how many solutions of the equation x^4-x^3*sinx-x^2*cosx=0?

Forum for the GRE subject test in mathematics.
cathy_liping
Posts: 9
Joined: Sat Oct 01, 2011 11:21 pm

how many solutions of the equation x^4-x^3*sinx-x^2*cosx=0?

Postby cathy_liping » Sat Oct 01, 2011 11:32 pm

how to solve this kind of problem? tks in advance :D

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: how many solutions of the equation x^4-x^3*sinx-x^2*cosx=0?

Postby owlpride » Sun Oct 02, 2011 12:48 am

There are 3 unique solutions (without counting multiplicities).

Notice that the left-hand side factors as x^2 * (x^2 - x sinx - cosx). The x^2 factor gives you a solution for x = 0. Let's focus on the other factor, f(x) = x^2 - x sinx - cosx. Notice that f is negative around a neighborhood of the origin and approaches + infinity as |x| gets large (since the x^2 term dominates). Hence f has at least two roots as its sign changes. Does it have any more?

Let's look at its critical points. f'(x) = 2x - x cosx = x * (2 - cosx) = 0 only at x = 0. (That's a local minimum of f because f' < 0 for x < 0 and f' > 0 for x > 0.) Since f has only a single critical point, it will have at most two roots (there's at least one critical point between any two roots). But we already found two roots. So we are done.

cathy_liping
Posts: 9
Joined: Sat Oct 01, 2011 11:21 pm

Re: how many solutions of the equation x^4-x^3*sinx-x^2*cosx=0?

Postby cathy_liping » Sun Oct 02, 2011 4:00 am

thank you very much!! now I think I conquer this kind of problems!

I can find the solution of another similar question by myself (cheers!!)

Quesion is y=logx and y=cx^4 have only one intersection, what is the value of c?
The answer is 1/4e.

let f(x)=logx-cx^4=0 which means f(x) has only one intersction with x-axis.
f'(x)=1/x-4cx^3=0
x^4=square1/4c
let c>0
x>square1/4c, f'(x)<0 f(x) go decrease
x<square1/4c, f'(x)>0, f(x) go increase
so f(x=square1/4c) must be 0 so that f(x) has only one intersection with x-axis.




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