The range of f(x)=a−|x|(a+1)−|x|⋅(a>0) is

f(x)=a−|x|(a+1)−|x|⋅(a>0) y2=a-x a+1-x (a+1)y2-xy2=a-x (a+1)y2-a=xy2-1 x=(a+1)y2-ay2-1 (y2-1)y2-aa+1≥0⇒y∈0,-aa+1∪1, ∞

The range of f(x)=a−|x|(a+1)−|x|⋅(a>0) is

f(x)=a−|x|(a+1)−|x|⋅(a>0) y2=a-x a+1-x (a+1)y2-xy2=a-x (a+1)y2-a=xy2-1 x=(a+1)y2-ay2-1 (y2-1)y2-aa+1≥0⇒y∈0,-aa+1∪1, ∞

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$The range of f (x) = \sqrt{\frac{a - | x |}{(a + 1) - | x |}} \cdot (a > 0) is$

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$[0, a]$

$[0, \infty] \cup [1 - \sqrt{\frac{a}{a + 1}} \cdot \sqrt{\frac{a}{a + 1}}]$

$[0, \sqrt{\frac{a}{a + 1}}] \cup (1, \infty)$

$[0, \sqrt{\frac{a}{a + 1}} + 1]$

answer is C.

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Detailed Solution

$f (x) = \sqrt{\frac{a - | x |}{(a + 1) - | x |}} \cdot (a > 0) y^{2} = \frac{a - |x|}{a + 1 - |x|} (a + 1) y^{2} - |x| y^{2} = a - |x| (a + 1) y^{2} - a = |x| (y^{2} - 1) |x| = \sqrt{\frac{(a + 1) y^{2} - a}{y^{2} - 1}} (y^{2} - 1) (y^{2} - \frac{a}{a + 1}) \geq 0$
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$\Rightarrow y \in [0, - \sqrt{\frac{a}{a + 1}}] \cup (1, \infty)$