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Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Fill in the blanks question  15

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Answer: f^{1}\left(\frac{\pi}{6}\right)=\frac{1+\sqrt{3}}{2}

Hint:

Given: f(x)=|\sin x-\cos x|

Solution:  f(x)=|\sin x-\cos x|

\begin{aligned} &f(x)=\left(\begin{array}{cc} \sin x-\cos x & 0 \leq x \leq \frac{\pi}{4} \\\\ -(\sin x-\cos x) & \frac{\pi}{4}<x \leq \frac{\pi}{4} \end{array}\right) \\\\ &f^{{}'}(x)=\left(\begin{array}{cc} \sin x+\cos x & 0 \leq x \leq \frac{\pi}{4} \\\\ -(\sin x+\cos x) & \frac{\pi}{4}<x \leq \frac{\pi}{4} \end{array}\right) \end{aligned}

\begin{aligned} f^{{}'}\left(\frac{\pi}{6}\right)=& \sin \frac{\pi}{6}+\cos \frac{\pi}{6} \\\\ &=\frac{1}{2}+\frac{\sqrt{3}}{2} \\\\ \end{aligned}

                    =\frac{\sqrt{3}+1}{2}

 

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