JEE MAIN - Mathematics (2005 - No. 6)
If $$f$$ is a real valued differentiable function satisfying
$$\left| {f\left( x \right) - f\left( y \right)} \right|$$ $$ \le {\left( {x - y} \right)^2}$$, $$x, y$$ $$ \in R$$
and $$f(0)$$ = 0, then $$f(1)$$ equals
$$\left| {f\left( x \right) - f\left( y \right)} \right|$$ $$ \le {\left( {x - y} \right)^2}$$, $$x, y$$ $$ \in R$$
and $$f(0)$$ = 0, then $$f(1)$$ equals
-1
0
2
1
Explanation
$$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {x + h} \right) - f\left( x \right)} \over h}$$
$$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|$$
$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0$$
$$ \Rightarrow f\left( x \right) = $$ constant
As $$f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0$$
$$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|$$
$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0$$
$$ \Rightarrow f\left( x \right) = $$ constant
As $$f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0$$
Comments (0)
