Question

The point of inflection for the curve $y=\frac{x+1}{x^2+1}$ is

• A. $(-2-\sqrt{3},\frac{1-\sqrt{3}}{4})$
• B. $(-2+\sqrt{3},\frac{-1-\sqrt{3}}{4})$
• C. $(-2+\sqrt{3},\frac{1+\sqrt{3}}{4})$
• D. $(1,1)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $(-2-\sqrt{3},\frac{1-\sqrt{3}}{4})$
C $(-2+\sqrt{3},\frac{1+\sqrt{3}}{4})$
D $(1,1)$

$y=\frac{x+1}{x^2+1}$
$\frac{dy}{dx}=\frac{-x^2-2x+1}{(x^2+1{)}^2}$
$\frac{d^{2}y}{d{x}^2}=\frac{2x^3+6x^2-6x-2}{(x^2+1{)}^3}$
For point of inflection $\frac{d^{2}y}{d{x}^2}=0$
$\Rightarrow 2x^3+6x^2-6x-2=0$
By inspection, $x=1$ is a solution.
$\Rightarrow (x-1)(x^2+4x+1)=0$
Hence the roots of the equation are $1,-2-\sqrt{3},-2+\sqrt{3}$.
At all the three points, the concavity changes. Hence, they are all points of inflection.
$f\left(1\right)=1$
$f(-2-\sqrt{3})=\frac{1-\sqrt{3}}{4}$
$f(-2+\sqrt{3})=\frac{1+\sqrt{3}}{4}$
Hence, options $A,C$ and $D$ are correct.