A) \[f(x)={{x}^{2}}+2,x\in (-\infty ,\infty )\]
B) \[f(x)=\left| x+2 \right|,x\in [-2,\infty )\]
C) \[f(x)=(x-4)(x-5),x\in (-\infty ,\infty )\]
D) \[f(x)=\frac{4{{x}^{2}}+3x-5}{4+3x-5{{x}^{2}}},\,\,x\in (-\infty ,\infty )\]
Correct Answer: B
Solution :
[b] The function \[f(x)={{x}^{2}}+2,x\in (-\infty ,\infty )\] is not injective as \[f(1)=f(-1)\]but \[1\ne -1.\] The function \[f(x)=(x-4)(x-5),x\in (-\infty ,\infty )\] is not one-one as \[f(4)=f(5),\] but \[4\ne 5.\] The function, \[f(x)=\frac{4{{x}^{2}}+3x-5}{4+3x-5{{x}^{2}}}x\in (-\infty ,\infty )\] is also not injective as \[f(1)=f(-1),but1\ne -1.\] For the function, \[f(x)=\left| x+2 \right|,x\in [-2,\infty ).\] Let \[f(x)=f(y),x,y\in [-2,\infty )\Rightarrow \left| x+2 \right|\] \[=\left| y+2 \right|\] \[\Rightarrow x+2=y+2\Rightarrow x=y\] So, f is an injection.
You need to login to perform this action.
You will be redirected in
3 sec
