A) \[i+{{(g(x))}^{n}}\]
B) \[1-g(x)\]
C) \[1+g(x)\]
D) \[1-{{(g(x))}^{n}}\]
E) \[{{(g(x))}^{n}}\]
Correct Answer: A
Solution :
\[\because \]\[y=f(x),\]then\[x={{f}^{-1}}(y)\] i,e., \[x=g(y)\] On differentiating, w. r. t.\[x,\]we get \[\Rightarrow \] \[1=g(y)\frac{dy}{dx}=g(y)\frac{1}{1+{{x}^{n}}}\] \[\Rightarrow \] \[g(y)=1+{{x}^{n}}=1+{{[g(y)]}^{n}}\] \[\Rightarrow \] \[g(x)=1+{{[g(x)]}^{n}}\]
You need to login to perform this action.
You will be redirected in
3 sec
