MATHEMATICS

@gordo......

Ans : B

GORDO BHAI EDITED THAT
PLEASE CHECK THAT AGAIN[1]

TPAPER KYA HAI????

manipal solve paper I, i m solving it too

NOPE

so is the comprehensiion solved?

Gordo bhai corrected[1]

Mani, are u sure about the options?
can u rewrite the 14th Q clearly for me?

but 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...

#14

GORDO THANX 4 TRYING THESE QUESTIONS
BUT I WILL BE GRATEFUL TO U

IF U GIVE THE WHOLE PROCEDURE

Ans : A

oye....Breaking rules !!!!!! [3][3][3] Rule 1 & 2 dekho..[3][3][3]

Let 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...

Comprehension
A class consists of n students. For 0 ≤ k ≤ n, let E_k 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(E_k) = 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(E_k) = 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(E_k) 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(E_k) 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)

both methods are correct :)

subash 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]

The solution of (x + y + 1) = 1 is

Ans : A

The distance between the origin and the tangent to the curve y = e^2x + x² drawn at the point x = 0, is
(A) 1/√5
(B) 2/√5
(C) 1/√3
(D) 2/√3

gud u are rite rkrish, someone post a new question