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Let N be the set of natural numbers and R be the set of all real numbers. If function f:N→Rf:N\to R defined by f(x)=4x2+12x+15f(x)=4x^2+12x+15, then f:N→range(f)f:N\to range(f) is

Solution

✅ Correct Option: 3

f(x)=(2x+3)2+6f(x)=(2x+3)^2+6 is strictly increasing for x≥1x\ge1, so distinct naturals give distinct images and ff is one-one.

Restricting the codomain to the range makes every image attained, so it is also onto, hence bijective.

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