IB DP Math MAA HL : IB Style Mock Exams – Set 10 Paper 1

Question

[Maximum mark: 6]
Let $f(x)=(x+2)^3$, for $x \in \mathbb{R}$.

(a) Find $f^{-1}(x)$               [3]

Let $g$ be a function so that $(f \circ g)(x)=27 x^6$.

(b) Find $g(x)$.               [3]

Answer/Explanation

(a) If we solve the equation $\left(f \circ f^{-1}\right)(x)=x$ for $y=f^{-1}(x)$, we get
$
\begin{aligned}
f\left(f^{-1}(x)\right) & =x \\
f(y) & =x \\
(y+2)^3 & =x \\
y+2 & =\sqrt[3]{x} \\
y & =\sqrt[3]{x}-2 \\
f^{-1}(x) & =\sqrt[3]{x}-2
\end{aligned}
$

(b) If we solve the equation $(f \circ g)(x)=27 x^6$ for $y=g(x)$, we obtain
$
\begin{aligned}
f(g(x)) & =27 x^6 \\
f(y) & =27 x^6 \\
(y+2)^3 & =27 x^6 \\
y+2 & =3 x^2 \\
y & =3 x^2-2 \\
g(x) & =3 x^2-2
\end{aligned}
$

Question

[Maximum mark: 22]
Let $f(x)=(x+1) e^{-2 x}, x \in \mathbb{R}$

(a) Find $\frac{\mathrm{d} f}{\mathrm{~d} x}$.           [2]

(b) Prove by induction that $\frac{\mathrm{d}^n f}{\mathrm{~d} x^n}=\left[n(-2)^{n-1}+(-2)^n(x+1)\right] e^{-2 x}$ for all $n \in \mathbb{Z}^{+}$.           [7]

(c) Find the coordinates of any local minimum and maximum points on the graph of $y=f(x)$. Justify whether any such point is a minimum or a maximum.                     [5]

(d) Find the coordinates of any points of inflexion on the graph of $y=f(x)$. Justify whether any such point is a point of inflexion.                    [5]

(e) Hence sketch the graph of $y=f(x)$, indicating clearly the points found in parts (c) and (d) and any intercepts with the axes.                   [3]

Answer/Explanation

(a) Using the product rule, we get
$
\begin{aligned}
\frac{\mathrm{d} f}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left[(x+1) e^{-2 x}\right] \\
& =e^{-2 x}-2(x+1) e^{-2 x} \\
& =[1-2(x+1)] e^{-2 x}
\end{aligned}
$

(b) Let $\mathrm{P}(n)$ be the following statement:
$
\frac{\mathrm{d}^n f}{\mathrm{~d} x^n}=\left[n(-2)^{n-1}+(-2)^n(x+1)\right] e^{-2 x} \quad\left[\text { for } n \in \mathbb{Z}^{+}\right]
$
Using mathematical induction, we will prove that $\mathrm{P}(n)$ holds for all $n \in \mathbb{Z}^{+}$.

Base case. For $n=1$, from part (a), we have
$
\frac{\mathrm{d} f}{\mathrm{~d} x}=[1-2(x+1)] e^{-2 x}
$
Hence $\mathrm{P}(1)$ holds.

Inductive hypothesis. We assume that $\mathrm{P}(k)$ holds for some $k \geq 1$. In other words, we assume that
$
\frac{\mathrm{d}^k f}{\mathrm{~d} x^k}=\left[k(-2)^{k-1}+(-2)^k(x+1)\right] e^{-2 x}\quad \quad\quad\quad\quad \text { (IH) }
$
for some $k \geq 1$.

Inductive step. To show that $\mathrm{P}(k+1)$ holds, we need to show that
$
\frac{\mathrm{d}^{k+1} f}{\mathrm{~d} x^{k+1}}=\left[(k+1)(-2)^k+(-2)^{(k+1)}(x+1)\right] e^{-2 x}
$
If we differentiate both sides of the equation (IH), we obtain
$
\begin{aligned}
\frac{\mathrm{d}^{k+1} f}{\mathrm{~d} x^{k+1}} & =\frac{\mathrm{d}}{\mathrm{d} x}\left[\left[k(-2)^{k-1}+(-2)^k(x+1)\right] e^{-2 x}\right] \\
& =\left[(-2)^k\right] e^{-2 x}-2\left[k(-2)^{k-1}+(-2)^k(x+1)\right] e^{-2 x} \\
& =\left[(-2)^k\right] e^{-2 x}+\left[k(-2)^k+(-2)^{k+1}(x+1)\right] e^{-2 x} \\
& =\left[(k+1)(-2)^k+(-2)^{(k+1)}(x+1)\right] e^{-2 x}
\end{aligned}
$
Hence $P(k+1)$ holds as well.

Conclusion. $\mathrm{P}(1)$ holds, and for every $k \geq 1$, if $\mathrm{P}(k)$ holds then $\mathrm{P}(k+1)$ also holds. Hence, by mathematical induction, we conclude that $\mathrm{P}(n)$ holds for all $n \in \mathbb{Z}^{+}$.

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