Question

The number of real solutions of the equation \(1 + |e^x - 1| = e^x(e^x - 2)\) is

• A. 1
• B. 2
• C. 4
• D. 8

Hint:

Check both cases for absolute value equations.

Answer 1

The Correct Option is A

To solve the equation \( 1 + |e^x - 1| = e^x(e^x - 2) \), we need to consider the nature of the absolute value function involved. The approach involves examining different cases for the absolute value expression and solving within those cases.

1.  First, note that the absolute value \(|e^x - 1|\) results in two cases:
   • Case 1: \(e^x - 1 \geq 0\) → \( |e^x - 1| = e^x - 1 \)
   • Case 2: \(e^x - 1 < 0\) → \( |e^x - 1| = 1 - e^x \)
2. For Case 1 (\(e^x \geq 1\)):
   1. Substitute \(|e^x - 1| = e^x - 1\) into the equation: \(1 + (e^x - 1) = e^x(e^x - 2)\)
   2. Simplifying, we get: \(e^x = e^{2x} - 2e^x\)
   3. This rearranges to: \(e^{2x} - 3e^x = 0\)
   4. Factorize the equation: \(e^x(e^x - 3) = 0\)
   5. This gives us the solutions: \(e^x = 0 \quad \text{or} \quad e^x = 3\)
   6. Since \(e^x > 0\) always, \(e^x = 0\) is not possible, so: \(e^x = 3 \Rightarrow x = \ln 3\)
3. For Case 2 (\(e^x < 1\)):
   1. Substitute \(|e^x - 1| = 1 - e^x\): \(1 + (1 - e^x) = e^x(e^x - 2)\)
   2. This simplifies to: \(2 - e^x = e^{2x} - 2e^x\)
   3. Rearranging gives: \(e^{2x} - e^x - 2 = 0\)
   4. Let \(y = e^x\), then the equation becomes: \(y^2 - y - 2 = 0\)
   5. Factor the quadratic equation: \((y - 2)(y + 1) = 0\)
   6. This gives solutions: \(y = 2 \quad \text{or} \quad y = -1\)
   7. Since \(y = e^x > 0\), \(y = -1\) is not valid. Hence, \(y = 2\).
   8. However, since we assumed \(e^x < 1\) in this case, \(y = 2\) does not fit, hence no solutions here.

We conclude that the only valid solution from both cases is \(x = \ln 3\). Therefore, the number of real solutions is:

1