A) 0
B) 1
C) \[(\alpha +\beta +\gamma ){{x}^{\alpha +\beta +\gamma -1}}\]
D) None of these
Correct Answer: A
Solution :
[a] we have, \[y=\frac{1}{1+\frac{{{x}^{\beta }}}{{{x}^{\alpha }}}+\frac{{{x}^{\gamma }}}{{{x}^{\alpha }}}}+\frac{1}{1+\frac{{{x}^{\alpha }}}{{{x}^{\beta }}}+\frac{{{x}^{\gamma }}}{{{x}^{\beta }}}}+\frac{1}{1+\frac{{{x}^{\alpha }}}{{{x}^{\gamma }}}+\frac{{{x}^{\beta }}}{{{x}^{\gamma }}}}\] \[=\frac{{{x}^{\alpha }}}{{{x}^{\alpha }}+{{x}^{\beta }}+{{x}^{\gamma }}}+\frac{{{x}^{\beta }}}{{{x}^{\alpha }}+{{x}^{\beta }}+{{x}^{\gamma }}}+\frac{{{x}^{\gamma }}}{{{x}^{\alpha }}+{{x}^{\beta }}+{{x}^{\gamma }}}\] \[=\frac{{{x}^{\alpha }}+{{x}^{\beta }}+{{x}^{\gamma }}}{{{x}^{\alpha }}+{{x}^{\beta }}+{{x}^{\gamma }}}=1\] \[\therefore \frac{dy}{dx}=0.\]
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