AP EAMCET · Maths · Application of Derivatives
The interval containing all the real values of \(x\) such that the real valued function \(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}\) is strictly increasing is
- A \((1, \infty)\)
- B \((0,1)\)
- C \((-\infty, 0) \cup(1, \infty)\)
- D \((-\infty, 0)\)
Answer & Solution
Correct Answer
(A) \((1, \infty)\)
Step-by-step Solution
Detailed explanation
\(f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}\) Clearly, \(f(x)\) is defined for \(x\gt0\) \(f^{\prime}=\frac{1}{2 \sqrt{x}}-\frac{1}{2} x^{-3 / 2}=\frac{1}{2 \sqrt{x}}\left(1-\frac{1}{x}\right)\) For \(f^{\prime}(x)\gt0\) We must have \(\left(1-\frac{1}{x}\right)\gt0 \Rightarrow x\gt1\)…
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