Question

Let $f\left(x\right)=\frac{x^4}{12}+\frac{(a+1)x^3}{6}+2x^2+5$ is concave upward for all $x$ then a can never lie in the interval

• A. $[6,7]$
• B. $[4,5]$
• C. $[1,2]$
• D. $[-3,-1]$

Answer 1

Answer: (A), (B)

$[6,7]$
$[4,5]$

$f^{\prime }\left(x\right)=\frac{x^3}{3}+\frac{(a+1)x^2}{2}+4x$
$f^{\prime \prime }\left(x\right)=x^2+(a+1)x+4$
For $f\left(x\right)$ to be concave upward
$f^{\prime \prime }\left(x\right)=x^2+(a+1)x+4\ge 0\forall x\epsilon R$
$D\le 0$
$\Rightarrow (a+1{)}^2-16\le 0$
$\Rightarrow (a+5)(a-3)\le 0$
$\Rightarrow -5\le a\le 3$