Let f: R -> R be defined as f(x) = (2x^2 - 3x + 2) / (3x^2 + x + 3). Then f is:

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q1 – JEE-MAIN 2026

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Home Maths Relations and FunctionsQuestion #368

Let

f : R \to R

be defined as

f (x) = \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3}

. Then

f

is:

Clue 1 of 3: Function Properties

To determine if a function is one-one (injective) or onto (surjective), you need to analyze its derivative for monotonicity and its range compared to its codomain, respectively.

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Step 1: Check for One-one (Injectivity)✦ Active

To check if $f (x)$ is one-one, we analyze its derivative $f^{'} (x)$ . The denominator $3 x^{2} + x + 3$ has a discriminant $1^{2} - 4 (3) (3) = - 35 < 0$ and a positive leading coefficient, so it is always positive. Thus, the domain of $f (x)$ is $R$ . Calculate $f^{'} (x)$ :

f^{'} (x) = \frac{(4 x - 3) (3 x^{2} + x + 3) - (2 x^{2} - 3 x + 2) (6 x + 1)}{(3 x^{2} + x + 3)^{2}}

Simplifying the numerator, we get $11 x^{2} - 11$ . So,

f^{'} (x) = \frac{11 (x^{2} - 1)}{(3 x^{2} + x + 3)^{2}} = \frac{11 (x - 1) (x + 1)}{(3 x^{2} + x + 3)^{2}}

Since $f^{'} (x)$ changes sign (e.g., $f^{'} (0) = - 11 / (3)^{2} < 0$ and $f^{'} (2) = 11 (3) / (17)^{2} > 0$ ), the function is not strictly monotonic over $R$ . Therefore, $f$ is not one-one.

Step 2: Check for Onto (Surjectivity)○ Expand

To check if $f (x)$ is onto, we find its range. Let $y = f (x)$ :

y = \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3}

Rearrange this equation into a quadratic in $x$ :

(3 y - 2) x^{2} + (y + 3) x + (3 y - 2) = 0

For $x$ to be real, the discriminant of this quadratic must be non-negative. Let $D_{x}$ be the discriminant:

D_{x} = (y + 3)^{2} - 4 (3 y - 2) (3 y - 2) \geq 0

This simplifies to:

(y + 3)^{2} - (6 y - 4)^{2} \geq 0

Using the difference of squares formula $A^{2} - B^{2} = (A - B) (A + B)$ :

((y + 3) - (6 y - 4)) ((y + 3) + (6 y - 4)) \geq 0

(- 5 y + 7) (7 y - 1) \geq 0

Multiplying by $- 1$ and reversing the inequality sign:

(5 y - 7) (7 y - 1) \leq 0

This inequality holds for $y$ values between the roots $\frac{1}{7}$ and $\frac{7}{5}$ . Thus, the range of $f (x)$ is $[\frac{1}{7}, \frac{7}{5}]$ . Since the codomain is $R$ and the range is a proper subset of $R$ , the function is not onto.

💡 Teacher's Secret Hint

Remember to consider the case where the coefficient of $x^{2}$ in the quadratic equation for $x$ is zero.

Step 3: Conclusion○ Expand

Based on Step 1, $f$ is not one-one. Based on Step 2, $f$ is not onto. Therefore, the function $f$ is neither one-one nor onto.

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Maths Question 1 – JEE-MAIN 2026

Clue 1 of 3: Function Properties

🧠 Full Solution Path