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the quadratic equation 3ax^2 +2bx+c=0 ha...

the quadratic equation `3ax^2 +2bx+c=0` has atleast one root between 0 and 1, if

A

at least one root in (0,1)

B

one root in (2,3) and ``
` the other in `(-2,-1)``

C

imaginary roots

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
Let `f(x) =ax^(3)+bx^(2)+ca+d" "...(i)`
`thereforef(0)=dand f(1)=a+b+c+d=d " "[becausea+b+c+=0]`
`therefore f(0)=f(1)`
f is continuous in the closed interval [0,1] and f is derivable in the open interval (0,1).
Also `f(0)=f(1)`
`therefore` By Rolle's theorem `f'(alpha)=0"for"0 lt alphalt1`
Now, `f(x)=3ax^(2)+2bx+c`
` impliesf'(alpha)=3aalpha^(2)+2balpha+c=0`
`therefore` Eq. (i) has exist atleast one root in the interval (0,1). Thus f'(x) must have root in the interval (0,1) or `3ax^(2)+2bx+c=0` lias root `in (0,1).`
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