Re: HSC 2013 3U Marathon Thread
Definately a 3U question:
(a) P'(x)=5x^4 -5c
To find any stationary points, solve P'(x)=0
5x^4 -5c=0
x^4 =c
x = ±∜c
But since c<0, ∜c is undefined and hence there are no stationary points.
P(0)=0-0+1
= 1
Also since c<0, we notice that P'(x)>0 for all real x, and hence P(x) is monotonic increasing.
By sketching a possible graph of P(x), we can clearly see that P(x) will have only one root, and that this root will be negative.
(b) The abscissae of the two stationary points are ∜c and -∜c.
P(∜c)=(∜c)^5 -5c(∜c) + 1
=c^(5/4) -5c^(5/4) + 1
= 1 - 4c^(5/4)
P(-∜c)=(-∜c)^5 -5c(-∜c) + 1
=-c^(5/4) +5c^(5/4) + 1
= 1 + 4c^(5/4)
By considering a graph of P(x), we can tell that there will be 3 distinct real roots if the product of the y coordinates is negative. i.e. one turning point is above the x-axis, one turning point is below the x-axis.
∴P(x) has three distinct real roots iff P(∜c).P(-∜c) < 0
P(∜c).P(-∜c)= [1 - 4c^(5/4)].[1 + 4c^(5/4)]
= 1 - 16c^(5/2)
= 1 - [4c^(5/4)]^2
< 0
1 - [4c^(5/4)]^2 < 0
[4c^(5/4)]^2 > 1
4c^(5/4) > 1
c^(5/4) > 1/4
c^5 > (1/4) ^4
c > (1/4) ^(4/5)
∴ P(x) has three distinct roots if and only if c > (1/4) ^(4/5).
fuuuu someone teach me how to use latex...