Let be defined as . Then f is : [2026]

Question

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 : [2026]

Smart Achievers · Accepted Answer

(4)

We have, $f (x) = \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3}$

$\Rightarrow \lim_{x \to \infty} f (x) = \lim_{x \to \infty} \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3} = \frac{2}{3}$

$\lim_{x \to - \infty} f (x) = \lim_{x \to - \infty} \frac{2 x^{2} - 3 x + 2}{3 x^{2} + x + 3} = \frac{2}{3}$

Since, $\lim_{x \to \infty} f (x) = \lim_{x \to - \infty} f (x) = \frac{2}{3}$ , the function must change direction to return to that value i.e., $\frac{2}{3}$ . This means, it fails the horizontal line test.

Thus, the given function is not one-one.

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

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

Since, $D \geq 0, {(y + 3)}^{2} - 4 {(3 y - 2)}^{2} \geq 0$

$\Rightarrow (7 y - 1) (5 y - 7) \leq 0 \Rightarrow y \in [\frac{1}{7}, \frac{7}{5}]$

Since, range of f(x) is a finite interval, f(x) is not onto.

Hence, f(x) is neither one-one nor onto.

Solution

Q.
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 : [2026]

1 both one-one and onto

2 one-one but not onto

3 onto but not one-one

4 neither one-one nor onto

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