A) \[{{g}^{-1}}(x)\]
B) \[x{{f}^{-1}}(x)-g({{f}^{-1}}(x))\]
C) \[x{{f}^{-1}}(x)-{{g}^{-1}}(x)\]
D) \[{{f}^{-1}}(x)\]
Correct Answer: B
Solution :
\[\int{f(x)dx}=g(x)\] \[\int{{{f}^{-1}}(x)}.1dx={{f}^{-1}}(x)\int{dx}-\int{\left\{ \frac{d}{dx}{{f}^{-1}}(x)\int{dx} \right\}dx}\] \[=x{{f}^{-1}}(x)-\int{x\frac{d}{dx}{{f}^{-1}}(x)dx}\] \[=x{{f}^{-1}}(x)-\int{xd\{{{f}^{-1}}(x)\}}\] Let \[{{f}^{-1}}(x)=t\] Þ \[x=f(t)\] and \[d\{{{f}^{-1}}(x)\}=dt\] \[=x{{f}^{-1}}(x)-\int{f(t)dt}=x{{f}^{-1}}(x)-g(t)=x{{f}^{-1}}(x)-g\{{{f}^{-1}}(x)\}\]. Trick : Put \[f(x)={{x}^{2}}\], then option is correct.
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