The domain of definition of the function f x - x -1-3- x isA- - 3B- 1-3C- -1-3-D- -1- -

The correct option is C [1, 3] f(x) = √(x 1)+√(3 x) For f(x) to be defined, (x 1) ≥ 0 ⇒ x ≥ 1 …… (1) and (3 x) ≥ 0 ⇒ 3 ≤ x …… (2) From (1) and (2), x ϵ [1, 3] ...

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Standard XII

Mathematics

Domain

The domain of...

Question

The domain of definition of the function $f (x) = \sqrt{x - 1} + \sqrt{3 - x}$ is

$[1, \infty]$

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$(- \infty, 3)$

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$(1, 3)$

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$[1, 3]$

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Solution

The correct option is D
$[1, 3]$

$f (x) = \sqrt{x - 1} + \sqrt{3 - x}$
For f(x) to be defined,
$(x - 1) \geq 0$
$\Rightarrow x \geq 1 \dots \dots (1)$
and $(3 - x) \geq 0$
$\Rightarrow 3 \leq x \dots \dots (2)$
From (1) and (2),
$x ϵ [1, 3]$

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The domain of definition of the function f(x)=√x−1+√3−x is

The correct option is D [1,3] f(x)=√x−1+√3−x For f(x) to be defined, (x−1)≥0 ⇒x≥1……(1) and (3−x)≥0 ⇒3≤x……(2) From (1) and (2), xϵ[1,3]

The domain of definition of the function $f (x) = \sqrt{x - 1} + \sqrt{3 - x}$ is

The correct option is D
$[1, 3]$

$f (x) = \sqrt{x - 1} + \sqrt{3 - x}$
For f(x) to be defined,
$(x - 1) \geq 0$
$\Rightarrow x \geq 1 \dots \dots (1)$
and $(3 - x) \geq 0$
$\Rightarrow 3 \leq x \dots \dots (2)$
From (1) and (2),
$x ϵ [1, 3]$