Q108 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
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
- Correct answerA$(1, \infty)$
- B$(0,1)$
- C$(-\infty, 0) \cup(1, \infty)$
- D$(-\infty, 0)$
Solution
$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}$
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 $\left(1-\frac{1}{x}\right)\gt0 \Rightarrow x\gt1$ So, $f(x)$ is strictly increasing in $(1, \infty)$
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 $\left(1-\frac{1}{x}\right)\gt0 \Rightarrow x\gt1$ So, $f(x)$ is strictly increasing in $(1, \infty)$
Hence, the correct answer is option: 'A'.