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: Use the condition for a decreasing function.
A function \(f(x)\) decreases for all values of \(x\) if \[ f'(x)\leq 0 \] for all \(x\).

Step 2: Differentiate the given function.
Given, \[ f(x)=\sqrt{3}\sin x-\cos x-2ax+b \] Differentiating with respect to \(x\), \[ f'(x)=\sqrt{3}\cos x+\sin x-2a \]

Step 3: Simplify the trigonometric expression.
We know that \[ \sqrt{3}\cos x+\sin x \] can be written in the form \[ R\cos(x-\alpha) \] Here, \[ R=\sqrt{(\sqrt{3})^2+1^2} \] \[ R=\sqrt{3+1}=2 \] Therefore, \[ \sqrt{3}\cos x+\sin x \] has maximum value \[ 2 \]

Step 4: Apply decreasing condition.
Since \[ f'(x)=\sqrt{3}\cos x+\sin x-2a, \] for \(f(x)\) to be decreasing for all \(x\), we need \[ \sqrt{3}\cos x+\sin x-2a\leq 0 \] for all \(x\).
The maximum value of \[ \sqrt{3}\cos x+\sin x \] is \[ 2 \] Hence, the greatest possible value of \(f'(x)\) is \[ 2-2a \] For the function to be decreasing for all \(x\), \[ 2-2a\leq 0 \] \[ 2\leq 2a \] \[ a\geq 1 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{a\geq 1} \]