Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 3 β€’ Question 10

The diagram below shows two function machines A and B. Both machines have the same input. Work out the range of values for which the output of A is greater than the output of B.

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Video Transcript

The diagram below shows two function machines A and B. Both machines have the same input. Work out the range of values for which the output of A is greater than the output of B.

The first thing we can do is take the diagram for A and B and turn them into expressions. For function 𝐴, if the input is π‘₯, π‘₯ times two is the first step. We write that as two π‘₯. We add one to the two π‘₯. We need to square everything that’s happened so far. That means we put brackets around the two π‘₯ plus one and square that. The output of function 𝐴 is two π‘₯ plus one squared.

Now for function 𝐡, if the input is π‘₯, we’re using the same variable for the input of both equations because we were told that both machines have the same input. The first thing that happens in function 𝐡 is to square the π‘₯-value, π‘₯ squared plus one. And then we need to multiply π‘₯ squared plus one times four.

We put brackets around π‘₯ squared plus one to indicate that the four is being multiplied by the π‘₯ squared and the one. The output of function 𝐡: four times π‘₯ squared plus one.

We want to consider when the output of function 𝐴 is greater than the output of function 𝐡. This means we need to consider when is two π‘₯ plus one squared larger than four times π‘₯ squared plus one.

To solve this, we’ll need to expand and multiply two π‘₯ plus one squared. We need to multiply two π‘₯ plus one times itself, starting with the first term, two π‘₯ times two π‘₯ plus two π‘₯ times one. Moving on to the second term, we have one times two π‘₯ and one times one. Two π‘₯ times two π‘₯ is four π‘₯ squared. Two π‘₯ times one is itself, two π‘₯. One times two π‘₯ equals two π‘₯, and one times one equals one. Two π‘₯ and two π‘₯ are like terms. They can be combined to equal four π‘₯. The expanded form of two π‘₯ plus one squared equals four π‘₯ squared plus four π‘₯ plus one.

We can also multiply this four across the brackets. On the right-hand side, four times π‘₯ squared equals four π‘₯ squared, and four times one equals four. We have a four π‘₯ squared term on both sides of the equation. If we subtract four π‘₯ squared from both sides of the equation, those terms cancel out.

Now we have four π‘₯ plus one is greater than four. And we subtract one from both sides. Four minus one is three. Four π‘₯ must be greater than three. And then we divide both sides by four. π‘₯ is greater than three-fourths. Any time π‘₯ is larger than three-fourths, the output of function 𝐴 will be greater than the output of function 𝐡.

Let’s do a quick test to see if this is true. Let’s solve both of these equations for π‘₯ equals one. One is larger than three-fourths. So we would expect the output of 𝐴 to be greater than the output of 𝐡. Two times one is two, plus one is three. Three squared equals nine. In function 𝐡, one squared equals one, plus one equals two. Four times two equals eight. Nine is larger than eight. The output of function 𝐴 was greater than the output of function 𝐡.

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