# C3 2016 4 Edexcel

(ai)

The point A has x-coordinate 0 so:

y = |4e^0 - 25| = 25

(aii)

The point B has y-coordinate 0 so:

|4e^{2x} - 25| = 0

So:

4e^{2x} = 25

Giving:

\displaystyle 2x = \ln \left(\frac {25} 4\right)

So:

\displaystyle x = \frac 1 2\ln \left(\frac {25} 4\right) = \ln \left(\frac 5 2\right)

(aiii)

As x \to -\infty, we have 4e^{2x} \to 0, so 4e^{2x} - 25 \to -25 and |4e^{2x} - 25| \to 25. So k = 25.

(b)

We have:

|4e^{2 \alpha} - 25| = 2 \alpha + 43

so either:

4e^{2 \alpha} - 25 = 2 \alpha + 43

or:

25 - 4e^{2 \alpha} = 2 \alpha + 43

Since \alpha > 0, we have 2 \alpha + 43 > 43. Note that since e^{2\alpha} > 0, we cannot have 25 - 4e^{2 \alpha} > 43, so we must have 4e^{2 \alpha} - 25 = 2 \alpha + 43. We then have:

4e^{2 \alpha} = 2\alpha + 68

So:

\displaystyle e^{2\alpha} = \frac \alpha 2 + 17

So:

\displaystyle \alpha = \frac 1 2 \ln \left(\frac \alpha 2 + 17\right)

as required.

We have:

\displaystyle x_1 = \frac 1 2 \ln \left(\frac 1 2 \times 1.4 + 17\right) = 1.4368 \text { (4dp)}

\displaystyle x_2 = \frac 1 2 \ln \left(\frac 1 2 \times x_1 +17\right) = 1.4373 \text { (4dp)}

(d)

We need to show that there is a root in the interval [1.4365, 1.4375). (these are the real numbers that are equal to 1.437 to 3 decimal places.

At x = 1.4365 we have:

|4e^{2x} - 25| - (2x + 43) = -0.112965\ldots

At x = 1.4375 we have:

|4e^{2x} - 25| - (2x + 43) = 0.0266965\ldots

Since |4e^{2x} - 25| - (2x + 43) is continuous and changes sign on [1.4365, 1.4375), there must be a root in this interval.

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