Question

The value of \(k\) such that the function \(f(x) = sinx-cosx-kx+b\) is strictly decreasing for all real \(x\) is

• A. \(k < 1\)
• B. \(k > 1\)
• C. \(k < \sqrt{2}\)
• D. \(k > \sqrt{2}\)

Hint:

f decreasing for all x needs f prime negative for all x; the maximum of cos x + sin x is sqrt2.

Answer 1

The Correct Option is D

Step 1: Differentiate:
\(f'(x)=\cos x+\sin x-k\).

Step 2: Condition for decreasing:
We need \(f'(x)<0\) for every real \(x\), so \(\cos x+\sin x<k\) for all \(x\).

Step 3: Find the maximum of cos x + sin x:
\(\cos x+\sin x=\sqrt2\sin\left(x+\dfrac\pi4\right)\), whose maximum value is \(\sqrt2\). So we need \(k>\sqrt2\). Option D.

Step 4: Why the other options are wrong.
\(k<1\), \(k>1\) and \(k<\sqrt2\) do not guarantee \(f'<0\) everywhere. For example \(k=1.2\) fails at \(x=\pi/4\), where \(f'=\sqrt2-1.2>0\).

Final Answer:
k must be greater than sqrt 2. \[ \boxed{\text{(D) }k>\sqrt2} \]