A) [? 1, 1]
B) [0, 1]
C) [?1, 0]
D) [?1,2]
Correct Answer: A
Solution :
Let \[f(x)=y=x+\frac{1}{x}\] Differentiating with respect to x, we get \[\frac{dy}{dx}=f'(x)=1-\frac{1}{{{x}^{2}}}\le 0\]\[\,\Rightarrow 1\le \frac{1}{{{x}^{2}}}\] or \[{{x}^{2}}\le 1\] Hence \[x\in [-1,\,1]\].
You need to login to perform this action.
You will be redirected in
3 sec
