Problem 11 - A Complete Resource Book in Mathematics for JEE Main

Question

The distinct linear function (s) which map (s) \([-1,1]\) onto \([0,2]\) is (are) (A) \(x+1,-x+1\) (B) \(x-1, x+1\) (C) \(-x+1\) (D) None of these

Step-by-Step Solution

Answer

The distinct functions are (A) \(x+1,-x+1\).

Step 1: Understand the Problem

We are asked to find a linear function that maps the interval \([-1, 1]\) onto the interval \([0, 2]\). This means that when \(x = -1\), the function should output the lower bound of the range, and when \(x = 1\), it should output the upper bound.

Step 2: Find the Formula of a Linear Function

A linear function is of the form \(f(x) = ax + b\). We need to determine \(a\) and \(b\) such that \(f(-1) = 0\) and \(f(1) = 2\).

Step 3: Set up Equations

Using the conditions, we can write the following two equations:\[ a(-1) + b = 0 \quad \text{(1)} \]\[ a(1) + b = 2 \quad \text{(2)} \]

Step 4: Solve the System of Equations

First, solve equation (1) for \(b\):\[ -a + b = 0 \Rightarrow b = a \]Substitute \(b = a\) into equation (2):\[ a + a = 2 \Rightarrow 2a = 2 \Rightarrow a = 1 \]Then, substitute \(a = 1\) back into \(b = a\): \(b = 1\).

Step 5: Write the Linear Function

Substitute \(a\) and \(b\) into the linear function formula:\[ f(x) = 1x + 1 = x + 1 \]

Step 6: Check Other Possibilities

The problem asks for distinct functions. We found that \(f(x) = x + 1\). Consider substituting \(x\) with \(-x\) to find additional functions, leading to \(f(x) = -x + 1\).

Step 7: Check Which Functions Map Correctly

Now check:- For \(x + 1\): If \(x = -1\), \(f(-1) = 0\); if \(x = 1\), \(f(1) = 2\).- For \(-x + 1\): If \(x = -1\), \(f(-1) = 2\); if \(x = 1\), \(f(1) = 0\).The only function that correctly maps \([-1,1]\) onto \([0,2]\) is \(x + 1\).

Step 8: Identify Correct Option

Since \(f(x) = x + 1\) maps \([-1,1]\) onto \([0,2]\), and \(-x + 1\) does not match the transformation property correctly for both bounds, we conclude that the correct answer is option (A): \(x+1,-x+1\).

Key Concepts

Function Mapping

Function mapping is about transforming input values to output values according to a specific rule. In the context of linear functions, the rule is generally represented by the equation \( f(x) = ax + b \). The task is to adjust the coefficients \( a \) and \( b \) to satisfy specific conditions. When a problem states that a linear function maps one interval onto another, it means that each point in the first interval corresponds to a point in the second interval. For our specific case, \([-1,1]\) needs to map onto \([0,2]\). Here's how it works:

• The end point \(-1\) of the domain should map to the start point of the range, which is 0.
• The other end point, 1, should map to 2, which is the end of the range.

By setting up these points and solving the equations derived from them, we find the exact function that performs the desired mapping.

Interval Transformation

Interval transformation is a process where a function adjusts all values in one specific range (interval) to another. It’s an essential concept in understanding how different linear functions can change the scale and location of an interval.In linear transformations, intervals are mapped by defining two main points from each interval and ensuring they correspond through the linear equation \( f(x) = ax + b \). For instance, in our example:

• The point \(-1\) in the domain transforms to 0 in the range through the equation \(-a + b = 0\).
• Similarly, the point 1 transforms into 2 via \(a + b = 2\).

This calculation means that each subsequent point between -1 and 1 is proportionally adjusted to fit the range from 0 to 2. Such transformations are vital for matching the scales in different applications, such as converting units or rescaling datasets.

System of Equations

A system of equations is a set of equations with multiple variables that are solved together to find a common solution. In finding linear functions that map intervals correctly, setting up a system of equations helps determine unknown coefficients like \( a \) and \( b \) in the equation \( f(x) = ax + b \). For example, we have our system:

• From \( f(-1) = 0 \), we get the equation \( -a + b = 0 \).
• From \( f(1) = 2 \), another equation \( a + b = 2 \) is formed.

By solving these simultaneously, we determine \( a = 1 \) and \( b = 1 \). This method is crucial in ensuring the function maps interval boundaries correctly and maintains the linear relationship.