JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 8)
Let f be a differentiable function from
R to R such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$
for all $$x,y \in $$ R.
If $$f\left( 0 \right) = 1$$
then $$\int\limits_0^1 {{f^2}} \left( x \right)dx$$ is equal to :
R to R such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$
for all $$x,y \in $$ R.
If $$f\left( 0 \right) = 1$$
then $$\int\limits_0^1 {{f^2}} \left( x \right)dx$$ is equal to :
1
2
$${1 \over 2}$$
0
Explanation
$$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$$
$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$
$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$$
$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$$ $$ \Rightarrow f'\left( x \right) = 0$$
$$ \Rightarrow f\left( x \right) = $$ constant
as $$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$$
$$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$$
$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$
$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$$
$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$$ $$ \Rightarrow f'\left( x \right) = 0$$
$$ \Rightarrow f\left( x \right) = $$ constant
as $$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$$
$$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$$
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