Question

Let $a$ and $b$ be the positive numbers such that the curve $y=a\ln x+b{x}^2+8x-2011$ has exactly one horizontal tangent. Then the smallest value of $b$ such that value of all inflection points of the curve are atmost $1,$ is

Answer 1

$y=a\ln x+b{x}^2+8x-2011$
$\Rightarrow y^{\prime }=\frac{a}{x}+2bx+8=0$ has exactly one root
$\Rightarrow 2b{x}^2+8x+a=0$ has exactly one root
From $D=0$
$\Rightarrow 64-8ab=0$
$\Rightarrow ab=8$
[One root cannot be negative because $a$ and $b$ are positive]
$y^{{}^{\prime \prime }}=\frac{-a}{x^2}+2b=0$
From $y^{\prime \prime }=0$
$\Rightarrow x^2=\frac{a}{2b}=\frac{4}{b^2}\le 1$
$\Rightarrow b^2\ge 4$
$\Rightarrow b\ge 2.(\because b>0)$
Hence, smallest value of $b$ is $2.$