A) \[a=4,\,\,b=2\]
B) \[a=4,\,\,b=-2\]
C) \[a=-2,\,\,b=4\]
D) None of these
Correct Answer: B
Solution :
\[\frac{d}{dx}\left( \frac{1+{{x}^{4}}+{{x}^{8}}}{1+{{x}^{2}}+{{x}^{4}}} \right)=\frac{d}{dx}(1-{{x}^{2}}+{{x}^{4}})\] \[\left( \because \,\,\frac{1+{{x}^{4}}+{{x}^{8}}}{1+{{x}^{2}}+{{x}^{4}}}=1-{{x}^{2}}+{{x}^{4}} \right)\] \[=0-2x+4{{x}^{3}}\] ... (i) but given that \[\frac{d}{dx}\left( \frac{1+{{x}^{4}}+{{x}^{8}}}{1+{{x}^{2}}+{{x}^{4}}} \right)=a{{x}^{3}}+bx\] ? (ii) On comparing Eqs. (i) and (ii), we get \[a=4,\,\,b=-2\]
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