Long Question 1 & 2

Posted on April 22, 2020 by user

Question 1:

A curve with gradient function

5 x - \frac{5}{x^{2}}

has a turning point at (m, 9).

(a) Find the value of m.

(b) Determine whether the turning point is a maximum or a minimum point.
(c) Find the equation of the curve.

Solution:
(a)

\begin{array}{l} \frac{d y}{d x} = 5 x - \frac{5}{x^{2}} \\ At turning point (m, 9), \frac{d y}{d x} = 0. \\ 5 m - \frac{5}{m^{2}} = 0 \\ \frac{5}{m^{2}} = 5 m \\ m^{3} = 1 \\ m = 1 \end{array}

(b)

\begin{array}{l} \frac{d y}{d x} = 5 x - \frac{5}{x^{2}} = 5 x - 5 x^{- 2} \\ \frac{d^{2} y}{d x^{2}} = 5 + \frac{10}{x^{3}} \\ When x = 1, \frac{d^{2} y}{d x^{2}} = 15 (> 0) \\ Thus, (1, 9) is a minimum point . \end{array}

(c)

\begin{array}{l} y = \int (5 x - 5 x^{- 2}) d x \\ y = \frac{5 x^{2}}{2} + \frac{5}{x} + c \\ At turning point (1, 9), x = 1 and y = 9. \\ 9 = \frac{5 {(1)}^{2}}{2} + \frac{5}{1} + c \\ c = \frac{3}{2} \\ Equation of the curve: y = \frac{5 x^{2}}{2} + \frac{5}{x} + \frac{3}{2} \end{array}

Question 2:

A curve has a gradient function kx²– 7x, where k is a constant. The tangent to the curve at the point (1, 3 ) is parallel to the straight line y + x– 4 = 0.

Find

(a) the value of k,

(b) the equation of the curve.

Solution:
(a)

y + x – 4 = 0

y = – x + 4

m = –1

f ’(x) = kx² – 7x

Given tangent to the curve at the point (1, 3) parallel to the straight line

k (1)² – 7 (1) = –1

k – 7 = –1

k = 6

(b)

\begin{array}{l} f' (x) = 6 x^{2} - 7 x \\ f (x) = \int (6 x^{2} - 7 x) d x \\ f (x) = \frac{6 x^{3}}{3} - \frac{7 x^{2}}{2} + c \\ 3 = 2 {(1)}^{3} - \frac{7 {(1)}^{2}}{2} + c at point (1, 3) \\ c = \frac{9}{2} \\ ∴ f (x) = 2 x^{3} - \frac{7 x^{2}}{2} + \frac{9}{2} \end{array}