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Difficulty: Very hardFunction Definitions, Evaluation, and Custom Operators

For all non-zero real numbers aa and bb, the custom operator Δ\Delta is defined by aΔb=a2−b2aba \Delta b = \frac{a^2 - b^2}{ab}. A function f(x)f(x) is defined for all x≠0x \neq 0 and x≠−1x \neq -1 by f(x)=xΔ(x+1)f(x) = x \Delta (x + 1). If kk is a positive constant such that f(k)=−712f(k) = -\frac{7}{12}, what is the value of (k+2)Δk(k + 2) \Delta k?

  1. A
    −1615-\frac{16}{15}
  2. B
    415\frac{4}{15}
  3. C
    56\frac{5}{6}
  4. 1615\frac{16}{15}Answer
  5. E
    32\frac{3}{2}

Answer

The value of (k+2)Δk(k + 2) \Delta k is 1615\frac{16}{15}.
Applying the custom operator to f(x)f(x) yields f(x)=x2−(x+1)2x(x+1)=−2x−1x2+xf(x) = \frac{x^2 - (x+1)^2}{x(x+1)} = \frac{-2x - 1}{x^2 + x}. Setting f(k)=−712f(k) = -\frac{7}{12} gives 2k+1k2+k=712\frac{2k + 1}{k^2 + k} = \frac{7}{12}, which simplifies to 7k2−17k−12=07k^2 - 17k - 12 = 0. Factoring (7k+4)(k−3)=0(7k + 4)(k - 3) = 0 and selecting the positive root gives k=3k = 3. Evaluating (3+2)Δ3=5Δ3=52−325⋅3=1615(3+2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \cdot 3} = \frac{16}{15}.

Step-by-Step Solution

1
Express f(x)f(x) using the definition of the custom operator Δ\Delta.
f(x)=xΔ(x+1)=x2−(x+1)2x(x+1)=x2−(x2+2x+1)x(x+1)=−2x−1x2+xf(x) = x \Delta (x + 1) = \frac{x^2 - (x + 1)^2}{x(x + 1)} = \frac{x^2 - (x^2 + 2x + 1)}{x(x + 1)} = \frac{-2x - 1}{x^2 + x}
Substitute a=xa = x and b=x+1b = x + 1 into aΔb=a2−b2aba \Delta b = \frac{a^2 - b^2}{ab}.
2
Set f(k)=−712f(k) = -\frac{7}{12} and solve for the positive constant kk.
−2k−1k2+k=−712  ⟹  2k+1k2+k=712  ⟹  12(2k+1)=7(k2+k)  ⟹  7k2−17k−12=0\frac{-2k - 1}{k^2 + k} = -\frac{7}{12} \implies \frac{2k + 1}{k^2 + k} = \frac{7}{12} \implies 12(2k + 1) = 7(k^2 + k) \implies 7k^2 - 17k - 12 = 0
Cross-multiply and rearrange into a standard quadratic equation.
3
Factor the quadratic equation to find the positive root kk.
(7k+4)(k−3)=0  ⟹  k=−47(7k + 4)(k - 3) = 0 \implies k = -\frac{4}{7} or k=3k = 3. Since k>0k > 0, k=3k = 3.
The question specifies that kk is a positive constant.
4
Evaluate (k+2)Δk(k + 2) \Delta k for k=3k = 3.
(3+2)Δ3=5Δ3=52−325×3=25−915=1615(3 + 2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \times 3} = \frac{25 - 9}{15} = \frac{16}{15}
Substitute a=5a = 5 and b=3b = 3 into the custom operator definition.

Key Concept

Function definition substitution, non-commutative custom operators, and quadratic solving.
Estimated Time:2m 30s
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Topics

Function Definitions, Evaluation, and Custom Operators

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