IIT JEE , AIEEE Forum - Mathematics , Algebra

man111

(1) if the expression x³-3x²-9x+p=0 has 2 equal root than thevalue of p =?

ans: -1 and other is +ve value.

(2) if x,y,z is choosen at random from whole no.(needed not to be different) than fimd the probability that x,y are even in x+y+z=11.

feynmann

let f( x ) be the given expn .

The equal value should satisfy f'( x ) = 0

which gives x^2 - 2x - 3 = 0

giving x = 3 , -1

put these values in first eqn to obtain the values of p .

It is not matching with the answers given .

LAMPARD

Is the ans. for 2nd one 0.0693????

suns_uday

for the 2nd bit::::::

x+y+z=11

the only possibilities of x,y are even iss::::

(x,y)=(2,2),,,,(2,4),,,(4,2),,(2,6),,,(6,2),,,(2,8),,,,(8,2),,,,(4,4),,(4,6),,,(6,4)

=10 ways

total possibilities ==¹¹⁺²C₂

there fore the answer is

==10x2 / 13x12== 5/39

suns_uday

Re:question.........

suns_uday

x³-3x²-9x+p=0

let the common root be "a" and the other be "b"

then 2a+b=3 .........(1)

and a²+2ab= -9........(2)

and a²b=-p => b= - p / a²............(3)

(1)=> 2a - p/a² =3 => a³-3a²-p =0 ....(4)

(2)=> a²-2p / a = -9 => a³+9a - 2p = 0...(5)

(4) & (5) must hav real roots and roots of both are same

(5) - (4) => 3a² +9a -p=0must hav real roots

there fore 9²-4(-p)(3) >= 0 => p >= -81/12

plzz check if any thing is wrong