A) odd and decreasing
B) even and decreasing
C) odd and increasing
D) even and increasing
Correct Answer: A
Solution :
Since, \[f(x)\] is decreasing odd function. \[\therefore \] \[f(-x)=-f(x)\,\,\,and\,\,\,f'(x)<0\] Now, let \[y={{f}^{-1}}(x)\] \[\therefore \] \[f(y)=x\] Then, \[f'(y)\frac{dy}{dx}=1\] \[\therefore \] \[\frac{dy}{dx}=\frac{1}{f'(y)}<0\] \[[\because f'(x)<0]\] \[\Rightarrow \] \[\frac{dy}{dx}<0\] \[\therefore \] \[{{f}^{-1}}(x)\]is decreasing function. Let \[z={{f}^{-1}}(-x)\] ??. (i) \[\therefore \] \[-x=f(z)\] \[\Rightarrow \] \[x=-f(z)\] \[=f(-z)\] \[[\because f\,\,is\,\,odd\,\,funcation]\] \[\therefore \] \[-z={{f}^{-1}}(x)\] \[\Rightarrow \] \[z={{f}^{-1}}(x)\] From Eq. (i), \[{{f}^{-1}}(x)=-{{f}^{-1}}(x)\] So,\[{{f}^{-1}}(x)\] is an odd function. Hence,\[{{f}^{-1}}(x)\]is an odd and decreasing function.
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