Question

$y=x^4+a{x}^3+\frac{3}{2}{x}^2+1$ is concave along the entire number scale then

• A. $\left|a\right|\ge 1$
• B. $\left|a\right|\le 1$
• C. $\left|a\right|\le 2$
• D. $\left|a\right|>2$

Answer 1

The correct option is C $\left|a\right|\le 2$

$y=x^4+a{x}^3+\frac{3}{2}{x}^2+1$
$y^{\prime }=4x^3+3a{x}^2+3x$
$y^{\prime \prime }=3(4x^2+2ax+1)$
Since, $y$ is concave for all $x$
Therefore, $y^{\prime \prime }>0$
$\Rightarrow 4x^2+2ax+1>0$
Since, $4>0$
Therefore, $b^2-4ac\le 0$ $\Rightarrow 4a^2-16\le 0$
$\Rightarrow \left|a\right|\le 2$
Ans: C