PR 3rd ed, Final test question 9

Forum for the GRE subject test in mathematics.
fullofquestions
Posts: 18
Joined: Sun Oct 07, 2007 1:49 pm

PR 3rd ed, Final test question 9

Postby fullofquestions » Fri Oct 26, 2007 12:11 pm

Question:
Let a be the smallest positive value of x at which the function f(x) = (cos x^2)(sin x^2) has a critical point. What is the value of f(a)?

The answer begins by using the double angle formula for the sine (sin(2x) = 2sinxcosx) to write the given function in the form f(x) = 1/2 * (sin(2x^2))

I can follow the problem after this point although I do not see how you get the rewritten form. I just need to have that clarified. Thanks.

mamal
Posts: 11
Joined: Tue Oct 30, 2007 3:02 am

Postby mamal » Tue Oct 30, 2007 3:21 am

The answer would be easy to follow!

Ok! look at this. Sin(2x)=2Sin(x)Cos(x). Now subtitute x^2 with U. Now we have the following relation.
f(x)=Sin(U)Cos(U) ==> f(x)=1/2*Sin(2U) ==> f(x)=1/2*Sin(2x^2)
Now since for Sin(U), the first critical point is licated at U=PI/2, we have,
2x^2=PI/2 ==> x=SQRT[PI]/2
That's All. :d

fullofquestions
Posts: 18
Joined: Sun Oct 07, 2007 1:49 pm

That was easy

Postby fullofquestions » Tue Oct 30, 2007 9:28 am

Thanks a bunch. I solved the problem another way but yours is more elegant and takes less steps. Better indeed.




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