Which of the following function is surjective | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$f (x) = x^4 + 2x^3 - x^2 + 1 	o A$ polynomial of degree even will always be into say
$f (x) = a_0x^{2n} + a_1 x^{2n-1} + a_2 x^{2n - 2} + .... + a_

Vedclass · Accepted Answer

$f : R 	o R f (x) = x^3 + 2x^2 - x + 1$

Vedclass · Answer

$f (x) = x^4 + 2x^3 - x^2 + 1 	o A$ polynomial of degree even will always be into say
$f (x) = a_0x^{2n} + a_1 x^{2n-1} + a_2 x^{2n - 2} + .... + a_{2n}$  
$\mathop {Limit}\limits_{x 	o \, \pm \,\infty } \,f (x) $

$=\mathop {Limit}\limits_{x 	o \, \pm \,\infty } \,$ $[x^{2n} $ $\left( {{a_0}\, + \,\frac{{{a_1}}}{x}\, + \,\frac{{{a_2}}}{{{x^2}}}\, + \,....\, + \frac{{{a_{2n}}}}{{{x^{2n}}}}} ight)$

 $= \left[ \begin{gathered}  \,\infty \,\,\,\,\,if\,{a_0} > 0 \hfill \  - \,\infty \,\,if\,{a_0}\, < 0 \hfill \ \end{gathered}  ight.$ 
Hence it will never approach $\infty / - \infty$ 
$f (x) = x^3 + x + 1 \Rightarrow f &lsquo;(x) = 3x^2 + 1$ - injective as well as surjective 
$f (x) =$$\sqrt {1 + {x^2}} \,$ - neither injective nor surjective 
$f (x) = x^3 + 2x^2 - x + 1\Rightarrow f &lsquo;(x) =3x^2 + 4x - 1 \Rightarrow D > 0$ 
Hence $f (x)$ is surjective but not injective.

Which of the following function is surjective but not injective

Similar Questions

The domain of the function $f(x) = \frac{{{{\sin }^{ - 1}}(x - 3)}}{{\sqrt {9 - {x^2}} }}$ is

The range of the polynomial $P(x)=4 x^3-3 x$ as $x$ varies over the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is

Let $f(x)=2 x^{2}-x-1$ and $S =\{n \in Z :|f(n)| \leq 800\}$ . Then value of $\sum_{n \in S} f(n)$ is . . . . .

Range of ${\sin ^{ - 1\,}}\left( {\frac{{1 + {x^2}}}{{2 + {x^2}}}} \right)$ is

Range of $f(x) = sin^{-1} (\sqrt {x^2 + x +1})$ is -