Question

The function $f\left(x\right)=x^2+bx+c$, where $b$ and $c$ are real constants describes

• A. One-One mapping
• B. Onto mapping
• C. Not one-one by onto mapping
• D. Neither one-one nor onto mapping

Answer 1

The correct option is D

Neither one-one nor onto mapping

Find the mapping of the given function

Given, $f\left(x\right)=x^2+bx+c$

Therefore $f\left(x\right)$ is a quadratic equation

Now,

$$\begin{gathered} y=f\left(x\right) \\ =x^2+bx+c \\ =x^2+2\frac{b}{2}x+\frac{b^2}{4}+c-\frac{b^2}{4} \\ ={\left(x-\frac{b}{2}\right)}^2+c-\frac{b^2}{4} \end{gathered}$$

So it is upward parabola.

For one element in co-domain this is exactly two pre-image. .

Hence by the definition of onto map it is not onto.

Now let $x_1,x_2\in x$

If it is one one then $f\left(x_1\right)=f\left(x_2\right)$ will imply $x_1=x_2$

$$\begin{gathered} \therefore f\left(x_1\right)=f\left(x_2\right) \\ \Rightarrow {x_1}^2+b{x}_1+c={x_2}^2+b{x}_2+c \\ \Rightarrow {x_1}^2-{x_2}^2=b\left(x_2-x_1\right) \\ \Rightarrow \left(x_1-x_2\right)\left(x_1+x_2\right)=b\left(x_2-x_1\right) \\ \Rightarrow x_1+x_2=-b \\ \Rightarrow x_1\ne x_2 \end{gathered}$$

Therefore, the function is not one-one

Hence, it is neither one-one mapping nor onto mapping.

Hence, the correct option is $\left(\mathrm{D}\right)$.