Q109 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics
Consider the function $f(x)=2 x^3-3 x^2-x+1$ and the intervals $Y_1=[-1,0], Y_2=[0,1]$, $r_3=[1,2], r_4=[-2,-1]$.
Then,
- A$f(x)=0$ has a root in the intervals $f_1$ and $f_4$ only
- B$f(x)=$ ohas a root in the intervals $f_1$ and $f_2$ only
- Correct answerC$f(x)=0$ has a root in every interval except in $f_4$
- D$f(x)=0$ has a root in all the four given intervals
Solution
We have,
$$
f(x)=2 x^3-3 x^2-x+1
$$
Let
$$
\begin{aligned}
& g(x)=\frac{x^4}{2}-x^3-\frac{x^2}{2}+x \\
& g(-1)=\frac{1}{2}+1-\frac{1}{2}-1=0
\end{aligned}
$$
and
$$
g(0)=0
$$
$\therefore f(x)=0$ has roots lie in $[-1,0]$.
Similarly, $g(0)=g(1)=0$
$$
\begin{aligned}
\therefore \quad f(x) & =0 \text { has roots lie in }[0,1] . \\
g(2) & =\frac{16}{2}-8-\frac{4}{2}+2=8-8-2+2=0 \\
g(1) & =g(2)=0
\end{aligned}
$$
But, $\quad g(-2 \neq 0$
$\therefore f(x)=0$ has roots in every interval except $I_4$.
Hence, the correct answer is option: 'C'.