If $f(x) = \begin{vmatrix}\sin x & \sin a & \sin b \\\cos x & \cos a & \cos b \\\tan x & \tan a & \tan b\end{vmatrix}$, where $0<a<b< \frac{\pi}{2}$, then the equation f'(x) = 0 has in the interval (a,b)
Answer & explanation
Correct answer: option 1
Here $f(a)=\begin{vmatrix}\sin a & \sin a & \sin b \\\cos a & \cos a & \cos b \\\tan a & \tan a & \tan b\end{vmatrix}=0$
Also f(b) = 0.
Moreover, as sinx, cosx, and tanx are continuous and differentiable in (a, b) for $0<a<b< \frac{pi}{2}$ therefore f (x) is also continuous and differentiable in [a, b].
Hence, by Rolle’s theorem, there exists a real number c in (a, b) such that f'(c) = 0.