Check whether the function f: R → R defined by f(x) = x³ + 5 is bijective — JEE Mathematics

A function is bijective if it is both injective (one-to-one) and surjective (onto).

1. Injective check: Let $f(x_1) = f(x_2)$.

$$x_1^3 + 5 = x_2^3 + 5 \implies x_1^3 = x_2^3 \implies x_1 = x_2$$

Since $x_1 = x_2$ is the only real solution, the function is injective.
2. Surjective check: Let $y = x^3 + 5$.

$$x^3 = y - 5 \implies x = (y - 5)^{1/3}$$

Since for every real number $y$, there exists a unique real number $x = (y - 5)^{1/3}$ such that $f(x) = y$, the range of the function is $\mathbb{R}$, which matches the codomain. Thus, the function is surjective.

Since it is both injective and surjective, it is bijective.

Answer: Bijective