A) [0,5]
B) [0,2)
C) (0,2)
D) None of these
Correct Answer: D
Solution :
Clearly, f(x) is defined, if \[25-{{x}^{2}}>0\]\[\Rightarrow \]\[-5<x<5\] Let\[y={{\log }_{5}}(25-{{x}^{2}}),\]then \[{{5}^{y}}=25-{{x}^{2}}\]\[\Rightarrow \]\[{{x}^{2}}-25-{{5}^{y}}\] \[\Rightarrow \]\[x=\pm \sqrt{25-{{5}^{y}}}\] For x to be real, we must have \[25-{{5}^{y}}\ge 0\] \[\Rightarrow \]\[{{5}^{y}}\le 25\]\[\Rightarrow \]\[y\le 2\] Also, \[y=f(x)\to -\infty \]as\[x\to \pm 5.\] Hence, range \[(f)=(-\infty ,2]\]
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