Model Test 1 Paper 1 Q2

Model Test 1 Paper 1 Q2

Postby skkwee on Wed Oct 31, 2007 1:45 pm

Question 2(a)

Method 1:
f(x) = kx -2

{\rm let } f(x)^{-1}= y

f(y) = x

ky - 2 = x

ky = x + 2

y =\frac{{x + 2}}{k}

f^{-1}(x) = \frac{{x + 2}}{k}


Compare the answer
f^{ -1} (x) = \frac{{x + 2}}{k}{\rm  with }f^{ - 1} (x) = \frac{{x + 2}}{5}


k = 5


Method 2:
f-1(x)=(x+2)/5

let f(x) = y
f-1(y) = x
(y + 2)/5 = x
(y + 2) = 5x
y = 5x -2

Substitute y with f(x) (because f(x) = y)

f(x) = 5x -2

Compare with f(x) = kx -2 (that given in the question), we get

k = 5
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Re: Model Test 1 Paper 1 Q2

Postby zacknistelrooy on Wed Jul 02, 2008 7:28 pm

thk :D :)
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