Let f and g be real functions- defined by f x - x -2 and g x -4- x 2- Findi f - g x ii f - g x iii fg x iv ff x v gg x vi f - g x

Clearly, f(x) = √(x+2) is defined for all x ϵ R such that x+2 ≥ 0, i.e., x ≥ 2 ∴dom (f) = [ 2, ∞) Again, g(x) = √(4 x^2) is defined for all x ϵ R such that 4 x ...

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Byju's Answer

Standard XII

Mathematics

Domain

Let f and g b...

Question

Let f and g be real functions, defined by $f (x) = \sqrt{x + 2}$ and $g (x) = \sqrt{4 - x^{2}}$ . Find

(i) $(f + g) (x)$

(ii) $(f - g) (x)$

(iii) $(f g) (x)$

(iv) $(f f) (x)$

(v) $(g g) (x)$

(vi) $(\frac{f}{g}) (x)$

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Solution

Clearly, $f (x) = \sqrt{x + 2}$ is defined for all $x ϵ R$ such that

$x + 2 \geq 0$ , i.e., $x \geq - 2$

$∴ dom (f) = [- 2, \infty)$

Again, $g (x) = \sqrt{4 - x^{2}}$ is defined for all $x ϵ R$ such that

$4 - x^{2} \geq 0$

But, $4 - x^{2} \geq 0 \Rightarrow x^{2} - 4 \leq 0 \Rightarrow (x + 2) (x - 2) \leq 0 \Rightarrow x ϵ [- 2, 2]$

$∴ dom (g) = [- 2, 2]$

$∴ dom (f) \cap dom (g) = [- 2, \infty] \cap [- 2, 2] = [- 2, 2]$

(i) $(f + g) : [- 2, 2] \to R$ is given by

$(f + g) (x) = f (x) + g (x) = \sqrt{x + 2} + \sqrt{4 - x^{2}}$

(ii) $(f - g) : [- 2, 2] \to R$ is given by

$(f - g) (x) = f (x) - g (x) = \sqrt{x + 2} - \sqrt{4 - x^{2}}$

(iii) $(f g) : [- 2, 2] \to R$ is given by

$(f g) (x) = f (x) . g (x) = (\sqrt{x + 2}) (\sqrt{4 - x^{2}})$

$= \sqrt{(x + 2)^{2} (2 - x)} = (x + 2) \sqrt{(2 - x)}$

(iv) $(f f) : [- 2, 2] \to R$ is given by

$(f f) (x) = f (x) . f (x) = (\sqrt{x + 2}) (\sqrt{x + 2}) = (x + 2)$

(v) $(g g) : [- 2, 2] \to R$ is given by

$(g g) (x) = g (x) . g (x) = (\sqrt{4 - x^{2}}) (\sqrt{4 - x^{2}}) = (4 - x^{2})$

(vi) ${x : g (x) = 0} = {x : 4 - x^{2} = 0} = {x : (2 - x) (2 + x) = 0} = {- 2, 2}$

$∴ dom (\frac{f}{g}) = dom (f) \cap dom (g) - {x : g (x) = 0}$

$= [- 2, 2] - {- 2, 2} = (- 2, 2)$

$∴ \frac{f}{g} : (- 2, 2) \to R$ is given by

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt{x + 2}}{\sqrt{4 - x^{2}}} = \frac{\sqrt{2 + x}}{(\sqrt{2 + x}) (\sqrt{2 - x})} = \frac{1}{(\sqrt{2 - x})}$

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