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Official PYQ · 1996Q2.1Sets, Relations and FunctionsMCQ1 mark

GATE CS 1996 Q2.1 — Sets, Relations and Functions

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Let RR denote the set of real numbers. Let f:R×R→R×Rf:R\times R \rightarrow R \times R be a bijective function defined by f(x,y)=(x+y,x−y)f(x,y) = (x+y, x-y). The inverse function of ff is given by

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Answer: C

To find the inverse function f−1f^{-1}, we set the output of the function f(x,y)f(x, y) to a new pair of variables, say (u,v)(u, v):

f(x,y)=(x+y,x−y)=(u,v)f(x, y) = (x+y, x-y) = (u, v)

This gives us a system of two linear equations:

x+y=u— (1)x−y=v— (2)\begin{align} x + y &= u \quad \text{--- (1)} \\ x - y &= v \quad \text{--- (2)} \end{align}

To solve for xx and yy in terms of uu and vv:

  • Adding equations (1) and (2): (x+y)+(x−y)=u+v  ⟹  2x=u+v  ⟹  x=u+v2(x + y) + (x - y) = u + v \implies 2x = u + v \implies x = \frac{u + v}{2}
  • Subtracting equation (2) from (1): (x+y)−(x−y)=u−v  ⟹  2y=u−v  ⟹  y=u−v2(x + y) - (x - y) = u - v \implies 2y = u - v \implies y = \frac{u - v}{2}

Thus, the inverse function is defined as:

f−1(u,v)=(u+v2,u−v2)f^{-1}(u, v) = \left( \frac{u+v}{2}, \frac{u-v}{2} \right)

Replacing the dummy variables (u,v)(u, v) with (x,y)(x, y) gives:

f−1(x,y)=(x+y2,x−y2)f^{-1}(x, y) = \left( \frac{x+y}{2}, \frac{x-y}{2} \right)

Therefore, the correct option is C.

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