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# Find the Domain and Range of the Real Valued Function: (Viii) F ( X ) = √ 9 − X 2 - CBSE (Science) Class 11 - Mathematics

ConceptCartesian Product of Sets

#### Question

Find the domain and range of the real valued function:

(viii)  $f\left( x \right) = \sqrt{9 - x^2}$

#### Solution

Given:

$f\left( x \right) = \sqrt{9 - x^2}$
$(9 - x^2 ) \geq 0$
$\Rightarrow 9 \geq x^2$
$\Rightarrow x \in \left[ - 3, 3 \right]$
$\sqrt{9 - x^2}$ is defined for all real numbers that are greater than or equal to – 3 and less than or equal to 3.
Thus, domain of f (x) is {x : – 3 ≤ x ≤ 3} or [– 3, 3].
For any value of x such that – 3 ≤ x ≤ 3, the value of f (x) will lie between 0 and 3.
Hence, the range of f (x) is {x: 0 ≤ x ≤ 3} or [0, 3].

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#### APPEARS IN

RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 3: Functions
Ex.3.30 | Q: 3.08 | Page no. 18
Solution Find the Domain and Range of the Real Valued Function: (Viii) F ( X ) = √ 9 − X 2 Concept: Cartesian Product of Sets.
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