9768 #65

Forum for the GRE subject test in mathematics.
YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

9768 #65

Postby YKM » Mon Aug 08, 2011 10:41 am

Dear all,

Can anyone show me the solution?

Thanks

ANDS
Posts: 47
Joined: Wed Aug 10, 2011 8:41 pm

Re: 9768 #65

Postby ANDS » Wed Aug 10, 2011 8:49 pm

For 65 you have:
p(-3): 27-9a+3b=c
p(2): -8-4a-2b=c
Equating the two, it simplifies down to 7+b=a.

We also have that p'(-3)<0; so 3x^2+2ax+b<0 ==> 27-6a=b<0 ==> 27+b<6a. Substituting in for a, 27+b<6(7+b)==> -15<5b ==> -3<b.

Now lets look at -8-4a-2b=c. We have that b>-3 ==> -2b <-6 ==> -2b <-14. Similarly, a>2 ==> -4a <-8. Adding the two we have -2b-8-4a < -22 ==> c <-22. Thus -27 is the only number that satisfies.

Hopefully this helps.

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: 9768 #65

Postby DDswife » Sat Aug 23, 2014 7:45 pm

3rd degree polynomials with first coefficient 1, and 2 known roots
c is the y i tercept, or p(0)

Going to the factorial decomposition of p

p = (x+3)(x-2)(x-k)

Plugging 0

c = 6k

Checking that p is decreasing in -3, coming from positive values, and unbounded in -oo, we observe that the last root k is less than -3

Hence, c <-18




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