36And loads and loads of thanks to both of u for ur help... :D
Please solve this questions....!!
I'll be grateful for your this act of kindness.....!!
and one more thing.... please do show the steps... :)
1. The number of pairs of reals (x,y) such that x = x2 + y2 and y = 2xy is
(A) 4 (B) 3
(C) 2 (D) 1 [Ans. is (A) 4]
2. Let P1 ,P2, P3, P4, P5 be five equally spaced points on the circumference of a circle of radius 1, centred at O. Let R be set of points in the plane of the circle that are close to O than any of P1 ,P2, P3, P4, P5. Then R is a
(A) Circular region (B) pentagonal region
(C) rectangular region (D) oval region that is not circular
3. The sides of a quadrilateral are all positive integers and three of them are 5, 20 and 20. How many possible values are there for the fourth side?
(A) 29 (C) 31
(C) 32 (D) 34
4. Let P(X) = 1 + x + x2 + x3 + x4 + x5. What is the remainder when P(x12) is divided by P(x)?
(A) 0 (B)6
(C)1+x (D)1 + x + x2 + x3 + x4.
5. In a triangle ABC the altitudes from B and C on to the opposite sides are not shorter than their respective opposite sides. Then one of the angles of ABC is
(A) 300 (B)450
(C) 600 (D)720
-
UP 0 DOWN 0 0 15
15 Answers
491) y=2xy
2xy-y=0
y(2x-1)=0
either
y=0
x=x2+y2
x=x2
x=0,1
or,
x= 1/2
x=x2+y2
y2=1/4
y=±(1/2)
soln.. (0,0), (1,0), (1/2,1/2) & (1/2, -1/2)
hence the answer!! [1]
49draw the circle, and point out the 5 points.
join the points to the centre.
use the perp bisector of the line joining 2 points is set of line equidistant from the two points.
u will find a pentagon.
49i get the 3rd one to be 43 ways..
none option matching..
please check the options once!
494) putting x=1 i get remainder 0
and according to the options, no other option gives remainder 0 except A
so i go for A! :)
495) it occurred to me at the sight of the problem that in case of isoceles right angled triangle, the altitudes to opposite side equals length of opposite side...hence satisfying the given conditions...
thus one of the angles is 45°
hence..(B) [1]
214) subho is wrong
the answer is 6
try rearranging P(x12)
there is another solution using complex number
w,w2,w3...w5 are roots of P(x)
we have w6=1
putting them on P(x12) - 6 we see it is a multiple of P(x)
so remainder is 6
36for the 4th one...!!
let me try some steps...!!
P(X) = 1 + x + x2 + x3 + x4 + x5
-> P(x) = 6(x6 - 1)/(x - 1) Now,
x6 - 1 is divisible by (x - 1) but what next ??????
39q3)
sum of 2 sides greater than 3rd sife
so 20,20,5
let diagonal (which contains the sides 20,20 on one side of it) be of length 'x'
its easy to see x ε [1,40)
let 4th side be y
5+y >x
so y ε [1,35)
so 34 values
71@ Rahul
-> P(x) = 6(x6 - 1)/(x - 1)
Isn't this divisible by 6 also since 6 is a factor?
71
@ Rahul ye kya hai???
Agar tum below average ho.. toh us hisab se toh ham to list mein bhi nahi aate!! Oh God!!
21QStN 4: http://www.targetiit.com/iit-jee-forum/posts/polynomials-17925.html
36There's gratitude for u all.....!!
Once Again,........ thanksssss...!!
@vivek-> Kucch jyada hi ho gaya, for ur 11th post