CAT 2020 Exam Shift 2 Solved Paper

© examsiri.com

Question : 57 of 76

Marks: +1, -0

The value of

\log _{a}(\;{a}/{b})+\log _{b}(\;{b}/{a})

, for

1< a ≤ b

cannot be equal to

$-0.5$
$1$
$0$
$-1$

Solution:

\log _{a}(\;{a}/{b})+\log _{b}(\;{b}/{a})

=\log _{a} a-\log _{a} b+\log _{b} b-\log _{b} a

=1-\log _{a} b+1-\log _{b} a \;\; [\log _{n} n=1]

since

(\log _{a} b+\log _{a} b) ≥ 2

∴

The above value is

≤ 0

.

∴ 1

can't be the answer.

© examsiri.com

Go to Question:

Prev Question

Next Question