JEE Main 2026 — Mathematics Algebra

Given $f(x) = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3}$f(x)=3x2+x+32x2−3x+2​

To check for one-one, let $f(x) = f(y)$f(x)=f(y)

$$\dfrac{2x^2 - 3x + 2}{3x^2 + x + 3} = \dfrac{2y^2 - 3y + 2}{3y^2 + y + 3}$$3x2+x+32x2−3x+2​=3y2+y+32y2−3y+2​

Cross-multiplying and simplifying, we get:

$$11x^2y - 11xy^2 - 11x + 11y = 0$$11x2y−11xy2−11x+11y=0

$$11xy(x - y) - 11(x - y) = 0$$11xy(x−y)−11(x−y)=0

$$11(x - y)(xy - 1) = 0$$11(x−y)(xy−1)=0

This implies $x = y$x=y or $xy = 1$xy=1.

Since $f(x) = f(1/x)$f(x)=f(1/x) for $x \neq 0$x=0 (for example, $f(2) = f(1/2) = \dfrac{4}{17}$f(2)=f(1/2)=174​), the function is many-one (not one-one).

To check for onto, let $y = \dfrac{2x^2 - 3x + 2}{3x^2 + x + 3}$y=3x2+x+32x2−3x+2​

$$(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$$(3y−2)x2+(y+3)x+(3y−2)=0

For $x \in \mathbb{R}$x∈R, the discriminant $D \ge 0$D≥0:

$$(y + 3)^2 - 4(3y - 2)^2 \ge 0$$(y+3)2−4(3y−2)2≥0

$$(y + 3 - 6y + 4)(y + 3 + 6y - 4) \ge 0$$(y+3−6y+4)(y+3+6y−4)≥0

$$(7 - 5y)(7y - 1) \ge 0$$(7−5y)(7y−1)≥0

$$(5y - 7)(7y - 1) \le 0$$(5y−7)(7y−1)≤0

$$y \in \left[\dfrac{1}{7}, \dfrac{7}{5}\right]$$y∈[71​,57​]

Since the range is a proper subset of the codomain $\mathbb{R}$R, the function is not onto.

Thus, $f$f is neither one-one nor onto.

Answer: neither one-one nor onto