JEE Main 2026 — 6 April, Evening Shift — Question 21

Let $f(x) = \frac{2x^2 - 3x + 2}{3x^2 + x + 3} = y$. Cross-multiplying: $2x^2 - 3x + 2 = y(3x^2 + x + 3)$. Rearranging: $(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$. For real $x$, discriminant $D \ge 0$: $(y+3)^2 - 4(3y-2)(3y-2) \ge 0$. Simplify: $y^2 + 6y + 9 - 4(9y^2 - 12y + 4) \ge 0$

$\Rightarrow -35y^2 + 54y - 7 \ge 0$. Multiply by -1: $35y^2 - 54y + 7 \le 0$. Factor: $(7y - 1)(5y - 7) \le 0$. Hence $y \in \left[ \frac{1}{7}, \frac{7}{5} \right]$, so range is a proper subset of $\mathbb{R}$ → not onto. Since $f(x)$ is a rational function with non-zero derivative sign changes,

it is many-one (non-monotonic). Thus $f$ is neither one-one nor onto.