Question:
If $y = | \cos\, x | + | \sin\, x |$, then $\frac{dy}{dx}$ at $x = \frac{2 \pi}{3}$ is
If $y = | \cos\, x | + | \sin\, x |$, then $\frac{dy}{dx}$ at $x = \frac{2 \pi}{3}$ is
Updated On: Jun 21, 2022
- $\frac{1 - \sqrt{3}}{2}$
- $0$
- $\frac{1}{2} ( \sqrt{3} - 1)$
- $None\, of\, these$
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The Correct Option is C
Solution and Explanation
We have, $y = | \cos\, x | + | \sin\, x |$
At $x = \frac{2 \pi}{3} , \cos \, x$ is -ve and $\sin \, x $ is +ve
$\therefore \:\:\:\: y = -\cos \, x + \sin \, x$
$\Rightarrow \:\: \frac{dy}{dx} = \sin \, x + \cos \, x$
$\therefore \:\: \frac{dy}{dx} \bigg|_{at \, x = 2 \pi / 3} = \sin \bigg( \frac{2 \pi}{3} \bigg) + \cos \bigg( \frac{2 \pi}{3} \bigg) $
$ = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} $
At $x = \frac{2 \pi}{3} , \cos \, x$ is -ve and $\sin \, x $ is +ve
$\therefore \:\:\:\: y = -\cos \, x + \sin \, x$
$\Rightarrow \:\: \frac{dy}{dx} = \sin \, x + \cos \, x$
$\therefore \:\: \frac{dy}{dx} \bigg|_{at \, x = 2 \pi / 3} = \sin \bigg( \frac{2 \pi}{3} \bigg) + \cos \bigg( \frac{2 \pi}{3} \bigg) $
$ = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} $
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.