IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Geometry-Identical representation of transformations
Topic :Geometry- Weightage : 21 %
All Questions for Topic : Volume and capacity (additional shapes),Enlargement around a given point,Enlargement by a rational factor,Gradients of perpendicular lines,Identical representation of transformations
Question
Question (a)
Let $f(x)=\ln x$ and $g(x)=x-6$.
Given that $h(x)=2 f(g(x))$, write down $h(x)$.
▶️Answer/Explanation
Ans:
1. First, we’re given two functions:
\(f(x) = \ln x\) and \(g(x) = x – 6\).
2. Now, we need to find \(h(x) = 2 f(g(x))\):
$\bullet$ Substitute the expression for \(g(x)\) into \(f(x)\):
\(f(g(x)) = \ln (x – 6)\).
$\bullet$ Multiply the result by 2:
\(2 \cdot \ln (x – 6)\).
So, the expression for \(h(x)\) is \(2 \ln (x – 6)\).
Question (b)
The $h(x)$ from part (a) is the transformation of $f(x)$ by:
$\bullet$ a vertical dilation with a scale factor $s$
$\bullet$ a horizontal translation of $t$ units to the right,
Write down the values of $s$ and $t$.
▶️Answer/Explanation
Ans:
In the given expression \(h(x) = 2 \ln (x – 6)\), we can identify the transformation of \(f(x) = \ln x\) as described. Let’s break down the transformation in terms of \(s\) and \(t\):
1. Vertical Dilation with Scale Factor \(s\):
The factor of 2 in front of \(\ln (x – 6)\) indicates a vertical dilation by a scale factor of \(s = 2\). This means that the function is stretched vertically by a factor of 2.
2. Horizontal Translation of \(t\) Units to the Right:
The term \(\ln (x – 6)\) involves \(x – 6\), which represents a horizontal translation of 6 units to the right. The value of \(t\) is \(6\).
So, the transformation is:
$\bullet$ Vertical dilation with \(s = 2\)
$\bullet$ Horizontal translation to the right by \(t = 6\) units.
Question (c)
Solve $h(x)=f(x)$.
▶️Answer/Explanation
Ans:
We are given that \(h(x) = 2 \ln (x – 6)\) and \(f(x) = \ln x\), and we need to find the values of \(x\) for which \(h(x) = f(x)\).
Setting up the equation and solving for \(x\):
\[2 \ln (x – 6) = \ln x\]
First, let’s simplify the equation by using properties of logarithms:
\[2 \ln (x – 6) = \ln x \implies \ln ((x – 6)^2) = \ln x\]
Now, we can remove the natural logarithm by taking the exponential of both sides:
\[(x – 6)^2 = x\]
Expand the left side:
\[x^2 – 12x + 36 = x\]
Rearrange the equation:
\[x^2 – 13x + 36 = 0\]
Now, we have a quadratic equation. To solve for \(x\), we can factor the quadratic or use the quadratic formula:
\[(x – 9)(x – 4) = 0\]
This gives us two possible solutions:
1. \(x – 9 = 0 \implies x = 9\)
2. \(x – 4 = 0 \implies x = 4\)
So, the solutions for \(x\) that satisfy \(h(x) = f(x)\) are \(x = 9\) and \(x = 4\). But ,the domain of the natural logarithm function is \(x-6 > 0\), ( argument of log must be positive ) so \(x\) cannot be 4.