Qualification > Math
[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !
islu_jf:
Hey its Oct/Nov 2001 P2
Question number one!.. i know how to do the whole thing, but i always get confused what to do with the result,
for ex.
if i got tanx=-1 limits are 0<x<360 (both equal sign but don't know how to write it)
x=-45
so then what should i do?
-45(-+)180 or 180(-+)-45?
and for 360/?
and if u can put the sinx part as well.. ??
plz help
elemis:
For tan keep adding/subtracting 180
islu_jf:
Okay implicit diff. ???
same paper question number 5 part two?
Vampire-Love4ever:
Q1- The function 3-2x-x^2 = a-(x+b)^2, find the value of a and b and hence find it's maximum value.
Q2- Express x^2+4x in the form a-(x+a)^2, stating values of a and b.
Q3- Express 2x^2-6x in the form a(x-b)^2+c stating the numerical values of a, b and c.
Arthur Bon Zavi:
--- Quote from: Vampire-Love4ever on December 27, 2010, 01:12:09 pm ---Q1- The function 3-2x-x^2 = a-(x+b)^2, find the value of a and b and hence find it's maximum value.
Q2- Express x^2+4x in the form a-(x+b)^2, stating values of a and b.
Q3- Express 2x^2-6x in the form a(x-b)^2+c stating the numerical values of a, b and c.
--- End quote ---
I do all these by a simple method, and if you don't understand, use the formula given in the book :
ax2 + bx + c = a ( x2 + b )
( ---x )
( a )
x2 + bx + c = (x2 + bx) + c = {(x + 1/2 b)2 - 1/4 b2} + c
(1)
-x2 - 2x - 3
-(x2 + 2x - 3)
-(x2 + 2x + 1 - 4)
-(x + 1)2 + 4 ---------> a = 4; b = -1; maximum value = 4.
(2)
x2 + 4x
(x2 + 4x)
(x2 + 4x + 4 - 4)
(x + 2)2 - 4 ----------> a= -4; b= -2
(3)
2x2 - 6x
2 (x2 - 3x)
2 (x2 - 3x + 2.25 - 2.25)
2 (x - 1.5)2 - 4.5 ------------> a= 2; b= 1.5; c= -4.5
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