CIE 2023 March 0606/22 Question 1/2 and solution

Question 1

On the axes below, sketch the graph of \( y = |4 \cos 2x| \) for \( 0 \leqslant x \leqslant \pi \), giving the coordinates of any points where the graph meets the axes. [3]

Solution

Step 1: Understand the function

The function to be graphed is

\[ y = |4 \cos 2x|, \]

where \( 0 \leq x \leq \pi \). This function consists of the absolute value of a scaled cosine function with frequency doubled.

Step 2: Analyze the function

The general cosine function, \( \cos 2x \), oscillates between \(-1\) and \(1\). Scaling it by \(4\) gives \(4 \cos 2x\), which oscillates between \(-4\) and \(4\). Applying the absolute value ensures that the negative values are reflected upwards, so \( y = |4 \cos 2x| \) oscillates between \(0\) and \(4\).

The period of \( \cos 2x \) is given by:

\[ \text{Period} = \frac{2\pi}{\text{frequency}} = \frac{2\pi}{2} = \pi. \]

Thus, within \(0 \leq x \leq \pi\), there will be two humps of the cosine graph due to the doubled frequency.

Step 3: Find key points

To sketch the graph, we determine where the function intersects the axes and other key points:

• When \( y = 0 \): This occurs when \( \cos 2x = 0 \), i.e.

  \[ 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \implies x = \frac{\pi}{4}, \frac{3\pi}{4}, \dots \]

  Within \( 0 \leq x \leq \pi \), the zeros are at \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \).

• Maximum values: The maximum value of \( y \) is \(4\), which occurs when \( \cos 2x = \pm 1 \). These happen at:

  \[ 2x = 0, \pi, 2\pi, \dots \implies x = 0, \frac{\pi}{2}, \pi, \dots \]

  Within \( 0 \leq x \leq \pi \), the maxima occur at \( (0, 4) \), \( \left(\frac{\pi}{2}, 4\right) \), and \( (\pi, 4) \).

Step 4: Sketch the graph

Below is the sketch of the graph \( y = |4 \cos 2x| \) for \( 0 \leq x \leq \pi \). Key points are labeled on the axes.

Question 2

DO NOT USE A CALCULATOR IN THIS QUESTION.

Expand and simplify $$\left(\frac{x \sqrt{11}}{2 \sqrt{3}-1}\right)^2,$$ giving your answer with a rational denominator. [4]

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Solution

Step 1: Write the expression

We are tasked to expand and simplify the following expression:

\[ \left( \frac{x \sqrt{11}}{2 \sqrt{3} - 1} \right)^2. \]

Our goal is to express the final result with a rational denominator.

Step 2: Rationalize the denominator

First, we rationalize the denominator of the fraction:

\[ \frac{x \sqrt{11}}{2 \sqrt{3} - 1}. \]

To do this, we multiply both the numerator and denominator by the conjugate of the denominator, \(2 \sqrt{3} + 1\):

\[ \frac{x \sqrt{11}}{2 \sqrt{3} - 1} \cdot \frac{2 \sqrt{3} + 1}{2 \sqrt{3} + 1}. \]

Step 3: Simplify the denominator

The denominator becomes:

\[ (2 \sqrt{3} - 1)(2 \sqrt{3} + 1) = (2 \sqrt{3})^2 - 1^2 = 4 \cdot 3 - 1 = 12 - 1 = 11. \]

Thus, the expression simplifies to:

\[ \frac{x \sqrt{11} (2 \sqrt{3} + 1)}{11}. \]

Step 4: Expand the numerator

Now expand the numerator:

\[ x \sqrt{11} (2 \sqrt{3} + 1) = 2x \sqrt{33} + x \sqrt{11}. \]

The expression becomes:

\[ \frac{2x \sqrt{33} + x \sqrt{11}}{11}. \]

Step 5: Square the entire expression

Now square the entire fraction:

\[ \left( \frac{2x \sqrt{33} + x \sqrt{11}}{11} \right)^2. \]

First, square the numerator:

\[ (2x \sqrt{33} + x \sqrt{11})^2 = (2x \sqrt{33})^2 + 2(2x \sqrt{33})(x \sqrt{11}) + (x \sqrt{11})^2. \]

Simplify each term:

\[ (2x \sqrt{33})^2 = 4x^2 \cdot 33 = 132x^2, \]

\[ 2(2x \sqrt{33})(x \sqrt{11}) = 4x^2 \sqrt{33} \sqrt{11} = 44x^2 \sqrt{3}, \]

\[ (x \sqrt{11})^2 = x^2 \cdot 11 = 11x^2. \]

Add these terms together:

\[ 132x^2 + 44x^2 \sqrt{3} + 11x^2 = 143x^2+ 44x^2\sqrt{3}=11x^2(13+4\sqrt{3}) \]

Thus, the squared numerator is:

\[ 11x^2(13+4\sqrt{3}). \]

The denominator squared is:

\[ 11^2 = 121. \]

Step 6: Final simplified result

Combine the squared numerator and denominator:

\[ \left( \frac{2x \sqrt{33} + x \sqrt{11}}{11} \right)^2 = \frac{11x^2(13+4\sqrt{3})}{121}=\boxed{\frac{x^2(13+4\sqrt{3})}{11}}. \]