x + y^2 = 17 17 > y + x^2 > 14 here x and y are real numbers greater than 1 then y-x will be

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03 Jan 2012 inequalities

x + y^2 = 17
17 > y + x^2 > 14
here x and y are real numbers greater than 1
then y-x will be

OPtion

1) more than 1
2) more than 2
3) less than 3
4) more than 5
5) more than 6
6) more than 7
7) equal to 6
8) equal to 7
9) equal to 8
10) none of these

Solution

According to the given condition
x + y^2 = 17
17 > y + x^2 > 14
so 17-17 < (x + y^2) - (y + x^2) < 17-14
so 0 < (y^2 - x^2) - (y - x) < 3
so 0 < (y-x)(y+x-1) < 3
so 0 < y-x < 3/(y+x-1)
so y-x will be less than 3

Correct Option: 3 Best Solution (3)

aman gupta   13 years AGO

subtracting the two given expressions we get
0 >(y-x)-(y-x)(y+x) >-3
0 >(y-x)[1-(y+x)] >-3
given that x and y are greater than 1 , so
y+x > 2
which implies
(y-x)[1-(y+x)]-3
this gives y-x

Like? Yes (3) | No   

P Sharath   13 years AGO

x+y^2=17 -------(1)
17 > y+x^2 >14 --(2)

eq (2)-(1)
0 > (y-x) + (x^2-y^2) > -3
=> 0 > (y-x) + (x-y)(x+y) >-3
taking y-x in common
=> 0 > (y-x)[1-(x+y)] >-3 ---(3)
dividing (3) by [1-(x+y)]
0 < (y-x) < -3/[1-(x+y)]
note: the sign changes because x and y are greater than 1 and quantity [1-(x+y)] is negative.

also as x and y are greater than 1 quantity
[1-(x+y)] < -1

hence the final equation is
0 < y-x < 3

Like? Yes (1) | No   

sridhar reddy   13 years AGO

given x and y are rational numbers greater than 1
and x+y^2=17...........(1)
and 14

Like? Yes (1) | No (2)