(a) k=0.5: 3 points; k=1: 5 points; k=1.2: 1 point
A1
For example: one meeting point when k is large, more when k is near 1, and the number is always odd
A1
(b) A meeting point on y=x has f(x)=x, that is x^3+x=k
M1
x^3+x is increasing and takes every real value, so this has exactly one solution
R1
(c) As for any f, (a,b) is a meeting point exactly when b=k-a^3 and a=k-b^3. Subtracting, b-a=(b-a)(a^2+ab+b^2), and a\ne b, so a^2+ab+b^2=1
M1A1
Adding, a+b=2k-(a^3+b^3)=2k-\left(s^3-3s(s^2-1)\right), using ab=s^2-1; so 2k=4s-2s^3, k=2s-s^3
A1
(d) b=s-a gives a^2+a(s-a)+(s-a)^2=1, hence a^2-sa+s^2-1=0
A1
They are real and different, so the discriminant s^2-4(s^2-1)=4-3s^2>0, which gives -\dfrac{2}{\sqrt3}<s<\dfrac{2}{\sqrt3}
A1
(e) Each s in this interval with 2s-s^3=k gives two meeting points (a,b) and (b,a). Five points need two such values of s
M1
On 0<s<\dfrac{2}{\sqrt3}, 2s-s^3 rises from 0 to \dfrac{4\sqrt6}{9} and falls back to \dfrac{4}{\sqrt3}-\dfrac{8}{3\sqrt3}=\dfrac{4\sqrt3}{9} at the end, and for s\le0 it is not positive
M1
Two values of s exactly when \dfrac{4\sqrt3}{9}<k<\dfrac{4\sqrt6}{9} (about 0.770<k<1.09)
A1
Note: the end s=\frac{2}{\sqrt3} is excluded because there a=b; at k=\frac{4\sqrt6}{9} the two values of s coincide.
Final answer
- (a) 3, 5, 1
- (e) \frac{4\sqrt3}{9}<k<\frac{4\sqrt6}{9}