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 ab=a+b2aba \star b = a + b - 2ab. If x1x^{-1} represents the inverse of an element xx under \star, what is the value of xx for which x1=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 ae=aa \star e = a, which gives a+e2ae=a    e(12a)=0    e=0a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0. Next, use the inverse definition xx1=e    x+x12xx1=0x \star x^{-1} = e \implies x + x^{-1} - 2x x^{-1} = 0. Substituting x1=3x^{-1} = 3 yields x+32x(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, ae=a    a+e2ae=a    e(12a)=0    e=0a \star e = a \implies a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0 for all a12a \neq \frac{1}{2}.
2
Set up the inverse equation for element xx.
xx1=0    x+x12xx1=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 x1=3x^{-1} = 3 into the inverse equation and solve for xx.
x+36x=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
Rate this question