2013 A-level H2 Mathematics (9740) Paper 1 Question 10 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
\frac{dz}{dx} = 3 - 2z

\frac{1}{3-2z} \frac{dz}{dx} = 1

\int \frac{1}{3-2z} dz = \int 1 dx

-\frac{1}{2} \mathrm{ln} |3-2z| = x + C

Since z > \frac{3}{2},

\mathrm{ln} ( 3 -2z )= -2x -2 C

3-2z = e^{-2x -2C}

2z = 3 - e^{-2x -2c }

z = \frac{3}{2} - Ae^{-2x} where A=\frac{e^{-2C}}{2}

(ii)
\frac{dy}{dx} = z

\frac{dy}{dx} = \frac{3}{2} - Ae^{-2x}

y = \frac{3}{2}x + \frac{Ae^{-2x}}{2} + D

(iii)
\frac{d^{2}y}{dx^{2}} = 2Ae^{-2x}

\frac{d^{2}y}{dx^{2}}= 2(\frac{3}{2} - \frac{dy}{dx})

\frac{d^{2}y}{dx^{2}}= 3 - 2\frac{dy}{dx}

\therefore, a=-2, b = 3

(iv)
Consider A = 0, then y = \frac{3}{2} x + D

Let D > 0, so y = \frac{3}{2}x  + 1 and D=0, then y = \frac{3}{2}x

\therefore, \mathrm{they~are~} y=\frac{3}{2}x ~\mathrm{and}~ y= \frac{3}{2} x + 1

Graph for 10iv

Graph for 10iv

KS Comments:

This is fairly good question. The first part explores students’ abilities to relate a 3-variables differential equation. In (iv), they are expected to understand the question first and set the arbitrary constants to the necessary conditions. Don’t forget to use the graphing calculator to draw the graphs!

Showing 2 comments
    pingbacks / trackbacks

    Leave a Reply to 2014 A-level H1 Mathematics (8864) Suggested Solutions | The Culture Cancel reply

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt
    0

    Start typing and press Enter to search