Short Questions (Question 14 – 16)

Posted on April 22, 2020 by user

Question 14:

Given y = x (6 – x), express

$y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18$ in terms of x in the simplest form.

Hence, find the value of x which satisfies the equation

$y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0$

Solution:

$\begin{array}{l} y = x (6 - x) = 6 x - x^{2} \\ \frac{d y}{d x} = 6 - 2 x \\ \frac{d^{2} y}{d x^{2}} = - 2 \\ ∴ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = (6 x - x^{2}) (- 2) + x (6 - 2 x) + 18 \\ = - 12 x + 2 x^{2} + 6 x - 2 x^{2} + 18 \\ = - 6 x + 18 \\ y \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + 18 = 0 \\ - 6 x + 18 = 0 \\ x = 3 \end{array}$

Question 15:

Find the coordinates of the point on the curve, y = (4x – 5)² such that the gradient of the normal to the curve is

$\frac{1}{8}$ .

Solution:

y = (4x – 5)²

$\frac{d y}{d x}$ = 2(4x – 5).4 = 32x – 40

Given the gradient of the normal is 1/8, therefore the gradient of the tangent is –8.

$\frac{d y}{d x}$ = –8

32x – 40 = –8

32x = 32

x = 1

y = (4(1) – 5)²= 1

Hence, the coordinates of the point on the curve, y = (4x – 5)² is (1, 1).

Question 16:

A curve has a gradient function of kx² – 7x, where k is a constant. The tangent to the curve at the point (1, 4) is parallel to the straight line y + 2x–1 = 0. Find the value of k.

Solution:

Gradient function of kx²– 7x is parallel to the straight line y + 2x–1 = 0

$\frac{d y}{d x}$ = kx²– 7x

y + 2x –1 = 0, y = –2x + 1, gradient of the straight line = –2

Therefore kx²– 7x = –2

At the point (1, 4),

k(1)² – 7(1) = –2

k – 7 = –2

k = 5