Mathematics - Application of Derivative Question with Solution | TestHub

The equation of the given curve is $\mathrm{y}=\frac{1}{\mathrm{x}-3},\mathrm{x}\ne 3$ The slope of the tangent to the given curve at any point $(\mathrm{x},\mathrm{y})$ is given by $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{(\mathrm{x}-3{)}^2}$ For tangent having slope $2,$ we must have $2=\frac{-1}{(x-3{)}^2}$ $$\Rightarrow 2(x-3{)}^2=-1\Rightarrow (x-3{)}^2=-\frac{1}{2}$$ which is not possible as square of a real number cannot be negative. Hence, there is no tangent to the given curve having slope 2 .