Solution
Following the process in the images provided:
The equation \( (\lambda x + \ln(x) + 1)(x + \ln(x) + 1) = x^2 \) is given. We divide both sides by \( x^2 \), assuming \( x \neq 0 \), which simplifies to:
\[ (\lambda + \frac{\ln(x) + 1}{x})(1 + \frac{\ln(x) + 1}{x}) = 1 \]
We then let \( t = \frac{\ln(x) + 1}{x} \) and consider the limits as \( x \) approaches 0 from the right and infinity to understand the behavior of \( t \):
\[ \lim_{x \to 0^+} t = \lim_{x \to 0^+} \frac{\ln(x) + 1}{x} = -\infty \]
\[ \lim_{x \to \infty} t = \lim_{x \to \infty} \frac{1/x}{1/x} = 0 \]
The derivative of \( t \) with respect to \( x \), \( \frac{dt}{dx} \), is found to be \( \frac{-\ln(x)}{x^2} \), and setting this equal to zero indicates that \( x = 1 \) is a critical point which corresponds to a local maximum on the graph of \( t \).
With \( t \) now in the equation, it is rewritten as:
\[ (\lambda + t)(1 + t) = 1 \]
\[ \lambda + \lambda t + t + t^2 = 1 \]
Let \( t_1 \) and \( t_2 \) be the roots of this quadratic in \( t \), and for exactly three values of \( x \) to satisfy the original equation, \( t \) must be in the range of \( t_1 < 0 < t_2 < 1 \) (because then for \(t_1\) there is one value of \(x\) and for \(t_2\) there are two distinct values of \(x\)). This condition ensures that the equation \( t^2 + (2\lambda + 1)t + \lambda - 1 = 0 \) has roots that lie within the valid range of \( t \) derived from \( x \).
Let \(g(x)=t^2 + (2\lambda + 1)t + \lambda - 1 \)
Considering the conditions for the roots and the quadratic formula, we conclude that \( \lambda \) must satisfy:
- \( g(0) < 0 \) which implies \( \lambda - 1 < 0 \) or \( \lambda < 1 \).
- \( g(1) > 0 \) which implies \( 2\lambda + 1 > 0 \) or \( \lambda > -\frac{1}{2} \).
- \( \frac{-\lambda + 1}{2} < 1 \) which simplifies to \( \lambda > -\frac{1}{2} \).
Combining these conditions, the range for \( \lambda \) is determined to be:
\[ -\frac{1}{2} < \lambda < 1 \]
