CMI integration problem

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  • #60258
    venkat jothi
    Participant

    source cmi entrance exam 2019

    please give a correct explanation for this problem. it is obtained from cmi BSc math entrance examination 2019 problem number 10.

    #60488
    Saumik Karfa
    Moderator

    Will get back to you within 24 hours

    #60719
    Saumik Karfa
    Moderator

    take $f(x)=1$ then $f(x)$ is continuous and the integral becomes $0$ which is less than $\frac14$, so the first option is not valid.

    We can see that for $f(x)=\frac12$ (constant function $\Rightarrow$ continuous) value of the integration is $\frac14$.

    Now what do you think ? Are there infinite values so that the value of the integral is $\frac14$

    #60990
    Saumik Karfa
    Moderator

    So have you tried it?

    3rd statement is not true :

    $f(x)(1-f(x))=f(x)-f(x)^2$

    $=\frac14 -(\frac14 -2.\frac12.f(x)+f(x)^2)$

    $=\frac14-(\frac12-f(x)^2)$

    Now let $g(x)=(1-f(x))^2\geq 0,$ then $ g$ is positive and continuous :

    then $\int_0^1 g(x)=0$ if and only if $g(x)=0$.

    Since $g$ is continuous and $g(x)=0$ everywhere then$f(x)=\frac12$

    Then $\int_0^1 f(x)(1-f(x)) \mathrm d x$

    $=\int_0^1 \frac14 \mathrm d x -\int_0^1 (\frac12 – f(x))^2\mathrm d x$

    $=\frac14-\int_0^1 (\frac12 – f(x))^2\mathrm d x$

    $=\frac14$ if and only if $f(x)=\frac12$

    • This reply was modified 1 month, 4 weeks ago by Saumik Karfa.
    • This reply was modified 1 month, 4 weeks ago by Saumik Karfa.
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