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NDA 2022
Exam
NDA
Year
2022
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2022 · Mathematics · Trigonometric Ratios & Identities

What is the maximum value of 3( x- x)+4( ^3 x..- ^3 x) ?

What is the maximum value of 3(sin⁡x−cos⁡x)+4(cos⁡3x−sin⁡3x)?3(\sin x-\cos x)+4\left(\cos ^3 x\right.\left.-\sin ^3 x\right) ?

Options

  1. A11
  2. B2\sqrt{2}
  3. C3\sqrt{3}
  4. D22

Correct answer

B. 2\sqrt{2}

Step-by-step solution

 Let f(x)=3(sin⁡x−cos⁡x)+4(cos⁡3x−sin⁡3x)f′(x)=3(cos⁡x+sin⁡x)−12sin⁡xcos⁡x(cos⁡x+sin⁡x)=3(cos⁡x+sin⁡x)(1−2sin⁡2x)=0⇒x=3π4,π6f′′(x)=3(−sin⁡x+cos⁡x)(1−2sin⁡2x)−3(cos⁡x+sin⁡x)(−4cos⁡2x)f′′(3π4)<0 and f′′(π6)>0∴f(x) is  maximum  at  x=3π4f(3π4)=3(12+12)+4(−122−122)=32−22=2\begin{aligned} & \text{ Let } f(x)=3(\sin x-\cos x)+4\left(\cos ^3 x-\sin ^3 x\right) \\ & f^{\prime}(x)=3(\cos x+\sin x)-12 \sin x \cos x(\cos x+\sin x) \\ & =3(\cos x+\sin x)(1-2 \sin 2 x)=0 \\ & \begin{array}{l}\Rightarrow x=\frac{3 \pi}{4}, \frac{\pi}{6} \\ f^{\prime \prime}(x)=3(-\sin x+\cos x)(1-2 \sin 2 x) \\ -3(\cos x+\sin x)(-4 \cos 2 x) \\ f^{\prime \prime}\left(\frac{3 \pi}{4}\right) < 0 \text{ and } f^{\prime \prime}\left(\frac{\pi}{6}\right) > 0\end{array} \\ & \therefore \quad f(x) \text{\text{ is }\text{ maximum }\text{ at } } x=\frac{3 \pi}{4} \\ & f\left(\frac{3 \pi}{4}\right)=3\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)+4\left(\frac{-1}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right) \\ & =3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}\end{aligned}
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