MCQ
$sin^px cos^qx $ નું એક મહત્તમ બિંદુ છે.
- A$x\, = \,{\tan ^{ - 1}}\,\sqrt {p/q} $
- B$x = {\tan ^{ - 1}}\sqrt {q/p} $
- C$x = 0$
- D$x = \pi /2$
$ \Rightarrow \,{\text{z}}\,\, = \,\,{\text{logy}}\,\, = \,\,{\text{plogsinx}}\,\, + \,\,{\text{qlogcosx}}\,\, $
$\Rightarrow \,\,\frac{{{\text{dz}}}}{{{\text{dx}}}}\,\, = \,\,p\cot x\,\, - \,\,q\tan x,$
$\frac{{{d^2}z}}{{d{x^2}}}\,\, = \, - p\cos e{c^2}x\, - \,\,q{\sec ^2}x$
હવે $\,\frac{{{\text{dz}}}}{{{\text{dx}}}}\,\, = \,\,0\,\, \Rightarrow {\tan ^2}x\,\, = \,\,p/q\,\,\, \Rightarrow \,\tan x\,\, = \,\,\sqrt {p\,/q} $
ઉપરાંત પછી $\frac{{{{\text{d}}^{\text{2}}}z}}{{d{x^2}}}$ ચોકકસ ઋણ છે.
તેથી, ${\text{x}}\, = \,{\tan ^{ - 1}}\,\sqrt {{\text{p/q}}} $ એ ${\text{z}}$ ના મહતમ બિંદુ છે. દા.ત. તે $\,\,{\text{y}}$ છે.
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