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Let $f(x)=x^{2}+a x+b$, where $a, b \in \mathrm{R}$. If $f(x)=0$ has all its roots imaginary, then the roots of $f(x)+f^{\prime}(\mathrm{x})+f^{\prime \prime}(\mathrm{x})=0$ are
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The correct answer is:
imaginary
$f(x)=x^{2}+a x+b$ has imaginary roots.
$$
\begin{array}{l}
\therefore \text { Discriminant, } D < 0 \Rightarrow a^{2}-4 b < 0 \\
f^{\prime}(x)=2 x+a \\
\Rightarrow f^{\prime \prime}(x)=2 \\
\text { Also, } f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0 ...(i)\\
\Rightarrow x^{2}+a x+b+2 x+a+2=0 \\
\Rightarrow x^{2}+(a+2) x+b+a+2=0 \\
\therefore x=\frac{-(a+2) \pm \sqrt{(a+2)^{2}-4(a+b+2)}}{2} \\
\quad=\frac{-(a+2) \pm \sqrt{a^{2}-4 b-4}}{2}
\end{array}
$$
Since, $a^{2}-4 b < 0$ $\therefore a^{2}-4 b-4 < 0$ Hence, Eq. (i) has imaginary roots.
$$
\begin{array}{l}
\therefore \text { Discriminant, } D < 0 \Rightarrow a^{2}-4 b < 0 \\
f^{\prime}(x)=2 x+a \\
\Rightarrow f^{\prime \prime}(x)=2 \\
\text { Also, } f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0 ...(i)\\
\Rightarrow x^{2}+a x+b+2 x+a+2=0 \\
\Rightarrow x^{2}+(a+2) x+b+a+2=0 \\
\therefore x=\frac{-(a+2) \pm \sqrt{(a+2)^{2}-4(a+b+2)}}{2} \\
\quad=\frac{-(a+2) \pm \sqrt{a^{2}-4 b-4}}{2}
\end{array}
$$
Since, $a^{2}-4 b < 0$ $\therefore a^{2}-4 b-4 < 0$ Hence, Eq. (i) has imaginary roots.
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