The number of possible integral values of the parameter k for which the inequality k2x2<(8k3)(x+6) holds true for all values of x satisfying x2<x+2 is?

A) 0
B) 1
C) 2
D) 3

1 Answer
Nov 27, 2017

0

Explanation:

x2<x+2 is true for x(1,2)

now solving for k

k2x2(8k3)(x+6)<0 we have

k(24+4x242+192x2x23x3x2,24+4x+242+192x2x23x3x2)

but

24+4x+242+192x2x23x3x2 is unbounded as x approaches 0 so the answer is 0 integer values for k obeying the two conditions.