A) \[3{{x}^{2}}+3\]
B) \[{{x}^{2}}-\frac{1}{{{x}^{2}}}\]
C) \[1+\frac{1}{{{x}^{2}}}\]
D) \[3{{x}^{2}}+\frac{3}{{{x}^{4}}}\]
Correct Answer: A
Solution :
\[fog(x)={{x}^{3}}-\frac{1}{{{x}^{3}}}\] Writing\[{{x}^{3}}-\frac{1}{{{x}^{3}}}\]using \[{{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b),\], we have \[{{\left( x-\frac{1}{x} \right)}^{3}}={{x}^{3}}-\frac{1}{{{x}^{3}}}-3x\cdot \frac{1}{x}\left( x-\frac{1}{x} \right)\] \[\Rightarrow \] \[{{x}^{3}}-\frac{1}{{{x}^{3}}}={{\left( x-\frac{1}{x} \right)}^{3}}+3\left( x-\frac{1}{x} \right)\] We have, \[f(g(x))={{x}^{3}}-\frac{1}{{{x}^{3}}}={{\left( x-\frac{1}{x} \right)}^{3}}+3\left( x-\frac{1}{x} \right)\] As\[g(x)=x-\frac{1}{x}\], this yields \[f\left( x-\frac{1}{x} \right)={{\left( x-\frac{1}{x} \right)}^{3}}+3\left( x-\frac{1}{x} \right)\] On putting\[x-\frac{1}{x}=t\], we get \[f(t)={{t}^{3}}+3t\] Thus, \[f(x)={{t}^{3}}+3x,\] and \[f'(x)=3{{x}^{2}}+3\]
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