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JEE Advanced Problems

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

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

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:

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