Question

Let a > 2, a $\in$ N be a constant. If there are just 18 positive integers satisfying the inequality $(x-a)(x-2a)(x-a^2)<0$ then which of the option(s) is/are correct?

• A. 'a' is composite
• B. 'a' is odd
• C. 'a' is greater than 8
• D. 'a' lies in the interval (3, 11)

Answer 1

The correct option is B 'a' is odd

$(x-a)(x-2a)(x-a^2)<0$

Solution for this inequality will be:

$x\in (-\infty ,a)\cup (2a,a^2)$

For positive integers:

$x\in [1,a)\cup (2a,a^2)$

Putting $a=5$ we have:

$x\in [1,5)\cup (10,25)$

No. of integral values of x for which x satisfies above inequality is 18 for $a=5$

Hence, $a=5$.