AP EAMCET · Maths · Application of Derivatives
Let \(f(x)=x^2+\frac{1}{x^2}\) and \(g(x)=x-\frac{1}{x}\) for \(x \in R-\{-1,0,+1\}\), then the local minimum of \(\frac{f(x)}{g(x)}\) is
- A \(-3\)
- B \(2 \sqrt{2}\)
- C \(-2 \sqrt{2}\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(f(x)=x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)+2, g(x)=x-\frac{1}{x}\) Let \(x-\frac{1}{x}=t\) \(f(x)=t^2+2, g(x)=t\) \(\begin{aligned} & \frac{f(x)}{g(x)}=t+\frac{2}{t}=h(t) \\ & h^{\prime}(t)=1-\frac{2}{t^2}\end{aligned}\) On putting \(h^{\prime}(t)=0\)…
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