JEE Main 2026 — Functions Question with Solution

Statement I: $f(x)=\frac{x}{1+|x|}$. For $x\ge 0$, $f(x)=\frac{x}{1+x}$ is strictly increasing. For $x<0$, $f(x)=\frac{x}{1-x}$ with $f^{\prime }(x)=\frac{1}{(1-x{)}^2}>0$, also strictly increasing. The function is continuous at $x=0$, so $f$ is strictly increasing on $\mathbb{R}$ and hence one-one. Statement I is true.
Statement II: $f(x)=\frac{x^2+4x-30}{x^2-8x+18}$. The denominator $(x-4{)}^2+2>0$ for all $x$.
$f^{\prime }(x)=\frac{-12(x^2-8x+14)}{(x^2-8x+18{)}^2}$
$f^{\prime }(x)=0\Rightarrow x=4\pm \sqrt{2}$. Since $f^{\prime }$ changes sign, $f$ is not monotonic and hence many-one. Statement II is true.
Both statements are true.