Q74 / 170 · Manipal (MET) 2019 · Mathematics
The value of $a(a \geq b)$ for which the sum of the cubes of the roots of $x^2-(a-2) x+(a-3)=0$ assumes the least value, is
- Correct answerA$3$
- B$4$
- C$5$
- DNone of the above
Solution
Let $\alpha$ and $\beta$ be the roots of the equation
$\begin{aligned}
& x^2-(a-2) x+(a-3)=0 \\
& \therefore \alpha+\beta=(a-2) \text { and } \alpha \beta=a-3
\end{aligned}$
Now, $\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$ $\begin{gathered} =(a-2)^3-3(a-3)(a-2) \\ =a^3-9 a^2+27 a-26=(a-3)^3+1 \end{gathered}$
It assumes the least value, if $(a-3)^3=0$ $\therefore \quad a=3$
Now, $\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$ $\begin{gathered} =(a-2)^3-3(a-3)(a-2) \\ =a^3-9 a^2+27 a-26=(a-3)^3+1 \end{gathered}$
It assumes the least value, if $(a-3)^3=0$ $\therefore \quad a=3$
Hence, the correct answer is option: 'A'.