The domain of fx-x-3-log 39-x is

The correct option is D [ 3,9) {8}f(x)=√(x+3)log_3(9 x) nbsp; f(x) is defined when nbsp; x+3≥0, 9 x gt;0, log_3(9 x) 0 ⇒ x ≥ 3, x 9, 9 x 1 ⇒ 9 x ≥ 3, ...

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Standard XII

Mathematics

Domain

The domain of...

Question

The domain of $f (x) = \frac{\sqrt{x + 3}}{{log}_{3} (9 - x)}$ is

R - {8}

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[- 3, 9)

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[- 3, \infty) - {8}

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[- 3, 9) - {8}

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Solution

The correct option is D $[- 3, 9) - {8}$
$f (x) = \frac{\sqrt{x + 3}}{{log}_{3} (9 - x)}$
$f (x)$ is defined when
$x + 3 \geq 0, 9 - x > 0, {log}_{3} (9 - x) \neq 0 \Rightarrow x \geq - 3, x < 9, 9 - x \neq 1 \Rightarrow 9 > x \geq - 3, x \neq 8 ∴ x \in [- 3, 9) - {8}$

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The domain of f(x)=√x+3log3(9−x) is

The correct option is D [−3,9)−{8}f(x)=√x+3log3(9−x) f(x) is defined when x+3≥0, 9−x>0, log3(9−x)≠0⇒x≥−3, x<9, 9−x≠1⇒9>x≥−3,x≠8∴x∈[−3,9)−{8}

The domain of $f (x) = \frac{\sqrt{x + 3}}{{log}_{3} (9 - x)}$ is

The correct option is D $[- 3, 9) - {8}$
$f (x) = \frac{\sqrt{x + 3}}{{log}_{3} (9 - x)}$
$f (x)$ is defined when
$x + 3 \geq 0, 9 - x > 0, {log}_{3} (9 - x) \neq 0 \Rightarrow x \geq - 3, x < 9, 9 - x \neq 1 \Rightarrow 9 > x \geq - 3, x \neq 8 ∴ x \in [- 3, 9) - {8}$