A) \[-2\]
B) \[-1\]
C) \[1\]
D) 2
Correct Answer: B
Solution :
Given, \[f(x)=1+\alpha x,\alpha \ne 0\] Let \[1+ax=y\] \[\Rightarrow \] \[\alpha x=y-1\] \[\Rightarrow \] \[x=\frac{y-1}{\alpha }=f(y)\] Since,\[f(x)\]is equal to its reciprocal \[\therefore \] \[\frac{y-1}{\alpha }=1+\alpha y\] \[\Rightarrow \] \[\frac{1}{\alpha }=\alpha \] and \[-\frac{1}{\alpha }=1\] \[\Rightarrow \] \[{{\alpha }^{2}}=1\] and \[\alpha =-1\] \[\Rightarrow \] \[\alpha =\pm 1\] and \[\alpha =-1\] \[\Rightarrow \] \[\alpha =-1\]
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