Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

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:

Options:

$a < 1$

$a ≥ 1$

$a ≥ \sqrt{2}$

$a < \sqrt{2}$

Correct Answer:

$a ≥ 1$

Explanation:

$f'(x)=\sqrt{3}\cos x+\sin x-2a=2(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x)-2a=2(\cos(x-\frac{π}{6})-a),f'(x)<0∀\,x∈R$ 

$⇒\cos(x-\frac{π}{6})=1⇒a≥1$