[shivam_01]

Originally Posted by hi.shivani

New Question Set:
1) For how many values of x
(x-1)(x+4)(x+2)/(x-3)(x+8](x+3)<0

2) For real values of x,the expression(x-c)(x-b)/(x-a) will assume all real values provided ;
a) a=<c=<b
b) b=<a=<c
c) b=<c=<a
d) a=<b=<c

3) If a^2+b^2+c^2=1;then ab+bc+ca lies in between (include)
a) -2 & 1 b) -1 & -2 c) 0 &1 d) NOT

4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5

5)The number of points (x,y) satisfies i) 3y-4x=20,ii)x^2+y^2=<16
is/are?

1) For how many values of x

(x-1)(x+4)(x+2)/(x-3)(x+(x+3) 0

number line..

--------(-)8-------(-)4---------(-3)------(-2)--------1-------3-----------



solution set is :

x in (-8,-4) U (-3,-2) U (1,3) in case it takes only integer values then the answer comes oout to be

-7,-6,-5,2

[shivam_01]

3) If a^2+b^2+c^2=1;then ab+bc+ca lies in between (include)
a) -2 & 1 b) -1 & -2 c) 0 &1 d) NOT

answer to this is [0,1] ??

[hi.shivani]

Originally Posted by shivam_01

eq. of the line joining the point (2,-1),(5,-3)
2x + 3y -1 = 0
(x1,4),(-2,y1)
x1 = -11/2 and y1 = 5/3

option (3) 2x+6y+1=0 satisfies i think u are wrong some where dude

i know i am behaving like a non-maths background student..but seriously i dont remember anything about co-ordinate geom...

so plz bear with my silly question:
from where does "x1 = -11/2 and y1 = 5/3" came..

[madhobi]

Originally Posted by shivam_01

draw a venn digram and check the answer
1) 181
2) 372

its not number but no vacuum cleaners, no refrigerators

[masoom]

is the ans of second is B..

[masoom]

Originally Posted by hi.shivani

i know i am behaving like a non-maths background student..but seriously i dont remember anything about co-ordinate geom...

so plz bear with my silly question:
from where does &quot;x1 = -11/2 and y1 = 5/3&quot; came..

see when we are given 2 points (x1.y1) and (x2,y2)..

the equation of line is...

(y-y1) = [(y2-y1)/(x2-x1)]*(x-x1)..

find equation of line from here...

then substitute the cordinates containing x1 and y1.. in the quation..

to get values of x1 and y1..

[shivam_01]

5)The number of points (x,y) satisfies i) 3y-4x=20,ii)x^2+y^2=<16
is/are?

x^2+y^2=<16 this is a circle whose radius is 4 and the line 3y-4x=20 intersects at one piont
Infact its is tangent so ans is 1 point ???

soln 2 :
(x-c)(x-b)/(x-a) = K
b+c>=2a
hence option 2 b=<a=<c

??

[shivam_01]

simply put the pints in the equation
like
eq. of the line joining the point (2,-1),(5,-3)
2x + 3y -1 = 0
(x1,4),(-2,y1)

2x1+3.4=1 x1 = -11/2 and
2*-2+3y1=1 so y1 = 5/3

option (3) 2x+6y+1=0 satisfies

masooom i am correect ??
i am now master in silly errors thats why asking

[masoom]

post edited...

[masoom]

Originally Posted by shivam_01

simply put the pints in the equation
like
eq. of the line joining the point (2,-1),(5,-3)
2x + 3y -1 = 0
(x1,4),(-2,y1)

2x1+3.4=1 x1 = -11/2 and
2*-2+3y1=1 so y1 = 5/3

option (3) 2x+6y+1=0 satisfies

masooom i am correect ??
i am now master in silly errors thats why asking

@shivam bhai..

line will be

(y+1)=-2/3(x-2)

3y+3=-2x+4...

2x+3y=1..

so this time i comitted a silly mistake