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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2021 · Mathematics · Application of Derivatives

For what value of θ , where 0 < θ < (π)/(2) , does θ+ θ θ attain...

For what value of θ\theta, where 0<θ<π20 < \theta < \frac{\pi}{2}, does sin⁡θ+sin⁡θcos⁡θ\sin \theta+\sin \theta\cos \theta attain maximum value?

Options

  1. Aπ2\frac{\pi}{2}
  2. Bπ3\frac{\pi}{3}
  3. Cπ4\frac{\pi}{4}
  4. Dπ6\frac{\pi}{6}

Correct answer

B. π3\frac{\pi}{3}

Step-by-step solution

 Let S=sin⁡θ+sin⁡θ⋅cos⁡θdsdθ=cos⁡θ+cos⁡2θ−sin⁡2θ=2cos⁡2θ+cos⁡θ−1=0=(cos⁡θ+1)(2cos⁡θ−1)=0∴cos⁡θ≠1∴cos⁡θ=12⇒θ=π3\begin{aligned} & \text{ Let } S=\sin \theta+\sin \theta \cdot \cos \theta \\ & \frac{d s}{d \theta}=\cos \theta+\cos ^2 \theta-\sin ^2 \theta \\ & =2 \cos ^2 \theta+\cos \theta-1=0 \\ & =(\cos \theta+1)(2 \cos \theta-1)=0 \\ & \therefore \cos \theta \neq 1 \therefore \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\end{aligned} Now, d2sdθ2=−4cos⁡θsin⁡θ−sin⁡θ(d2sdθ2)(x=π3)=−4×12×32−32<0∴\frac{d^2 s}{d \theta^2}=-4 \cos \theta \sin \theta-\sin \theta\left(\frac{d^2 s}{d \theta^2}\right)_{\left(x=\frac{\pi}{3}\right)}=-4 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} < 0\therefore function is maximum at x=π3x=\frac{\pi}{3}
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