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

The Correct Option is D

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