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

If 1/a, 1/H, 1/b are in AP, then (H+a)/(H-a) + (H+b)/(H-b) =

(1) 2

(2) 4

(3) 0

(4) 1

Solution:

Given 1/a, 1/H, 1/b are in AP.

2/H = 1/a + 1/b

2/H = (b+a)/ab

H/2 = ab/(a+b)

H = 2ab/(a+b)

(H+a) = [2ab/(a+b)] + a

= [2ab + a(a+b)]/(a+b)

= a[2b+a+b)/(a+b)

= a(a+3b)/(a+b)

(H-a) = [2ab/(a+b)] – a

= [2ab – a(a+b)]/(a+b)

= a[2b-a-b)/(a+b)

= a(b-a)/(a+b)

(H+a)/(H-a) = (a+3b)/(b-a) ..(i)

Similarly (H+b)/(H-b) = (3a+b)/(a-b)…(ii)

Add (i) and (ii)

(H+a)/(H-a) + (H+b)/(H-b) = [(a+3b)/(b-a)] + [(3a+b)/(a-b)]

= [(a+3b)/(b-a)] + [(-3a-b)/(b-a)]

= (2b-2a)/(b-a)

= 2(b-a)/(b-a)

= 2

Hence option (1) is the answer.

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