1. Tardigrade
  2. Question
  3. Mathematics
  4. The range of f(x) = 8 √2 sin√(π2/16) -x2 is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The range of $f(x) = 8 \sqrt{2} \sin\sqrt{\frac{\pi^{2}}{16} -x^{2}}$ is

Relations and Functions

Solution:

$ - \frac{\pi}{4} \le x \le \frac{\pi}{4 }\Rightarrow 0 \le \sqrt{\frac{\pi^{2}}{16} -x^{2 }}\le\frac{\pi}{4}$
$ \Rightarrow 8\sqrt{2}\sin \sqrt{\frac{\pi^{2}}{16 } -x^{2}}\in\left[0,8\right]$