There are exactly two distinct linear functions, ........., and ......... which map [-1, 1] onto [0,

Q: Wait, why do we only consider the endpoints? Can't the function map to other values?

Great question! Because the function is linear and we are mapping the interval $[-1, 1]$ *onto* $[0, 2]$, the function must be continuous and monotonic. If it didn't map the endpoints of the domain to the endpoints of the range, it would either miss part of the target interval or exceed it, failing the 'onto' condition.

Q: Why are there exactly two functions? Could there be a constant function?

A constant function would map the entire domain to a single point, not an interval. Since the range $[0, 2]$ has a length of 2, a constant function cannot cover it. Thus, the slope $a$ cannot be zero, leaving us with only the increasing and decreasing cases.

Q: How do I know if I should use $f(-1)=0$ or $f(-1)=2$?

You don't have to guess! You test both possibilities. One corresponds to a positive slope (increasing) and the other to a negative slope (decreasing). Both are valid linear functions that satisfy the mapping requirement.

Q: Is it possible for the function to be non-linear?

The problem specifically asks for 'linear functions'. If we allowed non-linear functions, there would be infinitely many ways to map one interval to another, but the constraint of linearity restricts our universe to just these two elegant solutions.

Q: What if the domain or range were open intervals like $(-1, 1)$?

The logic remains identical! Even with open intervals, the linear mapping must preserve the 'boundary' behavior. The endpoints define the transformation, regardless of whether they are included in the set.

Question

There are exactly two distinct linear functions, ........., and ......... which map $[-1, 1]$ onto $[0, 2]$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a line segment, specifically the interval $[-1, 1]$. Your goal is to stretch, shift, and potentially flip this segment so that it perfectly covers the interval $[0, 2]$.

When we talk about a linear function $f(x) = ax + b$ mapping one interval onto another, we are essentially asking: "How can I transform this specific piece of the number line to fit exactly over another?"

### The Power of Monotonicity

Because our function is linear, it possesses a beautiful property: **monotonicity**. A linear function is either strictly increasing or strictly decreasing. It never turns back on itself, and it never pauses.

This means that if we want to map the domain $[-1, 1]$ onto the range $[0, 2]$, the "start" of our domain must map to one end of the range, and the "end" of our domain must map to the other. There is no other way to cover the entire range without leaving gaps or going out of bounds.

### Case 1: The Ascending Path

Let us first explore the case where the function is strictly increasing. Here, the slope $a$ must be positive. To cover the range $[0, 2]$ as we move from $x = -1$ to $x = 1$, we must map the smallest input to the smallest output and the largest input to the largest output.

We set up our conditions:

$$f(-1) = a(-1) + b = 0$$

$$f(1) = a(1) + b = 2$$

This gives us a simple system of linear equations. By adding these two equations, we find $2b = 2$, which tells us $b = 1$.

Substituting this back into our first equation, we find $-a + 1 = 0$, which means $a = 1$. Thus, our first elegant solution is **$f(x) = x + 1$**. It is a simple, steady climb from $0$ to $2$.

### Case 2: The Descending Path

But what if we choose to flip the orientation? What if we map the start of our domain to the end of our range? This corresponds to a strictly decreasing function, where the slope $a$ is negative.

Now, we map the smallest input to the largest output and vice versa:

$$f(-1) = a(-1) + b = 2$$

$$f(1) = a(1) + b = 0$$

Again, we solve the system. Adding these equations yields $2b = 2$, so $b = 1$ once more.

Substituting $b = 1$ into the second equation, we get $a + 1 = 0$, leading to $a = -1$. Our second solution is **$f(x) = -x + 1$**. This function gracefully slides down from $2$ to $0$ as $x$ increases.

### The Elegance of the Result

We have found exactly two paths. One climbs, one descends. Both satisfy the requirement of mapping $[-1, 1]$ onto $[0, 2]$ perfectly.

In the world of JEE mathematics, it is easy to get lost in the algebra, but always remember the geometry. You are not just solving for $a$ and $b$; you are defining the only two linear bridges that can connect these two worlds.

Whether you choose **$f(x) = x + 1$** or **$f(x) = -x + 1$**, you have mastered the transformation. Keep this intuition close, and no linear mapping problem will ever intimidate you again.

Animated Solution for Mathematics - Functions: There are exactly two distinct linear functions, ........., and ......... which map $[- 1, 1]$ onto $[0, 2]$ .

Visualized Solution

Defining the Mapping Region

Properties of Linear Functions

Case 1: Strictly Increasing

Setting up Equations for Case 1

Solving for $b$

Solving for $a$

First Linear Function

Case 2: Strictly Decreasing

Setting up Equations for Case 2

Solving for $b$ (Case 2)

Solving for $a$ (Case 2)

Second Linear Function

Final Answer

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Power of Monotonicity

Case 1

Case 2

The Elegance of the Result

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } There are exactly two distinct linear functions, ........., and ......... which map [−1,1] onto [0,2].