A) \[[0,3]\]
B) \[[0,\sqrt{3}]\]
C) \[(-\infty ,\infty )\]
D) None of these
Correct Answer: B
Solution :
Given, \[f(x)=\tan \sqrt{\frac{{{\pi }^{2}}}{9}-{{x}^{2}}}\] For\[f(x)\]to be defined\[\frac{{{\pi }^{2}}}{9}-{{x}^{2}}\ge 0\] \[\Rightarrow \] \[-\frac{\pi }{3}\le x\le \frac{\pi }{3}\] Also, \[f'(x)=\left( {{\sec }^{2}}\sqrt{\frac{{{\pi }^{2}}}{9}-{{x}^{2}}} \right).\frac{1}{\sqrt[2]{\frac{{{\pi }^{2}}}{9}-{{x}^{2}}}}.(-2x)\] \[\Rightarrow \] \[f(x)\]is decreasing for\[0\le x\le \pi /3\]. \[\therefore \] \[f\left( \frac{\pi }{3} \right)\le f(x)\le f(0)\] \[\Rightarrow \] \[0\le f(x)\le \sqrt{3}\] \[\therefore \] \[f(x)\in [0,\sqrt{3}]\]
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