1. The no. of tangents to the curve y = cos(x+y) , |x| ≤ 2Π, that are parallel to the line x+2y=0 is ?
2. If 2008θ=Π, and 2 .\sum_{r=1}^{r=n} \sin(r\theta)\cos(r^2\theta) is an INTEGER , then find the least integral n ?
3. A circle passes through (-1,2) , (2,3) and (1,0) , find the intercept on y-axis.
N.B. Please mention the shortest method apart from solving the general equation of circle for g,f,c.
4. Area of the region given by \frac \pi4 \leq \arg\left(\frac{z+1}{z-1}\right)\leq\frac {3\pi}{4} is ?
262S = sin(n+n2)θ2 = 0 (only possible integer)
(n+n2)θ = kπ (k→integer)
(n+n2) = 2008k = 8*251*k
let n= 8t
8t+64t2 = 8*251*k
t(8t+1) = 251*k
now 251 being a prime number ,
either t = 251*p or 8t+1 =251*p
from first, t(min)=251
from second, p=1,2 gives no. soln
for p=3 , t(min) =94
thus t(min)=94
n(min) =752
262we dont need to think about whether S=1 or 0 or -1
just take the common case ,
(n+n2)θ = k*π/2 (k→integer)
and proceed likewise.
21rishabh how are you getting the points 3pi/2 and -pi/2 for question (1).
262Ah !! didnt see the '2'
then proceeding in the same way we get nmin=251
71It can't be 2007 nor 752. Please see the recent post #14.
71Note that a 2 has been provided to cancel off the two in denominator.
Now you have written that S = 0 (only integer). Why? I mean S = 1 is also a possibility or -1 (neglecting this one)
That would give n2+n = 1004 (taking the first possibility Î /2)
71How did you get it. after telescoping, I get S = sin(n2+n)θ
71@Swaraj Did the same way. Kindly answer the exact question where it asks the value of n . If n =-1 as what you say, it isn't correct.
@Rishabh . Thanks for the circle one. I had forgotten that.
11) 4.
slope of tangent = -1/2
now diff. the given curve and put y' = -1/2 to get sin(x+y) = 1.
now if sin(x+y) = 1 => cos(x+y) = 0
since y = cos(x+y) is the curve => y=0
so the points are (pi/2 , 0) (3pi/2 , 0) (-pi/2,0) (-3pi/2,0)
14) the lower bound is locus of major segment of a circle and the upper bound is the locus of minor segment of the same circle.
this means chord joining (-1,0) and (1,0) subtends pi/2 at centre so by using bit of geometry find coordinates of centre as (0,1) and radius = √2. so we are asked to find area of this circle which is 2pi
13) use family of lines,
(1,0) (-1,2) => (x-1)(x+1) + (y-0)(y-2) + a(x+y-1)
=> this circle shud pass thru (2,3) using this we get 'a' hence the equation. rest is just calc.
212)Ans=-1
Reduce the summation to sin((n2+n)θ)/2 by converting sinAcosB to 1/2(sin(A+B)+sin(A-B))