Let f : R → R be defined by f(x) = x^3 + x , where R is the set of re…

The correct answer is Option (3) → one-one and onto

One-one: let \(f(a) = f(b)\).

    \(a^{3} + a = b^{3} + b\)
    \(a^{3} - b^{3} + a - b = 0\)
    \((a - b)(a^{2} + ab + b^{2} + 1) = 0\)

The second factor is always positive, because

    \( \displaystyle a^{2} + ab + b^{2} = \left(a + \frac{b}{2}\right)^{2} + \frac{3b^{2}}{4} \ge 0 \)

So \(a - b = 0\), that is, \(a = b\).

Onto: for any real y, the equation \(x^{3} + x - y = 0\) is a cubic equation with real coefficients. Every such cubic has at least one real root x, and for that x, \(f(x) = y\).

Hence f is one-one and onto.