Test Index

Surds and Indices Practice Test 2

© examsnet.com

Question : 12 of 17

Marks: +1, -0

Given that

abc = 1

, then find the value of

(\frac{1}{1+a+b^{-1}}

+\frac{1}{1+b+c^{-1}}

+\frac{1}{1+c+a^{-1}})

.

ab
$\frac{1}{ab}$
1
0

Solution:

Concept:

The condition

abc=1

allows reciprocal substitution to simplify the given expression.

Explanation:

Given

abc=1

, we get

c^{-1}=ab

and

a^{-1}=bc

.

Let

S=\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}

.

For the second term,

1+b+c^{-1}=1+b+ab=b(1+a+b^{-1})

.

So

\frac{1}{1+b+c^{-1}}=\frac{b^{-1}}{1+a+b^{-1}}

.

For the third term,

1+c+a^{-1}=1+c+bc=c(1+b+ab)=cb(1+a+b^{-1})

.

Since

abc=1

,

cb=a^{-1}

, so

1+c+a^{-1}=a^{-1}(1+a+b^{-1})

.

Hence

\frac{1}{1+c+a^{-1}}=\frac{a}{1+a+b^{-1}}

.

Now

S=\frac{1+b^{-1}+a}{1+a+b^{-1}}

.

Therefore

S=1

.

Answer:

Option C:

1

© examsnet.com

Go to Question:

Prev Question

Next Question