Homework Help

# Respected Sir/Madam, Pls. help me out with Q.5 in the attached image.

user8235304 | Student, Grade 11 | Valedictorian

Posted June 13, 2013 at 5:41 PM via web

dislike 1 like

Pls. help me out with Q.5 in the attached image.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

Tagged with math, trigonometry

### 1 Answer |

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 14, 2013 at 3:09 PM (Answer #1)

dislike 2 like

Given `0<=a<=3,0<=b<=3` and `x^2+4+3cos(ax+b)=2x` has at least one solution, find a+b:

Rewrite as `3cos(ax+b)=-x^2+2x-4 ==> cos(ax+b)=-1/3x^2+2/3x-4/3` Now `-1<=cosalpha<=1` so:

`-1<=-1/3x^2+2/3x-4/3<=1`

`3>=x^2-2x+4>=-3`

Now `x^2-2x+4>=3` for all x, with equality at x=1.

(The graph is a parabola opening up: rewrite in vertex form to find the vertex. `x^2-2x+1-1+4 = (x-1)^2+3` so the vertex is at (1,3) )

Let x=1:

`1+4+3cos(a+b)=2`

`3cos(a+b)=-3`

`cos(a+b)=-1`

==> `a+b=pi` (With `a,b>=0` )

### Join to answer this question

Join a community of thousands of dedicated teachers and students.