Question

Define the functions \( f, g \) and \( h \) from \( \mathbb{R} \) to \( \mathbb{R} \) such that: \[ f(x) = x^2 - 1, \quad g(x) = \sqrt{x^2 + 1} \]

Consider the following statements:

• A. \( f \) is invertible.
• B. \( h \) is an identity function.
• C. \( f \circ g \) is not invertible.
• D. \( h \circ f \circ g = x^2 \). Then which one of the following is true?

Hint:

To check invertibility, verify whether a function is one-to-one (injective). Quadratic functions are not invertible over their entire domain unless restricted to a suitable interval.

Answer 1

The Correct Option is C

Step 1: Checking the invertibility of \( f \) 
The function \( f(x) = x^2 - 1 \) is a quadratic function. A function is invertible if it is one-to-one (injective). However, since quadratic functions are not one-to-one over \( \mathbb{R} \), \( f(x) \) is not invertible. 

Step 2: Checking if \( h(x) \) is an identity function 
The identity function is defined as \( I(x) = x \). However, the given function \( h(x) \) is: 

\[h(x) = \begin{cases} 0, & x \leq 0  x, & x>0 \end{cases}\]

 This does not satisfy \( h(x) = x \) for all \( x \), so it is not an identity function. 

Step 3: Checking if \( f \circ g \) is invertible 
\[ (f \circ g)(x) = f(g(x)) = f(\sqrt{x^2 + 1}) = (\sqrt{x^2 + 1})^2 - 1 = x^2 + 1 - 1 = x^2. \] Since \( x^2 \) is not a one-to-one function over \( \mathbb{R} \), \( f \circ g \) is not invertible. 

Step 4: Checking if \( h \circ f \circ g = x^2 \) 
\[ (h \circ f \circ g)(x) = h(f(g(x))) = h(x^2). \] From the definition of \( h(x) \), we get: 

\[h(x^2) = \begin{cases} 0, & x^2 \leq 0  x^2, & x^2>0 \end{cases}\]

 Since \( x^2 \geq 0 \) for all \( x \), we always have \( h(x^2) = x^2 \). Hence, \( h \circ f \circ g = x^2 \) holds true. Since statements (III) and (IV) are correct, the correct answer is (C).