If alpha_1

Question

(D) Let $f(x) = (x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6)$.We evaluate $f(x)$ at the given points:$f(\a

Vedclass · Accepted Answer

No real root in $(-\infty, \alpha_1)$

Vedclass · Answer

(D) Let $f(x) = (x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6)$.We evaluate $f(x)$ at the given points:$f(\alpha_1) = 0 + 3(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_4)(\alpha_1 - \alpha_6)$. Since $\alpha_1 $f(\alpha_2) = (\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3)(\alpha_2 - \alpha_5) + 0 = (+)(-)(-) = + > 0$.Since $f(\alpha_1)  0$,there is at least one real root in $(\alpha_1, \alpha_2)$.$f(\alpha_3) = 0 + 3(\alpha_3 - \alpha_2)(\alpha_3 - \alpha_4)(\alpha_3 - \alpha_6) = 3(+)(-)(-) = + > 0$.$f(\alpha_4) = (\alpha_4 - \alpha_1)(\alpha_4 - \alpha_3)(\alpha_4 - \alpha_5) + 0 = (+)(+)(-) = - Since $f(\alpha_3) > 0$ and $f(\alpha_4) $f(\alpha_5) = 0 + 3(\alpha_5 - \alpha_2)(\alpha_5 - \alpha_4)(\alpha_5 - \alpha_6) = 3(+)(+)(-) = - $f(\alpha_6) = (\alpha_6 - \alpha_1)(\alpha_6 - \alpha_3)(\alpha_6 - \alpha_5) + 0 = (+)(+)(+) = + > 0$.Since $f(\alpha_5)  0$,there is at least one real root in $(\alpha_5, \alpha_6)$.As $x 	o -\infty$,$f(x) 	o -\infty$. Since $f(\alpha_1) < 0$,we check the interval $(-\infty, \alpha_1)$. The function is a cubic polynomial with a positive leading coefficient. It crosses the x-axis at least once in each of the intervals $(\alpha_1, \alpha_2)$,$(\alpha_3, \alpha_4)$,and $(\alpha_5, \alpha_6)$. Since it is a cubic,it has exactly $3$ real roots. Thus,there are no roots in $(-\infty, \alpha_1)$.

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 :-

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