Let f(x) = x^2 + ax + b,where a, b in R. If f | vedclass

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(B) Given,$f(x) = x^2 + ax + b$ has imaginary roots. Therefore,the discriminant $D < 0$,which implies $a^2 - 4b < 0$. Now,we calculate the derivatives

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(B) Given,$f(x) = x^2 + ax + b$ has imaginary roots. Therefore,the discriminant $D Now,we calculate the derivatives: $f'(x) = 2x + a$ $f''(x) = 2$ Substituting these into the equation $f(x) + f'(x) + f''(x) = 0$: $(x^2 + ax + b) + (2x + a) + 2 = 0$ $x^2 + (a + 2)x + (b + a + 2) = 0$ The discriminant $D'$ of this new quadratic equation is: $D' = (a + 2)^2 - 4(1)(b + a + 2)$ $D' = a^2 + 4a + 4 - 4b - 4a - 8$ $D' = a^2 - 4b - 4$ Since $a^2 - 4b Thus,$D' Since the discriminant is negative,the roots of the equation $f(x) + f'(x) + f''(x) = 0$ are imaginary.

Let $f(x) = x^2 + ax + b$,where $a, b \in R$. If $f(x) = 0$ has all its roots imaginary,then the roots of $f(x) + f'(x) + f''(x) = 0$ are

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