Question:
If $f(x) = \sqrt{3}\sin x - \cos x - 2ax + b$ decreases for all $x \in \mathbb{R}$, then:
If $f(x) = \sqrt{3}\sin x - \cos x - 2ax + b$ decreases for all $x \in \mathbb{R}$, then:
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For $A\cos x + B\sin x$, maximum is $\sqrt{A^2+B^2}$.
Updated On: Jun 10, 2026
- $a \le 1$
- $a \ge 1$
- $a \le \frac{1}{2}$
- $a \ge \frac{1}{2}$
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The Correct Option is B
Solution and Explanation
\[
f'(x)=\sqrt{3}\cos x + \sin x - 2a
\]
For decreasing function:
\[
\sqrt{3}\cos x + \sin x \le 2a
\]
Maximum value:
\[
\sqrt{3^2 + 1^2} = 2
\]
\[
2a \ge 2 \Rightarrow a \ge 1
\]
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