Practicing Success
Let f(x) and g(x) be two real functions such that $f(x)=\frac{x}{|x|} g(x), x \neq 0$. If g(0) = g'(0) = 0 and f(x) is continuous at x = 0, then f'(0) is |
0 1 -1 non-existent |
0 |
We have, $f(x)=\frac{x}{|x|} g(x), x \neq 0 $ $=\left\{\begin{array}{rr}g(x), & x>0 \\ -g(x), & x<0\end{array}\right.$ It is given that g'(0) = g(0) = 0. This means that g(x) is differentiable and hence continuous at x = 0 and $\lim\limits_{x \rightarrow 0} g(x)=g(0)=0$ Now, $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{-}}-g(x)=-\lim\limits_{x \rightarrow 0} g(x)=0$ and, $\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0^{+}} g(x)=\lim\limits_{x \rightarrow 0} g(x)=0$ ∴ $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)$ So, f(x) is continuous at x = 0. ∴ $\lim\limits_{x \rightarrow 0} f(x)=f(0)$ $\Rightarrow f(0)=0$ Thus, we have $f(x)=\left\{\begin{array}{rr}g(x), & x>0 \\ 0, & x=0 \\ -g(x), & x<0 \end{array}\right.$ Now, (LHD of f(x) at x = 0) $=\lim\limits_{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$ $=\lim\limits_{h \rightarrow 0} \frac{f(-h)}{-h}$ $=\lim\limits_{h \rightarrow 0} \frac{-g(-h)}{-h}$ $=\lim\limits_{h \rightarrow 0} \frac{g(-h)-g(0)}{-h}$ [∵ g(0) = 0] $=-g'(0)=0$ Similarly, we have (RHD of f(x) at x = 0) = g'(0) = 0. Hence, f(x) is differentiable at x = 0 and f'(0) =0. |