If 0<a, b <5 and (x^2+5)/2=x-2 cos (a + bx) has at least one real roo…

We have,

$\frac{x^2+5}{2}=x-2 \cos (a + bx)$

$⇒x-\frac{x^2+5}{2}=2 \cos (a + bx)$

$⇒\frac{2x^2-x^2-5}{2}=2 \cos (a + bx)$

$⇒\frac{1}{2}[-(x^2 - 2x) −5] = 2 \cos (a + bx)$

$⇒\frac{1}{2}[-(x - 1)^2 −4] = 2 \cos (a + bx)$

$⇒-2-\frac{1}{2}(x - 1)^2 = 2 \cos (a + bx)$

Clearly, LHS = $-2-\frac{1}{2}(x - 1)^2 ≤- 2$ and, RHS ≥ -2.

Thus, the two sides are equal if each is equal to -2.

Now,

RHS = -2

$⇒ \cos (a - bx) = -1$

$⇒ \cos (a + bx) = \cos π$  

$⇒a + bx = 2n\, π±π$   ...(i)

and,

$LHS=-2 ⇒ x=1$  ...(ii)

Putting $x = 1$ in (i), we get

$a+b=2n\,π± π, n∈Z$

$⇒a+b=π, 3 π$  $[∵ 0<a+b<10]$

Hence, the greates value of $a + b$ is $3π$.