Qualification > Math
math doubts [NEW]
slvri:
--- Quote from: Z.J. on September 26, 2009, 08:26:44 pm ---yes slvri i did!
okay one more ques...
function f and g are defined by:
f(x): k-x k is a constant
g(x): 9/x+2 whr x nt = 2 both x E R
Find the values of k for whch the eqtn f(x)=g(x) has two equal roots and solve the equation f(x)=g(x) in these cases.
--- End quote ---
f(x)=g(x)
k-x=9/x+2
(k-x)(x+2)=9
kx-x2+2k-2x=9
x2+(2-k)x+9-2k=0
in this equation a=1, b=2-k, c=9-2k
the condition for the equation f(x)=g(x) to have two equal roots is b2-4ac=0
(2-k)2-4(1)(9-2k)=0
4-4k+k2-36+8k=0
k2+4k-32=0
k2+8k-4k-32=0
k(k+8)-4(k+8)=0
(k+8)(k-4)=0
k=-8 or k=4
when k=-8
x2+(2-k)x+9-2k=0
x2+(2--8)x+9-2(-8)=0
x2+10x+25=0
(x+5)2=0
x+5=0
x=-5
but when k=4
x2+(2-k)x+9-2k=0
x2+(2-4)x+9-2(4)=0
x2-2x+1=0
(x-1)2=0
x-1=0
x=1
there u go
zara:
function f and g are defined as
f(x): x2-2x
g(x):2x+3 both x E R
1) find the set of values of x for which f(x)>15
2)find the range of f...
im bad at this...had nly 2 or 3 classes of it...n didnt get much :-\
Ghost Of Highbury:
i guess the 1st one is -3>x>5
astarmathsandphysics:
--- Quote from: Z.J. on September 26, 2009, 08:49:01 pm ---function f and g are defined as
f(x): x2-2x
g(x):2x+3 both x E R
1) find the set of values of x for which f(x)>15
2)find the range of f...
im bad at this...had nly 2 or 3 classes of it...n didnt get much :-\
--- End quote ---
so
Complete the square
so so or so
slvri:
--- Quote from: Z.J. on September 26, 2009, 08:49:01 pm ---function f and g are defined as
f(x): x2-2x
g(x):2x+3 both x E R
1) find the set of values of x for which f(x)>15
2)find the range of f...
im bad at this...had nly 2 or 3 classes of it...n didnt get much :-\
--- End quote ---
its ok with practice u will get better :)
1) f(x)>15
x2-2x>15
x2-2x-15>0
x2-5x+3x-15>0
x(x-5)+3(x-5)>0
(x-5)(x+3)>0
(x-5))(x-(-3))>0
now theres a rule u need to keep in mind (i cant illustrate it here using a diagram so u will have to memorise it for the time being)
whenever an expression of the form(x-a)(x-b)>0 occurs where a<b then the solution is x<a and x>b
over here a is 5 and b is -3(5 is greater than -3)
so the solution is x<-3 and x>5
2)x2-2x
=(x2-2x+1)-1
=(x-1)2-1
now u can see that when x=1
f(1)=(1-1)2-1=-1
see for yourself that whatever value u put for x, the answer will always be greater than -1(-1 is the minimum value of f(x))
so the range of f(x) is f(x)>=-1
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