Previous year Question of jee mains
Chapter:- application and derivatives
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The correct option is (1)
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EXPLANATION.
⇒ f(x) = 4x³ - 3x²/6 - 2sin x + (2x - 1)cos x.
As we know that,
Differentiate the equation, we get.
⇒ f'(x) = (2x² - x) - 2cos x + 2cos x - sin(2x - 1).
⇒ f'(x) = (2x² - x) - sin x (2x - 1).
⇒ f'(x) = x(2x - 1) - sin x (2x - 1).
⇒ f'(x) = (x - sin x)(2x - 1).
As we know that,
⇒ x - sin x > 0 for x > 0.
⇒ x - sin x < 0 for x < 0.
⇒ f'(x) ≥ 0 = x ∈ (-∞, 0] ∪ [1/2, ∞).
⇒ f'(x) ≤ 0 = x ∈ [0, 1/2].
⇒ f(x) is increasing = [1/2, ∞).
Option [1] is correct answer.
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