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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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