IB Maths AA HL 2.7 Quadratic equations, inequalities and parameters Question Bank

Practise IB Maths AA HL 2.7 by solving parameterised quadratic equations and inequalities using discriminants and graph conditions.

Exam points

Derive parameter values for distinct, repeated or absent real roots using the discriminant.
Combine intersection and inequality conditions to determine a valid parameter interval.
Evaluate all algebraic branches and reject values that violate domain or graph constraints.

[Maximum number: 8]

The functions f and g are defined by

\begin{aligned} & f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \\ & g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \end{aligned}

(a)

Let h(x)=n f(x)+g(x) where $n \in \mathbb{R}, n>1$ .

[ 8 ]

(i)

By forming a quadratic equation in $\mathrm{e}^{x}$ , solve the equation h(x)=k, where $k \in \mathbb{R}^{+}$ .

[ 5 ]

$\quad(n+1) \mathrm{e}^{2 x}-2 k \mathrm{e}^{x}+(n-1)=0$ $\mathrm{e}^{x}=\frac{2 k \pm \sqrt{4 k^{2}-4\left(n^{2}-1\right)}}{2(n+1)}$ $x=\ln \left(\frac{k \pm \sqrt{k^{2}-n^{2}+1}}{n+1}\right)$

(ii)

Hence or otherwise show that the equation h(x)=k has two real solutions provided that $k>\sqrt{n^{2}-1}$ and $k \in \mathbb{R}^{+}$ .

[ 3 ]

for two real solutions, we require $k>\sqrt{k^{2}-n^{2}+1}$
and we also require $k^{2}-n^{2}+1>0$ $k^{2}>n^{2}-1$ $\Rightarrow k>\sqrt{n^{2}-1}\left(k \in \mathbb{R}^{+}\right)$

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SL 2.7—Quadratic equations and inequalities question 1

Question (a)

Question (i)

Question (ii)