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Find the Domain and Range of the Real Valued Function: (I) F ( X ) = a X + B B X − a - Mathematics

Find the domain and range of the real valued function:

(i) \[f\left( x \right) = \frac{ax + b}{bx - a}\]

 

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Solution

(i)
Given: 

\[f\left( x \right) = \frac{ax + b}{bx - a}\]

Domain of f : Clearly,  (x) is a rational function of x as

\[\frac{ax + b}{bx - a}\] is a rational expression.

Clearly, f (x) assumes real values for all x except for all those values of x for which ( bx-a) = 0, i.e. bx = a. 

\[\Rightarrow x = \frac{a}{b}\] 
Hence, domain ( f ) =\[R - \left\{ \frac{a}{b} \right\}\]
Range of f :
Let f (x) = y ⇒ (ax + b) = y (bx -a)
⇒ (ax + b) = (bxy -ay)
⇒ b + ay = bxy -ax
⇒ b + ay = x(by - a) 
 
\[\Rightarrow x = \frac{b + ay}{by - a}\] 
Clearly, f (x) assumes real values for all x except for all those values of x for which ( by - a) = 0, i.e. by = a.
\[\Rightarrow y = \frac{a}{b}\] 
Hence, range ( f ) =\[R - \left\{ \frac{a}{b} \right\}\] 
 
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.3 | Q 3.01 | Page 18
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