The domain of 1-log 10 1-x - x-2 is

The correct option is C ( 2≤ x f(x) = 1/log_10( 1 x )+√(( x+2 )) #160;For this function to be defined, (1 x) gt; 0,≠1 and (x+2) ≥0 ⇒x ...

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Standard XII

Mathematics

Definition of Functions

The domain of...

Question

The domain of $\frac{1}{{log}_{10} (1 - x)} + \sqrt{(x + 2)}$ is

(- 2 \leq x < - 1) \cup (0 < x < 1)

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(- 2 \leq x < 2)

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(- 2 \leq x < 1)

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(- 2 \leq x < 0) \cup (0 < x < 1)

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Solution

The correct option is C $(- 2 \leq x < 0) \cup (0 < x < 1)$
$f (x) = \frac{1}{{log}_{10} (1 - x)} + \sqrt{(x + 2)}$

For this function to be defined, $(1 - x) > 0, \neq 1$ and $(x + 2) \geq 0$

$\Rightarrow x < 1, \neq 0$ and $x \geq - 2$

Thus domain of $f (x)$ is $x \in [- 2, 0) \cup (0, 1)$

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Similar questions

Find the domain of $f (x) = \sqrt{x + 2} + \frac{1}{{log}_{10} (1 - x)}$

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Q. Let

f (x) = - 1 + | x - 1 |, 1 \leq x \leq 3

and

g (x) = 2 - | x + 1 |, - 2 \leq x \leq 2

, then (fog)(x) is equal to

Q. The domain of the function

f (x) = \frac{1}{{log}_{10} (1 - x)} + \sqrt{x + 2}

Q. The domain of the function

f (x) = \frac{1}{{log}_{10} (1 - x)} + \sqrt{x + 2}

, is

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The domain of 1log10(1−x)+√(x+2) is

The correct option is C (−2≤x<0)∪(0<x<1)f(x)=1log10(1−x)+√(x+2) For this function to be defined, (1−x)>0,≠1 and (x+2)≥0⇒x<1,≠0 and x≥−2Thus domain of f(x) is x∈[−2,0)∪(0,1)

The domain of $\frac{1}{{log}_{10} (1 - x)} + \sqrt{(x + 2)}$ is

The correct option is C $(- 2 \leq x < 0) \cup (0 < x < 1)$
$f (x) = \frac{1}{{log}_{10} (1 - x)} + \sqrt{(x + 2)}$

For this function to be defined, $(1 - x) > 0, \neq 1$ and $(x + 2) \geq 0$

$\Rightarrow x < 1, \neq 0$ and $x \geq - 2$

Thus domain of $f (x)$ is $x \in [- 2, 0) \cup (0, 1)$