Conder the function f(x)=1+2x+3x^2+4x^3 ...

Conder the function` f(x)=1+2x+3x^2+4x^3`
Let the sum of all distinct real roots ot f (x) and let t= |s| The function f(x) is

A

`(-1/4,0)`

B

`(-11,3/4)`

C

`(-3/4,-1/2)`

D

`(0,1/4)`

Text Solution

Verified by Experts

The correct Answer is:

C

Since `f'(x)=12x^(2)+6x+2`
Here `D=6^(2)-4.12.2=36-96=-60lt0`
`implies` Only one real root for `f(x)=0`
Also `f(0)=1,f(-1)=-2`
`implies` Root must lie in `(-1,0)`
Taking averate of 0 and (-1),f(-1/2)=1/4`
`implies` Root must lie in `(-1,-1/2)`
Similarly `f(-3/4)=-1/2`
`implies` Root must lie in `(-3/4,-1/2)`

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