If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{| x|-x}+\frac{1}{\sqrt{| x|-x}}\), then…
If \(f: A \rightarrow B\) is an onto function such that \(f(x)=\sqrt{| x|-x}+\frac{1}{\sqrt{| x|-x}}\), then \(A\) and \(B\) are respectively.
- \((-\infty, \infty),(0, \infty)\)
- \((-\infty, 0),[2, \infty)\)
- \((0, \infty),(2, \infty)\)
- \((-\infty, 0],(0, \infty)\)
Solution
We have,
\(f(x)=\sqrt{| x|-x}+\frac{1}{\sqrt{| x|-x}}\)
We know that, \(f(x)\) will be defined when
\(\begin{array}{lrl}
& & |x|-x > 0 \\
\Rightarrow & & |x| > x \\
\therefore & x \in(-\infty, 0)
\end{array}\)
Now,
\(\begin{aligned}
& f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}} \\
& =\sqrt{-2 x}+\frac{1}{\sqrt{-2 x}} \quad[\because x \in(-\infty, 0)] \\
& \therefore \quad f_{\min }=2 \sqrt{\sqrt{-2 x} \cdot \frac{1}{\sqrt{-2 x}}} \quad[\because A M \geq G M] \\
& =2 \\
& \therefore \quad A=(-\infty, 0) \text { and } B=[2, \infty)
\end{aligned}\)
\(\therefore \quad f(x) \in[2, \infty)\)
Asked in: AP EAMCET 2019 (22 Apr Shift 1)
Practice more Functions questions on Aicharya