Question

If 4^{4b - 3a} = 2^{2b + c} = 8^{c - a} and a + b + c = 11 , then find the value of \(4(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\)

• A. \(\frac{11}{9}\)
• B. \(\frac{9}{11}\)
• C. \(\frac{6}{5}\)
• D. \(\frac{5}{6}\)

Answer 1

The Correct Option is A

4^4b-3a = 2^2b+c = 8^c-a
2^2(4b-3a) = 2^2b+c = 2^3(c-a)
2^8b-6a = 2^2b+c = 2^3c-3a
2^8b-6a = 2^3c-3a
8b - 6a = 3c - 3a
8b = 3a + 3c ........ (1)
2^2b+c = 2^3c-3a
2b + c = 3c - 3a
3a + 2b = 2c ........ (2)
2^8b-6a = 2^2b+c
8b - 6a = 2b + c
6b = 6a + c ....... (3)
From (1) and (2):
8b = 2c - 2b + 3c => 10b = 5c => c = 2b =>b = c/2
From (1) and (3):
8b = 3a + 3 (6b - 6a) => 15a = 10b => 3a = 2b =>a = 2b/3 = c/3
a + b + c = 11
(c/3) + (c/2) + c = 11
c = 6, b = 3 and a = 2
Value of \(4[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}]\)
\(=4[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}]\)
\(=4[\frac{1}{6}+\frac{1}{18}+\frac{1}{12}]\)
\(=4[\frac{6+2+3}{36}]\)
\(=\frac{11}{9}\)
So, the correct option is (A) : \(\frac{11}{9}\)