11GORDO BHAI EDITED THAT
PLEASE CHECK THAT AGAIN[1]
this is a new thread in which all types of maths sums can be posted,
rules: 1) only 1 question at a time
2) no one start a new question unless the previous one is fully solved by someone and understood by everyone
3) every question must have options and must be of jee level.
4) this thread is only for seniors as they are the ones whu have covered the whole syllabus
5) question must be objective , solution must not be too long , but the question must be tricky( I HOPE EVERY ONE HAS SEEN JEE SOLUTIONS)
so guys i will start with a gud question:(easy one, but gud for start]
1)THE CIRCLE X2+Y2-4X-8Y+16=0 rolls up the tangent to it at (2+√3 , 3) by 2 units , assuming the x axis as horizontal , find the equation of the circle in the new position-
a)x2+y2-6x-2(4+√3)y+24+8√3
b)x2+y2-6x-2(2+√3)y+24+8√3
c)x2+y2-6x-2(4+√2)y+24+8√3
d)x2+y2-6x-2(4+√3)y+24+4√3
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31 Answers
1Mani, are u sure about the options?
can u rewrite the 14th Q clearly for me?
11@mani...
gud Q's.
koi #14 acchha se samjhao Pls !!! (Sry. gordo cud'nt understand although it may be simple)
1but can u give me a conformation on the answers...
the 3rd one ,
P(K)=CK (C constant of proportion)
we have summation (P(Ek))=1
or Cn(n+1)/2=1...or C=2/n(n+1)
again P(A)=summation { P(A/Ek)* P(Ek)}
or summ of [2/n(n+1)] r2/n... becoz {P(A/Ek) =k/n}
this way we get P(A)=(2n+1)/3n...
11GORDO THANX 4 TRYING THESE QUESTIONS
BUT I WILL BE GRATEFUL TO U
IF U GIVE THE WHOLE PROCEDURE
1answer is rite...ur way of solving it was different from mine...i did it dis way, by concidering the circle, (x-3)2+(y-3)2=4
the function is nothing but the distance of any pt on the circle from the origine...so max value=3√2 +2 and min is 3√2 -2..how did u like it??
its a one that i created myself...
1for the 13) Q is the B) option (2n+1)/3n....coz thats what im getting...
{3n+1/3n is any ways greater than 1}
11F(x) is the distance of (0,-3) from a pt. P whose Locus is :
y2 = 4 - (x - 3)2
(x - 3)2 + y2 = 22 ≡ Circle : centre (3,0) , r = 2
Let dist. of (0,-3) from the centre of the circle (3,0) be 'D'
D = 3√2
Fmax. = D + r = 3√2 + 2
Fmin. = D - r = 3√2 - 2
1oopssii! sorry for that...but try it out...ts a gud one still...
now for the earlier Q.
if P(Ek) =c, we realise that Ek is a group of mutually independent and exhaustive evens... so (n+1)C=1
or we see c=1/(n+1)
now P(A) =\sum_{i=0}^{n}{P(A/E_{i})P(E_{i})} so we have it as 1/2...
this cud have also been been understood by common sense, if all cases have equal probability , the any person has equal chances of passing...
1Let F(x)=\sqrt{x^{2} + [\sqrt{4-(x-3)^{^{2}}}+3]^{2}}
then the maximum and minimum values of F(x) are:
A)3√3 + 4 , 5√3 - 6
B)3√2 + 2 , 3√2 -2
C) 3√2 +4 , 3√2 +4
D) 3√3 +2 , 5√2 -6
try it out...
11Comprehension
A class consists of n students. For 0 ≤ k ≤ n, let Ek denote the event that exactly k student out of n pass in examination. Let P(Ek) = pk and let A denote the event that a student X selected at random pass in the examination.
11. If P(Ek) = C for 0 ≤ k ≤ n, then P(A) equals
(A) 1/2 (B) 2/3 (C) 1/6 (D) 1/(n+1)
12. If P(Ek) = C for 0 ≤ k ≤ n, then the probability that X is the only student to pass the examination is
(A) 3/4n (B) 2/(n+1)
(C) 2/n(n+1) (D) 3/n(n+1)
13. If P(Ek) is proportional to k for 0 ≤ k ≤ n, then P(A) equals
(A)3n/(4n+1) (B)3n+1/3n
(C) 1/n+1 (D) 1/n2
14. If P(Ek) is proportional to k for 0 ≤ k ≤ n, then the probability that X is the only student to pass the examination is
(A)3/n(n+1) (B) 6/n(n+1)(2n+1)
(C)1/n (D) 1/n(2n+1)
11subash make into dx/dy
then u get
\frac{\partial x}{\partial y}-x=y+1
its integrating factor is e-y
so multiplying both sides by it we get
e^{-y}\frac{\partial x}{\partial y}-e^{-y}x=y+1(e^{-y})
\frac{\partial (e^{-y} x)}{\partial y}=e^{-y}(y+1)
integrating on both sides we get
e^{-y}x = \int y+1/e^{y}
x=\int e^{-2y}y +e^{-2y}
x=e^{-2y}/(-2)-e^{-2y}y/(-2) -e^{-2y}/4
now u can simplify to get ur answer [1]
111. x = Cey - y - 2
2. y = Cex - x - 2
3. x = Cey + y + 2
4. y = Cex + x + 2
It gets way simpler with options try doing the more subjective way
= 1 is11At x=0
y = 1
y'(at (0,1)) = 2
Tngnt : y = 2x + 1
Dist of tngnt frm (0,0) = 1/√5
11The distance between the origin and the tangent to the curve y = e2x + x2 drawn at the point x = 0, is
(A) 1/√5
(B) 2/√5
(C) 1/√3
(D) 2/√3