MCQ
$f(x) = 1 + 2 sinx + 3cos^2x (0 < x < 2\pi /3) $ તો......
- ✓$x = \pi /2$ આગળ ન્યૂનત્તમ
- B$x\, = \,{\sin ^{ - 1}}\,(1/\sqrt 3 )$ આગળ મહતમ
- C$x = \pi /3 $ આગળ ન્યૂનત્તમ
- D$x = sin^{-1}(1/3) $ આગળ ન્યૂનત્તમ
$f'(x) = -2sinx + 6sin^2x - 6cos^2x $
$= -2sinx + 12sin^2x - 6$
હવે $ f'(x) =0 ==> cosx = 0 $ અને $sinx = 1/3$
અથવા $ x = \pi /2 \,\,\& \,\,x = sin^{-1} (1/3) $
તેથી $ f'(\pi /2) = -2 + 12 - 6 > 0$
${f}''\,\left( {{{\sin }^{ - 1}}\frac{1}{3}} \right)\,\, = \,\,\frac{{ - 2}}{3}\,\, + \,\,\frac{4}{3}\,\, - \,\,6\,\, < \,\,0$
$x = \pi /2$ આગળ ન્યૂનત્તમ છે.
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