Q41 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

Given that for any $n \in \mathbf{N}$ there exist an odd integer $q$ and a non-negative integer $r$ such that, $n$ can be written uniquely as $n=q \times 2^r$. Let $f: \mathbf{N} \rightarrow \mathbf{N} \times \mathbf{N}$ be function defined by $f(n)=\left(r+1, \frac{q+1}{2}\right)$. Then,

  • A
    $f$ is one-one but not onto
  • B
    $f$ is onto but not one-one
  • Correct answer
    C
    $f$ is a bijection
  • D
    only $f^{-1}(1,1)$ does not exist because $f$ is not a bijection

Solution

We have, $f(n)=\left(r+1, \frac{q+1}{2}\right)$ For one-one $\begin{aligned} & f\left(n_1\right)=f\left(n_2\right) \text { then } n_1=n_2 \\ & \because f\left(n_1\right)=\left(r_1+1, \frac{q_1+1}{2}\right) \Rightarrow f\left(n_2\right)=\left(r_2+1, \frac{q_2+1}{2}\right) \\ & \quad r_1+1=r_2+1 \text { and } \frac{q_1+1}{2}=\frac{q_2+1}{2} \\ & \quad r_1=r_2 \text { and } q_1=q_2 \\ & \because \quad f\left(n_1\right)=f\left(n_2\right) \end{aligned}$ Here, $f(n)$ is one-one $r$ is non-negative integer $\therefore \quad r+l \in \mathbf{N}$ $q$ is odd intege$ \because \quad \frac{q+1}{2} \in \mathbf{N} \Rightarrow f(n)=(N, N)$ $\therefore$ Range of $f(n)=N \times N$ $\therefore f(n) \text { is onto }$ Here, $f(n)$ is bijective

Hence, the correct answer is option: 'C'.

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