Practicing Success
The value of a so that the sum of the cubes of the roots of the equation $x^2-a x+(2 a-3)=0$ assumes the minimum value, is |
a = 1 a = 3 a = 0 none of these |
a = 3 |
Let $\alpha$ and $\beta$ be the roots of the given equation. Then, $\alpha+\beta=a$ and $\alpha \beta=2 a-3$ Let $S=\alpha^3+\beta^3$. Then, $S =(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$ $\Rightarrow S =a^3-3 a(2 a-3)$ $\Rightarrow S =a^3-6 a^2+9 a$ $\Rightarrow \frac{d S}{d a}=3 a^2-12 a+9=3\left(a^2-4 a+3\right)=3(a-1)(a-3)$ Clearly, $\frac{d S}{d a}=0$ when a = 1, 3. The changes in the signs of $\frac{d S}{d a}$ are shown in Figure. Clearly, S is minimum at a = 3. |