21OH OKI........
I GOT MA MIST.......
SORRY SIR [2]
KILL [48] [49] ME!!
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111 Answers
1sorry bhaiyya got it......same misconception ....!!!!!!!!will try to avoid this in jee[1]
1357tapan and akand
for triangle question
take a=2, b=√3 and c=1
this satisfies the given equation
Is this a right angled or not
21
1but this is d only triangle naa??????
tht condition must satisfy all right triangles naa???
1tapan......wats d mistake??? tell me also..........and then let bhaiyya kill u hehe
1paper 2 . Q 7 . what exactly does it mean . mughe which max and which min hi nahi pata chala
This I think will b my last Q on tiit for now. plz jaldi ki jiye
1so are all right triangles of d form √3a = 2b = 2√3c??????
no rite.....
so isnt d answer NONE OF THESE??
1for the triangle one u need to simplify it
a2 + b2 + c2 = ca+ab√3
= 2a (c/2 + b√3/2) = 2a (c cos 60 + b cos 30) .. assume right angled at A
=2a2
a2 + a2 = 2a2 .. holds true so right angled ..
1Q7 Let x,y be the real numbers satisfying the equation x2+2x+y2+2y-47=0
If the maximum and minimum values of x2+y2-4x-6y+13 are M and m respectively, then the numerical value of M-m is
Your Answer: 14
Correct Answer: none of these
Score: -1
Explaination :
x2+y2-4x-6y+13=(x-2)2+(y-3)2
P(2,3) lies inside the circle
let the distance between centre of circle and P be x
let M'=r+x and m'=r-x
M=M'2=r2+x2+2rx
m=m'2=r2+x2-2rx
M-m=4rx=4*7*5=140
1second eqn is a point na so max aur min kahaan se ???? [7] plz explain in detail
1i had a doubt in moment of inertia match the following .. question 3 a quarter circle .. isn't the moment of inertia MR2/2 ..
1being a circle .
the first one can have a max and min . but the second one is a point so the M-m should be 14 I thought
21SIR pl reply to # 27
i.e.
Q34 A point object O is placed at a distance of 30 cm from the pole of the convex mirror of focal length 20 cm as shown in the figure. Find the final image distance from the object.
Your Answer: Did Not Attempt
Correct Answer: 28 cm
Score: 0
Explaination :
1/u + 1/v =1/f
-1/30 +1/v=1/20
v=12 cm from pole
v'=42
V=42*2/3=28 cm
i din get the last step.......
only 12 cm length is outside the medium then Real depth, app depth concept shud b applied only to dat 12 cm ne Y wud ther b ne change in 30cm of glass slab ??
1357srinath
see it is properly mentioned that x,y be the real numbers satisfying the equation x2+2x+y2+2y-47=0 circle C
that means in the plane of xy, (x,y) always satisfy that equation
x2+y2-4x-6y+13
here same (x,y) is used. So whatever x,y taken here lies on the above circle C
x2+y2-4x-6y+13=(x-2)2+(y-3)2
which is distance between any point (x,y) (on circle C) and (2,3)
now you have to find the maximum and minimum distance
1357Yes Ankit option A is miswritten as MR2 which should have been MR2/2
1ok bhaiyya thank you . realsied my mistake ,[1]
thank you . ok bye see u later on sunday eveninng
1Q12 STATEMENT-I: a,b,c are unit vectors, then maximum value of |a-b|2+|b-c|2+|c-a|2 is 9
STATEMENT-II: a.b+b.c+c.a≤(-3/2)
Your Answer: a
Correct Answer: c
Score: -1
Explaination :
|a+b+c|2≥0
3+2(a.b+b.c+c.a)≥0
(a.b+b.c+c.a)≥(-3/2)
|a-b|2+|b-c|2+|c-a|2=2[(a.a)+(b.b)+(c.c)] -2(a.b+b.c+c.a)
=6-2(a.b+b.c+c.a)
1bhaiyaa...here in dat1 max value of gn expression
= 6-(a.b+b.c+c.a)
so dat its 6-(-1/2-1/2-1/2)
dat bcumz =9 naa..as θ becums 2π/3
1357are here statement II is incorrect
see it clearly
less than and greater than signs in explanation and statement
1I am not able to see the solutions, ie i am not able to scroll down the solutions, only the first page is viewable