|

Integral Calculus

Forum Expert
 Joined: 28 Feb 2007 Post: 2185
20 Mar 2008 18:26:59 IST
0 People liked this
4
665
Contrived Problem
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Integral Calculus

Ok one of my feeble attempts at making up a problem (Atleast I have not encountered one like this). Lets see how quickly this gets destroyed.

A function f(x) is such that 14 f(x) dx = 3. Find a function g(x) such that f(x)+g(x) has a zero in (1,4).

Blazing goIITian

Joined: 17 Jan 2008
Posts: 510
20 Mar 2008 19:30:28 IST
0 people liked this

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
20 Mar 2008 19:56:30 IST
0 people liked this

Don't assume f(x).

I am looking for a one-size-fits-all g(x) for all f(x) which satisfy the given condition

Hot goIITian

Joined: 26 Jan 2007
Posts: 156
20 Mar 2008 20:26:47 IST
2 people liked this

The Obvious solution is f(x) = -g(x).
But consider,
let g be any real valued,continuous  function that satisfies f + g =0 for at least one value of x (1.4)

Let H(x) =  ( f + g)dx

By LMVT for H over (1,4),

H'(c) = [H(4) - H(1)]/3

But H'(c) = f + g

if f + g =0 for at least one c  (1.4),

Then H'(c) = 0 for at least one c  (1.4),

implies , a solution for g can be got if ,<Though this is not the only solution>

H(4) - H(1) =0

implies [1][4]( f + g)dx = 0

implies

[1][4](g)dx = -3

Hence any function that satisfies [1][4](g)dx = -3

is one class of  solution.

example g(x) = (-2x/5)

That solves the problem!

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
20 Mar 2008 21:13:58 IST
0 people liked this

good. that was what i was looking for.

If ab f(x) = 0, then by mean value theorem, there exists c(a,b) so that f'(c) = 0

Another way to understand is that if ab f(x) = 0, then f(x) neither be +ve for all x in the interval nor -ve for all x. So there's got to be x in the interval for which it is zero.

 Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a avatar to have your pictures show up by your comment If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

## For Quick Info

Name

Mobile

E-mail

City

Class

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
 Altitude - 16545 m Post - 7958
2. Himanshu
 Altitude - 10925 m Post - 3836
3. Hari Shankar
 Altitude - 9960 m Post - 2185
4. edison
 Altitude - 10815 m Post - 7797
5. Sagar Saxena
 Altitude - 8625 m Post - 8064
 Altitude - 6330 m Post - 1979

Physics

Topics

Mathematics

Chemistry

Biology

Institutes

Parents Corner

Board

Fun Zone