Question

the distinct linear funtions which map[-1,1] onto[0,2] are _______

• A. f(n)=-x-1,g(x)=x-1
• B. f(x)=-x-1,g(x)=x+1
• C. f(x)=-x-1,g(x)=x+1
• D. f(x)=-x+1,g(x)=x+1

Answer 1

The Correct Option is D

To find the distinct linear functions that map the interval \([-1,1]\) onto \([0,2]\), we need to consider the general form of a linear function which is \(f(x) = ax + b\). For the function to map \([-1,1]\) to \([0,2]\), the endpoints of the domain must map to the endpoints of the range:

1. At \(x = -1\), \(f(x)\) should equal 0: \(a(-1) + b = 0\rightarrow -a + b = 0\rightarrow b = a\).
2. At \(x = 1\), \(f(x)\) should equal 2: \(a(1) + b = 2\rightarrow a + b = 2\).

We have a system of equations:

(1) \(b = a\)
(2) \(a + b = 2\)

Substituting \((1)\) into \((2)\):

• \(a + a = 2\rightarrow 2a = 2\rightarrow a = 1\)
• Therefore, \(b = 1\) from \((1)\).

This gives the linear function \(f(x) = x + 1\). To find the second function, consider reflection over the x-axis to maintain linearity and distinct mapping. Hence, we try \(f(x) = -x + 1\). Both functions \(f(x) = -x + 1\) and \(g(x) = x + 1\) satisfy the condition of distinct mappings and linear properties. 

Therefore, the functions are: \(f(x) = -x + 1, g(x) = x + 1\).