Q68 / 160 · TS EAMCET 2015 · Mathematics

If $f: N \rightarrow R$ is defined by $f(1)=-1$ and $f(n+1)=3 f(n)+2$ for $n \geq 1$, then $f$ is

  • A
    one-one
  • B
    onto
  • Correct answer
    C
    a constant function
  • D
    $f(n)>0$ for $n>1$

Solution

We have, $f: N \rightarrow R$ and $f(1)=-1$ Also, $f(n+1)=3 f(n)+2$ for $n \geq 1$ Now, $\operatorname{put} n=1$, we get $$ \begin{aligned} f(2) & =3 f(1)+2 \\ & =3(-1)+2=-1 \end{aligned} $$ Put $n=2$, we get $$ f(3)=3 f(2)+2=3(-1)+2=-1 $$ So, $f(n)=-1$ for all values of $n$. Hence, $f(n)$ is a constant function.

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

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