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Answer:$a=-2 \text { and } b=-5$

Hint:$\text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

Given:

$\text { The slope of the tangent to the curve } y=x^{3}+a x+b \text { at }(1,-6)$

Solution:

First we will find the slope of tangent, we get

$y=x^{3}+a x+b$

$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}(a x)+\frac{d}{d x}(b)$

$\frac{d y}{d x}=3 x^{3-1}+a \frac{d x}{d x}+0$

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

$\text { The slope of tangent to the curve } y=x^{3}+a x+b \text { at }(1,-6) \text { is }$

$\frac{d y}{d x(x=1, y=-6)}=3(1)^{2}+a$

$\Rightarrow 3+a \quad \rightarrow(1)$

$\text { The given line is } x-y+5=0$

$y=x+5 \text { is the form of equation of a straight line } y=m x+c$

$\text { where } \mathrm{m} \text { is the slope of line. }$

$\text { So, the slope of the line is } y=1 \times x+c$

$\text { so, the slope is } 1 \quad \rightarrow(2)$

$\text { Also the point }(1,-6) \text { lie on the tangent }$

$x=1 \& y=-6 \text { satisfies the equation }$

$y=x^{3}+a x+b$

$-6=(1)^{3}+a \times 1+b$

$-6=1+a+b$

$a+b=-7 \quad \rightarrow(3)$

$\text { Since the tangent is parallel to the line, from (1) \& (2) }$

$3+a=1$

$a=-2$

$\text { from (3) }$

$a+b=-7$

$-2+b=-7$

$b=-7+2$

$b=-5$

$\text { so, the value is } a=-2 \text { and } b=-5$

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