Question

If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4 < \alpha_5 < \alpha_6$, then the equation $(x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6) = 0$ has :-

• A. no real root in $(\alpha_5, \alpha_6)$
• B. no real root in $(\alpha_1, \alpha_2)$
• C. all roots are imaginary
• D. no real root in $(-\infty, \alpha_1)$

Answer 1

Answer: (D)

(D) no real root in $(-\infty, \alpha_1)$.

Answer 2

Explanation:
Let

$$f(x)=(x-\alpha_1)(x-\alpha_3)(x-\alpha_5)+3(x-\alpha_2)(x-\alpha_4)(x-\alpha_6)$$

Evaluate $f(x)$ at the points where one of the factors becomes zero:

1. At $x=\alpha_1$:

   $$f(\alpha_1)=0+3(\alpha_1-\alpha_2)(\alpha_1-\alpha_4)(\alpha_1-\alpha_6)$$

   Since $\alpha_1<\alpha_2, \alpha_4, \alpha_6$, each difference is negative so their product is negative. Thus, $f(\alpha_1)<0$.

2. At $x=\alpha_2$:

   $$f(\alpha_2)=(\alpha_2-\alpha_1)(\alpha_2-\alpha_3)(\alpha_2-\alpha_5)+0$$

   Here $(\alpha_2-\alpha_1)>0$ and $(\alpha_2-\alpha_3), (\alpha_2-\alpha_5)<0$. The product of one positive and two negatives is positive. Hence, $f(\alpha_2)>0$.
   By the Intermediate Value Theorem, there is a root in $(\alpha_1,\alpha_2)$.

3. Similarly, one finds a sign change (and hence a root) in $(\alpha_3,\alpha_4)$ and in $(\alpha_5,\alpha_6)$.

4. For $x<\alpha_1$, all factors $(x-\alpha_i)$ are negative. Thus, each product (being the product of three negatives) is negative. Their sum is negative, and no sign change occurs from $-\infty$ to $\alpha_1$.