Question

Difficulty: MediumBinary Operations

A binary operation ⋆\star defined on the set of real numbers R∖{12}\mathbb{R} \setminus \left\{\frac{1}{2}\right\} is given by a⋆b=a+b−2aba \star b = a + b - 2ab. If x−1x^{-1} represents the inverse of an element xx under ⋆\star, what is the value of xx for which x−1=3x^{-1} = 3?

  1. 35\frac{3}{5}Answer
  2. B
    25\frac{2}{5}
  3. C
    13\frac{1}{3}
  4. D
    −35-\frac{3}{5}

Answer

The value of xx is 35\frac{3}{5}.
First find the identity element ee using a⋆e=aa \star e = a, which gives a+e−2ae=a  ⟹  e(1−2a)=0  ⟹  e=0a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0. Next, use the inverse definition x⋆x−1=e  ⟹  x+x−1−2xx−1=0x \star x^{-1} = e \implies x + x^{-1} - 2x x^{-1} = 0. Substituting x−1=3x^{-1} = 3 yields x+3−2x(3)=0  ⟹  −5x+3=0  ⟹  x=35x + 3 - 2x(3) = 0 \implies -5x + 3 = 0 \implies x = \frac{3}{5}.

Step-by-Step Solution

1
Find the identity element ee under the operation ⋆\star.
e=0e = 0
By definition, a⋆e=a  ⟹  a+e−2ae=a  ⟹  e(1−2a)=0  ⟹  e=0a \star e = a \implies a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0 for all a≠12a \neq \frac{1}{2}.
2
Set up the inverse equation for element xx.
x⋆x−1=0  ⟹  x+x−1−2xx−1=0x \star x^{-1} = 0 \implies x + x^{-1} - 2x x^{-1} = 0
An element operating on its inverse yields the identity element e=0e = 0.
3
Substitute x−1=3x^{-1} = 3 into the inverse equation and solve for xx.
x+3−6x=0  ⟹  −5x+3=0  ⟹  5x=3  ⟹  x=35x + 3 - 6x = 0 \implies -5x + 3 = 0 \implies 5x = 3 \implies x = \frac{3}{5}
Substitute the given inverse value and perform simple linear algebraic rearrangement.

Key Concept

Identity and Inverse Elements in Binary Operations
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