Given that $f(x) = 2e^{x} - 5, \quad x \in \mathbb{R}$ (a) sketch, on separate diagrams, the curve with equation (i) $y = f(x)$ (ii) $y = |f(x)|$ On each diagram, show the coordinates of each point at which the curve meets or cuts the axes - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 3

For $y = |f(x)|$:

1. Determine intercepts:

   • Use the intercepts from part (i): $(\ln(\frac{5}{2}), 0)$ and $(0, -3)$ will reflect to $(0, 3)$ for the positive part.

2. Sketch curve:

   • The part of the curve below the x-axis will be reflected above the x-axis. The point $(0, -3)$ becomes $(0, 3)$.
   • The curve will approach $y = 5$ as $x \to -\infty$.
   • The asymptote remains at $y = 5$.

To find $x$ such that $f(x) = |f(y)|$:

1. Since $f(y)$ can be both negative and positive, check conditions:

   • When $f(x) \geq 0$, we have $f(x) = f(y)$; this occurs when $x \geq \ln(\frac{5}{2})$.
   • When $f(x) < 0$, this equality won't hold since it cannot equal its own negative. Hence take only the non-negative portion of $f(x)$.

2. Thus, the solution set is: $x \geq \ln\left(\frac{5}{2}\right)$.

To solve $|f(x)| = 2$:

1. Case 1: $f(x) = 2$:
   • Set $2e^x - 5 = 2$: 2e^x &= 7 \ e^x &= \frac{7}{2} \ x &= \ln\left(\frac{7}{2}\right) \end{align*}$$
2. Case 2: $f(x) = -2$:
   • Set $2e^x - 5 = -2$: 2e^x &= 3 \ e^x &= \frac{3}{2} \ x &= \ln\left(\frac{3}{2}\right) \end{align*}$$
3. Therefore, the solutions are: $x = \ln\left(\frac{7}{2}\right) \text{ and } x = \ln\left(\frac{3}{2}\right)$.