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Let F and G Be Two Real Functions Defined by F ( X ) = √ X + 1 and G ( X ) = √ 9 − X 2 . Then, Describe Function: (Viii) 5 8 - CBSE (Science) Class 11 - Mathematics

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Question

Let f and g be two real functions defined by \[f\left( x \right) = \sqrt{x + 1}\] and \[g\left( x \right) = \sqrt{9 - x^2}\] . Then, describe function: 

(viii) \[\frac{5}{8}\]

 

Solution

Given:

\[f\left( x \right) = \sqrt{x + 1}\text{ and } g\left( x \right) = \sqrt{9 - x^2}\]

Clearly,

\[f\left( x \right) = \sqrt{x + 1}\]  is defined for all x ≥ - 1.
Thus, domain (f) = [1, ∞]
Again,
 
\[g\left( x \right) = \sqrt{9 - x^2}\]   is defined for  9 -x2 ≥ 0 ⇒ x2 - 9 ≤ 0
⇒ x2 - 32 ≤ 0
⇒ (x + 3)(x - 3) ≤ 0
\[x \in \left[ - 3, 3 \right]\]
Thus, domain (g) = [ - 3, 3]
Now,
domain ( f ) ∩ domain( g ) = [ -1, ∞] ∩ [- 3, 3]    = [ -1, 3]
(viii)  \[\frac{5}{g}: \left[ - 3, 3 \right] \to \text{ R is defined by } \left( \frac{5}{g} \right)\left( x \right) = \frac{5}{\sqrt{9 - x^2}} .\]            {Since domain(g) = [ -  3, 3]}                                                          
 
 


 
 
 


 
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APPEARS IN

 RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 3: Functions
Ex.3.40 | Q: 4.8 | Page no. 38
Solution Let F and G Be Two Real Functions Defined by F ( X ) = √ X + 1 and G ( X ) = √ 9 − X 2 . Then, Describe Function: (Viii) 5 8 Concept: Functions.
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