Symbols & Functions: Level 3
1.
Let \(f(x) = x^2 – 5\). What is the value of \(f(f(f(1)))\)?
Solution (C):
1. Start from the inside: \(f(1) = (1)^2 – 5 = 1 – 5 = -4\).
2. Use this result: \(f(f(1)) = f(-4) = (-4)^2 – 5 = 16 – 5 = 11\).
3. Use this result: \(f(f(f(1))) = f(11) = (11)^2 – 5 = 121 – 5 = 116\).
2.
Let \( \nabla x = 3x – 3 \). If \( \nabla(\nabla k) = 15 \), what is the value of \(k\)?
Solution (A):
1. Let \(y = \nabla k\). The equation becomes \( \nabla y = 15 \).
2. Use the definition: \( 3y – 3 = 15 \).
3. \( 3y = 18 \implies y = 6 \).
4. Now we know \( \nabla k = 6 \).
5. Use the definition again: \( 3k – 3 = 6 \).
6. \( 3k = 9 \implies k = 3 \).
3.
Let \(x\#y = ax + by\). If \(1\#2 = 7\) and \(2\#1 = 8\), what is the value of \(5\#5\)?
Solution (E): This is a system of two linear equations.
1. \(1\#2 = 7 \implies a(1) + b(2) = 7 \implies a + 2b = 7 \).
2. \(2\#1 = 8 \implies a(2) + b(1) = 8 \implies 2a + b = 8 \).
3. Multiply the first equation by 2: \(2a + 4b = 14\).
4. Subtract the second equation from it: \((2a + 4b) – (2a + b) = 14 – 8 \implies 3b = 6 \implies b = 2\).
5. Substitute \(b=2\) into \(a + 2b = 7\): \(a + 2(2) = 7 \implies a + 4 = 7 \implies a = 3\).
6. The function is \(x\#y = 3x + 2y\).
7. \(5\#5 = 3(5) + 2(5) = 15 + 10 = 25\).
4.
Let \(x \text{ Ω } y = (x-y)^2\). If \((k \text{ Ω } 5) \text{ Ω } 2 = 49\), which of the following could be \(k\)?
Solution (D):
1. Let \(z = (k \text{ Ω } 5) = (k-5)^2\). The equation becomes \(z \text{ Ω } 2 = 49\).
2. Use the definition: \((z – 2)^2 = 49\).
3. Take the square root: \(z – 2 = 7\) or \(z – 2 = -7\).
4. Case 1: \(z – 2 = 7 \implies z = 9\).
5. Case 2: \(z – 2 = -7 \implies z = -5\).
6. Substitute back \(z = (k-5)^2\).
7. \( (k-5)^2 = 9 \implies k-5 = 3\) or \(k-5 = -3\). This gives \(k=8\) or \(k=2\).
8. \( (k-5)^2 = -5 \). A real square cannot be negative.
9. The possible values for \(k\) are 8 and 2. 8 is an option.
5.
If \( p \text{ & } q = p^2 + q^2 – 2pq \), for what non-negative value of \(q\) is \( p \text{ & } q = p^2 \) for all values of \(p\)?
Solution (C):
1. Set the two sides of the equation equal: \( p^2 + q^2 – 2pq = p^2 \).
2. Subtract \(p^2\) from both sides: \( q^2 – 2pq = 0 \).
3. Factor out \(q\): \( q(q – 2p) = 0 \).
4. This must be true for *all* values of \(p\).
5. If we choose \(q=2p\), it is not true for *all* \(p\) (e.g., if \(p=1\), \(q=2\), but if \(p=3\), \(q=6\)).
6. If we choose \(q=0\), the equation becomes \( 0(0 – 2p) = 0 \), which is \(0=0\). This is true for all values of \(p\).
6.
For a three-digit number `abc`, let \(f(abc) = 2^a \times 3^b \times 5^c\). What is the value of \( \frac{f(123)}{f(101)} \)?
Solution (D):
1. Evaluate the numerator, \(f(123)\):
\(a=1, b=2, c=3 \implies 2^1 \times 3^2 \times 5^3 = 2 \times 9 \times 125 = 18 \times 125 = 2250\).
2. Evaluate the denominator, \(f(101)\):
\(a=1, b=0, c=1 \implies 2^1 \times 3^0 \times 5^1 = 2 \times 1 \times 5 = 10\).
3. Calculate the ratio: \( \frac{2250}{10} = 225 \).
7.
For a 4-digit number \(abcd\), let \(f(abcd) = 3^a 5^b 7^c 11^d\). If \(m\) and \(n\) are 4-digit numbers and \(f(m) = 3 \times f(n)\), what is \(m-n\)?
Solution (A):
1. Let \(m = abcd\) and \(n = pqrs\).
2. \(f(m) = 3^a 5^b 7^c 11^d\).
3. \(f(n) = 3^p 5^q 7^r 11^s\).
4. \(f(m) = 3 \times f(n) \implies 3^a 5^b 7^c 11^d = 3 \times (3^p 5^q 7^r 11^s) = 3^{p+1} 5^q 7^r 11^s\).
5. By comparing exponents: \(a = p+1\), \(b = q\), \(c = r\), \(d = s\).
6. We need \(m-n = (1000a+100b+10c+d) – (1000p+100q+10r+s)\).
7. \( = 1000(a-p) + 100(b-q) + 10(c-r) + (d-s) \).
8. \( = 1000(1) + 100(0) + 10(0) + 0 = 1000 \).
8.
Let \(x \text{ ♦ } y = x^2 – y^2\). What is the value of \((10 \text{ ♦ } 9) \text{ ♦ } 8\)?
Solution (C):
1. First, evaluate the parenthesis: \((10 \text{ ♦ } 9)\).
2. \(10 \text{ ♦ } 9 = 10^2 – 9^2 = 100 – 81 = 19\).
3. Now substitute this result back: \(19 \text{ ♦ } 8\).
4. \(19 \text{ ♦ } 8 = 19^2 – 8^2 = 361 – 64 = 297\).
9.
Let \(x \text{ Ω } y = (x-y)^2\). If \(x \text{ Ω } (x-y) = 25\), what is \(y^2\)?
Solution (B):
1. We are given the equation \(x \text{ Ω } (x-y) = 25\).
2. Apply the definition of Ω: Let the first term be \(a=x\) and the second term be \(b=(x-y)\).
3. \( (a – b)^2 = 25 \).
4. Substitute back: \( (x – (x-y))^2 = 25 \).
5. Simplify the parenthesis: \( (x – x + y)^2 = 25 \).
6. \( (y)^2 = 25 \). The value of \(y^2\) is 25.
10.
For any four-digit number \(abcd\), let \(*abcd* = a^b + c^d\). What is \(*1234* + *2143*\)?
Solution (A):
1. Evaluate \(*1234*\): \(a=1, b=2, c=3, d=4\).
\(*1234* = 1^2 + 3^4 = 1 + 81 = 82\).
2. Evaluate \(*2143*\): \(a=2, b=1, c=4, d=3\).
\(*2143* = 2^1 + 4^3 = 2 + 64 = 66\).
3. Sum the results: \(82 + 66 = 148\).
Score: 0 / 10