = 4π x 1/4 = π \(\because\) Df(2-) \(\neq\) Df(2+) \(\therefore\) Function f(x) is not differentiable at x = 2. Hence, function f(x) is differentiable at x = 0 but not differentiable at x = 2. |
|