If $a_{1} < a_{2} < a_{3} < a_{4} < a_{5} < a_{6}$ , then the equation $(x - a_{1}) (x - a_{3}) (x - a_{5}) + 3 (x - a_{2}) (x - a_{4}) (x - a_{6}) = 0$ , has

Three real roots

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root lies in $(- \infty, a_{1})$

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no real root in $(a_{1}, a_{2})$

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no real root in $(a_{5}, a_{6})$

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Solution

The correct option is A
Three real roots

Explanation for the correct option:
Finding the value:
Let $f (x) =$ $(x - a_{1}) (x - a_{3}) (x - a_{5}) + 3 (x - a_{2}) (x - a_{4}) (x - a_{6}) = 0$
and $a_{1} < a_{2} < a_{3} < a_{4} < a_{5} < a_{6}$
As $x$ approaches $\infty$
$f (x) \to \infty f (a_{1}) = 3 (a_{1} - a_{2}) (a_{2} - a_{4}) (a_{1} - a_{6}) < 0$
Similarly
$f (a_{2}) > 0, f (a_{3}) > 0, f (a_{4}) < 0 f (a_{5}) < 0, f (a_{6}) > 0$
$f (x)$ changes sign in each interval $(a_{1} - a_{2}), (a_{3} - a_{4}) & (a_{5} - a_{6})$
Hence, option $(A)$ is the correct option.

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