Find the points on curve $y = x^{3} - 11 x + 5$ at which equation of tangent is $y = x - 11$ .

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Solution

Given equation of curve is $y = x^{3} - 11 x + 5$ .
Slope of tangent to the given curve is $\frac{d y}{d x}$ .
Therefore,
$\frac{d y}{d x} = 3 x^{2} - 11$

Also, slope of the tangent $y = x - 11$ is $1$ .
$\frac{d y}{d x} = 1$

$3 x^{2} - 11 = 1$
$3 x^{2} = 12$
$x^{2} = 4$
$x = \pm 2$
Then, when $x = 2$
Then, $y = (2)^{3} - 11 (2) + 5 = - 9$
When $x = - 2, y = 19$ .
Hence, the required points on the curve are $(2, - 9)$ and $(- 2, 19)$ .

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