Math Olympiad Problems for Grades 11 and 12: Problems with Solutions

Solution:
First, find the value of the inner function, [tex]f(2)[/tex].
Substitute [tex]x = 2[/tex] into the given function: [tex]f(2) = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3[/tex].
Now, substitute this result back into the function to find [tex]f(f(2))[/tex], which is [tex]f(3)[/tex]: [tex]f(3) = 2(3)^2 - 3(3) + 1 = 18 - 9 + 1 = 10[/tex].

Solution:
The given equation is [tex]\log_2(x) + \log_2(x - 2) = 3[/tex].
Using the product property of logarithms, combine the left side: [tex]\log_2(x(x - 2)) = 3[/tex].
Rewrite the logarithmic equation in exponential form: [tex]x(x - 2) = 2^3 \Rightarrow x^2 - 2x = 8[/tex].
Bring all terms to one side to form a quadratic equation: [tex]x^2 - 2x - 8 = 0[/tex].
Factor the quadratic equation: [tex](x - 4)(x + 2) = 0[/tex].
This gives two possible solutions: [tex]x = 4[/tex] or [tex]x = -2[/tex].
Since the problem states that [tex]x > 2[/tex], [tex]x = -2[/tex] must be rejected.
The fuel input level is [tex]x = 4[/tex].

Solution:
According to the Polynomial Remainder Theorem, if a polynomial [tex]P(x)[/tex] is divided by [tex](x - c)[/tex], the remainder is exactly [tex]P(c)[/tex].
In this case, dividing by [tex](x - 2)[/tex] means [tex]c = 2[/tex].
The theorem states that [tex]P(2) = 3[/tex].
Substitute [tex]x = 2[/tex] into the polynomial and set it equal to [tex]3[/tex]: [tex]P(2) = (2)^3 - k(2)^2 + 4(2) - 5 = 3[/tex]
[tex]\Rightarrow 8 - 4k + 8 - 5 = 3[/tex]
[tex]\Rightarrow 11 - 4k = 3[/tex]
[tex]\Rightarrow 4k = 8[/tex]
[tex]\Rightarrow k = 2[/tex].

Solution:
Let [tex]a[/tex] be the first term and [tex]d[/tex] be the common difference.
The formula for the [tex]n[/tex]-th term is [tex]a_n = a + (n - 1)d[/tex].
We are given the 3rd term ([tex]a_3[/tex]) and 7th term ([tex]a_7[/tex]): Equation 1: [tex]a + 2d = 14[/tex].
Equation 2: [tex]a + 6d = 26[/tex].
Subtract Equation 1 from Equation 2: [tex]4d = 12 \Rightarrow d = 3[/tex].
Substitute [tex]d = 3[/tex] back into Equation 1: [tex]a + 2(3) = 14 \Rightarrow a = 8[/tex].
To find the 10th minute ([tex]a_{10}[/tex]): [tex]a_{10} = a + 9d = 8 + 9(3) = 35[/tex].

Solution:
A 3-digit number is formed from the set [tex]\{1, 2, 3, 4, 5\}[/tex] without repetition.
For the number to be even, it must end in an even digit.
Step 1: Choose the Units digit.
From the set, the only even digits are [tex]2[/tex] and [tex]4[/tex] ([tex]2[/tex] options).
Step 2: Choose the Hundreds digit.
Out of [tex]5[/tex] digits, [tex]1[/tex] is used for the Units, leaving [tex]4[/tex] options.
Step 3: Choose the Tens digit.
Out of [tex]5[/tex] digits, [tex]2[/tex] are used, leaving [tex]3[/tex] options.
Total combinations [tex]= 4 \times 3 \times 2 = 24[/tex].