Find the value of f(x)
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Anonymous:
its x-f(x)
where did you find this question? is f(x) a polynomial in x ? or some other ? like exponential function
sir its not given
I will get the solution tomorrow -_-
it was given in my exam .
which exam
Coaching institution exam --- jee mains (type)
I got it ...
this is not regular course. It is JEE Advanced or even more ...
which institute is this ?
Answers
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ANSWERS: There are two functions. Irrational functions.
f(x) = [x + √(x² - 48) ] / 2 and [x - √(x² - 48) ] / 2
======== PROOF -=====
We are given a function f(x) , but the definition of f(x) is not known. We have to find that.
We are given that integral limits from x=0 to x= 1, of f(x) * (x - f(x)) dx is equal to 12. we have to find f(x) such that it is true.
Let us choose a function f(x) such that the product :
f(x) * [ x - f(x) ] = 12 always and is a constant.
Then the integrand is always constant. So after integration we get :
∫ 12 dx = 12 x .
Evaluated between x=0 and x= 1, we get the RHS , ie, 12.
====> So we need to find f(x) such that
f(x) * [ x - f(x) ] = 12
or, ( f(x) )² - x * f(x) + 12 = 0
This is a quadratic in f(x) . Ie., f(x) is a variable. x is a coefficient.
Roots are: f(x) = [ x + √{x² - 48 } ] / 2
This is our function. ANSWER.
==========
Verification:
![f(x)=\frac{x+\sqrt{x^2-48}}{2}\\\\x-f(x)=\frac{x-\sqrt{x^2-48}}{2}\\\\Product=f(x)*(x-f(x))=\frac{(x+\sqrt{x^2-48})(x-\sqrt{x^2-48})}{2*2}\\\\=\frac{x^2-(x^2-48)}{4}=12\\\\So \: \int \limits_{x=0}^{x=1} {f(x)*(x - f(x))} \, dx\\\\=\int \limits_{x=0}^{x=1} 12 \, dx=[12x]_0^1=12 \, \ \ DONE.......](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%2B%5Csqrt%7Bx%5E2-48%7D%7D%7B2%7D%5C%5C%5C%5Cx-f%28x%29%3D%5Cfrac%7Bx-%5Csqrt%7Bx%5E2-48%7D%7D%7B2%7D%5C%5C%5C%5CProduct%3Df%28x%29%2A%28x-f%28x%29%29%3D%5Cfrac%7B%28x%2B%5Csqrt%7Bx%5E2-48%7D%29%28x-%5Csqrt%7Bx%5E2-48%7D%29%7D%7B2%2A2%7D%5C%5C%5C%5C%3D%5Cfrac%7Bx%5E2-%28x%5E2-48%29%7D%7B4%7D%3D12%5C%5C%5C%5CSo+%5C%3A+%5Cint+%5Climits_%7Bx%3D0%7D%5E%7Bx%3D1%7D+%7Bf%28x%29%2A%28x+-+f%28x%29%29%7D+%5C%2C+dx%5C%5C%5C%5C%3D%5Cint+%5Climits_%7Bx%3D0%7D%5E%7Bx%3D1%7D+12+%5C%2C+dx%3D%5B12x%5D_0%5E1%3D12+%5C%2C+%5C+%5C+DONE....... "f(x)=\frac{x+\sqrt{x^2-48}}{2}\\\\x-f(x)=\frac{x-\sqrt{x^2-48}}{2}\\\\Product=f(x)*(x-f(x))=\frac{(x+\sqrt{x^2-48})(x-\sqrt{x^2-48})}{2*2}\\\\=\frac{x^2-(x^2-48)}{4}=12\\\\So \: \int \limits_{x=0}^{x=1} {f(x)*(x - f(x))} \, dx\\\\=\int \limits_{x=0}^{x=1} 12 \, dx=[12x]_0^1=12 \, \ \ DONE.......")
Hope it is clear. I have explained in great detail.
f(x) = [x + √(x² - 48) ] / 2 and [x - √(x² - 48) ] / 2
======== PROOF -=====
We are given a function f(x) , but the definition of f(x) is not known. We have to find that.
We are given that integral limits from x=0 to x= 1, of f(x) * (x - f(x)) dx is equal to 12. we have to find f(x) such that it is true.
Let us choose a function f(x) such that the product :
f(x) * [ x - f(x) ] = 12 always and is a constant.
Then the integrand is always constant. So after integration we get :
∫ 12 dx = 12 x .
Evaluated between x=0 and x= 1, we get the RHS , ie, 12.
====> So we need to find f(x) such that
f(x) * [ x - f(x) ] = 12
or, ( f(x) )² - x * f(x) + 12 = 0
This is a quadratic in f(x) . Ie., f(x) is a variable. x is a coefficient.
Roots are: f(x) = [ x + √{x² - 48 } ] / 2
This is our function. ANSWER.
==========
Verification:
Hope it is clear. I have explained in great detail.
do you know Integration too ? @durgapal
and also @addison ? you know Integration too ?
sir they are in 8,9 classes. they don't know
watching you answering is a great thing for us
Yes.
solly
sorry
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