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प्रश्न
At x= \[\frac{5\pi}{6}\] f(x) = 2 sin 3x + 3 cos 3x is ______________ .
विकल्प
0
maximum
minimum
none of these
उत्तर
none of these
\[\text { Given }: f\left( x \right) = 2 \sin 3x + 3 \cos 3x\]
\[ \Rightarrow f'\left( x \right) = 6 \cos 3x - 9 \sin 3x\]
\[\text { For a local minima or a local maxima, we must have }\]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 6 \cos 3x - 9 \sin 3x = 0\]
\[ \Rightarrow 6 \cos 3x = 9 \sin 3x\]
\[ \Rightarrow \frac{\sin 3x}{\cos 3x} = \frac{2}{3}\]
\[ \Rightarrow \tan 3x = \frac{2}{3} . . . \left( 1 \right)\]
\[\text { At x } = \frac{5\pi}{6}: \]
\[\tan 3x = \tan \frac{5\pi}{2}\]
\[ \Rightarrow \tan 3x = \tan \frac{\pi}{2}\]
\[\text { So,} \tan 3x \text { is not defined }. \left[ \tan 3x \neq \frac{2}{3} \text { is not satisfying eq } . \left( 1 \right) \right]\]
\[\text { Thus, }x = \frac{5\pi}{6}\text { is not a critical point } .\]
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