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Zorluk: ZorFunctions and Custom Symbol Operations
For all real numbers xx, the function ff is defined by f(x)=x22xf(x) = x^2 - 2x. The custom binary operation \otimes is defined for all real numbers aa and bb by ab=f(a+b)f(ab)a \otimes b = f(a + b) - f(a - b) If kk is a constant such that (k3)2=88(k \otimes 3) \otimes 2 = 88, what is the value of kk?
  1. A
    512\frac{5}{12}
  2. B
    1112\frac{11}{12}
  3. 2Cevap
  4. D
    176\frac{17}{6}
  5. E
    6

Cevap

The value of kk is 2.
Expanding the definition ab=f(a+b)f(ab)a \otimes b = f(a+b) - f(a-b) using f(x)=x22xf(x) = x^2 - 2x gives [(a+b)22(a+b)][(ab)22(ab)]=(a2+2ab+b22a2b)(a22ab+b22a+2b)=4ab4b=4b(a1)[(a+b)^2 - 2(a+b)] - [(a-b)^2 - 2(a-b)] = (a^2 + 2ab + b^2 - 2a - 2b) - (a^2 - 2ab + b^2 - 2a + 2b) = 4ab - 4b = 4b(a-1). Evaluating k3k \otimes 3 yields 12k1212k - 12. Substituting this into (12k12)2(12k - 12) \otimes 2 yields 4(2)(12k121)=96k1044(2)(12k - 12 - 1) = 96k - 104. Setting 96k104=8896k - 104 = 88 gives 96k=19296k = 192, so k=2k = 2.

Adım Adım Çözüm

1
Express the custom operation aba \otimes b in simplified algebraic terms.
ab=f(a+b)f(ab)=[(a+b)22(a+b)][(ab)22(ab)]=4ab4b=4b(a1)a \otimes b = f(a+b) - f(a-b) = [(a+b)^2 - 2(a+b)] - [(a-b)^2 - 2(a-b)] = 4ab - 4b = 4b(a - 1).
Expanding and canceling common terms simplifies the binary operation definition.
2
Evaluate the inner operation k3k \otimes 3.
k3=4(3)(k1)=12k12k \otimes 3 = 4(3)(k - 1) = 12k - 12.
Substitute a=ka = k and b=3b = 3 into the simplified operation formula 4b(a1)4b(a - 1).
3
Evaluate the outer operation (12k12)2(12k - 12) \otimes 2.
(12k12)2=4(2)[(12k12)1]=8(12k13)=96k104(12k - 12) \otimes 2 = 4(2)[(12k - 12) - 1] = 8(12k - 13) = 96k - 104.
Substitute a=12k12a = 12k - 12 and b=2b = 2 into 4b(a1)4b(a - 1).
4
Set the resulting expression equal to 88 and solve for kk.
96k104=88    96k=192    k=296k - 104 = 88 \implies 96k = 192 \implies k = 2.
Linear algebraic equation solving yields the value of kk.

Anahtar Kavram

Custom Binary Operations and Nested Function Evaluation
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