Question

Least integral value of $x$ satisfying the inequation $(x^2+1)<{(x+2)}^2<2x^2+4x-12$ is

• A. $-2$
• B. $-5$
• C. $2$
• D. $5$

Answer 1

The correct option is D $5$

First consider $x^2+1<(x+2{)}^2$

$\Rightarrow x^2+1<x^2+4x+4$

$\Rightarrow 4x>-3$

$\Rightarrow x>\frac{-3}{4}$

Now consider $(x+2{)}^2<2x^2+4x-12$

$\Rightarrow x^2>16$

$\Rightarrow x<-4;x>4$

Now consider $x^2+1<2x^2+4x-12$

$\Rightarrow x^2+4x-13>0$

$\Rightarrow x>\frac{-4+\sqrt{68}}{2}$ ; $x<\frac{-4-\sqrt{68}}{2}$

Least value satisfying this inequality is $x=5$