2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    Let f be a differentiable function from, R to R such that\[\left| f(x)-f(y) \right|\,\,\le \,\,2\,\,{{\left| x-y \right|}^{3/2}}\], 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: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) 2  

    B)                                                   \[\frac{1}{2}\]

    C)               0                                 

    D)               1

    Correct Answer: D

    Solution :

    \[\left| f\left( x \right)-f\left( y \right) \right|\le \,\,2{{\left[ x-y \right]}^{3/2}}\] \[\underset{y\,\to \,x}{\mathop{\lim }}\,\left| \frac{f(x)\,-\,f(y)}{x-y} \right|\,\,\le \,\,\underset{y\,\to \,x}{\mathop{\lim }}\,\,\,2\,{{\left| x-y \right|}^{1/2}}\] \[\Rightarrow \,\,\,\left| f'(x) \right|\,\,\le \,\,0\Rightarrow \,\,{f}'(x)\,\,=\,\,0\] \[\Rightarrow \,\,\,\,f\left( x \right)=constant\] \[as\text{ }f\left( 0 \right)=1\,\,\Rightarrow \,\,f\left( x \right)=1\] \[\int\limits_{0}^{1}{{{f}^{2}}(x)\,dx\,\,=\,\,1}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner