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If $f$ is a real valued differentiable function satisfying $\mid f(x)-f(y)\mid\leq (x-y)^2,x,y\in R$ and $f(0)=0$ then $f(1)$ equals

$(a)\;-1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;1$

1 Answer

$f'(x)=\lim\limits_{h\to 0}\large\frac{f(x+h)-f(x)}{h}$
$\mid f'(x)\mid=\lim\limits_{h\to 0}\mid\large\frac{f(x+h)-f(x)}{h}\mid$$\leq \lim\limits_{h\to 0}\mid \large\frac{(h)^2}{h}\mid$
$\Rightarrow \mid f'(x)\mid\leq 0\Rightarrow f'(x)=0$
$\Rightarrow f(x)$=constant
As $f(0)=0$
$\Rightarrow f(1)=0$
Hence (b) is the correct answer.
answered Dec 31, 2013 by sreemathi.v
 

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