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QJEEM19Ma9068: JEE Main 2019 Mathematics - Class 12 JEEM19Ma9068 Solution: Let f be a differentiable function from R to R such that…

Let f be a differentiable function from

R to R such that ∣f(x)−f(y)∣≤2∣x−y∣32,\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},    

for all   x,y∈x,y \in R.

If    f(0)=1f\left( 0 \right) = 1   

then    ∫01f2(x)dx\int\limits_0^1 {{f^2}} \left( x \right)dx   is equal to :

Multiple Choice Options

  • A 1
  • B 2
  • C 12{1 \over 2}
  • D 0

Step-by-Step Verified Solution

∣f(x)−f(y)∣≤2[x−y]3/2\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}

∣f(x)−f(y)x−y∣≤2∣x−y∣1/2\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}

lim⁡y→x∣f(x)−f(y)x−y∣≤lim⁡y→x2∣x−y∣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}}

⇒∣f′(x)∣≤0\Rightarrow \left| {f'\left( x \right)} \right| \le 0   ⇒f′(x)=0\Rightarrow f'\left( x \right) = 0

⇒f(x)=\Rightarrow f\left( x \right) = constant

as   f(0)=1⇒f(x)=1f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1

∫01f2(x)dx=1\int\limits_0^1 {{f^2}} \left( x \right)dx = 1

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