The domain of the function $y (x)$ given by $2^{x} + 2^{r} = 2$ for all $r \in (- \infty, 1)$ is:

(0, - 1]

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[0, 1]

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(- \infty, 0)

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(- \infty, 1)

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Solution

The correct option is C $(- \infty, 1)$
For domain of $x$ let's assume $r$ as a function of $x :$

Now,
equation
$2^{x} + 2^{r} = 2$

$=> r = l o g_{2} (2 - 2^{x})$

So,

$=> 2 - 2^{x} > 0$

$=> 2^{x} < 2$

$=> x < 1$

Therefore,solution is set form we have
$=> x ϵ (- \infty, 1)$

Foe domain of $r$ let's assume $x$ as a function of $r$

$=> x = l o g_{2} (2 - 2^{r})$ -------------------- $[2^{x} = l o g_{2} x]$

So,

$=> 2 - 2^{y} > 0$

$=> 2^{y} < 2$

$=> y < 1$

Therefore solution is set form we have $r ϵ (- \infty, 1)$

Hence,

$x ϵ (- \infty, 1)$ for $r ϵ (- \infty, 1)$

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