Question

If the tangent to the curve $y=x^3+\mathrm{ax}+b$ at (1, -6) is parallel to the line x - y + 5 = 0, then the value of a - b is .

Answer 1

Slope of the line x - y + 5 = 0 is 1. (Comparing with y = mx + c)
Therefore, Slope of the tangent to the curve $y=x^3+\mathrm{ax}+b$ is also 1.
$\frac{\mathrm{dy}}{\mathrm{dx}}=3x^2+a{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(1,-6)}=3+a\because \mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{curve}\mathrm{at}\left(1,-6\right)\mathrm{is}1.\therefore 3+a=1\Rightarrow a=-2\because \left(1,-6\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{given}\mathrm{curve}.\therefore -6=1-2+b\Rightarrow b=-5\therefore a-b=3.$