Bunuel wrote:
ajit257 wrote:
The operation ⊗ is defined for all nonzero numbers a and b by a ⊗ b = a/b – b/a. If x and y are nonzero numbers, which of the following statements must be true?
I. x ⊗ xy = x(1 ⊗ y)
II. x ⊗ y = -(y ⊗ x)
III. 1/x ⊗ 1/y = y ⊗ x
A. I only
B. II only
C. III only
D. I and II
E. II and III
Please can someone confirm the ans. I am not sure about the ans
Given: \(a@b=\frac{a}{b}-\frac{b}{a}\), for all nonzero numbers \(a\) and \(b\).
I. \(x@(xy) = x(1@y)\): \(LHS=x@(xy)=\frac{x}{xy}-\frac{xy}{x}=\frac{1}{y}-y\) and \(RHSx(1@y)=x(\frac{1}{y}-y)\) as you see LHS doen't equal to RHS;
II. \(x@y = -(y@x)\): \(LHS=x@y=\frac{x}{y}-\frac{y}{x}\) and \(RHS=-(y@x)=-(\frac{y}{x}-\frac{x}{y})=\frac{x}{y}-\frac{y}{x}\) --> LHS=RHS;
III. \((\frac{1}{x})@(\frac{1}{y}) = y@x\): \(LHS=(\frac{1}{x})@(\frac{1}{y})=\frac{y}{x}-\frac{x}{y}\) and \(RHS=y@x=\frac{y}{x}-\frac{x}{y}\) --> LHS=RHS.
Answer: E (II and III).
Hope it's clear.
Experts, can you use smart numbers to solve this? I set a = 2 and b=2 and then believed the unidentified operation was subtraction based on what made the left-hand side = right-hand side. However, then choices II and III did not work based on this. Are there any other ways to solve this then if you found the method above time consuming? Do you have to work from the left-hand side to right-hand side?
Also, for I. when you factor out the x on the left-hand side to arrive at (1/y - y), why does it just disappear? I did x(1/y - y)=x(1/y -y) and thought they were equal, but the expert above seems to do (1/y-y)=x(1/y - y).
Thank you in advance for your time.
RonTargetTestPrep