BITSAT 2013 Paper

The equation of the given curve is y = $\frac{1}{x-3}$ , x ≠ 3 The slope of the tangent to the given curve at any point (x, y) is given by $\frac{dy}{dx}$ = $\frac{-1}{(x-3)^2}$ For tangent having slope 2, we must have 2 = $\frac{-1}{(x-3)^2}$ ⇒ 2 $(x-3)^2$ = - 1 ⇒ $(x-3)^2$ = - $\frac{1}{2}$ 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.