KEAM 2011 Math Paper

Question : 108 of 120

Marks: +1, -0

\left|2x-3\right| < \left|x+5\right|,

x

lies in the interval

Solution:

\left|2x-3\right| < \left|x+5\right|

\Rightarrow\;\;\pm(2x-3) < \pm(x+5)

.....(i)

First we take positive sign and negative sign on both sides,

\Rightarrow\;\;2x-3 < x+5

\Rightarrow\;\;x<8

.....(ii)

-(2x-3)<-(x+5)

\Rightarrow\;\;2x-3>x+5

\Rightarrow\;\;x>8

.....(iii)

Now, we take negative sign in RHS and positive in LHS,

\Rightarrow\;\;2x-3<-x-5

\Rightarrow\;\;3x<-2

\Rightarrow\;\;x<-\frac{2}{3}

......(iv)

and we take negative sign LHS and positive in

\text{RHS}

\Rightarrow\;\;-(2x-3) < x+5

\Rightarrow\;\;-2x+3 < x+5

\Rightarrow\;\;-2<3x

\Rightarrow\;\;-\frac{2}{3} < x

.......(v)

From Eqs. (ii) and (v),

-\frac{2}{3} < x<8

\Rightarrow\;\;x\in\left(-\frac{2}{3},8\right)

From Eqs. (iii) and (iv),

x\in\left(-\infty,-\frac{2}{3}\right)\cup(8,\infty)

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