Question

let a>2 be a constant. if there are just 18 positive integers satisfying the inequality (x-a)(x-2a)(x-a²)<0, then find the value of a

Answer 1

Zeroes are a,2a,a^2

then, above♤

Answer 2

Answer: a = 5

Step-by-step explanation:

Here the given inequality,

(x-a)(x-2a)(x-a²)<0

⇒ (x-a) < 0 or (x-2a)(x-a²)<0

Since, x > 0,

Hence, the number of positive integers less than a ( from 0 ) = (a-1)

Also, The number of positive integers is less than a² and more than 2a = (a²-2a) - 1

Hence, the total positive integers that are satisfying the given inequality,

(a-1) + (a²-2a) - 1,

According to the question,

(a-1) + (a²-2a) - 1 = 18

⇒ a² - a - 2 = 18

⇒ a² - a - 20 = 0

⇒ a² - 5a + 4a - 20 = 0

⇒ a(a-5) + 4(a-5)=0

⇒ (a+4)(a-5) = 0

⇒ a = -4 ( Not possible )

Or a = 5

Hence, the value of a is 5.