Question 13
Let $$f:R^{2}\rightarrow R$$ be a real-valued function defined as $$f\left(0, y\right)=y+1 \text{and} f\left(x+1, y\right)=f\left(x, f\left(x, y\right)\right)+ x$$. What is the value of $$f(2, 2)$$?
Solution
Given : f(0,y) = y+1 and f(x+1,y) = f(x,f(x,y)) + x
To find : f(2,2)
f(x+1,y) = f(x,f(x,y)) + x
Let x=0
f(1,y) = f(0,f(0,y))
f(1,y) = f(0,y+1) (as f(0,y) = y+1 (given))
f(1,y) = (y+1)+1 = y+2
Hence, f(1,y) = y+2 $$\longrightarrow\ i$$
Now,
f(x+1,y) = f(x,f(x,y)) + x
Let x=1
f(2,y) = f(1,f(1,y)) + 1
f(2,y) = f(1,y+2) + 1 (from equation i)
f(2,y) = (y+2) + 2 + 1 = y+5
Put y=2
f(2,2) = 2+5 =7
$$\therefore\ $$ The required answer is B.
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