The number of integers in the domain of fx - -x-3-10-x-x-5 is

F(x)=√(x 3) √(10 x) √(x 5) nbsp; For square root to be defined, we get nbsp; x 3≥0, 10 x≥0, x 5≥0 ⇒ x≥3, 10≥ x, x≥5 ⇒ 10 ≥ x≥ 5 ∴ x=5,6,7,8,9,10 Hence, the ...

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Standard XII

Mathematics

Domain

The number of...

Question

The number of integers in the domain of $f (x) = \sqrt{x - 3} - \sqrt{10 - x} - \sqrt{x - 5}$ is

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Solution

$f (x) = \sqrt{x - 3} - \sqrt{10 - x} - \sqrt{x - 5}$
For square root to be defined, we get
$x - 3 \geq 0, 10 - x \geq 0, x - 5 \geq 0 \Rightarrow x \geq 3, 10 \geq x, x \geq 5 \Rightarrow 10 \geq x \geq 5 ∴ x = 5, 6, 7, 8, 9, 10$

Hence, the number of integers in the domain is $6.$

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The number of integers in the domain of f(x)=√x−3−√10−x−√x−5 is

f(x)=√x−3−√10−x−√x−5 For square root to be defined, we get x−3≥0, 10−x≥0, x−5≥0⇒x≥3, 10≥x, x≥5⇒10≥x≥5∴x=5,6,7,8,9,10 Hence, the number of integers in the domain is 6.

The number of integers in the domain of $f (x) = \sqrt{x - 3} - \sqrt{10 - x} - \sqrt{x - 5}$ is

$f (x) = \sqrt{x - 3} - \sqrt{10 - x} - \sqrt{x - 5}$
For square root to be defined, we get
$x - 3 \geq 0, 10 - x \geq 0, x - 5 \geq 0 \Rightarrow x \geq 3, 10 \geq x, x \geq 5 \Rightarrow 10 \geq x \geq 5 ∴ x = 5, 6, 7, 8, 9, 10$

Hence, the number of integers in the domain is $6.$