Q85 / 180 · COMEDK 2015 · Mathematics

If $y=|\cos x|+|\sin x|$, then $\frac{d y}{d x}$ at $x=\frac{2 \pi}{3}$ is

  • A
    $\frac{1-\sqrt{3}}{2}$
  • B
    0
  • Correct answer
    C
    $\frac{1}{2}(\sqrt{3}-1)$
  • D
    None of these

Solution

We have, $y=|\cos x|+|\sin x|$ At $x=\frac{2 \pi}{3}, \cos x$ is negative and $\sin x$ is positive. $\therefore \quad y=-\cos x+\sin x$ $\Rightarrow \frac{d y}{d x}=\sin x+\cos x$ $\left.\therefore \frac{d y}{d x}\right|_{\text {at } x=2 \pi / 3}=\sin \left(\frac{2 \pi}{3}\right)+\cos \left(\frac{2 \pi}{3}\right)$ $$ =\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2} $$

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

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