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Question

If $$2a+3b+6c=0$$, then atleast one root of the equation $$a{ x }^{ 2 }+bx+c=0$$ lies in the interval


A
(0,1)
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B
(1,2)
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C
(2,3)
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D
(1,3)
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Solution

The correct option is A $$\left( 0,1 \right) $$
Given, $$2a+3b+6c=0$$
Let $$f^{ ' }\left( x \right) =a{ x }^{ 2 }+bx+c$$
$$f\left( x \right) =\dfrac { a{ x }^{ 3 } }{ 3 } +\dfrac { b{ x }^{ 2 } }{ 2 } +cx+d$$
$$\Rightarrow f\left( x \right) =\dfrac { 2a{ x }^{ 3 }+3b{ x }^{ 2 }+6cx+6d }{ 6 }$$
Now, $$f\left( 1 \right) =\dfrac { 2a+3b+6c+6d }{ 6 } =\dfrac { 6d }{ 6 } =d$$
and $$f\left( 0 \right) =\dfrac { 6d }{ 6 } =d$$
$$\therefore f\left( 0 \right) =f\left( 1 \right) $$
$$\Rightarrow f^{ ' }\left( x \right) $$ will vanish atleast once between $$0$$ and $$1$$. (by Rolle's theorem)
Therefore, one of the roots of the equation $$a{ x }^{ 2 }+bx+c=0$$ lies between $$0$$ and $$1$$.

Mathematics

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