Injectivity and Surjectivity of Cubic Polynomial Function

To determine the nature of the function $f : [0, 3] \to [1, 29]$ defined by: $f(x) = 2x^3 - 15x^2 + 36x + 1$

we need to analyze its injectivity (one-one) and surjectivity (onto).

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Step 1: Checking for Injectivity (One-One)

To test whether the function is one-one, we find its derivative $f'(x)$: $f'(x) = \frac{d}{dx}\left(2x^3 - 15x^2 + 36x + 1\right)$ $f'(x) = 6x^2 - 30x + 36$ $f'(x) = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3)$

Now, let's analyze the sign of $f'(x)$ on the interval $[0, 3]$:

• For $x \in [0, 2)$, $f'(x) > 0 \implies f(x)$ is strictly increasing.
• For $x \in (2, 3)$, $f'(x) < 0 \implies f(x)$ is strictly decreasing.
• At $x = 2$ and $x = 3$, $f'(x) = 0$.

Since $f'(x)$ changes sign within the domain $[0, 3]$, the function $f(x)$ increases up to $x = 2$ and then decreases from $x = 2$ to $x = 3$.

Because $f(x)$ is non-monotonic on $[0, 3]$, there exist multiple values of $x$ that yield the same output $f(x)$. For instance, $f(x)$ takes values in the interval $[28, 29)$ twice (once in $[0, 2]$ and once in $[2, 3]$).

Hence, $f$ is not one-one.

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Step 2: Checking for Surjectivity (Onto)

To determine whether the function is onto, we find the range of $f(x)$ over the domain $[0, 3]$ and compare it with the codomain $[1, 29]$.

We evaluate $f(x)$ at the critical points and the endpoints of the domain $[0, 3]$:

• At $x = 0$: $f(0) = 2(0)^3 - 15(0)^2 + 36(0) + 1 = 1$

• At $x = 2$ (local maximum): $f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$

• At $x = 3$ (endpoint and local minimum): $f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$

Since $f(x)$ is a polynomial, it is continuous on the closed interval $[0, 3]$. By the Intermediate Value Theorem, $f(x)$ takes all values between its absolute minimum and absolute maximum.

• Absolute minimum value of $f(x)$ on $[0, 3] = 1$ (at $x = 0$)
• Absolute maximum value of $f(x)$ on $[0, 3] = 29$ (at $x = 2$)

Therefore, the Range of $f = [1, 29]$.

Since $\text{Range } f = [1, 29] = \text{Codomain}$, $f$ is onto.

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Conclusion

The function $f$ is onto but not one-one.

Correct Option: (B)