Question

If $a_1$ < $a_2$< $a_3$ < $a_4$ < $a_5$ < $a_6$, then the equation $(x-a_1)(x-a_3)(x-a_5)+2(x-a_2)(x-a_4)(x-a_6)$ = 0 has

• A. Three real roots
• B. One real root
• C. One real root in each interval ( ,), ( , ) and ( , )
• D. None of these

Answer 1

The correct option is C

One real root in each interval ( ,), ( , ) and ( , )

$f\left(x\right)=(x-a_1)(x-a_3)(x-a_5)+2(x-a_2)(x-a_4)(x-a_6)$ = 0

$a_1$ < $a_2$< $a_3$ < $a_4$ < $a_5$ < $a_6$

f( $a_1$ ) = 2 $(a_1-a_2)(a_1-a_4)(a_1-a_6)<0$

f( $a_2$ ) = 2 $(a_2-a_1)(a_2-a_3)(a_2-a_5)>0$

$\therefore$ at least one real root lies in ($a_1$ ,$a_2$ )

Similarly, at least one real root lies in each interval ($a_3$ , $a_4$ ) and ($a_5$ , $a_6$ )

But f(x) is cubic, therefore there are only three roots.

Hence, the equation f(x) = 0 has one real root in each interval ($a_1$ ,$a_2$ ), ($a_3$ , $a_4$ ) and ($a_5$ , $a_6$ )