Let f -2-3- - R and g -0-5- - R be such that f x - x 3- x - g x - x 3-2 x 2- Then the domain of f x - g x isA- 0-3-2B- 0-3-C- -2-3-D- -2-5-

The correct option is A (0,3] {2}g (x) is in the denominator. So, x^3 2x^20 ⇒ x^2 (x 2)≠ 0 ⇒ x≠ 0,2 D(f)=[ 2,3] and D(g)=[0,5] ∴f(x)g(x) exists if x∈ D(f)∩ D( ...

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Standard XII

Mathematics

Domain

Let f :[-2,3]...

Question

Let $f : [- 2, 3] \to R$ and $g : [0, 5] \to R$ be such that $f (x) = x^{3} - x, g (x) = x^{3} - 2 x^{2} .$ Then the domain of $\frac{f (x)}{g (x)}$ is

(0, 3]

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(0, 3] - {2}

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[- 2, 3]

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[- 2, 5]

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Solution

The correct option is B $(0, 3] - {2}$
$g (x)$ is in the denominator.
So,
$x^{3} - 2 x^{2} \neq 0$
$\Rightarrow x^{2} (x - 2) \neq 0$
$\Rightarrow x \neq 0, 2$

$D (f) = [- 2, 3]$ and $D (g) = [0, 5]$
$∴ \frac{f (x)}{g (x)}$ exists if
$x \in D (f) \cap D (g) - {0, 2}$
$\Rightarrow x \in [- 2, 3] \cap [0, 5] - {0, 2}$
$∴ x \in (0, 3] - {2}$

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Domain

Standard XII Mathematics

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Let f:[−2,3]→R and g:[0,5]→R be such that f(x)=x3−x, g(x)=x3−2x2. Then the domain of f(x)g(x) is

The correct option is B (0,3]−{2}g(x) is in the denominator. So, x3−2x2≠0 ⇒x2(x−2)≠0 ⇒x≠0,2 D(f)=[−2,3] and D(g)=[0,5] ∴f(x)g(x) exists if x∈D(f)∩D(g)−{0,2} ⇒x∈[−2,3]∩[0,5]−{0,2} ∴x∈(0,3]−{2}

Let $f : [- 2, 3] \to R$ and $g : [0, 5] \to R$ be such that $f (x) = x^{3} - x, g (x) = x^{3} - 2 x^{2} .$ Then the domain of $\frac{f (x)}{g (x)}$ is