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)
B
(1,2)
C
(2,3)
D
(1,3)

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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