Question

Let
ƒ : R → R
be defined as f ( x) = x -1 and
g : R - { 1, -1 } → R
be defined as
g(x) = $\frac{x²}{x² - 1}$
Then the function fog is :

• A. One-one but not onto
• B. Onto but not one-one
• C. Both one-one and onto
• D. Neither one-one nor onto

Answer 1

Answer: (D)

Neither one-one nor onto

Answer 2

The correct answer is (D) : Neither one-one nor onto
ƒ : R → R
be defined as
f(x) = x -1 and g : R - { 1, -1 } → R
be defined as
g(x) =$\frac{ x²}{x² - 1}$
Now fog(x)
=$\frac{ x²}{x² - 1}$ - 1 = $\frac{1}{x² - 1}$
∴ Domain of fog(x) = R - { -1, 1 }
And range of fog(x) = ( - ∞ , -1 ] ∪ (0, ∞)
Now ,
$\frac{d}{dx}$_ _ $(ƒog(x))$ = $\frac{-1}{( x² - 1 )²}$ . 2x =$\frac{ 2x}{( 1 - x² )²}$
∴ $\frac{d}{dx}$_ _ $(ƒog(x))$ > 0 for $\frac{2x}{(( 1 - x )(1 + x))²}$ > 0
⇒ $\frac{x}{(( x - 1)( x + 1))²}$ < 0
∴ x ∈ ( - ∞, 0 )
and
$\frac{d}{dx} (ƒog(x))$ < 0 for x ∈ ( 0, ∞ )
∴ fog(x) is neither one-one nor onto