In a $ riangle A B C$, angles $A, B, C$ are in $A P$. If $f(C)=\lim _{A ightarrow C} \frac{\sqrt{

Question

In a $	riangle A B C$, angles $A, B, C$ are in $A P$. If $f(C)=\lim _{A ightarrow C} \frac{\sqrt{3-4 \sin A \sin C}}{|A-C|}$, then $f^{\prime}\left(\frac{\pi}{12}ight)$ is equal to $\_\_\_\_$

Professor PC Thomas Classes · Answer

A, B, C are in AP.

$	herefore \quad 2B = A + C 	ext{ and } A + B + C = 180^\circ \quad 	herefore B = 60^\circ$

$	herefore \cos B = \frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac} \Rightarrow a^2 + c^2 = b^2 + ac \Rightarrow (a - c)^2 = b^2 - ac$

Or $|\sin A - \sin C| = \sqrt{\sin^2 B - \sin A \sin C}$

$\Rightarrow 2\cos\left(\frac{A+C}{2}ight)\left|\sin\left(\frac{A-C}{2}ight)ight| = \sqrt{\frac{3}{4} - \sin A \sin C} \Rightarrow 2\left|\sin\left(\frac{A-C}{2}ight)ight| = \sqrt{3 - 4\sin A \sin C}$

$	herefore \lim_{Aightarrow C} \frac{\sqrt{3 - 4 \sin A \sin C}}{|A - C|} = \lim_{Aightarrow C} \frac{2\left|\sin\left(\frac{A - C}{2}ight)ight|}{|A - C|}$

$\lim_{Aightarrow C} \left|\frac{\sin\left(\frac{A - C}{2}ight)}{\left(\frac{A - C}{2}ight)}ight| = |1| = 1 \Rightarrow f(x) = 1 \quad 	herefore f'(x) = 0$