The curves $y = e^{-2x}$ and $y = e^{x} - rac{1}{4}$ intersect at exactly one point as shown in the diagram. The point of intersection has coordinates $\left( \ln{2}, \frac{1}{4} \right)$. (Do NOT prove this.) (a) Show that the area bounded by the two curves and the $y$-axis, as shaded in the diagram, is $ \frac{1}{4} \ln{2} - \frac{1}{8} $. (b) Find the values of $k$ such that the curves $y = e^{-2x}$ and $y = e^{x} + k$ intersect at two points.

SSCE HSC Mathematics Advanced Absolute value functions: The curves $y = e^{-2x}$ and $y

The curves $y = e^{-2x}$ and $y = e^{x} - rac{1}{4}$ intersect at exactly one point as shown in the diagram - HSC - SSCE Mathematics Advanced - Question 32 - 2023 - Paper 1

Question 32

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The curves $y = e^{-2x}$ and $y = e^{x} - rac{1}{4}$ intersect at exactly one point as shown in the diagram. The point of intersection has coordinates $\left( \ln{2... show full transcript

Worked Solution & Example Answer:The curves $y = e^{-2x}$ and $y = e^{x} - rac{1}{4}$ intersect at exactly one point as shown in the diagram - HSC - SSCE Mathematics Advanced - Question 32 - 2023 - Paper 1

Step 1

Show that the area bounded by the two curves and the y-axis is $ \frac{1}{4} \ln{2} - \frac{1}{8} $.

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Answer

To find the shaded area between the curves, we will use the definite integral from $0$ to $\ln{2}$ :

$\text{Shaded Area} = \int_{0}^{\ln{2}} \left( e^{x} - \left( e^{-2x} + \frac{1}{4} \right) \right) dx$

Calculating the integral:

$= \int_{0}^{\ln{2}} \left( e^{x} - e^{-2x} - \frac{1}{4} \right) dx$

$= \left[ e^{x} + \frac{1}{2} e^{-2x} - \frac{1}{4}x \right]_{0}^{\ln{2}}$

Now, substituting the limits:

At $x = \ln{2}$ $x = ln 2$ :
- $e^{\ln{2}} = 2$
- $e^{-2 \ln{2}} = \frac{1}{4}$
- Substitute into the integral:

$= \left( 2 + \frac{1}{2} \cdot \frac{1}{4} - \frac{1}{4} \cdot \ln{2} \right)$

At $x = 0$ $x = 0$ :
- $e^{0} = 1$
- $e^{-0} = 1$
- Substitute into the integral:

$= \left( 1 + \frac{1}{2} - 0 \right)$

Combining both results:

$\text{Area} = \left( 2 + \frac{1}{8} - \frac{1}{4} \ln{2} \right) - \left( 1 + \frac{1}{2} \right) \implies \frac{1}{4} \ln{2} - \frac{1}{8}$

Step 2

Find the values of k such that the curves $y = e^{-2x}$ and $y = e^{x} + k$ intersect at two points.

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Answer

To find the values of $k$ , we set the equations equal:

$e^{-2x} = e^{x} + k$

Rearranging gives:

$e^{-2x} - e^{x} - k = 0$

Letting $y = e^{x}$ , we can rewrite this as:

$\frac{1}{y^{2}} - y - k = 0$

Multiplying through by $y^{2}$ yields:

$1 - y^{3} - ky^{2} = 0$

This is a cubic equation in $y$ , and for the equation to have two solutions, the discriminant must be greater than zero:

The discriminant $D > 0$ ensures there are multiple roots.
We also note that $1 + 4k > 0$ for the roots to be real. Therefore, we derive:

$k > -\frac{1}{4}$

Given the nature of the equations, we analyze the cubic nature and find that:

For two real solutions, we need $k < 1$ as well. Thus, the final answer is:

$-\frac{1}{4} < k < 1$

The curves $y = e^{-2x}$ and $y = e^{x} - rac{1}{4}$ intersect at exactly one point as shown in the diagram - HSC - SSCE Mathematics Advanced - Question 32 - 2023 - Paper 1

Worked Solution & Example Answer:The curves $y = e^{-2x}$ and $y = e^{x} - rac{1}{4}$ intersect at exactly one point as shown in the diagram - HSC - SSCE Mathematics Advanced - Question 32 - 2023 - Paper 1

Show that the area bounded by the two curves and the y-axis is \( \frac{1}{4} \ln{2} - \frac{1}{8} \).

Find the values of k such that the curves $y = e^{-2x}$ and $y = e^{x} + k$ intersect at two points.

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