Qualification > Queries
Pure maths 1 problem
Freaked12:
I am finding Question 5 part (i) October/November 2008 difficult to understand.
The function f is such that f(x) = a ? b cos x for 0? ? x ? 360?, where a and b are positive constants.
The maximum value of f(x) is 10 and the minimum value is ?2.
(i) Find the values of a and b.
Freaked12:
a - bcos x for O<= x <=360, where a and b are positive constants.
The max value of f(x) is 10 and the minimum value is -2
nid404:
simultaneous equations
a+b=10
a-b=-2
2a=8
a=4
b=10-4=6
Freaked12:
--- Quote from: nid404 on May 07, 2010, 04:52:15 pm ---simultaneous equations
a+b=10
a-b=-2
2a=8
a=4
b=10-4=6
but why is it a+b and a-b
Where did u get the hint from
--- End quote ---
nid404:
max value of cos x=1
since max value of f(x)=10
a+b=10 (cos x=1)
least value of cos x=-1
since min value of f(x)=-2
a-b=-2
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