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# p2 question on functionss watch

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1. Need some help on functions in P2 please...This question is Number 2 from June 2003 Paper (edexcel)

f: x -> x^2 - 2x + 3 0 less the or eqal to x wich is less then or equal to 4
g : x -> ax^2 + 1 where a is a constant

a) Find the range of f
b) given that gf(2) = 16 find the value of a.

a) i thort taht u shud only put in f(0) and f(4) into the equation but that ent right....y not? how do u work it out..

b) firstly u put in f(2) into x^2 - 2x =3 and get f(x) = 3
then i thort u use the f(x) = 3 and put that into ax^2 = 1 to get 3^2a + 1 = 16....but that isnt right either..help plsss
Need some help on functions in P2 please...This question is Number 2 from June 2003 Paper (edexcel)

f: x -> x^2 - 2x + 3 0 less the or eqal to x wich is less then or equal to 4
g : x -> ax^2 + 1 where a is a constant

a) Find the range of f
b) given that gf(2) = 16 find the value of a.

a) i thort taht u shud only put in f(0) and f(4) into the equation but that ent right....y not? how do u work it out..

b) firstly u put in f(2) into x^2 - 2x =3 and get f(x) = 3
then i thort u use the f(x) = 3 and put that into ax^2 = 1 to get 3^2a + 1 = 16....but that isnt right either..help plsss
You should really post your maths queries now in the maths sub-forum, at the top of this page. You'll get answered a lot faster.

a)
f(x) = x² - 2x + 3, 0 <= x <= 4
g(x) = ax² + 1, a const

I don't know if P2 covers max and min of functions, but if it does this is how.

f'(x) = 2x - 2
at f'(x) = 0
2x - 2 = 0
x = 1
===
This is a minimum (obvious since you have a quadratic/parabola)

f(x=1) = 1² -2.1 + 3
f(1) = 2
fmin = 2
======

To find the max,

f(0) = 0 - 0 + 3
f(0) = 0, end of range of x
=====

f(4) = 4² - 2.4 + 3
f(4) = 11, top of range of x
======

so, by comparing results,

fmax = 11
======

range of f is
2 <= f <= 11
=========

b)
g(f(x)) = a(x² - 2x + 3)² + 1
g(f(2)) = a(2² - 2.2 + 3)² + 1
g(f(2)) = a(3)² + 1
g(f(2)) = 9a + 1

then,

9a + 1 = 16
9a = 15
a = 5/3
=====

That looks like what you would have gotten!

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Updated: June 13, 2004
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