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Mathematics Question on Continuity and differentiability

If y=∣cos⁡ x∣+∣sin⁡ x∣y = | \cos\, x | + | \sin\, x |, then dydx\frac{dy}{dx} at x=2π3x = \frac{2 \pi}{3} is

A

1−32\frac{1 - \sqrt{3}}{2}

B

00

C

12(3−1)\frac{1}{2} ( \sqrt{3} - 1)

D

None of theseNone\, of\, these

Answer

12(3−1)\frac{1}{2} ( \sqrt{3} - 1)

Explanation

Solution

We have, y=∣cos⁡ x∣+∣sin⁡ x∣y = | \cos\, x | + | \sin\, x |
At x=2π3,cos⁡ xx = \frac{2 \pi}{3} , \cos \, x is -ve and sin⁡ x\sin \, x is +ve
∴    y=−cos⁡ x+sin⁡ x\therefore \:\:\:\: y = -\cos \, x + \sin \, x
⇒  dydx=sin⁡ x+cos⁡ x\Rightarrow \:\: \frac{dy}{dx} = \sin \, x + \cos \, x
∴  dydx∣at x=2π/3=sin⁡(2π3)+cos⁡(2π3)\therefore \:\: \frac{dy}{dx} \bigg|_{at \, x = 2 \pi / 3} = \sin \bigg( \frac{2 \pi}{3} \bigg) + \cos \bigg( \frac{2 \pi}{3} \bigg)
=32−12=3−12= \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2}