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The value of a in order that $f(x)=\sqrt{3}\sin x-\cos x-2ax+b$ decreases for all real values of x, is given by: |
$a < 1$ $a ≥ 1$ $a ≥ \sqrt{2}$ $a < \sqrt{2}$ |
$a ≥ 1$ |
$f'(x)=\sqrt{3}\cos x+\sin x-2a=2(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x)-2a=2(\cos(x-\frac{π}{6})-a),f'(x)<0∀\,x∈R$ $⇒\cos(x-\frac{π}{6})=1⇒a≥1$ |
