Mathematics | TS EAMCET 2019 | Discussion Page

Consider the function $f(x)=2x^{3}-3x^{2}-x+1$ and the intervals $I_{1}$ =[-1,0], $I_{2}$ = [0,1], $I_{3}$ =[1,2], $I_{4}$ =[-2,-1]

Then,

A) f(x) =0 has a root in the intervals

$I_{1}$ and

$I_{4}$ only

B) f(x) =0 has a root in the intervals

$I_{1}$ and

$I_{2}$ only

C) f(x) =0 has a root in every interval except in

$I_{4}$

D) f(x)=0 has a root in all the four given intervals

Option C

We have,

$f(x)=2x^{3}-3x^{2}-x+1$

Let

$g(x)=\frac{x^{4}}{2}-x^{3}-\frac{x^{2}}{2}+x$

$g(-1)=\frac{1}{2}+1-\frac{1}{2}-1=0$

and g(0)= 0

$\therefore$ f(x)=0 has roots lie in [-1,0]

Similarly , g(0)=g(1)=0

$\therefore$ f(x)=0 has roots lie in [0,1].

$g(2)=\frac{16}{2}-8-\frac{4}{2}+2=8-8-2+2=0$

g(1)= g(2)=0

But , g(-2)≠ 0

$\therefore$ f(x)=0 has roots in every interval except $I_{4}$

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