A) (a) 1
B) (b) \[-1\]
C) (c) \[\sqrt{2}\]
D) (d) \[-\sqrt{2}\]
Correct Answer: B
Solution :
[b] \[\therefore f(x)=\frac{\alpha .x}{x+1}\] \[x=\frac{\alpha .\frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1}=\frac{{{\alpha }^{2}}x}{\alpha x+x+1}x=\frac{{{\alpha }^{2}}x}{(\alpha +1)x+1}\] \[\Rightarrow (a+1){{x}^{2}}+(1-{{a}^{2}})x=0\] \[\{{{x}^{2}}+(1-\alpha )x\}=0\] If \[\alpha +1=0\] \[\therefore a=-1,\forall =x\] Hence, option [b]is correct.You need to login to perform this action.
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