(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.