Question:
The domain of definition of $f(x)=\sqrt{4 x-x^{2}}$ is
(a) R − [0, 4]
(b) R − (0, 4)
(c) (0, 4)
(d) [0, 4]
Solution:
(d) [0, 4]
Given:
$f(x)=\sqrt{4 x-x^{2}}$
Clearly, f (x) assumes real values if
$4 x-x^{2} \geq 0$
$\Rightarrow x(4-x) \geq 0$
$\Rightarrow-x(x-4) \geq 0$
$\Rightarrow x(x-4) \leq 0$
$\Rightarrow x \in[0,4]$
Hence, domain (f )= [0, 4].
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