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Q.

Let f:N→N where N is set of natural numbers be a  function such that f(x+y)=f(xy)∀x≥4 and y≥4, then

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a

f(4)=f(5)

b

f(8)=f(9)

c

f(7)=f(8)

d

f(5)=f(6)

answer is C.

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Detailed Solution

f(1+f(1))=4f(1)⇒ f(1+4)=4×4⇒ f(5)=16 Now f(5+f(5))=4f(5)⇒ f(5+16)=4×16⇒f(21)=64
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