A LevelMathsCambridge (CIE)Pure 3FlashcardsAlgebra & FunctionsModulus Functions

Modulus Functions (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

Exam code: 9709

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Cards in this collection (5)

  • Two non-parallel straight lines meet once. Why can an equation involving a modulus have more solutions than that?

    Because the modulus reflects part of a graph upwards, and the reflected piece can cross the other graph as well.

    That produces an intersection the unreflected lines never had.

  • To solve \left|\text{f} \left(x\right)\right| = \left|\text{g} \left(x\right)\right|, solve both \text{f} \left(x\right) = \text{g} \left(x\right) and \text{f} \left(x\right) = \_\_\_\_\_\_.

    To solve \left|\text{f}\left(x\right)\right| = \left|\text{g}\left(x\right)\right|, solve both \text{f}\left(x\right) = \text{g}\left(x\right) and \text{f}\left(x\right) = - \text{g}\left(x\right).

    The two cases arise because each side can take either sign once the modulus is removed.

  • What should you do before solving a modulus equation algebraically?

    Sketch both graphs, including the reflected parts, and locate the intersections.

    The sketch tells you how many solutions there should be, and therefore which algebraic answers to keep.

  • True or False?

    Every solution of x - 4 = 2 x - 5 is also a solution of \left|x - 4\right| = 2 x - 5.

    False.

    x = 1 satisfies the first, but \left|1 - 4\right| = 3 while 2 \left(1\right) - 5 = - 3, so it fails the second.

  • How do you solve \left|\text{f} \left(x\right)\right| = \text{g} \left(x\right), with a modulus on one side only?

    Solve both \text{f}\left(x\right) = \text{g}\left(x\right) and - \text{f}\left(x\right) = \text{g}\left(x\right).

    Then test each answer in the original equation, because a modulus can never be negative, so any root making \text{g}\left(x\right) negative has to be thrown out.

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