A) \[\frac{-1}{3}\]
B) \[\frac{-1}{2}\]
C) \[3\]
D) \[6\]
Correct Answer: A
Solution :
[a] Let \[D=\frac{d}{dx}\left( \frac{g(x)}{g\left( g(x) \right)} \right)\] at \[x=4\] \[={{\left. \frac{g\left( g(x) \right)g'(x)-g(x).g'\left( g(x) \right).g'(x)}{{{\left( g\left( g(x) \right) \right)}^{2}}} \right]}_{x=4}}\] Now, \[f(x)={{x}^{3}}+3x+4\] \[\Rightarrow \,\,\,f'(x)=3{{x}^{2}}+3>0\] Clearly, \[f(x)\] is an increasing function. Now, \[f(0)=4\] \[\Rightarrow \,\,\,\,{{f}^{-1}}(4)=g(4)=0\] Also, \[g\left( f(x) \right)=x\] \[\therefore \,\,\,g'(f(x))f'(x)=1\] \[\Rightarrow \,\,g'(f(0))f'(0)=1\] \[\Rightarrow \,\,g'(4)=\frac{1}{f'(0)}=\frac{1}{3}\] \[f(-1)=0\Rightarrow {{f}^{-1}}(0)=g(0)=-1\] \[\therefore \,\,\,D=\frac{g\left( g(4) \right)g'(4)-g(4)g'\left( g(4) \right)g'(4)}{{{\left( g\left( g(4) \right) \right)}^{2}}}\] \[=\frac{g(0).\frac{1}{f'(0)}-0}{{{\left( g(0) \right)}^{2}}}=\frac{\left( -1 \right)\times \frac{1}{3}}{1}=\frac{-1}{3}\]
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