A) \[(1,\infty )\]
B) \[(2,\infty )\]
C) \[(0,\infty )\]
D) \[[1,\infty )\]
Correct Answer: D
Solution :
[d] Let \[f(x)=y={{x}^{2}}+2x+2={{(x+1)}^{2}}+1\] \[y-1={{(x+1)}^{2}}\Rightarrow x+1=\sqrt{y-1}\] \[x=\sqrt{y-1}-1\] Since, \[y-1\ge 0\therefore y\ge 1\] \[\therefore \] range is \[\left[ 1,\infty \right)\].
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