6. Application of derivatives — ગણિત ધોરણ 12 સાયન્સ — Question

$= 3^{3cos2x }3^{4sin2x} = 3^{3cos2x + 4\,sin 2x}$

જ્યારે $3cos2x + 4sin2x $ ન્યૂનત્તમ હશે

ત્યારે $y $ ન્યૂનત્તમ હશે.

$z = 3cos2x + 4sin2x $ તો $3 = rcos\alpha ; 4 = rsin\alpha $ મૂકો.

$ r^2 = 3^2 + 4^2 = 25 r = 5$

${\text{tan}}\,\alpha \, = \,\,\frac{4}{3}\,\, $ $\Rightarrow \,\alpha \,\, = \,\,{\tan ^{ - 1}}\,\,\frac{4}{3}\,\,\,\,$

$\therefore \,z\,\, = \,\,r\sin \,(2x\, + \,\alpha )$

$ = \,\,5\sin \,\left( {2x\, + \,{{\tan }^{ - 1}}\frac{4}{3}} \right)\,\,\, $

$\Rightarrow \, - 5\,\, \leqslant \,\,z\,\, \leqslant \,\,5$

ન્યૂનતમ $\,\, = \,\,{\text{ - 5}}\,\, \Rightarrow $ ન્યૂનતમ ${\text{y}}\,\, = \,\,{{\text{3}}^{{\text{ - 5}}}}\, = \,\,\frac{1}{{243}}$