Moderate -0.8 This is a straightforward simultaneous equations problem requiring substitution of fractional powers. Setting the equations equal gives $x^{2/3} - 1 = x^{3} + 1$, which simplifies to a quadratic in $x^{1/3}$. The algebraic manipulation is routine and the question requires only standard technique with no problem-solving insight, making it easier than average but not trivial due to the fractional indices.
Elim y to form 3-term quad eqn in x1/3(or u or y or even x)
(M1
(=0 ) (=0 )
Expect x2/3 −x1/3 −2 or u2 −u−2 etc.
Answer
Marks
Guidance
x1/3 (or u or y or x) = 2, −1
*A1
Both required. But x = 2,‒1 and not then cubed or cube rooted scores A0
Cube solution(s)
DM1
Expect x = 8, ‒ 1. Both required
(8, 3), (‒1,0)
A1)
OR
Answer
Marks
Guidance
Elim x to form quadratic equation in y
(M1
y+1=( )2
Expect y−1
Answer
Marks
Guidance
y2 −3y=0
*A1
Attempt solution
DM1
Expect y = 3, 0
(8, 3), (‒1,0)
A1)
Total:
4
Question
Answer
Marks
Question 3:
3 | EITHER
Elim y to form 3-term quad eqn in x1/3(or u or y or even x) | (M1 | (=0 ) (=0 )
Expect x2/3 −x1/3 −2 or u2 −u−2 etc.
x1/3 (or u or y or x) = 2, −1 | *A1 | Both required. But x = 2,‒1 and not then cubed or cube rooted scores A0
Cube solution(s) | DM1 | Expect x = 8, ‒ 1. Both required
(8, 3), (‒1,0) | A1)
OR
Elim x to form quadratic equation in y | (M1 | y+1=( )2
Expect y−1
y2 −3y=0 | *A1
Attempt solution | DM1 | Expect y = 3, 0
(8, 3), (‒1,0) | A1)
Total: | 4
Question | Answer | Marks | Guidance
Find the coordinates of the points of intersection of the curve $y = x^{\frac{2}{3}} - 1$ with the curve $y = x^{\frac{1}{3}} + 1$. [4]
\hfill \mbox{\textit{CAIE P1 2017 Q3 [4]}}