3oh yaa....both will be integers....
ok thnk u sir...
21 Answers
106mathie which graph.. (1) or (2)? (2) im perfectly sure.... 1 hr baitha ha us par sirf d2y/sx2 nikaalne mein...
21thnx asis!!!! got ur pt.
write man! hadnrt read post 10 earlier jus saw ur graf!!!
write write !!! only int, int satusfy!!!
106bhaiyya thats what i have written in post #10.... isnt that correct?
3Nishant sir...
is graph 1 as drawn here correct ???
I feel, the only solution is (x,0), where x is any integer....
62bhargav.. i think you have slighty gone wrong!
at x=0 cosx/cos2x=1!!
as x-> pi/4 LH Limit
cosx-> 1/√2
while cos 2x- > 0+
so the limit will tend to + infinity!
106mathie put x=Ï€ in the graph equation.
u get y=-1/1 which is definitely below... not above...
392nd one, wat i am getting is that ∩ will bne exactly above and concave upwards and those half lines at extremes should be mirror images{i mean downwards in lower half}
21for 0<x<1 same as kal ka graf!! xy = x+ y;
the function is periodic with integers......
39asish are u sure with the graph? check once, kahiin kuch upar niche toh nahiin ho gaya.
106graph 1.
{x}{y} = {x}+{y}
==> {y} = {x}/({x} -1)
Now, {x}-1 is always negative as {x}<1
So, {y} ≤ 0 {y}=0 when {x} =0
But{y} is always ≥0
Combining these two,
we get {y}=0 and {x}=0
i.e. the graph is only all integer points...
106maine bahut kuch likhaa tha par accidentally cancel press kar diya... :(
im not going to rite much now again.
just graph is symmetrical abt y-axis.
dy/dx=sinx(1+2cos2x)/cos22x... depends upon sinx.
d2y/dx2 is very complex.. i cant write...
at x=2nÎ , y=1
at x=(2n+1)Î /2, y=0
at x=(2n+1)Î , y=-1
discontinuity at cos2x=1/2
if cos2x = 1/2- then y=-∞
if cos2x = 1/2+ then y=∞
bada khatarnaak graph hai
by periodicity rest of graph is decided.
21no sir da graf is 4 first part, u askd naa how wud it luk!!!
2nd part graf is cont decreasing!!!!!!!!!!
21part 2...
let cosx = u;
y = u/(2u^2-1)
so discont at u^2 = 1/2;
range of u [-1,1];
112)
we have to remember the graph is not valid for x=nÎ /2 + 2nÎ
at x=0 y=1
at x=Î /2 y=0
