MAT Reasoning about Solutions Practice

  1. By factorising or otherwise, determine how many real solutions each of these equations havefor , and what those solutions are. is a constant.
    The number of solutions may depend on the constant. In such cases, detail how the number of solutions varies as the constant varies.
  2. (the log here is base 10)
  3. By using the discriminant, determine the range of values of for which the equation has one or more real solutions.
  4. ,
  5. ,
  6. What is the minimum and maximum number of roots these graphs have?
  7. By thinking about the shape of these graphs and considering their turning points, determine the range of for which the graph has the specified number of solutions.
  8. 1 solution
  9. 2 distinct solutions
  10. 3 distinct solutions
  11. Given that , then for :
  12. 0 solutions
  13. 1 solution
  14. 2 distinct solutions
  15. 3 distinct solutions
  16. 4 solutions
  17. The following questions involve solving equations with exponentials:
  18. Solve
  19. Solve
  20. How many solutions does have?
  21. [MAT 2010] Given a positive integer and a real number , consider the following equation in ,
    Which of the following statements about the equation is true?
    (a) If then the equation has no real solution for some values of .
    (b) If is even, then the equation has a real solution for any given value of .
    (c) If then the equation has (at least) one real solution .
    (d) The equation never has a repeated solution for any given values of and .

ANSWERS

  1. –
  2. Just the one. Factorising:
    So , which has no solutions, or , which gives .
  3. It has 6 roots, but 4 of them are repeated.
    This gives us , , .
    The worst thing you could possibly do is write and then divide both sides by . Doing so loses you the solution.
  4. 3 distinct solutions.
    So solutions are , and .
  5. Factorising, we get , so .
    If , then there will be one solution for in each interval.
    If or , there will be no solutions for .
    If , then there will be two solutions for in each interval.
  6. Factorising:
    The first factor doesn’t give any solutions since has no real solutions, but and .
  7. Like part (d), spotting again that this is a squared expression:
    So and .
    Then .
    The equation thus has 2 solutions provided , and no solutions otherwise.
  8. The key here is spotting that it’s a binomial expansion:
    We have an inflection point (i.e. triple-repeated root) when
    .
  9. –
  10. or (you first need to factorise out the ).
  11. so
    Thus or . Then by consider the graph of , we get: or .
  12. So or
    Using the discriminant on the first gives us and on the second . Combining we get .
  13. Note that for all the polynomials where the highest power is even, there can be a global minimum or maximum, and thus the graph need not cross the x-axis. However for the odd polynomials, the value of the polynomial varies from to and thus must cross the x-axis at some point.
  14. Min: 0, Max: 2
  15. Min: 1, Max: 3
  16. Min: 0, Max: 4
  17. Min: 1, Max: 5
  18. –
  19. Differentiating we get the turning points and .
  20. If there’s one solution, the x-axis either cuts the line above the local maximum or below the local minimum.Soor .
    This gives us
  21. We get two distinct solutions when the x-axis touches the local maximum and crosses the line later, or equivalently with the minimum. This gives us
  22. This occurs when the x-axis is inside the horizontal strip between the maximum and minimum. We get .
  23. The y-values of the two minima is . The y-value of the maximum is . Then by using the shape of the quartic:
  24. No solutions when , because the minima will be above the y-axis.
  25. It’s not possible to have exactly 1 solution.
  26. 2 distinct solutions when (the two minima will be touching the y-axis)
  27. 3 distinct solutions when (when the x-axis crosses the quartic twice and touches the maximum)
  28. 4 solutions when (i.e. the x-axis is between the maximum and minima)
  29. Use a similar approach to that within the C2 A Level module:

  30. or
    The first is not possible because an exponential function always gives a positive value. So is the only solution.
  31. Let for convenience:
    So , or
    The first is not possible, the second gives us and the third is .
  32. When there’s clearly no solutions. If , then the modulus function changes from a negative to a positive value, so use . The discriminant is negative, so there are no solutions.
  33. If then we have a cubic. We know that cubics always have a solution, so it can’t be (a).
    If is even, then the highest power of our polynomial is even, and we will always have a minimum (given the coefficient of the highest-order term is positive). We know from Q3 that there may be no solutions, and since we can use to shift the graph up and down, there will always be some for which the x-axis never cuts the graph.
    This leaves (c) and (d) so far.
    For (c), we know we will always have a solution when is odd. If is even, then if for example, we have a graph which approaches as , and definitely crosses the x-axis as it has roots . If increases, this shifts the graph down. We can see that however much we increase and shift the graph down, it will always still cross the x-axis. So the answer is (c).
    For (d), it’s not too difficult to find a counter-example. When (the simplest case where might have a repeated root), then . Using the discriminant to get a repeated root, we find that , and the resulting would be .