**P148. Show that there is no real constant c > 0 such that \((\cos\sqrt{x+c}=\cos\sqrt{x})\) for all real numbers \((x\ge 0)\).**

**Solution:**

If the given equation holds for some constant c>0 then,

f(x) = \((\cos\sqrt{x}-\cos\sqrt{x+c}=0)\) for all \((x\ge 0)\)

\((\Rightarrow 2\sin\frac{\sqrt{x+c}+\sqrt{x}}{2}\sin\frac{\sqrt{x+c}-\sqrt{x}}{2}=0)\)

Putting x=0, we note

\((\Rightarrow\sin^2\frac{\sqrt{c}}{2}=0)\)

As \((c\not=0)\)

\((\sqrt{c}=2n\pi)\)

\((\Rightarrow c=4n^2\pi^2)\)

We put n=1 and x=\((\frac{\pi}{2})\) to note that f(x) is not zero.**Hence no c>0 allows f(x) =0 for all \((x\ge 0)\). (proved)**