NIMCET 2019 Question Paper with solutions

© examsiri.com

Question : 11 of 120

Marks: +1, -0

Solution set of inequality

\log _{3}(x+2)(x+4)+\log _{\;{1}/{3}}(x+2)<\;{1}/{2} \log _{√{3}} 7

is

(-2, -1)
(-2, 3)
(-1, 3)
(3, ∞)

Solution:

We have,

\log_{3}(x+2)(x+4)+\log_{1/3}(x+2)<\;\;{1}/{2}\log_{√{3}}7

⇒\log_{3}(x+2)(x+4)-\log_{3}(x+2)<\log_{3}7

⇒\log_{3}\;\;{(x+2)(x+4)}/{x+2}<\log_{3}7

⇒\log_{3}(x+4)<\log_{2}7

⇒(x+4)<7

and

x+2 > 0

x < 3

and

x > -2

∴ x∊ (-2,3)

© examsiri.com

Go to Question:

Prev Question

Next Question