To determine the values of \(\lambda \) for which the function \(f(x)=\sin x-\cos x-\lambda x+C\) is always decreasing, we need to find the values of \(\lambda \) for which the derivative of \(f(x)\) is less than or equal to zero for all real \(x\).
The derivative of \(f(x)\) with respect to \(x\) is:
\(f^{\prime }(x)=\cos x+\sin x-\lambda \)
For \(f(x)\) to be a decreasing function for all real values of \(x\), we must have:\(f^{\prime }(x)\le 0\)\(\cos x+\sin x-\lambda \le 0\) that is
\(\cos x+\sin x\le \lambda \)
Note that for an expression of the form \(a\cos x+b\sin x\), the maximum value is \(\sqrt{a^{2}+b^{2}}\).
Here, \(a=1\) and \(b=1\). So the maximum value of \(\cos x+\sin x\) is:\(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
Now the condition \(\cos x+\sin x\le \lambda \) for all real \(x\).implies that \(\lambda \) must be greater than or equal to the maximum possible value of \(\cos x+\sin x\).
Therefore, we must have:\(\lambda \ge \sqrt{2}\)