IB DP Math MAA SL : IB Style Mock Exams – Set 3 Paper 2

Question

Let $f(x)=x^3$ and $g(x)=2 x-16$, for $x \in R$.

(a) Find $\rm{g^{-1}(x)}$.

(b) Solve the equation $(g \circ f)(x)=0$.

Answer/Explanation

Ans:

(a) To find the inverse of $g(x)=2x-16$, we need to solve for $x$ in the equation $y=2x-16$ in terms of $y$. We have:

\begin{align*}
y &= 2x-16 \\
y+16 &= 2x \\
\frac{y+16}{2} &= x
\end{align*}

Therefore, $g^{-1}(x) = \frac{x+16}{2}$.

(b) The composition $(g \circ f)(x)$ means we apply the function $f$ first and then apply the function $g$ to the result. Therefore,

$$(g \circ f)(x) = g(f(x)) = g(x^3) = 2x^3 – 16$$

To solve $(g \circ f)(x) = 0$, we need to find the values of $x$ such that $2x^3 – 16 = 0$. We can factor this equation as:

\begin{align*}
2x^3 – 16 &= 2(x^3-8) \\
&= 2(x-2)(x^2+2x+4) \\
\end{align*}

The quadratic factor $x^2+2x+4$ has no real roots, so the only real root of the equation $2x^3-16=0$ is $x=2$.

Therefore, the solution to $(g \circ f)(x) = 0$ is $x=2$.

Question

Let A(2, -3, 5) and B(-1, 1 , 5).

Find
(a) the distance between A and B
(b) the distance between O and B
(c) the coordinates of the midpoint M of the line segment [AB]
(d) the coordinates of point C given that B is the midpoint of [AC].

▶️Answer/Explanation

Sol:

(a) The distance between points A and B is given by the distance formula:
\begin{align*}
AB &= \sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2} \\
&= \sqrt{(-1-2)^2+(1-(-3))^2+(5-5)^2} \\
&= \sqrt{3^2+4^2+0^2} \\
&= \sqrt{9+16} \\
&= \sqrt{25} \\
&= 5
\end{align*}

Therefore, the distance between A and B is 5 units.

(b) Let O(0,0,0) be the origin. The distance between O and B is also given by the distance formula:
\begin{align*}
OB &= \sqrt{(x_B-x_O)^2+(y_B-y_O)^2+(z_B-z_O)^2} \\
&= \sqrt{(-1-0)^2+(1-0)^2+(5-0)^2} \\
&= \sqrt{1+1+25} \\
&= \sqrt{27} \\
&= 3\sqrt{3}
\end{align*}

Therefore, the distance between O and B is $3\sqrt{3}$ units.

(c) The coordinates of the midpoint M of the line segment [AB] is given by the midpoint formula:
\begin{align*}
M &= \left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2},\frac{z_A+z_B}{2}\right) \\
&= \left(\frac{2+(-1)}{2},\frac{-3+1}{2},\frac{5+5}{2}\right) \\
&= \left(\frac{1}{2},-1,5\right)
\end{align*}

Therefore, the midpoint of the line segment [AB] is $\left(\frac{1}{2},-1,5\right)$.

(d) Let C(x,y,z) be the coordinates of point C. Since B is the midpoint of [AC], we can use the midpoint formula to find the coordinates of C:
\begin{align*}
B &= \left(\frac{x_A+x_C}{2},\frac{y_A+y_C}{2},\frac{z_A+z_C}{2}\right) \\
(-1,1,5) &= \left(\frac{2+x}{2},\frac{-3+y}{2},\frac{5+z}{2}\right) \\
\end{align*}

Solving for x, y, and z, we get:
\begin{align*}
x &= -4 \\
y &= 5 \\
z &= 5
\end{align*}

Therefore, the coordinates of point C are (-4, 5, 5).

Scroll to Top