Q55 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
$\alpha, \beta, \gamma$ are the roots of the equation $8 x^3-42 x^2+63 x-27=0$ If $\beta \lt \gamma \lt \alpha$ and $\beta, \gamma, \alpha$ are in geometric progression, then the extreme value of the expression $\gamma x^2+4 \beta x+\alpha$ is
- A$\frac{3}{4}$
- B3
- Correct answerC$\frac{3}{2}$
- D$\frac{21}{4}$
Solution
Given equation : $8 x^3-42 x^2+63 x-27=0$...(i)
$\alpha, \beta, \gamma$ are root of eqn. (i)
$\gamma^2=\beta \alpha(\beta, \gamma, \alpha \text { in G.P. })$
Sum of root, $\alpha+\beta+\gamma=\frac{-b}{a}=\frac{42}{8}$...(ii) $\alpha \beta \gamma=\frac{-c}{a}=\frac{27}{8}$....(iii) $\alpha \beta+\beta \gamma+\gamma \alpha=\frac{63}{8}$....(iv)
From (iii) and $\gamma^2=\beta \alpha \Rightarrow \gamma^3=\frac{27}{8} \Rightarrow \gamma=\frac{3}{8}$ On solving (ii), (iii) by putting values of $\gamma=\frac{3}{8}$ in (ii) and (iii), we get $\beta=\frac{3}{4}, \alpha=3$ Now extreme value of $\gamma x^2+4 \beta x+\alpha$ is $\frac{-D}{4 a}$ i.e. $\frac{-\left(16 \beta^2-4 \gamma \alpha\right)}{4 \times \gamma}=\frac{-(9-18)}{4 \times \frac{3}{2}}=\frac{9}{6}=\frac{3}{2}$.
Sum of root, $\alpha+\beta+\gamma=\frac{-b}{a}=\frac{42}{8}$...(ii) $\alpha \beta \gamma=\frac{-c}{a}=\frac{27}{8}$....(iii) $\alpha \beta+\beta \gamma+\gamma \alpha=\frac{63}{8}$....(iv)
From (iii) and $\gamma^2=\beta \alpha \Rightarrow \gamma^3=\frac{27}{8} \Rightarrow \gamma=\frac{3}{8}$ On solving (ii), (iii) by putting values of $\gamma=\frac{3}{8}$ in (ii) and (iii), we get $\beta=\frac{3}{4}, \alpha=3$ Now extreme value of $\gamma x^2+4 \beta x+\alpha$ is $\frac{-D}{4 a}$ i.e. $\frac{-\left(16 \beta^2-4 \gamma \alpha\right)}{4 \times \gamma}=\frac{-(9-18)}{4 \times \frac{3}{2}}=\frac{9}{6}=\frac{3}{2}$.
Hence, the correct answer is option: 'C'.