If the tangent to the curve

Solution:

Given:

$x-y+5=0$

$\Rightarrow y=x+5$

$\Rightarrow \frac{d y}{d x}=1$\

Now,

$y=x^{3}+a x+b \quad \ldots(1)$

$\Rightarrow \frac{d y}{d x}=3 x^{2}+a$

Slope of the tangent at $(1,-6)=$ Slope of the given line

$\Rightarrow\left(\frac{d y}{d x}\right)_{(1,-6)}=1$

$\Rightarrow 3+a=1$

$\Rightarrow a=-2$

On substituting $a=-2, x=1$ and $y=-6$ in eq. (1), we get

$-6=1-2+b$

$\Rightarrow b=-5$

$\therefore a=-2$ and $b=-5$