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Miscellaneous Exercise · Q8

Q.Let f={(1,1),(2,3),(0,−1),(−1,−3)}f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\} be a function from Z\mathbb{Z} to Z\mathbb{Z} defined by f(x)=ax+bf(x) = ax + b, for some integers a,ba, b. Determine a,ba, b.

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Use any two points from the given set to form simultaneous equations in aa and bb; solving reveals the linear function is f(x)=2x−1f(x) = 2x - 1, so a=2a = 2 and b=−1b = -1.

A function defined by f(x)=ax+bf(x) = ax + b is a linear function, completely determined by two parameters: the slope aa and the yy-intercept bb. When we're given specific input-output pairs, we can substitute them into the function's formula to create equations that pin down these unknowns.

The set f={(1,1),(2,3),(0,−1),(−1,−3)}f = \{(1, 1), (2, 3), (0, -1), (-1, -3)\} tells us that f(1)=1f(1) = 1, f(2)=3f(2) = 3, f(0)=−1f(0) = -1, and f(−1)=−3f(-1) = -3. Since the function has the form f(x)=ax+bf(x) = ax + b, any two of these pairs will suffice to find aa and bb. The remaining pairs then serve as a check.

Let me use the simplest pair first: (0,−1)(0, -1).

  1. Extract bb directly from f(0)=−1f(0) = -1:

    Substituting x=0x = 0 into f(x)=ax+bf(x) = ax + b gives:

f(0)=a⋅0+b=bf(0) = a \cdot 0 + b = b

Since f(0)=−1f(0) = -1, we immediately have b=−1b = -1.

  1. Find aa using another point, say (1,1)(1, 1):

    Substitute x=1x = 1 and f(1)=1f(1) = 1 into f(x)=ax+bf(x) = ax + b:

1=a⋅1+b=a+b1 = a \cdot 1 + b = a + b

We already know b=−1b = -1, so:

1=a−1  ⟹  a=21 = a - 1 \implies a = 2

  1. Verify with the remaining points: …

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