Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be defined by \( f(x) = \frac{x}{\sqrt{1+x^2}} \), then \( (f \circ f)(x) \) is

• A. $\frac{x}{\sqrt{1+x^2}}$
• B. $\frac{x}{\sqrt{1+3x^2}}$
• C. $x$
• D. $1$

Hint:

In composition, simplify denominator carefully—common mistake area.

Answer 1

The Correct Option is B

Concept: Composition of function: $(f \circ f)(x) = f(f(x))$

Step 1: Substitute $f(x)$ into itself.
\[ f(f(x)) = f\left(\frac{x}{\sqrt{1+x^2}}\right) \]

Step 2: Apply definition of $f$.
\[ = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1+x^2}}\right)^2}} \]

Step 3: Simplify denominator.
\[ = \frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1 + \frac{x^2}{1+x^2}}} \] \[ = \frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{\frac{1+2x^2}{1+x^2}}} \]

Step 4: Final simplification.
\[ = \frac{x}{\sqrt{1+2x^2+x^2}} = \frac{x}{\sqrt{1+3x^2}} \] Conclusion:
$(f \circ f)(x) = \frac{x}{\sqrt{1+3x^2}}$