MCQ
$[0, 2\pi ]$ માં $x + sin2x $ ની એક મહત્તમ કિંમત?
- A$\frac{{2\pi }}{3}\, + \,\frac{{\sqrt 3 }}{2}$
- B$\frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}$
- C$\frac{\pi }{3} + \frac{{\sqrt 3 }}{2}$
- D$\frac{\pi }{3} - \frac{{\sqrt 3 }}{2}$
$==> f'(x) = 1 + 2 cos2x $
$==> f'(x) = -4sin2x$
હવે $ f'(x) = 0 ==> cos2x = -1/2$
$==> 2x = 2\pi /3, 4\pi /3, ….$
$==> x = \pi /3, 2\pi /3$
પરંતુ ${f}{\text{''(}}\pi {\text{/3)}}\, = \,{\text{ - 4}}\,{\text{(}}\sqrt {\text{3}} /2) < \,0$
$\therefore {\text{ x = }}\pi {\text{ /3 }}$ આગળ ${\text{ }}{f}{\text{(x)}}$ મહતમ છે અને તેની એક મહતમ કિમત
$ = \,\,\pi /3\, + \,\,\sin (2\pi /3)\,\, $
$\Rightarrow =\,\pi /3\,\, + \,\sqrt 3 /2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.