11bhaiyya one query
i started both tests but dint complete now when i click on the tests a message comes out that ive already submitted. I want to complete the test
can something be done?
The results are out
please post your queries here only
For others test is still open
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111 Answers
21P1
Q5 a2+b2+c2=ca+ab√3
then the triangle is
I took a,b,c as 3,4,5 a pyhta triplet...........
the equation dusnt hold..... [2]
11
1bhaiyya one doubt..........
in d above question i took......
S=1/3+1/3+7/27+5/27......
as
S=2/3+12/27+.....
so
S=2/3/(1-6/9)
=2/3/3/9
=2...
so i got 2.....is dere any mistake???? (I KNO THERE IS)...waahh
21I KNOW I M WRONG HERE..........
Can sum1 pl. explain me da explanation.........
Q3 The number of different matrices that can be with the numbers 1,2,3,4 each having four elements is
Your Answer: 44
Correct Answer: 3.44
Score: -1
Explaination :
four places can be filled up in 4*4*4*4=44 places
matrices possible are 2X2, 4X1 and 1X4
so total no. of ways=3.44
11@tapanmast thiis might help
http://targetiit.com/iit_jee_forum/posts/matrices_4592.html
21Q13 A function f(x) is defined as f(x)=x2sin(1/x), x≠0 and f(0)=0
STATEMENT-I: f'(0)=0
STATEMENT-II: f'(x)=2xsin(1/x)-cos(1/x)
Your Answer: d
Correct Answer: b
Score: -1
Explaination :
f'(x)=2xsin(1/x)-cos(1/x)
f'(0)=lim(h→0)[f(0+h)-f(0)]/h=[h2sin(1/h)]/h
lim(h---->0) hsin(1/h)=0
shud it not be 1 ??
1look tapan.......
the diff types of matrices wid 4 elements are... 2X2,4X1,1X4.....
and each type can have 4 diff types....
so 44+44+44
=3.44
1tapan......
lim(h---->0) hsin(1/h)=0
lim(x-->0) sinx/x=1
is basically an extension of d L hospitals rule....
I THINK u cant apply d same to lim(h---->0) hsin(1/h)=0
as when u try 2 differentiate ull get more complex things...
so lim(h---->0) hsin(1/h)=0
=0.sin(1/h)
=0 is correct i guess
1
how in d world can this be of a right angled triangle???????
ive tried substituting 3,4,5 and 1,1,√2
but nothing came...........and moreover i cant understand d solutiion.......
please sum1 explain........
1repeating my doubt
x2+y2-2|x|-2|y|+4 ≤0
what does this equation represent ? [7]
acc to me this is null set
as
it will be
(x ±1) 2+(y ± 1)2 +2 <= 0
[7][7]
1dude philip u cant take it in tht way................
REMEMBER ABOUT DOMAINS AND RANGES..............
|x| is not always ±x...
1357P1
Q5 a2+b2+c2=ca+ab√3
then the triangle is
I took a,b,c as 3,4,5 a pyhta triplet...........
the equation dusnt hold.....
who said this ....every right angle triangle satisfies this equation
The question was given a condition and you have to prove it is right angled
you all are doing ulta
21But sir if it is a true statement then it shud satisfy ulta conditions also ne.......
Wat i thot was takin 3,4,5 as a pytha triplet, LHS is RATIONAL but RHS is NOT, then how cud it b a rite triangle ??
21Yeah!! Sir, after seeing the soltn i realised that in the soltn it is a = 5........
but SirJEE wats wrong in # 21 ???
Rationality must hold ne, in any order of a,b,c
1bhaiyya the same doubt for me as akand i also got dat 1 to be two....is dat wrong bhaiyya>?
1357@akand
from the two terms that you got
how do you came to know that it is in GP and not in AP??
1given
x2+y2-2|x|-2|y|+4 ≤0
for ist quadrant:
(x - 1) 2+(y - 1)2 +2 <= 0
for 2nd
(x + 1) 2+(y - 1)2 +2 <= 0
for 3rd
(x + 1) 2+(y + 1)2 +2 <= 0
for 4th
(x - 1) 2+(y + 1)2 +2 <= 0
now whats wrong in here [7][7]
i really dont understand i had to do all this it wasn't my problem clear straightaway
help me out here
1if so why??cant we add and den sum it upto infinite terms ???[7]
pl see post 5 its da same doubt
21Q34 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 ??
1hehe bhaiyya......thts a mistake i agree.........
JUST imagine........in d tensed exam hall, all i cud see was its GP at d first look.....then i got stuck as i cudnt find a common ratio i tried desperately for everything..... at last i added them.........and voila..... i got those two terms and in a second i applied d GP formula and waa i got d answer as 2......
and when i looked at d options i cud see 2 innocently sitting there......and in a jiffy i marked it and was happy...........
and then didnt even think about this question........ waaaaaahhhhhhhh
I HOPE I DONT DO THIS IN D JEE......... im sorry bhaiyya
21YEAH philip!!!
same doubt.......
all the four wer cumin NULL SETS.......
so i wrote the ans as area of jus da 1 given circle.......
1357@philip you re right
there was a little mistake in the question
instead of x2+y2-2|x|-2|y|+4 ≤0
it should have been x2+y2-2|x|-2|y|-4 ≤0
sorry for the mistake
21But sir, in dat case ANS WUD HAV bn the area of SOLE circle not considering the NULL circles........