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Problems of the Day 2

  1. Let the number of ways in which 3 positive integers \( f_1, f_2, f_3 \) can be selected such that \( f_1f_2f_3 = 2310 \) is \( \lambda \), then if \( f_1 < f_2 < f_3 \), then value of \( \lambda \) is ____.

    Solution:

    To solve for the number of ways to select positive integers \( f_1, f_2, f_3 \) such that \( f_1f_2f_3 = 2310 \) and \( f_1 < f_2 < f_3 \), first factorize 2310 to get its prime factors:

    \[ 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \]

    This factorization is analogous to distributing 5 distinct objects, corresponding to the primes 2, 3, 5, 7, 11, into three distinct boxes representing \( f_1, f_2, f_3 \). An empty box would indicate that it contains the number 1.

    Since \( f_1 < f_2 < f_3 \), and all three must be distinct, this is ensured by ensuring that:

    a. No box is empty.

    b. Exactly one box is empty.

    Thus, these 5 objects (the prime factors) can be distributed into 3 boxes according to the following distributions:

    1. \( 3, 1, 1 \) distribution (and its permutations 1, 3, 1 and 1, 1, 3)
    2. \( 2, 2, 1 \) distribution
    3. \( 4, 1, 0 \) distribution

    The total number of ways is then calculated by:

    1. For the \( 3, 1, 1 \) distribution, the total number of ways to distribute the objects is given by the multinomial coefficient divided by the arrangement number to account for the order of \( f_1 < f_2 < f_3 \):

      \[ \frac{5!}{3!1!1!} \times \frac{3!}{2!} \times \frac{1}{3!} = 10 \]

    2. For the \( 2, 2, 1 \) distribution, the number of ways is:

      \[ \frac{5!}{2!2!1!} \times \frac{3!}{2!} \times \frac{1}{3!} = 15 \]

    3. For the \( 4, 1, 0 \) distribution, the number of ways is:

      \[ \frac{5!}{4!1!} \times 3! \times \frac{1}{3!} = 5 \]

    4. For the \( 3, 2, 0 \) distribution, the number of ways is:

      \[ \frac{5!}{3!2!} \times 3! \times \frac{1}{3!} = 10 \]

    Adding all the valid distributions together gives us the total number of distributions for which \( f_1 < f_2 < f_3 \):

    \[ 10 + 15 + 5 + 10= 40 \]

    Hence, the number of ways to choose \( f_1, f_2, f_3 \) such that they are distinct and their product is 2310 is 40.

  2. If the equation \( (\lambda x + \ln x + 1)(x + \ln x+1) = x^2 \) has exactly three solutions, then determine the range of \( \lambda \).

    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 \( t \).

    Graph of t vs x

    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:

    1. \( g(0) < 0 \) which implies \( \lambda - 1 < 0 \) or \( \lambda < 1 \).
    2. \( g(1) > 0 \) which implies \( 2\lambda + 1 > 0 \) or \( \lambda > -\frac{1}{2} \).
    3. \( \frac{-\lambda + 1}{2} < 1 \) which simplifies to \( \lambda > -\frac{1}{2} \).

    Combining these conditions, the range for \( \lambda \) is determined to be:

    \[ -\frac{1}{2} < \lambda < 1 \]