Practicing Success
Let $f(x)=\min \{1, \cos x, 1-\sin x\},-\pi \leq x \leq \pi$. Then, f(x), is |
not continuous at $x=\pi / 2$ continuous but not differentiable at x = 0 neither continuous nor differentiable at $x=\pi / 2$ none of these |
continuous but not differentiable at x = 0 |
We have, $f(x)=\min \{1, \cos x, 1-\sin x\}$ $\Rightarrow f(x)= \begin{cases}\cos x, & -\pi / 2 \leq x \leq 0 \\ 1-\sin x, & 0<x \leq \pi / 2 \\ \cos x, & \pi / 2<x \leq \pi\end{cases}$ We find that $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{-}} \cos x=1$ $\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0} 1-\sin x=1$ and, $f(0)=\cos 0=1$ ∴ $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$ So, f(x) is continuous at x = 0. Now, (LHD at x = 0) = $\left(\frac{d}{d x}(\cos x)\right)_{x=0}=0$ (RHD at x = 0) = $\left(\frac{d}{d x}(1-\sin x)\right)_{x=0}=-1$ ∴ (LHD at x = 0) ≠ (RHD at x = 0) Hence, f(x) is not differentiable at x = 0. However, it is continuous there at. At $x=\pi / 2$ also, f(x) is continuous but not differentiable. |