AMC 8 Daily Practice - New Patterns

Complete problem set with solutions and individual problem pages

Problem 6 Easy

Define a new operation:   a \odot b = \begin{cases} a^{2} - b & \text{if } a \leq 0 \\ -a + b & \text{if } a > 0 \end{cases}.   For example, -2 \odot 4 = (-2)^{2} - 4 = 0, and 2 \odot 3 = -2 + 3 = 1. If x \odot 1 = -\frac{3}{4}, find the value of x.

  • A.

    -\frac{1}{2}

  • B.

    \frac{1}{2},-\frac{3}{4}

  • C.

    \frac{7}{4}

  • D.

    -\frac{1}{2},\frac{1}{2}

  • E.

    -\frac{1}{2},\frac{7}{4}

Answer:E

To find x, we consider the two cases based on the definition of \odot:

Case 1: x \leq 0

If x \leq 0, then x \odot 1 = x^{2} - 1 = -\frac{3}{4}  ， x = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}

Since x \leq 0 in this case, we discard x = \frac{1}{2} and retain x = -\frac{1}{2}.

Case 2: x > 0

If x > 0, then x \odot 1 = -x + 1 = -\frac{3}{4}, x = \frac{7}{4} > 0, this solution is valid.

Thus, the values of x are:   \boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{\frac{7}{4}}

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