Respected Sir/Madam,

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

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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` )

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