Let $f:(0,1)\rightarrow \mathbb{R}$ is injective function, then choose the correct options-
(a) Range of $f$ must contains rationals.
(b) Range of $f$ must contains irrationals.
(c) Range of $f$ must contain both rationals and irrationals.
(d) $f$ is onto.
Explanation: (b) is correct.
(a) There exists a bijective function $f$ from $(0,1)$ to $\mathbb{Q}^c\subset \mathbb{R}$ (as both are uncountable and have the same cardinality), which has no rationals in its range.
(b) If the range of $f$ does not contain irrationals, then the range set will be countable being a subset of $\mathbb{Q}$, and thus $(0,1)$ will be countable, a contradiction.
(d) Take $f(x)=x$, it is one-one function from $(0,1)$ to $\mathbb{R}$ which is not onto.