Question

Consider the function $f\left(x\right)=x^2+ax+b$ where $a,b\in R$. If $f\left(x\right)=0$ has all its roots imaginary, then the roots of $f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=0$ are

• A. real and distinct
• B. imaginary
• C. equal
• D. rational and equal

Answer 1

The correct option is B

imaginary

Explanation for the correct option.

Step 1. Find the condition for imaginary roots.

As the roots of the equation $f\left(x\right)=0$ are imaginary, so the value of discriminant of $x^2+ax+b=0$ is less than $0$. Thus $b^2-4ac<0$. So,

$$\begin{gathered} a^2-4\times 1\times b<0\left[\mathbf{Here}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{b}\mathbf{=}\mathit{a}\mathbf{,}\mathbf{}\mathit{c}\mathbf{=}\mathit{b}\right] \\ \Rightarrow a^2-4b<0 \end{gathered}$$

Now, for the function $f\left(x\right)=x^2+ax+b$ the first derivative is given as $f\text{'}\left(x\right)=2x+a$ and the second derivative is given as: $f\text{'}\text{'}\left(x\right)=2$.

Step 2. Find the nature of roots for the equation.

The equation $f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=0$ can be written as:

$$\begin{gathered} x^2+ax+b+2x+a+2=0 \\ \Rightarrow x^2+\left(a+2\right)x+\left(a+b+2\right)=0 \end{gathered}$$

Now, the value of the discriminant of the equation is given as:

$\begin{array}{rcl}D& =& {\left(a+2\right)}^2-4\times 1\times \left(a+b+2\right)\left[\mathbf{Here}\mathbf{,}\mathbf{}\mathit{a}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{b}\mathbf{=}\mathit{a}\mathbf{+}\mathbf{2}\mathbf{,}\mathbf{}\mathit{c}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathbf{2}\right]\\ & =& a^2+4a+4-4a-4b-8\\ & =& a^2-4b-4\end{array}$

Now, as $a^2-4b<0$, so $a^2-4b-4<0$ is also true.

So for the equation $f\left(x\right)+f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=0$the value of the discriminant is less than zero and so the roots of the equation are imaginary.

Hence, the correct option is B.