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:\(\lambda \ge \sqrt{2}\)
Given: \(f(x) = \sin x - \cos x - \lambda x + C\)
For \(f(x)\) to be decreasing for all real values of x, we must have \(f'(x) \le 0\) for all x.
First, find the derivative of \(f(x)\) with respect to x:
\(f'(x) = \frac{d}{dx}(\sin x - \cos x - \lambda x + C)\)
\(f'(x) = \cos x + \sin x - \lambda\)
Now, we need \(f'(x) \le 0\) for all x:
\(\cos x + \sin x - \lambda \le 0\)
\(\cos x + \sin x \le \lambda\)
The maximum value of \(\cos x + \sin x\) can be found by rewriting it in the form \(R\cos(x - \alpha)\), where \(R = \sqrt{1^2 + 1^2} = \sqrt{2}\). Thus, the maximum value is \(\sqrt{2}\).
Therefore, \(\cos x + \sin x \le \sqrt{2}\)
So, we must have \(\lambda \ge \sqrt{2}\) for the inequality \(\cos x + \sin x \le \lambda\) to hold for all x.
Correct Answer: \(\lambda\ge\sqrt{2}\)
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