Test Index

Sets, Relations and Functions

© examsnet.com

Question : 2 of 65

Marks: +1, -0

If

y=f(x)=\frac{a x+b}{a x-a}

f(y)

$-x$
$2 x$
$x$
$x^{2}$

Solution:

Concept:

Substitute

y = f(x)

into

f

and simplify.

Explanation:

Given

y = f(x) = \frac{ax+b}{ax-a}

.

Compute

f(y) = \frac{a y + b}{a y - a}

.

Substitute

y

:

f(y) = \frac{a \cdot \frac{ax+b}{ax-a} + b}{a \cdot \frac{ax+b}{ax-a} - a}

.

Combine numerator and denominator over common denominator

(ax-a)

:

Numerator:

\frac{a(ax+b) + b(ax-a)}{ax-a} = \frac{a^2x+ab + abx - ab}{ax-a} = \frac{a^2x + abx}{ax-a}

.

Denominator:

\frac{a(ax+b) - a(ax-a)}{ax-a} = \frac{a^2x+ab - a^2x + a^2}{ax-a} = \frac{ab + a^2}{ax-a}

.

Thus

f(y) = \frac{a^2x + abx}{ab + a^2} = \frac{ax(a+b)}{a(a+b)} = x

, provided

a \neq 0

and

a+b \neq 0

.

Answer:

x

© examsnet.com

Go to Question:

Prev Question

Next Question