Question

if all the solutions of the inequality x² -6ax + 5a²<=0 are also the solutions of inequality x²- 14x + 40<=0 then find the number of possible integral values of a.​

Answer 1

Answer:

if x square -6ax + 5 bada hoga bada and equal to zero are also the solution of inequality x square - 14x + 40 is bigger and equal to zero then

x ≤0

and, a = 50 , 10

Answer 2

Step-by-step explanation:

• So let p(x) = x^2 – 6ax + 5a^2
•                   = x^2 – ax – 5ax + 5a^2
•                   = x(x – a) – 5a (x – a)
•                    (x – a) (x – 5a) = 0
•                  Or x = a,5a
• So let q(x) = x^2 – 14 x + 40
•                   = x^2 – 10 x – 4x + 40
•                 So x(x – 10) – 4(x – 10) = 0
•                       (x – 10)(x – 4) = 0
•                             So x = 10,4
• Now q(m) ≤ 0 and q(5m) ≤ 0
•       Or q(x) ≤ 0
• So we get m^2 – 14m + 40 ≤ 0
•          Or (m – 10)(m – 4) = 0
•               Or m = 10,4
• So we have 4 ≤ m ≤ 10-------1
• Also 2q(5m) ≤ 0
• So m^2 – 14(5m) + 40
•     (5m)^2 – 14(5m) + 40
•        25 m^2 – 70 m + 40 ≤ 0
•        25m^2 – 50m – 20m + 40 ≤ 0
•          25 m(m – 2) – 20(m – 2) ≤ 0
•            So (m – 2) (25m – 20) ≤ 0
•             Or m – 2 = 0  25 m – 20 ≤ 0
•             Or m = 2, m = 4/5
•        Now 4/5 ≤ m ≤ 2 -------- 2
• So from 1 and 2 on integration we get 0. Therefore the number of possible values is 0.

Reference link will be

https://brainly.in/question/28524191