If a and 4a-3b-2c have same sign- Then- ax2-bx-c-0- a- 0 cannot have both roots belonging to

The correct option is C (1,2) a and 4a+3b+2c have same sign f( x ) =a x ^ 2 +bx+c=0,a≠ 0 As a and 4a+3b+2c have same sign a+4a+3b+2c ...

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Standard XII

Mathematics

Proof of Intermediate Value Theorem

If a and ...

Question

If $a$ and $4 a + 3 b + 2 c$ have same sign. Then, $a x^{2} + b x + c = 0, (a \neq 0)$ cannot have both roots belonging to

(- 1, 2)

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(1, 2)

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(- 2, 1)

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Solution

The correct option is C $(1, 2)$
a and $4 a + 3 b + 2 c$ have same sign
$f (x) = a x^{2} + b x + c = 0, a \neq 0$
As a and $4 a + 3 b + 2 c$ have same sign
$a + 4 a + 3 b + 2 c < 0$ (or) $a + 4 a + 3 b + 2 c > 0$
$5 a + 3 b + 2 c < 0$ (or) $5 a + 3 b + 2 c > 0$
$(a + b + c) + (4 a + 2 b + c) < 0$ (Or) $(a + b + c) + (4 a + 2 b + c) > 0$
$f (1) + f (2) < 0$ (Or) $f (1) + f (2) > 0$
$∴ f (x)$ cannot have both roots belonging to $(1, 2)$

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If a and 4a+3b+2c have same sign. Then, ax2+bx+c=0,(a≠0) cannot have both roots belonging to

The correct option is C (1,2)a and 4a+3b+2c have same signf(x)=ax2+bx+c=0,a≠0As a and 4a+3b+2c have same signa+4a+3b+2c<0 (or) a+4a+3b+2c>05a+3b+2c<0 (or) 5a+3b+2c>0(a+b+c)+(4a+2b+c)<0 (Or) (a+b+c)+(4a+2b+c)>0f(1)+f(2)<0 (Or)f(1)+f(2)>0∴f(x) cannot have both roots belonging to (1,2)

If $a$ and $4 a + 3 b + 2 c$ have same sign. Then, $a x^{2} + b x + c = 0, (a \neq 0)$ cannot have both roots belonging to