Question

Let $ a > 2,a \in N $ be a constant. If there are just 18 positive integers satisfying the inequality $ \left( {x - a} \right)\left( {x - 2a} \right)\left( {x - {a^2}} \right) < 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

Hint: Find the zeroes of the LHS considering ‘a’ and ‘x’ are positive integers, a>2. Find the limits of x so the values of the inequality will be negative(less than zero). There are 18 positive integers which satisfy the above given inequality, which gives the final ‘a’ value.

Complete step-by-step answer:
We are given that ‘a’ is greater than 2 and ‘a’ belongs to natural numbers.
Let us consider the inequality $ \left( {x - a} \right)\left( {x - 2a} \right)\left( {x - {a^2}} \right) < 0 $
\[\left( {x - a} \right)\left( {x - 2a} \right)\left( {x - {a^2}} \right)\] , is a polynomial
Zeroes of the above polynomial are as follows
  $
  x - a < 0 \to x < a \\
  x - 2a < 0 \to x < 2a \\
  x - {a^2} < 0 \to x < {a^2} \\
  $
Which means the limits of x will be $ \left( { - \infty ,a} \right) \cup \left( {2a,{a^2}} \right) $
But the x values which satisfy the above inequality are positive integers.
So the limit of x becomes $ \left[ {1,a} \right) \cup \left( {2a,{a^2}} \right) $ .
So from the limit the values of x are $ \left( {1,2,3,.....,a - 1} \right) + \left( {2a + 1,2a + 2,2a + 3,.....,{a^2} - 1} \right) $
No. of values from $ 1 \to a - 1 = a - 1{\text{ }} $
No. of values from $ 2a + 1 \to {a^2} = {a^2} - \left( {2a + 1} \right) $
 $ = {a^2} - 2a - 1 $
As given in the question the total number of solutions of ‘x’ for the above inequality is 18.
Therefore,
 $
  a - 1 + {a^2} - 2a - 1 = 18 \\
  {a^2} - a - 2 - 18 = 0 \\
  {a^2} - a - 20 = 0 \\
  $
We have to find the solutions for the above equation
 $
  {a^2} - a - 20 = 0 \\
  {a^2} - 5a + 4a - 20 = 0 \\
  a\left( {a - 5} \right) + 4\left( {a - 5} \right) = 0 \\
  \left( {a - 5} \right)\left( {a + 4} \right) = 0 \\
  a = 5,a = - 4 \\
  $
We have got two values i.e. a=5, a=-4
As given $ a > 2,a \in N $ , ‘a’ must be a positive number
So the value of ‘a’ is 5.
Therefore, from among the options given in the question option B and D are correct as a=5 is an odd number and ‘a’ lies in the interval (3, 11).
So, the correct answer is “Option B AND D”.

Note: Zeros of a polynomial are the values where the polynomial becomes zero. Inequalities can also be solved by adding, subtracting other terms. Multiplying and dividing can change the direction of the inequality.