21Sir can u pl. bried a lil on ur diag in # 59......... i m strill not gettin it [2]
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111 Answers
1its ok sir, i am clicking on the graph and i am getting the soluion
by the way match the following are for 4 marks?
and negative marking for match the following??
211PAPER II
i am talking about match following of maths.
and what is the marking of match the following in physics? since it is not given in the solution
1357no negative marking for match the following
but actully, I gave it in Mathematics and chemistry (which will be removed)
For total yes it is four marks only
there no bonus this time for all correct
1ok sir, and for physics are six marks given for all correct in match the following?
1357
see if this diagram helps
As seen in the diagram OI'=42 cm
now see the slab
The object seems to be at I' which is get refracted to form image at point I
If you still do not understand this
then lets go with basic
At the slab let the angle of incidence be i and refraction be r
μsini=sinr
(3/2)(H/OI')=H/OI
OI=(2/3)OI'
1sir till when will be the negative marking be removed ??(FOR MATCH THE FOLLOWING)
1YA I AM TALKING ABOUT NEGATIVE MARKING GIVENN, WHEN WILL IT BE REMOVED??
AND MY TOTAL BECOMES 43 IN PAPER ii WITHOUT REMOVAL OF NEGATIVE MARKING
1sir can u display the names of the top ten people of both the tests
1i attended all the questions.... but frm qno82 it shows dat "did not attempt" in P-II
1Paper II question 6 is cancelled , then wont every one get marks there?
my answer was correct then why is it cancelled?
21Can sum1 pl. give the complete balance equation 4 ferus oxalate + Kmno4
115Fe2+ + MnO4- +8H+ ------> 5Fe3+ + Mn2+ + 4H2O
5C2O42- + 2MnO4- + 16H+ ------> 2Mn2+ + 10CO2 + 8H2O
1Q6 This is being cancelled
find the area bounded by the curves
x2+y2=4
x2+y2-2|x|-2|y|+4≤0
If want to solve then second circle equation is
x2+y2-2|x|-2|y|-4≤0
Your Answer: 8Ï€-16
Correct Answer:
Score: 0
Explaination :
x2+y2-2|x|-2|y|+4≤0
is not a real circle
this is being cancelled
For
this circle solution is
x2+y2-2|x|-2|y|-4≤0
brown area is to be found
area of yellow + area of brown+area of yellow=2*2=4 units
area of yellow + area of brown=Ï€r2/4=Ï€
area of brown 2Ï€-4
area will be in all 4 quadrants A=4(2π-4)=8(π-2)=8∩-16
why is this cancelled , my answer was correct[12]
1heeh check dis out.......
percentile 100?????
and my mark (84) isnt in d rank list waaaaahhhhhh
1357is it really showing 100%
your mark was in the rank list, but it was not updated (as some marks of your has increased or decreased)
1357x2+y2-2|x|-2|y|+4≤0
It is being canceled because the second circle doesn't satisfy for any x and y
The equation x2+y2-2|x|-2|y|-4≤0 was misprinted as x2+y2-2|x|-2|y|+4≤0
that why it is cancelled
1the answer in the solution and my answers are same, so i must get a correct
11Q7 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
my doubt is the circle 1 is a point circle
so please explain wat this ques is all about ?????????
11Q13 STATEMENT-I: The number of common tangents that can be drawn to the circle x2+y2=a2 and hyperbola x2-y2=a2 is two
STATEMENT-II: y=mx±a√(m2-1) is a tangent to the hyperbola. For common tangent |a√(m2-1)|/√(m2+1)=a.
Your Answer: d
Correct Answer: a
Score: -1
Explaination :
the tangents are x±a=0
i thought there were4 tangents vich could me drawn 4m a circle to a hyperbola
help me inm case of misconception
11Q26∫-∩∩ {sinx}dx :where {.} is the fractional part
Refer Prev Ques
Your Answer: d
Correct Answer: a
Score: 0
I DID
x=[x]+{x}
value daalien
mera to 0 aa raha tha and aa raha hai
please help in case of mistake!!!!!!
11please help in this 1
For a triangle ABC with angles A, B and C, such that 6A=3B=2C Match the following
28) tan(B/2)tan(C/2)+tan(C/2)tan(A/2)+tan(A/2)tan(B/2)
29) 4sin(A/2)sin(B/2)sin(C/2)
30) 4cos(A/2)cos(B/2)cos(C/2)
31) cot(A/2)cot(B/2)cot(C/2)
A) (√3-1)/2
B) 1
C) (3+√3)/2
D) √3(2+√3)
Your Answer: b
Correct Answer: b
Score: 1.00
Explaination :
A=30°, B=60° and C=90°
tan(B/2)tan(C/2)+tan(C/2)tan(A/2)+tan(A/2)tan(B/2)=1 for any angles
4sin(A/2)sin(B/2)sin(C/2)=cosA+cosB+cosC-1
4cos(A/2)cos(B/2)cos(C/2)=sinA+sinB+sinC
cot15°=2+√3
YEH PUTTING KA IDEA DIMAAG MEIN NAHIN AAYA
IS THERE A WELL METHOD TO SOLVE THIS 1 ???