Let a,b,c∈R and α,β be the roots of ax2+bx+c=0 such that α 1 then 1+ca+ba=.

We have af(−1)<0 and af(1)<0⇒ a(a−b+c)<0 and a(a+b+c)<0⇒ a21−ba+ca<0 and a21+ba+ca<0⇒ 1−ba+ca<0 and 1+ba+ca<0⇒ 1+ba+ca<0Hence, the result

Let a,b,c∈R and α,β be the roots of ax2+bx+c=0 such that α 1 then 1+ca+ba=.

We have af(−1)<0 and af(1)<0⇒ a(a−b+c)<0 and a(a+b+c)<0⇒ a21−ba+ca<0 and a21+ba+ca<0⇒ 1−ba+ca<0 and 1+ba+ca<0⇒ 1+ba+ca<0Hence, the result

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Let $a, b, c \in R and α, β$ be the roots of $a x^{2} + b x + c = 0$ such that $α < - 1 and β > 1$ then $1 + \frac{c}{a} + |\frac{b}{a}|$ =.

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$< 0$

$\geq 0$

$\leq 0$

$> 0$

answer is D.

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Detailed Solution

We have

$\begin{array}{l} a f (- 1) < 0 and a f (1) < 0 \\ \Rightarrow a (a - b + c) < 0 and a (a + b + c) < 0 \end{array}$

$\begin{array}{l} \Rightarrow a^{2} (1 - \frac{b}{a} + \frac{c}{a}) < 0 and a^{2} (1 + \frac{b}{a} + \frac{c}{a}) < 0 \\ \Rightarrow (1 - \frac{b}{a} + \frac{c}{a}) < 0 and (1 + \frac{b}{a} + \frac{c}{a}) < 0 \\ \Rightarrow (1 + |\frac{b}{a}| + \frac{c}{a}) < 0 \end{array}$