MCQ
Which of the following function is surjective but not injective
- A$f : R \to R$ $f (x) = x^4 + 2x^3 - x^2 + 1$
- B$f : R \to R$ $f (x) = x^3 + x + 1$
- C$f : R \to R^+ f (x) =$ $\sqrt {1 + {x^2}} \,$
- ✓$f : R \to R f (x) = x^3 + 2x^2 - x + 1$
$=\mathop {Limit}\limits_{x \to \, \pm \,\infty } \,$ $[x^{2n} $ $\left( {{a_0}\, + \,\frac{{{a_1}}}{x}\, + \,\frac{{{a_2}}}{{{x^2}}}\, + \,....\, + \frac{{{a_{2n}}}}{{{x^{2n}}}}} \right)$
$= \left[ \begin{gathered} \,\infty \,\,\,\,\,if\,{a_0} > 0 \hfill \\ - \,\infty \,\,if\,{a_0}\, < 0 \hfill \\ \end{gathered} \right.$
Hence it will never approach $\infty / - \infty$
$f (x) = x^3 + x + 1 \Rightarrow f ‘(x) = 3x^2 + 1$ - injective as well as surjective
$f (x) =$$\sqrt {1 + {x^2}} \,$ - neither injective nor surjective
$f (x) = x^3 + 2x^2 - x + 1\Rightarrow f ‘(x) =3x^2 + 4x - 1 \Rightarrow D > 0$
Hence $f (x)$ is surjective but not injective.
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$(A)$ $-2$ $(B)$ $\frac{-2}{3}$ $(C)$ $2$ $(D)$ $\frac{2}{3}$