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Difficulty: HardBinary Operations

A binary operation ⊙\odot defined on the set of real numbers R\mathbb{R} is given by a⊙b=a+b+kaba \odot b = a + b + kab, where kk is a non-zero constant. If the inverse of 22 under ⊙\odot is −4-4, what is the value of (3⊙1)−1(3 \odot 1)^{-1}?

  1. −523-\frac{52}{3}Answer
  2. B
    134\frac{13}{4}
  3. C
    −285-\frac{28}{5}
  4. D
    −133-\frac{13}{3}

Answer

The value of (3⊙1)−1(3 \odot 1)^{-1} is −523-\frac{52}{3}.
First, the identity element is determined by solving a⊙e=aa \odot e = a, which gives a+e+kae=a  ⟹  e=0a + e + kae = a \implies e = 0. Next, using the inverse property x⊙x−1=0x \odot x^{-1} = 0, we get x−1=−x1+kxx^{-1} = \frac{-x}{1 + kx}. Given 2−1=−42^{-1} = -4, substituting gives −21+2k=−4\frac{-2}{1 + 2k} = -4, leading to k=−14k = -\frac{1}{4}. Evaluating 3⊙13 \odot 1 yields 3+1−34=1343 + 1 - \frac{3}{4} = \frac{13}{4}. Finally, applying the inverse formula to 134\frac{13}{4} gives −1341−1316=−523\frac{-\frac{13}{4}}{1 - \frac{13}{16}} = -\frac{52}{3}.

Step-by-Step Solution

1
Find the identity element ee under the operation ⊙\odot
e=0e = 0
By definition of identity element, a⊙e=a  ⟹  a+e+kae=a  ⟹  e(1+ka)=0  ⟹  e=0a \odot e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0 for all real numbers aa.
2
Derive the formula for the inverse x−1x^{-1} of an element xx
x−1=−x1+kxx^{-1} = \frac{-x}{1 + kx}
By definition of inverse, x⊙x−1=e  ⟹  x+x−1+kxx−1=0  ⟹  x−1(1+kx)=−x  ⟹  x−1=−x1+kxx \odot x^{-1} = e \implies x + x^{-1} + kxx^{-1} = 0 \implies x^{-1}(1 + kx) = -x \implies x^{-1} = \frac{-x}{1 + kx}.
3
Use the given inverse condition 2−1=−42^{-1} = -4 to find the constant kk
k=−14k = -\frac{1}{4}
Substituting x=2x = 2 into the inverse formula gives −21+2k=−4  ⟹  −2=−4(1+2k)  ⟹  −2=−4−8k  ⟹  8k=−2  ⟹  k=−14\frac{-2}{1 + 2k} = -4 \implies -2 = -4(1 + 2k) \implies -2 = -4 - 8k \implies 8k = -2 \implies k = -\frac{1}{4}.
4
Evaluate the operation 3⊙13 \odot 1
3⊙1=1343 \odot 1 = \frac{13}{4}
Using the operation definition with k=−14k = -\frac{1}{4}: 3⊙1=3+1+(−14)(3)(1)=4−34=1343 \odot 1 = 3 + 1 + \left(-\frac{1}{4}\right)(3)(1) = 4 - \frac{3}{4} = \frac{13}{4}.
5
Calculate the inverse of 134\frac{13}{4} under ⊙\odot
−523-\frac{52}{3}
Using the inverse formula y−1=−y1+kyy^{-1} = \frac{-y}{1 + ky} for y=134y = \frac{13}{4}: y−1=−1341+(−14)(134)=−1341−1316=−134316=−134×163=−523y^{-1} = \frac{-\frac{13}{4}}{1 + \left(-\frac{1}{4}\right)\left(\frac{13}{4}\right)} = \frac{-\frac{13}{4}}{1 - \frac{13}{16}} = \frac{-\frac{13}{4}}{\frac{3}{16}} = -\frac{13}{4} \times \frac{16}{3} = -\frac{52}{3}.

Key Concept

Binary Operations: Finding Identity Elements, Unknown Parameters, and Inverse Elements
Estimated Time:2m 0s
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Topics

Binary Operations

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