Question

\[ \text{If } y = |\cos x - \sin x| + |\tan x - \cot x|, \text{ then } \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{3}} + \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{6}} = \]

• A. 1
• B. -1
• C. 2
• D. 0

Hint:

When differentiating absolute value functions, check the sign of the expression inside to choose the correct case. Use one-sided evaluations if needed.

Answer 1

The Correct Option is D

We are given:
\[ y = |\cos x - \sin x| + |\tan x - \cot x| \] Let:
\[ f(x) = |\cos x - \sin x|, \quad g(x) = |\tan x - \cot x| \] We need:
\[ f'(x) + g'(x) \text{ at } x = \frac{\pi}{3} \text{ and } x = \frac{\pi}{6} \] Step 1: Analyze \( f(x) = |\cos x - \sin x| \)
- At \( x = \frac{\pi}{3} \), \( \cos x = \frac{1}{2}, \sin x = \frac{\sqrt{3}}{2} \Rightarrow \cos x - \sin x = \frac{1 - \sqrt{3}}{2} < 0 \Rightarrow f(x) = -(\cos x - \sin x) = \sin x - \cos x \)
\[ \Rightarrow f'(x) = \cos x + \sin x \Rightarrow f'\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{\sqrt{3}}{2} \] - At \( x = \frac{\pi}{6} \), \( \cos x = \frac{\sqrt{3}}{2}, \sin x = \frac{1}{2} \Rightarrow \cos x - \sin x = \frac{\sqrt{3} - 1}{2} > 0 \Rightarrow f(x) = \cos x - \sin x \)
\[ \Rightarrow f'(x) = -\sin x - \cos x \Rightarrow f'\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2} \] Step 2: Analyze \( g(x) = |\tan x - \cot x| \)
- At \( x = \frac{\pi}{3} \), \( \tan x = \sqrt{3}, \cot x = \frac{1}{\sqrt{3}} \Rightarrow \tan x - \cot x > 0 \Rightarrow g(x) = \tan x - \cot x \)
\[ g'(x) = \sec^2 x + \csc^2 x \Rightarrow g'\left(\frac{\pi}{3}\right) = 4 + \frac{4}{3} = \frac{16}{3} \] - At \( x = \frac{\pi}{6} \), \( \tan x = \frac{1}{\sqrt{3}}, \cot x = \sqrt{3} \Rightarrow \tan x - \cot x < 0 \Rightarrow g(x) = -(\tan x - \cot x) = \cot x - \tan x \)
\[ g'(x) = -(\sec^2 x + \csc^2 x) \Rightarrow g'\left(\frac{\pi}{6}\right) = -\left(\frac{4}{3} + 4\right) = -\frac{16}{3} \] Step 3: Add the derivatives
\[ \left(f' + g'\right)_{x = \frac{\pi}{3}} = \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) + \frac{16}{3} \] \[ \left(f' + g'\right)_{x = \frac{\pi}{6}} = \left( -\frac{1}{2} - \frac{\sqrt{3}}{2} \right) - \frac{16}{3} \] Add both:
\[ \left(f' + g'\right)_{\frac{\pi}{3}} + \left(f' + g'\right)_{\frac{\pi}{6}} = 0 \] \[ \boxed{0} \]