MCQ
$f(x) = 2sinx + cos2x, 0 \leq x \leq 2\pi$ મહત્તમ કયાં છે ?
- A$x = \pi /2$
- B$x = 3\pi /2$
- ✓$x = \pi /6 $
- Dક્યાયં નહી
${{f'}}(x)\,\, = \,\,2\,\cos \,x\,\, - \,\,2\,\sin \,2x$
$ \Rightarrow \,\,{{f'}}(x)\,\, = \,\,0\,\,\,\,\,\,\,\, \Rightarrow \,\,2\cos \,x\,(1\,\, - \,\,2\,\sin \,x)\,\, = \,\,0\,\,\,\,\,\,\, \Rightarrow \,\,x\,\, = \,\,\frac{\pi }{6}\,,\,\,\frac{\pi }{2}$
હવે, ${{f''}}{\text{(x)}}\,\, = \,\,{\text{ - }}\,{\text{2}}\,{\text{sin}}\,{\text{x}}\,\,{\text{ - }}\,\,{\text{4cos}}\,{\text{2x}}$
$ \Rightarrow \,\,\,{{f''}}\left( {\frac{\pi }{{\text{6}}}} \right)\,\, = \,\,{\text{ - 2}}\,{\text{.}}\,\frac{{\text{1}}}{{\text{2}}}\,\,{\text{ - }}\,\,{\text{4}}\,{\text{.}}\,\frac{{\text{1}}}{{\text{2}}}\,\, = \,\,{\text{ - 3}}\,\, < \,\,{\text{0}}$
$ \Rightarrow \,\,{{f''}}\left( {\frac{\pi }{{\text{2}}}} \right)\,\,\, = \,\,{\text{ - }}\,{\text{2}}\,\,{\text{ - }}\,\,{\text{4}}\,{\text{( - 1)}}\,\, = \,\,{\text{2}}\,\, > \,\,{\text{0}}$
તેથી $\,{\text{x}}\,\, = \,\,\frac{\pi }{{\text{6}}}\,$ આગળ $\,{f}{\text{(x)}}$ મહતમ છે.
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વિધાન $2$:બિંદુ $A(1,0,7)$ અને $B(1,6,3)$ ને જોડતો રેખાખંડનેા રેખા $\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}$ દ્વિભાજક બને.