A cubic function $f$ is defined for $x \in \mathbb{R}$ as
$$f : x \mapsto x^3 + (1 - k^2)x + k,$$
where $k$ is a constant - Leaving Cert Mathematics - Question 3 - 2012
Question 3
A cubic function $f$ is defined for $x \in \mathbb{R}$ as
$$f : x \mapsto x^3 + (1 - k^2)x + k,$$
where $k$ is a constant.
(a) Show that $-k$ is a root of $f$.
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Worked Solution & Example Answer:A cubic function $f$ is defined for $x \in \mathbb{R}$ as
$$f : x \mapsto x^3 + (1 - k^2)x + k,$$
where $k$ is a constant - Leaving Cert Mathematics - Question 3 - 2012
Step 1
Show that $-k$ is a root of $f$.
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Answer
To show that −k is a root of f, we evaluate f(−k):
f(−k)=(−k)3+(1−k2)(−k)+k
Calculating this gives:
f(−k)=−k3−(1−k2)k+k
Combining the terms:
=−k3−k+k3+k=0
Thus, we have shown that −k is indeed a root of f.
Step 2
Find, in terms of $k$, the other two roots of $f$.
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Answer
Since −k is a root of f, we know by the Factor Theorem that (x+k) is a factor of f(x). We perform polynomial long division to find the other factor.
The result of the division is:
x3+(1−k2)x+k=(x+k)(x2−kx+1)
Thus, the other two roots of f are the solutions to the equation:
x2−kx+1=0
Applying the quadratic formula:
x=2k±k2−4
So the other two roots of f are:
k+2k2−4andk−2k2−4
Step 3
Find the set of values of $k$ for which $f$ has exactly one real root.
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Answer
From the solution to part (b), we see that f has exactly one real root if and only if k2−4<0. This is equivalent to:
k2<4
Thus, the set of values of k is:
−2<k<2
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