The function f : [0, 3] → [1, 29], defined by f(x) = 2x³ - 15x² + 36x + 1, is (a) one-one and onto (

Q: Wait, why do we have to check the critical points AND the boundaries for the range?

Great question! Because the function is not monotonic, it doesn't just go from $f(0)$ to $f(3)$. It climbs to a peak and then drops. You must capture the absolute maximum and minimum, which could happen at the 'peaks' (critical points) or the 'edges' (boundaries). By checking all three, you ensure you haven't missed the true range.

Q: Is there a shortcut to check if it's one-one without calculus?

While you could try to solve $f(x_1) = f(x_2)$, that is algebraic torture for a cubic! Calculus is the elegant path. By checking the derivative, you are essentially asking: 'Does the slope ever change direction?' If it does, the function fails the horizontal line test. Calculus turns a complex algebraic problem into a simple sign-analysis task.

Q: I got the range as $[1, 29]$. Does this mean it's onto?

Exactly! The definition of an 'onto' function is that the range must equal the codomain. Since your calculated range $[1, 29]$ matches the given codomain $[1, 29]$ perfectly, every element in the target set is 'hit' by the function. You've nailed it!

Question

The function $f : [0, 3] 	o [1, 29]$, defined by $f(x) = 2x^3 - 15x^2 + 36x + 1$, is

Options:

(A) one-one and onto

(B) onto but not one-one

(C) one-one but not onto

(D) neither one-one nor onto

Sigma Percentile · Accepted Answer

### The Anatomy of a Cubic Function

Imagine you are standing on the graph of the function $f(x) = 2x^3 - 15x^2 + 36x + 1$. You are tasked with determining its nature: is it one-one? Is it onto?

In the world of JEE Advanced, we don't just guess; we investigate. We treat the function like a living entity, and to understand its behavior, we must look at its 'velocity'—its derivative.

### Phase 1: The Derivative as a Compass

To see if a function is one-one, we need to know if it is strictly monotonic. If it is always climbing or always falling, it is one-one. But if it turns, it repeats values.

Let us calculate the derivative to find these turning points:

$$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) = 6x^2 - 30x + 36$$

This derivative is the heartbeat of our curve. To find where the curve pauses or turns, we set $f'(x) = 0$.

Factoring out the $6$, we get:

$$6(x^2 - 5x + 6) = 0$$

This simplifies beautifully to $6(x-2)(x-3) = 0$. Our critical points are $x=2$ and $x=3$.

Notice that both points lie within our domain of $[0, 3]$. This is the 'aha!' moment. Because the derivative changes sign at $x=2$, the function increases until $x=2$ and then decreases until $x=3$.

It wiggles! Therefore, it fails the horizontal line test. It is **not one-one**.

### Phase 2: The Treasure Hunt for the Range

Now, we must determine if the function is onto. For a function to be onto, its range must perfectly cover the codomain $[1, 29]$.

We need to find the absolute minimum and maximum values of $f(x)$ on the interval $[0, 3]$. We check the boundaries and the critical points:

1. At the start, $f(0) = 2(0)^3 - 15(0)^2 + 36(0) + 1 = 1$.

2. At the peak, $f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$.

3. At the end, $f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$.

Looking at these values, the lowest point is $1$ and the highest point is $29$. Since the function is continuous, it sweeps through every value between $1$ and $29$.

Thus, the range is $[1, 29]$.

### The Final Synthesis

We have discovered that the range is exactly equal to the codomain $[1, 29]$. This means the function is onto.

However, because it 'turned' at $x=2$, it is not one-one. We have successfully dissected the function, proving it is **onto but not one-one**.

You have just mastered the art of function analysis!

Animated Solution for Mathematics - Functions: The function $f : [0, 3] \to [1, 29]$ , defined by $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1$ , is

Select Answer:

Visualized Solution

Defining the Function and Domain

Strategy for One-One

Finding the Derivative $f^{'} (x)$

Factoring the Derivative

Analyzing Monotonicity

Conclusion on One-One

Strategy for Onto

Evaluating Boundary and Critical Points

Calculating $f (0)$

Calculating $f (2)$

Calculating $f (3)$

Determining the Range

Final Conclusion

The Sigma Insight: Classification of Functions

The Anatomy of a Cubic Function

Phase 1

Phase 2

The Final Synthesis

Similar Questions

The function $f : R \to R, f (x) = \frac{x ^{2} + 2 x - 15}{x ^{2} - 4 x + 9}, x \in R$ is

Let a function $f : N \to N$ be defined by $f (n) = ⎩ ⎨ ⎧ 2 n, n - 1, \frac{n + 1}{2}, n = 2, 4, 6, 8, \dots n = 3, 7, 11, 15, \dots n = 1, 5, 9, 13, \dots$ then, $f$ is

If $f : [0, \infty) \to [0, \infty)$ , and $f (x) = \frac{x}{1 + x}$ then $f$ is

A function $f$ from the set of natural numbers to integers defined by $f (n) = {\frac{n - 1}{2}, - \frac{n}{2}, when n is odd when n is even$ is

The function $f : (- \infty, \infty) \to (- \infty, 1)$ , defined by $f (x) = \frac{2 ^{x} - 2 ^{- x}}{2 ^{x} + 2 ^{- x}}$ is:

Let function $f : R \to R$ be defined by $f (x) = 2 x + sin x$ for $x \in R$ , then $f$ is

For real $x$ , let $f (x) = x^{3} + 5 x + 1$ , then

Let $f (x) = {- a x + a if - a \leq x \leq 0 if 0 < x \leq a$ where $a > 0$ and $g (x) = (f (∣ x ∣) - ∣ f (x) ∣) /2$ . Then the function $g : [- a, a] \to [- a, a]$ is

Let the function defined in column I have domain $(- \frac{π}{2}, \frac{π}{2})$ and range $(- \infty, \infty)$ .

List-I

List-II

If the functions $f (x)$ and $g (x)$ are defined on $R \to R$ such that $f (x) = {0, x, x \in rational x \in irrational$ ; $g (x) = {0, x, x \in irrational x \in rational$ then $(f - g) (x)$ is

Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f:[0,3]→[1,29], defined by f(x)=2x3−15x2+36x+1, is