Q229 / 270 · NDA 2022 (Phase 2) · Mathematics

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

  • A
    $1$
  • Correct answer
    B
    $\sqrt{2}$
  • C
    $\sqrt{3}$
  • D
    $2$

Solution

$\begin{aligned} & \text { Let } f(x)=3(\sin x-\cos x)+4\left(\cos ^3 x-\sin ^3 x\right) \\ & f^{\prime}(x)=3(\cos x+\sin x)-12 \sin x \cos x(\cos x+\sin x) \\ & =3(\cos x+\sin x)(1-2 \sin 2 x)=0 \\ & \begin{array}{l}\Rightarrow x=\frac{3 \pi}{4}, \frac{\pi}{6} \\ f^{\prime \prime}(x)=3(-\sin x+\cos x)(1-2 \sin 2 x) \\ -3(\cos x+\sin x)(-4 \cos 2 x) \\ f^{\prime \prime}\left(\frac{3 \pi}{4}\right) \lt 0 \text { and } f^{\prime \prime}\left(\frac{\pi}{6}\right) \gt 0\end{array} \\ & \therefore \quad f(x) \text { is maximum at } x=\frac{3 \pi}{4} \\ & f\left(\frac{3 \pi}{4}\right)=3\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)+4\left(\frac{-1}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right) \\ & =3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'B'.

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