Question

Find the co-ordinates of the point (s) on the curve $y=\frac{x^2-1}{x^2+1},x>0$ such that tangent at these point (s)have the greatest slope.

• A. $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{2}}).$
• B. $(\frac{1}{\sqrt{3}},-\frac{1}{2}).$
• C. $(\frac{1}{3},-\frac{4}{5}).$
• D. $(\sqrt{3},\frac{1}{2}).$

Answer 1

The correct option is B $(\frac{1}{\sqrt{3}},-\frac{1}{2}).$

$y=\frac{x^2-1}{x^2+1}=\frac{x^2+1-2}{x^2+1}=1-\frac{2}{x^2+1}$

$\Rightarrow y=1-\frac{2}{x^2+1}$

S=slope $=\frac{dy}{dx}=\frac{4x}{{(x^2+1)}^2}$

For maximum and minimum of S, $\frac{dS}{dx}=0$

$\frac{dS}{dx}=4\frac{{(x^2+1)}^2.1-x.2(x^2+1).2x}{{(x^2+1)}^4}$

$\frac{dS}{dx}=4\frac{x^2+1-4x^2}{{(x^2+1)}^3}$

or $\frac{dS}{dx}=4\frac{1-3x^2}{{(1+x^2)}^3}=0$

$\therefore x=\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$

Now consider change of sign at $x=\frac{1}{\sqrt{3}},\frac{dS}{dx}$ changes from +ve to -ve.

Hence S has maximum at $x=\frac{1}{\sqrt{3}}$.

When $x=\frac{1}{\sqrt{3}}$ the value of $y=-\frac{1}{2}$

Hence the point is $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{2}}).$