Question

If \(f(x)=\sqrt{3\sin x-\cos x-2ax+b}\) decreases for all values of \(x\), then

• A. \(a\geq 1\)
• B. \(a=1\)
• C. \(a\leq 1\)
• D. \(a<1\)

Hint:

For a function to be decreasing for all \(x\), check that \[ f'(x)\leq 0 \] for every \(x\). Also, the maximum value of \[ A\cos x+B\sin x \] is \[ \sqrt{A^2+B^2}. \]

Answer 1

The Correct Option is A

Step 1: State the condition for a function to be decreasing everywhere.
f(x) decreases for all x if and only if f'(x) ≤ 0 for all x.

Step 2: Differentiate the given function.
f(x) = √3 sin x - cos x - 2ax + b → f'(x) = √3 cos x + sin x - 2a.

Step 3: Find the maximum value of the trigonometric part.
√3 cos x + sin x can be expressed as R cos(x - α) with R = √((√3)² + 1²) = 2. Its maximum value is 2.

Step 4: Impose the decreasing condition.
For f'(x) ≤ 0 for all x, the maximum possible value of f'(x) must be ≤ 0. The maximum is 2 - 2a ≤ 0 → a ≥ 1.

Step 5: Final conclusion.
The required condition is a ≥ 1.