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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

The maximum value of the expression sin θ cos^ 2 θ ( θ ∈[0, π]) is

The maximum value of the expression sinθcos2θ(∀θ∈[0,π])issin \theta cos^{2}\theta \left(\forall \theta \in\left[0, \pi\right]\right) is

Options

  1. A23\dfrac{2}{3}
  2. B23\dfrac{2}{\sqrt{3}}
  3. C233\dfrac{2}{3\sqrt{3}}
  4. D13\dfrac{1}{\sqrt{3}}

Correct answer

C. 233\dfrac{2}{3\sqrt{3}}

Step-by-step solution

As sin2θ+cos2θ=1,sin^{2}\theta + cos^{2}\theta = 1,

⇒sin2θ+cos2θ2+cos2θ2=1⇒ sin^{2}\theta + \dfrac{cos^{2}\theta}{2} + \dfrac{cos^{2}\theta}{2} = 1

Applying AM≥GM,AM \ge GM, we have

sin2θ+cos2θ2+cos2θ23≥(sin2θcos4θ4)13\dfrac{sin^{2}\theta + \dfrac{cos^{2}\theta}{2} + \dfrac{cos^{2}\theta}{2}}{3} \ge {\left(\dfrac{sin^{2}\theta cos^{4}\theta}{4}\right)}^{\dfrac{1}{3}}

⇒(13)3≥sin2θcos4θ4⇒ {\left(\dfrac{1}{3}\right)}^{3} \ge \dfrac{sin^{2}\theta cos^{4}\theta}{4}

⇒(sinθ⋅cos2θ)2≤427⇒ \left(sin \theta ⋅ cos^{2}\theta\right)2 \le \dfrac{4}{27}

⇒sinθ⋅cos2θ≤233⇒ sin \theta ⋅ cos^{2}\theta \le \dfrac{2}{3\sqrt{3}}

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