The function f(x) = x^{2} + bx + c, where b a | vedclass

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

Question

(D) The given function is $f(x) = x^{2} + bx + c$. This is a quadratic function representing a parabola. For a function to be one-to-one, $f(x_1) = f(

Vedclass · Accepted Answer

neither one-to-one nor onto mapping

Vedclass · Answer

(D) The given function is $f(x) = x^{2} + bx + c$. This is a quadratic function representing a parabola. For a function to be one-to-one, $f(x_1) = f(x_2)$ must imply $x_1 = x_2$. Here, $f(x) = (x + \frac{b}{2})^2 + (c - \frac{b^2}{4})$. Since the square term is always non-negative, $f(x_1) = f(x_2)$ does not necessarily imply $x_1 = x_2$ (e.g.,$f(x) = x^2$ gives $f(1) = f(-1) = 1$), so it is many-to-one. Furthermore, the range of this function is $[c - \frac{b^2}{4}, \infty)$, which is not equal to the codomain $\mathbb{R}$ (assuming the codomain is $\mathbb{R}$), so it is not onto. Therefore, the function is neither one-to-one nor onto.

The function $f(x) = x^{2} + bx + c$, where $b$ and $c$ are real constants, describes:

Similar Questions

The function $f: (-\infty, \infty) \rightarrow (-\infty, \infty)$ defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is :

Are $f$ and $g$ both necessarily onto,if $g \circ f$ is onto?

Let $a > 1$ and $0 < b < 1$. If $f: R \rightarrow [0, 1]$ is defined by $f(x) = \begin{cases} a^x, & -\infty < x < 0 \\ b^x, & 0 \leq x < \infty \end{cases}$,then $f(x)$ is

The number of bijective functions $f: Z \rightarrow Z$ such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$ is:

For real $x,$ let $f(x) = x^3 + 5x + 1,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module