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If `x=2` is a real root of cubic equation f (x) = `3x^(3)+bx^(2)+bx+3=0`. Then the value of lim_(xrarr(1)/(2)) (e^(f(x)-1)/(2x-1))`

A

`(27)/(4)`

B

`-(27)/(4)`

C

`-(27)/(8)`

D

`(27)/(8)`

Text Solution

Verified by Experts

The correct Answer is:
C

`3x^(3)+bx^(2)+bx+3=3(x^(3)+1)+bx(x+1)=0`
has roots `- 1,2,(1)/(2)` so,
`underset(xrarr(1)/(2))(lim)((3x^(3)+bx^(2)+bx+3)/(2x-1))=(3(x-1)(x-2)(x-(1)/(2)))/(2(x-(1)/(2)))`
`(3)/(2).(3)/(2).(-3)/(2)=-(27)/(8)`
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