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

$\therefore \,\,\,\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\,\, = \,\,\cos x\, - \,\sqrt 3 \,\sin x$

 $\frac{{{d^2}y}}{{d{x^2}}}\, = \, - \sin x\, - \,\sqrt 3 \,\cos x$

મહતમ કે ન્યૂનતમ માટે $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\,\, = \,\,0\,\,$

$\therefore \,\,\cos x\,\, = \,\sqrt 3 \,\sin x$$\Rightarrow \,{\rm{tanx}}\,\, = \,\,\frac{{\rm{1}}}{{\sqrt {\rm{3}} }}\,\, \Rightarrow \,x\, = \,{30^ \circ }$

જ્યારે $\frac{{{{\rm{d}}^{\rm{2}}}y}}{{d{x^2}}}\,\, < \,\,{\rm{0}} $ માટે $\,{\rm{x}}\, = \,{\rm{3}}{{\rm{0}}^ \circ }$

$(3)$ સાચો જવાબ છે