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$f\left ( x \right )= ax^{2}+bx+c\; \left ( a,b,c\: \epsilon \, R \right )$ . If $a+b+c> 0$ and $a-b+c< 0$ then $f\left ( x \right )= 0$ will have a root in

Option 1)
$\left ( 0,1 \right )$

Option 2)
$\left ( -1,1 \right )$

Option 3)
$\left ( -1,0 \right )$

Option 4)
$\left ( 1,2 \right )$

Answers (1)

best_answer

$f\left ( x \right )=ax^{2}+bx+c$

$f\left ( 1 \right )=a+b+c\: > \: 0$ (given)

$f\left ( -1 \right )=a-b+c\: < \: 0$ (given)

$\therefore \: f\left ( -1 \right )$ and $f\left ( 1 \right )$ are of opposite sign , So

$f\left ( x \right )=0$ for one $x\, \epsilon \, \left ( -1,1 \right )$

$\therefore$ Option (B)

f (α ) & f (β) are of opposite signs -

Here, $f\left ( x \right )= ax^{2}+bx+c$

$f\left ( x \right )= 0$ for one $x\in \left ( \alpha ,\beta \right )$

- wherein

Option 1)

$\left ( 0,1 \right )$

This is incorrect

Option 2)

$\left ( -1,1 \right )$

This is correct

Option 3)

$\left ( -1,0 \right )$

This is incorrect

Option 4)

$\left ( 1,2 \right )$

This is incorrect

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