Question

Let $(a-1){\left(x^2+\sqrt{3}x+1\right)}^2-(a+1)\left(x^4-x^2+1\right)\le 0\forall x\in R$, then which of the following is/are correct?

• A. $a\in \left[-\frac{1}{\sqrt{3}},\frac{4}{\sqrt{3}}\right]$
• B. Largest possible value of a is $\frac{2}{\sqrt{3}}$
• C. Number of possible integral values of a is 3
• D. Sum of all possible integral values of a is ' 0 '

Answer 1

Answer: (B), (C), (D)

$(a-1){\left(x^2+\sqrt{3}x+1\right)}^2-(a+1)\left[{\left(x^2+1\right)}^2-(x\sqrt{3}{)}^2\right)\le 0$
$\text{ or }(a-1){\left(x^2+\sqrt{3}x+1\right)}^2-(a+1)\left[x^2+x\sqrt{3}+1\right)$
$\left(x^2-x\sqrt{3}+1\right)\le 0$
$\left(x^2-x\sqrt{3}+1\right)[(a-1)\left(x^2+\sqrt{3}x+1\right)-(a+1)$ $\left(x^2-x\sqrt{3}+1\right)]\le 0\forall x\in R$
$\Rightarrow -2\left(x^2+1\right)+2a\sqrt{3}x\le 0$
$\Rightarrow x^2-a\sqrt{3}x+1\ge 0\forall x\in R$
$\Rightarrow 3a^2-4\le 0(D\le 0)$
$\Rightarrow a\in \left[-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right]$

$\Rightarrow \text{ Number of possible integral value of a is }\{-1,0,1\}$
$\Rightarrow 3$
and sum of all integral values of a is $-1+0+1=0$