Author Topic: tsfx 2008 exam 1 (Read 1718 times) Tweet Share

unknown id · « **on:** November 06, 2008, 06:10:12 pm »

are the solutions to 10.b) wrong by any chance?

Exam: http://www.tsfx.com.au/vic/unit4-2008-MM-schools-exam-1-FINAL.pdf

Solutions: http://www.tsfx.com.au/vic/unit4-2008-MM-schools-exam-1-SOLUTIONS-FINAL.pdf

i get $\frac{0.1sin(1) + 0.2cos(1)}{e^2}$ while the answers say $\frac{0.2cos(1) - 0.2sin(1)}{e^2}$

unknown id · « **Reply #1 on:** November 06, 2008, 06:33:29 pm »

this is my working:

$f(x+h) - f(x)$

$= hf'(x)$

$= -0.1 \times f'(1)$

$= -0.1 \times \frac{-sin(1)-2cos(1)}{e^2}$

$= 0.1 \times \frac{sin(1)+2cos(1)}{e^2}$

$= \frac{0.1sin(1)+0.2cos(1)}{e^2}$

/0 · « **Reply #2 on:** November 06, 2008, 06:35:56 pm »

Quote from: unknown id on November 06, 2008, 06:33:29 pm

this is my working:

$f(x+h) - f(x)$

$= hf'(x)$

$= -0.1 \times f'(1)$

$= -0.1 \times \frac{-sin(1)-2cos(1)}{e^2}$

$= 0.1 \times \frac{sin(1)+2cos(1)}{e^2}$

$= \frac{0.1sin(1)+0.2cos(1)}{e^2}$

I got too

unknown id · « **Reply #3 on:** November 06, 2008, 06:37:17 pm »

^ phew, i was panicking for a moment there!

shinny · « **Reply #4 on:** November 06, 2008, 08:33:22 pm »

Yeah I remember seeing that too...they screwed up a sign somewhere

Tea.bag · « **Reply #5 on:** November 06, 2008, 08:38:17 pm »

got that too

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Author Topic: tsfx 2008 exam 1 (Read 1718 times) Tweet Share

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