# inequality

• April 3rd 2011, 11:02 AM
jiboom
inequality
the question is simplY:

solve for 0<x<10

ln(2x+1)<3 cos x

{it is only woth 4 marks, part one, which was worth 2 marks,was what 2 transformations take
ln(x) to ln(2x+1) so i cant see there is to much work here but i cant get it.}

I can see i need to find the points of intersection, but i must be missing something here.

The mark scheme is no help,it just states the answer.
• April 3rd 2011, 02:28 PM
pickslides
Have you graphed each of these functions?

What do you get?
• April 4th 2011, 06:17 AM
jiboom
yes,graphed them so i just need points of intersection, but how to find them is the problem
• April 4th 2011, 01:44 PM
pickslides
Fair problem to have.

You need to use technology (spreadsheet or graphic calculator) to find the intersection of these functions, if you can't use those, a numerical method like the bisection or newton's method will help.
• April 8th 2011, 09:44 AM
jiboom
Quote:

Originally Posted by pickslides
Fair problem to have.

You need to use technology (spreadsheet or graphic calculator) to find the intersection of these functions, if you can't use those, a numerical method like the bisection or newton's method will help.

this is what i guessed, but as i say its only 4 marks,compared to the 2 marks for the 2 transformations in part (1).

it gives no starting values and i need 4 numbers (0 being the easy on to find)

on another paper have same sort of question when need to solve e^(-x)-x+1=0, once again mark scheme gives no suggestion as to how the answer is to be found.
• April 10th 2011, 02:00 PM
pickslides
I can tell you there are 3 points of intersection between $\displaystyle \ln (2x+1)$ and $\displaystyle 3\cos x$ on this interval.

I suggest when using a choosen numerical method, use the starting points as x=1,5 & 7.