The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field.

The Computer Laboratory

Practice Paper 3 Question 16

Give an example (i.e. equation) of a non-constant polynomial function of the smallest degree \(f(x),\) such that all of the following hold: (a) it has at least one inflexion point, (b) all inflexion points have \(y\)-coordinate equal to \(0,\) (c) all its roots are real, and (d) it is symmetric with respect to the \(y\)-axis.

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

Warm-up Questions

  1. Evaluate \(\frac{d}{dx}{(x^2-3)^3}.\)
  2. Find the roots of the polynomial \(x^2-7x+12.\)
  3. Find the inflexion point of the curve \(y=x^3-3x^2+5x-1.\)

Hints

  • Hint 1
    What does \(f\) being symmetrical about the \(y\)-axis imply about its degree \(n?\)
  • Hint 2
    Does the fact that \(f\) has a point of inflexion restrict the values that \(n\) can take?
  • Hint 3
    How can one express a polynomial in terms of its roots?
  • Hint 4
    If you arbitrarily choose one root of the polynomial, what conditions does that set on other roots?
  • Hint 5
    Once you have arbitrarily selected a pair of roots, how can you select the other pair to satisfy the inflexion point condition?

Solution

Since \(f\) is symmetrical about the \(y\)-axis, it must be of even order \(n.\) At an inflexion point, the second derivative changes sign. For \(n \leq 2,\,\) \(f''(x)=0\) and so we deduce \(n > 2.\) Hence, the smallest possible degree is \(n=4.\) If the \(y\)-coordinate equals \(0\) at all the inflexion points, then every inflexion point is a root. If all roots are real, then the polynomial may be expressed as a product of real monomials, i.e. \(f(x)=(x-a)(x-b)\cdots.\) The symmetry of the function also means that the roots must be symmetrical, so \(f(x)=(x-a)(x+a)(x-b)(x+b)\)\(=(x^2-a^2)(x^2-b^2).\)

We can arbitrarily choose a pair of symmetric roots, so let \(b=1.\) We then find the second derivative of \(f,\) which is \(f''(x)=12x^2-2a^2-2.\) The roots of this derivative, \(\pm\sqrt{(a^2+1)/6},\) are points of inflexion. In order to satisfy the condition that all inflexion points are roots of \(f,\) we equate these to \(\pm a\) and solve to get \(a=\frac{1}{\sqrt5}.\) Therefore \(f(x)=(x^2-1)(x^2-1/5)\) in this case.

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.foo[email protected]. Please do not write to this address regarding general admissions or course queries.