IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: I'm a mistake - legalize abortion! on May 11, 2010, 09:07:15 pm
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ok can anybody help me?q.no.10 of M/J 2009
The function f is defined by f : x ? 2x2 ? 12x + 13 for 0 ? x ? A, where A is a constant.
(i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants.
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry.
(iii) When A has this value, find the range of f.
The function g is defined by g : x ? 2x2 ? 12x + 13 for x ? 4.
(iv) Explain why g has an inverse.
(v) Obtain an expression, in terms of x, for g?1(x).
thanks in advance
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Sorry internet is slow right now..Paper is being downloaded.
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hello anybody there who can help?? i would be thankful :D
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ok can anybody help me?q.no.10 of M/J 2009
The function f is defined by f : x ? 2x2 ? 12x + 13 for 0 ? x ? A, where A is a constant.
(i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants.
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry.
(iii) When A has this value, find the range of f.
The function g is defined by g : x ? 2x2 ? 12x + 13 for x ? 4.
(iv) Explain why g has an inverse.
(v) Obtain an expression, in terms of x, for g?1(x).
thanks in advance
Nid to your rescue :)
I hope im not late
i) completing the square method 2(x-3)2-5
ii) the line of symmetry is given by -b/2a or you can simply derivate and equate to 0
using the first method
-(-12)/ 2X2= 3
x=3 (line of symmetry passes through x=3)
second method
dy/dx= 4x-12
line of symmetry will pass through the maxima or minima as it's a quadratic
4x-12=0
x=3
u can use any method
iii) Just substitute 3 in the eqn
u get y= -5
now to decide whether this is max or min
sub 3 in d2y/dx2 = 4x
4X3=12 >0 hence minima
that mean all values of y will be greater than -5
y>=-5 is the range
iv) g has an inverse because it is a 1-1 function. It has a given set of domain.
v) y=2x2-12x-13
We wrote this is the form (a-b)2+c
2(x-3)2-5
y=2(x-3)2-5
make x the subject
y+5=2(x-3)2
y+5
----= (x-3)2
2
/2} = (x-3))
/2} +3)
in terms of x, so u swap the y with x
g-1(x)=
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"nid to the rescue." lol(superman)
thank u so much miss "nidman"
I hope u r not mad at me anymore
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superwoman :P
Nooo...I'm not :D
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@nid why dont u change ur name to "nicole nidman"? ;D Sounds like a nice name to me.... :P :P
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no, I like to be original :P
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(ii) Only a few candidates realised that because y = f(x) has a line of symmetry at x = 3, then A = 6.
y is it so in the june 2009 q10?
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didnt get ur questio n :P u sayin why is the ER say so......... or wat
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the domain is
0<=x<=A
you found x=3 is line of symmetry
to the left of this line x is only uptil 0 difference of 3 units. To the right as well it should be at the same difference to have a reflection along x=3. Hence maximum value x can is 6
Hope ur getting me
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hmm i actualy got that...... :P
but i didnt understand why ifraha18 was quoting from the Examiner Report and asking why many pppl didnt get it correct lolz...
;D ;D ;D
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WHICH PAPER U HAVE
??? ???
ANSWER ME
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??? ??? ??? ???
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but line of smmetry comes 6... acc to mark scheme..
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line of symmetry is at x=3
Value of A is 6