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:\(\lambda \ge \sqrt{2}\)
A function $f(x)$ is decreasing for all real values of $x$ if its derivative $f'(x) \le 0$ for all $x \in \mathbb{R}$.
Given $f(x) = \sin x - \cos x - \lambda x + C$. Differentiating with respect to $x$:
$$f'(x) = \cos x + \sin x - \lambda$$We require $f'(x) \le 0$, which implies:
$$\cos x + \sin x - \lambda \le 0$$ $$\cos x + \sin x \le \lambda$$We know that the expression $\sin x + \cos x$ can be written as $\sqrt{2} \sin(x + \frac{\pi}{4})$. The range of this expression is $[-\sqrt{2}, \sqrt{2}]$. For the inequality to hold for all $x$, $\lambda$ must be greater than or equal to the maximum value of the expression $\sin x + \cos x$.
$$\text{Max}(\sin x + \cos x) = \sqrt{2}$$ $$\text{Therefore, } \lambda \ge \sqrt{2}$$Final Answer: \lambda \ge \sqrt{2}
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