Question

The interval in which the inequality $\sqrt{x^2+3x+2}<1+\sqrt{x^2-x+1}$ holds good is

• A. $(-\infty ,-2]\cup (-1,\frac{3}{4})$
• B. $x\in (-\infty ,-2)\cup [-1,\frac{-1+\sqrt{13}}{6}]$
• C. $(-\infty ,-2)\cup [-1,0]$
• D. $-2\le x\le -1$

Answer 1

The correct option is B $x\in (-\infty ,-2)\cup [-1,\frac{-1+\sqrt{13}}{6}]$

$\sqrt{(x+1)(x+2)}<1+\sqrt{{(x-\frac{1}{2})}^2+\frac{3}{4}}$

Now,$(x+1)(x+2)\ge 0$ and $\sqrt{x^2-x+1}\ge \frac{3}{4}$

$\Rightarrow x\le -2\cup x\ge -1\therefore x\le -2$ and $x\ge -1$

Again,$x^2+3x+2<1+x^2-x+1+2\sqrt{x^2-x+1}\Rightarrow 2x<\sqrt{x^2-x+1}\forall x\le -2\cup x\ge -\mathrm{1...}\left(A\right)$

Squaring$($if $x\ge 0)$ both sides of $2x<\sqrt{x^2-x+1},$

We get $3x^2+x-1<0;$

$3(x^2+\frac{x}{3}-\frac{1}{3})<0x^2+2\cdot \frac{x}{6}+\frac{1}{36}-\frac{1}{36}-\frac{1}{3}<0\Rightarrow {(x+\frac{1}{6})}^2-\left(\frac{\sqrt{13}}{6}\right)<0\Rightarrow |x+\frac{1}{6}|<\frac{\sqrt{13}}{6}\left|x\right|<\frac{-1+\sqrt{13}}{6}...\left(B\right)$

By (A) and (B) we get,$\frac{-1-\sqrt{13}}{6}\le x<\frac{-1+\sqrt{13}}{6}$