Question

The coordinates of the point on the curve $(x^2+1)(y-3)=x$ where a tangent to the curve has the greatest slope are given by

• A. $(\sqrt{3},3+\sqrt{3}/4)$
• B. $(-\sqrt{3},3-\sqrt{3}/4)$
• C. $(0,3)$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $(\sqrt{3},3+\sqrt{3}/4)$
C $(0,3)$

Solving for y the given equation we have

$y=3=\frac{x}{x^2+1}\Rightarrow \frac{dy}{dx}=\frac{1-x^2}{{(1+x^2)}^2}=f\left(x\right)$

Now $f^{\prime }\left(x\right)=\frac{-2x(3-x^2)}{{(1+x^2)}^2}$

For extremum of $f\left(x\right)$, we have $f^{\prime }\left(x\right)=0\Rightarrow x=0,x=\pm \sqrt{3}$

At $x=0,f^{\prime }\left(x\right)$ changes sign from positive to negative

$\Rightarrow f\left(x\right)$ has maxima at $x=0$.

Thus the required point is $(0,3)$