1any responses to the above ?
Find the maximum attained by \int_0^1 x^2 f(x) - x f^2(x) dx
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1!! :D ... trying again....
btw, my second step is wrong i suppose ...dat z' one ... ??
i saw jus now.....
341sorry, i have been preoccupied with some work. but yes, rohan is quite right. The integrand is maximum at every point when f(x) = x/2
1I meant to say this :
consider the graph of the function
x2f(x)-xf2(x)=G(x)
put f(x)=y
now for a particular x this value will will depend only on y
it will be maximum when dG(x)/dy=0 ( partial )
that is when
x2-2xy=0
=> y=x/2
thus at each x the value of this quantity will be maximum when y=x/2
thus if y=x/2 the quantity will be the maximum possible for all x from 0 to 1
thus then the area will also be maximum(as it will be maximum possible at every point)
341rohan, why dont you actually write out the steps.
Ok I(f) = \int_0^1 x^2 f(x) - x f^2(x) \ dx
Now \frac{\partial I}{\partial f} = 0
What next?
21Rohan : Can u pl. show sum more steps dude i didnt get much......... and iska matlab kya hai??
"if are is to be maximum if"
thus are will aso be maximum in that case
1and one error in the last line it will be partial differentiation ..w.r.t f(x)
1i am talking for each definite x that varies from 0 to 1 read my soln carefully
13u cant hav x fixed it varies .... ye integral hai mere bhai
1I think i have an easier method
if we maximise the given expression fixing x (i.e for a fixd value of x)
then we see that we get an expression which only depends on f(x) (as we have fixed x)
now if are is to be maximum if we can maximise this value then we will get f(x) in terms of x and thus the maximum possible value of this expression for any x
thus are will aso be maximum in that case
thus we diffrentiate it partially w.r.t x
we get f(x)=x/2 thus the ans as 1/16
341This is the method. I flicked this one from goiit where i posted this solution about a month back.
Here, the stumbling block is f(x), because it can be just about any function. Then it strikes that you encounter a similar situation when you are looking for the global minimum or maximum of a quadratic
So there you isolate the constant and push the unknown x into a perfect square
Same thing is done here for f(x). Because we know how to integrate a polynomial in x. So we push f(x) into some perfect square and we are rid of it
Problems in JEE will use some crossover techniques like this. When you play around with the expression, your brain will trigger some response that we have encountered this situation before. follow such hunches and you will be through!
13prophet sir any alternate soln for this problem plzz...
sumthin including more of calculus than makin perfect sq....
we generlly dont do this in calculus
66Let g(x)=x^2f(x)-xf^2(x). Complete the square in f(x) as follows:
g(x)=-x\left(f^2(x)-2\cdot f(x)\cdot \dfrac{x}{2}+\dfrac{x^2}{4}-\dfrac{x^2}{4}\right)
\Rightarrow\ g(x)=\dfrac{x^3}{4}-x\left(f(x)-\dfrac{x}{2}\right)^2\leq \dfrac{x^3}{4}
since 0 ≤ x ≤1.
Therefore,
\dfrac{x^3}{4}-g(x)\geq 0
\Rightarrow\ \int_0^1\left(\dfrac{x^3}{4}-g(x)\right)\mathrm{d}x\geq 0
\Rightarrow\ \int_0^1\dfrac{x^3}{4}\ \mathrm{d}x -\int_0^1g(x)\ \mathrm{d}x\geq 0
\Rightarrow\ \int_0^1g(x)\ \mathrm{d}x\leq \dfrac{1}{16}
Accordingly, the maximum value of \int_0^1 (x^2f(x)-xf^2(x))\ \mathrm{d}x is \dfrac{1}{16}. The equality is attained for f(x)=\dfrac{x}{2}.
11sir ek faaltu ques
aapne topic ka naam Find Minimum poocha hai
and in the ques r asking type max value
wats the dilemma [7][7][7]
341I was doing just that when i got disconnected.
Its not second derivative. its the square of the function.
1solving the above diff eqn :
we get ..
f2(x) - xf(x) = x2e-3x+C
=>x f2(x) - x2f(x) = x3e-3x+C
so we get
z = x3e-3x+C
1have got a diff eqn..
let z= x2f(x) -xf2(x)
z' = x2f'(x) + 2xf(x) - f2(x) - x.2f(x).f'(x) = 0 [for maxima or minima]
f'(x) = [f2(x) - 2xf(x)]/[x2 - 2xf(x)]
let f(x)=y
so we have: dy/dx = [y2 - 2xy]/[x2-2xy] ... homogeneous..
am i doing anything worthful ?
11i did it wrt x
but i understand now that im wrong f(x) and x are variables
341RHS is a function of what variable?
You are differentiating w.r.t to what variable?
11but i agree with you in the example
this cant be done the normal way then?
11but since the limits are given to be constants
differentiating the entire thing would always give zero
341how do you prove that?
If f(x) = 0, the integral is 0. If f(x) = 1, then it is -1/6. So, the integral is not constant