Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

I was solving this question : $I = \int_0^1xf(x)\,dx = \frac{1}{6}$ $J = \int_0^1 (f(x))^2\,dx = \frac{1}{12}$ $f\left( \frac{1}{2} \right) = ?$ Here $f(x)$ is continuous. So, to solve this I tried to get an integral that contained both the given integrals. So, I assume a parameter $t$ and : $\int_0^1 (f(x) - tx)^2\,dx =0 $ $\int_0^1(f(x))^2,dx -2t\int_0^1xf(x)\,dx +t^2\int_0^1x^2\,dx =0$ Putting in the values and solving, I got : $t = \frac{1}{2}$. So then, $(f(x) - 0.5x)^2$ is always positive, so in order for the integral I assumed to evaluate to $0$, the function had to be $0$. $(f(x)-0.5x)^2 =0$ $f(x) = 0.5x$ $f(0.5) = 0.25$ This was correct according to the answer key, but my doubt is that : There can be another function $g(x) \neq tx$ which also satisfies the given conditions, and $g(0.5) \neq 0.25$. So, how can we prove that either $tx$ is the only function satisfying the given conditions, or that for every possible $g(x)$, $g(0.5)$ will have to be 0.25 ?

I was solving this question : 
 
$I = \int_0^1xf(x)\,dx = \frac{1}{6}$
 
$J = \int_0^1 (f(x))^2\,dx = \frac{1}{12}$
 
$f\left( \frac{1}{2} \right) = ?$
 
Here $f(x)$ is continuous.
So, to solve this I tried to get an integral that contained both the given integrals. So, I assume a parameter $t$ and  : 
 
$\int_0^1 (f(x) - tx)^2\,dx =0 $
 
$\int_0^1(f(x))^2,dx -2t\int_0^1xf(x)\,dx +t^2\int_0^1x^2\,dx =0$
 
Putting in the values and solving, I got : $t = \frac{1}{2}$.
 
So then, $(f(x) - 0.5x)^2$ is always positive, so in order for the integral I assumed to evaluate to $0$, the function had to be $0$.
 
$(f(x)-0.5x)^2 =0$
 
$f(x) = 0.5x$
 

$f(0.5) = 0.25$
 
This was correct according to the answer key, but my doubt is that :
 
There can be another function $g(x) \neq tx$ which also satisfies the given conditions, and $g(0.5) \neq 0.25$. So, how can we prove that either $tx$ is the only function satisfying the given conditions, or that for every possible $g(x)$, $g(0.5)$ will have to be 0.25 ? 

Grade:12th pass

2 Answers

Arun
25763 Points
one year ago
Dear student
 
Question is not understandable. Please check and repost the question with an attachment. I will be happy to help you
Vikas TU
14149 Points
one year ago
Dear student 
Question is not clear 
Please upload an image.
We will happy to help you.
Good luck 
Cheers 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free