InterviewSolution
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InterviewSolution
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The function `f:R->[-1/2,1/2]` defined as `f(x)=x/(1+x^2)` isA. surejective but not injectiveB. neither injective nor surjectiveC. invertibleD. injective but not surjective |
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Answer» Correct Answer - A Injectivity: Let `x,y in R` be such that f(x)=f(y) `Rightarrow (x)/(1+x^(2))=(y)/(1+y^(2))` `Rightarrow x-y+xy(y-x)=0` `Rightarrow (x-y)(1-xy)=0` `Rightarrow x=y or xy=1` Putting x=2 in xy=1, we obtain y=1//2 `therefore f(2)=f((1)/(2))=(2)/(5)` So, f is not injective. Surjectivity: `"Let" y in [-1//2, 1//2]` such that `f(x)=y` `Rightarrow (x)/(1+x^(2))=y Rightarrowy+x^(2)y Rightarrow x^(2)y-x+y=0 Rightarrow x=(1+sqrt(1-4y^(2)))/(2y)` Clearly, `x in R` for all `y in [-1//2,1//2]` except y=0 We observe that f(0)-0. So, all `y in [-1//2,1//2]` has its pre-image in R. Hence, f is surjective. |
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