NIMCET 2011 Question Paper with solutions

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Question : 28 of 120

Marks: +1, -0

Solution set of the 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:

Given,

\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<0

⇒ \log _{3} {(x+2)(x+4)}/{(x+2) 7}<0 × 1 ≠-2

⇒ \log _{3} {(x+4)}/{7}< 0, x ≠-2

⇒ {x+4}/{7}<1, x ≠-2

⇒ x<3, x ≠-2

∴ -2 < x < 3 \;\;\;\; [ ∵ x<-2 \log (x+2)

is not defined

]

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