Test Index

Indices

© examsnet.com

Question : 57 of 86

Marks: +1, -0

If

a^{2}+b^{2}=45

a b=18,

\frac{1}{a}+\frac{1}{b}

is

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{1}{2}$
None

Solution:

Concept:

Use

(a+b)^2 = a^2 + b^2 + 2ab

and

\frac{1}{a}+\frac{1}{b} = \frac{a+b}{ab}

.

Explanation:

Given

a^2+b^2 = 45

and

ab = 18

.

Compute

(a+b)^2 = a^2+b^2 + 2ab = 45 + 2(18) = 45+36 = 81

.

Thus

a+b = \pm 9

.

Now

\frac{1}{a}+\frac{1}{b} = \frac{a+b}{ab} = \frac{\pm 9}{18} = \pm \frac{1}{2}

.

Among the options, only positive

\frac{1}{2}

appears (assuming

a

and

b

are positive).

Answer:

\frac{1}{2}

© examsnet.com

Go to Question:

Prev Question

Next Question