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Mathematics · 2018
JEE Main 2018 · 2018-04-08 Offline · Q73
Let f(x)=x^2+1/x^2 and g (x)=x-1/x, x ∈ R -{-1,0,1}. If h (x)=(f(x))/(g (x)), then the local minimum value of h (x) is:
Let $\displaystyle f(x)=x^2+\frac{1}{x^2}$ and $\displaystyle \mathrm{g}(x)=x-\frac{1}{x}$, $\displaystyle x \in \mathbf{R}-\{-1,0,1\}$. If $\displaystyle \mathrm{h}(x)=\frac{f(x)}{\mathrm{g}(x)}$, then the local minimum value of $\displaystyle \mathrm{h}(x)$ is :
Official answer
(4)
$\displaystyle 2 \sqrt{2}$
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.