6673 edexcel gce decision mathematics d1 advanced advanced subsidiary 2 (3)

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1. The solution to the quadratic equation x2−5x+6=0x^2 - 5x + 6 =
0x2−5x+6=0 is
A. x=1,6x = 1, 6x=1,6
B. x=−1,−6x = -1, -6x=−1,−6
C. x=2,3x = 2, 3x=2,3
D. x=−2,−3x = -2, -3x=−2,−3
Answer: C) x=2,3x = 2, 3x=2,3
Rationale: Factoring the quadratic equation x2−5x+6=0x^2 - 5x + 6 =
0x2−5x+6=0, we get (x−2)(x−3)=0(x - 2)(x - 3) = 0(x−2)(x−3)=0, so
x=2x = 2x=2 and x=3x = 3x=3.


2. The limit of lim⁡x→01−cos⁡(x)x2\lim_{x \to 0} \frac{1 -
\cos(x)}{x^2}limx→0x21−cos(x) is
A. 12\frac{1}{2}21
B. 0
C. ∞\infty∞
D. 1
Answer: A) 12\frac{1}{2}21
Rationale: Using the Taylor expansion for cos⁡(x)\cos(x)cos(x),
lim⁡x→01−cos⁡(x)x2=12\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} =
\frac{1}{2}limx→0x21−cos(x)=21.

,3. The roots of the equation x2+6x+9=0x^2 + 6x + 9 = 0x2+6x+9=0
are
A. x=−3,−3x = -3, -3x=−3,−3
B. x=3,3x = 3, 3x=3,3
C. x=−3,3x = -3, 3x=−3,3
D. x=3,0x = 3, 0x=3,0
Answer: A) x=−3,−3x = -3, -3x=−3,−3
Rationale: The equation x2+6x+9=0x^2 + 6x + 9 = 0x2+6x+9=0
factors as (x+3)2=0(x + 3)^2 = 0(x+3)2=0, so x=−3x = -3x=−3 is the
repeated root.


4. The equation of the normal to the curve y=x2+2xy = x^2 +
2xy=x2+2x at x=1x = 1x=1 is
A. y=−2x+3y = -2x + 3y=−2x+3
B. y=2x−1y = 2x - 1y=2x−1
C. y=2x+1y = 2x + 1y=2x+1
D. y=−x+2y = -x + 2y=−x+2
Answer: A) y=−2x+3y = -2x + 3y=−2x+3
Rationale: The slope of the tangent at x=1x = 1x=1 is
f′(1)=2x+2=4f'(1) = 2x + 2 = 4f′(1)=2x+2=4. The slope of the normal
is the negative reciprocal, −14-\frac{1}{4}−41. Using the point
(1,3)(1, 3)(1,3), the equation of the normal is y−3=−14(x−1)y - 3 = -
\frac{1}{4}(x - 1)y−3=−41(x−1), which simplifies to y=−2x+3y = -
2x + 3y=−2x+3.

, 5. If y=sin⁡(x)+cos⁡(x)y = \sin(x) + \cos(x)y=sin(x)+cos(x), what is
dydx\frac{dy}{dx}dxdy?
A. cos⁡(x)−sin⁡(x)\cos(x) - \sin(x)cos(x)−sin(x)
B. cos⁡(x)+sin⁡(x)\cos(x) + \sin(x)cos(x)+sin(x)
C. cos⁡(x)−sin⁡(x)\cos(x) - \sin(x)cos(x)−sin(x)
D. sin⁡(x)+cos⁡(x)\sin(x) + \cos(x)sin(x)+cos(x)
Answer: A) cos⁡(x)−sin⁡(x)\cos(x) - \sin(x)cos(x)−sin(x)
Rationale: Differentiating y=sin⁡(x)+cos⁡(x)y = \sin(x) +
\cos(x)y=sin(x)+cos(x), we get dydx=cos⁡(x)−sin⁡(x)\frac{dy}{dx}
= \cos(x) - \sin(x)dxdy=cos(x)−sin(x).


6. The second derivative of f(x)=3x3−5x2+2x−1f(x) = 3x^3 - 5x^2 +
2x - 1f(x)=3x3−5x2+2x−1 is
A. 18x−1018x - 1018x−10
B. 18x−518x - 518x−5
C. 9x2−109x^2 - 109x2−10
D. 9x2−59x^2 - 59x2−5
Answer: A) 18x−1018x - 1018x−10
Rationale: The first derivative of f(x)=3x3−5x2+2x−1f(x) = 3x^3 -
5x^2 + 2x - 1f(x)=3x3−5x2+2x−1 is f′(x)=9x2−10x+2f'(x) = 9x^2 -
10x + 2f′(x)=9x2−10x+2. The second derivative is f′′(x)=18x−10f''(x)
= 18x - 10f′′(x)=18x−10.