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2024

Problems of the Day (29 May 2024)

  1. There are three dice A, B, and C, each with six faces. Die A has three faces marked with 2, two faces with 4, and one face marked with 6. Die B has one face marked with 1, three faces marked with 2, and two faces marked with 3. Die C has two faces marked with 1 and four faces marked with 2. When the three dice are thrown randomly once, let \( E \) be the event of getting a sum of the numbers appearing on top faces equal to 8. Find \(P(E)\).

    Solution:

    \( n(E) = \text{coefficient of } x^8 \text{ in } \)

    \( (3x^2 + 2x^4 + x^6)(x + 3x^2 + 2x^3)(2x + 4x^2) \)

    \( = \text{coefficient of } x^4 \text{ in } \)

    \( (3 + 2x^2 + x^4)(1 + 3x + 2x^2)(2 + 4x) \)

    \( = \text{coefficient of } x^4 \text{ in } \)

    \( (3 + 2x^2 + x^4)(2 + 10x + 16x^2 + 8x^3) \)

    \( = 2 \times 16 \)

    \( = 32 \)

    \( \therefore P(E) = \frac{32}{6^3} = \frac{4}{27} \)

  2. Let \( f(x) = \frac{\sin x + \sin 3x + \sin 5x + \sin 7x}{\cos x + \cos 3x + \cos 5x + \cos 7x} \)

    Find the fundamental period of f.

    Solution:

    \( f(1) = \frac{(\sin x + \sin 7x) + (\sin 3x + \sin 5x)}{(\cos x + \cos 7x) + (\cos 3x + \cos 5x)} \)

    \( = \frac{2 \sin 4x \cos 3x + 2 \sin 4x \cos x}{2 \cos 4x \cos 3x + 2 \cos 4x \cos x} \)

    \( = \frac{2 \sin 4x (\cos 3x + \cos x)}{2 \cos 4x (\cos 3x + \cos x)} \)

    \( = \tan 4x\frac{\cos 2x \cos x}{\cos 2x \cos x} \)

    So the given function is essentially equal to \( \tan 4x \) only. It looks like that the fundamental period is \( \frac{\pi}{4} \). When we look closely we find out that whenever \( \cos 2x \cos x = 0 \) then \( f \) is not defined. This results in a fundamental period \( \pi \) as can be seen on the graph.

  3. Let \( A \) be a square matrix of order 3 satisfies the matrix equation \( A^3 - 6A^2 + 7A - 81I = 0 \). The the value of \( \text{det}(\text{adj}(I - 2A^{-1})) \) is equal to ....

    Solution:

    Observe that \( \text{adj}(I - 2A^{-1}) \) = \( |I - 2A^{-1}|^2 \)

    \(= \frac{|A - 2I|^2}{|A|^2} \)

    The characteristic equation of \( A \) is given as \( \lambda^3 - 6\lambda^2 + 7\lambda - 8 = 0 \)

    \( \Rightarrow |A| = -(-8) = 8 \)

    \( A^3 - 6A^2 + 7A - 8I = 0 \)

    \( \Rightarrow (A - 2I)(A^2 - 4A - I) - 10I = 0 \)

    \( \Rightarrow (A - 2I) \left((A - 2I)^2 - 5I\right) - 10I = 0 \)

    \( \Rightarrow X(X^2 - 5I) - 10I = 0 \)

    Where \( X = A - 2I \)

    \( \Rightarrow X^3 - 5X - 10I = 0 \)

    Which is the characteristic eqn of \( X \)

    Thus \( |X| = |A - 2I| = -10 \)

    \( \therefore \text{det}(\text{adj}(I - 2A^{-1})) = \frac{|A - 2I|^2}{|A|^2} = \frac{100}{64} = \frac{25}{16} \)

The Importance of Honesty in Mathematical Problem-Solving 🧮

In the journey of solving mathematical problems, honesty plays a pivotal role, often more than we initially appreciate. Whether you're a student facing a tricky calculus problem, a teacher trying to impart clear concepts, or a professional using statistics to make decisions, the integrity with which one approaches the problem-solving process can significantly impact the results and, importantly, the understanding of the subject matter.

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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:

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

Circle of Apollonius

The Circle of Apollonius is defined as the locus of a point \(P\) which moves in such a way that the ratio of its distances from two fixed points, \(A\) and \(B\), is a constant \(\lambda\), with \(\lambda \neq 1\). This geometric construct is named after the ancient Greek geometer Apollonius of Perga, who extensively studied these circles among other conic sections.

How to plot a region correctly

In JEE Mains and Advanced problems, especially when it comes to calculating areas bounded by curves, one of the initial challenges students often face is correctly plotting the region in question. This can be particularly true for complex shapes or when dealing with inequalities that define regions not immediately intuitive to visualize. If you find yourself grappling with how to plot these regions effectively, you're not alone. The following discussion aims to shed light on a systematic approach that can simplify this process, making it not just manageable but also insightful.