Problems of the Day (29 May 2024)

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:
\( n(E) = \text{coefficient of } x^8 \text{ in } \)
\( (3x^2 + 2x^4 + x^6)(x + 3x^2 + 2x^3)(2x + 4x^2) \)
\( = \text{coefficient of } x^4 \text{ in } \)
\( (3 + 2x^2 + x^4)(1 + 3x + 2x^2)(2 + 4x) \)
\( = \text{coefficient of } x^4 \text{ in } \)
\( (3 + 2x^2 + x^4)(2 + 10x + 16x^2 + 8x^3) \)
\( = 2 \times 16 \)
\( = 32 \)
\( \therefore P(E) = \frac{32}{6^3} = \frac{4}{27} \)

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.

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 \( X = A  2I = 10 \)
\( \therefore \text{det}(\text{adj}(I  2A^{1})) = \frac{A  2I^2}{A^2} = \frac{100}{64} = \frac{25}{16} \)