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Real-valued function f(2, 2)

XAT 2026 · Quantitative Ability & Data Interpretation · Medium · Functions

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Let f:R2→Rf : \mathbb{R}^2 \to \mathbb{R} be a real-valued function defined as f(0,y)=y+1f(0, y) = y + 1 and f(x+1,y)=f(x,f(x,y))+xf(x + 1, y) = f(x, f(x, y)) + x. What is the value of f(2,2)f(2, 2)?

  1. A.

    6

  2. B.

    7

  3. C.

    None of the other options is correct

  4. D.

    5

  5. E.

    4

Answer

B

Explanation

Given f(0,y)=y+1f(0, y) = y + 1. First find f(1,y)f(1, y): f(1,y)=f(0,f(0,y))+0=f(0,y+1)+0=(y+1)+1=y+2f(1, y) = f(0, f(0, y)) + 0 = f(0, y + 1) + 0 = (y + 1) + 1 = y + 2.

Now find f(2,y)f(2, y): f(2,y)=f(1,f(1,y))+1=f(1,y+2)+1=(y+2+2)+1=y+5f(2, y) = f(1, f(1, y)) + 1 = f(1, y + 2) + 1 = (y + 2 + 2) + 1 = y + 5.

For y=2y = 2: f(2,2)=2+5=7f(2, 2) = 2 + 5 = 7.

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