A) \[g\,(x)={{\left( \frac{b-{{x}^{1/3}}}{a} \right)}^{1/2}}\]
B) \[g\,(x)=\frac{1}{{{(a{{x}^{2}}+b)}^{3}}}\]
C) \[g\,(x)={{(a{{x}^{2}}+b)}^{1/3}}\]
D) \[g\,(x)={{\left( \frac{{{x}^{1/3}}-b}{a} \right)}^{1/2}}\]
Correct Answer: D
Solution :
\[f\,(x)={{(a{{x}^{2}}+b)}^{3}}=y\]\[\Rightarrow \]\[x=\pm {{\left( \frac{{{y}^{\frac{1}{3}}}-b}{a} \right)}^{\frac{1}{2}}}\] \[\therefore \] \[g\,(x)={{\left( \frac{{{x}^{1/3}}-b}{a} \right)}^{1/2}}\]
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