Post by: zara on September 26, 2009, 03:00:03 pm
can summon plz explain?
Post by: astarmathsandphysics on September 26, 2009, 03:18:32 pm
Post by: slvri on September 26, 2009, 03:28:38 pm
multiply by x4
18+x2=4x4
4x4-x2-18=0
4(x2)2-x2-18=0
u can see that this is a quadratic equation in x2
factorising we have
4x4-9x2+8x2-18=0
x2(4x2-9)+2(4x2-9)=0
(4x2-9)(x2+2)=0
either 4x2-9=0 or x2+2=0
4x2=9 or x2=-2
x2=9/4 (x2=-2 has no real roots, complex roots are +-sqrt(2)i)
so from the first one
x=sqrt(9/4)
x=+-(3/2)
hope this helped zara :)
Post by: nid404 on September 26, 2009, 03:40:07 pm
Post by: zara on September 26, 2009, 05:36:28 pm
multiply by x4
18+x2=4x4
4x4-x2-18=0
4(x2)2-x2-18=0
u can see that this is a quadratic equation in x2
factorising we have
4x4-9x2+8x2-18=0
x2(4x2-9)+2(4x2-9)=0
4(x2-9)(x2+2)=0
either 4x2-9=0 or x2+2=0
4x2=9 or x2=-2
x2=9/4 (x2=-2 has no real roots, complex roots are +-sqrt(2)i)
so from the first one
x=sqrt(9/4)
x=+-(3/2)
the ones in red i didnt get it....plz elaborate... :-\
y is 4x2-9=0? wont it be x2-9=0?
Post by: zara on September 26, 2009, 05:42:11 pm
The function h is defined by
h:x > 6x-x2 for x>=3
Express 6x-x2 in the form a-(x-b)2, where a and b are positive constants.
Post by: zara on September 26, 2009, 05:46:22 pm
g is defined by: g:x >6x-x2 for x E R
Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9.
i did the expression part...buh im confused in the second part...how to show it? :-\
n yea one more thing....what does this mean "x E R"?
i don't remember our sir explaining this to us....n its der in our worksheet!
Post by: Ghost Of Highbury on September 26, 2009, 05:48:56 pm
the first one...u multiply it by x^4 so that u can simplify it and solve easily...as the divisor can be cancelled..
wen u multiply x^4 with 18/x^4 + 1/x^2 = u get => 18 (as the x^4 gets cancelled) + x^4/x^2 = 18 + x^2
next...
(4x^2 - 9)(x^2 + 2) = 0
this means that either (4x^2 - 9) = 0...or (x^2 +2) = 0....as something HAS to be multiplied by 0 to get a 0.
so 4x^2 - 9 = 0
4x^2 = 9..
and so on...
hope that helped :)
Post by: Ghost Of Highbury on September 26, 2009, 05:50:39 pm
PS: wud try to solve thos.e
Post by: zara on September 26, 2009, 05:55:43 pm
tht helped!
n yea sure go ahead solve it!
its nt for specific people...whoever can is free to do! =)
Post by: Ghost Of Highbury on September 26, 2009, 06:00:48 pm
therefore a = 9
b = 3
Post by: zara on September 26, 2009, 06:12:55 pm
x^2 + 6x = 9 - (x-3)^2im lost adi..:\
therefore a = 9
b = 3
i need the steps if possible?!
Post by: Ghost Of Highbury on September 26, 2009, 06:17:04 pm
x2 + nx = (x + n/2)2 - (n/2)2
n = 6
therefore n/2 = 3
here its -x2 + 6x = (x-3)2 - (3)2 = -x2 + 6x = 9 - (x-3)2
Post by: Ghost Of Highbury on September 26, 2009, 06:30:31 pm
Post by: astarmathsandphysics on September 26, 2009, 07:28:11 pm
the general rule
x2 + nx = (x + n/2)2 - (n/2)2
n = 6
therefore n/2 = 3
here its -x2 + 6x = (x-3)2 - (3)2 = -x2 + 6x = (x-3)2 - 9
Should be (x+3)2-9
Post by: Ghost Of Highbury on September 26, 2009, 07:43:36 pm
9 - (x-3)2 is correct right?
(i edited it now)
(x+3)^2 - 9 doesnt give 6x - x^2
Post by: slvri on September 26, 2009, 08:12:11 pm
18/x4 + 1/x2 =4sory i made a mistake there
multiply by x4
18+x2=4x4
4x4-x2-18=0
4(x2)2-x2-18=0
u can see that this is a quadratic equation in x2
factorising we have
4x4-9x2+8x2-18=0
x2(4x2-9)+2(4x2-9)=0
4(x2-9)(x2+2)=0
either 4x2-9=0 or x2+2=0
4x2=9 or x2=-2
x2=9/4 (x2=-2 has no real roots, complex roots are +-sqrt(2)i)
so from the first one
x=sqrt(9/4)
x=+-(3/2)
the ones in red i didnt get it....plz elaborate... :-\
y is 4x2-9=0? wont it be x2-9=0?
and for the first one, u multiply each term in the equation by x^4 to turn it into a expression that u can factorise easily
Post by: zara on September 26, 2009, 08:14:59 pm
sory i made a mistake thereyea Thanks...adi expalined it...=)
and for the first one, u multiply each term in the equation by x^4 to turn it into a expression that u can factorise easily
Post by: slvri on September 26, 2009, 08:15:52 pm
yea Thanks...adi expalined it...=)i hope u understood :)
Post by: zara on September 26, 2009, 08:26:44 pm
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.
Post by: slvri on September 26, 2009, 08:36:00 pm
yes slvri i did!f(x)=g(x)
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.
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
Post by: zara on September 26, 2009, 08:49:01 pm
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 :-\
Post by: Ghost Of Highbury on September 26, 2009, 08:58:12 pm
Post by: astarmathsandphysics on September 26, 2009, 08:58:47 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 :-\
so
Complete the square
Post by: slvri on September 26, 2009, 09:02:11 pm
function f and g are defined asits ok with practice u will get better :)
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 :-\
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
Post by: Ghost Of Highbury on September 26, 2009, 09:05:46 pm
Post by: slvri on September 27, 2009, 12:52:26 am
Post by: slvri on September 27, 2009, 12:55:54 am
Post by: Ghost Of Highbury on September 27, 2009, 07:13:46 am
Post by: nid404 on September 27, 2009, 07:15:29 am
Should be (x+3)2-9
yeah it should be
Post by: Ghost Of Highbury on September 27, 2009, 07:20:05 am
if the answer is (x+3)2 -9 ...it simplifies to x2 + 6x + 9 -9 = x2 + 6x
but the question says that it shud be -x2 + 6x
for this the answer should be...
9 - (x-3)2 which simplifies to 9 - (x2 - 6x + 9)
==> 9 - x2 + 6x - 9 = -x2 + 6x
???
Post by: nid404 on September 27, 2009, 07:34:54 am
the question says....."Express 6x-x2 in the form a-(x-b)2"...
if the answer is (x+3)2 -9 ...it simplifies to x2 + 6x + 9 -9 = x2 + 6x
but the question says that it shud be -x2 + 6x
for this the answer should be...
9 - (x-3)2 which simplifies to 9 - (x2 - 6x + 9)
==> 9 - x2 + 6x - 9 = -x2 + 6x
???
yeah right.....the question says 6x-x2...
yup ur right...infact it says express it in the form c-(a+b)2
Post by: Ghost Of Highbury on September 27, 2009, 07:38:06 am
9 - (x-3)^2
therefore a = 9
b = 3
Post by: nid404 on September 27, 2009, 07:38:58 am
it says express it in the form of a - (x-b)^2
9 - (x-3)^2
therefore a = 9
b = 3
whatever...it just says express it in that form :P
Ur right
Post by: Ghost Of Highbury on September 27, 2009, 07:48:09 am
Post by: nid404 on September 27, 2009, 07:59:38 am
yah...try the mininimum point wala q..i'm not gettingd answer for that....
There r too many questions and answers...I'm getting confused....can you post the question again....that will make it easier
Post by: nid404 on September 27, 2009, 08:07:27 am
Post by: slvri on September 27, 2009, 08:12:07 am
slvri has answered it correctly....it says find the values of x for which f(x) is greater than 15hey nid.......need any other q's solved?
Post by: nid404 on September 27, 2009, 08:13:48 am
hey nid.......need any other q's solved?
Well not for now....actually adi did not get the the domain for the f(x)>15 thingy....so I was trying to explain to him why you were correct...hope he got it
Post by: Ghost Of Highbury on September 27, 2009, 08:14:35 am
" f is defined by: f:x > 3x-2 for x E R
g is defined by: g:x >6x-x2 for x E R
Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9."
Post by: slvri on September 27, 2009, 08:25:13 am
ya slvri...this one...this q wasnt answered...gf(x)=g(f(x))=g(3x-2)
" f is defined by: f:x > 3x-2 for x E R
g is defined by: g:x >6x-x2 for x E R
Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9."
=6(3x-2)-(3x-2)2
=18x-12-(9x2-12x+4)
=18x-12-9x2+12x-4
=-16+30x-9x2
=-16-(-30x+9x2)
=-16-(9x2-30x)
=-16-9(x2-(10/3)x)
=-16-9((x)2-2(x)(5/3)+(5/3)2)+9(5/3)2
=-16+9(5/3)2-9(x-5/3)2
=-16+25-9(x-5/3)2
=9-9(x-5/3)2
now u can see that when x=5/3
gf(5/3)=9-(5/3-5/3)2=9-0=9
and whatever value u put for x, the value of gf(x) will always be less than or equal to 9
so the maximum value of gf(x)=9
shown
Post by: Ghost Of Highbury on September 27, 2009, 08:29:11 am
Post by: slvri on September 27, 2009, 08:40:29 am
can we find out the maximum point by differentiation?yes u can assuming this is an a level math or an igcse add math q(but not if this is an igcse math q)
Post by: nid404 on September 27, 2009, 08:56:50 am
gf(x)=16+30x-9x2
when u differentiate
-18x2+30
x=-5/3
substitute in the equation you get max value of gf(x) as 9
Post by: slvri on September 27, 2009, 09:03:25 am
differentiating becomes much easierwhatever way u like
gf(x)=16+30x-9x2
when u differentiate
-18x2+30
x=-5/3
substitute in the equation you get max value of gf(x) as 9
Post by: nid404 on September 27, 2009, 09:07:51 am
whatever way u like
You can do that right??
Post by: slvri on September 27, 2009, 09:10:14 am
You can do that right??yup
Post by: astarmathsandphysics on September 27, 2009, 10:04:25 am
yeah..then..how can it be less than -5 and greater than 3 at the same time??
x<-5 OR x>3
NOT x<-5 AND x>3
Post by: slvri on September 27, 2009, 10:13:18 am
x<-5 OR x>3whoops.....meant to say that both of these are the possible solution sets
NOT x<-5 AND x>3
Post by: Ghost Of Highbury on September 27, 2009, 02:03:52 pm
Post by: nid404 on September 27, 2009, 02:07:45 pm
ok...so the final answer is ..... x>5 and x<-3
plot it and check it if you r still not sure
Post by: Ghost Of Highbury on September 27, 2009, 03:39:36 pm
Post by: coolcar221 on October 02, 2009, 07:48:18 pm
find the set of values of k for which y=kx-4 intersects the curve y=x^2-2x at two distinct points
Post by: astarmathsandphysics on October 02, 2009, 08:03:52 pm
http://www.astarmathsandphysics.com/a_level_maths_notes/C1/a_level_maths_notes_c1_using_the_discriminant_to_find_the_number_roots_of_a_quadratic_curve.html
Post by: astarmathsandphysics on October 02, 2009, 08:11:09 pm
find the set of values of k for which y=kx-4 intersects the curve y=x^2-2x at two distinct points
so
Post by: simba on October 13, 2009, 11:52:01 am
Q: f(x): 3-2sinx , for 0<x<360 (its greater than or equal & less than or equal)
find the range of f
Post by: slvri on October 13, 2009, 11:55:34 am
guyz i need help........ok..........u know that the sine function is always between -1 and 1 (max 1 and min -1). so when sinx=1, f(x)3-2sinx=3-2(1)=3-2=1(min)
Q: f(x): 3-2sinx , for 0<x<360 (its greater than or equal & less than or equal)
find the range of f
and when sinx=-1, f(x)=3-2(-1)=3+2=5(max)
so 1<=f(x)<=5
Post by: simba on October 13, 2009, 12:16:44 pm
Post by: simba on October 13, 2009, 05:27:17 pm
Post by: slvri on October 13, 2009, 05:31:39 pm
i gt a question bt nt abt fuctions.............in june 06 .....num.3 ..how do i get the rate in decimal rather than the percentage ????um can u post the question here?
Post by: simba on October 13, 2009, 05:35:52 pm
(i) the grant given in 2011
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive
Post by: astarmathsandphysics on October 13, 2009, 06:24:14 pm
Post by: astarmathsandphysics on October 13, 2009, 09:46:09 pm
(i) the grant given in 2011
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive
i)
ii)
Post by: simba on October 13, 2009, 10:59:57 pm
Post by: Eamyzz on October 14, 2009, 01:03:44 am
anythng else? =)
Post by: simba on October 14, 2009, 11:28:09 am
(i) find the first term of the progression
will someone solve it :)
Post by: simba on October 14, 2009, 11:49:52 am
Post by: slvri on October 14, 2009, 11:59:25 am
the second term of a geometric progression is 3 and the sum to infinity is 12the nth term of a GP is given by an=arn-1
(i) find the first term of the progression
will someone solve it :)
for the second term n=2
ar2-1=3
ar=3
a=3/r
the sum to infinity of a GP is given by S=a/(1-r)
a/(1-r)=12
a=12(1-r)
put a=3/r
3/r=12(1-r)
3=12r(1-r)
3=12r-12r2
12r2-12r+3=0
4r2-4r+1=0
(2r)2-2(2r)(1)+(1)2=0
(2r-1)2=0
2r-1=0
r=1/2
a=3/(1/2)=6
is this the correct answer?
Post by: Eamyzz on October 14, 2009, 04:39:09 pm
Post by: simba on October 14, 2009, 04:44:15 pm
will someone help :???
Post by: Ghost Of Highbury on October 14, 2009, 04:55:09 pm
guyz another question ....num.4 in nov.07
will someone help :???
A level maths --> paper no?
Post by: simba on October 14, 2009, 04:56:56 pm
Post by: badboy on October 14, 2009, 05:11:41 pm
each year a company gives a grant to charity. The amount given each year increases by 5% of its value in the preceding year. The grant in 2001 was $5000. Find :
(i) the grant given in 2011
(ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive
i)
ii)
ur answers r wrong i) 8144 ii)71034
Post by: slvri on October 14, 2009, 05:43:36 pm
(i)the grant given in 2011 should be the 11th term
ur answers r wrong i) 8144 ii)71034
the 11th term is given by a11=arn-1=5000*1.0511-1=5000*1.0510=$8144
(ii)total amount is the sum of the money from 2001 to 2011
so the sum from terms 1 to 11 is S11
=a(1-rn)/1-r
=5000(1-1.0511))/1-1.05
=5000(1-1.0511)/-0.05
=$71034
hope this helped :)
Post by: astarmathsandphysics on October 14, 2009, 07:15:36 pm
Post by: astarmathsandphysics on October 14, 2009, 07:20:58 pm
I will look now
Post by: astarmathsandphysics on October 14, 2009, 07:31:40 pm
ii)
iii)
Post by: sameer210394 on October 19, 2009, 09:05:47 am
shudnt -3>x>5
abbey Adi, u and the guy wrote the same answer and arguing? ???
Post by: Ghost Of Highbury on October 19, 2009, 09:33:24 am
abbey Adi, u and the guy wrote the same answer and arguing? ???
yea lol...v realized it later.. :P :P
Post by: helllife on October 29, 2009, 05:06:48 pm
1) Find the sets of values of k for which the roots of the equation x^2+kx-k+3=0 are real and of the same sign.
2)Given that y=px^2+2qs+r, shows that y will have same sign for all real values of x if and only if q^2<pr.
can anyone explain this?
Post by: astarmathsandphysics on October 29, 2009, 05:19:41 pm
k^2+4k-12=(k+6)(k-2)>0 so k>6 or k<-2 and k^2>k^2-4*1*(-k+3) so-k+3>0 so k<3
so k>6 or k<-2 anf k<3ie k<-2
")b^2-4ac<0 since y=0 has no roots
(3q)^2-4pr<0 so 4q^2<4pr so q^2<pr.
Post by: zara on October 31, 2009, 11:42:13 am
Q1)Given that x=sin-1(2/5), find the exact value of
(i) cos2x
(ii)tan2x
Post by: zara on October 31, 2009, 11:45:10 am
1/cos(theta) + tan(theta) = cos(theta)/1-sin(theta)
Post by: Ghost Of Highbury on October 31, 2009, 11:49:56 am
sin-1 2/5 = 23.5
cos2 x = 0.84
and
tan2x = 0.19
just substitute and find it
---
next
(1/cos x + sinx/cosx) = (1+sinx/ cos x) * (1-sinx/1-sinx) = 1-sin2x/cosx - sinxcosx = cos2x/cosx-sinxcosx
= cos2x / cosx(1-sinx) = cosx/1-sinx
x = theta
LHS = RHS
hence proved
Post by: zara on October 31, 2009, 11:55:29 am
[sin(theta) - cos(theta)]/[sin(theta)+cos(theta)] =6sin(theta)/cos(theta),
find the exact values of tan(theta).Hence find (theta) for 0<=(theta)<=360
Post by: zara on October 31, 2009, 12:02:08 pm
first find xadi how u got cos2=0.84? n same for tan?
sin-1 2/5 = 23.5
cos2 x = 0.84
and
tan2x = 0.19
just substitute and find it
---
next
(1/cos x + sinx/cosx) = (1+sinx/ cos x) * (1-sinx/1-sinx) = 1-sin2x/cosx - sinxcosx = cos2x/cosx-sinxcosx
= cos2x / cosx(1-sinx) = cosx/1-sinx
x = theta
LHS = RHS
hence proved
Post by: Ghost Of Highbury on October 31, 2009, 12:04:33 pm
Post by: zara on October 31, 2009, 12:16:21 pm
find cos x --> square the answerhow?
Post by: Ghost Of Highbury on October 31, 2009, 12:23:14 pm
how?
calculator?
Post by: zara on October 31, 2009, 12:52:32 pm
lol thts wht hppns ven im doin many things at a time...
now i get the whole thing....Thanks...
Post by: zara on October 31, 2009, 12:53:44 pm
Q3)Given that the equationsome one solve this ques plz...
[sin(theta) - cos(theta)]/[sin(theta)+cos(theta)] =6sin(theta)/cos(theta),
find the exact values of tan(theta).Hence find (theta) for 0<=(theta)<=360
Post by: falafail on October 31, 2009, 02:17:01 pm
some one solve this ques plz...
(http://i33.tinypic.com/308wg0p.jpg)
i believe this is how it's done.
feel free to ask if you need me to explain anything, or to correct me if i'm wrong XD
oh, the picture's too small. here: http://i33.tinypic.com/308wg0p.jpg
Post by: nid404 on October 31, 2009, 02:28:14 pm
By the way if you don't want to confuse urself bt taking tan2theta and then substituting
assume tan theta=x and then solve. Get value for x then sub tan theta in it
and yeah use periodic property to get the answer within the given range
Post by: zara on October 31, 2009, 06:15:38 pm
Post by: zara on October 31, 2009, 06:19:57 pm
Post by: Ghost Of Highbury on October 31, 2009, 06:24:18 pm
x-1 = x2 - 5x -8
simplifies to
-x2 + 6x + 7 =0
x = -1 x = 7
y = -2 y= 6
------------------------------
y = -2x2 + qx + p
substitute the values
x = -1 ; u get p=q
substitute x=7 and 'q' for 'p' (because they are equal) ; u get q = 13
so the curve is
-2x2 + 13q + 13
@Z.J - equation of a circle?..or area
Post by: zara on October 31, 2009, 06:27:12 pm
Post by: Ghost Of Highbury on October 31, 2009, 06:32:44 pm
yea aadi its equation of a circle as mentioned in my worksheet.:\
ok..check this
http://www.analyzemath.com/CircleEq/CircleEq.html
so the answer wud be
By the way--did u understand the other question?
Post by: nid404 on October 31, 2009, 06:33:01 pm
yeah It's
r=8cm....let centre be (h,k)
let point P be a point on the circumference of a circle with coordinates (x,y)
then
(x-h)2 + (y-k)2=r2
(x-h)2+(y-k)2=16(42)
I assume centre is (0,0) since it is not given
assuming....h and k become 0
x2+y2=16
equation is x2+y2-16=0
Post by: nid404 on October 31, 2009, 06:35:31 pm
x2+y2+2gx+2fy+c=0
where centre is (-g,-f) and radius is root of (g2+f2-c)
@adi...how do u use the square root sign??
Post by: Ghost Of Highbury on October 31, 2009, 06:38:17 pm
equation of the circle has the general form
x2+y2+2gx+2fy+c=0
where centre is (-g,-f) and radius is root of (g2+f2-c)
@adi...how do u use the square root sign??
its latex
click that "pi" symbol next to "sub" "sup" and others..
for square root type ... sqrt(type here) in the latex form.. i.e after clicking "pi"
like this
Post by: zara on October 31, 2009, 06:40:53 pm
first find where they intersectwait....hows p=q?
x-1 = x2 - 5x -8
simplifies to
-x2 + 6x + 7 =0
x = -1 x = 7
y = -2 y= 6
------------------------------
y = -2x2 + qx + p
substitute the values
x = -1 ; u get p=q
substitute x=7 and 'q' for 'p' (because they are equal) ; u get q = 13
so the curve is
-2x2 + 13q + 13
@Z.J - equation of a circle?..or area
im stuck..
Post by: Ghost Of Highbury on October 31, 2009, 06:43:44 pm
wait....hows p=q?
im stuck..
wen u substitute x=-1 ad y=-2 in the equation
y = -2x2 + qx + p
u get
-2 = -2 -q + p
-2 + 2 = p - q
p-q =0
p=q
Post by: astarmathsandphysics on October 31, 2009, 06:48:28 pm
r=4 if the diameter is 8
Post by: nid404 on October 31, 2009, 06:54:32 pm
Post by: zara on October 31, 2009, 07:00:05 pm
Post by: astarmathsandphysics on October 31, 2009, 07:04:35 pm
Post by: falafail on October 31, 2009, 07:13:09 pm
thanks aadi nid falafail n astar!
anytime ^_^
Post by: Ghost Of Highbury on November 05, 2009, 07:47:57 am
thanks aadi nid falafail n astar!
ur welcome
Post by: zara on November 14, 2009, 06:30:00 pm
n if its -ve then its minima n if +ve then maxima?
Post by: slvri on November 14, 2009, 06:32:14 pm
If coordinates are (2,-63); is the y coordinate considered for deciding the maxima or minima?nope..........have u done differentiation in p1 yet or not?
n if its -ve then its minima n if +ve then maxima?
Post by: zara on November 14, 2009, 06:32:45 pm
Jane's bank balance put $1000 into a savings bank account for her on each birthday from her 10th to her 18th. The account pays interest at 6% for each complete year that the money is invested. How much money is in the account on the day after her 18th birthday?
Post by: zara on November 14, 2009, 06:35:08 pm
nope..........have u done differentiation in p1 yet or not?yes i have done it...
sorry its nt the coordinate....it says if (2,-63) is a minimum point, then whats the minima?
Post by: astarmathsandphysics on November 14, 2009, 06:38:49 pm
Post by: slvri on November 14, 2009, 06:39:13 pm
another question...ok this is a difficult q.......can u plz tell me the ans first bcuz itll take me a few mins to solve it?
Jane's bank balance put $1000 into a savings bank account for her on each birthday from her 10th to her 18th. The account pays interest at 6% for each complete year that the money is invested. How much money is in the account on the day after her 18th birthday?
Post by: zara on November 14, 2009, 06:49:02 pm
ok this is a difficult q.......can u plz tell me the ans first bcuz itll take me a few mins to solve it?i dunno the answer...
the question is from a worksheet.
Post by: zara on November 14, 2009, 06:49:47 pm
Post by: slvri on November 14, 2009, 06:51:45 pm
Post by: vaan on November 14, 2009, 06:55:50 pm
another question...
Jane's bank balance put $1000 into a savings bank account for her on each birthday from her 10th to her 18th. The account pays interest at 6% for each complete year that the money is invested. How much money is in the account on the day after her 18th birthday?
this question is easy...buts its very time consumin.....
Post by: zara on November 14, 2009, 07:03:17 pm
well the minima is (2,-63).....minima is just another word for minimum pointno slvri..
in my book its written
The value of the function at minimum point is known as minima.
so like im confused now?!
Post by: slvri on November 14, 2009, 07:03:24 pm
on eleventh bday 1000 is put and on 18th bday it becomes 1000(1.06)7 bcuz for 7 years interest is put on it(12th to 18th bdays)
and so on till on 18th bady 1000 is put but it stays 1000 bcuz thats the end of the time period
this forms a geomtric progression 1000+1000(1.06)+1000(1.06)2+.....+1000(1.06)8
a=1000, r=1.06, n=9(NOT 8!)
S9=1000(1.069-1)/(1.06-1)=$11491.32
is this correct?
Post by: slvri on November 14, 2009, 07:04:51 pm
Post by: zara on November 14, 2009, 07:06:55 pm
oh sory........then -63 is the minima cuz -63 is the value of the function when the value of x is 2ayte! so if it was +63 then we say its the maxima?
Post by: slvri on November 14, 2009, 07:12:02 pm
its not necessary that a minima is always negative....it can be positve tooo..
and a maxima can be negative
Post by: zara on November 14, 2009, 07:15:20 pm
no....if theyve said itz a minimum point then 63 will still be a minima......ohkay...thnkx..
its not necessary that a minima is always negative....it can be positve tooo..
and a maxima can be negative
Post by: astarmathsandphysics on November 14, 2009, 08:08:23 pm
?It will be
it is a geometric sequence with first term 1000 and common raio 1.06 and 9 terms
Post by: astarmathsandphysics on November 14, 2009, 08:14:17 pm
n if its -ve then its minima n if +ve then maxima?
It is not y that has to be negative for a min.
and
explained here
http://www.astarmathsandphysics.com/a_level_maths_notes/C1/a_level_maths_notes_c1_maxima_and_minima_the_second_differential_criterion.html
Post by: zara on November 15, 2009, 06:16:46 pm
Post by: astarmathsandphysics on November 20, 2009, 08:22:31 pm
(i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3]
(ii) State the value of A for which the graph of y = f(x) has a line of symmetry. [1]
(iii) When A has this value, find the range of f.
i)2x^2-12x + 13=a(x + b)^2 + c=ax^2+2abx+ab^2+c
2x"2=ax^2 so a=2
-12x=2abx=4bx so b=-3
13=ab^2+c=2*(-3)^2+c so c=-5
f(x)=2(x-3)^2-5
ii)A=6
iii)max value of f(x) is f(6)=13
range of f is [-5,13]
Post by: zara on December 10, 2009, 03:35:12 pm
Hence express the infinite recurring decimal 0.108108108...as a fraction in its lowest terms.
Post by: vanibharutham on December 10, 2009, 03:41:57 pm
Find the sum of the infinite series 1/103+1/106+1/109+... expressing ur answer as a fraction in its lowest terms.
Hence express the infinite recurring decimal 0.108108108...as a fraction in its lowest terms.
We find that r = 0.001, and a = 0.001
S to infinity = a / 1 - r
=====> 0.001 / 1 - 0.001
=> 1 / 999
Since
1/ 999 = 0.001001001001001 etc....
2/999 must equal 0.002002002002 etc
Hence 108/999 must equal 0.108108108108108
In its simplest form 108/999 = 4/37
:)
Post by: zara on December 10, 2009, 03:46:13 pm
Post by: zara on December 10, 2009, 03:52:37 pm
Post by: @d!_†oX!© on December 10, 2009, 04:52:54 pm
A = P ( 1 + r/100 )n
6000=P(1+2/100)24
thus, P= $3730.328928
and because the monthly payments are equal, thus each monthly payment= 3730.328928/24 = $155.43
:) :)
Post by: zara on December 20, 2009, 03:36:31 pm
i got rong answer prolly cause one of my limit is rong and thts whr im confused, can summon plz do n lemme kno?
Post by: SGVaibhav on December 20, 2009, 06:02:47 pm
im talking about the normal maths which everyone takes
do they have integration, differentiation and limits ???
if they have, what all do they have?
Post by: astarmathsandphysics on December 20, 2009, 07:00:39 pm
Post by: cooldude on December 21, 2009, 04:23:34 pm
if anyone uses Pure mathematics 1 by Hugh Neill & Douglas Quadling can u plz turn over to page 253 Ex.16D Q4.
i got rong answer prolly cause one of my limit is rong and thts whr im confused, can summon plz do n lemme kno?
if u still want the answer, here it is (i dont know how to use the integral sign, sorry)
ive attached a drawing, (please refer to that while reading the explanation)
first of all we'll find the area K+L by integration, we then subtract the area L from this area to find the area K.
first of all we find the area K+L by integration----->>>
the integral of the function (2x-5)^4---> [(2x-5)^5)/10]
the limits will be b/w Q and R, we can find the co-ordinate of Q by equating the equation of the curve to 0.
then the area K+L comes out to be---->
=> 24.3-0=24.3
we then find the area L. (we first have to find the equation of PR though).
since f(x)=(2x-5)^4
f'(x)=[4(2x-5)^3]*2
therefore the gradient of the tangent=216 at (4,81)
then the equation of PQ becomes---> y=216x-783, we find out the x-coordinate of R by equating the equation of PQ to 0.
then we find the area L-->
0.5*0.375*81=15.1875
therefore K=24.3-15.1875=9.1125
hope i helped
p.s. sorry about the bad handwriting in the drawing, so just for reference Q(2.5,0), P(4,81), R(4,0), S(3.625,0)
Post by: SGVaibhav on December 21, 2009, 04:53:04 pm
i wanted to know what all they teach in AS level mathematics EDEXCEL
im talking about the normal maths which everyone takes
do they have integration, differentiation and limits ???
if they have, what all do they have?
ok but then i wanted to know what all they teach in limits, differentitaion and integration
only the topic names of content...
Post by: zara on December 22, 2009, 01:18:42 pm
if u still want the answer, here it is (i dont know how to use the integral sign, sorry)thanks a lot!!
ive attached a drawing, (please refer to that while reading the explanation)
first of all we'll find the area K+L by integration, we then subtract the area L from this area to find the area K.
first of all we find the area K+L by integration----->>>
the integral of the function (2x-5)^4---> [(2x-5)^5)/10]
the limits will be b/w Q and R, we can find the co-ordinate of Q by equating the equation of the curve to 0.
then the area K+L comes out to be---->
=> 24.3-0=24.3
we then find the area L. (we first have to find the equation of PR though).
since f(x)=(2x-5)^4
f'(x)=[4(2x-5)^3]*2
therefore the gradient of the tangent=216 at (4,81)
then the equation of PQ becomes---> y=216x-783, we find out the x-coordinate of R by equating the equation of PQ to 0.
then we find the area L-->
0.5*0.375*81=15.1875
therefore K=24.3-15.1875=9.1125
hope i helped
p.s. sorry about the bad handwriting in the drawing, so just for reference Q(2.5,0), P(4,81), R(4,0), S(3.625,0)
Post by: cooldude on December 23, 2009, 02:04:12 pm
thanks a lot!!
np