Find the point on the curve y = x³ - 11x + 5 at which the equation of tangent is y = x - 11.

OR

Using differentials, find the approximate value of √49.5.

Given; y = x³ - 11x + 5 and the equation of tangent is y = x - 11

∴ Slope of the tangent is the coefficient of x [y = mx + c] = 1

From the first equation;

From (i) and (ii)

⇒ 3x² − 11 = 1

⇒ 3x² = 12

∴ x = ±2

⇒ y = x − 11 = −13, −9

∴ The required points are (2,−9) and (−2,−13)

OR

Given; √49.5

Let f(x) = √x and f(x + δx) = √(x + δx)

Here; x = 49 and δx = 0.5

We know;

⇒f(x + δx) = δx f^’(x) + f(x)

Here;

= 7.036