Home » Ask & Discuss » Mathematics. » Trignometry « Back to Discussion

Trignometry

Hot goIITian

Joined:	23 Mar 2008
Post:	159

30 Mar 2008 21:57:51 IST

0 People liked this

321

Functions

None

sin(e^X)=2^X+2^-x then find the number of real solutions?

Share this article on:

Comments (4)

Akhil

Blazing goIITian

Joined: 17 Jan 2008

Posts: 1451

30 Mar 2008 22:00:41 IST

Like 0 people liked this

RHS minimum value=2 at x=0
LHS max value=1
so no solution

Anand Hegde

Blazing goIITian

Joined: 12 Jun 2007

Posts: 1078

30 Mar 2008 22:01:45 IST

Like 0 people liked this

$\text{By AM GM, we get}\\ \\ \frac{2^{x}+2^{-x}}{2} \ge \sqrt{2^{x}.2^{-x}}\\ \\ 2^{x}+2^{-x} \ge 2\\ \\ \text{RHS is always greater than or equal to 2, but LHS is always less than or equal to 1. Hence no solution}$

Anand Hegde

Blazing goIITian

Joined: 12 Jun 2007

Posts: 1078

30 Mar 2008 22:02:15 IST

Like 0 people liked this

oh sorry akhil....dint know you were at it.

ankit aggarwal

Blazing goIITian

Joined: 10 Feb 2008

Posts: 303

1 Apr 2008 19:57:33 IST

Like 0 people liked this

no solution because sin(e^X) has the range from [-1,1]
whereas 2^X+2^{-x has the least value of 2.}

Quick Reply

Reply

	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