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my friend i check ans is squt5
this is "AIEEE" 2007 question
friend bobby ask good question of maxima $minima
tell me about true answer
sorry at place k put t
x+y-z=0 ;
z=x+y;
Max(3x-y-3z);
put the valu of z
Max(-4y)=.........(d)
x^2+ 2y^2=1............(A)
lets assume trigonometric variable which setisfied (A)
x=sin(k),z=cos(t)/squt(2);
y=[cos(t)/squt(2) -sin(t)];
From (d)
Max(-4y)=max (4sin(t)-2*squt(2)cos(t));
solve like previous question
obtain squt(24),
"TELL ME FRIEND about process ans may be wrong due to calculation mistake"
why all of you are going so large my friend
x^2+y^2=1
let x=sin(a);
y=cos(a);
x+2y=sin(a)+2cos(a)=squt(5)sin(a+t);
hence maximum valu=squt(5), where t=tan^-1(2)
in such type of problem of inequality i.e
A.M>=G.M
maxima $ minima exist at point where equality will hold i.e "if x=y=z"
you can assume x=y=z=k;
given that x+y+z=1
hence 3k=1
k=1/3
put the valu $obtain the answer
my friend this trick is quite comfortable if you are preparing for "I.I.T"
hello my friend in my openion arctan(x)=tan^-1(x)
tan inverse x
above function can be written as arctan[2(y/x)/(1-y^2/x^2)]
now let y/x=tanb
F(x)=arctan[2tanb/(1-tan^ 2b)]
F(x)=arctan[tan2b]
F(x)=2b
F(x)'=2*b'
where b=tan inverse y/x
hello bob you are absolutely wrong
to diffrentiate acute obtuse angle bisector take one line and one bisector
"LET tan(x) is angle between one line and one bisector if [tan(x)<=1]then bisector is acute angle opther wise obtuse angle bisector"
hello friend in my openion you are wrong in frist question
who will responsible for genration of trigonometry
please tell me about initial steps am not test you friend actually i want to know
because i want to work on this field
ALSO WRITE MATHMATICAL STEPS OF "GENRATION"
hiiii friends please try to ans a ques.
if X=0+i;
then what is valu of log(X)
where i represents a complex number
hi my friend cosx-2sinx cosx/cos^2x-sin^2x+sinx-1
cosx(1-2sinx)/(-sin^2x-(1-cos^2x)+sinx)
cosx(1-2sinx)/sinx(1-2sinx) {wherex!=n *pi}
=cotx
hi kfarnan i think at valu of x where cos x & cot x =0 then that enequality will't valid
if valid then
{sec^2x-tan^2x}=1,proved
dear friend anakin in my openion you do mistake in inquality
~if log b (a)>c
if base of log is {.}
then inequality will change
if>=1 then as it remain
hello JaneFairfax i think that you are intresested in trigo.
but i challange you in trigo. calculus algebra. coordinate pleaseeeeeeeee
accept my challenge and ask question from me
second partial derivative of a given function Y=F(x) exist if there will be no sharp turn in the graph of F'(x)
let y=f(x,z)
then y'=f(x+h,z)-f(x,z)/h when lim'h'tends to 0 is partial derivative because x is variable
similerly find derivative of y treating z as variable because of
derivative is limiting process so for y'' repeat above process
what is range of 1/(3-5cosx)
my openion about this is
Mathmatics is the perfect science
friend i ask one question related to range yesterday very intresting
try to solve it
2 or 3 days before i ask a problem what is eaist way to find area of triangle i given sides as 1-(x)^1/2 ,5-(x)^1/2 ,7-(x)^1/2 friend all of you not thing about question
here triangle will not be form
triangle will form if (sum of smallest side )>=biggest side
my friends i want to help
all of you know that
A=abc/4R
hence 4RA=abc
A.M>=G.M
(a+b+c )/3>=(abc)^1/3
now put the valu of abc=4RA
hence you can get the answer
if you want solution of sin30=1/2 then search on google
"why sin30=1/2" am allready given solution on that
if a,b,c are sides of a triangle & A be the area of triangle R will be radius of
circumcircle of that triangle
p.t (a+b+c)>=3{4RA}^1/3
if A is area of triangle &a,b,c are sides & R is radius of circumcircle which circumscribed the triangle
then prove that (a+b+c)>=3(4RA)^1/3
why sino=0