Question

Statement 1 : The function \(f:\mathbb{R}\to\mathbb{R}\) defined by \[ f(x)=\frac{x}{1+|x|} \] is one–one.
Statement 2 : The function \(f:\mathbb{R}\to\mathbb{R}\) defined by \[ f(x)=\frac{x^2+4x-30}{x^2-8x+18} \] is many–one.
Which of the following is correct?

• A. Both Statements are correct
• B. Both Statements are false
• C. Statement 1 is false and Statement 2 is correct
• D. Statement 1 is correct and Statement 2 is false

Hint:

A function is one–one if it is {strictly monotonic}. Rational functions with same degree numerator and denominator are usually {many–one}.

Answer 1

The Correct Option is A

Statement 1 Analysis:
\[ f(x)=\frac{x}{1+|x|} \] Write it piecewise: \[ f(x)= \begin{cases} \dfrac{x}{1+x}, & x\ge 0\\ \dfrac{x}{1-x}, & x<0 \end{cases} \]

For \(x\ge 0\): \(f'(x)=\dfrac{1}{(1+x)^2}>0\) \(\Rightarrow\) strictly increasing. \
For \(x<0\): \(f'(x)=\dfrac{1}{(1-x)^2}>0\) \(\Rightarrow\) strictly increasing. 
Also, \[ \lim_{x\to 0^-}f(x)=0=\lim_{x\to 0^+}f(x) \] Thus, the function is strictly increasing on \(\mathbb{R}\), hence one–one
. \[ \Rightarrow Statement 1 is correct.
\] Statement 2 Analysis:
\[ f(x)=\frac{x^2+4x-30}{x^2-8x+18} \] Rewrite numerator using denominator: \[ x^2+4x-30=(x^2-8x+18)+12x-48 \] \[ f(x)=1+\frac{12x-48}{x^2-8x+18} \] Since the function is a rational function of degree \(2/2\), it is not strictly monotonic
on its domain. Indeed, \[ f(2)=\frac{4+8-30}{4-16+18}=\frac{-18}{6}=-3 \] \[ f(6)=\frac{36+24-30}{36-48+18}=\frac{30}{6}=5 \] Also, multiple distinct \(x\)-values can give the same \(f(x)\), hence the function is many–one
. \[ \Rightarrow Statement 2 is correct.
\] 
Final Conclusion:
Both statements are correct. \[ \boxed{\text{Option (1)}} \]