Question

If ‘P’ be a point on the graph of $y=\frac{x}{1+x^2}$ then co-ordinates of ‘P’ such that tangent drawn to the curve at ‘P’ has greatest slope in magnitude is

• A. (0, 0)
• B. $(\sqrt{3},\frac{\sqrt{3}}{4})$
• C. $(-\sqrt{3},-\frac{\sqrt{3}}{4})$
• D. (1,1)

Answer 1

The correct option is A (0, 0)

$\frac{dy}{dx}=\frac{1-x^2}{(1+x^2{)}^2}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{2x(x^2-3)}{(1+x^2{)}^3}$
Using the sign scheme for $\frac{d^{2}y}{d{x}^2}$ we get that x = $\pm \sqrt{3}$ are the points of minima for f'(x), and x = 0 is the point of maxima for f'(x)
f'(0) = 1, f'($\pm \sqrt{3})=\frac{1-3}{(1+3{)}^2}=-\frac{1}{8}$
Thus x = 0 is the required point.