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If $2a+3b+6c= 0,$ then at least one root of the equation

$ax^{2}+bx+c= 0,$ lies in the interval

Option 1)
$\left ( 2,3 \right )$

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

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

Option 4)
$\left ( 1,3 \right )$

Answers (1)

best_answer

As we learnt in

Rolle's Theorems -

Let f(x) be a function of x subject to the following conditions.

1. f(x) is continuous function of $x:x\epsilon [a,b]$

2. f'(x) is exists for every point : $x\epsilon [a,b]$

3. $f(a)=f(b)\:\:\:then\:\:f'(c)=0\:\:such \:that\:\:a<c<b.$

-

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

$f\left ( x \right )=\frac{ax^{3}}{3}+\frac{bx^{2}}{2}+cx$

now $f\left ( 0 \right )=0$ and

$f\left ( 1 \right )=\frac{a}{3}+\frac{b}{2}+c$

$\Rightarrow \frac{2a+3b+6c}{3}=0$

$\therefore f\left ( 1 \right )=f\left ( 0 \right )$ So $f{}'\left ( x \right )=0$ having at least one root between (0,1) by Rollers Rule.

Option 1)

$\left ( 2,3 \right )$

Incorrect option

Option 2)

$\left ( 1,2 \right )$

Incorrect option

Option 3)

$\left ( 0,1 \right )$

Correct option

Option 4)

$\left ( 1,3 \right )$

Incorrect option

Posted by

prateek

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