Question

# If set of values of ' $$a$$ ' for which $$f(x)=a x^{2}-(3+2 a) x+6$$ $$a \neq 0$$ is positive for exactly three distinct negative integral values of $$x$$ is $$(c, d],$$ then find the value of $$\left(c^{2}+4 |d|\right)$$.

Solution

## $$f(x)=a x^{2}-(3+2 a) x+6$$$$=(a x-3)(x-2)$$Here, roots of the equation $$f(x)=0$$ are 2 and $$\dfrac{3 }{ a},$$ and $$f(0)=6$$.$$f(x)$$ should be positive for exactly three negative integral values of $$x$$Therefore, graph of $$f(x)$$ must be a downward parabola passing through $$x=2$$ and $$x=3 / a$$ and $$-4 \leq \dfrac{3}{a}<-3$$$$\therefore \quad a \in\left(-1,-\dfrac{3}{4}\right]$$$$\therefore \quad c=-1, d=-\dfrac{3}{4}$$$$\Rightarrow \quad c^{2}+4|d|=1+3=4$$Maths

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