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Question
The distinct linear functions that map [−1, 1] onto [0, 2] are
Options
\[f\left( x \right) = x + 1, g\left( x \right) = - x + 1\]
\[f\left( x \right) = x - 1, g\left( x \right) = x + 1\]
\[f\left( x \right) = - x - 1, g\left( x \right) = x - 1\]
None of these
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Solution
Let us substitute the end-points of the intervals in the given functions. Here, domain = [-1, 1] and range =[0, 2]
By substituting -1 or 1 in each option, we get :
Option (a):
\[f\left( - 1 \right) = - 1 + 1 = 0 \text{ and }f\left( 1 \right) = 1 + 1 = 2\]
\[g\left( - 1 \right) = 1 + 1 = 2 \text{ and }g\left( 1 \right) = - 1 + 1 = 0\]
So, option (a) is correct.
Option (b):
\[f\left( - 1 \right) = - 1 - 1 = - 2 \text{ and }f\left( 1 \right) = 1 - 1 = 0\]
\[g\left( - 1 \right) = - 1 + 1 =0 \text{ and }g\left( 1 \right) = 1 + 1 = 2\]
Here, f (-1) gives -2
\[\not\in \left[ 0, 2 \right]\]
So, (b) is not correct.
Similarly, we can see that (c) is also not correct.
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