Question

Exhaustive set of values of $a$ for which the function $f\left(x\right)=x^4+a{x}^3+\frac{3x^2}{2}+1$ will be concave upwards along the entire real line is

• A. $[-1,1]$
• B. $[-2,2]$
• C. $[0,2]$
• D. $[0,4]$

Answer 1

The correct option is B $[-2,2]$

$f\left(x\right)=x^4+a{x}^3+\frac{3x^2}{2}+1$
$\Rightarrow f^{\prime }\left(x\right)=4x^3+3a{x}^2+\frac{6x}{2}$
$\Rightarrow f^{\prime \prime }\left(x\right)=12x^2+6ax+3$
as $f\left(x\right)$ is concave upwards along the entire real line,
So, $f^{\prime \prime }\left(x\right)\ge 0,\forall x\in R$
$\Rightarrow 12x^2+6ax+3\ge 0$
Hence,
$D=(6a{)}^2-144\le 0$
$\Rightarrow a^2-4\le 0$
$\Rightarrow a\in [-2,2]$