KVPY 2020-21 SB SX Exam Previous Papers

© examsiri.com

Question : 10 of 120

Marks: +1, -0

Let

{x}, {y}

be real numbers such that

{x}>2 {y}>0

and

2 \log (x-2 y)=\log x+\log y

.
Then the possible value (s) of

{x}/{y}

is 1 only
are 1 and 4
is 4 only
is 8 only

Solution:

\log (x-2 y)^{2}=\log (x y)

\⇒(x-2 y)^{2}=x y

\⇒({x-2 y}/{y}^{2})={x}/{y}

\⇒({x}/{y})^{2}-5({x}/{y})+4=0

\⇒{x}/{y}=1,4

\⇒{x}/{y}=4 \;\;\;\; (\as {x}/{y}>2)

© examsiri.com

Go to Question:

Prev Question

Next Question