Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If a, b, c are three distinct positive real number which are in HP, then (3a-2b)/(2a-b) + (3c + 2b)/(2c - b) is

(1) greater than or equal to 10

(2) less than or equal to 10

(3) only equal to 10

(4) None of these

Solution:

Given a, b, c are in HP.

So b = 2ac/(a+c)

b(a+c) = 2ac…(i)

(3a-2b)/(2a-b) + (3c + 2b)/(2c – b) = (6ac+4bc-3ab-2b2+6ac+4ab-3bc-2b2)/(4ac-2bc-2ab+b2)

= (12ac+b(a+c)-4b2)/(4ac-2b(a+c)+b2)

Substitute (i)

= (12ac+2ac-4b2)/(4ac-4ac+b2)

= (14ac – 4b2)/b2

= 14ac/b2 – 4

Substitute 2ac/b = a+c

= 7(a+c)/b – 4

We know 2/b = 1/a + 1/c

7(a+c)/b – 4 = 7(a+c)(½) (1/a + 1/c)- 4

= (7/2)(1+ a/c + c/a + 1) – 4

= (7/2)(2+ a/c + c/a) – 4

= 7+(7/2)(a/c + c/a) – 4

= 3 + (7/2)(a/c + c/a)

Using AM relation on a/c , c/a

((a/c)+(c/a))/2 ≥ √((a/c)(c/a))

(a/c)+(c/a) ≥ 2

So 3 + (7/2)(a/c + c/a) ≥ 3+(7/2)2

≥ 10

Hence option (1) is the answer.

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