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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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