If x-x2-3 x-1-x2-4 x-3- then domain of fx is

The correct option is D ℝ { 1, 3} nbsp;f(x)=x^2 3x+1x^2+4x+3 =x^2 3x+1(x+1)(x+3) Since, f(x) is not defined at x= 1, 3. Therefore, domain =ℝ { 1, 3} ...

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Standard XII

Mathematics

Domain

If x=x2-3 x+1...

Question

If $f (x) = \frac{x^{2} - 3 x + 1}{x^{2} + 4 x + 3},$ then domain of $f (x)$ is

R - {- 1, 3}

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R

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R - {1, 3}

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R - {- 1, - 3}

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Solution

The correct option is D $R - {- 1, - 3}$
$f (x) = \frac{x^{2} - 3 x + 1}{x^{2} + 4 x + 3}$
$= \frac{x^{2} - 3 x + 1}{(x + 1) (x + 3)}$

Since, $f (x)$ is not defined at $x = - 1, - 3.$
Therefore, domain $= R - {- 1, - 3}$

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Domain

Standard XII Mathematics

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If f(x)=x2−3x+1x2+4x+3, then domain of f(x) is

The correct option is D R−{−1,−3} f(x)=x2−3x+1x2+4x+3 =x2−3x+1(x+1)(x+3) Since, f(x) is not defined at x=−1,−3. Therefore, domain =R−{−1,−3}

If $f (x) = \frac{x^{2} - 3 x + 1}{x^{2} + 4 x + 3},$ then domain of $f (x)$ is

The correct option is D $R - {- 1, - 3}$
$f (x) = \frac{x^{2} - 3 x + 1}{x^{2} + 4 x + 3}$
$= \frac{x^{2} - 3 x + 1}{(x + 1) (x + 3)}$

Since, $f (x)$ is not defined at $x = - 1, - 3.$
Therefore, domain $= R - {- 1, - 3}$