Tagged: integral calculus
- March 24, 2020 at 10:31 pm #60258venkat jothiParticipant
please give a correct explanation for this problem. it is obtained from cmi BSc math entrance examination 2019 problem number 10.March 26, 2020 at 8:44 pm #60488
Will get back to you within 24 hoursMarch 27, 2020 at 9:10 pm #60719
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 $\frac14$March 29, 2020 at 8:05 pm #60990
So have you tried it?
3rd statement is not true :
$=\frac14 -(\frac14 -2.\frac12.f(x)+f(x)^2)$
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 $f(x)=\frac12$
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