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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: (Iii) F G - Mathematics

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प्रश्न

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: 

(iii) f g

उत्तर

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


    
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अध्याय 3: Functions - Exercise 3.4 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 3 Functions
Exercise 3.4 | Q 4.3 | पृष्ठ ३८

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