CAT 2023 Slot 1 Solved Paper

© examsiri.com

Question : 46 of 66

Marks: +1, -0

If

x

and

y

are positive real numbers such that

\log _x (x^2+12) =4

and

3 \log _y x=1

, then

x+y

equals

20
68
10
11

Solution:

Given,

\log _x (x^2+12) =4

\;⇒x^2+12=x^4

\;⇒ x^4-x^2-12=0

\;⇒x^4-4 x^2+3 x^2-12=0

\;⇒x^2 (x^2-4) +3 (x^2-4) =0

\;⇒ (x^2-4) (x^2+3) =0⇒

since,

{\x}

is a positive real number (given)

⇒ {\x}=2 .

Now, Given

3 \log _y x=1

\;⇒ \log _y x=\;{1}/{3}

\;⇒x=y^{\;{1}/{3}}

\;⇒y=x^3⇒{\y}=8

\;⇒ {\x}+ {\y}=2+8=10

© examsiri.com

Go to Question:

Prev Question

Next Question