Answer

Correct Answer : c ) 14

Considering the given expression

\({1\over a}+{1 \over b}+{2 \over ab}={2 \over 3}\)

\(⇒a+b+2={2 \over 3}ab\)

\(⇒ 3a + 3b + 6 = 2ab\)

\(⇒ 3b + 6 = a (2b – 3)\)

\(⇒{3 \over 2}(2b-3)+{9 \over 2}+6=a(2b-3)\)

\(⇒(2b-3)(a-{3 \over 2})={21 \over 2}\)

\(⇒ (2b – 3)(2a – 3) = 21 …. (1)\)

As 21 can be be expressed as 1 × 21 or 3 × 7

Now, comparing equation (1) with these values we get

On considering (1 × 21 = 21)

2b – 3 = 1

⇒ b = 2

Similarly,

2a – 3 = 21

⇒ a = 12

And, on considering (3 × 7 = 21)

2b – 3 = 3

⇒ b = 3

And, 2a – 3 = 7

⇒ a = 5

Thus, integral values of (a, b) will be (2, 12), (12, 2), (3, 5), and (5, 3)

Thus, maximum possible sum of possible values of a and b = 2 + 12 = 14

Hence, (c) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA etc.

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