Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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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: (Iv) F G - Mathematics

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: 

(iv) \[\frac{f}{g}\]

 
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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]
(iv) \[\frac{f}{g}: \left[ - 1, 3 \right] \to \text{ R is given by } \left( \frac{f}{g} \right)\left( x \right) = \frac{f\left( x \right)}{g\left( x \right)} = \frac{\sqrt{x + 1}}{\sqrt{9 - x^2}} = \sqrt{\frac{x + 1}{9 - x^2}}\].
 
 


 
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.4 | Q 4.4 | Page 38
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