Q55 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
With respect to the roots of the equation $3 x^3+b x^2+b x+3=0$,
match the items of List-I with those of List-II.
- AA-V, B-III, C-I, D-II
- BA-IV, B-I, C-II, D-III
- CA-V, B-II, C-III, D-I
- Correct answerDA-IV, B-II, C-V, D-III
Solution
If $b=9$ this equation becomes $3 x^3+9 x^2+9 x+3=0$ no change of sign and all coefficient are +ve So, all roots are negative
$\therefore(\mathrm{A}) \rightarrow($ iv $)$
If $b=-3$ then equation becomes $\begin{aligned} & 3 x^3-3 x^2-3 x+3=0 \Rightarrow x^3-x^2-x+1=0 \\ & \Rightarrow x^2(x-1)-1(x-1)=0 \Rightarrow\left(x^2-1\right)(x-1)=0 \\ & \Rightarrow x=1,1,-1\end{aligned}$ Two roots are positive $\therefore(C) \rightarrow(v)$ Hence option (4) is correct.
If $b=-3$ then equation becomes $\begin{aligned} & 3 x^3-3 x^2-3 x+3=0 \Rightarrow x^3-x^2-x+1=0 \\ & \Rightarrow x^2(x-1)-1(x-1)=0 \Rightarrow\left(x^2-1\right)(x-1)=0 \\ & \Rightarrow x=1,1,-1\end{aligned}$ Two roots are positive $\therefore(C) \rightarrow(v)$ Hence option (4) is correct.
Hence, the correct answer is option: 'D'.