Answer only one of the following two alternatives. OR The curve C has equation y = (x² + λ x - 6λ²)/(x + 3) where λ is a constant such that λ ≠ 1 and λ ≠ -3 .
- (i)Find dy/dx and deduce that if C has two stationary points then -3/2 < λ < 1 .[5]
- (ii)Find the equations of the asymptotes of C.[3]
- (iii)Draw a sketch of C for the case \( 0 < \lambda < 1 \).[3]
- (iv)Draw a sketch of C for the case \( \lambda > 3 \).[3]
Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- [M1A1] Differentiates y wrt x. dy/dx = ((x + 3)(2x + λ) - (x² + λ x - 6λ²))/((x+3)²) = (x² + 6x + 3λ + 6λ²)/((x+3)²)
- [M1A1] Uses discriminant and factorises. dy/dx = 0 has distinct roots if discriminant > 0 . 6² - 4(1)(3λ+6λ²) > 0 36 - 12λ - 24λ² > 0 ⇒ 3 - λ - 2λ² > 0 (3 + 2λ)(1 - λ) > 0
- [A1] Obtains result. (No equality.) ⇒ -3/2 < λ < 1 (AG)
- [M1] Uses division to obtain the form for recognition of oblique asymptote. (x² + λ x - 6λ²)/(x + 3) = x + (λ - 3) + (9 - 3λ - 6λ²)/(x + 3)
- [B1] (Tolerate error on remainder term.) ⇒ asymptotes are: x = -3
- [A1] and y = x + λ - 3 .
- [B1] For \( 0 < \lambda < 1 \): Axes and asymptotes.
- [B1] Upper branch with minimum below x-axis.
- [B1] Lower branch with maximum.
- [B1] For \( \lambda > 3 \): Axes and asymptotes.
- [B1] (n.b. oblique asymptote has positive y intercept.) Left-hand branch.
- [B1] Right-hand branch.
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