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  • If <img alt="\mathrm{f(x)=\left|\begin{array}{ccc}\sin x & \sin a & \sin b \\ \cos x & \cos a & \cos b \\ \tan x & \tan a & \tan b\end{array}\right|, \quad where 0<a<b

If \mathrm{f(x)=\left|\begin{array}{ccc}\sin x & \sin a & \sin b \\ \cos x & \cos a & \cos b \\ \tan x & \tan a & \tan b\end{array}\right|, \quad where 0<a<b<\frac{\pi}{2}}, then the equation \mathrm{f^{\prime}(x)=0} has, in the interval \mathrm{(a, b)}
 

Option: 1

at least one root
 


Option: 2

at least 4 roots
 


Option: 3

no root
 


Option: 4

none of these


Answers (1)

best_answer

\mathrm{f(a)=\left|\begin{array}{ccc} \sin a & \sin a & \sin b \\ \cos a & \cos a & \cos b \\ \tan a & \tan a & \tan b \end{array}\right|=0}

Also \mathrm{f(b)=0}
Moreover, as \mathrm{\sin x, \cos x, and \: \tan x} are continuous and differentiable in \mathrm{(a, b)} for \mathrm{0<a<b<\frac{\pi}{2}, therefore\: f(x)} is also continuous and differentiable in \mathrm{[a, b].} Hence, by Rolle's theorem, there exists a real number \mathrm{c} in \mathrm{(a, b)} such that \mathrm{f^{\prime}(c)=0}

Hence option 1 is correct.

Posted by

Devendra Khairwa

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