C3 June 2007 Q3

EdexcelOld spec10 marksDifferentiation

3. A curve \(C\) has equation\[y = x^2\mathrm{e}^x.\]

(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), using the product rule for differentiation. (3)

(b) Hence find the coordinates of the turning points of \(C\). (3)

(d) Determine the nature of each turning point of the curve \(C\). (2)

Mark scheme (a)

Scheme	Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^2\mathrm{e}^x + 2x\mathrm{e}^x\)	M1,A1,A1
	(3)

Notes

Generous M for attempt at \(f(x)g^{\prime}(x) + f^{\prime}(x)g(x)\)
1^st A1 for one correct, 2^nd A1 for the other correct.
Note that \(x^2e^x\) on its own scores no marks

Mark scheme (b)

Scheme	Marks
If \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\), \(\mathrm{e}^x(x^2 + 2x) = 0\) setting (a) \(= 0\)	M1
\([\mathrm{e}^x \neq 0]\) \(x(x + 2) = 0\) \((x = 0)\) or \(x = -2\)	A1
\(x = 0,\ y = 0\) and \(x = -2,\ y = 4\mathrm{e}^{-2}\ (= 0.54\ldots)\)	A1ft
	(3)

Notes

1^st A1 (\(x = 0\)) may be omitted, but for
2^nd A1 both sets of coordinates needed; f.t only on candidate’s \(x = -2\)

Mark scheme (c)

Scheme	Marks
\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = x^2\mathrm{e}^x + 2x\mathrm{e}^x + 2x\mathrm{e}^x + 2\mathrm{e}^x\) \(\left[= (x^2 + 4x + 2)\mathrm{e}^x\right]\)	M1, A1
	(2)

Notes

M1 requires complete method for candidate’s (a), result may be unsimplified for A1

Sketch of y = x squared e to the x: maximum to the left of the y-axis, minimum at the origin, x-axis asymptote to the left — Mark scheme (d)

Scheme	Marks
\(x = 0,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\ (=2)\) \(x = -2,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\ \left[= -2\mathrm{e}^{-2}\ (= -0.270\ldots)\right]\) M1: Evaluate, or state sign of, candidate’s (c) for at least one of candidate’s \(x\) value(s) from (b)	M1
\(\therefore\) minimum \(\therefore\) maximum	A1 (cso)
	(2)
	(10 marks)

Notes

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Alternative